A guest post by Dr. Weinstein in response to a post by Dr. Spencer.
EFFECT OF ATMOSPHERIC MASS AND CO2 ON GROUND TEMPERATURE
Leonard Weinstein, ScD
January 17, 2012
The issue has been raised about the effect of the mass of an atmosphere on surface temperature. It can be shown that if no optically absorbing gases, aerosols, or clouds are present in the atmosphere, that the average surface temperature will be determined by albedo, absorbed surface solar radiation (ignoring small radioactive heating effects), and outgoing thermal radiation at the ground level, and I will not discuss that issue here. The present discussion only considers an atmosphere with greenhouse gases, and for simplicity only looks at the effect of CO2 as a greenhouse gas. Only long time average global average values of temperature are considered, and only at long term constant CO2 levels (i.e., transient responses are ignored).
Some simplifications are made here, as the complete analysis is complex, and requires accurate experimentally measured data values and assumptions that are not well settled. The main simplification I make is the ground temperature sensitivity of the Earth atmosphere to increases in atmospheric CO2 levels. Values from less than 0.5 C/doubling to over 4 C/doubling have been suggested as the result of CO2 increase plus all feedback effects, However I am only describing the CO2 effect independently here, and this has been shown in most studies to give a surface increase of about 1.2 C/doubling of CO2, ignoring all other effects.
I use here is a mean virtual temperature, Tv~250 K that is based on an average temperature between sea level and approximate average location of outgoing radiation to space. This is an approximation, but its exact value has little effect on the comparison shown later. In addition, I use the wet lapse rate (as found in our lower to mid Troposphere) of -6.5 C per km height, even though I ignored the feedback effect of water vapor and clouds in the atmosphere to simplify the analysis.
In an atmosphere, the height from ground to a particular pressure level can be found from the following equation:
The value of H, which differs somewhat for different lapse rates, is called the scale height, and is the height where the pressure decreases by a factor of 1/e. I am using here the value of H ~29.3Tv for Earth’s atmosphere (based on the actual measured average atmospheric gas properties and Earth’s gravity), so combining this with the value of Tv selectedgives H~7.33 km. Changes in this value would be small enough for different assumptions that it would not change the basic result shown here.
I now examine two simplified cases:
- The case of a surface pressure of 1 bar (Earth’s actual value), with present amounts of CO2 (390 ppm), and with the effect of other greenhouse gases, aerosols, or clouds having a constant effect that is independent of atmospheric mass or changes in CO2, and assuming the same albedo as at present.
- The case of 2 bars surface pressure, with the same total amount of CO2 as case 1, but with an added equal amount of a mix of N2 and or O2, so that the average specific heat and molecular weight of the atmospheric gases are the same for 1 and 2. The greenhouse heating effect of other greenhouse gases, clouds, and aerosols are considered to be exactly the same as case 1 to separately show the effect of CO2 alone, and the albedo is still the same as in case 1.
The total effect of the present amount of CO2 alone on an increase in temperature above the no-greenhouse gas for case 1 is not accurately established, with estimates for CO2 alone from 5 C to 15 C as compared to the 33 C estimated total greenhouse effect with all gases, clouds, and aerosols. Since some of the CO2 absorption and radiation wavelengths overlap some of the water vapor wavelengths, the effect of CO2 in the presence of water vapor is even less addition than if considered alone. I am examining the effect of only CO2 here. I use an estimated value of the total CO2 effect of 10 C for the present amount as being reasonable (the exact amount is not important as long as is significantly larger than the effect of one doubling). If case 1 has the same mass atmosphere as the present atmosphere, except the concentration of CO2 was 0.5 times that of the present (195 ppm), this would have resulted in a reduction of surface temperature of 1.2 C for the lower concentration, ignoring feedback. Case two does has half the concentration of case 1, but also has twice the atmospheric mass, so the total mass of CO2 is the same for both case 1 and 2, and the only difference is atmospheric mass (and corresponding thickness) of the total atmosphere. The question is: what does this do to surface temperature?
The atmosphere is considerably thicker for case 2 than case 1 due to having twice the mass of gases, and this raises the altitude of some of the (assumed well mixed) CO2 a considerable amount. A simple analogy to see the effect is that if a thin unmixed layer of CO2 containing all the present CO2 mass in the present atmosphere were forced to lie close over the surface, and most of the atmosphere above it had none, the greenhouse gas effect would only raise the location of outgoing radiation a short distance above the surface. Multiplying the average outgoing altitude by the lapse rate would result in surface temperature increasing only a fraction of the 10 C presently possible for mixed atmospheric CO2. While the gases would mix eventually up into the atmosphere, this point shows the effect of altitude of the greenhouse gas as also being a factor.
The equation for the relation between pressure and height for p1/p2=2 gives a value of (Z2-Z1)=5.08 km. Thus the pressure at 5.08 km for case 2 matches the surface pressure for case 1. The fact that a 0.5 change in CO2 would only change surface temperature 1.2 C implies that it only changes the average location of outgoing radiation by 1.2/6.5= 0.18 km if that were the only factor considered. However, the total change of 10 C possible for all of the CO2 alone implies the average altitude of outgoing radiation to space for all the CO2 alone was about 10/6.5=1.54 km. This is nearly an order of magnitude larger than the change due to a 0.5 change of CO2 (i.e., it is the result of the exponential response).
We thus have case 2 with only 0.5 the CO2as case 1 in the lower 5.08 km of atmosphere, but where it has the same total mass of the entire case 1 atmosphere. However, we have on top of that, additional atmosphere with the same total mass of atmosphere as all of case 1, and also with 0.5 the CO2 as all of case 1. This upper layer would be as thick as the entire present case 1 atmosphere. If the upper layer absorbed and radiate all portions of wavelengths absorbed and radiated in the lower 5.08 km, this upper portion alone would have a location (for CO2 alone) 1.36 km above the 5.08 km level where outgoing radiation occurred. The actual solution of the resulting average altitude would require a full radiation analysis, and is not as high as that oversimplified version. However, it is clear that a thicker atmosphere, even without increasing total greenhouse gases over the thinner case, would have increased surface temperature due to the increased average altitude of outgoing radiation. It is also true that it is not the mass or pressure of the atmosphere alone that causes the increase, it is the combination of average altitude of outgoing radiation and lapse rate, and the increase in mass of atmosphere would raise the average location of outgoing radiation by virtue of thickening the total atmosphere. The final increase in surface temperature is the product of average outgoing altitude (including from the ground, greenhouse gases, clouds and aerosols), and lapse rate.
the Air Vent 
I left this comment at Roy’s thread:
Dr. Spencer,
There are multiple effects to remove heat from the planet surface. Only one effect to space.
For communication purposes, let’s say we have a gas which blocks/emits zero IR yet has mass and we increase the mass of the atmosphere to 10x using only a special non- IR absorbing/emitting gas.
Of course in this 10x world, the same amount of the heat from the sun is still surface absorbed. The heat removal mechanisms from the surface are obviously not totally IR and include conduction and convection. Assume for this concept that convection doesn’t change from this huge change in thickness.
The greenhouse effect is a delay in energy release from earth to space. If surface heat has any increased delay in reaching emission altitude, temps go up. Consider that increased conduction time causes the same sort of delay as increased emission time.
The sun in this model would hit the surface of our planet with the same intensity, yet IR would not be emitted or absorbed by our special gas – with a non-zero energy content. Conduction would obviously create a delay in release of energy from the surface to emission altitude – space.
While I don’t know if the paper you discussed above is correct, it does seem that the addition of insulating and energy storing materials deserves a proper mathematical treatment analyzing the delay in heat transport.
To continue the concept, the paper referenced discussed “pressure” as the primary motivator of surface temp. Pressure in their model is certainly not correct as a primary independent variable. Were Earth two times the mass and the same diameter, the atmosphere would be denser, pressure would double, but this planet would not likely fall on the pressure/temp curve of the paper above.
Pressure is a proxy for heat transport, as are greenhouse gases. It would take something new and convincing for me to see either as unimportant in result.
These “delays” are a mathematically one-time phenomenon, as once equilibrium has been reached, the flow out again matches flow in. So it becomes necessary to get specific about just how long the delay is. I have seen no persuasive calculations of this.
Leonard – please can you tell us your preferred reference for this number?:
“However I am only describing the CO2 effect independently here, and this has been shown in most studies to give a surface increase of about 1.2 C/doubling of CO2, ignoring all other effects.”
Thanks.
I’m not confident what the situation Leonard describes really does in net effect, but it does seem that the increased conduction/radiation ratio would do something non-zero to the energy transport.
Curious said,
There are numerous sources for values between 1 C and 1.2 C as the rise ideally due to doubling CO2 alone effect (ignoring all feedbacks). I selected the high end to be conservative. An example was given by Christopher Monckton (but I don’t have the reference handy). I could look some up if you insist, but it is so commonly used by both sides that it would be a good assignment for you to find.
Jeff Id,
It is not the thermally insulating or heat storage properties of the atmosphere that raises the location of outgoing radiation, it is only the optical properties. The optically partially insulating gases move the average location of outgoing radiation to a higher altitude than a greenhouse gas free equivalent atmosphere, and the lapse rate does the rest. In order for this to heat the ground more, the lapse rate has to be reasonably large, and sufficient atmospheric mixing (buoyancy, convection, evaporation and re-condensation upwards, wind, etc.) insures this. The only effect of the more massive atmosphere, with a given amount of greenhouse gas, that is important here is the increase in total height of the atmosphere.
Brian,
The effect of the optically absorbing gas is not to produce a significant temporal delay, but results in a transfer of the effective outgoing location of radiation to space to a higher altitude. It is in equilibrium for my analysis, so will not change with added time.
“The effect of the optically absorbing gas is not to produce a significant temporal delay”
I think you would find that the delay caused by the absorption/emission with added CO2 is exactly equal to the increased atmospheric temperature. The emission altitude rising represents a time change in energy transport from the surface to average emission altitude.
Jeff,
There is a delay of the average transit time of ground (and atmospheric) absorbed energy to the average level of outgoing radiation. However, some of the radiation is directly radiated from the ground (through the atmospheric “window”) with no delay. Most is carried from ground to higher altitude by convection of heated air (buoyancy) and evaporation followed by condensation at higher altitude (which take some finite time). Thermal radiation absorbed and radiated by the atmosphere also transports some of the energy to higher altitude (this process is much faster than convection). Each process has different average altitudes of final radiation to space, and different transit times, but an average location of all outgoing radiation is all that is needed to find temperature increase. The time scale of this delay is not directly the cause of the temperature increase (the average time scale can be different with different mixes of atmosphere, and result in the same final effective altitude). Different atmospheres could have different portions of radiation and convection, but only the average altitude of outgoing radiation combined with the lapse rate determine the increase in temperature. The only thermal properties of the atmosphere needed to know are the average specific heat (Cp) and the gravity, and also assurance that there is sufficient mixing to maintain the lapse rate. The fact that there is some delay time is related to the process, but is not a direct cause. In an average state situation, energy in equals energy out, and once the atmosphere goes through transient response to average state balance, it does not change temperature or energy level further. Note that I only am talking about long term average values.
Re: Leonard Weinstein (Jan 22 16:20),
But that’s exactly the “lag” whose energy buildup determines the temperature increase. Power x Time = Energy.
The average surface temperature is primarily raised because convection, conduction, evaporation, ocean currents, etc. carry energy away from the hot area (the tropics) and heat the colder areas.
It is all about cooling the hot spot that radiates at T^4 and warming the cold areas. The total energy going into the system equals the total energy going out, but a surface with heat evenly distributed will on average be much warmer than a surface with a hot spot.
Jeff,
To make my point more clear, consider two cases with greatly different atmospheres and greenhouse gas concentrations so that the altitude of outgoing radiation is greatly different, and surface temperatures are also greatly different. As long as all of the solar energy was absorbed by the surface and the albedo is the same, the outgoing energy is exactly the same long term, and is equal to incoming. There is no change in energy transport from the surface to average emission altitude. There is a lag in the time of absorption to time of final emission at altitude, but no steady state change in heat transfer rate. This is despite the fact the surfaces may have temperature differences. I may be wrong, but I think the lag can vary even for the same temperature increase but different composition and mass atmospheres. If you think otherwise, can you show any justification.
Genghis,
It turns out you are exactly wrong. Hot spots radiate far more than cold areas (T^4), so spreading the heated surface energy out before transporting it up to the outgoing layer LOWERS the average surface temperature. It is true convection and other processes spread the energy out, but not enough to LOWER the temperature much, so that the greenhouse gas effect does most of what it is claimed to do.
“There is no change in energy transport from the surface to average emission altitude. There is a lag in the time of absorption to time of final emission at altitude, but no steady state change in heat transfer rate. This is despite the fact the surfaces may have temperature differences.”
I think we agree on this. I believe Brian was referring to the time delay of individual energy quanta absorption to release. The transfer rate is power not energy so there is no disagreement there. The power energy concept is often confused in discussions due to loose terminology. Power (flow rate) doesn’t change from IR absorption, energy/temp does.
Leonard,
Everything that diffuses the energy throughout the system, increases the average temperature of the system, precisely because of the T^4 law. Radiation primarily loses energy (radiates it out in accordance to the T^4 law) and doesn’t transport the energy from the hot spots to the colder areas.
Great post Dr. Weinstein! Sometimes one puts their thoughts out to see if they are really understanding the subject so here I go.
The post is also apropos to the famous (or infamous to some) donnybrooks over the climate of Venus where you had one side saying it was all about a “runaway” greenhouse effect due to the atmosphere being 97% CO2 and the opposing faction claiming the surface temperatures were due to the huge mass of the atmosphere / high surface atmospheric pressure (about 93 bar). Observing that the Venusian atmosphere from the 1 bar altitude (about 50 km, 75C) up closely resembles that of Earth’s troposphere in temperature and lapse rate even though the Venusian atmosphere is mostly CO2. Gases with radiative properties are necessary emit from the upper troposphere and thus develop a temperature lapse rate profile in the troposphere but the dominant factor in determining the planetary surface temperature once a lapse rate structure is established is the mass of the atmosphere. A surface temperature of 462C yields a lapse rate of about 7.7C/ km between the surface and the 1 bar altitude.
Atmospheric gases with radiant properties are more akin to a catalyst enabling the existence of the temperature lapse rate structure rather than being a primary temperature driver. This is manifest by the theoretical logarithmic temperature response to increases in greenhouse gases where CO2 has done most of its heavy lifting from zero to 200 ppm and has little more temperature influence at higher concentrations. So in the Venus climate kerfuffle, the answer is somewhere in the middle with both the necessary presence of greenhouse gases and the high atmospheric mass. What does not appear to be required for the high Venusian temperatures is the high concentration of greenhouse gases vs. non greenhouse gases.
Leonard Weinstein,
“Thermal radiation absorbed and radiated by the atmosphere also transports some of the energy to higher altitude (this process is much faster than convection).”
Except part of the energy is not going up in radiative transfer. Each extra stop on the way out increase the time for all to get out. Only a very small proportion goes straight out.
How much is the delay again??
Robert,
You are correct in that the greenhouse gas establishes the altitude of energy balance with the solar energy, and the lapse rate does the rest. You can call it a catalyst or not, but it is a direct result of the optical absorption properties of the gases in the atmosphere. This is what I said in the Venus discussion at Science of Doom’s site, and what I am saying here. The only specific point I was trying to make here was that making the atmosphere taller (by increasing it’s mass) increases that outgoing average altitude even with the same total amount of greenhouse gas as the lower mass atmosphere.
Genghis,
The total energy in and out are no different for all cases at equilibrium. When you lower the energy of some “hot” locations, you move that exact amount of energy to “cooler” locations. Using the T^4 law for the redistributed energy fluxes give a LOWER average surface temperature (but equal total energy flux). Break it into 2 parts and try it with energy and then average the area temperatures to see.
Jeff,
I now see your point. I had already said that there are transfer lags, although the radiation (direct photon transfer lag) is very small, and it is related to, but it is not uniquely proportional to the temperature increase (due to the much slower convection lags, which can have a range of values even for a final radiation out height the same). It is proportional for a given atmospheric mix and mass, but the point I make is that given the temperature rise, you cannot find a unique lag for all combination of gases and altitudes of effective outgoing radiation. I thought you were implying there was a direct unique relationship, but it is not a cause and effect.
Genghis,
I apologize. You are correct. I did a two area case and found spreading the energy out does result in a higher average temperature. I had read elsewhere the opposite, and did not bother to check. This indicates that changes in air patterns or ocean currents can increase a simple geometric temperature average for a given total energy level in and out and atmosphere composition. However, I also did calculations it for small changes as seen on Earth, and it would not have a large effect for those cases.
Leonard,
Lets take two square meters the first radiating 400 watts (289.8 K) and the second radiating 200 watts (243.7K). Their average temperature is 266.75K.
Now lets average their radiation at 300 watts, we get 269.7K which is almost 3K warmer.
Kuhnkat,
Look at the case where the gas is transparent to solar energy, but almost totally opaque to outgoing thermal radiation at reasonable densities. There is radiation going in all directions in the gas, but the net radiation heat transfer is nearly zero for that case. The photon radiative delay is not even a significant player for that case, as thermal convection time constants are the totally dominating factor in raising the energy up. What is the point of looking at interior processes, when the absorbed surface energy and altitude to eventually leave the atmosphere are all that matter. In that case, regardless of the optical resistance to photon transfer, the surface temperature will be still determined by the lapse rate (which is only a function of Cp, g, and the fact of adequate mixing), and the altitude of outgoing radiation.
So an imaginary surface at this height has the same atmospheric mass above it as the P1 case, but 50% of the CO2, and hence should be 1.2C lower than the P1 surface case.
The P2 surface is 6.5*5.08=33.02C hotter than this imaginary surfacem so doubling the atmospheric mass with no additional GHE adds 33.02C.
Something fishy here…
5 “There are numerous sources for values between 1 C and 1.2 C as the rise ideally due to doubling CO2 alone effect (ignoring all feedbacks). I selected the high end to be conservative. An example was given by Christopher Monckton (but I don’t have the reference handy). I could look some up if you insist, but it is so commonly used by both sides that it would be a good assignment for you to find.”
Leonard – you misundertand, I wasn’t requesting a lesson on googling, I was requesting your reference supporting your work. Sure there are lots of claims re: global avg. temp response to CO2 concentrations in the atmosphere and I could spend many a while googling and reading them. What I wanted to know however is your preferred reference so I can evaluate its worth. If you don’t have one, fine – just say so.
No comment yet, I just want be informed on the various posts
Steveta_uk,
The heights are not just additive. You have to do a complete radiation analysis to get the new average outgoing height. I only show that increasing the height of the total atmosphere has to increase the outgoing height but the increase depends on several details such as direct radiation to space from the ground, and amount of radiation blocking. If the radiation transmission resistance (“blocking”) is high enough, doubling the atmospheric mass would double the level of outgoing radiation. However, for the case in question, there are contributions directly from ground to space, and the modest CO2 level, and change in dilution, would result in outgoing radiation from throughout the atmosphere. Averaging this is not as simple as for the optically very opaque atmosphere.
Curious,
I have run across several examples of the simple ideal increase for doubling CO2, but did not record the sites, so do not have a reference on hand. If I ever expanded this post (which would also have to have some quantitative numbers) I would look up suitable references.
23 – equals “I don’t have one”. Ok.
I want to make a comment on an extreme case here.
If the atmospheric mass were increased by about a factor of four with O2 and N2, and there were the same total mass of CO2 as present, but water vapor and clouds were considered, the surface temperature would approach the boiling point of the surface water due to the height of outgoing radiation (mainly due to increased water vapor and higher clouds). Under these conditions the continual increase in water vapor at higher temperature and greater atmospheric mass would continually increase the total mass and therefore height of the atmosphere, and a true runaway greenhouse gas effect would occur. All of the oceans would evaporate, and the surface temperature become several hundred C.
However, if the mass of the atmosphere did not increase, but the percent CO2 continually increased, the surface temperature would not increase very much, as the height of outgoing radiation would be limited by the mass of atmosphere, and the lapse rate would not change much. There cannot be a runaway greenhouse effect with a limited height atmosphere regardless of composition.
This implies that there has to be greenhouse gases to heat above the transparent gas case, but mass and thus thickness of the atmosphere is the actual limiting factor for the magnitude of the greenhouse gas effect.
Curious,
What is your point here? If it bothers you, I will get references. I know there is some at Science of Dooms site. The exact level was not the point of this post, and the level did not matter except as order of magnitude.
26 – The point is you have written a post including this assertion:
“However I am only describing the CO2 effect independently here, and this has been shown in most studies to give a surface increase of about 1.2 C/doubling of CO2, ignoring all other effects.”
When asked for your preferred reference instead of simply saying “I don’t have one, nor can I cite the studies I refer to” you brush it aside with “I refer you to google”. Wow – big deal, your audience can use google to see if it can find the information to support LW’s post and they can treat it as an “assignment”. Nice approach. IMO the assignment is yours to provide refs for the arguments you are advancing but it is clear we differ on this.
Curious Said,
Since you make an issue of this here are a few sources sources. It took 2 minutes to find them and selected only 3 of many more. I did a search on: Textbook sensitivity to a doubling of CO2 concentration where temperature feedbacks are absent or sum to zero:
http://chriscolose.wordpress.com/2009/10/08/re-visiting-cff/ (1 C per doubling)
http://wattsupwiththat.com/2011/12/28/sense-and-sensitivity-2/ (1.2 C per doubling)
http://scienceofdoom.com/2010/02/19/co2-an-insignificant-trace-gas-part-seven-the-boring-numbers/ (1.1 C per doubling)
I am still very curious why you made an issue of this since it is so easily found and not a major point of question.
Curious, 1.2° C/doubling is a standard textbook calculation. It’s well below the the level where anybody is going to be able to publish a study just to demonstrate it. Leonard could perhaps have just stated it that way instead.
The closest you will get is papers that include a derivation of this result (or related quantities), but it would be an ancillary result to the main point of the paper.
In most “real world” papers, they include information like the atmospheric profile, effect of oceans, the variation in constituents with height (esp H2O), and as you can imagine, as you do so, your number changes accordingly.
The biggest relatively straightfoward correction to this simplistic calculation is including moisture and associated latent heat. This is probably the classic paper on that,. Lal and Ramanathan, 1984.
The so called greenhouse effect is an artifact of a failed experimental configuration, first perpertrated by John Tyndall over 100 years ago and repeated endlessly ever since.
Carbon dioxide, by its molecular weight and atomic structure, expands more rapidly under heat than does an average sample of air. This causes increased pressure in the sealed container holding the carbon dioxide sample, relative to the container holding just air. Increased pressure results in the temperature of the carbon dioxide to rise higher than the air in the other container. That is the so called greenhouse effect.
HOWEVER, when the pressure in the two vessels are allowed to equate, such as by providing a hole in the top for the excess gas to escape to the outside air, then the temperature in each vessel rises to a lower level than before and the temperature in both containers remains the same. Bye Bye, greehouse effect. This has also been demonstrated on a number of occasions, but with far less noise and propaganda than the false greehouse effect.
Now, when we turn to the real atmosphere, we see that gravity draws more air molecules towards the surface and leaves fewer higher up, where in any event, there is more room for them to expand as the diameter of the atmosphere keeps increasing, the further up you go from the surface, so pressure is lower for both reasons.
More molecules near the surface means higher pressure than further up. Higher pressure, as already demonstrated, means higher temperature near the surface and lower temperature further up. However, there is no net increase in temperature or energy for the whole atmosphere, only a redistribution.
Let me quote: “Adiabatic changes in temperature occur due to changes in pressure of a gas while not adding or subtracting any heat”.
Now please go back and read Nikolov and Zeller’s papers with a fresh mind.
28 – yeah, you’re right – who needs concise traceable references eh?
Guess you know those ones inside out and have bottomed em out loads of times? How did you feel about James McC’s comments? Seemed reasonable to me and SoD agreed but I haven’t got the time to bottom it out where it went next. No doubt you followed through on it so if you could spare a bit more of your google fu to help out it’d be appreciated.
Oh and btw there’s good news (or so some claim) for bean eaters:
http://www.mensfitness.com/nutrition/what-to-eat/beans-beans-dont-make-you-fart
29 Thanks Carrick – response appreciated. Lal and Ramanathan looks a good paper from the brief read I’ve just had – I’ll revisit at some point. If you have a ref to a standard textbook calc. of the 1.2degC per CO2 doubling that would also be appreciated. I looked at the sites LW quotes and I didn’t find a complete and continuous treatment on any of them.
Ausie Dan,
You are still confusing the temperature gradient caused by the adiabatic lapse rate, which does cause the lower level of the atmosphere to be warmer than the higher levels, and the cause of an absolute temperature level, which is determined by the location of where the energy in and out balances. You also confuse adiabatic compression and expansion with greenhouse effects. I do not know the level of gas dynamics or thermodynamics you have had, but it requires understanding of these subjects.
Leonard Weinstein,
I was not questioning your overall presentation, only the statement of radiation being faster than convection. Convection moves a large amount of energy slowly. Radiation moves a small amount extremely fast but not all in the upward direction. I do not pretend to know the exact figures.
Leonard,
I know enough physics to understand what you are saying.
What I am saying is that those who believe in AGW do suffer such confusion.
The whole hypothesis of the so-called “greenhouse effect” falls apart for the simple reason that backradiation from a cooler atmosphere cannot warm an already warmer surface. As DeWitt Payne mentioned on the third thread on Backradiation on SoD, gases only start to absorb radiation when the source starts to get warmer than the gas. This confirms the work of Prof Claes Johnson “Computational Blackbody Radiation” which is summarised on the Radiation page of my site http://climate-change-theory.com
DeWitt then asked a question as to how a laser can cut metal even when it gets very hot. A laser is hardly acting like a blackbody which emits frequencies based on its temperature. Laser beams are induced radiation, not spontaneous for a start. Laser cutting involves laser light (with higher frequencies than those emitted by the material being cut) producing melting or burning because of the intensity resulting from the focussing process. The region being cut is also cooled somewhat by a blowing process. The burning could be compared with the use of a magnifying glass to focus the Sun’s light on a combustible material. Similar questions have been asked about microwave ovens.
Everyone should remember that Johnson is writing about blackbody radiation – not lasers or microwave ovens. You won’t get much spontaneous microwave radiation (let alone laser radiation) in all that “backradiation.” Any microwave radiation would only warm water and a few other substances to a depth of about 3 or 4 cm.
Inspired by Claes Johnson and in support of Doug Cotton, I compared one-stream and two-stream implementations of a single slab and of multi-layer models of the atmosphere. I found that two-stream models of heat flow, although giving athe same temperature distribution as the one-stream model, give in a symptomatic way spurious absorption.
Two-stream descriptions of heat flow and thereby back-radiation , as is used by IPCC authors, should be avoided.
Click to access IR-absorption.pdf
Leonard,
I notice that you did not comment on the main thrust of my post, namely that the so called greenhouse effects of H2o, CO2 etc, are merely artifacts of differential expansion under heat, in a confined container.
When unconstrained they all expand at the same rate.
there is no greenhouse effect.
Curious:
ScienceOfDoom seems to me to about as complete as you will get. Can you point to where the lack of continuity that you perceive to be located in this article? (In general, anything more detailed would probably sit in an exercise for the student.)
Curious, I know you’re looking for a source for that 1.2°C/doubling CO2 number. As I pointed out above, the precise number depends on the assumptions of the model. This report provides a much more extensive treatment of the problem circa 2003.
AusieDan said
January 24, 2012 at 1:28 am
Leonard,
I notice that you did not comment on the main thrust of my post, namely that the so called greenhouse effects of H2o, CO2 etc, are merely artifacts of differential expansion under heat, in a confined container.
When unconstrained they all expand at the same rate.
there is no greenhouse effect.
Aussie,
I believe it would be reasonable to consider gravitation to be an elastic confinement. If atmosphere was unconstrained there should be no pressure gradient. It isn’t rigid as employed in the ideal gas law. I would be interested in Doctor Wienstein’s repsonse as well.
38 – Carrick – have a look at the reference LW quoted at SoD. Do a “ctrlF” search for “james” and you should find this inserted note from SoD:
“[added note, James McC kindly pointed out that my calculation of temperature is wrong and so maybe it is too simplistic to use this method when there is an absorbing and re-transmitting atmosphere in the way. I abused this approach myself rather than following any standard work. All errors are mine in this bit – we’ll let it stand for interest. See James McC’s comments in About this Blog)”
In turn this leads to this response from SoD to James McC on the “About this blog thread”:
******
“scienceofdoom on March 18, 2010 at 11:02 pm
James McC:
I am one of those tiresome people who don’t understand scientific concepts till I have worked out the maths. Usually I get it all wrong, but then I worry about it until I understand why.
I’m the same. It keeps nagging at me. I “accept” it, but then have to go back to it, and finally 3 months later I figure it out. Or if I don’t I’m never really sure I understood it right..
On the IPCC definition:
I was a bit puzzled by the reference to the tropopause..
It’s just that a standard definition is needed for comparisons. There were lots of different calculations done over a couple of decades, all arriving at slightly different values, but all using different definitions (and also using different CO2 concentrations and ignoring some less important bands like CO2 absorbing solar radiation in the 4um band).
It could be defined at the top of the stratosphere if everyone did the same thing.
But basically the main effective radiative emissions from the surface out to space take place from within the troposphere. So when we think about balancing incoming radiation with outgoing radiation adding “forcing” above the troposphere is fine.
And so..
If I understand it correctly, what we mean by RF in plainer language is..
Your plainer language is correct. And it’s really about being able to use it for 0-dimensional models. (1-d being when each layer up through the atmosphere is separately considered).
And now onto your calculations. Thanks for pointing out my sloppy work. (And check I inserted “^” in the right places in your post)
I was pleasantly surprised when my (incorrect) back of envelope calculation came out so close to the properly calculated number so threw it in.
Your analysis seems spot on. I said in part seven “It’s a rough and ready approach. It’s not quite right, but let’s see what it churns out.” So you’re not really at odds with the experts as their number comes out of a big computer using all the formulae.
Note that your calculation for current conditions won’t be quite right because CO2 is 1.7W/m^2 but total GHG effect is 2.4W.m^2. And you can’t throw the 2.4W/m^2 into the ready reckoner of 5.35ln(C/Co) because it’s only valid for CO2 (it might happen to be correct but I wouldn’t assume so).
Therefore current temps calculated from CO2 increase alone – using the Stefan Boltzmann ready reckoner – are too low by some amount.
But future temps calculated in the same way from CO2 alone – as you have done – show 0.7′C compared with 1.2′C as calculated.
There is enough of a difference to wonder why. More digging is required.”
******
And then returning to the original thread SoD inserted this note after his no feedback case calculation:
“[End of dodgy calculation that when recalculated is not close at all. More comments when I have them].”
Checking dates of posts and comments doesn’t immediately show where the trail goes next and whether or not SoD and James reconciled their views. Please don’t misunderstand – I have a lot of respect for the work SoD does and the transparent approach he takes to things. I’ve only extracted the key items on this dialogue to show the problem with LW’s use of it as a reference in this instance. Perhaps there is a resolution available but it is not on the page LW pulled out the hat with google.
Re: Christopher Monckton – his “trademark” approach is to use the IPCC numbers which in turn AFAIK do not have a transparent bottomed out derivation of temp. sensitivity to CO2. This is something which I tried to find a longwhile ago and failed, the search for which took me via CA where SteveMc agreed he had not found it either and he had in fact requested that the IPCC include a standard derivation in their ARs as a reference point. AFAIK this was not done and IMO it is something which should be revisited as a requirement for the forthcoming IPCC report which is in progress now.
Chris Colose gave a good treatment as far as he went. In the comments there were several good contributions and Patrick027 added a lot of valuable detail – I would say of the three refs, for someone with some time and knowing what holes they were looking to fill this was the best reference.
I think I’ve made my point about how I feel about LWs attitude to these search results for “Textbook sensitivity to a doubling of CO2 concentration where temperature feedbacks are absent or sum to zero” as references.
39 Carrick – thanks again. I’ve not got time to look at that in detail but from a quick look at the introduction and the closing session remarks I’m not expecting it to provide a single derivation for CO2 sensitivity. As you rightly say – assumptions and proposed model are what count.
My 42 – re: the IPCC AR5 – I’ve just seen Lucia is reporting David Appell’s ZOD hosting:
http://rankexploits.com/musings/2012/zods-new-location/
http://www.davidappell.com/ZODS/
“The atmosphere is considerably thicker for case 2 than case 1 due to having twice the mass of gases, and this raises the altitude of some of the (assumed well mixed) CO2 a considerable amount”
I think this assumption is incorrect.
http://www.mikeblaber.org/oldwine/chm1045/notes/Gases/Mixtures/Gases06.htm
The pressure exerted by a particular component of a mixture of gases
Dalton’s Law of Partial Pressures:
Pt is the total pressure of a sample which contains a mixture of gases
P1, P2, P3, etc. are the partial pressures of the gases in the mixture
Pt = P1 + P2 + P3 + …
So the partial pressure of CO2 would be the same regardless of whether total pressure for atmosphere is 1 atmosphere or 2 atmospheres.
OK, sounds reasonable Leonard, so long as the adiabate is supported by heating from absorption of solar energy at the ground.
If higher atmospheric pressure (with the same total weight of CO2 in the atmosphere) would be expected to result in a warmer surface, then is the effect very large? For example, it is likely oxygen concentration varied considerably over the past 500 million years…. ~15% to ~35% (http://www.pnas.org/content/96/20/10955.long), which would have produced changes in total atmospheric pressure of ~ -5% to +15% compared to today. How much impact on surface temperature would a 15% increase in total pressure have, all else being equal?
Curious Said,
It is clear you have spent a lot of time on the 1.2 C per doubling issue. I do not have a better source of this number than I already gave you. If you find later that there is a better number (and source), I would appreciate if you made it available. Unless the value is very greatly different from 1.2, it does not affect the point I was trying to make here, that the mass of atmosphere is as major a factor to the greenhouse effect as gas composition (as long as there is some greenhouse gases).
Aussie Dan.
I am less knowledgeable in many other fields than gas dynamics, and do not (generally) throw poorly understood comments at experts in those fields where I am far less expert (but I admit I have on some occasions). I am quite expert in gas dynamics, and understand about compressing or heating confined or free expanding gases. I don’t mean to offend you, but I would guess from your comments that you have a very limited understanding of these subjects, and are making statements that have no relationship to the validity of the greenhouse gas effect.
Steve Fitzpatric,
The greenhouse gas heating depends on both the mass of the atmosphere and details of the greenhouse gases, water vapor, clouds, albedo, etc. Just increasing the atmospheric mass if all else is constant would heat at less than a direct linear amount. However, if all greenhouse factors also increased in direct amounts as the atmosphere mass increased, the temperature increase would likely be about linear with mass. That is, the difference would be less than 15% increase in the 33 C present if all else were the same mass as before, and near 15% increase in the 33 C (about 5 C) if water vapor and cloud and CO2 were near constant concentration.
Curious #42 just remember all models are wrong, all real-world calculations have incorrect assumptions.
If you wanted to do the zero-dimensional version of the planet, the way Science of Doom is going about it, is completely, preposterously wrong. And yet he gets the same order of magnitude as other estimates that are more careful. Why?
To understand how this could be, imagine you have an exact quantity that you are expanding in order of approximations:
F = F0 + eps * (F1 + F2 + … + Fn) + eps^2 * (S1 + S2 + … + Sm) + …
F0 is the superconducting Earth with no atmosphere approximation.
“eps” tracks the relative magnitude of the different corrections you are making (in the end you set eps=1).
The F1, F2, ,… Fn are the “n” first order corrections you need to make to reach an expression for F which is correct to first order if F. Similarly the S1, S2, … are the “m” second order corrections and so on..
Suppose, I make a single correction to the zeroth order F = F0 approximation, like this,
F = F0 + eps F1
The thing to note is this is no more accurate of a result than F0. In fact even this:
F = F0 + eps * (F1 + F2 + … + Fn-1)
is no more accurate of a result than F0.
When you are computing estimates you have to include all relevant corrections of the same order. In fact, I’ve seen nonsense like this:
F = F0 + eps^2 S1
and the person coming away saying “look! I’ve corrected it to second order because i including a second order correction!”.
Um no they didn’t. They’re just stupid. 😉
Doing things like “well let’s including the rotation of the Earth”, “let’s allow the atmosphere to respond to forcings” etc doesn’t help you, unless you’ve nailed them all.
Fundamentally what SOD was wrong, but so was James Mc. There is a way to do the zero-th order equation “correctly”, but it’s way more complicated than the toy models that get used. And as long as the toy models are employed, even the “classical” greenhouse gas number 1.2°C/doubling, including agreement among the different sources, is mostly a statement they’ve used similar assumptions, not that there is any really deep meaning to the 1.2°C/doubling. (That is there will never be a fourteen digit accurate version of this 1.2°C/doubling of CO2).
It’s not a physical number, it can’t be measured, all it just rough order of magnitude, what you get from GHG forcings without throwing in any of the higher order corrections.
Anyway, sorry for this long-winded meta-comment, I hope it is useful in helping you sort some of this out.
47 – “If you find later that there is a better number (and source), I would appreciate if you made it available.”
Leonard – I’m tempted to say “Treat it as an assignment”…. however I hope at the least we are now on the same page wrt to making statements without knowing your references.
I spent some time earlier reading the WG1 ZOD Ch8 that David Appell has kindly provided and I was dismayed at the level of abstraction that is being aimed for with the Global Warming Potential and the Global Temperature Change Potential metrics, without having the essential and foundational CO2 sensitivity derivation explicitly provided. The reference they quote is
“Ramaswamy, V., et al., 2001: Radiative Forcing of Climate Change. Climate Change 2001: The Scientific Basis.
54 Contribution of Working Group I to the Third Assessment Report of the Intergovernmental Panel on Climate
55 Change, Cambridge University Press”
which I have yet to bottom out.
48 Carrick – thank you – yes your comment is useful and I appreciate you putting it together.
I’m guilty of not having nailed this issue to my satisfaction and I know from following the blogs that there will be guys out there who have. Putting that aside my big beef is the lack of a foundational reference which one can point to which specifies assumptions, boundaries and method to come up with a consistent and traceable solution. The worth of that solution can then be evaluated in the different contexts where it is applied.
Given WG1 Ch8 is called “Chapter 8: Anthropogenic and Natural Radiative Forcing” I find it incredible that such a reference is not centre stage. After all AFAIK tell if the answer is roughly = nil, the whole AR is just recycling in waiting. Maybe I’ll be pleasantly surprised by Ramaswamy et al! 🙂
In reply to #42 from “Curious”.
Nir Shaviv has a derivation of your 1.2 C here:
http://www.sciencebits.com/OnClimateSensitivity
I myself would guess it would be closer to your 0.7 C figure.
A doubling of CO2 would supposedly increase the wattage flux at surface by 3.7 watts.
{(390.7 + 3.7)/(current 390.7 surface flux)}^0.25 * 287 -287 is closer to 0.7 C. I think what Shaviv is calculating is the effect if the flux we got from the sun increased by a factor of
239/235, rather than an additional 3.7 watt greenhouse effect at earth’ surfaces
curious said
January 24, 2012 at 2:33 pm
‘I spent some time earlier reading the WG1 ZOD Ch8 that David Appell has kindly provided and I was dismayed at the level of abstraction that is being aimed for with the Global Warming Potential and the Global Temperature Change Potential metrics, without having the essential and foundational CO2 sensitivity derivation explicitly provided. The reference they quote is
“Ramaswamy, V., et al., 2001: Radiative Forcing of Climate Change. Climate Change 2001: The Scientific Basis.
54 Contribution of Working Group I to the Third Assessment Report of the Intergovernmental Panel on Climate
55 Change, Cambridge University Press”’
Curious: I don’t think the Ramaswamy reference will be much help. It is Chapter 6 ‘Radiative forcing of climate change`of the 2001 IPCC WG1 report, so is itself only a secondary source. However it does define the term ‘radiative forcing’ and provide estimates of the effects of various forcing agents in W/m2. As far as I can see that chapter does not discuss the surface temperature response to radiative forcing.
There is a brief mention of the temperature response to a doubling of CO2 in IPCC AR3 WG1 (2001) Chapter 1 The climate system: an overview, p 93, where we read “…the radiative forcing corresponding to a doubling of the CO2 concentration would be 4 W/m2. To counteract this imbalance the temperature of the surface-troposphere system would have to increase by 1.2° C (with an accuracy of +/- 10%), in the absence of other changes.” No literature reference is given to back up this statement.
I think you are right to want to track down the derivation of this widely quoted value. Good luck!
Leonard: can you comment on this comment by ‘Phil.’ on WUWT, in Willis’ post on the ‘Mystery of Equation 8’?
“Agreed, the obvious reason for the pressure dependence of Earth and Venus is the pressure dependence of their atmospheric absorption/emission due to pressure broadening of the absorber/emitter spectral lines, far from falsifying the Greenhouse theory this result is what would be expected. The GHE is amplified by pressure (no surprise to anyone who understands the physics)!”
Coldish,
If the atmospheric mass is unchanged, and other greenhouse effects unchanged except for increasing CO2, this does effectively broaden the absorption lines, by virtue of the fact that the absorption lines are not zero width, but have a finite width with weak edges (due to doppler shift and other causes). There is a mean length to 50% absorption in the wavelength regions where the lines dominate, and the numbers of layers of this length to TOA is like a number of layers of 50% transmitting optical filters. Increasing concentration just effectively shortens each 50% path to encounter an equal number of absorbing molecules. the shorter path thus increases the number of effective filter layers. The increased number of absorbers thus absorbs more of the weak edges in the lines and can be described as filter absorption line broadening. This is the cause for an approximate log effect of increasing concentration. This is not pressure broadening, which is an effect to increase the width of the emission lines from surrounding gas interaction.
Increasing mass of the atmosphere, and thus increasing pressure, does cause a small amount of pressure broadening, but it is small compared to the effect of the increase in total atmosphere height, thus raising the average greenhouse gas altitudes. The total optical absorption may be unchanged, but the altitude to TOA increases, and lapse rate does the rest.
All real cases are very complex, due to unequal illumination, clouds, etc., but despite that both Venus and Earth closely follow the relationship of an increase in surface temperature over non-greenhouse calculation equal to the effective lapse rate times the average altitude to outgoing radiation.
While CO2 may have some part to play in Earth’s surface temperature, Leonard’s 1.2 Kelvin/doubling sounds rather high to me but my calculations are crude compared to his. The contribution due to clouds and water vapor is beyond me and probably beyond everyone else too.
With regard to the discussions on “Science of Doom” about the planet Venus, the situation is totally different. On Venus it is absurd to suggest that CO2 has any effect on surface temperature as a result of its absorption spectrum. Thanks to the 100% cloud coverage upward radiation cannot get directly into space until you get above the clouds.
While I respect Leonard Weinstein, I find Rodrigo Caballero (University College Dublin) very illuminating (hat tip to DeWitt Payne):
Click to access PhysMetLectNotes.pdf
While I have some quibbles with Nikolov & Zeller owing to their ignoring the effect of water vapor on planet Earth, they have done a great job predicting planetary surface temperatures in terms of atmospheric mass rather than chemical composition. Who needs RTEs except above the clouds?
Gallopingcamel,
The CO2, and other trace greenhouse gases are so dense on Venus beneath the clouds, that they are nearly as opaque to outgoing radiation as the clouds. In fact, only a small amount of solar radiation reaches the surface, and is required to be raised back up.Thus cloud blockage is not a big factor for radiation conduction. The main source of heating is direct absorption directly within the atmosphere (much of it in the thick cloud layer). The pressure is dropping above the clouds until outgoing radiation can leave. The temperature would probably be about the same without the clouds if the albedo would be the same, but the clouds mainly control the albedo. Without clouds, the solar energy to the ground would be a major location of absorbed energy, and mainly convection would raise the energy up. The only factors determining the surface excess temperature are the lapse rate (which is close to the adiabatic lapse rate) times the average outgoing altitude in the atmosphere (however the energy got into the atmosphere).
Jeff Condon,
“If I use a supercooled laser emitting at 15um will warm CO2 absorb some of the energy or will it send a telegram to the laser and request information on its temperature? I’m thinking telegram must be the answer.”
Exactly what PUMPS the CO2 in the atmosphere?? Where is the MAGNIFYING Glass or power supply to concentrate the energy?? Your response is silly
Leonard Weinstein said
January 24, 2012 at 11:36 pm
“Increasing mass of the atmosphere, and thus increasing pressure, does cause a small amount of pressure broadening, but it is small compared to the effect of the increase in total atmosphere height, thus raising the average greenhouse gas altitudes. The total optical absorption may be unchanged, but the altitude to TOA increases, and lapse rate does the rest.”
WRONG! !You’re ignoring Dalton’s law. The partial pressure of CO2 at any height, 100 meters, 1 kilometer, 2 kilometers would be the same as it is now if the amount of CO2 stayed the same, regardless of whether we have 1 or 2 or 3 atmospheres of pressure, when the rest of the atmosphere consists of non greenhouse gases.
Gallopingcamel,
I looked at Rodrego’s notes. This is a super reference and lays out a lot of useful material. If you examine his comments on greenhouse effect, he totally agrees with my basic contentions. A main point to consider, especially on Venus and to a slightly lesser effect on Earth, is that radiation heat transfer is relatively small. The greater the amount of greenhouse gas, the smaller the relative radiative to convective heat transfer ratio. Convective heat transfer dominates, so optically opaque clouds do not make that big a difference as long as albedo is the same.
Alan D McIntire,
You are correct. I need to think about that more. In a real case, increasing atmospheric mass would result in an increase in water vapor content and raised clouds, so they would be the main source of increased temperature at fixed total CO2 level. Thanks for catching that.
Jeff (1st comment) said: “The greenhouse effect is a delay in energy release from earth to space. If surface heat has any increased delay in reaching emission altitude, temps go up.”
Not so. As Prof Claes Johnson has shown, no radiation from a (cooler) atmosphere can add thermal energy to a (warmer) surface. This is further confirmed by others – see links on my site at http://climate-change-theory.com . Surface “heat” (“thermal energy” I guess you mean) does not have to reach emission altitude as it can transfer to the atmosphere by diffusion (molecular collision) and evaporation, the proportion of such transfer easily increasing if necessary, and probably over 50% anyway. Who cares anyway if the atmosphere 1,000 metres or more above our heads is a little less cold? The surface and the air we breathe will not, and cannot be warmed as a result. Any so-called back radiation will not be absorbed by the surface and converted to thermal energy because its frequency is below the cut-off frequency, as determined by Wien’s Displacement Law – see Claes Johnson “Computational Blackbody Radiation.”
The final plot at the foot of my Home page shows the rate of increase of SST has actually been declining slightly since 1880, and projections would indicate only about 0.5 deg.C increase by 2100, this being entirely due to a 900 to 1100 year natural cycle which may well pass a maximum in the next 50 to 200 years. There is absolutely no evidence in that plot of any effect of carbon dioxide, and such would be a physical impossibility anyway.
Dr Weinstein:
This article is yet another one perpetrating that -18 deg.C figure (calculated using S-B and average radiative flux over 24 hours) supposedly then resulting in “the 33 C estimated total greenhouse effect with all gases, clouds, and aerosols.”
The only true “blackbody” is the complete “Earth+atmosphere” system because it is in fact surrounded by space, as is required for the S-B Law to apply. In other words, a blackbody must be totally insulated from its surrounds so that no thermal energy can enter or exit via any means other than radiation.
The internal surface/atmosphere interface is nothing remotely like a perfect blackbody. Some thermal energy “sinks” deeper into the oceans and below the surface every sunny morning, and some exits at night, or maybe not all of it until the next local winter. Probably over 50% of the remaining thermal energy exits by evaporation and diffusion (molecular collision) followed by convection once the energy is into the first (say) 1mm of the atmosphere. So there is less than half (maybe less than 30%) of the energy left to radiate. In fact, in calm conditions, there is very close thermal equilibrium between that 1mm of air and the surface, as anyone can observe. Hence S-B gives a result which is very close to zero radiative flux, and would even be zero when temperatures are equal.
So, all your calculations of that -18 deg.C figure are garbage. No doubt you’ll find that sort of temperature somewhere up in the atmosphere as a kind of weighted mean – but what would that tell you anyway about surface temperatures? Furthermore, you would get a much lower figure (like about -100 deg.C) if you did separate calculations for day and night and then averaged the results. The temperature gradient is mostly caused by pressure having an effect on the rate at which warm air rises by convection.
The truth of the matter is that carbon dioxide has absolutely no effect and the so-called atmospheric greenhouse effect caused by so-called backradiation is a physical impossibility as I explain with all due evidence on my site http://climate-change-theory.com – see also my post above.
Dave C.;
You might as well give up carrying the torch for Claes. This is the only response you’re ever going to get. His conflation of mass absorption and quantum absorption has failed to persuade anyone (except you and a very few others). Extraordinary claims, etc.
Brian H –
Except that physical evidence supports Claes computations, which no one has been able to prove wrong. When you warm an emitter surrounded by a gas, and observe the spectrum you will see no absorption until the emitter is warmer than the temperature of the gas. Explain that, Brian H, by any other physics than that which Professor Claes Johnson (who I suspect knows somewhat more than yourself about all this) has derived. The same result has also be deduced by another author too now – see the first para at http://climate-change-theory.com and, furthermore, the temperature gradient plot at the foot of my home page confirms absolutely no anthropogenic effect any time since 1880. Read also my post to Dr Weinstein above which was posted simultaneously with yours.
Anyone:
Meanwhile I await anyone’s explanation as to why the above gas does not absorb when it is cooler than the emitter.
Leonard – thank you for your responses and for the original post. I’ll also look at Caballero’s paper. I can’t help thinking that someone ought by now to have published some semi-quantitative estimates of the atmospheric mass effect you so clearly (but qualitatively) describe. Or are we (you) at the threshold of new knowledge?
Brian H
I simulated a one slab atmosphere with the one-stream approach and with a two-steam approach with the so-called back-radiation.
I found that the two-stream approach gives an absorption which is twice as big than the the correct absorption in the one-stream approach.
I conclude that the two-stream approach with the so-called back-radiation should be avoided, because the calculated absorption is a factor 2 too big.
Click to access IR-absorption.pdf
Well, there’s double counting of the energy buried somewhere, in there, for my money. Every emission cools the source as much as it warms the target on absorption. It’s a mobile zero! 😉
63 – Doug – I think you have a typo:
“When you warm an emitter surrounded by a gas, and observe the spectrum you will see no absorption until the emitter is warmer than the temperature of the gas.”
compared to
“Meanwhile I await anyone’s explanation as to why the above gas does not absorb when it is cooler than the emitter.”
??
So, just to get things clear in my mind (re LW at 54 and 59, Alan M at 57): adding N2, O2 to the atmosphere will not result in pressure broadening of CO2, but the mean emission height will be raised. Is that correct?
Coldish,
My basic initial writeup has been falsified by Alan M (57). However, this writeup made unrealistic assumptions about all other factors being unchanged. They would change (especially water vapor total content). It would be the water vapor that would have to increase for a doubled atmosphere that would be the cause of increased greenhouse gas effects by raising the mean emission height. Pressure broadening is not a major factor for Earth. It is a bigger factor for Venus. That does not change the final fact that the greenhouse effect on temperature is the combination of mean emission height and lapse rate. It is only how you get the mean emission height that is wrong in my post.
“My basic initial writeup has been falsified by Alan M (57).”
I think you are too quick to declare Alan’s statement correct. Because of the increased thickness of the case 2 atmosphere, the basic volume assumptions of Dalton’s law In Alan’s example need to be adjusted.
If you consider the partial pressure of CO2 at ground level, it would be exactly the same in case 1 and 2. But in a thickened case 2 atmosphere, it is simple to imagine a point near the top where a well mixed gas would propel CO2 molecules above the maximum physical altitude of the case 1 atmosphere. We have more total height and the same amount of CO2 so emission altitude does increase and the partial pressure does not match at each physical altitude.
“However, it is clear that a thicker atmosphere, even without increasing total greenhouse gases over the thinner case, would have increased surface temperature due to the increased average altitude of outgoing radiation.”
To be clear, the increased average emission altitude doesn’t necessarily prove out this statement for me. For case 2, there is greater mean path between emission/absorption, the absorbing molecules exist in a higher density environment on average and are more likely to conduct energy than in Case 1. They are also more likely to be energized by conduction so does that make them less receptive to IR? I get all confused at that point.
Curious,
The answer is more simple of course. Doug is wrong. Another rehash of the silly backradiation misunderstanding. Every time I read it, the person on the other end has little basic understanding of physics. Doug has added a new one where a cold emission cannot be absorbed by a warmer gas. Very interesting.
If I use a supercooled laser emitting at 15um will warm CO2 absorb some of the energy or will it send a telegram to the laser and request information on its temperature? I’m thinking telegram must be the answer.
Photons are photons and warm or cold emitters of photons are clueless as to the other emitters condition. More energy trapped/delayed is more heat period. The examples given at his link are silly in the extreme.
63 and 72 – I think it would be worth Doug very carefully describing the experimental set up he has in mind (or has observed) for his assertion 1):
1) “When you warm an emitter surrounded by a gas, and observe the spectrum you will see no absorption until the emitter is warmer than the temperature of the gas.”
2) “Meanwhile I await anyone’s explanation as to why the above gas does not absorb when it is cooler than the emitter.”
In 67 I was highlighting what I read as a turnaround in claim?
It might be my reading, but it in sentence one Doug appears to be claiming that when the emitter is warmer than the gas, absorbtion is observed – ie the gas is cooler than the emitter and it absorbs. In sentence two he appears to be claiming when the gas is cooler than the emitter the gas it does NOT absorb. Maybe it is just in the way he has chosen to pose his question but I would like to be clear.
“Coldish said
January 25, 2012 at 8:15 am
So, just to get things clear in my mind (re LW at 54 and 59, Alan M at 57): adding N2, O2 to the atmosphere will not result in pressure broadening of CO2, but the mean emission height will be raised. Is that correct?”
No, I think that’s bakcwards. The mean emission height will NOT change (look up Dalton’s Law), but there MAY be pressure broadening of the CO2 lines.
Jumping to a slightly different topic, note that O2 makes up about
21 % of our atmosphere, and has been relatively stable for millions of years, yet O2 is very unstable and wouldn’t exist at all without life. I suspcet that besides controlling the amount of O2, life has had a significant part in controlling the amount of CO2 in the atmosphere.
My 73 – Apologies – editing error in my second last sentence:
“In sentence two he appears to be claiming when the gas is cooler than the emitter the gas it does NOT absorb.”
should read:
“In sentence two he appears to be claiming when the gas is cooler than the emitter it does NOT absorb.”
@Jeff Condon 72
I have done a numerical experiment.
A one-slab atmosphere, modelled with a one-way heat flow as well as with a two-stream heat flow.
It turned out that the temperature distribution is equal for the two cases, but the absorption of the slab and re-emission – because we are in steady state conditions –
is in the one way heat flow correct but in the two-steamheat flow doubled.
Do not talk about sending telegrams.
Try to convince me that the one way heat flow model, with the correct absorption, is not correct.
While I say that the two-stream heat flow with an absorption twice as big giving rise to back-radiation, is not correct.
In the same paper I compare also a stack of completely opaque layers with both the one-way heat flow and the two-stream heat flow.
The latter gives in a synptomatic way huge absorptions.
My conclusion is that the two-stream modelling with back-radiation should be avoided, because the model shows spurious absorption.
Click to access IR-absorption.pdf
It is the merit of Claes Johnson to have given a possible explanation of the fact that there is only a one-way traffic of heat radiation.
Look also in the paper of Matthias Kleespies, giving an overview of radiation theory since 200 years.
Click to access History-of-Radiation.pdf
My numerical experimentation -algebraical experimentation because you do not need a computer for it- is a conformation of the one-way traffic of heat radiation.
You are free to imagine a “telegram” exchange.
Alan #74,
“The mean emission height will NOT change”
Hummm… I see your point. When the ratio of P1/P2 in Leonard’s equation reaches the same value for the higher total mass atmosphere as for the lower one (the same height above the surface), the total mass of CO2 above that altitude will be identical to the lower pressure atmosphere…. so the infrared transparency to space will be identical at all altitudes in either atmosphere, and the effective emission height will not change so long as the total mass of CO2 is constant.. But pressure broadening of CO2 absorption bands would appear to be significant (http://www.heliosat3.de/e-learning/remote-sensing/Lec7.pdf), so there will almost certainly have to be at least some increase in emission height and surface temperature for the higher total atmospheric mass. Calculating exactly how much looks complicated.
WRT O2/CO2. Yes, the level of CO2 in the atmosphere would appear to have a fairly strong long term negative feed-back mechanism via plants and photosynthetic algae that end up sequestering carbon by fossilization (coal, petroleum, natural gas). In the shorter term, net accumulation in short-term carbon stores (living and dead plant matter, peat bog accumulation, soil carbon) also appears significant, but probably less important than ocean absorption of CO2.
Jeff and Steve,
You are both right that there would be some increase in outgoing height, due to pressure broadening, and due to deviation from a simple log pressure variation at the very top region of the atmosphere. However these are not the issue I initially brought up, so I agreed the basic argument was flawed.
Brian H:
You said: “Every emission cools the source as much as it warms the target on absorption.”
This is now proven incorrect by Claes Johnson and JWR in his linked paper above. For more information study my page http://climate-change-theory.com/RadiationAbsorption.html which contains links to “Computational Blackbody Radiation” etc.
In due course, following the publication of my book on this, there will be a substantial reward offered to any university Physics Dept which can empirically prove Claes wrong regarding the fact that no absorption resulting in conversion to thermal energy occurs when the receiving body has a cut-off frequency significantly below the frequency of the emission being received.
Jeff Condon:
You wrote: “… warm or cold emitters of photons are clueless as to the other emitters condition.”
This is incorrect. Blackbodies detect the frequency of the emission being received.
You have obviously not read my page and the linked papers http://climate-change-theory.com/RadiationAbsorption.html
I really don’t have time to copy it all here and don’t know if you have a genuine interest in seeking the truth of the matter, but some others might …
Briefly: the example of the warmer gas not absorbing any emission from a cooler source (as proven by spectroscopy – so you cannot argue against this) is very relevant, because the warmer surface of the Earth does not absorb any emission from the cooler source, namely the atmosphere. By “absorb” I mean convert the energy in the radiation to thermal energy. It may either immediately re-emit it with exactly the same frequency spectrum and intensity, or merely deflect it. Either way, the net effect is equivalent to reflection except for the angles involved.
The “information” is contained in the frequency which, if you know tertiary physics (Wien’s Displacement Law) is proportional to absolute temperature. If the frequency received is significantly lower than the frequency being emitted then there is no conversion to thermal energy. If you have a really good knowledge of computational physics then you should be able to follow the proof in the linked paper on my site and also that just written by JWR – see his posts above. If not, then you will just have to consider the empirical evidence that this is the case.
Anyone:
See my posts above first.
I challenge anyone to show me any experiment which demonstrates that “backradiation” at night has any effect on the surface, either warming it or slowing the rate of cooling. This basic glossed-over assumption implicit in the official IPCC explanation of the “greenhouse effect” (see their website) has never been proven empirically – because it can’t be, because it is a physical impossibility.
I just put forward this simple challenge and, until someone proves me (and Professor Claes Johnson) wrong on this, I rest my case.
Doug Cotton,
Hans Schreuder has some observations which would appear to support your position:
http://climaterealists.com/index.php?id=9004
Backradiation doesn’t manage to melt frost even when the air above and ground below is warm enough!!!
“Try to convince me that the one way heat flow model, with the correct absorption, is not correct.”
I don’t understand one way radiant heat flow in your model.
Two equal surfaces at 30 and 40C in a vacuum with energized IR absorbing molecules will emit their photons. The photons go to the opposite surface and are absorbed. The photons do not care if the surface they impinge on is hotter or colder – excepting extreme cases. They energize the surface they hit and don’t care what the temperature of the object was which generated them.
If the surface they hit is warmer and therefore emitting more photons of IR, so what the photon still hit right? The net flow addition guarantees that energy is traveling from hot to cold, but the energy is traveling in both directions. This is not a 2nd law violation because NET power is still in the correct orientation.
—
Doug,
See 82.
79 Doug – please can you describe the experimental setup which you envisage to prove/disprove your views? Please can you give a definition of “significantly below”? I looked at your linked page but didn’t find it – if it is there please point me to a reference. Thanks.
Obvious correction to final paragraph to Jeff Condon:in my post above …
In due course, following the publication of my book on this, there will be a substantial reward offered to any university Physics Dept which can empirically prove Professor Claes Johnson wrong regarding the fact that no absorption resulting in conversion to thermal energy occurs when the receiving body has a cut-off frequency significantly ABOVE the frequency of the emission being received.
Jeff
You are still wrong in saying “The photons do not care if the surface they impinge on is hotter or colder”
I suggest you read the paper by Claes and my summary thereof on the “Radiation” page on my site http://climate-change-theory.com
Please understand that I do not have the time to keep posting brief explanations here just because you have not bothered to study what has been explained in much greater detail there.
PS Jeff (and others)
When you have studied what I have referred you to you will then be able to come back here and explain why, as shown empirically by spectroscopy, a gas will not absorb spontaneous emission from a source which is significantly cooler than itself, but will absorb when the source becomes warmer than itself.
Likewise, the Earth’s surface does absorb high energy radiation from the Sun (and converts such energy to thermal energy) but it does not absorb and convert to thermal energy any radiation from significantly cooler layers of the atmosphere. The frequency of the radiation (and hence its energy) has to be above the cut-off frequency for the surface, such cut-off frequency being determined by Wien’s Displacement Law. This is now proven computationally and, at least for gases, empirically.
Show me an experiment which shows otherwise! For example, two metal plates on the ground at night with one shielded from all that backradiation – compare the temperatures. Why hasn’t the IPCC arranged such an experiment? They probably have, but it didn’t support their hoax.
Jeff,
… Photons are photons and warm or cold emitters of photons are clueless as to the other emitters condition. More energy trapped/delayed is more heat period. The examples given at his link are silly in the extreme. …
Just as an example that physics is about as settled a science as climatology, Dr. Tom Phipps, Jr., has advocated the “telegram” approach as more consistent with quantum mechanics. He goes so far as to suggest that no photon is emitted that is not guaranteed a receiver. A Feynman diagram would require a time-reversed particle that communicates between receiver and emitter.
to Jeff Condon 82
You are not coming away with such an answer.
Look in the paper fig1a and fig 1b.
In 1a the correct SB from warm to cold is implemented.
In 1b your claims, what I call Prevost-Fourier type of source terms, the the slab is emitting back from warm to cold.
Look to the results.
What is the slab absorbing and emitting immediately?
I have given you the answers for both implementations, and I conclude that the approach which you continue to defend gives the wrong value for the absorption.
If you want to continue the discussion, study the two figures 1a and 1b.
I hope that at least those who follow the blog, see the algebraical proof that two-stream heat flow with therby
so-called back-radiation gives spurious absorption.
I’m a lowly engineer, so please be patient. As I understand things, two bodies having different “temperatures” are generally adjacent, and each emits photons having energies generally corresponding to the respective bodies’ temperatures. Each incident photon may be absorbed, transmitted or reflected by the receiving body.
If I understand you correctly, photons from the cooler body are either reflected or transmitted by the warmer one, in all cases. This suggests to me that the impinging photons must not have sufficient energy to raise the energy level of ANY of the recipient body’s molecules by a “quanta”, suggesting that ALL of said body’s molecules (ok, atoms?) are at or above the energy of the impinging, lower energy photon. If my memory hasn’t suffered too much in the last 50 years, I seem to recall that we’re talking PROBABILITIES of any given energy state, not absolutes. I don’t buy it.
On the other hand, in the wavelength world, I interpret your words as suggesting that at wavelengths longer than infrared, total reflection will take place,
Can EM energy having wavelengths well below those of IR cause a body to heat up? Diathermy? Microwave ovens? What am I missing?
“has to be above the cut-off frequency for the surface,”
Doug, the cutoff ‘frequency’ of CO2 is very low and not related to its temp except through broadening of the absorption spectra. I would like to see your spectroscopy reference because it doesn’t make any sense.
I read your link earlier and it has mistakes which make it difficult to follow.
Kuhnkat,
I don’t know what you mean. Are you asking me to explain the concept of back-radiation or the basic greenhouse effect? My answer was supposed to be silly because the concept presented is silly.
Mark F,
Yup. That is exactly right.
#90 They go through equations come to different answers with two different models and conclude one model is right and the other wrong having given no evidence either way. This article is by no means statistical proof of anything.
Jeff Condon 95
Again, you are a bad loser.
It is clearly said and shown that both models give the same temperature distribution.
But when you look in more detail it is seen that the one-way heat flow model gives a coherent absorption,
and the two-stream heat model gives an absorption twice as high.
It has nothing to do with ststistics. Do not make your case more rediculous.
96 typing errors
It has nothing to do with statistics. Do not make your case more ridiculous.
JWR,
I am not losing, the argument is too small to care about but you are not reading carefully. They point out that back radiation increases the insulation factor in their flow – that is all. Then they jump to conclude it can’t be right. Unscientifically, I saw no proof or even any evidence presented as to why back radiation cannot be right. You said it was a statistical proof – it is not.
There are so many devices which wouldn’t work if the claims made at the end of this thread were true. Imagine a thermal imaging camera viewing ice. The silly claims of that paper would be proof that the image sensor would see only black unless it were cooler than the source. There are thousands of operational devices which prove this wrong.
Here is one which doesn’t incorporate a cooled sensor. Ask yourself how come it works….
Here are the specs for the camera. Note that it can somehow detect temperatures to -20C.
http://www.infraredcamerasinc.com/Thermal-Cameras/Fix-Mounted-Thermal-Cameras/ICI7320_Pro_fix-mounted_thermal_camera.html
Jeff,
I bet you faked that thermal imaging clip just to fool people who know back radiation is impossible. 😉
Convincing people that all objects radiate continuously is only possible if they are inclined to improve their understanding of how the world works. Many show clearly they are not so inclined; to continue trying is then just tilting at windmills.
Steve,
I never have learned that lesson but assuming that we have good people, we may see a change of thinking. The problem is so simple to me. Photons emit from warm sources. They do not interact in any manner which changes their direction, and then they hit something else. Cold or warm, they still hit. They don’t turn around, in Earth temp examples there is no saturation of absorbers.
As you know, the second law of thermodynamics in particles is based on probability across a LOT of interactions. On an individual basis, temps can and do move in the wrong direction. It ain’t that kind of law.
Anyway, I’ve done it again and derailed an otherwise interesting concept.
JWR:
’tis better to be a bad loser, yet win—than lose, yet not know that you lost and really suck at physics to boot.
Jeff,
I understand the desire to help people get past their limitations of understanding (heck I worked as a technical consultant for many years!). But connecting thermodynamics and radiative transfer to the underlying statistics of large ensembles (which is what you have to do to really understand what is going on) may be out or reach for many, at least until they study some statistical mechanics. All hail Boltzman!
You were not the one who derailed the thread. Besides, I think Leonard had already accepted that the whole approach was mistaken… no warming from a thicker atmosphere with the same total weight of CO2, except from secondary effects like pressure broadening of absorption bands.
Thanks Steve. I am still not certain about the net effect of non-absorbing gasses but am convinced that it shouldn’t be very big.
Jeff, Re: your thermal imaging camera clip.
John N-G had an interesting post some time ago on what makes a scientific theory believable (http://blog.chron.com/climateabyss/2011/11/the-crackpot-einstein-scale/).
John’s conclusion: scientific theory that is relied upon for successful technology (like your infrared camera) is the most certain of science…. everything else in science is less certain. An interesting analysis I think.
It looks like the critics have vanished. I hope they come around again.
Physics is an unforgiving beast.
I read John’s article and it was well done.
Jeff –
You write: ” the cutoff ‘frequency’ of CO2 is very low and not related to its temp except through broadening of the absorption spectra”
This would be totally contrary to Wien’s Displacement Law.
Do I have to provide you even with links to Wikipedia / Wien’s Displacement Law ?
When are you going to read my Radiation page and Prof Claes Johnson’s paper?
When are you going to explain why a gas does not absorb spontaneous (blackbody) radiation from an emitter which is cooler than itself?
When are you going to explain why dew can remain frozen all day long in the shade (on ground that is above freezing point and in air that is likewise) without being melted by all that back radiation which shows so clearly in, for example, Trenberth’s energy diagrams.
When are you going to explain how the Earth cooled for 30 years from mid 1938 to mid 1968 despite increasing carbon dioxide levels and despite a long-term (since Little Ice Age) natural rising trend of (currently) 0.06 deg.C per decade?
I’m not going to waste time answering your questions if you don’t display some degree of willingness to read (and learn) with an open mind. Nor am I interested in reading papers which ultimately depend on false assumptions about backradiation or whatever.
Doug,
“This would be totally contrary to Wien’s Displacement Law.”
No it isn’t. WTF???
“When are you going to read my Radiation page and Prof Claes Johnson’s paper?”
Did.
“When are you going to explain how the Earth cooled for 30 years from mid 1938 to mid 1968 despite increasing carbon dioxide levels and despite a long-term (since Little Ice Age) natural rising trend of (currently) 0.06 deg.C per decade? ”
Did you ask me to do that before? Sorry, I missed it. Natural variance is higher than Climate science approves of.
“I’m not going to waste time answering your questions if you don’t display some degree of willingness to read (and learn) with an open mind. Nor am I interested in reading papers which ultimately depend on false assumptions about backradiation or whatever.”
The great thing about the internet is that you never know to whom you are talking. I take it that you won’t be willing to provide the single reference I requested? I rarely request references here. Perhaps 10 in 3 years and 40K comments.
Please Dr. Doug, don’t waste your valuable time providing evidence for your incredibly extreme claims.
Doug:
LOL.
You don’t have an open mind, nor that you know anything that you could teach us.
107 Doug,
No need to read any papers, just look at the IR camera clip. Just about all objects radiate continuously in the infrared, as the IR camera clip shows (perfect reflectors don’t radiate, but that is a whole other issue). Few who know anything about science are going to bother with reading a paper which claims to refute the operation of well known, widely used technology. Frost can remain in shaded areas (in winter, I presume) during the day, even if the air goes somewhat above freezing, because those areas are losing sufficient heat via infrared emission that they can’t get above freezing with the energy they absorb from all sources (including conduction, radiation, and convection). That does not mean that there is no back radiation from the atmosphere, only that there is enough net loss of radiation (total absorbed less total emitted) to keep the surface from warming to the melting point.
Jeff Condon
“I read your link earlier and it has mistakes which make it difficult to follow.”
Which link and which :”mistakes” ?
Just for your information, not only do I have a degree in Physics, but I have engaged in ongoing study and tertiary tutoring of Maths and Physics for over 4 decades, and written comprehensive Maths tutoring software for all secondary levels.. I have also done other post graduate courses and studied climate science privately.
Steve:
Personally I think classic thermodynamics is a waste of time, that people should start with the statistical theory, and learn the generalizations of classic thermodynamics only in the context of what this means on a microscopic scale.
IMO, if you don’t understand the connection, you really don’t understand the modern theory of thermodynamics.
Thus if you don’t understand that radiative transfer involves the emission of a thermal photon from one atom or molecule and later reabsorption by another (which may or may not have a higher kinetic energy than the first), but are still willing to prattle on about e.g. Clausius’s statement of the Second Law, then you’ve show me you don’t know anything except a bunch of mumbledigoop that you read in a paper some place.
Jeff Condon, Carrick, Steve Fitzpatrick
We are living on different continents, and I went to sleep yesterday night.
It is funny to see that IR-camera looking to the ice cubes in a cup of coffee.
But nobody said that IR-cameras do not work.
What I did was to apply SB in two different ways between two plates with or without
holes defined by a cross-section f. (f=1 fully opaque, f=0 fully parent)
Plate 1 is loaded by a heat flux q. The plates are supposed “black”.
fig 1a) Applying SB on plate 1 and plate 2 as a pair.The emission by plate 1 is already taking
into account the presence of plate2, and no heat flux from plate 2 to plate 1.
There is no back-radiation.
Of course plate 2 is emitting to outer space at zero K.
fig 1b) Plate 1 emits if it were looking to zero K.
The net flux is obtained by subtracting the flux from plate 2
in the direction of plate 1 and defined if plate 2 were looking to zero K.
That is the so-called back-radiation.
Plate 2 is also emitting on the other side a flux to outer space at zero K.
The resulting temperatures are the same for the two cases and for all cross-sections f.
One could conclude that the two implementations are equivalent!
We look to the heat which is absorbed by plate 2 and re-emitted inmediately
since we assume steady state conditions.
In the paper is shown that in the implementation A plate 2 absorbs fq/(2-f) and in
implementation B plate 2 absorbs 2fq/(2-f), twice as big.
For a completely opaque plate 2 with f=1,
plate 2 absorbs q in case A and 2q in case B.
And the value q for implementation A is correct.
Up to now nobody has claimed that it is not correct.
There remains only one conclusion:
The implementation B is not correct.
And the only difference between A and B is that what is called
back-radiation is include in the fux emitted by plate !.
Doug:
That’s really scary.
If you can’t spot the problems with Claes Johnson’s paper, either you are lying or you are a poser.
“Just for your information, not only do I have a degree in Physics, but I have engaged in ongoing study and tertiary tutoring of Maths and Physics for over 4 decades, and written comprehensive Maths tutoring software for all secondary levels.. I have also done other post graduate courses and studied climate science privately.”
If you want to play resume, you have come to a bad spot.
Lets say that the crowd is something bigger here and you aren’t too old to learn.Degree’s don’t prove correctness and that is a silly game to play here. Were Galleleo’s credentials the proof over everyone else?
Jeff:
I don’t need a link. What he actually knows is more important. And he left some pretty good rat droppings here.
My guess is high school teacher in physical sciences.
Wow Carrick!! Where did that compilation come from?
Sorry Leonard for your entertaining thread.
Doug- after that pile of complete bovine scatology, you may reply to me with a link only. You have stated your opinion with a fervence that any muslim would take pride in. Id Out.
Jeff It’s from his web site.
I couldn’t get past the first paragraph that I linked above without guffawing in laughter.
“Professor Johnson’s mathematical analysis satisfies (and actually provides a theoretical proof for) the various laws of radiation, and also explains the ultra violet catastrophe without having to resort to photons.”
It just gets better with each reading. Jeez!!!
A physical experiment that nearly everyone can do:
Stand in front of the refrigerator with your eyes closed about 3 ft away.
Put on headphones and play Pink Floyd the Wall at top volume – take care to not hurt your ears.
Have someone else hold the door to the top section of the “cold box” open by only a couple of inches at the widest point – nearly closed.
Have your special someone open the top door of the cold box slowly.
Note any change in temperature of your facial skin.
Close the door.
—-
What happened?
—
Extra credit – Does the radiant effect of the moon heat or cool the Earth?
Jeff Condon,
“Does the radiant effect of the moon heat or cool the Earth?”
when the side of the moon that is facing the earth is hotter than the earth or when the side of the moon facing the earth is colder than the earth?
HAHAHAHAHAHAHAHAHAHAHAHAHA
Leonard @56 & 58
When it comes to Venus you are right about the lower atmosphere being opaque to radiation around 15 microns and Caballero discusses this on page 133. In particular you will note that when an atmosphere is opaque to upward IR radiation the effective radiating surface is raised.
You don’t need RTEs if they make only minor contributions to moving the heat around. We seem to agree that the abiabatic lapse rate can explain the high temperature on the surface of Venus and it would make little difference if the CO2 was replaced by Nitrogen or even Helium. It is the mass of the atmoshere and the gravitational field that creates the temperature gradient in the lower atmosphere. Hence my statement “Who needs RTEs”.
When it comes to planet Earth the problem is more complex owing to the massive amounts of water in the lower atmosphere. Cloud cover is not 100% so IR radiation can be lost into space through direct radiation. While I am skeptical about your 1.2 Kelvin per doubling of CO2 I do concede that the radiative properties of CO2 could have some effect on Earth even though they have none on Venus until you get above the clouds.
I don’t think we have a “failure to agree” except on the magnitude of the contribution of CO2 to Earth’s surface temperature. While I can’t refute your calculations, the ice core records show that atmospheric CO2 concentrations are driven by global temperatures. In the 9th grade we learn that the solubility of gases in water falls with rising temperatures so you don’t need a supercomputer to understand why the oceans exude CO2 when temperatures rise.
Re Doug Cotton # 107
“When are you going to explain why a gas does not absorb spontaneous (blackbody) radiation from an emitter which is cooler than itself?”
Actually it does absorb radiation from an emitter that is cooler. The amount it absorbs is determined by the probability that there will be a given number of photons with the specific energy to match a specific absorbion line in the absorbing gas. If the emitting gas is cooler than the absorber there will only be a small number, and if it is greatly cooler the number will be tiny, but NOT zero. It will be far less than the number of photons absorbed by the cold gas that are supplied by the hot gas. That is a basic principle of quantum mechanics. As has been pointed out by others it is not the same as the NET transfer that occurs which is simply the difference between the photon transfer in one direction minus the other direction. On a macro scale the NET transfer of photons is from hot to cold. The number of photons that are able to do this “magic” is given by the Boltzman distribution.
Damn. My microwave oven has stopped working!
Apparently some Swedish Professor has convinced the food that it can only absorb radiation from a source hotter than it is, so as soon as the food gets slightly warmer than the machine, it sends all the microwaves back to where they came from and just sits there luke warm.
Very annoying.
123 🙂
Steveta_uk
Microwaves (in radio frequency bands) are a red herring here because they are not emitted spontaneously in blackbody emission. For a start microwaves only heat water molecules in the food, not the food itself, and only to the boiling point of water. They are a very special case of radiation specifically generated by electric power of course, not by spontaneous conversion of thermal energy. So too are laser beams which are induced emission, not spontaneous emission.
Now consider the real world cases I have suggested, such as why frost in shade on the ground is not melted by all that backradiation hitting it all day long. If it can’t melt frost, then it can’t warm the oceans. And why doesn’t a gas absorb radiation from a cooler emitter? Only Prof Johnson’s conclusion can explain both, I suggest.
125 Doug – I thought you might raise that as an objection to Steveta_uk. So please can you respond to my 73 and 84?
Please take the time to put together a consistent paragraph on exactly what your experimental set up would be (or is) wrt to your claims on gas emission/absorbtion. Thanks.
Terry is getting close, for at least he seems to realize that some of the photons will not get absorbed. If you had read Prof Johnson’s paper you would note that he refers to the distribution being strongly attenuated, as it indeed is both above and below the peak. So I usually say that the temperatures have to be significantly different to ensure negligible absorption. But indeed they usually are significantly different in most parts of the troposphere which is about -20 deg.C at 14,000 feet and goes down to below -60 deg.C at the tropopause. So there is no significant absorption of backradiation
As JWR has explained above, you get spurious results if you treat it all as two-way radiation. I really only believe a small amount of radiation from hot spots in the atmosphere actually goes towards the surface, nothing like 100 W/m^2.
Suppose you have a blackbody radiating away and you determine its temperature with an IR thermometer. Now reflect most of that radiation back to it. Does that stop it cooling because net radiation is zero, or does it then radiate nearly twice as much? If the latter, the thermometer would say it is warmer. Is it really? The IPCC effectively says it almost stops it cooling. But does it? The answer, according to Johnson, would be neither because it does not absorb the backradiation and convert it to thermal energy.
Doug – I found Terry’s paragraph informative too, especially in the context of Carrick’s comments wrt statistical thermo. My thermo understanding is from a mech eng. perspective and it made me realise I should get a stat. thermo. text and have a read.
However at the moment I am interested to try to bottom out the experiment which would prove your claim:
“When you warm an emitter surrounded by a gas, and observe the spectrum you will see no absorption until the emitter is warmer than the temperature of the gas.”
AFAICT the experiment you describe above (127) is not quite analogous:
“Suppose you have a blackbody radiating away and you determine its temperature with an IR thermometer. Now reflect most of that radiation back to it. Does that stop it cooling because net radiation is zero, or does it then radiate nearly twice as much? If the latter, the thermometer would say it is warmer. Is it really? The IPCC effectively says it almost stops it cooling. But does it? The answer, according to Johnson, would be neither because it does not absorb the backradiation and convert it to thermal energy.”
This seems more akin to me to a demonstration of the power of insulation. My understanding is emission is a rate of transmission hence in this example I would expect your black body to reduce temp at a slower rate when you reflect some of its radiation back to it. Or if it being held at a constant temp. by a power source I would expect it to require less power to maintain the same temp.
#120
Oops. Don’t forget that the moon is blocking a moon sized chunk of 4 degree Kelvin outer space. Want to try again?
Jeff,
“Oops. Don’t forget that the moon is blocking a moon sized chunk of 4 degree Kelvin outer space. Want to try again?”
Please compute the difference between the difference of the temperature of the earth between the moon blocking that large 4k “radiative surface” and it not being blocked. When you have done that please show me some experimental evidence that actually shows the difference you postulate!! HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA
Jeff – I’m sure you will find my links to Professor Claes Johnson’s paper “Computational Blackbody Radiation” on the Radiation page of my site http://climate-change-theory.com You can always chat with him on his blogs.
Best you read it in full I suggest because you obviously have a lot to learn and a lot of preconceived ideas to shed. His understanding approach may well help you better than my brief summary. Perhaps when you have over 40 years of physics behind you (as I have) you will be more open minded.
Regarding those photons, he explains why it is not necessary to postulate a particle nature for photons – everything can be explained by assuming EM radiation has wave nature only. Einstein was always uneasy about those particles.
Jeff
The Moon has a “day” about 27 Earth days long and its surface warms up over those days to a maximum around 120 deg.C. Certainly at full Moon (and probably for a few days either side of full Moon) I would say yes its radiation would be capable of warming the Earth’s surface to some small extent because the frequencies would be above Earth’s surface cut-off frequencies, especially at night. The extent of warming may be so small it would be hard to measure, but in theory there would be some.
Curious and others
In writing “… I would expect your black body to reduce temp at a slower rate when you reflect some of its radiation back to it… ” you show me you still don’t understand Johnson’s result. Please at least study my summary on my Radiation page or read his whole paper. You’ll also find posts of mine on his blogs which may help. The reflected radiation does not have frequencies greater than the original radiation and so it will have no effect. You are still thinking like Trenberth et al, not like Johnson. It’s your choice whether you accept Trenberth’s pseudophysics or Johnson’s real physics which is compatible with the real world – like that frost on the ground. Trenberth’s backradiation would melt it for sure – at least he would think so.
132 – Doug – I certainly don’t claim complete knowledge. However from our exchange I’ve concluded you don’t have an expermiment in mind or in practice wrt:
“When you warm an emitter surrounded by a gas, and observe the spectrum you will see no absorption until the emitter is warmer than the temperature of the gas.”
I just wrote a long reply and it disappeared when I posted it. Normally I copy to clipboard first, but forgot to because it’s after midnight and I’m tired.
The experiment was posted by DeWitt Payne in the third article on Backradiation at Science of Doom (where I’ve been censored because they have no answers and cannot prove Claes wrong.)
Please just read everything on both my Home page and the Radiation page, including the footnote http://climate-change-theory.com – if interested in natural cycles, read my original site http://earth-climate.com
Cheers
Doug (in Sydney)
134 – I’ll look up your SoD reference – if you have a comment timestamp you can quote that would be useful.
Tangentially this looks interesting and the paper is free access:
26 January 2012 Last updated at 03:25
‘Cloaking’ a 3-D object from all angles demonstrated’
http://www.bbc.co.uk/news/science-environment-16726609
Click to access 1367-2630_14_1_013054.pdf
DC, from your linked page:
This exposes an enormous hole in your understanding of “greenhouse theory”.
Do you think that proponents of the “back radiation” concept actually believe that the surface gets warmer from back radiation?
As far as I know nobody has ever suggested this – they have suggested that the rate of cooling may be reduced and so the temperature at a given moment may be higher than it would otherwise be. That’s a very different concept to the one you are rejecting in the above quoted paragraph.
Curious, Terry has described the picture as completely correctly as you can without invoking math (without dealing with pressure and thermal doppler broadening, which for a “beginners discussion” we can ignore, but when discussing the not so aptly named greenhouse gas effect, we can’t, since these affect the line widths as you move vertically through the atmosphere). Doug’s suggestion that he was “getting close” (as if anything Terry said was inconsistent with what Jeff, Steve or I said) is another snorter.
In terms of stat mech approaches, I have a thermo book at work taught from that perspective. I’ll link it when I get in. (I was going to work at home until I remembered I’ve got some calibrations I have to start this morning.)
I seriously don’t know why so many professors try and couch everything in the same language as the old, obsolete classical theory. For example heat is sometimes called a “process variable”, it refers in that context to the heat energy exchanged during one complete cycle of a thermodynamic process. Why do we need a separate word for it? What’s wrong with heat energy? Heat itself is a caloric theory concept, antiquated and confusing, the way it gets used both in classic thermo and by common day layman usage (e.g., the idea that “heat flows”). The idea we need to keep that same obsolete/lay term and use it in one special place, or that there is a need for it as a separate word to be used in exactly one context is just silly.
Doug:
To be fair, reading SOD’s comments, your comments were deleted because they violated the blog commenting policy.
Why didn’t you link to the original thread on your website? That seems very odd.
Steveta_uk:
Well this is the rub I think. People get confused because there is a subtlety here. If you put a blanket over a heater, that reduces the rate of heat loss near the heater, and the temperature of the air trapped by the heater increases from what it would be, where it allowed to freely convect:
Something got warmer in the process of impeding the loss of heat energy exchange. What these guys seem to be arguing is we are claiming the reason the air heats up is because the cold object is warming the warmer object. What is really happening is the warmer object is cooling less rapidly, so in equilibrium, it has a higher net temperature, with the extra heat energy coming from the warmer object, not the colder one.
This statement is true regardless of mechanism or even whether you are talking about rate of heat energy exchange (via any mechanism…. including convection, conduction and radiation) or some a rate associated with any other quantity, such as the rate at which water drains in the bottom of your sink.
Gedanken experiment time: turn on your bathroom faucet and let it run. Measure the level of the water in the bottom of your sink (assume that the opening of the faucet constricts the flow enough so you end up with a finite height above the bottom of the sink). Now constrict the drain so it has 1/2 it’s original area.
What happens? The level of water in the sink rises due to a new higher water level. (The flow rate through the pipe is proportional to the pressure at the opening, in turn proportional to the height of the column of water, and inversely proportional to the area of the constriction.)
The GHG effect in the atmosphere works the same way. Having a GHG present impedes the loss of heat energy from the surface of the Earth into space. Just as the constriction in the sink doesn’t imply that there is now water flowing back up the pipe from below but rather eventually comes from the faucet itself, the extra heat energy that warms the air near the surface comes from the extra heat energy trapped near the surface because it can’t freely radiate into space.
The GHG effect is a mechanism for slowing the loss of heat. That’s all. In order to describe it correctly, you need to use language such as Terry has used. The old classic thermodynamics doesn’t even have a concept of radiation heat exchange, it’s language and outmoded concepts are totally useless for this problem.
Jeff Condon,
What do you think your IR photo proved?? I am too stupid to even grasp that.
134 Doug – is this the comment by DeWitt you had in mind (in response to Jack Frost)?:
*****
DeWitt Payne January 11, 2012 at 12:31 am
It’s not SB that applies here, it’s Beer-Lambert. The absorption of surface radiation by the atmosphere for a path length of 1 m is quite small on average. You will get some absorption at the peak of the CO2 absorption, but since the temperature difference is only 1 degree, the emission of CO2 will almost completely make up for the absorption. It would require an extremely precise measurement to tell the difference.
If you have an IR absorbing gas in a cell and put a black body emitter at the same temperature as the gas at one end of the cell, you won’t see absorption or emission lines or bands regardless of path length. You only get absorption if the emitter has a temperature higher than the gas, which is how IR spectrophotometers work. You only get emission above the Planck curve of the emitter if the emitter is at a lower temperature than the gas.
http://scienceofdoom.com/2010/07/31/the-amazing-case-of-back-radiation-part-three/#comment-15197
*****
137 and 139 – Carrick – thanks if you have ref. I’ll check it out.
A quick comment on the whole AGW theory/debate, as I must put this down now, is the lack of clear analysis or comment on the diurnal aspects and impacts of the radiation fluxes incoming and outgoing. I have seen some but IMO it appears to get lost in the simple “solar constant intercepting disc IN and SB radiating bb sphere OUT” model.
JWL, or anyone.
In figure 1a, I cannot for the life of me figure out how the input energy to the sheet can be affected by t2 if the claim is that it cannot. How did t1 get informed of the existence of t2. t is theta.
f(t1-t2) into the ‘slab’ from the left.
Leonard,
“Cold or hot, the Moon heats the Earth. In Both cases it is very small effect. It is due to two effects. Reflected sunlight and thermal radiation.”
Yes, I have been told that and shown computations. What I have not been shown are experiments that would show this conclusively.
Curious the book is Bohren and Albrecht, Atmospheric Thermodynamics.
Regarding diurnal effects, how about this?
Doug,
As others pointed out, you are confusing heat transfer, which has to be hot to cold, with radiation energy transfer, which is bi-directional, with a net transfer hot to cold. Backradiation is not heating the warm ground, it is radiation insulation allowing the location of outgoing radiation to be raised in altitude, and thus the lapse rate heats the ground.
Gallopingcamel,
It is true that warmer water outgases CO2 if the concentration above the water is constant, but if the concentration is higher, the amount absorbed can go the other way. It all depends on both temperature and concentration in the atmosphere. I think the small water temperature increase would not be enough to outgas in the presence of a 100 ppm level increase above the water. Nevertheless, the amount changing in the water, is very small, is buffered, and not a problem.
The issue of replacing CO2 on Venus with N2 and O2 may not change the result below the clouds, but now the effect above the clouds would change, and there would be some temperature drop. It depends on if the outgoing level is at the clouds or above, and it is somewhat above. Also, the lapse rate in CO2 is much larger than in N2 and O2 (change in Cp).
Kuhnkat,
Cold or hot, the Moon heats the Earth. In Both cases it is very small effect. It is due to two effects. Reflected sunlight and thermal radiation.
142.Jeff Condon said
January 26, 2012 at 11:41 am
JWL, or anyone.
In figure 1a, I cannot for the life of me figure out how the input energy to the sheet can be affected by t2 if the claim is that it cannot. How did t1 get informed of the existence of t2. t is theta.
f(t1-t2) into the ‘slab’ from the left
Hello Jeff,
Now we can discuss.
When I followed the discussion on yes/no back-radiation, at various blogs, I started to make a model.
One layer, two layer…and I developed the model for N layers.
All based on the net heat from warm to cold.
I got coherent results, without back-radiation.
But I became puzzled when I checked the claim of Miskolczi, Aa=Ed and I found that he and other K&T authors used an absorption which is 2 to 3 times as big as the one I obtained with a 20 layer model of semi-transparent atmosphere, which I checked to have converged, meshwise.
So in order to understand what was going on, I wrote down the equations for the one slab model of a semi-yransparent atmospher, for a one way flow of heat desription and for the two-stream description of heat flow.
I found the same temperature distribution.
But the two-stream model with back-radiation gives an absorption twice as big as the one-way heat flow model.
And the absorption of the one-way heat flow model, fig 1a, is correct.
Now you are wondering how plate 1 knows that plate 2 is colder?
The same question as Claes Johnson is asking himself.
If you find the papers from Claes to abstract, please read the paper of Matthias Kleespies, a history of 200 year radiation theory.
Click to access History-of-Radiation.pdf
Matthias Kleespies concludes that there is no experimental confirmation for the particle theory of Planck, nor for the wave therie of Johnson.
With my comparison for a one slab atmosphere , according to the two-stream theory(particle theory) and the wave theory with a one-way heat flow.
I give people arguments to reconsider things:
The one-way heat flow model gives the correct absorption, the two-stream heat flow model -the so-called back-radiation model-
gives in a synptomatic way too big values for the absorption, which is re-emitted immediately in steady state conditions.
I do not need to know how plate 1 knows that it is warmer than plate 2, there is only one formulation with the correct absorption!
That does not mean that I study the Galerkin approach of Claes Johnson with interest.
With regards,
Jef Reynen
JWR,
What I was asking was very specific does not require references, links or anything else.
Can you answer my question to explain your equation?
It is fairly simple to estimate but your slight giggle tells me that the physics won’t convince you.
I notice that nobody has managed to reply to the IR camera demonstration.
Jeff Condon hand waves,
“It is fairly simple to estimate but your slight giggle tells me that the physics won’t convince you.”
I did not ask for an estimate. Read what I asked for again. Yes I am still giggling because you are funny. You continuously spout opinions, admittedly accepted by others of AUTHORITY, which you have no way of proving.
I haven’t called anyone stupid.
The IR sensor of the camera is room temp, yet it receives IR from objects colder than itself. The IR causes an energy increase/decrease of the sensor despite sometimes being cooler than the
objectsensor the energy is striking.It is back radiation from a cooler source 100% confirmed for all to see.
142.Jeff Condon said
January 26, 2012 at 11:41 am
JWL, or anyone.
In figure 1a, I cannot for the life of me figure out how the input energy to the sheet can be affected by t2 if the claim is that it cannot. How did t1 get informed of the existence of t2. t is theta.
f(t1-t2) into the ‘slab’ from the left.
Do you mean the equation f(t1-t2)?
Look into the parer eq (1)
JWR
Look into the paper eq (1)
Jeff Condon,
“It is back radiation from a cooler source 100% confirmed for all to see.”
Does it HEAT or slow the cooling of the sensor??
Jeff Condon,
Why does the picture have black spots on the ice cubes if the camera is good down to -20c??
Limitation in translating the picture??
You obviously haven’t even THOUGHT about the fact that the ice cubes will be covered with a warmer layer of melted water or condensed water which is mostly what is being measured!! You really need to be more rigourous if you expect to be believed by those who are biased against your story.
Anonymous,
I’m not quite that slow. Thank god. As the video progresses you can see the camera shifts its dynamic range such that the table (at room temp) drifts to ever warmer colors. Can you see any colors which are cooler than the room temp table?
Jeff Condon,
“I’m not quite that slow. Thank god. As the video progresses you can see the camera shifts its dynamic range such that the table (at room temp) drifts to ever warmer colors. Can you see any colors which are cooler than the room temp table?”
Jeff, the fact that this camera appears to show the ice cubes as being cooler than the table in no way supports your theory that IR can warm or slow the heating of a warmer body, that is how smart you are. It does do a nice job of exhibiting the effect that Einstein received the Nobel prize for.
http://www.physicsforums.com/showthread.php?t=454821
(snicker)
Leonard (144)
You had better read what Prof Claes Johnson has proven, because you clearly have no understanding thereof. There is a brief summary I have written at http://climate-change-theory.com/RadiationAbsorption.html I am not confusing anything, thank you. With over 40 years of physics behind me, I have a quite clear understanding and acceptance of Johnson’s “Computational Blackbody Radiation.” Until you also have such an understanding there is no point in my trying to explain to you why, for example, a cold Moon does not transfer any thermal energy to a warmer Earth surface.
IR detector:
Obviously, photons from the ice are affecting the elements of the sensor. As the ice has a temperature greater than absolute zero, there must be photons being emitted. There will be SOME photons from other sources that are reflected, but IR photography and thermography technologies recognize and deal with that.
Is this a photo-electric effect, that is, does the absorption of a photon cause an electrical signal to be generated by a sensor element? Does the electric or magnetic field of a passing photon affect the conductivity (well, gotta be absorption and a rise in energy state) or other property of the sensor element? I can’t think of a way to detect a photon without extracting some kinetc or other energy therefrom.
Or maybe the ice cubes are sucking photons out of the sensor elements. Who knew?
Doug,
How does the camera work?
#160 – You are aware that you have linked to an article proving my point right?
FINAL POST TO EVERYONE
Curious (141) is right
Yes, AGW “science” is flat Earth science and similarly flawed.
It “forgets” that the fourth root of the average of some numbers is nothing like the average of the fourth roots of those numbers.
It takes no account of solar energy seeping into and out of the surface each day and night for example. (That’s why it gets wrong answers when applied to the Moon.)
It “thinks” the Earth’s surface is a blackbody when it is not remotely so, because it is not insulated from the atmosphere or sub-surface layers.
It thinks backradiation slows the rate of cooling and/or affects the lapse rate, whereas thermodynamics affects the lapse rate far more than radiative transfer – see my post on the lapse rate thread at WUWT..
It forgets about the thermal inertia of the whole huge amount of thermal energy under the surface down to the core.
It almost forgets about thermodynamics altogether in its obsession with radiative transfer theory, which it has wrong anyway.
And it has no empirical evidence to back it up.
You need to follow posts on WUWT as they discuss serious stuff, not fun stuff.
Never in the history of mankind have more people been bluffed in more ways by so few.
Cheers everyone
Come on over to http://wattsupwiththat.com/
“You need to follow posts on WUWT as they discuss serious stuff, not fun stuff.”
What is the difference?
Doug:
Even given your complete lack of any meaningful insight beyond that of a high school teacher, this kind of whoring/shilling for Johnson is really … kind of down there in the swamp.
WGAFF what you think? You don’t even believe photons exist as quantized particles.
Jeff, not that I’m disagreeing with your point about radiation, but couldn’t a camera that observed Claesian physics reproduce that video? I might be wrong, but wouldn’t the camera need to distinguish between two objects, where each object is colder than the sensor? eg if the ice was floating in water that was itself colder than the sensor?
166,
If you wait until part way into the video, you can see the table top (room temp) color shift to ever brighter rendering as the camera scale changes. If you observe the edges of the still very dark colored ice, you can see there are far cooler colors around the edge of the ice than the room temp table.
If the warm sensor is not absorbing photons from the cooler objects, it would not be able to discern these colors. Also, the manufacturer seems to believe they can image minus 20 C with the same camera.
Re Doug Cotton # 158
“I have a quite clear understanding and acceptance of Johnson’s “Computational Blackbody Radiation.” Until you also have such an understanding there is no point in my trying to explain to you why, for example, a cold Moon does not transfer any thermal energy to a warmer Earth surface.”
Sorry to say but wrong again. If it didnt transfer any energy back then you would not be able to see the dark side of it that occurs from earthshine. All photons possess energy whichever end of the spectrum they are from visible or IR.
143 – Carrick – Thanks for the ref. Interesting reviews on Amazon UK. I’ll see if I can get a library copy. Your diurnal link is dead for me?
+++++
156 – Anonymous – Looks like minus 20degC is fairly std. capability – have a google.
This one must work when warmer than the subject as they say you can only operate down to minus 10degC!:
http://www.aikencolon.com/Fluke-TiR4FT-TiR4-FT-IR-Infrared-Thermal-Imager_p_865.html – data sheet on the right.
btw aren’t the colours irrelevant – they could be inverse mapped if you preferred?
+++++
163 – Doug – thanks for the vote of confidence but is that for finding the right link to DeWitt at SoD? Or for the diurnal comment? If the comment I highlighted from DeWitt is the one you meant; for clarity, I would still like you to explain exactly what you think it demonstrates. Thanks.
2nd try on the diurnal link.
#163,
“FINAL POST TO EVERYONE”
Is that a promise? Let’s hope so. You are very lost my friend; the shrill ravings and aggression do not make you any less so. Please, for your own sake, and everyone else’s sanity, learn some basic radiative physics.
170 – thanks Carrick, works. Just had a quick read – looks good. Will read again when feeling sharper!
Curious, Bohren definitely got the dander up of people who want to retain poorly defined concepts and crapped-up terminology.
IMO, a pretty fair percentage of classic thermodynamics language is pure unadulterated bovine excretion The language is often borrowed from common language (already confusing enough), then borrowed a second, a third and so forth… each time to try and get it to loosey-goosey fit into the actually correct theory, which comes from statistical mechanics.
I noticed that there were a fair number of criticisms of the book for its disdain of the differential. I happen to agree with Bohren here and am taking the opportunity to launch off into a lecture on Parmenides and his illusion of change argument, er no wrong power point slides. Er, here it is…
The differential as it was introduced in thermodynamics was done as an attempt to describe changes in physical systems while trying to pretend that we can ignore the passage of time during the change in state variables. Which is a crock.
In the real world, the time scale over which the change occurs matters greatly. They had to make the argument to ignore the time variable classically because, well absent statistical mechanics, they had no way to say whether a system remained in thermodynamic equilibrium or not, nor even a technically accurate way of defining when that occurred. So they were left with a lot of handwaving about the process happening “very slowly” so philosophy-of-science level arguments on equilibrium thermodynamics can be retained. Introduce statistical mechanics, and we can calculate how fast you have to change state variables before the system is no longer in equilibrium, and we can even tell you want it means for a system to be in equilibrium, etc.. And the artificial and awkward language of classical thermodynamics can simply be abandoned and just flushed down the toilet (which IMO it should be).
As a technical example, if you compress air in a chamber using a piston (assume sinusoidal variation in pressure), the rate at which the compression is done changes the answer.
(We can work out how fast you have to run the piston before the system is no longer in thermodynamic equilibrium, in a statistical mechanical sense, this involves computing the microscopic relaxation time for the system, see e.g. here.. For “typical” pressures and densities of a gas, this is a very tiny number, typically 10^-8 to 10^-10 seconds, depending on the gas.)
As you compress or rarify the air in the chamber, the air in the chamber heats up (or cools off). The walls of the chamber in the real world, have a finite heat conductance, so if you perform the compression (and rarifaction) of the air in the chamber slowly enough, the excess heat energy from compression will be conducted through the wall of the container, and the pressure change will occur effectively at constant temperature (that is “isothermal”). In this case, the relationship between the volume of the chamber and its pressure is:
(P + Delta P) (V + Delta V) = P V => Delta P/P = –Delta V/V.
(Ignoring second order.)
If you do it rapidly enough, we can assume no heat is exchanged between the tank and its environment… so the process can be treated as isoentropic, in which case you can show P V^gamma = constant holds, where gamma is the ratio of specific heat capacities at constant pressure to constant volume, and instead you have:
(P + Delta P) (V + Delta V)^gamma = P V^gamma => Delta P/P = – gamma \Delta V/V.
And for intermediate time scales, you have to include the rate of heat loss into the system (still doable analytically, at least for certain geometries).
Anyway the point of all this is using exact differentials may or may not be a consistent mathematical construct (that’s a bit too meta for this blog), but regardless, it is a useless concept here. The duration of the process does matter: Not only for a laboratory scale problem like I’ve described, but for atmospheric processes as well, and actually for almost any real world problem as well.
You can’t up simply ignore dt so that your crapped up 19th century “philosophy of science” arguments using outmoded thermodynamics language and obsolete and contradictory concepts can be shoe-horned into fitting into the framework of statistical mechanics. And with that, I’ll try and “play nice” with all of the old professors and their protégées, who still insist that there’s nothing wrong with their outmoded ways of thinking. 😉
I should mention the above discussion assumes the rate of change of volume by the piston is being done slowly enough that the pressure can be assumed to be constant throughout the chamber. At higher frequencies this will no longer be the case but is still amenable for certain geometries to an exact solution.
I want to thank all who participated in this discussion. I learned some things, and am now corrected in a major area. Some of the participants have shown wisdom, and some ignorance. The important thing is to try to learn, not just dig in when evidence is shown to refute positions.
IIRC, the full moon adds about .01 W/m2 to the nitetime “sky”. Or was it .1 W/m2? Whatever — it’s insignificant, but measurable.
Good discussion here. From an engineer’s perspective, some people seem to forget that the bitter cold of outer space is always present & anything that decreases the heat-loss of an object to that ubiquitous cold sink (like insulation or GHGs) ultimately affects its temperature. It’s a matter of cooling less, not “heating up”.
http://wattsupwiththat.com/2012/01/29/making-things-matters/#comment-878941
So now we’re going to leave links to comments rather than actual comments?
Jeff Condon,
I was just wondering why I didn’t see any blurring from the water vapor that should have been around the top of the coffee. Any comment?? Doesn’t prove anything but would have been nice to see it as it would have carried your basic argument much better. That GHG’s emit substantial IR.
Oh, and no it does not prove your point. You really aren’t that smart if you think an effect in one class of materials proves a similar effect in other materials. I am wondering if you now expect me to believe that DLR causes eddy currents in materials heating them.
“You really aren’t that smart if you think an effect in one class of materials proves a similar effect in other materials.”
I can promise you I’m not that smart. However, I do study math and science with care. If you have an effect of absorbing photons in one class of materials from cold to hot and the math predicts and explains that effect, on what evidence would you determine the superposition of light doesn’t exist for another class of materials?
It really is basic and standard physics we are talking about. What is your theory as to what happens to the photons emitted by the cooler body?
Jeff Condon,
can you give me an example of ir or visible absorption that knocks an electron out of place without being intensity dependent?? That is a dummies description of the photoelectric effect. Doesn’t sound like it has anything to do with varying energy levels or vibration etc. unless it is a harmonic thingy. If we had been discussing the photoconductive effect (also known as photoconductivity or photoresistivity), the photovoltaic effect, or the photoelectrochemical effect, I would have to concede.
I leave the theorizing to the smart people unless they can’t whup up some good observational evidence to back it up.
Kuhnkat, why don’t you tell us why you think the mechanism for thermal photon emission and absorption should be radically different in “other materials” and why it “isn’t smart” to assume it is the same mechanism.
Otherwise maybe you should just stay quiet and let us suspect you’re an idiot, rather than opening your mouth and removing any doubt.
carrick,
i would much rather make sure there is absolutely no doubt in anyone’s mind.
We don’t have to wait for the climate in the next few years to tell us who’s right and who’s wrong. Physics all along has been telling us. The warmists just don’t understand physics.
The most glaring mistake they make is in saying the atmosphere has warmed the surface (like a blanket) from -18C to +15C. The first figure is a theoretical temperature (call it small t) which is only related to the intensity of radiation via the S-B law which only relates to perfect blackbodies. Such blackbodies are usually other bodies in space which are perfectly insulated by space so there is no heat loss by conduction. In contrast the Earth’s surface is continually losing heat to the first millimetre of the atmosphere by diffusion (see Wikipedia “Heat Transfer” second paragraph) and also into the depths of the Earth’s crust or the oceans. So there is less energy left to radiate.
The actual temperature (call it capital T) is a totally different entity without direction for a start. So you cannot just subtract and get T – t = 33 deg.C, because t is not a real temperature. Without carbon dioxide and its colleagues, thermal energy would still diffuse from the surface into the atmosphere, greatly reducing the radiation, as it does. In fact the net radiation from the surface is probably less than 25 W/m^2, so what value of little t would that give you? Very cold I assure you.
This is why an IR thermometer cannot calculate temperature by measuring the intensity of the radiation and using S-B law. It can only do so by measuring the frequency and using Wien’s Displacement Law which says absolute temperature is proportional to the peak frequency.
So, given the major fallacy in the warmists “science” when they calculated and widely promulgated that 33 degree “difference” between apples and oranges, what confidence could we possibly have in any other deductions of theirs? They are also wrong in assuming radiation from a cold atmosphere can warm an already much warmer surface.
The atmosphere cools the Earth by reducing the amount of incident solar radiation which gets through. Hop out of a spacecraft and see how hot you feel in the sun’s rays. But radiation “temperature” is a very different thing from ambient temperature, both in space and, for example, at the top of a high mountain where the Sun’s rays might feel like 40C but the actual temperature of the air might be -15C.
It is important to remember that radiation is a measure of energy (Watts) transferred through a unit cross section (one square metre) and it is thus a vector with both magnitude and direction, nothing like a temperature. The only “connection” with temperature can be made if a true blackbody is emitting it, that body is not also losing thermal energy by conduction, diffusion, convection, evaporation or any other means. If it does lose energy in such ways then, at the very least, you would need much more information before making any inferences about its temperature.
Yes, the whole Earth plus atmosphere system looks like a blackbody from outer space and some average radiating temperature could be calculated by remembering that it is a spinning sphere, not a flat disk as warmists treat it as being. But whatever temperature is calculated is merely an average temperature somewhere in the atmosphere.
Doug,
So how does the camera work.
Doug Cotton:
What the….
Radiation isn’t a measure of anything, it’s a phenomenon. The units of energy is joules not Watts. Irradiance —the term you were scratching for— is, like temperature, a positive-definite scalar and is measured in units of power/unit area (e.g., W/m2). The irradiance I is given by I = |<S >|; where S is the Poynting vector and < … > the time average of that quantity.
Whatever else it is, irradiance is definitely no vector “in both magnitude and direction” (if you have magnitude and direction you are already a scalar).
If you get something so basic so wrong, “what confidence could we possibly have in any other deductions” of your?
Latest science of doom blah blah blah vs equations:
Sound familiar?
Kuhnkat:
Well…
E = h nu
The frequency nu of the light is related to the energy E of the photon, so no Jeff probably can’t give you an example. Of course this same equation is integrally related to the derivation of Planck’s Law of Radiation—it’s needed to solve the ultraviolet catastrophe–and from this Wien’s Displacement Law and the Stefan-Boltzman Law can be derived.
It all ties together in the end, and depends on the microscopic picture involving the emission and absorption of photons (unless you are Claes Johnson or his clone Doug Cotton, in which case there are no such things as photons and all doubt has really been removed, not just in jest).
Fair enough. Fair enough.
Thanks for the info Carrick. I have read Claes’ stuff and find his writing interesting, so, remove the doubt. Any hope of a link to solving the 2 slot with photons?
Carrick: Yes I accept that I should have used the words irradiance (or radiative flux), and “power” instead of energy and perhaps “Euclidean vector” as here http://en.wikipedia.org/wiki/Euclidean_vector – such having magnitude and direction.
My only comment is that most readers of climate articles would rarely see some of these words, let alone know their meaning. I suggest climatologists are just as much at fault using the terms “heat content” or “backradiation” as I’m sure you’d agree.
So what – you had no trouble knowing what I meant.
Now how about addressing …
(1) the issue of very invalid physics used by climatologists when they “determine” that -18 deg.C figure ignoring the fact that the surface is not insulated as a blackbody should be.
(2) The now proven fact that a body (such as the surface) cannot convert to thermal energy any of the energy carried by radiation which has a frequency significantly below its cut-off frequency, that being the peak frequency (proportional to the body’s absolute temperature) as determined by Wien’s Displacement Law. I remind you that there is no published experiment showing otherwise or in any way proving that backirradiance can warm the surface.
It is not a matter of counting scientists.
What matters is who is applying correct physics, mathematics or whatever, Nothing else matters.
It is incorrect physics to treat the Earth’s surface as if it were radiating as a blackbody in space would do. The surface is not insulated from its surrounds, namely the atmosphere and sub-surface crust, deep ocean waters etc. It transfers most of its energy by diffusion, conduction, convection, evaporation and chemical processes. Hence there is not much left for radiation.
Hence it is incorrect physics to apply Stefan-Boltzmann calculations to the surface as is done in the development of the AGW hypothesis when calculating so-called sensitivity and the infamous -18 degrees C figure.
Hence the AGW hypothesis is not grounded in correct physics. Need I make the final statement?
Kuhnkat,
“can you give me an example of ir or visible absorption that knocks an electron out of place without being intensity dependent?”
Phosphors in white LED’s are a good example. Since you requested visible or IR, often a blue light is emitted to a phosphor which absorbs and re-emits to yellow-red. It is an absorption downconversion. Only one photon is required to emit one photon of lower energy. More photons are emitted when it is brighter but the process knocks an electron out of place one photon at a time. My understanding is that the surrounding material around the individual atom offsets the amount of jump experienced by the electron and changes the output wavelength on decay.
Jeff Condon,
thank you. So IR is not generally involved in the photoelectric effect, only visible or higher. I believe Wikipedia tells us that alkaline metals are sensitive to visible, the metals in gener to blue and non metallic to ultraviolet. Now, do we agree that this happens in quanta. (sorry about my terrible terminology, I am ignorant)That is, if the photon is not the right amount of energy then it is simply reemitted apparently?
The distribution of numbers of photons in each wavelength (i.e., individual energy of each photon) is temperature dependent, but a wide range of wavelengths occur at any given temperature. It is only the individual photon wavelength (i.e., energy) that matters as to whether it is absorbed or not in a gas with multiple atoms (i.e, with molecular vibration energy states). There are only a limited number of possible excitation states in the molecular vibration, so a limited absorption and emission spectra, but it covers part of the Earth’s temperature range. Photons are emitted and absorbed, and the extra vibration energy is excited and de-excited by mode redistribution (collisions with other atoms), so there is continual passing of radiation in all directions in greenhouse gases for suitable gas temperatures. There is upward and back radiation which is absorbed. However the net balance (i.e., heat transfer) is always hot to cold. That is the difference between back radiation being present but back radiation not heating the warmer ground. The increase in ground temperature in the presence of greenhouse gases is due only to the lapse rate (always present if there is sufficient convective mixing) and the average altitude of outgoing radiation to space. Back radiation is a result of the gas above the ground having suitable excitation states, but it is not the cause of the warmer ground except for the special case of the ground being colder than the air. Thus greenhouse gases do cause a warmer Earth than without, but not due to back radiation.
I implore those who do not understand this to quit making statements that do nothing but show their lack of understanding.
I need to add that gases without suitable wavelength absorption and radiation ranges (i.e., non-greenhouse gases) will not absorb or radiate photons, and thus are only heated by conduction and convection. N2 and O2 do have some wavelengths of interaction, but they are almost totally outside of the the thermal wavelengths from Earth, so are effectively non greenhouse gases. Gases, unlike solids or liquids are not like black or gray bodies in their radiation sensitivity. The non-greenhouse gases can’t radiate to space, so can’t raise the altitude of outgoing radiation. Clouds can, but that is a separate issue altogether from the present discussion.
kuhnkat:
Which part are you looking for? The basic two slit is a famous experiment that proved the wavelike nature of light. It doesn’t require quantum mechanics for it to work. (The mathematical formula for the interference pattern is a standard calculation). It only requires that light is “wave-like” to compute.
There is a nice link here on the interference of individual particles, where it is easy enough to understand that the wavefunction has not been “collapsed”, and so even for a photon, the interference reduces to an interference in the probability that it will get detected on the back screen.
The single-particle two-slit experiment turned into one of the battlegrounds between Bohr and Einstein over quantum mechanical theory (Bohr wins this round).
Leonard:
think you have your physics a bit off there. 1/2 of the thermal photons emitted at any elevation, end up as downwelling photons. Eventually these lead to a warmer ground because they heat the air around them. The convective lapse rate ends up dominating but it’s because the radiative lapse rate is too steep, and this radiation forces additional convection.
See this figure.
It is the interplay between infrared radiation (including backradiation) and the convective instability of the atmosphere that leads to an elevated temperature in the lapse rate and yes, a warmer surface temperature. Interestingly what convective instability does is reduce the efficiency of the GHG effect, because it results in more heat being carried back up to the top of the atmosphere than might otherwise have been transported.
You can see that latter point in the figure I linked, just shift the lapse rate for radiation so the temperature is the same at the top of the atmosphere… the ground would end up being hotter if convective instability weren’t present.
(The source for the figure is Manabe and Strickler, 1964.)
Leonard #139,
It sounds like you are suggesting that increased CO2 would not narrow the ‘atmospheric window’ and so never restrict direct flux of photons from the surface to space (under clear sky conditions). That is simply not correct. Adding more CO2 most certainly will narrow the window and most certainly will reduce the rate of direct radiation to space from the surface when the sky has less than 100% cloud cover. And the narrowing of the atmospheric window does indeed manifest itself as additional ‘back radiation’ from the added CO2, which can be confirmed by measuring the spectrum of that back radiation. This will happen independent of whether or not the atmosphere is locally following the theoretical lapse rate. As I have noted several times before, there is nothing magical about the lapse rate; it represents only the boarder/transition between conditions where only radiant flux controls the local temperature and conditions where both radiant flux and convective transfer jointly control the local temperature. Increasing CO2 reduces direct radiant loss to space both at the surface and at all levels of the atmosphere.
While there are many exceptions (all types of temperature inversions), it is also true, of course, that in most places most of the time the atmosphere does follow the lapse rate reasonably well (or the lower moist lapse rate, depending on local humidity), but that does not mean that the lapse rate 100% “controls” the local surface temperature, it just means that the flow of energy from the surface to space is most of the time at least in part due to convection. If you increase CO2, you will always reduce the direct radiant energy flow, which will result in an increase in convective flow (and a concurrent increase in surface temperature) in order to maintain energy balance at the surface. Having less energy flow via direct radiation and more via convection requires a greater temperature difference across the atmosphere… the surface must warm relative to the top of the troposphere to increase convective flow. It is important to differentiate between cause and effect: the requirement of flow of energy from the surface to space to maintain energy balance is what leads to convection (and the observed lapse rate), not the lapse rate which “controls” the surface temperature. Just as in every physical system, the local temperature is always controlled by the flows of energy in and out, including radiation and convection.
Sorry, that should have been: Leonard #193.
Carrick,
Great reference! I hope Leonard reads it.
Thanks, Steve. And that was an excellent exposition in #199. Hopefully people read and understand that too.
Carrick,
The adiabatic lapse rate does require sufficient mixing, and I previously stated that. Given sufficient mixing, the size of the adiabatic lapse rate is independent of the radiation effects (notice I said adiabatic). The lapse rate of Earth and Venus are in fact the adiabatic value on average with condensation effects included (wet lapse rate for Earth). As long as convection can result in sufficient mixing (buoyancy plus wind), this will hold. Since the thermal effect of the radiation heat transfer is restricted by buoyancy adjusting the temperature to maintain the lapse rate to the adiabatic lapse rate by the air currents (more thermal heating, more buoyancy), it is only the Cp, g, and presence of sufficient mixing that determines the air temperature GRADIENT (note, this is only so for adiabatic lapse rate conditions, but these are on average present on Earth). The level locks the level of the entire profile, and this occurs by the balance at the effective average outgoing location.Thus the back radiation, on average, does NO HEATING OF THE GROUND. It is purely a result, not cause as I stated.
Steve,
You put claims in my comment that I did not have. The higher concentration of CO2 would add optical thickness to the atmosphere, resulting in raising the level of outgoing radiation. I clearly said the average outgoing level, along with the lapse rate determined the heating. A greater concentration does close the window slightly (but not completely), and requires slightly more relays up to exit. It does this by more fully saturating the absorption lines which were broadened by doppler shifting due to thermal interaction. I never claimed otherwise. Also I said that all comments were “on the average”, and stated there were local variations. However, thermal heating by radiation is self compensated by buoyancy and wind mixing so that if it is the dominate heat transfer up or a small net transfer, it does not matter, and has no significant effect on ground temperature ON AVERAGE AND IF THERE IS SUFFICIENT MIXING.
Leonard, I think you need to think this through a bit more carefully and review what you said originally.
All I’m going to say.
Leonard #202,
“Back radiation is a result of the gas above the ground having suitable excitation states, but it is not the cause of the warmer ground except for the special case of the ground being colder than the air. Thus greenhouse gases do cause a warmer Earth than without, but not due to back radiation.”
This part of your earlier comment is the part you are mistaken about. Increased CO2 in the atmosphere reduces the width of the atmospheric window, and so always restricts heat loss from the surface (and every other height!). The reduction of radiative loss from the surface is independent of whether there is a temperature inversion (no convective transport) or no inversion (there is convective transport). Please consider that even in the presence of “an inversion”, we are not necessarily talking about the atmosphere above actually being warmer than the surface… it can be much colder than the surface but still non-convective, because it is not as cold as you would expect from the theoretical lapse rate. In other words, even “an inversion” is not necessarily an actual increase in temperature with height, only less of a decrease in temperature with height than you would calculate from the lapse rate. It is an inversion of “potential temperature” not an inversion of sensible temperature. The “colder” CO2 molecules above still absorb infrared…. that absorption takes place independent of the local temperature.
I remain astounded that people are so willing to reject radiative physics; everything radiates, all the time.
“The level locks the level of the entire profile, and this occurs by the balance at the effective average outgoing location.Thus the back radiation, on average, does NO HEATING OF THE GROUND. It is purely a result, not cause as I stated.”
I’m starting to agree with carrick that the terminology is worthless. Is this statement claiming that back-radiation does not add energy to the ground?
It clearly does so what is the negotiation?
Leonard Weinstein – the distribution is not wide as you claim – it is strongly attenuated – see http://en.wikipedia.org/wiki/Wien's_displacement_law
Incoming solar SW radiation does not overlap significantly with upward LW radiation as you no doubt know.
Stefan-Boltzmann Law cannot be correctly applied to the Earth’s surface as it is nothing like a true blackbody. losing energy by several processes other than radiation.
Any radiation from a cold atmosphere cannot be converted to thermal energy by a warmer surface. Physics says so, and you cannot produce empirical evidence to the contrary.
I usually find those who fall for the IPCC bluff about that -18 deg.C temperature really don’t understand this area of physics.
If anyone wants to learn, come and ask questions on WUWT.
Doug,
How does the camera work. I’m tired of asking. You have stated your physics knowledge is superior to all here, please answer my ignorant question.
Carrick – it is you who have your physics more than a bit off. Radiation from a colder part of the atmosphere cannot warm a layer that is significantly warmer. The energy transfers from warm to cold, not from cold to warm. Nor can that radiation warm the warmer surface as I have been telling you all over and over Wow. What a blunder!
Go read Prof Claes Johnson’s “Computational Blackbody Radiation” or my summary on my Radiation page at http://climate-change-theory.com
I suppose you too fell for the bluff that the Earth’s surface radiates like a blackbody. Did you ever learn any physics about blackbodies?
Jeff The IR camera makes use of Wien’s Displacement Law which says peak frequency is proportional to absolute temperature. Hence, measuring frequency allows it to calculate to temperature of an emitting surface. The camera then artificially colors different areas of the image according to the frequencies detected. http://en.wikipedia.org/wiki/Wien's_displacement_law
Thanks for finally answering. 200 questions later.
How can the camera measure peak frequency of thermal emissions which are cooler than the sensor? Did the sensor not receive photons from the cooler source? and did those photons not add energy to the sensor?
Doug,
If the cooler source emits at ALL, what happens to the photons directed at the warmer source. I’m very weary of the games you are playing. Tell me what happens to the photons. Do they bounce off, do they not emit in the first place, do they magically vainish?
I’m growing very impatient with the aloofness. You have polluted this thread with claims you cannot back up.
Prove me wrong and since you are the big dog, don’t link to papers to answer simple questions, use your own words. You won’t like me when I’m impatient.
Carrick – if you want to believe stuff from 1964 about lapse rates good luck to you. They are affected by pressure, not carbon dioxide, being a function of how fast warm air rises. None of these processes can transfer energy back to the surface.
It can’t happen and it isn’t. The world is following a natural increase (part of a 1000 year cycle) which was warming by 0.06 deg.C/decade a hundred years ago and has now reduced to 0.05 deg.C/decade – meaning it should be half a degree warmer by 2100. But wait, there’s more. After 2200 it will cool for 500 years.
Jeff – Prof Claes Johnson answers your question quite clearly in his discussion of resonators. Think of it like a stone hitting a catapult and being fired out again. The end result is like deflection or reflection as far as energy and frequency are concerned – there is no change – the radiation eventually finds its way to space and is never converted to thermal energy by the surface.
See the footnote on my Radiation page for more detail on this process.
Others – please look first for answers on my website as I really don’t have time now to answer any more questions.
Doug ” Radiation from a colder part of the atmosphere cannot warm a layer that is significantly warmer”.
For the last time (literally, you’re on /ignore with me until you grow a brain stem), the colder part of the atmosphere does not warm the warmer part, it causes it to cool less slowly.
” if you want to believe stuff from 1964 about lapse rates good luck to you”;
We still use planetary law that was developed by Newton, which is considerably older than 1964. It’s not when it was written, it’s whether it is correct or not.
Bye.
Doug #208,
It is clear that you do not understand the real issue. I suggest you read Read Willis’s ‘steel greenhouse’ article (at WUWT) for a start.
The question is not a net transfer of energy from colder to warmer, it is a reduction of net transfer from the warmer surface to colder space, because the atmosphere (with infrared absorbing GHG gases) stands in the path. Even very cold CO2 molecules in the upper troposphere absorb and re-emit photons, and that re-emission is in random directions. Do all the photons absorbed by CO2 in the upper troposphere get re-emitted toward the surface of the Earth? Heck no, but some of them do. Those that are emitted in the direction of space are of course lost, but those which are emitted downward balance part of the original upward flux, thus reducing the net upward flux. There is not (and can never be) a net radiative flux from cold to warm. But that is never claimed when people talk about re-radiation in random directions. The net flow is always form the warmer surface to the colder CO2 in the atmosphere (I repeat, the net flow is always from the warmer surface to the colder CO2 in the atmosphere!). What is claimed is that the net radiative flow from the Earth’s surface toward space is reduced by GHG’s absorbing and re-emitting photons. The cold upper troposphere is a lot warmer (~220K) than space (about 2.75K), so the net flow from the surface is automatically less because the (relatively) warmer atmosphere with GHG gases stands between the surface and space, partially absorbing and re-emitting surface radiation. Just like Willis’s steel greenhouse.
Carrick,
yes it is famous and I have read a number of short explanations of it. Except for those that um, fork from accepted science none of them gave an explanation that got rid of waves. I am looking for an explanation that can do that since so many people seem fixated on photons only.
Kuhnkat, don’t know if this will help, but the photon has the same wave-particle duality of any other quantum particle.
So even the single photon still has a wave function, which acts just like the classic wave obtained by the superposition of many photons. It doesn’t act like a particle unless you make a measurement that localizes the position. As I mentioned Bohr and Einstein had a bit of a tiff over the single-photon two slit experiment (which had long before been performed by Taylor circa 1909).
The problematic part for the photon, as I mentioned above, is it is wholly a relativistic particle, whereas Schrödinger and other classic formations are entirely non-relativistic. (Somehow many people who study these equations don’t grok that, and spend an inordinate amount of time worrying about causality in an acausal theory. Now that’s what I call “quantum weirdness”!)
The place you really see a breakdown of the classic quantum mechanical formulation (beyond the obvious issues with causality, the irreconcilability of simultaneity with special relativity etc) is the interaction of a photon with an electric field, where individual photons can be created and destroyed in free space (and you even end up with violations energy in these interactions via creation and destruction of virtual pairs, which is allowed as long as the timescale is short enough it isn’t observable and other really weird stuff like that).
From Pierrehumbert himself.
Click to access PhysTodayRT2011.pdf
“Coupled vibrational and rotational states are the key players in IR absorption. An IR photon absorbed by a molecule knocks the molecule into a higher-energy quantum state. Those states have very long lifetimes, characterised by the spectroscopically measurable Einstein A coefficient. For example, for the CO2 transitions that are most significan in the thermal IR, the lifetimes tend to range from a few milliseconds to a few tenths of a second. In contrast, the typical time between collisions for, say, a nitrogen-dominated atmosphere at a pressure of 10^4 Pa and temperature of 250K is well under 10^-7s. Therefore, the energy of the photon will almost always be assimilated by collisions into the general energy pool of the matter and establish a new Maxwell-Boltzmann distribution at a slightly higher temperature. That is how radiation heats matter in the LTE limit.”
This experimentally shown result is IMO an extremely important one that is never discussed. A GHG at sea level atmospheric temperature and pressure rarely gets a chance to radiate. Its energy absorbed into the atmosphere within about 10m of the earth’s surface.
The fact that the Boltzmann distrubution of energies applies to the atmosphere (and GHGs within it) is irrelevent if the energy is reassorbed into the pool of energy held by the atmosphere before it actually radiates. There are many thousands of collisions that occur between it acquiring sufficient energy and it radiating.
I dont think there is much back radiation seen at the earth’s surface at all.
IMO Pierrehumbert fluffs his logic in the statement that follows in his paper
“According to the equipartition principle, molecular collisions maintain an equilibrium distribution of molecules in higher vibrational and rotational states. Many molecules occupy those higher energy states, so even though the lifetime of the excited states is long, over a moderately small stretch of time a large number of molecules will decay by emitting photons.”
And the reason I believe he fluffed it is because he has to assume that its the same actual molecules in those higher energy states whereas I believe they’re constantly changing and no one molecule has its energy state high enough for long enough to radiate. Its a massive assumption to switch to averages and claim they radiate at that point. The individuals dont radiate for a “long time” for a reason and I’m assuming that reason doesn’t disappear when talking averages.
As one moves higher into the atmosphere, the temperature and pressure drops and the collisions become less frequent and radiation will increase until at some particular altitude and temperature and GHG concentration, radiation becomes predominant.
Hence IMO there is a sound physical reason why increased pressure (ie more non-GHG atmosphere) will be warmer.
Doug writes “Radiation from a colder part of the atmosphere cannot warm a layer that is significantly warmer. The energy transfers from warm to cold, not from cold to warm. Nor can that radiation warm the warmer surface as I have been telling you all over and over Wow. What a blunder! ”
If you look at individual molecules in the two bodies, the distribution of energies says that there are some in the warmer body with less energy than some in the cooler body. There will be an overlap and the closer the temperatures of the two bodies, the more the overlap..so even if it were true that the hotter body somehow couldn’t accept energy from the cooler body on average, individually that logic breaks down.
Carrick – choose what you wish to believe – it is quite clear on my website that it does not cause the surface to cooler more slowly, any more than reflected solar insolation would do so. Reason: backradiation (if it even exists) cannot be not converted to thermal energy. Radiation is quite different from thermal energy. Something which is not thermal energy cannot affect the amount of thermal energy anywhere unless and until it is converted to thermal energy. For the last time, study what Johnson says.
Tim: If the temperatures are very close (which is not normally the case) near resonance can allow conversion to thermal energy, but the distributions are strongly attenuated (ie narrow) and the temperatures are very rarely close enough. Read Prof Johnson’s note. It’s all in there.
Tim the process of exciting to a higher state happens only when there is near resonance. Conversion to thermal energy usually happens by a different process usually.solar.insolation
Doug writes at his website “Why would it when that coffee was not warmed by the “mirrors” on the inside of the flask? Only direct Solar radiation warms the oceans and land surfaces….Just because thermal energy (“heat” energy) is slowed down in its exit from the Earth’s surface to space, this does not mean that it will somehow bank up and make it hotter at the surface. This did not happen to our coffee at 97 deg.C. All it means is that the thermal energy takes longer to escape to space.”
I think you’re really confused. Much of what you say is true about back radiation being unable to heat that which caused it in the first place but then you seem to go off the rails. If you’re looking at a thermos for an analogy, then you need to periodically add energy to it as happens in the day/night cycle.
Imagine if every few hours you took out a cup of water from the thermos and added back a boiling cup of water. Eventually an equilibrium will be established and the thermos will have an average temperature. Now if the thermos somehow got better (maybe you shined it up or something) then after a further few cycles of refilling, the contents of the thermos is going to be hotter isn’t it because it cooled slower and residual energy remained. This is more like the reasoning used for GHG warming.
Doug writes “Conversion to thermal energy usually happens by a different process usually.solar.insolation”
Dont think so Doug. IR energy is absorbed by a GHG molecule which spins faster and/or vibrates faster or whatever and then an O2 or N2 molecule bumps into it and the energy is subsequently shared between the two molecules.
Occasionally the GHG molecule may end up with more energy and the non-GHG molecule less but more often than not, the GHG molecule loses energy to the non-GHG molecule. This is how GHG molecules heat the atmosphere through LW radiation.
I agree that the sun’s SW radiation provides the energy to the ocean or land surface in the first place. Nobody disputes that.
Kuhnkat,
“thank you. So IR is not generally involved in the photoelectric effect, only visible or higher.”
Who said that? There is also an effect called upconversion but it does rely on multiple photons and is less efficient. Light is light and it doesn’t matter what frequency of it you are discussing. There is no magic threshold where it stops behaving like a photon.
“That is, if the photon is not the right amount of energy then it is simply reemitted apparently?”
If the photon is not of the correct energy it can be transmitted or even reflected (although both are also EM resonant processes). If it is to be re-emitted, it must first be absorbed.
Jeff Condon,
you sounded as if you thought there might be an IR photoelectric effect. I am no expert so would be interested in information on that.
You say the photon must be absorbed before it is reemitted. How do we know it was absorbed and not reflected. Even if it is “absorbed” if it is immediately reemitted it doesn’t change anything.
For that matter I have read where particles will not absorb energy if it doesn’t match the energy states of the particle in general, not just with the photoelectric effect. Is that true? What are the broad details there if you or Carrick have some time. Or is it they absorb and reemit what is not used to change levels? It seemed to me this would be an obvious example of absorbing photons with no change in energy.
Carrick,
thank you for the info on the two-slit. i am trying to reach an understanding of what the current accepted knowledge is of how it all works. Having so many people seem to think there was no wave effect when I read even electrons apparently have wave properties keeps me a touch confused as to what and who should be listened to. Of course, I realize that just because someone is misled in an area like this doesn’t necessarily mean they don’t have other areas right.
Jeff Condon,
Wiki, AbsoluteAstronomy, PhysicsForums… all have pretty much the same info that the photoelectric effect is only from shorter wavelengths from visible through ultraviolet. Just do a search on Photoelectric effect and you will see what i was reading on the first two pages.
Steve and others;
Any radiation heading for the surface (at some angle in practice) has absolutely no effect on the surface. It does not get converted to thermal energy and so cannot affect the rate of thermal energy leaving the surface. It is merely immediately radiated out again with the same frequency and intensity, never having been converted to thermal energy. How could it possibly affect radiation coming out at different angles from other molecules? When you shine two torches towards each other, but not directly – just so the beams cross – they have no effect on each other’s beams. This I suggest would be a close analogy if backradiation even exists.
The surface does not need to radiate at all to lose heat – it can do so by diffusion, conduction, convection, evaporation and chemical processes. The surface does not act like a blackbody because it is not surrounded by a vacuum or insulated from losses by these other means.
I suspect that most radiation actually starts in the atmosphere, not the surface. But then I also suspect that any backradiation is extremely small compared with upward radiation, because I do not believe radiation has an equal probability of going towards warmer areas than towards cooler areas due to the higher energy and density of molecules “blocking” it in the warmer direction. If there are numerous captures and re-emissions, then even a slightly higher probability than 50% will, in the limit, ensure the vast majority heads for cooler regions. There are no experiments to my knowledge which demonstrate backradiation warming something, or slowing its rate of cooling.
But whatever happens, the end result (if any gets to the surface) as far as energy and rates of cooling are concerned is just the same as if it had been reflected by a mirror. A mirror neither warms nor cools more slowly when it reflects IR radiation.- it, like the surface, is not affected at all because the radiating energy is never converted to thermal energy. You can only add and subtract like things such as thermal energy. Radiation does not cancel out other radiation as there are different angles involved for a start. The transfer of all thermal energy is in one direction, and the reason it only takes place in one direction is because only the cooler body “receives” it and converts it back to thermal energy.
Hopefully this will help all to understand why an atmospheric greenhouse effect resulting from radiation is a physical impossibility.
Doug,
You sent me to Claes Johnson and a footnote again. WTF? At least you are claiming that the photon is magically reflected. See how easy that was to write! So if that is the case, how come the camera sensor isn’t “reflecting” the 8 micron photon the camera uses to measure the ice. You should be smart enough after 40 years of physics to know that the Planck curve covers that range for room temp — or else the camera couldn’t measure it!
“Others – please look first for answers on my website as I really don’t have time now to answer any more questions.” –DOUG listen!! You haven’t answered mine.
How does that physics denying camera work?
You provided a link to Wein’s displacement law which only explains the shape in the emission spectrum with temperature. You have absolutely ignored the flat fact of physics that in order to measure light you need to absorb light. Silly rule that! But the god of physics is funny about her rules. Since the detector of the camera is warmer than the source, it cannot absorb according to your ‘it bounces off’ theory. This is a real theory right? Or is there some reason that it sometimes can be absorbed and sometimes not?
Another hilarious fact is that if it bounces off, half the energy bounces back toward the ground, delaying the travel from ground to space. Lets call it Doug’s backradiation. Now I assume that even though ground emits a wide range of frequencies, that you would state the magic backradiation would reflect again from the same temperature ground back upward, never interacting thermally with anything because everything going upward is cooler so it is either reflective or non-absorptive?
So many physics rules are violated. Here is one which directly refutes what I call the magic reflection theory of light.
Your interpretation of Weins displacement rule means that an object slightly warmer than another object – very slightly warmer – only has the cute tip of higher frequency energy to transfer and all of radiative transfer physics fails.
Two different temperature plates have an energy emission of cT^4. The absorbed energy according to me of the cooler plate is the emission of the first minus the emission of the second. Between them they have a common side so the rest of the world thinks the radiative energy transfer is 1/2 cT1^4 – 1/2 cT2^4. I’m not going to do the math right now for you but it is easy to see that this equation fails COMPLETELY if the only energy available for absorption is the little sliver of higher frequency light at the tip of Planck’s curve.
But wait Jeff! Planck’s curve never goes to zero!!! http://en.wikipedia.org/wiki/Planck%27s_law
WTF! If it never goes to zero, then at what threshold does the magical reflection happen?!!
Oh man!!
steve fitzpatrick 198
“Just as in every physical system, the local temperature is always controlled by the flows of energy in and out, including radiation and convection.”
I believe that a more accurate statement is that “changes” in the local temperature are controlled by the flow of energy. It is not possible to calculate the temperature of a body based on flow alone. If we have a net flow of zero we know the temperature is constant but we have no idea of the temperature. If we know that we have a flow of 100 Joules/sec entering a body with a thermal mass of 100 Joules/deg we know the temperature will rise at a rate of 1º/sec. We still have no information on the actual temperature.
If we are to calculate a temperature where the energy flows sum to zero we need additional information. At least one of the energy flows will have to be a known function of the actual temperature. It is the temperature difference between the current value and the equilibrium value that drives the energy flow.
It is therefore quite reasonable to claim that the temperature causes the flow rather than saying the flows cause the temperature. In practice, the flow and the temperature difference vary simultaneously so the question of causality is mainly philosophical. We have an equation linking the flow and the temperature difference so if we know one we can always calculate the other.
I rather like the view expressed by Leonard as the undoubted presence of a lapse rate avoids all the difficulties involved in the calculation of layer by layer temperatures. One cannot calculate the radiation exiting at the top of the atmosphere without that knowledge. There is also a relatively simple relationship between the effective height of the radiation, the effective temperature, the lapse rate and the surface temperature. Whatever arguments one may have about cause and effect it seems a much easier way to link the presence of CO2 to effective height and, via the lapse rate, to surface temperature.
All,
I think there are many comments being made due to less optimum choice of words resulting in misunderstandings. Radiation is always bi-directional. I said so several times. It is the net balance that determines heat transfer, and this is always hot to cold. The absorption of radiation by the atmosphere acts as a radiation insulation, but ON AVERAGE convection compensates for radiation heat transfer to maintain the lapse rate that would be present for any level of radiation heat transfer. ON AVERAGE the only effect of radiation heat transfer being SOMEWHAT restricted is to raise the level of outgoing radiation to space. This increase in level along with the lapse rate is what heats the AVERAGE surface. Back radiation does transfer energy to the surface, but it is ON AVERAGE always exceeded by upward radiation, and thus there is NO average heat transfer by radiation to the surface. Obviously there are local exceptions, due to the face that there is not always complete atmospheric mixing, but it is close to mixes on average.
An informative (but not exact) example for the effect is an electrically heated object in a vacuum. Putting a layer of insulation on it results in a higher temperature at the bottom of the insulation, since the heat loss is slowed by the insulation, but energy is continually being added. The object temperature increases until the radiation from the exterior of the insulation matches the power into the object. You would not say in that case that the insulation is heating the object by back conduction. That is exactly analogous to claiming back radiation is heating the surface.
Leonard:
What you are missing is that the lapse rate doesn’t act as a form of heat transfer from the top of the atmosphere to the bottom, so it can’t be an explanation of the warming of the surface. I fact when you account for convective heat loss (which is real, because the environmental lapse rate is rarely or never the adiabatic lapse rate, and hence must be diabatic), you find that you end up with less surface warming than if the atmosphere weren’t allowed to convect.
Also, nobody here (sensible anyway) is saying that back radiation is heating the surface, the back radiation simply impedes the loss of heat energy from the surface. It is not a source of heat energy, because it is a cooler object, and net heat energy must flow towards it. The source of heat energy is the surface of the earth (which presumably has stored solar radiation energy), and the warming comes from the surface heat energy, not from the back radiation. If the surface were at 0K, there’d be no warming.
That said, I think you need to revisit this statment:
That is the statement that Steve and I both were really reacting to.
TimTheToolMan, sorry I don’t have time to comment on all of what you wrote today. Meeting day. Let’s just stick with this one:
Simply because the mean free path is short near the surface, doesn’t mean there isn’t an upwards gradient in thermal photons. There is, because the density decreases with elevation, and even more importantly, once you’re out of the boundary layer (at night time about 200-m above the surface) the water vapor concentration in the atmosphere dramatically drops.
Nobody said the infrared photons were treated as if they were free.
Doug Cotton #227 and other comments,
“Any radiation heading for the surface (at some angle in practice) has absolutely no effect on the surface”
My friend, you are utterly confused about all this. Infrared wavelength photons reflected by the surface? Never turned into heat? No effect? Look at the infrared absorption spectrum of some common substances… like liquid water (~70% of Earth’s surface), which absorbs ~100% of 14 micron wavelength light (the primary emission frequency for CO2) within 1 mm of the surface. The water molecules DO NOT KNOW the temperature of the CO2 molecule which emitted the photon (all 14 micron wavelength photons are created equal… they have exactly the same energy content, which is determined by their wavelength!). And why do you think black-body emission/absorption has anything to do with having a vacuum surrounding an object? The presence or absence of convection and thermal conduction has nothing to do with black-body radiation.
You continue to confuse the net flux of photons (which is always form warmer to cooler) with absolute emission rate. They are not the same. The absolute emission rate per unit surface area goes as c* T^4 (T being the absolute temperature, c is a constant). The net flux per unit surface area between two black-bodies in the form of parallel plates of infinite extent goes as c*{(T1)^4 – (T2)^4)}, where T1 and T2 are the absolute temperatures of the two plates. If the plates are at the same temperature, the net flux is zero (of course), but the two plates continue to exchange photons, because they both continue to emit in proportion to T^4. Black-bodies always emit photons continuously. More complicated materials are not perfect black-bodies, but their emission continues to be proportional to ~T^4, and most common materials (like those that form the surface of the Earth) are pretty close to perfect black-bodies in the infrared.
This physics is 100+ years known, broadly confirmed, and widely used in common applications (like Jeff’s IR camera video shows). A lot of people have spent time trying to explain the basics of radiative flux to you on this thread. Seems to me they have been wasting their time. There are lots of good, reasoned technical arguments in support of the view that climate science has consistently and substantially overstated the influence of increasing GHG forcing on Earth’s surface temperature, but a gross error in the basic physics of black-body radiation is not one of them. If you want to make a contribution to the technical dialog (and I suspect you do), then do yourself a favor and try to learn the basics. It will take some effort, but that is better than continuing to make outrageous “scientific” claims and (as my Japanese friends sometimes say) frankly speaking, continuing to embarrass yourself.
230 – Leonard – please can I check what constitutes “Putting a layer of insulation” an object in a vacuum?
Steve,
“The water molecules DO NOT KNOW the temperature of the CO2 molecule which emitted the photon ”
You are taking away my fun. 😉 I was going to write next that apparently the time-invariant and otherwise identical photons obviously have a tag on their butts indicating the temperature of the emission source.
Jorge #229,
Of course the heat loss from an object is a function of its temperature (radiative, convective, and conductive losses), and you need to know that function explicitly to calculate the object’s temperature based ont he energy flow into the object. This is obvious, and I would never suggest otherwise.
“I rather like the view expressed by Leonard as the undoubted presence of a lapse rate avoids all the difficulties involved in the calculation of layer by layer temperatures. ”
You may like it, but it doesn’t actually avoid those difficulties. The existence of a lapse rate equal to the theoretical rate depends on radiative flux upward being not sufficient to maintain energy balance at the surface in the absence of convection; if the energy flux to the surface is low enough, the temperature change with altitude will not follow the theoretical lapse rate at all (AKA an inversion). In addition, the lapse rate is not constant… it depends on other factors like atmospheric moisture, so no difficulties are avoided.
I think Doug’s “missing heat” is possibly an assumption that CO2 molecules are the ONLY ones involved. The IR camera uses photon sensors having much lower energy gaps (if that’s the mechanism used for detection) than CO2,
In other words, the IR sensor elements can absorb low-energy photons, which CO2 molecules cannot.
Jeff #235,
Yes, but a further complication is that you would need a Maxwell Demon sitting on each potentially absorbing molecule to read the tags and decide whether to absorb or reflect based on the tag values. 😉
I can’t spend any more time on this fun-stuff today. Take care.
MarkF,
Besides some conflict with Dougs theory, the IR sensor
itis tuned to higher energy photons.Jeff Condon and Mark,
the camera specs claim a spectral range of 7-14 um. OOPS, misses the two big CO2 bands, but, that is definitely in the upper IR.
DC #227
Equally, an electric grill (broiler) doesn’t need to radiate IR and red light to lose heat – it can do so by diffusion, conduction, convection, evaporation and chemical processes.
So even though it doesn’t need to do this, it strangely continues to radiate anyway – I wonder why?
Carrick,
It is absorbed solar energy that is heating the ground, not the lapse rate. However, it is the lapse rate and location of outgoing radiation that determines what the resultant surface temperature needs to adjust to, to match incoming solar and outgoing thermal radiation. Since the lapse rate is essentially constant for adding trace amounts of CO2, only the altitude is changing. If something changed the lapse rate, that would be a factor. I am assuming the bottom of the atmosphere (on average) is about the same as the ground temperature. This is not exactly true always, but is on average valid. If the ground temperature changed from increased sunlight or from more greenhouse gases, the ground heats the air by conduction, convection, evaporation/condensation, and radiation/absorption. Since the CP and g stay the same, the result has to be a raising of the location in the atmosphere where radiation to space (average) balances incoming absorbed radiation. I repeat it is not the lapse rate heating the ground and atmosphere, it is the Sun. It is also not backradiation.
234, Curious,
Vacuum is conductive and convective isolated, but not radiation isolated. Putting a low thermal conductivity (insulator) layer on an internally heated object that had a high emissivity surface (e.g., styrofoam or aerogel on say Iron), would slow heat transfer to the exterior surface of the insulating layer, that then has to eliminate the building heat by radiation to surrounding vacuum. Note I said internally heated object, so energy keeps being added. The result is that the surface of the heated object has to become hotter to conduct the energy to be radiated out. i.e, there has to be a temperature gradient in the insulator to push the energy out.
A consequence of the magic reflection theory of IR.
Build a metal box at room temperature with all sides closed. IR is emitted inside the box from wall to wall, never being absorbed. In a short time the energy inside the box builds to enormous levels while bouncing around, if you forget to open the box, it would gain mass from that energy as heat from the environment piled into the center of the box. If you opened it, you had better be warmer than the box or you may absorb gigajoules of energy. If you didn’t open it, a substantial fraction of the energy in the universe would end up inside the box. Insta-black hole.
Think of the magic power storage unit we have just created. Truly amazing stuff.
—
Another question, in reflection the photon interacts with the electrons in the atom and by resonance the field is flipped and the particle changes direction. Of course that means there is interaction with the electrons so of course being an imperfect world, some heat is created. In this perfect no-interaction system, how does the EM field get reversed? Is it through interaction with the electrons? Is the reflection diffuse or specular?
or is this all simply physically false.
Carrick,
If almost all of the solar energy was absorbed in the atmosphere going down (but a small amount made it to the surface), and almost all the upward thermal radiation absorbed in a very short distance, and also if the atmosphere were sufficiently mixed to maintain the adiabatic lapse rate, it would be the adiabatic compression heating of downward mixing atmosphere and possible conduction from atmosphere to ground (until the ground matched the lower atmosphere) that would primarily assure the surface was hot enough to nearly match the surface temperature to the surface value on the adiabatic lapse rate as controlled by the location of outward going radiation times lapse rate. The assumption of being sufficiently mixed is critical here, and may not occur if no radiation directly reached the surface (a local inversion may excessively damp mixing, as buoyancy would not help). However, even Venus, which has a relatively small solar insolation at the surface, has a surface temperature and adiabatic lapse rate that satisfy that condition closely, so mixing is adequate even there. Earth is different due to the atmosphere absorption of solar energy going in being low relative to ground absorption.
Discussions about less than perfect mixing, local temperature inversions or reduced lapse rates, and condensation effects, may be valid separate points, which I had agreed to, but I qualified my discussion by specifically stating I was looking at the simplified average case.
Carrick “Simply because the mean free path is short near the surface, doesn’t mean there isn’t an upwards gradient in thermal photons. There is, because the density decreases with elevation”
Hence my comment further down in that same post…
“As one moves higher into the atmosphere, the temperature and pressure drops and the collisions become less frequent and radiation will increase until at some particular altitude and temperature and GHG concentration, radiation becomes predominant.”
The important point I’m trying to make is that I dont believe there is much GHG induced DLR reaching the earth’s surface (and by implication any heating is largely by conduction) and that the density of the atmosphere plays a direct role in how the GHGs operate particularly with respect to setting the altitude of effective outbound radiation (and by implication the surface temperature through standard theory)
Leonard: No, no, no. Firstly the lapse rate is primarily (say, 70%) determined by pressure. Secondly, you need to come to grips with Hansen’s error and other maters on my (updated) Home page,http://climate-change-theory.com Energy absorbed by carbon dioxide is very quickly re-emitted by one means or another, and even if the atmosphere warms a bit, such warming cannot affect the surface by backradiation. Yet warming by backradiation is the official IPCC view.
Hey Doug, My physics profs usually respected my input and I theirs – this was a long time ago. I corrected more than one back then. Unmix your teacher-student mentality. My 25 years of hard physics application tell me that old dogs can indeed learn new tricks but the old dog must listen first.
Almost forgot. I hate to admit it but I was wrong. i stated that the IR camera was based on the photoelectric effect and it isn’t. Apparently it is the pyroelectric or ferroelectric effect which does operate in the IR range. At least that settles one bit of confusion.
Jeff said ” there is interaction with the electrons so of course being an imperfect world, some heat is created.”
This is garbage. Photons don’t have mass and don’t cause friction. If the frequency of the radiation is significantly below the peak emission frequency of the receiving body then there is absolutely no conversion of the coherent radiated energy to thermal energy because it is all re-emitted with exactly the same frequencies and intensities, thus leaving no energy behind.
Doug Cotton #245,
You are still utterly lost. As they say, you can lead a horse to water, but you can’t make him drink. Like Carrick, I’ve now added you to my do-not-respond list. Good luck with your adventures in insanity.
Doug, you need to turn in your physics credentials. Light does in fact cause heat in mirrors. It also creates pressure on objects even in vacuum.
Read on my friend, but lecture no more.
And who said physics isn’t fun!
Jeff: See my (updated) Home page http://climate-change-theory.com
PS My physics professors (including Harry Messel, Julius Sumner-Miller and Verner VonBraun) came from the days prior to Hansen’s huge mistake in 1981.
#251,
Doug,
As an aeronautical engineer, I would absolutely love to read your stories about them, if you want to write about your experiences, I will post them here. This little blog has more views than a lot of town newspapers and the education level is very high so you would find it hard to locate a better audience. That said, my god man!, resumes for correctness? Is that what they taught you? I think not.
I’m glad you went to a nice place and learned from awesome people. VonBraun would turn you inside out on radiation as we have because the concepts are basic and he would tell you EXACTLY the same things. The difference is that you would have listened to them rather than recite names and links. Physics is a harsh mistress even when you are older. It doesn’t care if you are famous for janitorial services or relativity.
Real is real my friend and that is our world.
Jeff and others.
Physics must be verified by empirical evidence.
I am still waiting for just one of you to show me just one documented simple experiment such as with metal plates at night receiving backradiation and getting warmer, or cooling slower.
Without such empirical evidence there is no point discussing any greenhouse effect.
And what have you to say about Hansen’s obvious error: http://climate-change-theory.com ?
Jesus dude, C.A.M.E.R.A!!!
HINT: There is a plate inside it!!
Answer the question!!!
EVERYONE:
Any radiation from a cooler atmosphere does not have its coherent energy converted to incoherent thermal energy when it meets a significantly warmer surface. Only thermal energy can affect other thermal energy. Only solar SW radiation is converted to thermal energy when it meets the surface.
You can only add or subtract like things. There is no new thermal energy created so no “slowing of the rate of cooling” is caused by any backradiation. See my Radiation page at http://climate-change-theory.com
So the official IPCC explanation of the GHE is debunked and there is no experiment supporting their claim that warming happens or the rate of cooling is slowed.
Jeff I have explained above how an infra red camera works by measuring frequency and converting that to temperature using Wien’s Displacement Law. . Its “plate” is not warmed by the radiation.
Doug,
You have explained nothing. Tell me how a 15 micron photon from an ice cube is not absorbed by the earth yet a 15 micron laser is?
Your sophistry has reached my limit of tolerance. Your aloofness has exceeded it. Show me some thought or stop pretending.
PS How would the camera form an image if it had a physical plate which was warmed by different amounts in different places? How would it measure the temperature in each pixel (sensor) so to speak? Conduction would spread the heat out, diffusing the image anyway. What it can and does measure in each sensor is the frequency of radiation received. Note the reference to “temperature calculations” here http://coolcosmos.ipac.caltech.edu/image_galleries/ir_zoo/lessons/background.html
PPS Visible light has far higher frequencies (and so far higher energy) than IR radiation. Theoretically it does warm the CCD of a digital camera, but the camera does not measure the temperature increase in each pixel – instead it measures the frequency (just like an IR camera) and converts that frequency to colors.
A laser is not a blackbody emitting spontaneous radiation with frequencies relating to its physical temperature. It is fed by electricity which produces stimulated (induced) emission which very different from spontaneous emission. http://en.wikipedia.org/wiki/Stimulated_emission
Um, Doug? Stop digging. Now.
#258 Doug,
You just stated that the radiation is not received at all, it is reflected. Focus on your claims that this is impossible or in lieu of that, why energy is received for the camera.
#259, the digital camera is sensitive to 8 micron wavelengths, not visible, read the datasheet. Besides that, I’m pretty sure that ice cubes don’t light up in the dark to my eyes. Claims of visible light sensitivity are inaccurate. Also, the measurement of any light sensitive camera – no matter the wavelength – requires absorption.
#260, what is the difference between a 15 micron laser photon and a 15 micron blackbody photon?
Are we getting the point yet?
Mark,
Doug has 40 years of physics background, I’m sure he can hold his own with a mere Aeronautical Engineer. – yes I’m that tired.
I did not say it was reflected.
The digital camera I take photos with – http://douglascotton.com – is sensitive to visible light.
You have not produced evidence for what you say about 15 micron lasers etc, let alone any warming by actual backradiation.
BTW your ice cube would have to be about -80 deg.C to have a peak frequency of 15 microns http://www.calctool.org/CALC/phys/p_thermo/wien
Maybe this is Jeff’s “15 micron laser” (LOL)
variable angle manual ellipsometer L117F300 specifications
Alignment: Built-in axis of rotation of incident arms is in the sample plane. Incidence angles are easily set with no need for alignment prisms. The robust frame maintains alignment for precise and reproducible results.
Method of Measurement: Nulling type ellipsometer using high quality calcite prisms
Detector: Solid State
Incidence Angle: 30°, 45°, 50°, 55°, 60°, 65°, 70°, 75°, 80°, 90° (prealigned and easily set via detents)
Light Source: HeNe 6328 Laser gives less than 1 mW output on sample
Beam Diameter: 1mm diameter ( 1 x 3mm on wafer @ 70°)
Microspot (Optional): 15 micron diameter (15 x 52 micron on wafer @ 70°)
Induced emission is very different from spontaneous emission because the intensity is such that photons arrive more quickly than the rate at which they can be re-emitted. So the molecule is still excited when the next photon arrives.
When laser beams are generated two identical photons are emitted in the same direction as the incoming photon because the intensity is high enough to produce this induced emission..
I acknowledge that there are lasers up to about 16 microns – some used for detection of weak concentrations of chemicals. http://cqd.eecs.northwestern.edu/research/qcl.php If they cause warming it is because photons are arriving while the molecules are still excited. This rarely happens in nature. Johnson’s paper only relates to blackbody radiation.
in any event, the majority of lasers emit in the visible and UV range.
Here is Wiki’s version of the IR Camera explanation:
http://en.wikipedia.org/wiki/Thermographic_camera
Kuhnkat: The Wikipedia article is incorrect on one important point. The IR camera does inot measure intensity. It measures frequency (or wavelength) – that’s why its specifications have a wavelength range. Frequency is proportional to absolute temperature. Intensity tells you nothing about wavelength. See what the manufacturers themselves say: http://www.brickhousesecurity.com/about-night-vision-ir-cameras.html
To all: Just read my posts and others on http://wattsupwiththat.com
The key issue to start with is Hansen’s blunder about that 33 degree difference supposedly due to you know what. Well, you think you know. If Hansen was wrong, then AGW crumbles. See my (updated) Home page: http://climate-change-theory and similar info by others on WUWT: http://wattsupwiththat.com/2012/01/31/jim-hansens-balance-problem-of-0-58-watts/#more-55750
#242.
Jeff, this metal box experiment won’t work as you describe, because clearly you don’t read EXACTLY what Doug says, or you’d see the flaw in your logic.
Doug has said over and over and over and over (you get the idea) again that a cold photon cannot be absorbed by a surface that is significantly warmer – 12 times, in fact, if you care to count them
Despite being asked more than once (twice, in fact) he fails to respond to anyone who asks exactly what significantly is. But as all the walls of your box are about the same temp, we can assume that they can accept the photons from the opposing wall.
So a different thought experiment for Doug.
Image two steel sheets of 1m^2 each, in a vacuum, separated by 10cm. The sheets are in lab conditions, so are about 15C to start. Both have electrical heating elements attached to the rear surface and temperature measuring devices. The whole apparatus has been left to stabilise.
Heat up sheet A to 100C. Sheet B will show a temperature increase. From here on keep the power provided to the heater on sheet A constant.
Next, bring sheet B up slowly to 100C. What happens to temp A? Is 90C significantly cooler than 100C? How about 95C? At exactly what temperature in B does temp A react?
Let sheet B cool to say 50C. Let it stabilise, and measure temp A. Does it drop?
Next drop sheet B to the floor of the chamber (overall 15C, remember). What happens to temp A?
Steveta_uk: Thanks for the support. When I first read Jeff’s box experiment I felt his statement about energy building up indefinitely to “a substantial fraction of the energy in the universe” was so absurd as not to warrant a reply. It cannot build up faster than the rate at which it is input. In practice the box would burn or explode. (Put an electric radiator inside such a box and try it if you don’t believe me.)
Could a radiator contribute “a substantial fraction of the energy in the universe ” (LOL)
But remember, no wall on the box is a perfect insulator, so each wall will lose some energy by conduction and then radiation and diffusion to the outside world.
Now, regarding “significantly” well, once again you need to read Johnson’s computations. As he explains, each distribution is strongly attenuated. This means the graphs are narrow. The amount of overlap could be calculated from formulae, but I’m not getting into that here. The effect reduces as the amount of overlap reduces. imagine two squashed up normal distributions moving apart from each other.(though they are not strictly the shape of a normal distribution) and you will understand that the overlapping area reduces rapidly. The peak frequency (proportional to absolute temperature) is easily observed and in fact measured by IR cameras.
In general, with typical differences between atmospheric temperatures and surface temperatures as they are, there is negligible conversion to thermal energy. And whatever is happening has probably happened for a billion years or more.
Doug Cotton,
you made an assertion about the IR camera. Please provide links to back it up.
PS Once again, the walls in Jeff’s box are not themselves blackbodies. They will absorb radiated energy while the source of the energy is warmer, but they also diffuse thermal energy to the air inside the box and conduct it to their outside surfaces. But the air can get as hot as the source of the energy (which I will assume is an electric jug heating element) because energy is transferred to it by diffusion, not all by radiation. Probably before that happens, an equilibrium point will be established when the outside surfaces will be radiating the same total amount of energy per second as the element is producing – just as the inside of your car only warms up to a certain temperature when the Sun delivers energy to the seats etc inside, which warm the air, which warms the windows, which conduct the energy to their outside surfaces, which then radiate it away, establishing equilibrium. Note that the walls probably have have a larger surface area than the element, so they can disperse all the energy with a lower power per unit area than the element.
Why do I waste my time explaining these red herrings? I’m going to keep to climate matters now, and I suggest you focus on Dr James Hansen’s 1981 blunder where he assumed the Earth’s surface was a blackbody.
See http://climate-change-theory.com for all your answers, including an explanation of Hansen’s error and even one about how IR cameras work (on the Radiation page)..
Doug,
No energy balance? No realization that you have no energy balance. No recognition that your theory violates the very concept of reflection. No recognition that your new description violates your old one. I have repeatedly proven your statements wrong and repeatedly asked questions which you are unable to answer. In fact, your replies are so disconnected that they indicate that you are not even aware that your argument has been completely shredded.
Your “teachers” are laughing somewhere but I’m just irritated.
A mirror will still warm to the temperature of the room it’s in.
And your theory produces ““a substantial fraction of the energy in the universe” does it? You really don’t understand the point made by myself and Prof Claes Johnson, do you?
Steveta tried to point out your error, but you are obviously too narcissistic to ever admit your misunderstanding.
Goodbye – your future posts will be ignored by myself and others I suspect. Try arguing against other experts on WUWT.
Steveta,
“be absorbed by a surface that is significantly warmer – 12 times, in fact, if you care to count them”
I am fully aware of what he wrote and have asked several times myself when the photon can be absorbed and when it cannot. The magic box was my last ditch attempt to get him to admit that one 15micron photon is the same as any other. There are no tags on their butts to indicate where they came from. He has been unable to reply even one time to that question. It gets right to the heart of the argument he makes. So since he can’t reply to the obvious question other than ‘signficantly warmer’, I gave him the stupid box – which again he failed to explain.
The whole concept is quite impossible and several others here have tried repeatedly to explain to Doug that he is seriously mixed up. I don’t believe that he has studied numerical physics at all at this point.
Doug,
I’m sorry you cannot see reason. Your aloofness is beyond tolerable anyway. My guess is that our conversation cost me 1 reader but the boredom of it has long ago chased most of the PhD’s out of the room. You really didn’t realize who you were talking with. I will go back to running my lighting company now.
Jeff Condon,
sorry to go off in another direction, but:
“The magic box was my last ditch attempt to get him to admit that one 15micron photon is the same as any other. There are no tags on their butts to indicate where they came from.”
So, how do they know what altitude the radiation originates from when they are measuring from the ground or TOA?!?!?! This is basically what I thought. Whether the 15 micron is true BB or from a dipole molecule from a collision or spontaneous, how do we know??
Kuhnkat,
I’m not sure the minimum energy threshold for the photoelectric effect.
The process of reflection/transmission itself is an interesting one but it does require interaction which vibrates electrons. This vibration, creates heat sometimes so my questions to Doug were intended to get him to consider how come this cannot create heating in one case where in others it does. If a mirror emits at 15um and a laser strikes that mirror, in my world the laser adds some energy to that mirror’s surface. Where does his magic theory say we get a perfect reflection because the high energy physics world could really really use that kind of revalation.
Your window in your home absorbs very little visible light because it doesn’t match the energy states of the particle. UV light is absorbed though. You can see the same kind of effect in the absorption bands of CO2. Black paint doesn’t let much visible light get away though.
In the case of the camera, my point was that the absorption did occur. Energy was indeed added from a cooler source to a warmer one. Doug also failed to explain that.
Jeff Condon,
The original photoelectric effect has a limit somewhere between visible and near infrared based on the online articles. The pyroelectric and ferroelectric effect that appears in crystals is in the far infrared.
When the particle reemits the photon the vibration is gone. If it can reemit the photon the vibration which you claim cannot have affected anything or there wouldn’t be the energy to reemit the photon. Of course, if there are other absorptions before this has had time to happen…
I’m amazed that Doug seemed to think I was supporting his argument. He completely ignored the heated plates section of my post (#274).
Here is another link
http://www.glolab.com/pirparts/infrared.html
Kuhnkat,
Reflection is sometimes a complex process. (Pun intended.) 😉
There are two contributions to reflection, both of which arise from the interaction of the photon’s oscillating electric field with the electrons in the surface they strike. Specular reflection from a ‘transparent’ material depends on both the angle of incidence and the refractive indexes of the materials that meet at the surface. Reflection from metallic surfaces has an additional component which is related to the high electrical conductivity of the metal… which translates essentially to an extremely high optical absorption coefficient (photons can’t penetrate the surface even a tiny fraction of their wavelength without being absorbed). The high electrical conductivity of metals means there are relatively “free” conductance electrons, which respond to the photon’s electric field by moving to generate an opposite (in a phase sense) oscillating electromagnetic field, which heads off into space opposite the original angle of approach of the photon… an oscillating electromagnetic field is just another photon! All materials have a “real” refractive index (which you probably are familiar with) and an “imaginary” refractive index, which describes the material’s rate of absorption of photons with depth. Both values are dependent on wavelength. Describing a material’s optical properties requires the use of the complete refractive index, which is a complex number. Aluminum is remarkably high in reflectivity, especially in the infrared spectrum, which is why you often see insulation that is faced with an aluminized surface coating. Conservation of energy demands that a perfectly reflective surface can’t emit as a blackbody at all…. if it did, it would cool continuously via loss of IR photons, so an aluminized surface mostly loses or gains heat by convection, not radiation… which is useful if you want to reduce heat flow..
A fun web page for anyone who wants to see how the optical properties of materials influence reflection is: http://refractiveindex.info/?group=METALS&material=Aluminium, and associated pages.
Steve,
thank you for your patience in instructing me in the details of reflection.
I’ve read some bizarre theoretical physics regarding reflections – I think is was in Brian Greene’s book “The Elegant Universe” – where the ‘wave’ properties of reflection are discussed, and in line with a two-slit experiment, the idea is that virtual photons travel via every possible path from a source via a mirror to your eye, and all paths except the traditional linear reflection cancel out via destructive interference, hence you see a perfect image. As I said, bizarre stuff.
Steveta,
It does explain the two slit experiment but without having read the book the concept is expandable to particles of matter as well. That means particles/atoms etc are all standing waves in spacetime. Unstable constituent particles are missing portions of their waveform and decay into various types of EM waves.
I really would have fun discussing that but am not much of an expert.
You all need to understand what Johnson is explaining, together with my explanation of “significantly” above which he explains similarly prior to the quote below. Then you can answer your own questions about metal plates Steveta.
None of this has anything to do with pyroelectric effects (in which you would expect warming because the human body is warmer than the device) nor has it anything to do with reflection. The IR camera works differently and nothing is warmed – just the frequency is measured. Only lasers with wavelengths shorter than about 9 microns would warm anything in the human body, as the calculations show. The vast majority of lasers have much shorter WL – wonder why?
All I ever said was that the end result was the same as reflection so far as energy is concerned. The mere resonating vibration is a once only up and down thing because the radiation is immediately re-emitted. There is no energy loss because it leaves the surface with the same spectrum and the same energy, so how could any energy have been converted to thermal energy? Thermal energy doesn’t just appear just because radiation hits a molecule. The energy in radiation has to be converted back to thermal energy by a specific process which is not like friction, because a photon has no mass.
Some of you make the same mistake as Hansen in overlooking diffusion, evaporation etc.
None of you has discussed Hansen’s error – presumably because you now realize he was wrong as explained right at the top of my Home page http://climate-change-theory.com
What is the point of discussing any GHE when Hansen was wrong from the very start of his argument?
Here’s an excerpt from Johnson which it seems most have been to lazy to read, so I’ll spoon feed you just this once more …
Bye
.
Claes Johnson couldn’t explain the camera either.
Fact: Spectroscopy proves a warmer gas does not absorb spontaneous emission from a cooler source
Question: Why?
Fact: If there is any backradiation it is not even enough to melt frost (which is in the shade) in 8 hours, yet it is supposed to be a quarter as powerful as the Sun at noon
Question: Why?
Fact: The atmosphere cools faster than the surface at night – see Nahle’s experiment in Sept 2011.
Question: Why?
Fact: An infrared camera measures frequency. Its sensors do not have to be warmed by IR radiation
Evidence: Read the manufacturer’s description of how it works linked in one of my posts above
http://www.brickhousesecurity.com/about-night-vision-ir-cameras.html says …. ” Night Vision Security Cameras are extremely sensitive to these thermal-infrared emissions and amplify them, converting the measurement of the invisible wavelength into an image on your monitor. ”
Obviously wavelength is calculated from a measure of frequency.
“Fact: Spectroscopy proves a warmer gas does not absorb spontaneous emission from a cooler source”
Question: Why will you not provide a source for your statement?
Fact: If there is any backradiation it is not even enough to melt frost (which is in the shade) in 8 hours, yet it is supposed to be a quarter as powerful as the Sun at noon
Question: Why do you cite a useless observation without values?
Fact: The atmosphere cools faster than the surface at night – see Nahle’s experiment in Sept 2011.
Question: So what?
Fact: An infrared camera measures frequency. Its sensors do not have to be warmed by IR radiation
Question: If its sensor cannot absorb light, how can it measure light frequency?
Same questions as before Doug. Same non-answers. Pay attention to the questions if you wish to make a point.
Doug if you want to suggest a theory, might I suggest you start with an accepted fact or set of observations and then logically progress from that to your point? At the moment you’re just blankly stating that radiation from a cooler body cannot be absorbed by a warmer body.
Where is any evidence that is the case?
Please also note that attempts to point me/us to another website will be taken as “no evidence” because you ought to be able to spell out the argument starting with a known fact or observations in just a few words. If your idea is “too complex” to cleanly state then its probably wrong.
By comparing #287 with #289 others will realize how little Jeff Condon reads and understands.
And, no, a sensor does not need to absorb and convert radiation to thermal energy in order to measure its frequency, any more than you need to be run over by passing vehicles in order to count them.
And still no one can refute my point that Hansen made a huge error, hence the AGW hypothesis crumbles.
Tim
My book does that – I’m just finishing off the appendices.:
If in 1981 Jim Hansen wanted to “suggest a theory” he should have started with correct physics (rather than his huge blunder) and empirical proof that backradiation warmed the surface or slowed its cooling.
Read my site folks! http://climate-change-theory.com – note also the Radiation page.
Tim
I’m not blandly stating that – I have linked the proof (derived from accepted facts) both on my Radiation page of my site and in earlier posts here.
\
We really are going around in circles here, and I have about seven other forums I’m posting on, so I can’t spend more time on a few people here. Over 30,000 have visited my site, so why don’t you guys?
Doug Cotton, B.Sc (Physics), B.A.(Econ), Dip.Bus.Admin
(engaged in full time climate theory research)
Finally, Tim, it is not a new theory that radiated thermal energy only passes from warmer to cooler bodies.. Any process engineer and many contributors to forums like http://wattsupwiththat.com are very much aware of it.
What’s a “new theory” and a wrong one, is that there is backradiation slowing the cooling rate and increasing the warming rate each day and night..
It’s not a new theory that the Earth’s surface transfers thermal energy by diffusion, conduction, convection, evaporation and chemical processes.
What’s a new theory and a wrong one is that it acts like a blackbody which only transfers energy by radiation.
.
“And, no, a sensor does not need to absorb and convert radiation to thermal energy in order to measure its frequency, any more than you need to be run over by passing vehicles in order to count them.”
See Doug, that is where you and others have missed the amazing nature of the camera example I have given you. I’ve been waiting for you to finally address some specifics of your claim but apparently the idiots who built the camera, used a sensor called a microbolometer. If you read the two links I have given YOU as I have read yours, you would know already that this is a very special kind of sensor which actually measures its own temperature change.
From Wikipedia:
A microbolometer is a specific type of bolometer used as a detector in a thermal camera. Infrared radiation with wavelengths between 7.5-14 μm strikes the detector material, heating it, and thus changing its electrical resistance. This resistance change is measured and processed into temperatures which can be used to create an image. Unlike other types of infrared detecting equipment, microbolometers do not require cooling.
You are in a perfect trap now. You have admitted that the sensor needs to absorb, yet now claim it doesn’t “need” to turn into thermal energy.
So now how does that silly camera work?
“No amount of experimentation can ever prove me right; a single experiment can prove me wrong.” – Albert Einstein.
Can I suggest a revision to your book?
Human body temperature corresponds to about 9.3 microns – well within the 7.5 to 14 micron range of the camera.
Clearly the camera sensor would be cooler than body temperature. It may be attached to the outside of a plane and so cooler than the surface it is photographing.
It will be limited in some circumstances, but there is a spread to the distribution. so it is not a sharp cut-off as Johnson explains.
Jeff’s point does not disprove Johnson’s “Computational Blackbody Radiation”
PS As on my site, January 2012 was 0.09 deg.C cooler than the mean for the whole period since January 1979.
hehe.
What about -20 C?
I read the “cutoff” apparently you don’t follow Johnsons math well.
PS
These IR cameras are used at night. Even at eye height the air is usually cooler than the surface at night, so the sensor would be too. Hence it will easily form an image of the land profile and any human or animal even when using it at eye height..
“These IR cameras are used at night. ”
WTF? It is an ice cube!
Doug, it directly refutes your claims. End of story. I have used bolometer cameras – for fun mostly but comeon man – you were so cocky that you wrote how little I understand while obviously not even bothering to read my single link.
Care to revise your statements counselor?
Actually it supports my claims.
Read about the limitations: http://www.dias-infrared.de/pdf/p020.pdf
Why does it only read down to -20 deg.C when 14 microns corresponds to -66 deg.C?
The reason is that 14 microns is its physical limit and it probably would read such if it were at that temperature itself.
However, at typical cool temperatures at night – say around 0 deg.C, the value of -20 C is about as far as the near resonance extends – remember I said “significantly different”
So the very fact that it is not going to be able to be pointed from the surface to a cloud at say -40 deg.C and measure its temperature (even though it should detect 13 to 14 microns) shows that the radiation from the significantly cooler cloud does not warm its sensor..
I knew it would not be able to measure -40 deg.C even before I read this. However, a more expensive IR camera that measures frequency can do so.
QED
Bye
Doug writes “Finally, Tim, it is not a new theory that radiated thermal energy only passes from warmer to cooler bodies.. Any process engineer and many contributors to forums like http://wattsupwiththat.com are very much aware of it”
Isn’t it possible you’ve misinterpreted the accepted theory that the *net* radiation goes from hot to cold and imposed your own additional restriction that *no* radiation goes from cold to hot? And that in fact there is no actual evidence for what you’re suggesting?
PS And even if the camera were at, say, 20 C then I would expect it to read 0 C from your ice cube as this would be within the range of not being significantly different. But It may not read -20 C very well, if at all, if it were much over 20 deg.C itself though.
The vast majority of the atmosphere from which 15 micron carbon dioxide radiation is supposed to come, is obviously significantly cooler than -20 deg.C – in fact 15 microns corresponds to -79.965 deg.C. The atmosphere gets down to about -100 deg.C at the mesopause.
Tim; I did not say “no radiation goes from hot to cold.”
You completely miss the point of Prof Claes Johnson’s computations. If you don’t want to read it that’s your choice, but don’t waste my time showing myself and others that you don’t understand the very basis of it that conversion to thermal energy only takes place when the received radiation has frequency above the cut-off frequency of the receiving surface, as determined by Wien’s Displacement Law.
correction: I did not say “no radiation goes from cold to hot”
Doug writes “Tim; I did not say “no radiation goes from hot to cold.””
I never said you did. I specifically wrote “and imposed your own additional restriction that *no* radiation goes from cold to hot?”
Cold to hot. Isn’t that what you’re saying doesn’t happen? No radiation from cold to hot?
Doug writes “correction: I did not say “no radiation goes from cold to hot””
Then precisely stated, what ARE you saying?
Doug,
First, you have chosen a different camera. Why? I gave you the datasheet!
The camera I gave you is sensitive to 7um but all of that is moot.
the Johnson “cutoff” of a -20C ice cube is far beyond 14um. Its 14 um energy should not affect a room temp sensor by your favorite theory. It seems that you need to read Jonson’s claims more carefully as well as what I have written.
Actually Johnson seems to ignore several consequences of his own math but we are nowhere near discussing math at this point.
See correction posted a minute before your comment: frankly you make me so angry I can hardly concentrate.
Now how about addressing the rest of the post #304 pointing out your misunderstanding, and that of Hansen et al?
Doug writes “Now how about addressing the rest of the post #304 pointing out your misunderstanding, and that of Hansen et al?”
But your claim here has nothing to do with climate. You appear to be suggesting there is something fairly fundamentally happening with the radiation and its ability to be absorbed. If its not simply stated to be “no radiation from cold to hot”, then what exactly is your theory?
Jeff – the Johnson cut-off is determined by Wien’s Displacement Law and is the peak frequency of radiation by a blackbody at any particular temperature. You don’t appear to understand that. There’s a calculator (linked above and on the Radiation page of my site) showing -20 deg.C has peak frequency -11.4468 microns, nothing like 14 microns which is about -80 deg.C as I said. What’s your problem?
Tim – you won’t understand until you read Johnson or my site.
Doug,
So 11 is above 7, are you sure that means the camera can’t detect below -20c or is it the thermal noise that the idiotic scientists state which causes the problem. We aren’t worried about the peak frequency of the -20 C source though, we are worried about the peak frequency of the sensor right? The sensor is “significantly” higher temperature than the source so it cannot absorb?
Anyway, after you recent claim that an uncooled bolometer cannot measure more than -20, I looked around for almost 5 minutes. Sweat pouring off my fingers, I can provide this link:
Click to access 16414861.pdf
How does 100 Kelvin sound?
Good thing I didn’t write a book.
Doug writes “conversion to thermal energy only takes place when the received radiation has frequency above the cut-off frequency of the receiving surface, as determined by Wien’s Displacement Law.”
By “conversion to thermal energy”, you really mean “absorbed” dont you? AFAIK there are only two other options, reflected or ignored entirely. What exactly do you mean by cut-off frequency?
Doug writes “you won’t understand until you read Johnson or my site.”
I started reading your site and it was very confused. Again, I haven’t heard from you any actual evidence. Are there any predictions this theory makes successfully for example?
Doug,
I should mention that to be fair, I didn’t bother to calculate the cutoff frequency. I just assumed they would overlap. I’m a little surprised they don’t but it doesn’t matter to my point.
You are now faced with an uncooled bolometer which can and does measure to 100K or -173C using a sensor with a possible temperature of 50C.
Is that ‘significantly’ different?
If not, what is!
It stinks being wrong Doug. It stinks worse when you are loud and wrong.
Your blog has 30,000 views, this post has 2000 by itself.
Jeff writes “You are now faced with an uncooled bolometer which can and does measure to 100K or -173C using a sensor with a possible temperature of 50C. ”
To be fair a bolometer does measure local temperature, not local temperature *increase* per se. Net radiation will still be away from the bolometer in that case.
#318, NET radiation is from warm to cold. Doug’s claim is that if the temperature difference is arbitrarily great enough no energy can go the other way.
I’ve given him several examples now which prove that radiation travels both ways and the difference can be used to calculate the net. Now his theory has been flatly proven wrong, right before his eyes. Unfortunately, Doug has become so entrenched in the thing that he has written a book and will NOT see any evidence which contradicts his belief.
Jeff,
The sensor elements in the microbolometer are in fact cooling/heating relative to the sensor back-plate (held at a controlled temperature) due to net emission or net absorption to/from the imaged object. I mean, the germanium optics work in both directions. So while there certainly are photons being absorbed by each sensor element, there are also photons being emitted by each element back through the optics, mapping back onto the object that is being imaged. And just like all black-body radiation, it is the net flux that changes the temperatures of the individual elements, which must be always from warmer to cooler. If part of the imaged object is warmer than the sensor, then there is a net flux from that warmer part of the object to the area of the sensor where the image of that part of the object forms. And visa-versa for parts of the object that are colder than the sensor.
The net photon flux can be in either direction for each pixel (each sensor element)… the net for each pixel can be either negative or positive. But there is still a continuous photon flux in both directions, of course! 😉
I agree with you Jeff
“I’ve given him several examples now which prove that radiation travels both ways and the difference can be used to calculate the net.”
He’s yet to give even one clear example where radiation isn’t flowing in both directions. Its his claim therefore its his responsibility to do so…
Tim
I am not saying what you state I am.
I am saying what is on my Radiation page.
Steve,
Yup. Isn’t it simpler when reality takes hold? LED’s can detect light. Lighting the LED doesn’t stop that effect until the crystal is saturated.
TTTM,
He is completely cornered. Were I his lawyer, I can make a few more sophistic arguments but would be looking for a settlement.
Doug “I am saying what is on my Radiation page.”
If you cant state it concisely in a few lines, and give an experimental result that supports it and refutes the opposing view then its probably not correct.
Jeff has not produced evidence of his linked camera actually measuring below -20 deg.C, as far as I can see in his linked article.
Please quote the section or page number and the actual words for all to see.
But the article he links does discuss using the camera on a space mission with temperatures down to -20 degrees C for the sensor. So at those temperatures it could possibly measure down to some of the coolest parts of the atmosphere I would guess. So what?
The camera with similar technology that I linked to (in #301) was specific about its ability to form images of subjects but only if they were at a minimum of -20 deg.C (see table on last page), which proves my point. It can’t do so for clouds at -40 deg.C if viewed from a surface at, say, 0 deg.C.
The camera’s inability to measure below -20 deg.C confirms what Johnson says because radiation from such significantly cooler clouds is not converted to thermal energy in a sensor which is at about the temperature of the surface.
Tim
“A few words” – OK read the bold type in #304
A proof? That’s a different matter – so read “Computational Blackbody Radiation” by Claes Johnson, well published Professor of Applied Mathematics whose work extends that of Einstein and Planck in particular.
Doug claims “The camera’s inability to measure below -20 deg.C confirms what Johnson says because radiation from such significantly cooler clouds is not converted to thermal energy in a sensor which is at about the temperature of the surface.”
No it doesn’t. You’re going to have to do a LOT better than that. Ever heard the statement “extrordinary claims require extrordinary evidence” ?
Douglas,
You have got to read the words. You have quoted the thermal limits the sensor is being tested to, not the temperature range it can measure.
Your quote sir:
The technical requirements of the microbolometer array flow directly from the science and measurement requirements
of MERTIS to achieve excellent science at Mercury. The MPO spacecraft will orbit Mercury in an eccentric orbit with a
periherm of 400 km and an apoherm of 1500 km. MERTIS measures thermal infrared radiation emitted from the surface
of Mercury, the magnitude of which varies greatly over an orbit; Mercury has the largest diurnal temperature variation
of any planet in the Solar System, ranging from 100 K to 700 K.
MERTIS, with another 1 minute of hard sweat, returned this link.
http://solarsystem.dlr.de/TP/MERTIS_en.shtml
You may find it interesting that the other uneducated fools think that they will be able to measure 200K with a 50:1 signal to noise ratio.
I suggest that after you write your book, you call the ESA before they spend a billion dollars on a launch in 2012.
Doug writes ““A few words” – OK read the bold type in #304”
“conversion to thermal energy only takes place when the received radiation has frequency above the cut-off frequency of the receiving surface, as determined by Wien’s Displacement Law.”
I’m going to assume “conversion to thermal energy” simply means absorbed, and there is no “cut-off” frequency inherent in Wien’s Displacement Law. What is the basis for the cut-off frequency? That is, afterall the crux of this theory. You cant simplify your statement without explaining that part.
Jeff @#325 and TTTM:
Why do you have trouble comprehending simple English like “using the camera on a space mission with temperatures down to -20 degrees C for the sensor.”
It is my article which gives specifications for “measured temperatures” from -20 deg.C.
I asked you to quote such specifications for yours and you didn’t. You just brought in another red herring about measuring Mercury temperatures from space. So what?
Now answer the question. Produce the evidence that one of these microbolometers can measure temperatures significantly below -20 deg.C when the sensor is at typical Earth surface temperatures, not in space – the quoted frequency range does not prove that they can do so at Earth surface temperatures, even if they can from space which would be understandable.
I repeat over and over: the fact that they can’t measure more than perhaps 20 to 40 degrees below their sensor temperature (ie significantly below) proves my point and provides the proof TTTM wants.
Now, either of you provide proof of the relatively new theory of the 1980’s that backradiation can slow the rate of cooling and increase the rate of warming of the surface.
The fact is that all such experiments trying to support the IPCC claim have failed and thus been hushed up.
Tim at #329
“I’m going to assume “conversion to thermal energy” simply means absorbed, and there is no “cut-off” frequency inherent in Wien’s Displacement Law.”
You assume wrongly. Radiation can be “absorbed” and immediately emitted at the same frequencies and intensities, without any loss of energy – in other words, without any conversion to thermal energy. That’s what blackbodies do. That’s 100+ year old physics.
The cut off frequency is the same as the peak frequency (as you would know if you had read Johnson) which is certainly in Wien’s Displacement Law which says it is proportional to absolute temperature.
Whilever you refuse to read Johnson and/or my Radiation page you just keep wasting my time.
Now how about discussing Hansen’s error – see my post #61 and Home page at http://climate-change-theroy.com
Doug,
I gave you exact quotes from the paper as to what MERTIS will measure. I even linked to another article saying the same thing. MERTIS is only designed to operate with its own physical temp between -20 and 50C yet they anticipate a 50:1 S/N ratio for measurements of 200k which is -73 C by my calculations and it has the ability to measure even further down to 100K or -173C.
“MERTIS will map 5-10% of the surface with a spatial resolution higher than 500m. The flexibility of the instrumental setup will allow to study the composition of the radar bright polar deposits with a S/N ratio of >50 for an assumed surface temperature of 200K.”
Dude, read the links. You have been PROVEN wrong and you have gone so nuts that you won’t read the words.
Here is another – completely different bolometer sensor designed to measure to 220K
http://jemeuso.riken.jp/en/about14.html
It is real, face it.
Sorry typo: http://climate-change-theory.com which reads …
In 1981 NASA’s Dr James Hansen made a huge mistake: he assumed that the Earth’s surface acts like what physics calls a “blackbody.” But a blackbody has to be surrounded by space, or totally insulated so that it cannot lose “heat” (which should be called thermal energy) by conduction or other means to its surrounds. The whole Earth plus atmosphere system does indeed act like a blackbody when viewed from outer space where all that can be detected is its electromagnetic (EM) radiation. But the surface is continually transferring heat to the atmosphere and the crust beneath by various means including diffusion, conduction, convection, evaporation and chemical processes. We can deduce the mean temperature of the whole system, but not the surface.
As this article explains, Hansen deduced (because of the above wrong assumption) that there is an unexplained 33 degree warming of the surface which he claimed must be due to an atmospheric “greenhouse effect” all caused by water vapour, carbon dioxide, methane and a few other gases which do indeed absorb the infra-red (IR) radiation coming from the surface. But that is not a reason for warming the surface.
So this “greenhouse effect” is based entirely upon the assumption that so-called “backradiation” from a cold atmosphere is able to increase the temperature of a surface which is warmer than the source of the radiation. This is a physical impossibility as is proven theoretically by Prof Claes Johnson and empirically by Prof. Nasif Nahle in documents linked below. This paper also proves that the “two stream” heat flow concept is wrong.
Consider “backradiation” created by the silver reflective lining on the inside of a vacuum flask filled with coffee at, say, 97 deg.C. The reflected radiation will not raise that 97 degree temperature at all. Radiation which is received from a source which is cooler is rejected by being re-emitted with exactly the same frequency and energy that it had before it arrived. Only radiation from a warmer source can lead to warming of a cooler body. Imagine two metal plates at different temperatures placed close to each other in a vacuum. They will each tend towards a temperature which is between the initial two temperatures. At no stage will the cooler one cause the warmer one to get hotter still. A mirror held above the ground at night will send back to the surface more than twice as much backradiation as all greenhouse gases combined. According to those energy diagrams there is about a quarter as much radiation coming out of the surface, even at night, as there is coming into the surface from the Sun when it is directly overhead. So will it make the surface warmer if reflected back? Why would it when that coffee was not warmed by the “mirrors” on the inside of the flask? Only direct Solar radiation warms the oceans and land surfaces. I postulate that virtually all IR radiation which is generated in the atmosphere undergoes multiple capture and re-emission processes, including collisions with surface molecules which also re-emit it. Eventually it finds the only escape route, namely to space.
Even if some small amount of IR radiation from a cooler part of the atmosphere does reach the surface it will not even slow the rate of cooling. Why would it affect radiation coming from different molecules at different angles? If it were converted to thermal energy then, yes, it could slow the rate of cooling – but it is not converted. If some thermal energy is “trapped” in the atmosphere it cannot affect the temperature of the surface. And, if the atmosphere does warm in some hot spots, it will increase its rate of radiation in order to cool back again.
Another link:
Click to access Poster%20-%20Imagers%20-%20Katayama.pdf
Parameter Specification
Size < 100mm x 150mm x 200mm
Mass < 3kg
Detector Uncooled infrared detector
Wavelength 8 to 12μm
Number of pixels 640 x 480
Spatial resolution < 200m @600km (< 0.33 mrad)
Field of View 12° x 9°
Frame rate 30Hz
High gain: 180K to 340K
Dynamic range
Low gain: 180K to 400K
NEdT 0.2 K@300K
Oh sh#$, 180K……..
I’m done for a bit Doug. I am a little sorry that your concept was broken but your arrogance helped get past that.
I will answer later if you are a bit more respectful of the rest of us but changing the subject isn’t going to work.
#333 Jeff:
“they anticipate a 50:1 S/N ratio at 200k which is -73 C by my calculations and to measure even further down to 100K or -173C.”
I call your bluff. Show me your calculations
Then show me evidence of measuring below -20 deg.C when the sensor is at surface temperatures.
Why would your linked camera which uses the same technology have such superior capabilities to the one I linked which specifically quotes a “measured temperature” range with a minimum of -20C?
Don’t play games with me, son, no one pulls wool over my eyes.
Show me your calculations
200 kelvin minus 273 Kelvin (carry the 1) = -73 Celcius
LMFAO literally. Thanks Doug.
The sensor is only intended to operate between -20 and 50 as stated in the paper. Read it yourself.
I was finished blogging tonight until you asked me to calculate for you. hahahaha.
good stuff.
Kuhnkat,
We don’t know anything about where the photon originated, including deep space. We only know that by probability there is a predicted amount from the air which will swamp other sources, and the values match up very well.
Kuhnkat,
Nothing about what I’ve written means that the warming, which does happen, is measurable or dangerous. Sure we measured warming, who can 100% prove it wasn’t caused by H20?
There are many arguments against AGW, backradiation is not one of them.
Jeff at #336
You might bluff others but not me.
This camera is for use on satellites. With sensor temperature at TOA about 200K or less it is hardly surprising that it could measure 180K which is not significantly colder.
I asked for realistic minimums from realistic surface temperatures. My specifications for my linked camera gave realistic values for what is, after all, the same technology.
I suspect that none of these cameras form images of subjects much more than a mere 20 to 30 degrees below the temperature of their sensors, which is within the range of not being significantly different.
The fact that they don’t display significantly cooler objects proves my point. QED
“If it can reemit the photon the vibration which you claim cannot have affected anything or there wouldn’t be the energy to reemit the photon. ”
Now you’ve got it. If it re-emits at the same frequency, no change, no heat. The answer to why things aren’t always that perfect is only a step away.
“With sensor temperature at TOA about 200K ”
The Mercury sat sensor does not EVER go below -20C.
Kuhnkat,
look to absorption band widening as an interesting way that a photon might be re-emitted at a different level. Also, consider what happens if the electron jumps to a shell which in a crystal allows it to move away from the nucleus.
Jeff Condon,
“Also, consider what happens if the electron jumps to a shell which in a crystal allows it to move away from the nucleus.”
not much of that pyroelectronic crystal laying around. Which of the possible modes of broadening actually apply to the surface? Would we be able to mostly ignore doppler for instance? Resonance broadening certainly wouldn’t apply except for self radiation which doesn’t appear to be much of a subject even though it would affect the SB results.
When a photon is absorbed by the surface why would it necessarily be reemitted even close?? With the dynamic state the energy could be used to fire off a different photon couldn’t it? The warmers point to the chunks in the spectra as proof that energy isn’t making it out of the system. If that much wasn’t getting out it wouldn’t take years to know it!! Obviously the energy is getting shifted to other bands for radiation as the energy balance is pretty close.
Yes I waved on your suggestions what to look at. I really don’t know enough to evaluate the jumps.
Kuhnkat, pyroelectric just means that that you have a transducer that produces electricity in response to heat or vise versus (usually). It doesn’t specify the mechanism. Solid state pyroelectric effect is one example. Electro-mechanical is another (e.g., the piezoelectric bimorph, one of my favorite toys). Biological pyroelectric works using electro-chemistry (it’s used in various types of biological sensors). You can even use the pyroelectric effect to generate x-ray radiation in crystals. (And that doesn’t seem compatible with your claim about the pyroelectric effect appearing in crystals in the far infrared.)
Article on crystals, pyroelectric effect and X-ray generation.
I’ll leve it there, because if I said everything you could do with it, it would just sound like sci-fi, but it’s not.
Douglas Cottonmouth would be interesting if he sounded like something besides one of those poorly written scripts for Data on STTNG (you know where he launches off into some silly-sounding sciencey explanation of something.)
Carrick,
the camera is said to be utilizing the pyroelectronic effect. The camera senses 7-14 um. Doesn’t that imply that the pyroelectronic effect in specific crystals is sensitive to far infrared??
Carrick,
here is one of the articles I glanced at looking for the info on the camera:
Wikipedia also lists it. Probably where Douglas got the quote about body temps.
OOOPS, just realized I used the term Far Infrared when it should have been Long Wave IR. Shoulda stuck with the numbers.
Jeff @340
“They anticipate” is hardly scientific proof that they will succeed.
I don’t consider such hypothetical untested guesses as valid evidence: they are probably making the same mistake as Hansen.
What is relevant is what such cameras can do at Earth surface temperatures. Can they form images of clouds at -40 deg.C for example? At least one manufacturer admits they can’t, and obviously that would be tested from actual use of the product. Your space camera is not yet tested up there I gather.
Why does the camera I quoted only measure down to -20 deg.C when it can work with longer WL up to 14 microns which should measure -66 deg.C?
You have to admit you have no empirical evidence of such cameras measuring significantly lower temperatures than their sensors, whereas there would appear to be evidence that they don’t at surface temperatures, which is what is relevant to the greenhouse effect.
Dr Roy Spencer’s plot continues its downward curved trend with the January 2012 figures just out.
January 2012 was 0.19 deg.C BELOW the mean for the whole satellite period since 1979.
“You have to admit you have no empirical evidence of such cameras measuring significantly lower temperatures than their sensors, whereas there would appear to be evidence that they don’t at surface temperatures, which is what is relevant to the greenhouse effect.”
I just gave you 3 separate links to cameras which do exactly that. What is wrong with you?
Here’s the plot: http://climate-change-theory.com/latest.jpg
Notice the good fit of that curved trend line. This is showing the superimposed ~60 year cycle. The underlying trend has been decreasing from 0.06 deg C / decade to 0.05 deg.C / decade using data since either 1880 or 1900, it makes no difference.
The year 2100 should see an increase in the ~1,000 year cyclic trend of about 0.4 to 0.5 deg.C with a long term maximum before 2200 about another 0.3 deg.C higher, after which the world should see ~500 years of cooling to a Little Ice Age scenario.
#351 Jeff: None of them spells out that they are measuring with the sensor at surface temperatures I suspect, but quote me the page or table number to save my time and others’.
I showed you a link with a specification table that proved one camera does not go below -20C and you could not answer why without using Johnson’s result.
I’m still waiting for an actual experiment measuring what actual backradiation from the atmosphere actually can do to warm an actual object at surface temperature.
All this hypothetical stuff about what cameras in space are “anticipated” to do is not science. I want to see photos of cold clouds (~-40 C) or similar taken with one of these cameras from the Earth’s surface. Not much to ask, surely.
There seems to be a sub-discussion on measuring cold surfaces with long wave IR cameras. Unfortunately there seems to be a lot of misunderstanding going on. There are four basic types of detectors for UV, visible light, or IR. These are: Photomultiplier tubes containing a photocathode which emits electrons when illuminated, the electrons are then amplified by a chain of dynodes photoelectric, where photons of suitable energy cause electrons to be emitted, and the electrons are accelerated to cause a cascade amplification of electrons, which is measured. This is only useful for fairly high energy input photons (generally visible or UV), and uses the photoelectric effect. However, this is not what is used on long wave IR cameras. Photoresistors or Light Dependent Resistors (LDR) which change resistance according to light intensity are usable at longer wavelengths but have some limitations. Photovoltaic cells or solar cells which produce a voltage and supply an electric current when illuminated are often used for IR cameras. All of the above have stray signals if the detector is not considerably cooled, and this occurs as noise. The problem with this is not that a signal is not detected, but that signal to noise is poor. The final type is an array of optical detectors that are effectively thermometers, responding purely to the heating effect of the incoming radiation, such as pyroelectric detectors, Golay cells, thermocouples and thermistors, but the latter two are much less sensitive. Please don’t keep arguing about photoelectric effect detectors for IR.
“None of them spells out that they are measuring with the sensor at surface temperatures I suspect, but quote me the page or table number to save my time and others’. ”
You are wrong, as I have shown above but I am done being your librarian.
“I showed you a link with a specification table that proved one camera does not go below -20C and you could not answer why without using Johnson’s result.”
This was your first explanation of what YOU consider significant – stated in reverse order from what Johnson writes (source vs sensor). I replied in minutes with a colder measuring room temp sensor. One which exceeds your new – and completely arbitrary – definition of significant.
“All this hypothetical stuff about what cameras in space are “anticipated” to do is not science”
The 2012 camera had better be tested and installed already or they won’t get it launched. Are you really this far gone that you are going to deny the cameras work? What a friggin joke. The papers are years old and written by people building the instruments.
I am absolutely certain that these sorts of devices are in operation right now but don’t want to look for them. I’m not the guy writing a book on imaginary physics.
You have demonstrated the #1 strongest version of scientific denial I have witnessed in 25,000 comments here.
I forgot to specifically mention in 354 that only the Photomultiplier type detector uses the photoelectric effect. An electron has to be emitted from the surface to space away from it to be the Photoelectric effect.
Yes Leonard, I am quite aware of how the photelectric effect works.
But we were talking about Microbolometers which are relevant to climate because they depend on warming of the sensors and then detecting a resulting change in electrical resistance.
I’m sure the space camera “works” in some sense. And I would expect it would be operating either at temperatures inside or outside the spacecraft. Neither would be -20 deg.C I suspect, so neither of us really know what they expect. Obviously they would have no problem photographing the surface, and clouds would show up somehow as they would block the surface image. So I don’t think the images would prove or disprove the point.
So I come back to Earth and say, if one microbolometer can’t measure below -20 deg.C when its sensor is at surface temperatures, then why would another brand be all that different. If you haven’t looked up specifications in detail so that you can quickly point me to them, as I did for you, then you have presented no evidence. I scanned all your links but missed seeing what I was looking for. I am not just being lazy.
Obviously we will have to agree to disagree about Johnson’s result, but it is well known among process engineers according to one such blogger.
If it were not true, then you could set up a radiating body (such as an electric radiator) with much more intensity (power) than one such as a small light bulb. The radiator would send more radiative flux towards the light bulb, so would it make it glow brighter?
“, as I did for you”
Yeah right.
What a joke.
Link at my #301 last page – table of specifications – second line reads …
“Measured temperature range -20 C to 500 C (two ranges)”
What’s the “joke” ?
And if you have played with such a camera, how come you haven’t thought to try taking a photo of some clouds?
These are the questions you avoid answering ….
(1) Why do the specs say it can’t take images of objects colder than -20 deg.C.
(2) Question in bold @ $359
(3) Question about any other experiment (not based on cameras) showing warming effect of backradiation.
(4) Question about Hansen’s blunder assuming Earth’s surface is a blackbody even though it’s not insulated from its surrounds above and below.
I throw these 4 questions open to anyone.
Jeff In your link* Table 2, page 2 says it has a range 8 to 12 microns.
Well 12 microns corresponds to -31.67 deg.C.
So it also has a limit. I’m not going to argue between -20 and -31.67 deg.C
The point is, neither can photograph clouds that are, say, -40 deg.C and which are supposed to send backradiation that is supposed to warm the surface, yet doesn’t warm the sensors.
Good material for my book thanks – I’ll make more enquiries.
* http://www.icsoconference2008.com/cd/pdf/Poster%20-%20Imagers%20-%20Katayama.pdf
Kuhnkat
You’re still not quite understanding what Prof Johnson and I have been saying.
Your statements …
(1) “With the dynamic state the energy could be used to fire off a different photon couldn’t it?”
No, not if the frequency was significantly below cut-off frequency, because it is not converted to thermal energy.
(2) “Obviously the energy is getting shifted to other bands for radiation as the energy balance is pretty close.”
Only the SW from the Sun gets converted to thermal energy which may exit hours or even months later in winter, for example. So the broad IR spectrum that is emitted comes from that energy warming all the elements in the surface.
Everyone:
Jeff has now pointed us to a cheaper type of IR camera that does indeed depend on the sensors being physically warmed by the IR radiation, but, from what Claes Johnson determined, we would not expect such cameras (when at surface temperature themselves) to register radiation from sources colder than about -20 to -30 deg.C which is what their specifications do in fact state. But so-called backradiation should come from areas of the atmosphere that are colder than that, so, as I have been saying all along, such backradiation does not get converted to thermal energy and thus has no effect on the surface temperature, whether such is warming or cooling. QED
Douglas Cotton,
I was conversing with Jeff and others. I have read and find Claes work interesting, but, am not discussing it until it is further along. I do not have the math and physics skills to argue it anyway.
Doug,
You see the camera system designed to measure -173C in a direct quote from the ESA using a camera that never drops below -20 and you make up false information to ignore the facts in front of your face.
How does the camera work Doug?
BTW, you have gone against Claes Johnsons theory also.
“The point is, neither can photograph clouds that are, say, -40 deg.C and which are supposed to send backradiation that is supposed to warm the surface, yet doesn’t warm the sensors.”
So these clouds at say 6000m altitude and -40C cannot warm the surface due to the “significantly” warmer ST, at say 10C?
But I assume they could warm CO2 in the air about 100m below the clouds, which is at perhaps -39C? Not significantly warmer, is it?
And I assume that air could warm the air about 100m below, which is at perhaps -38C?
And I assume this effect can continue all they way down to air at 100m elevation, which is at perhaps 19C, so is clearly warmer than the surface?
Douglas,
The radiation and absorption from solids can be blackbody, or have some lesser levels (the absorptivity and emissivity can be different from 1), but in all real material cases, even though the peak in energy emission is temperature dependent, a very wide range of absorption occurs (for a black body, all incoming photons are absorbed). The net energy transfer to or from a material depends on both its temperature and the temperature of objects it is facing. Using black body as a simplification, the relation is simply net energy transfer, E=sigma(Thot^4-Tcold^4). i.e., it depends on both. For a fixed initial temperature object, the result would be to either cool or heat it, depending which surface is hotter. In both directions you would get a signal. Cooling the detector only reduces thermal noise, to improve signal to noise. It is not to detect or not detect.
Kuhnkat:
Absolutely. In your original comment, it seemed like you were saying that pyroelectric in crystals worked only at longer wavelengths than piezoelectric. Yes you can use it in IR too.
Leonard Weinstein #370,
No need to explain it again; he has seen the same explanation several times before. He didn’t believe the other explanations, nor will he believe yours. Just lost, totally and utterly lost.
Stevea_uk, you’ve nailed the problem of course, and I suspect without reading Cotton’s website.
Once he admits that a photon emitted from a colder object can be absorbed by a warmer object, “significantly warmer” or not, then you have backradiation and a GHG effect, due to absorption and reemission by GHGs .
Steve,
I’m not sure lost is the right word. I was thinking about a personality so unable to admit error last night that they would resort to not being able to read numbers. I’m not generally afraid of people but Doug is a little frightening. Psychology is a messy sport and people crack for all kinds of weird reasons. He can see the numbers, but refuses to discuss them and instead drifts to a position ever closer to reality while completely ignoring that he has now violated both his own original claims and observed reality. Instead of recognizing the problem, he says it is good material for his book and he will make “enquiries”.
I don’t think a theory has been smashed any more thoroughly here before yet on he trundles as though the wall doesn’t exist. It is the same effect with the ‘top’ paleos of Climate Science. At least they realize their work is very weak when discussing between themselves, Doug has no idea yet.
“Stevea_uk, you’ve nailed the problem of course, and I suspect without reading Cotton’s website.”
Carrick, I tried – really I tried – but the logical discontinuities really confused me. I simply could not work out what he was trying to argue.
I suspect that Professor Claes Johnson would be just as mystified.
Jeff, did you notice that Doug has drifted from an absolute statement about radiative transfer between cooler and warmer objects to now saying that perhaps -20C is only border-line significantly cooler than +20C, but -40C is definitely significantly cooler.
Steveta,
Yup. It bothers me that he hasn’t figured that out. Your point in 369 pretty well stuck a fork in the thing.
Steveta: @369
The flaw in your argument is this: There is only a small probability of slight warming when the temperature difference is, say, 30 degrees. Normally you only expect warming when the source is warmer, but the distributions can still overlap a bit when temperatures diverge.
What you are suggestion is a series of events each of which has a low probability, Hence the probability of all happening is far smaller still.
Doug Cotton:
Congratulations!
You’ve just admitted a GHG exists. Now you’re just quibbling over how large it is. So are we and so is everybody else.
So that means, welcome to the club.
As TimTheToolMan alluded to above, the path length is pretty short for a thermal photon in the absorption bands (so you are definitely getting substantial absorption by a “not significantly warmer” molecule for downwelling LWR in the GHG absorption bands), so you do need to include absorption/reemission to get the right answer. If you look at the multilayer models, in the simplest ones all of the LWR is assumed to be absorbed by each successive layer. See Figure 1 of TTTM’s reference. See this for a more complete treatment, and for extra points explain to us what the terms in the integrand are and what their origins are. With 40 years of experience, if you see problems with this treatment (originally developed by some nut named Chandrasekhar, who slipped on a banana peel and accidently won a Nobel prize for it along the way), you can awe us by pointing out in detail the problems with them.
I suggest you now admit you were completely wrong (we are all lifetime members of that club too, so no shame in that).
Leonard @370
You miss the point altogether because you have not read the first two pages at http://climate-change-theory.com nor understood Claes Johnson “Computational Blackbody Radiation”
Only the cheaper IR cameras that depend on actual warming of their sensors help to confirm Johnson’s conclusions.
Of course the original, more expensive ones that measure frequencies will handle much lower temperatures.
Here’s some interesting reading ….
http://wattsupwiththat.com/2012/02/03/monckton-responds-to-skeptical-science/#more-55890
Come and argue with myself and others on http://wattsupwiththat.com
And, one day, get around to reading the first two pages at http://climate-change-theory.com – you will learn something I promise.
“Only the cheaper IR cameras that depend on actual warming of their sensors help to confirm Johnson’s conclusions.”
The friggin -173K sensor is an uncooled bolometer that never has been even tested below -20C!! Since it is slated to orbit Mercury, I wonder if ‘cheaper’ was the reason.
Jesus, I’ve never met anyone so much in denial of the facts in front of them. Try and learn instead of lecture.
I’m sorry your pet theory doesn’t work, are you really surprised that your non-mathematical approach would overturn a century of physics?
Why would you keep linking to Wattsupwiththat? We all know the site. I guest post there on occasion.
Carrick @ 382
The problem with the argument has been outlined in my post above. A series of only slightly probable events has an absolutely negligible probability. That’s the probability of any GH warming. That’s why there is no effect seen above the long-term nature 1000 year cyclic trend currently heading for a maximum within 200 years, and rising at about 0.5 degrees per century.
Jeff:
There’s no statement in the specs saying it will actually measure such temperatures and I simply don’t believe them that it would. You just deduced that it would, but the specs don’t actually say that. They could not measure such unless the camera is attached outside the spacecraft so the sensor (and camera) go well below the -20 deg.C figure. I doubt that they have measured 100K temperatures on Earth with the camera at -20 deg.C. That would simply be impossible because of Johnson’s proof and other experiments now being completed by Prof Nahle this year which I can’t discuss here.
To all:
Be patient – you will all see numerous experiments within the next year confirming that Johnson is right. Meanwhile, Keep an eye on that frost in the shade and compare how long similar frost takes to melt in the Sun which is supposed to have about 4 times the power.
Bye
http://wattsupwiththat.com
http://climate-change-theory.com
An interesting segment of a recent debate: (See WUWT)
Jon Cook: “Monckton also repeats a myth … that most climate sensitivity estimates are based on models, and those few which are based on observations arrive at lower estimates. The only study which matches Monckton’s description is the immensely-flawed Lindzen and Choi (2009).”
Reply: I am not sure what qualifications Mr. Cook has to find Professor Lindzen’s work “immensely flawed”. However, among the numerous papers that find climate sensitivity low are Douglass et al. (2004, 2007) and Coleman & Thorne (2005), who reported the absence of the projected fingerprint of anthropogenic greenhouse-gas warming in the tropical mid-troposphere; Douglass & Christy (2009), who found the overall feedback gain in the climate system to be somewhat net-negative; Wentz et al. (2007), who found that the rate of evaporation from the Earth’s surface with warming rose thrice as fast as the models predicted, implying climate-sensitivity is overstated threefold in the models; Shaviv (2005, 2011), who found that if the cosmic-ray influence on climate were factored into palaeoclimate reconstructions the climate sensitivities cohered at 1-1.7 C° per CO2 doubling, one-half to one-third of the IPCC’s central estimate; Paltridge et al. (2009), who found that additional water vapor at altitude (caused by warming) tends to subside to lower altitudes, allowing radiation to escape to space much as before and greatly reducing the water vapor feedback implicit in a naïve application of the Clausius-Clapeyron relation; Spencer and Braswell (2010, 2011), who found the cloud feedback as strongly negative as the IPCC finds it positive, explicitly confirming Lindzen & Choi’s estimated climate sensitivity; Loehle & Scafetta (2011), who followed Tsonis et al. (2006) in finding that much of the warming of the period 1976-2001 was caused not by us but by the natural cycles in the climate system, notably the great ocean oscillations; etc., etc.
Doug,
Uncoold bolometer MERTIS with a sensor temperature never below -20C:
#328
MERTIS measures thermal infrared radiation emitted from the surface of Mercury, the magnitude of which varies greatly over an orbit; Mercury has the largest diurnal temperature variation of any planet in the Solar System, ranging from 100 K to 700 K.
#333
“MERTIS will map 5-10% of the surface with a spatial resolution higher than 500m. The flexibility of the instrumental setup will allow to study the composition of the radar bright polar deposits with a S/N ratio of >50 for an assumed surface temperature of 200K.”
Your answer — All this hypothetical stuff about what cameras in space are “anticipated” to do is not science.
I gave other examples of UNCOOLED sensors:
Another link:
Click to access Poster%20-%20Imagers%20-%20Katayama.pdf
Parameter Specification
Size < 100mm x 150mm x 200mm
Mass < 3kg
Detector Uncooled infrared detector
Wavelength 8 to 12μm
Number of pixels 640 x 480
Spatial resolution < 200m @600km (< 0.33 mrad)
Field of View 12° x 9°
Frame rate 30Hz
High gain: 180K to 340K
Dynamic range
Low gain: 180K to 400K
NEdT 0.2 K@300K
In exchange for my links, facts and effort, you respond with stupid crap about cameras not seeing clouds below -40. False claims that nothing was specific about the sensor dynamic range. I would call them lies if I didn't fully believe you are pushing a shopping cart full of cans on the weekends. You have violated Johnson's own math without even recognizing it. You have attempted to weasel out at every corner by deception rather than reason. You have even called the ESA instrument about to launch to mercury – "NOT SCIENCE!!" Doug, you have lost your point so badly that it is astounding you would pretend to ignore it.
Nearly 50k comments and after this exchange, I'm ready for TCO to come back.
Guys, I have to cut down on time spent on forums, so I’m just going to copy posts I write (or include links) on WUWT. I’ll not be looking here for responses, only on WUWT, so if you have a response written in a suitable polite form for that well known site (which has had 100 million views) then come on over ….
Here are a couple written last night and today …
http://wattsupwiththat.com/2012/02/03/monckton-responds-to-skeptical-science/#comment-883884
and
Oh yes, our friend Doctor Hansen, I presume. The one who, back in 1981 thought the earth’s surface acted like a blackbody. It must be looking a bit whiter down there now, is it Jimmy?
Anyway, the story goes that he calculated that the surface temperature would have been -18 degrees C because, just simplifying it all to a flat disk shaped Earth to keep the arithmetic easy, and just bunging in all the radiation from the Sun that gets through the atmosphere (which is thus cooling the planet) then, well then that good old reliable Stefan-Boltzmann stuff he once copied down from the blackboard (without really understanding) proved that -18 degree figure must be the case. And we have another 33 degrees to blame on mankind.
The only trouble is that a bit down the track it was decided to add on some evaporation and conduction and chemical processes and whatever other ways all that “heat” could get out of the surface. So all that just kinda went into the atmosphere, got double counted and then got radiated away to space by those helpful carbon dioxide molecules and their cronies, and, well … we still didn’t have enough of that lovely radiation.
So let’s invent a bit of (what’s a good name?) “backradiation” – no “Downwelling Long Wave Radiation” sounds more impressive than low frequency or low energy IR radiation. Yes DLW will be a good name for those proclaiming the science of doom.
But seriously, though, now that we have these things called IR cameras, some with that enchanting name of “microbolometers” we find that their sensors don’t actually get warmed by clouds that are at around -30 degree C or colder. What is the backradiation doing? If it can’t warmer a tiny sensor, how is it going to warm the oceans, or even melt a bit of frost all day long?
Seems like it’s not getting converted to thermal energy when it comes from a much colder source than the warm surface. And I guess if it’s not thermal energy, then it’s not going to affect the temperature of the surface, whether it’s warming or cooling.
And of course we can guess that the IPCC tried the obvious experiment to try to show backradiation warming the surface, but it failed so miserably it wasn’t even worth writing a confidential email about … hush, hush!
Prof Claes Johnson will help explain all this pretty basic physics known by many for a long time. You’ll find a summary here: http://climate-change-theory.com/RadiationAbsorption.html
Cool off and relax all!
Jeff at #386 – Yes the specs you quote read only to 12 microns – look up my previous post pointing this out. It is just below -30 deg.C as I keep saying – not your 100K. Read your “evidence” and my posts a bit more carefully, son.
Doug Cotton:
LOL. Not “negligible probability” if it is used in commercial applications.
Your argument, which you are correct you outlined using words above, is refuted by experience, data, theory, commercial instrumentation, and about everything else, including sanity.
You should admit that your model is in error and your infatuation with Claes Johnson, who you obviously lack the skill and training to recognize as a charlatan, is also in error.
We know you won’t, because you’ve said too much to back down. My suggestion:
Anonymity is your friend here. Choose a new alias, a new blog host, and start over. I think your reputation is completely ruined by your inability to admit your own gaffes at this point, and see no way for you to salvage it.
“Jeff at #386 – Yes the specs you quote read only to 12 microns – look up my previous post pointing this out. It is just below -30 deg.C as I keep saying – not your 100K. ”
They have built the instrument to measure 100K which is minus 173c.
They have claimed a 50:1 noise ratio to minus 73c
They intend to launch the thing this year.
I have given you other cameras with uncooled sensors of similar performance.
What is wrong with your eyes?
I suspect it’s not his eyes but his living-in-an-ivory-tower-and-not-used-to-being-challenged pride that’s getting him in trouble now.
You’ve really rung his ideas through the shredder. I don’t think even he’s dense enough to not be aware of how badly mangled his ideas have gotten. You’ve more patience on this matter than I admit I ever want.
Copy of my post on WUWT awaiting approval. Replies there please, not here
http://wattsupwiththat.com/2012/02/03/friday-funny-how-to-do-climate-graphs/#more-55904
Sharperoo:
Actually it’s more like 1,000 years over which natural trends tend to cancel out. Superimposed on a (roughly) 1,000 year natural cycle there is also a 60 year cycle, so 30 year trends are about the worst you could cherry pick.
The plot at the foot of my Home page http://climate-change-theory.com shows just why. It is a plot of the gradient of 30 year trends calculated on a moving basis each month.
It clearly shows just how much the gradient changes, but it also shows a cyclic pattern in such changes, so, by using an “axis trend” for the apparent, though incomplete sinusoidal pattern, we see a decline in the underlying 1,000 year trend’s rate of increase.
I hope I am making myself clear. The green line effectively takes out ENSO cycles and the 60 year cycle and tells us about the underlying 1,000 year trend. That trend was increasing at about 0.06 degree C per decade between 1900 and 1930, whereas it is now increasing at the slower rate of about 0.05 degree C per decade, and should continue to decrease, passing through zero in about 200 years, maybe less. When it does so we should see a 1,000 year maximum in the underlying trend which would be about 0.8 deg.C above the current level in the year 2200. After that it should decline for 500 years.
This method of fitting an “axis trend” overcomes accusations of cherry picking because, even if the data started 30 years earlier, the green line would be very close to the same. More importantly, we can be confident it can be extrapolated at least another 30 years, probably somewhat more.
Of course the 60 year trend will add superimposed short-term maxima and minima, meaning net level or slightly declining temperatures until about 2028, then rises for the next 30 years.
Jeff at 390:
Well they are contradicting themselves if your interpretation is right.
Remember the other brand of the same technology (and there are only two manufacturers of such) has specs down to -20 deg.C and said 14 microns. This brand says 12 microns which would be just cooler than -30 deg.C using Wien’s Displacement Law. So the specs are in close agreement, but the comment about S/N ratio appears to be way out of the range of the quoted specs. I will make more enquiries on this point, but at this stage, seeing that both specs are similar, I’m going with the specs.
Carrick writes “As TimTheToolMan alluded to above, the path length is pretty short for a thermal photon in the absorption bands ”
Yes, but its not just the path length that is short. As far as I can see the actual radiation from the GHG molecules in the lower atmosphere should be very low too. Hence the Kiehl Trenberth energy budget diagram showing considerable (324W) back radiation being absorbed by the earths surface is (I suspect) wrong because most GHG induced “back radiation” starts much higher in the atmosphere and never makes it anywhere near the ground.
Drat… that was me above rather than some anonymous poster who agrees with me 😉
#394: Anon
You are of course right about there really being little backradiation. Even the initial IR radiation is greatly overstated because about 50% to 70% of the energy is transferred from surface to atmosphere by other means, leaving far less to radiate. Thus the surface is not radiating anything like a blackbody and so that 33 degree figure is a joke.- see http://climate-change-theory.com
Copy of new post on WUWT – please reply there, not here ….
Joel Shore says:
February 3, 2012 at 5:44 pm
Of course a lot of people don’t yet believe what Claes Johnson, a well published Professor of Applied Mathematics has written in his Computational Blackbody Radiation which extends the work of Einstein and Planck in particular, and solves a dilemma which troubled Einstein all his life. It’s common for such scientists to be disbelieved in the early stages.
Maybe I wouldn’t believe it either if the climate records actually did show a human footprint, or if IR cameras like those microbolometers did actually form images of objects more than 30 degrees below zero. But, according to their manufacturers’ specifications, any IR radiation from significantly cooler sections of the atmosphere apparently doesn’t warm their sensors, which is exactly what Prof. Johnson would predict – and exactly why backradiation doesn’t warm the surface.
So how about you go and argue with him on his site if you think you can find fault with his assumptions, computations and/or conclusions? Come back and let us all know if you think you are the first in the world to do so.
You might also like to put up your argument as to why you think the Earth’s surface acts like a blackbody, and thus why my arguments on my Home page are considered faulty by yourself. No doubt you’ll find all your answers on SkS – or will you?
If you want to read a brief summary, or get a link to his document, you know where to look: http://climate-change-theory.com/RadiationAbsorption.html
But let’s agree to talk facts and figures on this site, shall we? Everyone here has heard enough comments from people like yourself in the past – just echoing the IPCC and SkS. What’s your motive – to push an agenda, or to seek the real truth?
Tim, you have to remember that the layer near the ground get heated up by upwards and downwards welling radiation, so it’s the reemission that is responsible for the DWR eventually making it to the ground.
Think about figure 1 of your reference.
Can’t resist sorry:
Doug “Everyone here has heard enough comments from people like yourself in the past – just echoing the IPCC and SkS.”
LMAO x 10^10^10.
Jeff echoes the IPCC or SkS???
HAHHAHAHAHAHAAHHAHA
Doug is obviously getting PO’d by the a$$ kicking he’s getting from Jeff and CO.
Just friggin’ admit you were totally off base, may be pick another anonymous name, and keep blogging. No point to continue when your entire thesis has been ripped to shreds and used as the liner for a cat liter box. Seriously give it up. You’re wrong, you can’t even maintain a consistent argument, your whole theology you practice is bankrupt. It’s over.
Copy of my post just now on WUWT. Please reply there …
Re Sharperoo’s “30 year trends” ….
Well, here are a few 30 year trend rates calculated each 10 years with the rate of increase per decade against each. Which would you like to cherry-pick Sharperoo? Data source is linked at the foot of http://climate-change-theory.com where you’ll see a plot including all the intermediate moving monthly values of such 30 year trends between the dates shown below.
Jan 1900 to Dec 1929: 0.020 deg.C/decade
Jan 1910 to Dec 1939: 0.086 deg.C/decade
Jan 1920 to Dec 1949: 0.096 deg.C/decade
Jan 1930 to Dec 1959: 0.036 deg.C/decade
Jan 1940 to Dec 1969: 0.002 deg.C/decade
Jan 1950 to Dec 1979: 0.016 deg.C/decade
Jan 1960 to Dec 1989: 0.064 deg.C/decade
Jan 1970 to Dec 1999: 0.113 deg.C/decade
Jan 1980 to Dec 2009: 0.084 deg.C/decade
So, according to you, any of these 30 year trends would have sufficed at the time to indicate “the” long-term trend. I’ll settle for the green “trend of the trends” line showing 0.06 reducing to 0.05 deg.C / decade. I had drawn that before calculating the arithmetic mean of the above 9 values for which I then got 0.057 deg.C / decade.
So take you pick anywhere around half a degree per century.
Any good reason for any advance on that?
“The kicking from Jeff” ????? You’d better try to understand that in #386 Jeff quoted specs “Wavelength 8 to 12μm” and that 12μm is equivalent to -31.67 deg.C which is nothing like Jeff’s -173 deg.C – and yet he himself copied the specs without even reading them it would seem. I have no time for dealing with such careless posts.
Three of my posts make the same point: #365, #388 and now #400.
I hope you can understand that I barely have the patience to have to point out Jeff’s misunderstanding three times now as well as Carrick’s lack of understanding and perhaps others’ too.
I have now put this useful information which Jeff has linked on my website as it supports Johnson’s result very nicely. Thank you, even though you didn’t mean to help.
It’s not hard to see that none of you rejoice at realizing the world is not warming any more than half a degree per century, and that the end is in sight within 200 years when 500 years of cooling will start..
You really don’t want to rejoice in this good news, do you? Not even to check it out a bit more carefully just in case you can come to understand it and believe the truth.
You really don’t want to know the good news, do you?
I genuinely wonder why?. But then I don’t know what vested interests you might have in this, do I? I have none I can assure you, either way. All I do is spend my own money and time trying to help others to understand for the good of mankind – nothing else.
Prof Claes Johnson has a wonderful paper on “Computational Blackbody Radiation” in which he explains precisely why you will never be able to convince Doug Cotton that he is deluded. Simply put, the information you are providing to him is below his cut-off frequency so he cannot absorb it.
I kid you not – here it is in the words of the professor himself:
And we’re supposed to take Johnson seriously?
Carrick “Tim, you have to remember that the layer near the ground get heated up by upwards and downwards welling radiation, so it’s the reemission that is responsible for the DWR eventually making it to the ground.”
I have. Perhaps you missed my point on your hurried initial reading. Please reread the post at #219
I cant see how there is much downwards radiation at that temperature and pressure at all as per Pierrehumberts own description and my reasoning applied to it.
So the long wave radiation comes up from the earth’s surface and is absorbed within about 10m and then transfers its energy to the atmosphere. Very little of the GHGs then radiate at all because they never get the chance.
There is a chance I dont understand the nature of the delay between the gaining of the required energy and the actual radiating of the GHGs and if you’re aware of something relevent please let me know.
Copy of post at http://wattsupwiththat.com/2012/02/03/monckton-responds-to-skeptical-science/#comment-884135
Lord Monckton & HenryP:
No one seems to have published a simple experiment showing backradiation warming something.
How easy it would be to have two metal plates insulated from the ground sitting there, one absorbing backadiation at night and the other shielded from the backradiation. Just compare the pair.
I find it really hard to believe that someone has not done such an obvious test, or, if not, why the IPCC would not have requested it? The whole thing looks very suspicious, but then, what doesn’t in this field?. Surely Lord Monckton could arrange such a simple experiment.
The specifications for those microbolometer IR cameras whose sensors are not warmed by radiation from a cloud at -35 deg.C for example, should be an encouragement.
Carrick is wrong again: “get heated up by upwards and downwards welling radiation”
No it doesn’t get heated by downwelling (low frequency and thus low energy) radiation from a (significantly) colder layer.
I think Johnson’s little analogy on the side (obviously taken out of context) applies quite well to many of you here. You just can’t come to grips with a fact of physics that upsets your preconceived ideas, can you?
TTTM ” if you’re aware of something relevent please let me know”
Sure, the upward radiation warms the cooler layers above.
The downward radiation does not warm the warmer layers below but gets scattered and eventually goes to space, or back up to warm cooler layers above where it originated.
.
Doug,
They specifically state their measurement temperature range with an uncooled far-warmer sensor. They specifically state 100Kelvin in one example, 200 Kelvin in another and 180 Kelvin in a third. They are measuring those temps in the ten micron range using a BLACKBODY sensor.
This means the PHOTONS from a cooler source are being absorbed by the WARMER blackbody.
You replied on this thread very much inconsistently meaning to me that you recognize the problem. Your latest reply is simply to say it is below the cutoff frequency so it does not exist. Except that you can see absolute proof in every single one of the examples presented here that it does.
Backradiation is real.
End of story my friend.
——-
The real battle regarding global warming is in the magnitude caused by man, whether it is a problem, and whether a problem should be addressed. These questions have not been adequately answered and my contention is that most scientists know that.
Tim:
Why wouldn’t they reradiate it? If they absorb a photon at that frequency, they certainly can reemit it (just isotropically).
Are you worried about the GHG losing energy through collisions?
Remember they can gain energy too via collisions. So on average that is a wash, other than what they radiate is going to look black-bodyish multiplied by their corresponding emission spectra, and photon number doesn’t end up being conserved as a result (we just have to conserve energy).
(Getting the absorption and emission spectra right as a function of height is the real trick.)
You can also “cheat” and use use the known, measured lapse rate of temperature, then you can ignore the amount of absorption and remission going on, and just treat each layer as contributing a certain number of photons to the surface. This is where formulas, like Inamdar and Ramanathan come from:
E.g., see this section.
They write Fc = epsilon sigma T_s^4 – G_a.
Here G_a is the backwards welling radiation absorbed by the surface.
The full paper can be found here.
The confusion is that you appear to be ignoring absorption and remission, but that gets “imprinted” on the lapse rate, so you really aren’t.
Left the link out for Inamdar and Ramanathan, sorry.
Carrick asks “Why wouldn’t they reradiate it? If they absorb a photon at that frequency, they certainly can reemit it (just isotropically)”
But the relevent part of Pierrehumberts quote is
“for the CO2 transitions that are most significant in the thermal IR, the lifetimes tend to range from a few milliseconds to a few tenths of a second. In contrast, the typical time between collisions for, say, a nitrogen-dominated atmosphere at a pressure of 10^4 Pa and temperature of 250K is well under 10^-7s.”
So the reason (I believe) the photons aren’t emitted (or at least are only rarely emitted) is that well before the GHG gets to reemit it, the energy is absorbed back into the energy pool of the atmosphere through any one of thousands of collisions that happen to it.
Carrick and TimTheToolMan,
the amount of RADIATION (please forget the made up term BackRadiation) would seem to be quite variable. As TTTH points out the mean free path is very short at the surface and there will be collisions before spontaneous emission. Carrick is right that GHG’s can gain and lose energy through collision. Isn’t it both depending what part of the diurnal cycle we are in??
When the ground is warming the GHG’s should be at a higher energy level due to absorption and be losing more energy than gaining through collision as the conduction is very slow. The GHG’s would warm the air on balance until the insolation drops off and we reverse the situation with the GHG’s gaining energy from the warmed atmosphere on balance. As we go higher in the atmosphere the situation would be less extreme, but, through the stratosphere does it ever get to where the emissions will happen more often than collisions?
TTTM,
I hope you don’t mind the interruption. You are thinking about the collisions correctly but the energy transfer of the collisions goes both ways. Therefore there is a non zero probability of a CO2 or H20 molecule transferring and receiving energy yet the emission times you quote are also probability based. The average may be milliseconds but the event could happen in the first nanosecond or even a tenth of a second later.
Jeff writes “Therefore there is a non zero probability of a CO2 or H20 molecule transferring and receiving energy”
Yes of course but that doesn’t change the conclusion. If the net transfer of energy is towards the non-GHG atmosphere (and it is) then the probability of the non-GHG atmosphere imparting sufficient energy on the GHG for it to radiate is lower than the probability of transfer in the other direction. The fact there are so many collisions before radiation implies a reduction in probability it will radiate.
It is the reduction in probability I’m looking at and if it takes, say 0.1s on average to radiate (Pierrehumbert said “lifetimes tend to range from a few milliseconds to a few tenths of a second.”) then that means 10^6 collisions on average reducing that probability at sea level.
As far as I can see the probability will become very low…but I’m willing to be convinced otherwise if I’ve missed something.
TTTM,
The conduction of GH gasses to other gasses is nearly balanced by the conduction back to the GH. The time of the transfer doesn’t change the number of energized molecules. The low absorption bands of CO2 and H20 guarantee that there is always IR emission despite the fast conduction rate.
Jeff writes “The conduction of GH gasses to other gasses is nearly balanced by the conduction back to the GH.”
Nearly, but not equal. Thats why they’re GHGs…because they additionally turn the earth’s radiating LW energy into thermal energy in the atmosphere. I’m quite sure there is a net transfer of energy from GHG to non-GHG and hence I’m quite sure GHGs on average have more energy than non-GHGs.
Jeff writes “The time of the transfer doesn’t change the number of energized molecules.”
Not the number, no. But as per my reasoning above I do believe it does change the probability and hence the rate at which GHGs at sea level radiate. So I dont know what you mean by your statement “The low absorption bands of CO2 and H20 guarantee that there is always IR emission despite the fast conduction rate.”
I’m hoping that if my logic is faulty then it ought to be reasonably straightforward to state why.
I wrote “I’m quite sure there is a net transfer of energy from GHG to non-GHG ” But I need to qualify that with “at sea level” . Obviously with increasing altitude things change.
“I’m hoping that if my logic is faulty then it ought to be reasonably straightforward to state why.”
“As far as I can see the probability will become very low”
I don’t think your logic is faulty. However, the fact that we are in a temperature stable system is a good indicator that the probability of eventual radiation is reasonably balanced.
Jeff writes “I don’t think your logic is faulty. However, the fact that we are in a temperature stable system is a good indicator that the probability of eventual radiation is reasonably balanced.”
Dont assume I’m trying to overturn the world with this line of reasoning. Conduction and convection are perfectly good forms of energy transport and the atmosphere at sea level and earth’s surface itself are nearly the same temperature due to the absorbed upwards longwave radiation (as well as any conductive effect).
What I’m trying to understand is precisely how the DLR looks at the earth’s surface and I’m not yet convinced there is much. Or indeed that much is needed to match reality. Obviously with increasing altitude the radiation increases in line with this reasoning and inline with reality 🙂
Kuhnkat writes “Isn’t it both depending what part of the diurnal cycle we are in?”
Possibly. I think its fair to say that the earth’s surface (but more obviously the ocean) is always radiating energy that the GHGs will be absorbing. The way I see it, the sea level GHGs will radiate more when the atmospheric temperature increases to a value greater than the surface induced warming would have given it. I think these will be the exceptions rather than the rule.
Tim, conduction is very poor in a gas so not much heat is carried away by that mechanism, except at the interface between the ground and the atmosphere. This is an operational statement, not something you can necessarily deduce by logic alone.
I’ve been a bit remiss in not calculating all of these rates myself. One of the points of confusion for me is in translating the quantum mechanical change of state of the molecule upon absorption into a classical interpretation of an increase in kinetic energy. If anybody has some comments on that, I’d be interested.
Indeed Carrick you are correct in this regarding conduction: “not much heat is carried away by that mechanism, except at the interface between the ground and the atmosphere”
Now, this surely means that not all the available energy is radiated by the surface, right?
So how can we apply Stefan-Boltzmann calculations to get that infamous -18 deg.C when we don’t have all the energy being radiated – meaning the surface is not acting like a blackbody, so the answer is wrong.
So the attribution of that 33 degrees to so-called greenhouse gases is also wrong because the 33 is calculated using the wrong 18 degree figure – right?
So Dr Hansen was either ignorant or bluffing.
Copy of my post on http://wattsupwiththat.com/2012/02/03/monckton-responds-to-skeptical-science/#comment-884767
MartinGAtkins:
Babsy is correct, but you should note that you started talking about emitted power, whereas Babsy was questioning if there was any empirical evidence that radiation from a cooler atmosphere can transfer thermal energy to a warmer surface. You change the subject to emitted power.
To explain, as I probably need to, if you shine two identical torches at a mirror it will indeed emit twice the power of one torch. But if there is perfect (100%) reflection there is no energy to be transferred, and thus no warming of the mirror.
The facts of life are that radiation from a (significantly) cooler source has a lower peak frequency than the peak frequency of the warmer receiving surface. Such radiation is scattered and does not lose any energy and so does not alter the thermal energy in the surface, whereas radiation which has a higher peak frequency (eg SW solar) is converted to thermal energy.
My backyard experiment (February 5 & 6, 2012)
I shielded a small section of my backyard with a car windscreen shade (silver on each side) which I suspended at an angle of about 45 degrees so that it would not interfere with convection loss and would reflect away upward radiation from the ground. I used a digital thermometer with a metal spike which I inserted into the ground, or held in the air just above the ground for the ambient readings. The “shielded” ground readings were taken under the shade about 20cm from where it came down to the ground, whilst the “unshielded” readings were taken in an open area about 2m away.
Below are the results (temperatures in deg.C) …
time unshielded shielded ambient
21:33 23.3 23.1 22.1
05:34 21.7 21.7 17.7
(a) I found no evidence of “backradiation” slowing the rate of cooling.
(b) My results agreed with those of Prof Nahle (Sept 2011) showing that the air was cooler than the surface and also cooled faster than the surface.
self snip…
What is your expected temperature for each model if correct?
Dr Roy Spencer’s backyard experiment is mentioned above but, in my view, is flawed, as explained in an earlier post.
So I went out yesterday and bought a $15 thermometer with a metal spike and conducted my own backyard experiment here in Sydney. I estimate that I shielded about three-quarters of any backradiation (probably roughly equivalent to removing all carbon dioxide and methane plus a bit of WV) and I found no evidence that any backradiation was slowing the rate of cooling of the ground. Just before sunrise the temperatures of the shielded and unshielded ground were each 21.7 deg.C.
Got $15 to try it yourself? The thermometer might be useful for your next barbecue anyway.
I’ve posted the readings at the foot of this page on my website http://climate-change-theory.com/RadiationAbsorption.html
There’s at least one obvious problem with your experimental design:
I wouldn’t try to measure the temperature of the soil, because there are too many confounding variables:
A better experiment would be to measure the temperature of two objects not in good thermal conduct with the ground. A couple of medium-sized garden pots full of dry soil, suspended at an elevation of 1-m above the surface would probably work for a zeroth order experiment.
I’d also trade in your car windshield shade for a good thermal blanket (the type with a shiny surface on both sides). Or if you really want to be a hero, go out and buy a couple of pieces of sheet metal, and sandwich a piece of foam board insulation between them.
What you should find is that the temperature of the covered object is slightly warmer than the exposed object. Both of these should be cooler than the temperature of the soil under them. (All of this assumes night-time measurements.)
You also need to buy separate thermometers for each object you want to measure, and to start by calibrating them (by putting them all in the soil for about an hour to the same depth, then measuring their temperatures).
And after you do, then tell us why the cameras work.
Jeff: You seem to be unable to read posts above which talk about the cameras. Try to improve your understanding of what they do and their limitations as per their specs. It’s all on my website if you need further explanation.
Carrick
Maybe pots sunk into the ground would be an improvement. But if suspended they will receive radiation from the ground.
I am not hoping to exclude all backradiation, but have excluded more than all that which CO2 contributes.
You could always try some experiment yourself – it is repeatable. I just wanted to keep the materials simple so people can try it. I was careful to measure exactly the same spots on the ground.
There certainly is a difference in sunlight today, so backradiation should have had roughly a quarter the effect of the Sun.
For new readers, the reasons why energy in backradiation does not get converted to thermal energy in the surface are explained by Prof Claes Johnson Computational Blackbody Radiation
Doug:
No that’s not a problem. It’s much better to have them in radiative contact with the ground than contact through conductance. Over the time scales of interest, the ground is pretty much isothermal over the scales you’re looking at (assuming homogeneous surface).
Just a request, can you stop with the link whoring? thanks.
Of course it’s a problem if they receive upward radiation from the ground. How are you going to know that there is no effect from back radiation when the upward radiation would obviously warm them?
The whole point of the experiment is to demonstrate what happens to the ground itself. I may try two wide mouthed vacuum flasks (filled with dry soil) with their lids off sitting in holes in the ground – that’s my best suggestion. It would stop conduction from nearby areas of the ground. Then I would shield one of the flasks, perhaps with two windscreen shields at maybe only 30 degrees. These screens are quite opaque enough.
What’s to stop you (or others here) doing likewise?
You are not correct about the time scales. Conduction underground will not be effective over 2m in the space of one night I would suggest.
8-hours conduction time for 2-m horizontally? On what planet would that be on?
As a point of fact, it’s quite well know in meteorology that the surface layer responds quite rapidly to temperature, amd it’s also know that the rate at which heat diffuses is a very strong function of the amount of soil moisture. Without knowing whether you irrigate on your property, whether there is a grass surface, or any other factors, it’s impossible to put very accurate numbers to it, which is why I described this as a “confound” above. But certainly 1/4-m per hour or whatever you were suggesting for a horizontal diffusion rate was extremely low ball, if for no other reason that the surface layer is extremely porous, so air molecules does flow through it. If you are heating one area, you’ll actually get convection of air through the soil in response to it.
We study soil physics at my lab, so I have some insight into the problem, while admitting my own understanding isn’t perfect.
Doug, I don’t understand the point of your experiment.
You are trying to shield some measured object from DLR, I assume, to show that despite this the temperature does not drop any faster than without the shield, and thus show that DLR is not altering the measured temperature.
But the shield itself is warmer than the atmosphere you are trying to block, so it is emits any IR itself, it proves nothing. By using a silvered surface, you remove almost all emitted IR but must introduce reflected IR instead, which is almost certainly from a source warmer than the atmosphere you are trying to shield.
So I don’t see why this would give the result you are looking for.
I’d prefer this experiment.
Create a pair of tall thin boxes from polystyrene foam, as per Roy Spenser’s experiment, place digital thermometers in base, cover with cling film/saran/clingy (depending on country) to reduce mixing air, and point at the sky on a warm clear dry night.
Take regular readings from both, and verify they both lose heat in the same way.
Next night (assuming warm/clear/dry again) place a tray supported a distance above one of the boxes, and fill with ice. Repeat the measurements. I would expect the cold try to mask DLR, so that box will be cooler.
I have all the required apparatus, but not the weather, so can’t do this myself at present.
Carrick:
It is not a matter of taking 8 hours to conduct horizontally at all.
The temperatures drop simultaneously (as I observed with some intermediate readings) which indicates the heat loss is by way of diffusion, convection and radiation all acting the same at each point, not a result of horizontal conduction at all. There will be absolutely no horizontal conduction when the temperatures are already the same. Horizontal conduction is not the reason for the uniform temperatures.
When the weather fines up later in the week I’ll switch the position of the shield which should cancel out other effects.
Backradiation from a colder atmosphere cannot be converted to thermal energy when it strikes a warmer surface. My results are not in conflict with this. If you wish to disagree with this, then go and find an error in the computational derivation of this. I don’t perceive any fault in the mathematics.
Steveta:
If you read my experiment description the shield was angled so as to reflect most radiation from the ground away roughly horizontally. Even if it had been reflected back by a horizontal shield it would have no effect. The air was always cooler than the ground, so any radiation from the air (even close to the ground) would have no effect on the rate of cooling. The angle was primarily to ensure convection worked. This is not a case of two errors cancelling out so the result is identical for each point.
All should understand that the backradiation is supposed to be about a quarter as strong as the Sun at noon. You know there would be a big difference in temperatures of shaded ground compared with that in direct sunlight. Try walking on sand on a hot day, then stand on sand in the shade.
Why not do the experiment yourself and interchange the shielding, even half way through the night. Try using identical wide necked vacuum flasks filled with soil and with the lids off and buried with only the open tops exposed.
I tip that any such experiments will show that Prof Claes Johnson is right.
.
Douglas Cotton,
“…so any radiation from the air (even close to the ground) would have no effect on the rate of cooling. ”
That is what you are supposed to be proving. Your experiment must show that it is true, not assume it.
Steveta: Regarding your ice, it will cause the box to get cooler even if you turned everything upside down..
The reason is not because it is blocking backradiation. The reason is simply that the ice will absorb more radiated and convected energy from the box before equilibrium is reached. This does not mean that the box will cool faster – it will just keep cooling longer than it would have otherwise. This is not a good analogy anyway, because some energy goes into phase change.
Doug,
You have not explained the cameras at all. You have actually (insanely) denied their specifications. You even took the time to write that the cameras are “unscientific”.
If a camera can measure at 200K (-73C) with a 50:1 dynamic range using a sensor of -20C to 50C temp, how does it work? If that same camera is intended to measure to 100K, how does that work?
Stop referring me to your scatological link and explain, in your own words, how these “unscientific” observations exist in your theory or admit it is crap. Science first here Douglas. Hint: When models don’t fit observation, you don’t usually chuck observation.
Well Michael Mann does but normal scientists don’t.
A blackbody at 100K emits at about 29 microns. The camera specs say 8 to 12 microns. Didn’t you notice the difference which I pointed out. Go ask them why. Obviously you need an answer from them… I just suspect they have made a mistake.
How could theirs be so much better than their competitors who probably overestimate the range anyway when they say 7.5 to 14 microns.
Neither specs go anywhere near 29 microns.
“I just suspect they have made a mistake.”
Baby steps Doug. At least you are beginning to recognize that observations don’t match your theory. You do recognize that they do match the standard physics model which allows superposition of electromagnetic radiation as well as back-radiation right?
Douglas:
If the horizontal diffusion time is short, it is a potential reason why they are the same. It’s a “confounding factor”. and indicates a poor experimental design.
Your explanation of they are already the same temperature and therefore there is no horizontal conduction is completely circular reasoning. The whole point of the measurement was to test whether they were the same or not, and horizontal conduction of heat would have made them similar enough that your temperature sensor wouldn’t be able to distinguish them. But you already “knew” the answer so failure to confirm due to confound is not a defect in your experiment.
Geesh.
(This al assumes that your windshield reflector is at all efficient in blocking IR, which I doubt…. it’s designed to block visible radiation not IR. That’s why I suggested using a thermal blanket, which we know is designed to block IR.,)
Again, this is something you must test experimentally.
My claim is your fundamental flaws are conceptual, and this can only be tested experimentally. You can’t prove anything by rigorous mathematics, and when you repeated claim that Johnson “proved” anything, that demonstrates your own lack of understanding of the scientific method, in addition to everything else.
You are very confused about the angle of your shield. You are apparently trying to ‘mask’ IR from the atmosphere, so you place it at 45 degrees so it’s not reflecting IR from the ground. Exactly where do you think the shield is pointing? There is no direction possible that will produce less incoming IR than an unshielded test.
Steveta_uk: There is no direction possible that will produce less incoming IR than an unshielded test.
And the real problem arises here. If he does configure it properly, an object under the barrier will be warmer under the barrier than outside of it.
Douglas, as a point of curiosity can you link the make and model of the thermometer you’re using?
To all of you, I suggest you read http://www.slayingtheskydragon.com/en/blog/185-no-virginia-cooler-objects-cannot-make-warmer-objects-even-warmer-still?showall=1
Carrick poses a circular argument above which barely warrants discussion. Horizontal diffusion about 15 cm underground he assumes to lead to simultaneous thermal equilibrium at two point 2m apart. Really? Then we would expect vertical “diffusion” to do the same, so the whole underground for many meters down would change temperature up and down every day and night, just like the surface, would it? Go do some caving!
He also makes statements like “the object under the barrier will be warmer ..” without a slither of evidence..
You too should study No Virginia cooler objects cannot make warmer objects even warmer still linked above – which rubbishes Roy Spencer’s contrary claim and yours.
Douglas,
To you, I suggest you STOP reading nonsense which you obviously don’t have the wherewithal to parse. It is costing you IQ points every time you read it. When observations contradict a theory, the theory must be revised. Fortunately, this revision was done long before the new theory ever sprung up.
A – One physics model having cold to hot backradiation makes sense in all cases and matches observation.
B – One model having no absorbed cold to hot backradiation does not match observation.
Hmmmm.. Tough call.
It is you who does not have the “wherewithal to parse.” Pick any fault in this excerpt:
“Can Energy “Flow Uphill”? Spencer says “In the case of radiation, the answer to that question is, “yes”. While heat conduction by an object always flows from hotter to colder, in the case of thermal radiation a cooler object does not check what the temperature of its surroundings is before sending out infrared energy. It sends it out anyway, no matter whether its surroundings are cooler or hotter. Yes, thermal conduction involves energy flow in only one direction. But radiation flow involves energy flow in both directions.”
“I say “In the case of radiation, the answer to that question is, “no”. While heat conduction by an object always flows from hotter to colder, in the case of thermal radiation a cooler object does not check what the temperature of its surroundings is before sending out infrared energy, but the receiving absorber does and will only absorb radiation hotter than itself, otherwise it reflect, transmits, or scatters it. It sends it out (emits) anyway, no matter whether its surroundings are cooler or hotter.” Yes, thermal conduction involves energy flow in only one direction. But radiation flow involves energy flow in both directions, but actual transfer in only one from hot to cold, just like conduction and convection.”
“Spencer offers “But, if ANY flow of energy “uphill” is totally repulsive to you, maybe you can just think of the flow of IR energy being in only one direction, no, it emits in all directions, but with it’s magnitude being related to the relative temperature difference between the two objects. OK. The result will still be the same: The presence of a cooler object can STILL cause a warmer object to become even hotter.” Nope, if you agree the magnitude of actual transfer is related to T hot **4 – T cold **4 then you cannot then assert the contrary. Contradiction is not allowed.”
Douglas:
The comments I made above had to do with the horizontal, not vertical, diffusion rates. And at the time you agreed with them, before you realized this invalidated your argument, then you came back with this lame nonsense.
Here is what I said again:
It has to do with the fact that the porosity in the surface layer is high, allowing air to flow through the soil when there is a horizontal temperature gradient. As I said, it’s well known in meteorology that the horizontal diffusion rate is much larger than the vertical diffusion rate.
Want a reference?
That’s because it’s my hypothesis for what happens!. This is something that you test experimentally, not by providing links to random websites and claiming this answers the question for you. (That by the way is the definition of “link whore.”)
Sorry that was me.
Douglas,
The comment is so full of flaws that it is literally hard to read. I hate anti-scientific drivel. Literally. Anyone here can tell you that I have made mistakes and admitted them. I disagree with myself almost as often as with others but on this you are really not in touch with your basic physics lessons. Very basic ones as far as I can tell.
“While heat conduction by an object always flows from hotter to colder”
On an individual molecule basis this statement is completely, demonstrably and [self snip] FALSE!! And this is where so many go wrong. On a net flow basis, this is true but individual molecules transfer energy on a quantum basis, the flow of energy is a probabilistic transfer of energy which en-bulk is always from hot to cold.
“in the case of thermal radiation a cooler object does not check what the temperature of its surroundings is before sending out infrared energy, but the receiving absorber does and will only absorb radiation hotter than itself, otherwise it reflect, transmits, or scatters it”
Here we are making the claim that the receiving object ‘checks the temperature’ of the transmitting object, yet we know that all photons of the same wavelength are the same. This is entirely different from Johnson’s claims and amazingly weird. So the Earth checks the Moon’s temp and refuses its 15um radiation yet Venusian 15um radiation is accepted. Stupid at best but again it was completely refuted at YOUR recent link by one Dr. Spencer for the same reasons the cameras I have shown, invalidate your claim here. The link made the same claim that the ‘frequency’ is measured and completely ignored that the photon must be absorbed to measure the frequency. A writer who is very ignorant of basic physics of sensors.
See that is where the argument breaks down. We can prove that energy is still transferring in both directions because we can measure it. Your link lost the argument with Spencer as badly as you have lost here, and still is equally incapable of realizing it.
“But radiation flow involves energy flow in both directions, but actual transfer in only one from hot to cold, just like conduction and convection.””
If the flow is in both directions, and there is no evidence that absorption doesn’t occur in both directions, why the irrational assumption in the face of massive evidence to the contrary? Spencer’s IR sensor for example. Conduction is both directions, convection even happens in both directions as some hot molecules are zooming against the flow, why not radiation?
—–
Sensors are observations. You are ahead of Johnson and your link in that you realize that if the sensors I’ve identified actually work, it is a contradiction of your theory. You admit this implicitly because you disagree now that they exist. The scientists have made a mistake!!
If you will admit that this bolded observation is correct, I will spend a few minutes to find a functioning example of your impossible sensor and you will need to change theories.
Do we have a deal?
Ah yes, we will need a functional cold-hot delta T which you consider significant. It is an oddity in your argument which results from a misinterpretation by yourself of Johnsons work.
Well you gave no indication of the depth of what you refer to as the surface layer. Of course within a few millimeters it will be affected by the air which itself will diffuse horizontally quiet quickly. But my metal probes went a bit deeper than that.
I can see that I will just have to do the experiment using the two wide necked vacuum flasks and maybe more windscreen shades when I have a chance to buy such. The result will be the same. Go try it yourself, as you probably won’t believe my figures.
Meanwhile read the emails etc in the above link.
Also, this statement in particular applies to the surface layer: “As I said, it’s well known in meteorology that the horizontal diffusion rate is much larger than the vertical diffusion rate.”
There are technical reasons I would expect it to be true in general, even for deeper layers, but it wouldn’t have anything to do with the mechanism provided above (porosity).
I believe it has to do with alpha'(z)/alpha(z) being large compared to the depth of penetration, where alpha(z_ = K(z)/rho(z) c_s(z) (all quantities assumed to be horizontally stratified), rho(z), with K(z) = heat conduction of the soil, rho(z) is the density of the soil, and c_s(z) is the specific heat capacity of the soil (it varies with depth, mostly because the moisture content varies with depth).
At least for simple model, like K(z)/K'(z) = lambda = constant, the solution to the heat equation should be analytic. Somebody solve this for a point-like heat source on the surface, and get back to us with this. 😉
Douglas:
I shouldn’t have to for somebody with 40 years of science experience.
It’s a well defined concept already.
This is the place where I remind you of the plumber’s motto: When in a hole, quit digging.
The layer is typically about 6-8″ deep.
Link whore alert.
Jeff I never said the energy was not transferring (as radiated energy) in both directions. I said, like Johnson said, that it is not converted to thermal energy when it meets a (significantly) warmer surface.
You don’t seem to believe that the receiving body can first detect the frequency (hence temperature of the emitting body) then “decide” whether to scatter the radiation or converts its energy to thermal energy.
I suggest that your disbelief is not warranted.
For example, you say “Here we are making the claim that the receiving object ‘checks the temperature’ of the transmitting object, yet we know that all photons of the same wavelength are the same”
But wavelength implies frequency, and the absolute temperature of the emitting source is proportional to frequency.
Another quote …
“I accept radiation detector surfaces need not be colder than the incident radiation to detect and measure cold radiation. My eyes see ice. My eyes do not re-radiate ice light. Penzias & Wilson detected CMBR = 3.7 K in 1964 with a radio telescope in a warmer NJ. Radio telescopes do not re-radiate CMBR. I suppose warm IR thermometers can indeed measure radiating temperature of colder refrigerators, without absorbing refrigerator radiation and warming further. What does that have to do with whether warm radiators get warmer from cold re-radiation?
Maybe this quote will help you all to understand and change your incorrect beliefs about cold warming hot …
“Spencer is dead wrong about the 2nd Law. He makes the classic mistake of equating incident radiation with conversion to heat (as if reflection, scattering, and transmission don’t exist), which causes him to contradict himself: Yes, the surface radiates no matter what’s out there, so how can it “know” that the net heat transfer will end up being hot to cold?”
“Jeff I never said the energy was not transferring (as radiated energy) in both directions. I said, like Johnson said, that it is not converted to thermal energy when it meets a (significantly) warmer surface.”
This is different from the claims made in the link you provided in #448. I believe you asked me to find errors in that. This is why links are a terrible way to communicate ideas. I will assume that the link is now a dead reference which should not have been added to the discussion.
“You don’t seem to believe that the receiving body can first detect the frequency (hence temperature of the emitting body) then “decide” whether to scatter the radiation or converts its energy to thermal energy.”
This is absolutely correct. Two photons of identical frequency can be emitted by ‘significantly’ different temperature objects by every theory of physics we live by. There is no way to tell these photons apart. An IR laser can cut material of any temperature, even though the material gets so hot it turns to plasma – another good exercise requiring your explanation. Photons have no tags on their butts.
“But wavelength implies frequency, and the absolute temperature of the emitting source is proportional to frequency.”
False,
Temperature does not correspond to a frequency but rather to a range of frequencies.
How does an infrared laser cut through white hot metal?
By your theory, Infrared lasers work by back-radiation. Oh shoot, it’s usually a CO2 molecule used in industry. 😀
Define significant difference, I will provide a new example of a sensor which absorbs radiation exceeding your difference.
Carrick,
I can’t find any AVHRR data which is current enough.
How many people do you need to hear about who believe what I do ?????
“I do not believe in the “standard absorption/emission physics”, promoted by GHG theorists, because it has been easily falsified on many grounds by professional physicists, chemists and engineers as outside science. Study Gerlich & Tscheuschner, “Falsification of the Atmospheric CO2 Greenhouse Effects Within the Frame of Physics”, Jan 2009 (arxiv.org/abs/0707.1161) and Kramm & Dlugi, “Scrutinizing the atmospheric greenhouse effect and its climatic impact”, Dec 2011 “
“How many people do you need to hear about who believe what I do ?????”
Doug,
The blog is called “noconsensus”.
I take it you don’t want the challenge? Are you worried that I will find another example of a functional instrument invalidating your claims?
Give me a range – consistent with your theory and I will find the device which refutes it.
====
If you don’t want that challenge, how does an infrared laser heat steel to plasma states?
I honestly cannnot imagine how that works under your model.
Jeff: Now I really do know that you do not know physics.
You have just made a statement which is the complete opposite of the well-known Wien’s Displacement Law clearly stated and explained in Wikipedia.
I said (in accord with that law) ““the absolute temperature of the emitting source is proportional to frequency.”
You said “False” and Wien’s Displacement Law says true.
Now explain to Wikipedia authors why they are wrong. http://en.wikipedia.org/wiki/Wien%27s_displacement_law
,
Jeff Your reference to lasers proves nothing. We are talking about natural spontaneous emission in the atmosphere and surface. In such cases the peak frequency is proportional to the absolute temperature of the source as per Wien’s Displacement Law.
Radiation created using electricity (or another energy source) and an oscillator is totally different, whether radio waves, microwaves, lasers, x-rays or whatever. Just about any frequency can be generated by a machine http://www.geo.mtu.edu/~scarn/teaching/GE4250/EM_wave_lecture.pdf
#465 Read the answer more carefully. I specifically avoid conflict with Wein’s rule.
#466 Seriously, I don’t understand how IR photons from a laser can be absorbed by a far hotter source. Are you saying the IR laser photons are intrinsically different from the same frequency of blackbody photons?
PS The vast majority of lasers do in fact have high frequencies which correspond to high temperatures, usually in the visible light spectrum. You can focus sunlight to burn something, just like laser light is focused. However, I would hope they use something a little less powerful when operating on the retina of my eye.
PPS Anyone who doesn’t think that the frequency of visible light corresponds to high temperatures would do well to remember that the Sun’s peak frequency is in the visible light spectrum.
Jeff: I wrote 468 and 469 before reading your 467.
Have you never seen a display of laser lights – beams in the visible light spectrum? Are you so ignorant of physics that you didn’t know what I said in the above two quotes.
I trust you do now understand that just about any frequency can be generated as a highly focused laser beam. It only has to be in the visible light spectrum to be far hotter than the hot steel it is cutting by melting the steel.
The plain fact is that I made a statement quoting Wien’s Displacement Law and you said my statement was false. You did not “avoid conflict” with what you incorrectly call Wien’s Rule.
I just wanted to point out for the record that Doug has been arguing in a dishonest fashion. This is probably obvious to the causal reader of this thread, but here is an example:
1) I pointed out here that Doug’s experimental design was flawed because he placed the two temperature sensors in the soil, and that they would be at nearly the same temperature (within measurement error of his 0.1° C resolution device) regardless of whether his set up was any good.
2) Doug responded with the assertion: “You are not correct about the time scales. Conduction underground will not be effective over 2m in the space of one night I would suggest.”
3) I then explained the physics of why horizontal conduction is different than vertical
That last was a warning to Doug against trying to B.S. his way through this. I have people I can ask, some of them are world experts on the thermodynamic properties of soils, if I get stuck, Doug probably doesn’t.
3) Doug responded with the non-sequitur “It is not a matter of taking 8 hours to conduct horizontally at all.” Apparently he was under the impression that Aristotlean physics is still in vogue on this blog (that is we should ignore what our senses tell us because they are flawed in deference to our perfect wisdom).
Circular logic exercise 101 as I and Stevea_uk pointed out separately.
4) Doug then tries to using Stalin-esque tactics by accusing me of what he did, namely circular reasoning (without actually being able to point to anything I said that was remotely circular):
5) To which I point to my previous physics explanation and pointed out that he seemed to have accepted the fact of horizontal conduction mattering until it dawned on him finally it undermined his experiment.
6) To which Doug replies “Well you gave no indication of the depth of what you refer to as the surface layer. Of course within a few millimeters it will be affected by the air which itself will diffuse horizontally quiet quickly. But my metal probes went a bit deeper than that.”
7) But of course I realized that the probe was deeper than a few millimeters, and Doug certainly was aware that I realized this too, which makes this a deceptive statement on his part, designed to distract from his many many problems with basic science concepts as I pointed out here.
I think we have somebody who shamelessly refuses to admit when they’ve made an error, no matter how slight.
I won’t even go into the much larger list of self-contradictory arguments that Doug has made as we have advanced through this thread on blackbody radiation.
My suggestion to Doug is, “keep digging, if you don’t mind deep holes.” (And having Jeff repeatedly hand you your ass.)
Do you trust German engineering and science? I have always respected such.
It’s not surprising their press is now revealing the CO2 lies … their scientists won’t be bluffed by the pseudo science that is AGW, any more than I will be.
http://notrickszone.com/2012/02/06/body-blow-to-german-global-warming-movement-major-media-outlets-unload-on-co2-lies/
“PS The vast majority of lasers do in fact have high frequencies which correspond to high temperatures, usually in the visible light spectrum.”
So you are claiming that CO2 industrial lasers make “plasma’ with visible light?
#472=distraction.
oh god, you have to write “Doug’s distraction” or else he will take it wrong and not pay attention to the point.
==
Doug,
You clearly haven’t worked with lasers in your lifetime. You would know that the power spectrum of most lasers is astoundingly flat outside of the single emission band (multi-frequency devices do exist). This happens for a lot of reasons not including my intended trap of the laser comment — lasing! I did enjoy irony of the non-absorbing guy not noticing that the light amplification occurs through similar processes to his non-absorbing theory. Except of course that for the really big C02 systems, the energizing light (often of RF frequency) is absorbed by the CO2 and re-emitted at an even higher frequency. Some lasers use higher frequencies for energization, others use far lower ones. A violation of Doug’s first law.
As you may guess, the industrial stuff is all about emitting as much 10um light as possible. I guess those guys are anti-science too.
Doh!
#471 Was stuck in the spam trap due to links. Enumerated references between here and there are offset.
Thanks Jeff. Nonetheless, like a fine, well worth the wait. 😛
grrr.. “Like a fine wine, well worth the wait.”
Consider what is happening in just one country Germany because of the AGW hoax …
“Dr. Guenter Keil’s report focusses in detail on the amazing absurdities of Germany’s Renewable Energy Feed-In Act and the country’s utopian Energy Transformation. The government, through intrusive meddling and ballooning bureaucracy, has maneuvered Germany’s energy supply system into a vicious death spiral: the more the government intervenes, the greater the mess becomes. And the greater the mess becomes, the more the government intervenes! Dr. Keil concludes:
“Germany’s energy transformation has already failed. For Germans, the outlook is bleak. …the planned mismanagement is heavily damaging the economy and will fail spectacularly some years later because its economic and social costs will have become unbearable. The question remaining open is how many billions of euros will have to be destroyed before a new energy policy (a new energy transformation?) picks up the shattered pieces.”
“Consider what is happening in just one country Germany because of the AGW hoax …
Consider what is happening when those who claim knowledge of the ‘hoax’ don’t understand the basic arguments. People with motives and poor understanding do more damage to their own cause than had they not spoken at all.
#480, Yep that is the basic problem here. By arguing poorly, Doug comes off as a quack, and threatens to paint anybody skeptical of catastrophic AGW into the same corner as him.
One must fight doubly hard against those who may hold the same fundamental beliefs as ours about CAGW, but get their science so totally, totally wrong.
#481, I hope that we differentiate ourselves from the Mannian style brow beating by holding the forum in public.
Doug may confuse some people, but he cannot set back the science by being incorrect. He can only make it seem less clear than it is to the uninformed. They were already uninformed so meh..
Education is a process, not an event.
In the meantime, I read math every day to expand my own understanding. I study business, news, climate, datasets, papers. Every day, something new. The Internet is a wonderful tool.
Let’s see a show of hands for those who already have learned everything!
Jeff Condon,
“Let’s see a show of hands for those who already have learned everything!”
I learned everything already. Just ask me a question!!
Oh wait! Dang, I forgot it AGAIN!! Guess I have to start over.
What a lot of waffle in the last few comments without a strain of physics or logic..
Jeff can’t even apologize for getting his physics wrong about Wien’s Displacement Law and the fact that lasers with high frequencies can of course heat and melt metal, seeing that their radiated temperature can be equivalent to that of the Sun.
No wonder he didn’t understand why the specifications of that microbolometer IR camera.(8 to 12 microns) limited its temperature reading range – he didn’t understand the relevance of Wien’s Displacement Law.
I really don’t have time to give free online tuition in basic tertiary physics here on this site.
Doug:
What a twat.
You aren’t even competent to explain F=ma. You’d probably explain how it was really F=mv.
So very much time wasted on one person’s misunderstandings.
Two in the one day! I’ve just encountered another AGW proponent on WUWT who, like Jeff here, was ignorant of Wien’s Displacement Law. I wonder how many other AGW proponents are in the same boat. No wonder their conclusions are false about the effect of backradiation.
Here’s my response on WUWT …
Gary shows his lack of knowledge of physics when he says “Which raises the question of how his roof can tell the difference between a “cool” radio wave and a “hot” radio wave of the same wavelength”
Wien’s Displacement Law states that the peak frequency is proportional to the absolute temperature of the emitter. http://en.wikipedia.org/wiki/Wien's_displacement_law
I trust Gary at least knows the connection between wavelength and frequency and has thus now learnt a bit of basic physics, namely that you cannot have “cool” and “hot” radiation of the same wavelength.
It must worry you a touch that so many Germans agree with me …
“Today Germany’s national tabloid Bild (which has a whopping circulation of 16 million) devoted half of page 2 on an article called:.
“”THE CO2 LIES … pure fear-mongering … should we blindly trust the experts?”
“That’s what Germany’s leading daily Bild (see photo) wrote in its print and online editions today, on the very day that renowned publisher Hoffmann & Campe officially released a skeptic book – one written by a prominent socialist and environmental figure.”
(source: WUWT)
More quackery from Doug. Obvious CAGW plant.
Speaking Douglas Cotton and quackery…
The fact that Carrick apparently could not see that Jeff’s knowledge of physics was lacking (and he never previously corrected Jeff but instead supported whatever Jeff said, even when he was wrong about the camera and lasers, for example) just shows me that neither of them really understands basic thermodynamics and Radiative Transfer Theory. Sure Carrick might have known more than myself about surface soil composition and conductivity and I was willing to learn. Even before that I had spoken about a refinement to my experiment using wide necked vacuum flasks filled with soil and with their lids removed.
But anyone who knew their physics would not mistakenly refer to Wien’s “Rule” instead of the more relevant Wien’s Displacement Law. They would understand that if a cooler body really could warm a warmer one then the warmer one would re-radiate at higher frequency effectively increasing the frequency (LOL) and then other cooler bodies all around would absorb this and radiate it back again, and the process repeat for ever. (Wow – free energy!)
Clearly Jeff (like Gary on WUWT) was ignorant of WDL, even though I have mentioned it several times, both here and on my websites. It is critical to understanding what happens with radiation.
So this error about cooler warming warmer, and the error in the calculations of that 33 degree of warming (from that -18 deg.C) are the two major errors of the AGW hypothesis. One error would be enough to debunk it, but we have two at least.
“Obvious CAGW plant”
Well the “plant” (or whatever it is) has sprouted 16 million seeds – and that’s just the start.
Expect to see this series of articles (and the new book released today) both translated into English etc etc and reaching hundreds of millions real soon now.
Keep a brave face!
The last nail in the AGW coffin is looking up at a big hammer.
This reply of mine on WUWT may be a clearer explanation of what Prof Johnson, myself and others are now saying …
In general, spontaneously emitted radiation has an attenuated frequency distribution with a peak frequency which is proportional to the absolute temperature, quantified in Wien’s Displacement Law. When it strikes another body (eg Earth’s surface) its coherent energy will be converted to incoherent thermal energy only if the surface is emitting (or would emit) at a peak frequency less than that of the radiation. In other words, the surface has to be (significantly) cooler than the source of the radiation, usually the Sun.
So the Earth’s surface converts incident high frequency solar radiation to thermal energy, but reflects or scatters low energy (low frequency) radiation from a cooler atmosphere, the energy therein not being converted to thermal energy and thus not affecting the temperature of the surface.
For mathematical proof see Prof Claes Johnson’s Computational Blackbody Radiation.
You also mentioned IR spectroscopy which does in fact provide evidence for my point. You will find it common knowledge that a gas will not been seen to be absorbing radiation from a cooler emitter when the radiation is analysed by spectroscopy. It is only when the emitter is warmed and its temperature starts to exceed that of the gas that we do in fact start to see absorption lines.
On WUWT:
Manfred says:
February 6, 2012 at 11:53 pm
Doug Cotton says:
February 6, 2012 at 9:40 pm
…Yes there will be slight cooling in the near future until about 2028, but we should still expect another 0.4 to 0.8 degree rise in the long-term trend before it passes a maximum some time in the next 50 to 200 years. Let’s not be accused of alarmist claims ourselves….
—————————————
Could you elaborate on that or give a link ?
___________________________
This has been discussed in previous posts – you should also see the WUWT article by Nicola Scafetta regarding 60 year cycles and the correlation with planetary orbits.
Briefly, there appears to be a ~1,000 year roughly sinusoidal cycle I will call the long-term trend. Yes there is some correlation with Solar activity, but it may be due to a common cause and there does appear to be a lag of about 50 years for Earth’s climate to follow. In the short term the correlation is not strong and the (superimposed) 60 year cycle is observed. This is discussed on my original site http://earth-climate.com
At the foot of the Home page of my newer site http://climate-change-theory.com is a revealing plot of the gradient of 30 year trends calculated monthly starting Jan 1900-Dec 1930. This shows a rate of increase (green line) of about 0.06 deg.C/decade back then, but this rate has now decreased to about 0.05 deg.C/decade, and, if the trend is sinusoidal and approaching a maximum within 200 years or so, I postulate that the gradient could be down to zero by about then, thus indicating a maximum in the long-term (~1000 year) cycle. Meanwhile, the next maximum in the 60 year cycle is expected in 2058 as it is actually about 59.6 years if caused by the Jupiter/Saturn resonance cycle which has that periodicity.
Linked is a chart of laser wavelengths – note the majority in the visible and UV high energy / high frequency ranges ….
http://www.lexellaser.com/techinfo_wavelengths.htm
Douglas Cotton,
“Linked is a chart of laser wavelengths – note the majority in the visible and UV high energy / high frequency ranges ….”
You forget that even if ALL of them but one appeared to prove your point just one would DISPROVE it. Please stop with the poor logic.
PS Here again is the link to the calculator to convert laser wavelengths to radiated temperature equivalents.
http://www.calctool.org/CALC/phys/p_thermo/wien
Doug’s attempted distraction makes me think ho doesn’t actually read the items he links to.
I hate to ask, Jeff, but do you ever end up banning commenters?
Steveta,
“I hate to ask, Jeff, but do you ever end up banning commenters?”
I have nobody on a ban list. A couple of times I’ve put people in moderation. Usually for lies, or completely random critique. I put up a long post once that took 8 hours to write (I used to do more technical posts) and TCO had a critique up in less than 1 minute. I told him I was going to moderate his critiques because he obviously hadn’t even read, he vanished after that.
Doug has resorted to fabricating errors by me and Carrick now. He obviously recognizes that his pet theory is full of holes because he is avoiding the most difficult questions.
Douglas,
You are fun to tease.
“But anyone who knew their physics would not mistakenly refer to Wien’s “Rule” instead of the more relevant Wien’s Displacement Law.”
Somehow I knew that one would get you.
Know how?
Because you are grasping at straws and avoiding any difficult questions. It is obvious that you are not attempting honest discussion. I actually laughed when I deleted “law” and replaced it with “Rule”. Actually, I put in ‘rule of thumb’ but decided that was going too far. I didn’t want you to explode.
Sorry for the tease Doug but you are so busy declaring victory without noticing it is dead, you sound like the black knight in Monte Python. http://www.youtube.com/watch?v=2eMkth8FWno
A CO2 laser turns steel into plasma emitting 10um light not UV, not visible, 10um. Answer me specifically – How does that device work in your theory?
If you cannot explain the laser cutting metal with infrared light, your theory dies again.
Unfortunately Prof Johnson does have a sort of answer to this. He distinguishes between coherent and incoherent light waves (as he doesn’t seem to believe in photons) and appears to think that emissions from hotter to cooler objects are “coherent” but not the other way round.
Since the CO2 laser output is coherent, it’s allowed to excite atoms at any temperature. It’s just incoherent cool radiation that cannot.
So you have to distinguish between a coherent source, such as a laser, and an incoherent source, such as Prof Johnson.
Steveta,
Coherent has specific meaning in lasers. Output from any blackbody is not in any way coherent light. Output from an industrial laser is not known for its coherence either. Is there a quote from Johnson which is more complete so that we can distinguish what he is saying.
Jeff, did my reply (a minute ago) get lost, or do you filter anything with a link to Prof Johnson in it?
Jeff:
I think it’s really because his understanding is so poor, that he’s forced to this sort of sophistic rhetoric. He really isn’t capable of following our arguments, or even the badly flawed theories of Claes Johnson, but rather decides on what is true based on what fits into his world view (which is driven by political beliefs rather than observation).
Jeff:
If you want to make light coherent from a blackbody, all you need to do is pass it through a pinhole aperature.
There isn’t anything magical about coherency.
Here Johnson explicitly disavows the existence of the photon:
Here’s a link where Johnson explains his theory of phlogistons.
Yes, he literally is arguing Greek versions of science.
For extra credit count how many times Doug endorses Claes Johnson’s work on this thread.
I get 42 times (I used a script to calculate it).
I also get that he linked to his own website 31 times, a new record for link-whoring I believe.
Carrick,
I don’t disagree because I used to do holography (with lasers) on a consulting basis. Metrology in particular. I was waiting for the next magic theory from Doug.
Carrick, no fair!
Johnson’s phlogiston theory is there to debunk the photon theory – he equates the two, so this isn’t there to support ancient Greek theory, it’s to debunk Planck/Einstein.
Even he isn’t that daft.
Fair enough. He just doesn’t believe in photons, or probably quantum mechanics by extension .
I’m not sure I can endorse the “even he isn’t that daft” statement however.
Seems to me you can make an “everything is a wave… there are really no particles” argument, but that still doesn’t change the continuous emission of photons(electromagnetic waves) continuously from a blackbody, independent of temperature. Of course, there is still that ultraviolet catastrophe thing to deal with…. if everything is a wave. OK, so they are a special kind of wave with only certain energy levels allowed… that behave like particles. Wait, but that suggests they are really particles, but, but, if particles don’t really exist, then are not particles, and they must be waves… wait….. 😉
Steveta_UK,
” Wait, but that suggests they are really particles, but, but, if particles don’t really exist, then are not particles, and they must be waves… wait….. ;-)”
but, neither particles or waves explains the 2 slit experiment fully….
I think we are at the edge of the known and until more knowledge is gathered the arguments will continue. I don’t believe Claes bothers to deny Quantum Mechanics. Why deny a DESCRIPTION of what is happening that appears to work without knowing all the dirty details of what is happening?
It would be interesting to see Claes’s theory on what gamma rays are. How the High Intensity Gamma Ray Source works would be a side bar on that discussion.
When you get to the place where there are so many effects that are known, and physical devices that use it on a practical level perhaps as many as a million times a day, it becomes a very dubious tack to try and just dismiss all of that over what amounts to disagreement on policy (e.g., climate change mitigation vs adaptation).
I can guarantee you that all of this skepticism from uninformed laypeople on back radiation wouldn’t be there if there wasn’t an IPCC pushing a policy direction they don’t agree with.
Imagine that the story that was being pushed by the IPCC was: “AGW will be mostly beneficial and harmful side effects relatively economically adapted to, so this world can look to a productive next century, both industrially and agriculturally. However it must be recognized, that the benefits of this CO2 increase will be short lived, as once the economies inevitably shift away from fossil-fuel based industries, we can expect CO2 over the centuries that follow to return to their preindustrial level. Unless well planned for, this could lead to a global disaster of unprecedented proportions, if society does not prepare for the coming harsher global environment that follows this golden age of CO2.”
Oddly, this version of the story is probably more likely to be true than the one promulgated by the IPCC (I think it’s close to the median of possible outcomes).
My implied question there is “Would these same ‘skeptics’ be arguing against back radiation then?” were this the IPCC “front page story”?
Carrick,
on your IPCC Front Page story. There would have been no need of an IPCC and they definitely would not have become involved in all the BS that they did if the story were that innocous. I agree that ~99.99999999% of the people not directly involved in Climate Science would have been totally uninterested in Radiation (why the term Backradiation?? It is no different from any other with the same attributes?!?!?!) from the atmosphere if the IPCC hadn’t tried to blame everything on one minor cause for a whole range of issues.
Kuhnkat,
CO2 lasers (ironically enough) are designed to emit in the IR range and to my knowledge are the most common for industrial cutting. Doug is toast again.
I take it you are more convinced now?
Jeff Condon,
uh no, I am not more convinced now. Apparently what Doug is willing to argue separates quickly what I or Claes or a number of other people believe or are willing to argue.
I would point out that the fact a range of crystalline material absorbing energy in a particular range of frequencies when heated is only peripherally connected to the IR range impinging on the surface of the earth. You may be right, but, the IR camera, pyroelectrical and ferrolectrical effects still does not clearly PROVE it any more than microwave ovens and lasers in the 10 um range do.
I should have said ‘usually’ in the IR range. I know that gasses can be tuned to different wavelengths but am not enough of an expert to know what other bands CO2 might be tuned to.
Lord Monckton:
” I hope shortly to be in a position to demonstrate formally that climate sensitivity is unarguably little more than one-third of the IPCC’s central estimate.”
Firstly, the original calculations of sensitivity were, I understand, based on an assumption that the Earth’s surface would have been -18 degrees C but for so-called greenhouse gases and water vapour.
However, in using Stefan-Boltzmann’s Law to calculate that -18 C figure one would have to have assumed that the Earth’s surface acted like a blackbody. A blackbody absorbs and emits only radiated energy, whereas at least half of all energy which transfers between the Earth’s surface and the first millimeter of the atmosphere does so by diffusion (molecular collision) and evaporation. Hence less than half the energy remains for radiation and, obviously, this greatly affects the calculation of that -18 degree C figure which, in a nutshell, is totally meaningless.
Secondly, spectroscopy proves that a gas does not absorb radiation from an emitter which is cooler than itself. Hence, the lower atmosphere does not absorb radiation coming from cooler carbon dioxide molecules above it. The same applies for solids and liquids in the Earth’s surface as Claes Johnson, a well-published Professor of Applied Mathematics has proved.
Hence an atmospheric greenhouse effect supposedly due to “backradiation” is a physical impossibility.
I would like to suggest that it would be more productive to focus on these two solid facts of physics than to spend all your days arguing about whether Arctic or Greenland ice is melting, or whether the world is warming – which the 500 year trend is, but only by about 0.05 deg.C / decade, and only for another 50 to 200 years at the most.
Kuhnkat @512
You don’t produce any evidence that even one such laser does in fact demonstrate that any solid is converting to thermal energy any emission with a frequency which corresponds to a radiated temperature significantly less than the current temperature of the solid. You are clutching at straws.
The laws of physics cannot be broken, and if they seem to be then there is another explanation.
If they could actually cut steel with low frequency lasers, then why bother to produce high frequency ones?
Whatever you people say about Johnson, he is only disputing the particle nature of photons. He is not saying that there cannot be “packets” of wave nature radiation. It is not necessary for his mathematics to assume that the radiation is continuous rather than in packets. The trouble is, none of you has enough background in applied mathematics and physics to follow and understand and acknowledge the veracity of his calculations.
But you cannot explain by any other “physics” why spectroscopy shows that a warmer gas does not absorb radiation from a (significantly) cooler emitter.
How about trying to explain this last sentence, or prove it incorrect? That is my final challenge to you all.
Douglas Cotton,
please quit embarrasing yourself before I tell Claes on you:
http://www.laserk.com/newsletters/whiteCO.html
” A kilowatt of radiation at 10 microns, focused down to its diffraction limit, is a power density of 1 gigawatt per square cm. However, it is more practical to think of focusing the power down to 100 times the diffraction limit which is 0.040 in. Because most materials absorb at 10 microns, considerable interest has been shown in CO2 lasers for cutting applications. This means cutting such things as paper, cardboard, plastics, glass, quartz, wood (tree surgery), meat, flesh (bloodless surgery), metal, and rock. Except for the last two items, the cutting process is just what one might imagine if a small intense flame were to burn through the material. In a sense, the process is very similar to electron-beam cutting and welding, but it is much simpler to use. For metal, the initial absorption depends on the surface condition, and the cutting depends strongly on the size and thermal conductivity of the material. Thin-sheet stock can be cut, but molten metal tends to freeze out on the edges giving a rough cut. Thick metal, however, tends to form a molten pool which must be removed by some means if further penetration is to occur. It was discovered that irradiation of hard rocks with power densities of the order of 200 W per square cm caused the surface to become white hot. The resulting mechanical stress would cause a general weakening of the rock; in some cases, it caused the rock to crumble. At higher power densities, the rock would melt, relieving the stress locally. Further research into the comminution of hard materials is under way. Most materials are opaque at 10 microns, and any problem which requires controlled surface heating or burning might find a potential solution with the CO2 laser.”
Carrick and Jeff, please note the power densities. I just don’t see that anywhere in our atmosphere. 8>)
“But you cannot explain by any other “physics” why spectroscopy shows that a warmer gas does not absorb radiation from a (significantly) cooler emitter.”
Reference please.
Any reference.
Any at all.
anything.
any.
?
Further to my response to J Fischer above …
Suppose you somehow placed a small metal marble-sized ball inside a hollow soccer ball-sized metal sphere and then sucked all air out to form a vacuum inside. Now, let’s assume the small ball was a few degrees hotter than the surrounding sphere. Further assume that the outer sphere is large enough so that there is much more radiative flux coming from it than from the smaller ball. This would be due to its greater surface area which would more than compensate for its cooler temperature.
So, we have a net radiative flux going from the cooler sphere to the warmer small ball inside it.
Will the small ball start to get warmer or start to cool?
Physics says that the flow of thermal energy can only be from hot to cold. But we have net radiative flux going from cold to hot. Hence the small ball must be rejecting (scattering and reflecting) the cooler radiation from the larger sphere. The large sphere will however absorb and convert to thermal energy the warmer radiation from the small ball. They each “detect” the temperature of the other because they detect the peak frequency and that frequency is proportional to the absolute temperature – see: http://en.wikipedia.org/wiki/Wien's_displacement_law
The significance of this fact of physics is that a warmer Earth surface does not convert radiation from a cooler atmosphere to thermal energy. So the radiative atmospheric greenhouse effect is debunked.
Kuhnkat,
So I/we’ve beaten Doug’s argument to death. I agree that Claes doesn’t agree with Doug either, however my reading of Claes leaves me with a difficult to reconcile proposition of a CO2 laser cutting into literally anything not specifically tuned to reflect/transmit regardless of temp. Even those highly pure substances often wear out.
Can you enlighten me on your concept?
Actually, I’m quite confused as to why a bolometer doesn’t prove any concept. Perhaps that would be a good place to start.
Jeff Condon,
“Actually, I’m quite confused as to why a bolometer doesn’t prove any concept. Perhaps that would be a good place to start.”
First, the camera uses a microbolometer. Second, please narrow the statement down to a particular concept, not just any. Of course it doesn’t prove ANY concept.
To everyone: Consider a patch of rock being warmed by the Sun in the morning. The IPCC says backradiation will add more thermal energy, so it must warm faster. (It is not just a matter of backradiation slowing the cooling rate – it must be consistent in whatever it does. Either it adds thermal energy or it doesn’t.)
At some time soon after noon the Sun will bring the rock to a maximum temperature before it starts to cool towards evening. When at that maximum will the backradiation cause it to warm more? How could it, becuase that would be transferring thermal energy from a cold source to a warmer body. It is simply against the laws of physics. It simply cannot and does not happen. Yet the IPCC “explanation” of the GHE says it does.
Prof Johnson has proven why it doesn’t in his Computational Blackbody Radiation. The GH theory is debunked.
kuhnkat:
Hence the conundrum. If you admit the story can be (and is most likely actually) this innocuous, indeed even a rather pleasing picture, no funding.
Because it is downward pointing and it is the term that accounts for the totality of the classic GHG effect.
Pretty sure he doesn’t believe in QM either. He argues for a classic picture of radiation instead. Do you actually take his dreck seriously??? Please say no.
What’s left to argue?
Either you accept that there are these thingies called infrared photons, and they can be absorbed and emitted by “radiatively active” molecules, and there is a GHG effect, or you say there aren’t these thingies called infrared photons, in which case you gotta lotta ‘xplaining t’do.
Carrick
No conundrum. A group decided they wanted to push a particular story and fely that being less than honest would work better. i refer you to examples such as the Nuclear Winter excesses that were promulgated by scientists and possibly the gubmint to drive nuclear disarmament. It could never work but they tried anyway.
Carrick, is there a difference other than vector in the radiation that a CO2 molecule emits?? This is part of the disinformation with the use of the term backradiation. This was carried by the many fellow travelers of the Climate types. When I first tried to find out what the heck the deal was I first came to the conclusion that they were saying that CO2 somehow reflected the radiation back down to the earth. That many layman’s explanations used this excessively to get their point across. Most of us were fooled into thinking that was what was happening. When I actually started thinking about what little I knew about physics I quickly came to the conclusion that it was BS because something heated radiates in all directions. Gases do not reflect in a preferred direction without some ionizing or electromagnetic effect. BackRadiation is a PROPAGANDA term. What hits the earth is radiation no different from what is emitted up, horizontally etc. It is even less than half as can be easily seen by drawing the geometry without the earth as an infinite plane.
Do I take Claes’ “dreck” seriously?? I don’t take YOUR “dreck” seriously. You can’t PROVE it anymore than he can. You can only refer me to things that other peole have done that are generally accepted that would APPEAR to prove it. I would also remind youu that mathematics is excellent at DESCRIBING things, NOT creating them. It is also excellent at being misused to describe things that are not real, as many scientists have found to their chagrin after years of chasing their tails because they thought their math was somehow reality and PROVED something.
What do I think of Quantum Mechanics?? I think it is the result of geniuses trying to reach past their limited technology to understand something they couldn’t figure out how to measure yet. I think it is what you get when you can’t think of anything else it could be. The experiments show that whatever it is has both classic wave and particle dynamics. Instead of deciding that they don’t understand they did their best and decided on a duality. It works as far as it goes. We apparently are waiting for the next genius that can take us past the current dead end we have ended up in.
All you have to do is look at the work at the LHC to realize that we are still a bunch of blind folded monkeys feeling an elephant while wearing metal gloves.
“What’s left to argue? Either you accept that there are these thingies called infrared photons, and they can be absorbed and emitted by “radiatively active” molecules, and there is a GHG effect, or you say there aren’t these thingies called infrared photons, in which case you gotta lotta ‘xplaining t’do.”
Sorry, I make up my own rules and games. I am not required to bend to yours. Yes, you have a lot of splainin’ to do about how those PARTICLES can KNOW when a remote slit is open or not and whether they are being observed, plus some other stuff I don’t understand.
“Carrick and Jeff, please note the power densities. I just don’t see that anywhere in our atmosphere. 8>)”
Kuhnkat,
Exactly right, so….
do very very slightly lesser amounts transfer some energy? Yes
what about half the power density?
etc.
Edited for clarity.
Carrick,
I agree with you agian. Getting tired of it, but Claes make some very extreme claims decorated (that is the right word) with “non-supporting” algebra that contradict the existence of quantum mechanics.
The algebra seems fine to me – I didn’t reproduce it – but the conclusions are unsupported.
It’s like JWR who wrote an entire article using the same algebra two ways and decided one was better than the other even though they said the same thing.
A microbolometer is a small bolometer which has pixels that physically warm from infrared radiation. These examples were designed to absorb energy (blackbody) in IR ranges around 12um. In the examples I’ve given, the sensor is far warmer than the source. Backradiation!!
The sensor pixels change temp according to the backradiation. Each pixel actually changes temp to detect the image!! This was the original point of the ice image which was then expanded to far colder and higher cost interplanetary camera examples.
100% physical proof of the mechanism in action.
Doug recognized the problem and said not “significantly lower temp” to which I gave an example of very much lower temp.
It is absolutely, beyond a doubt, backradiation in action on a physical object that would not function without the transfer of energy from cold to warm. Standard physics confirmed!
Of course the net power is in the other direction.
Carrick,
Actually I bet that if someone bothered to take the time to use Claes cutoff frequency, they could confirm that his power transfer between two plates followed a different relationship than Planck. I can’t imagine a more dry and useless waste of my time but it seems simple enough. Once that was confirmed, there would be more experimental evidence of contradiction. Unfortunately, that wouldn’t convince people like Doug either. Steve F has the right idea I’m afraid.
kuhhkat:
This is pure nonsense. It’s a quantity that is mathematically exact and physically well defined.
It’s the Ga term.
End of story. (Really.)
People can take liberties when they try and construct English language explanations for laypeople, but it’s important to remember that the underlying theory is written in the language of physics and mathematics, and for that, there is no wiggle room. There is an exact, right way to express the result. I’m not responsible and refuse responsibility for lame explanations given by idiots that you decide may be deceiving you. What crap they spew is not my monkey, I simply refuse to accept that scenario, sorry.
Once again, to reenforce, you simply have no idea what you are talking about here. The idea that it is propaganda is the only propaganda being perpetuated here.
I think you could learn something by spending a bit more time on the history of science. We don’t end up with results that get used millions of times a day, then suddenly realize it’s all nonsense. What happens when we extend the edges of science, we realize the rules that we used in another area, no longer apply, and everything gets dumped out, and we start over.
For example, F = ma is an approximation to Einstein’s special relativity, but it works really well on Earth at the velocities and distances we humans typically work at. We don’t have to give it up simply for most problems because it doesn’t apply to say accelerator physics.
As long as you recognize you’re just playing games in your own mind and not dealing with reality that’s fine.
However, there’s only a single reality, and it’s governed by an underlying rationality. Unlike games in your mind, there is a single right answer, starting from a given set of precepts, and all other answers are wrong. No “yes Billy that is a good answer. Does anybody have another one?” allowed here.
And what I am trying to tell you is, if you accept the existence of IR photons and the atomic theory of gases, there is an inescapable train of rational thought that allows no other conclusion than the GHG effect is real.
That isn’t all on me, fortunately. 😉
They “know” they’ve been observed, when their wavefunction collapses (and by “observed”, it doesn’t have to be observed by a sentient being! Key point.). You really should go back and read through the exchanges between Bohr and Einstein, it’ll give you a much deeper understanding that I could possibly communicate here.
Carrick,
since you insist on being a propagandist, please point me to the first use of the term Backradiation in the literature where it was defined as a unique form of radiation identifiable as separate from other radiation.
You did a lot of arm waving and presented no actual useable facts to establish this term as anything other than how I have identified it.
Carrick,
I did not state, nor do I believe, that a PERSON needs to observe to cause the collapse of the wave or whatever the correct terms are. I understand that it can be instruments that have been set up or that it is the normal function for it to collapse at interaction, observation being an interaction. Yes I do need to reread it a few times and find some other sources with different viewpoints.
Of course, this does bring on interesting questions. When they are doing experiments at the LHC, would there be any difference in the result if there were NOT the instrumentation present to record?? Are we creating a result by observation or recordings that is not the same as an in the wild result?? Does Quantum Mechanics give you the tools to compute these differences? The question also extends to observations in the atmosphere and surface.
Let me take the opportunity to revise and extend my remarks. 😉 Let this sentence say:
“What happens when we extend the edges of science, we realize the rules that we used in another area, no longer apply, but everything DOESN’T GET dumped out, and we HAVE TO start over.”
Newton’s F=ma is my favorite example of good science “proven” wrong.
Einstein proved this equation wrong. We are still proving Einstein’s equations right but F=ma is amazingly good for nearly everything we deal with in daily experience. It will never be thrown out. Nor will Quantum Mechanics, although it will be expanded on and revised. It hopefully will be re-written in more conceptually simple terms, but relativity is beyond most already.
In the end, quantum mechanics will not be any more wrong than Newton was because it is also based on observation.
Jeff, I am one of those few people who can say I actually analyzed (and published!) data that required using corrections from Einstein’s General Relativity to get the answer right.
By that doesn’t mean that F = G m1 m2 /r^2 isn’t a great approximate that works in most cases.
Interestingly, you can get the perturbative general relativistic version by including the self-enegy, using E=mc2 and adding that additional mass and so forth.
I can’t harp on this enough though: When we have science that reaches the level that millions to billions of dollars are being earned from it, chances are future changes to the theory won’t sweep everything away that we think is true now…. all of the framework of radiation physics is empirically based on literally billions (if not trillions by now) of confirmations.
Will the theory evolve?
Certainly.
As it turns out the interface between the quantum mechanical description of a gas and the classic (“kinetic”) theory is incomplete: There are classical results that still cannot be derived exactly (though to be fair, not that many people are spending any time on the problem). But the point is, it will evolve.
The same thing happened with Maxwell’s equations.
When we went to the the kinetic theory we realized the “bulk” versions were only an approximation (there are higher order corrections even classically). Then when quantum electrodynamics was developed additional corrections were found.
Guess which version of the theory we use when designing radar antennas? The classic version.
Funny that. What happened to all of these scientists fooling themselves by playing with math equations?
Carrick,
your example of using Classical theory for designing antenna is excellent. Classical deals with the wave properties better than QM. That is the direction I am wandering. To try and make my belief clear though, it would appear that QM does explain MORE than Classical does so takes precedence. It does not appear to explain some of the things Classical does which is why the smart guys compromised on the duality I believe. I obviously do not understand all the details of this.
“Funny that. What happened to all of these scientists fooling themselves by playing with math equations?”
Science of Doom has made a post on that topic. I read it today and was suspicious that this endless thread may have motivated it. He mentions that those most skeptical of the basics are the worst with equations. I don’t agree with the social studies aspects of his post but the basic point made sense.
Jeff writes “Newton’s F=ma is my favorite example of good science “proven” wrong.”
I dont think enough people have a good enough grasp on reality and how its described to understand the difference between F=ma which is useful and mostly correct for everything vs a climate model which might be somewhat useful and could easily be wrong about everything in 100 years.
Climate models are smashmatics at their best.
I would say climate models are more useful in helping us understand the past than predicting the future.
Finally, something we can disagree on.
Too many arbitrary DOF for use IMO. The Hadley cell section of CAM3 really poisoned them for me.
I didn’t tell you what I thought they were useful for.
You can use them to “infill” missing data (like ECMWF or NCEP), that has proven to be pretty successful…. used like this, they are just a version of “back-cast” weather modeling.
Given the temperature record, you can use them to solve the inverse problem for the total forcing as a function of time (and if you make an assumption about the climate sensitivity of CO2 you can use that to compute anthropogenic aerosol forcings, sort of by default). Of course you don’t need a full AOGCM for that, which is good, because that means it’s a relatively robust result.
I think if you start with just non-anthropogenic forcings you can deduce when anthropogenic forcings started becoming important (that’s the point at which the known non-anthropogenic forcings can no longer be used to describe past climate behavior). Probably doable without an AOGCM again, so again relatively robust.
Infilling – ok for short records.
Total forcing, not as impressed. Too many assumptions in the feedbacks. I really have a problem with the cloud feedback in models. The robustness of the result would require a complete post by yourself to convince me. 😀
Actually, that might be what it takes at this point.
Last paragraph….. again, the cloud feedbacks drive the sensitivity. We don’t really know that part. At least I don’t.
When we talk about clouds, we get into semantics as to what is a forcing and what is a feedback. 😉
If the feedback is changing over time, it’s what I call a “parametric forcing”. Indeed, CO2 is an example of that. So would be cloud feedback, were it to vary over time.
So it’s a definitional thing really, but point #2 is almost self-referentially true, if you use my definitions.
I grant you point #3 is harder to prove: that is, when anthropogenic forcings became measurable.
Still useful to look at and see what we can learn though, just as it’s useful to criticism attempts to look at it (like the IPCC efforts, which suggest AGW began circa 1970).
I should have said “net anthropogenic forcings became measurable”.
Obviously we can establish that CO2 has increased from pre-industrial days. What we don’t know nearly as well is by how much aerosol (direct and indirect) forcings increased in lock step with this. The IPCC suggests they were roughly equal and opposite until circa 1970, when pollution controls allowed CO2 to dominate.
(Interestingly, one explanation for the current plateau in temperature is the invocation of increased pollution from China and India.)
“(Interestingly, one explanation for the current plateau in temperature is the invocation of increased pollution from China and India.)”
I know, and if you visit the current planetary manufacturing center, it isn’t an unreasonable point. Brown sky’s, combined with poorly mixed CO2, Northern hemisphere exceeding the south, on and on.
The evidence for warming by CO2 isn’t exactly missing. However, the evidence for some unpredicted feedbacks isn’t either.
Have you seen my latest post. New things are interesting after pedantic beating of backradiation.
Yes, Smokey, yours is similar to my “thought experiment” posted at 4:35pm today. But I don’t think we can expect any warmists to come to logical conclusions about thermal energy appearing to flow only from warmer to cooler bodies or gases. You see, the IPCC told them that photons (without mass or momentum) carry thermal energy along with them and spill it wherever they happen to crash into something with all the momentum they don’t have. Besides, it seems that only words, not science, flow from warm things.
These warmists don’t understand that radiated energy is something totally different from thermal energy. Thermal energy gets shared around by molecular collisions, whereas radiated energy has to go through a process of being converted to thermal energy by cooler molecules which are able to do so.
It’s like the sound of your voice which can be converted and broadcast as radio waves, but then may or may not be converted back to sound by something it strikes, like a radio receiver tuned to the right frequency. Ah, frequency! That’s what it’s all about. Does the radiation have a frequency above the cut-off?
But the warmists thought the surface was blind to frequency and they have never heard of resonance and near resonance, or indeed anything much to do with physics.
Doug Cotton,
“You see, the IPCC told them that photons (without mass or momentum) carry thermal energy along with them and spill it wherever they happen to crash into something with all the momentum they don’t have.”
Then you also disagree with the physicists that light causes recoil when emitted and transfers momentum on impact??
$533 K: Either you say “there are these thingies called infrared photons, and they can be absorbed and emitted by “radiatively active” molecules, and there is a GHG effect, or you say there aren’t these thingies called infrared photons, in which case you gotta lotta ‘xplaining t’do.””
No, it’s not an either/or case as you describe it. It simply doesn’t matter if the radiation is continuous or in “packets.” But have you ever thought about how many wavelengths would be in a packet, and why more than one wave if each was due to a single energy jump? And what if the energy jumps are more frequent than the frequency? Or could that not happen? Or is the radiation the sum of the contributions of many molecules, just like a wave in the ocean?
You all need to seriously consider my “thought experiments” above to understand that the IPCC concept of backradiation is contrary to physics because it has to imply transfer of thermal energy from cold to hot. This simply does not happen.
#518: K:
“I don’t believe Claes bothers to deny Quantum Mechanics.”
That is correct – he wrote on SoD that he only rejects the “new” statistical mechanics which is very dubious. He certainly accepts standard quantum mechanics.
Nor does he reject photons in wave form.
Jeff: The info about gases not absorbing has been discussed before and was on SoD. I actually trust DeWitt Payne on that one – wouldn’t you?
#525 K: “Because most materials absorb at 10 microns”
What about the materials that don’t absorb? Why is it so? The IPCC said anything on the surface would absorb IR radiation from the cold atmosphere which has much less energy than 10 micron radiation.
A surface at +16.6 deg.C has a peak frequency of 10 microns, so what’s the big deal? There is still a margin of at least another 30 degrees to play with anyway, so human temperatures would be OK in the outer regions of the distribution.
You people really need to use the calculator I linked you to, or click ‘converted” on my Radiation page if you lost the link.
But it’s good to see you find out for yourself that CO2 lasers work around 10 microns, not 15 microns as Jeff first implied.
So how many lasers have you found cutting things with 15 micron radiation????. I’m all ears.
Doug Cotton,
here are a couple 16 micron lasers:
http://www.freepatentsonline.com/4276517.html
http://www.freepatentsonline.com/4136317.html
Now, what you have to do is decide whether you are going to argue that a 16 micron laser is impossible or whether for physical reasons engineers don’t use them due to the excessive energy requirements compared to a 10 micron laser doing the same job. What’s it gonna be big boy??
#544 Jeff: ” the cloud feedbacks drive the sensitivity. We don’t really know that part. At least I don’t.”
Don’t bother looking. Water vapour also does not add thermal energy to a warmer surface. Why would it, when carbon dioxide doesn’t and can’t and never will be able to.
#420 TTTM “the atmospheric temperature increases to a value greater than the surface induced warming would have given it.”
This would be an exceptionally rare event due to turbulent weather for example.
Normally the atmosphere even just a few meters from the ground is cooler than the ground in calm conditions when there is no major change in the surface conditions in the near vicinity, such as a change from land to ocean.
By what process are you imagining it would get warmer? Hopefully you are not assuming it absorbs IR radiation from colder areas higher up. Even when it absorbs IR from the surface it cannot warm above the surface temperature as a result, as indeed you imply.
So folks, this is yet another example of Johnson’s “Law” in operation.
It seems no one on any of the four forums I posted the above two “thought experiments’ on has been able to show any fault in the logic. Yet strangely they accept the IPCC hypothesis!
Now everyone, if you accept my conclusion in #519 (about the marble inside the soccer ball) then you are implicitly accepting Johnson’s “Law” and thus rejecting the GH effect. Speak now or forever hold your peace.
Today’s thought experiment.
You can make a large cone-shaped funnel with reflective walls on the inside lined with aluminium foil. Make a hole in the end
and place just below it some water in a container with a much smaller cross sectional area than the mouth of the funnel. You can then capture the Sun’s rays and heat up the water, maybe boiling it .
So, turn the funnel upside-down, place it on the ground and capture the IR radiation from the ground. Place water (cooler than the surface) in the small container just above the hole in the end of the funnel, leaving room for air to escape. Remember that the funnel is capturing radiation from a larger surface area than the water’s area, so the water cannot radiate as much away as it receives, and so it warms.
Will the water at the top of the funnel ever get warmer than the ground? If not, how does it “know” when to stop warming?
What is it then doing with all the radiation which is still hitting it?
Mydogsgotnonose says (on WUWT):
February 8, 2012 at 1:09 am
“The real GHG warming is at second phases, cloud droplets, black carbon etc. The former gives increased convection so accelerates precipitation thus reducing relative humidity, the physical origin of Miskolczi’s observation]”
____________________________________________________
Indeed yes, Mydog … Just what the ‘Slayers’ and I have been saying. Increased convection goes with reduced relative humidity, and vice versa. So, as we know, the moist adiabatic lapse rate is only about two-thirds of the dry adiabatic lapse rate, primarily because of the release of latent heat in the formation of rain drops which then carry the energy back downwards, warming air and sometimes the ocean and land as well
Apart from turbulent weather, this is the only way thermal energy can go downwards in the troposphere, because it cannot do so by convection or radiation – just by those rain drops that “keep falling on my head … ”
Elementary my Dear Watson.
Doug, I was wondering if you could explain why the inner surface of your football was firing all of it’s IR precisely at the little marble?
Because if it isn’t, then your argument about it being bigger is irrelevant, as it is equally a bigger target for itself.
“Jeff: The info about gases not absorbing has been discussed before and was on SoD. I actually trust DeWitt Payne on that one – wouldn’t you?”
Doug, trust has nothing to do with ‘science’ as you should know. You have misinterpred Claes Johnsons work here already among a vast number of other mistakes. I’m far more concerned about your interpretation than your source. If you cannot provide a reference, I must assume it doesn’t exist.
I don’t recall saying that a CO2 laser must emit at 15um. Why does it make you think the problem is any different? The IR beam is turning metal into plasma (and every temperature inbetween) so I would call that ‘absorption’ by a colder body.
Looking forward to your answer — if you can make one up.
Copy ….
Roy, I am sorry to hear about your daughter and wish her a full recovery.
I note that you have not had time to respond to my post on the thread about atmospheric pressure in which I pointed out that the -18 deg.C figure is meaningless because the Earth’s surface is not insulated and thus does not act like a blackbody. It loses thermal energy (perhaps more than 50%) by diffusion, conduction, convection, evaporation and chemical processes and so there is far less energy left for radiation than was inserted in the S-B equation to obtain that figure.
But secondly, I join with Prof Claes Johnson in saying that backradiation cannot be converted to thermal energy in a warmer surface. Now I know you say you are just considering net radiation, but there is no physical way in which an approaching “ray” of radiation can somehow reduce or cancel out a part of an exiting ray which would usually be at a different angle anyway.
The only way you could get any cancelling effect, thus slowing the rate of cooling for example, would be if the radiated energy is first converted to thermal energy within the surface. Only thermal energy can be added to, or subtracted from, other thermal energy to get a net effect.
Johnson is saying that radiation below the cut-off frequency (ie, basically radiation from a cooler source) will be merely reflected, transmitted or scattered without ever being converted to thermal energy.
Now you would agree that reflection could not cause a greenhouse effect, but he is saying that the end result (energy-wise) is exactly the same. Scattering is the same as reflection except that the angles are different.
Now, over the years, relative humidity has been decreasing while temperatures have been rising. I say that decreasing relative humidity causes the adiabatic lapse rate to increase because it is drier. That in turn means emission is taking place at lower temperatures and is thus reduced in intensity, so there is a net positive radiative balance as a result. The only way the extra thermal energy left in the atmosphere gets back to the surface is due to the precipitation being warmer, or less cold – less snow and warmer rain.
So there you have it. There is no greenhouse effect due to carbon dioxide or any backradiation. Only water vapour affects the radiative balance because of its unique capacity to affect the lapse rate (due to release of latent heat) and to carry the thermal energy to the surface via precipitation.
Doug,
Not only are you incredibly wrong, you are now copying replies from other blogs to this thread. This is spam and will be snipped in the future if you are not replying to someone here. You should realize by now that your knowledge does not exceed those who are asking you questions. The very concept that you haven’t figured that out is incredible by itself.
Please stop copying other comments here and answer the specific questions people have asked or admit your failure and go away.
Until Doug can provide a rational answer for how an IR laser works to heat an object, I think he’s snookered.
He doesn’t appear to reject the photon theory of light, or if he does, he won’t admit to it, in spite of the 40+ times he’s referenced and yes, even praised, Claes Johnson’s work.
Photons are photons, they don’t get labels for where they came from. Either an object can absorb a photon, or it can’t. If it can, the story is over. The IR photon comes from whatever source, travels through space, and gets absorbed by the material in question.
It is only the average over a large ensemble of particles that reproduces the 2nd law of thermodynamics. Microscopically, all processes involving photon exchange are reversible.
Concepts like “classical wave coherency” apply to ensembles of photons, not to individual ones, so Claes Johnson’s “I’m throwing about 90% of the science that makes things work the way they do just so I don’t have back-radiation” approach is about the only “viable” one, even if he’s created a gazillion more new questions with no obvious answers than the few questions he was trying to address originally.
#563 – I’d almost agree with everything there, except that laser cooling also works. That’s how they get to teeny-weeny fractions of a degree above absolute 0, by hitting a molecule at exactly the right moment so that the photon reflection reduces the Brownian motion.
So the idea that resonance has some impact is not implausible – just wrong in the GHG theory instance.
I wonder how Doug copes with the concept that a photon can hit something very very cold, and REDUCE its temperature?
#559 Steveta Well, you guys are the ones who believe in backradiation, so there’s a lot of radiation bouncing around at all angles and being back radiated inside the large sphere, so sooner or later most of it is going to strike that poor little marble, just as if the whole inside surface of the sphere were a mirror. You either accept or reject the notion of backradiation having any effect.
Jeff – you show me evidence of a warm gas absorbing radiation from a colder source then. You show me evidence of backradiation warming anything.
You say “You have misinterpred Claes Johnsons work here already” but you don’t say where or how – ie you provide no evidence for this statement, nor most of what you say.
You claimed that 15 micron lasers cut steel, but in fact they don’t and now we see that lasers rarely have longer WL than 10 microns which is equivalent to about atmospheric temperature.
So you are all talk without evidence, yet again.
Now I ask you a question:
How come the relative humidity (worldwide mean) has declined (long-term) as in the paper referred to on WUWT yesterday?
This implies that water vapour feedback is negative – quite a problem for the IPCC hypothesis I suggest.
#563 Carrick asks for “a rational answer for how an IR laser works to heat an object”
Amswer: You select a laser which has a high enough frequency. Those lasers which have frequencies in the visible light spectrum could potentially heat an object to about the same temperature as the Sun. This is established by Prof Johnson’s computations. What’s your problem? What WL laser were you thinking of? Your question was far from specific.
Uh, Dougie Pooh,
the paper I linked on the 10.6 micron laser that YOU say is room temperature does cut steel and granite. Wanna try again big boy??
#563 Carrick:
You say” “The IR photon comes from whatever source, travels through space, and gets absorbed by the material in question.”
Prof Johnson says that is not true when the frequency of the radiation is below the cut-off frequency of the material in question,
If you wish to argue about this, then do so on one of Prof Johnson’s blogs, not with myself. (You’ll find some posts of mine there.)
$563 Carrick
You say “all processes involving photon exchange are reversible.”
Prof Johnson says this is not true. The econversion of coherent radiated energy to incoherent thermal energy is not reversible.
Again I say, go and argue with him.
Just exactly what questions do you think Claes has not answered. He has certainly explained why thermal energy when transferred by radiation can only appear to move from warmer to cooler bodies. Surely that rules out backradiation transferring thermal energy to the surface.
You show me any evidence – just one experiment in all history, that actually shows backradiation increasing thermal energy in the surface.
What happens in the morning when the Sun is warming the ground – does backradiation help it to warm faster? Show me evidence in an experiment with some ground shielded and some exposed to backradiation.
I will be waiting for you to link me to just one experiment anywhere in this whole wide world in which the relative humidity has been decreasing for over a century.
Doug,
“you show me evidence of a warm gas absorbing radiation from a colder source then.”
What do you think happens to steel gas right before it turns into “plasma” under a 10um industrial CO2 laser.
“How come the relative humidity (worldwide mean) has declined (long-term) as in the paper referred to on WUWT yesterday?”
Because you won’t address the problems with your theory.
[snip] – I don’t snip Doug. Perhaps 3 in the past 12 months.
Linking to links for questions not asked is also not permitted. We need you to focus. Like a laser. On the actual questions.
Johnson has not thrown away 90% of all physics. Planck dreamed up a particle nature of radaition in desperation to try to explain the UV catastrophe. Einstein was uncomfortable with the particle idea. Johnson’s brilliant mind led him to be able to explain the UVC using only a wave nature for radiation, which of course it is, because it has a wave length. It does not have mass or momentum, like a particle, now does it? Nor is it a packet of thermal energy.
Indeed it is not thermal energy itself which travels along at the speed of light any more than it is the sound of your voice that does so over thousands of Km or miles when transmitted by radio waves.
Yes, Johnson also rejects “statistical mechanics” which has a lot of flaws, but he does not reject quantum mechanics or any other standard physics.
#571 Jeff –
You say “What do you think happens to steel gas right before it turns into “plasma” under a 10um industrial CO2 laser.”
OK. Show me the specifications of such a laser and some documentation of it being used for such
Why the hell would anyone use such a low energy laser to cut steel when there are far more powerful ones available as in the list I linked somewhere above.
Why would they build lasers which emit in the visible light spectrum? Just for a light show?
You would only have to look up Wikipedia to learn that laser radiation has some very unique qualities. Note the last paragraph below where it explains the difference between it and thermal radiation. Not all photons are the same! Maybe the “temporal coherence” somehow creates a greater warming effect – I really don’t know. But I would suggest that the only relevant experiments relating to Johnson’s work on blackbody radiation should in fact be ones which use what he is talking about, namely spontaneous emission resulting from the temperature of the emitter, not the induced (stimulated) emission of a laser beam which is generated by electrical energy input and has nothing to do with the temperature of the machine emitting it..
“A laser is a device that emits light (electromagnetic radiation) through a process of optical amplification based on the stimulated emission of photons. The term “laser” originated as an acronym for Light Amplification by Stimulated Emission of Radiation.[1][2] The emitted laser light is notable for its high degree of spatial and temporal coherence, unattainable using other technologies.
Spatial coherence typically is expressed through the output being a narrow beam which is diffraction-limited, often a so-called “pencil beam.” Laser beams can be focused to very tiny spots, achieving a very high irradiance. Or they can be launched into a beam of very low divergence in order to concentrate their power at a large distance.
Temporal (or longitudinal) coherence implies a polarized wave at a single frequency whose phase is correlated over a relatively large distance (the coherence length) along the beam.[3] A beam produced by a thermal or other incoherent light source has an instantaneous amplitude and phase which vary randomly with respect to time and position, and thus a very short coherence length.”
“Why the hell would anyone use such a low energy laser to cut steel when there are far more powerful ones available as in the list I linked somewhere above.”
Do you even know the definition of energy and power? Seriously, I’m not certain you do.
To answer what I assume is your question, they use long wavelength, high power, CO2 lasers because they have ~10% efficiency – and they cut like crazy.
The next surprise for you was supposed to be that the scientists were stupid enough to energize CO2 lasers with even longer ‘radio waves’ but after waiting this long to make that simple point, it isn’t worth it.
Jeff #571
Once again you avoid my question about why relative humidity is declining. This is highly significant, because the IPCC thinks water vapour has a positive feedback, amplifying their assumed effect of CO2 by about three. Now we see from real world data that it has a negative effect. There goes 2 out of 3 dgerees of their assumed warming, bring us back to only 1 degree of warming by 2100. I can live with that, as the long-term trend should increase about half a degree and the 60 year cycle add another half degree temporarily, but probably around 2058 and again 60 year after that. But it will be natural warming.
So answer my question in a somewhat more scientific manner please, or I will assume that you agree with the above.
“Once again you avoid my question about why relative humidity is declining. This is highly significant,”
It isn’t significant until we are able to converse on simple science topics. I have often written that I don’t know how much warming was/will be caused by CO2. Other than ‘some’, I’m as skeptical as they come.
You are very loud with your opinions so It is time you learned.
#576: I asked you for evidence – links to laser manufacturers’ websites with specs including microns. I don’t need to know the power – I need the actual WL, not a generalisation like “long wavelength” For example, 15 microns is equivalent to only two thirds the absolute temperature that 10 microns generates, ie about 90 degrees less. From an enquiry I made to a local manufacturer,their CO2 lasers cannot be used for cutting steel http://www.rayjetlaser.com/en-US/your_success_with_rayjet/applications/Pages/Laser_applications.aspx
Well Dougie Pooh,
since you have shown yourself incompetent at even doing simple web searches along with understanding simple physical concepts here is a link to an industrial CO2 laser you can buy for about $65k that operates at either 10.6 or 9 microns for cutting STEEL!
http://www.alkras.com/al1500/
Doug,
A link to an obscure concept of gas non-absorption is one thing but I flatly refuse to type in CO2 lasers for you into google. You obviously realize the problem from your answers, any reader can do it themselves so you are on your own.
I call bullshit dude. I also don’t swear on this blog but you have now exceeded any possible version of reality.
#578
I give you evidence that the IPCC has made a huge mistake about WV feedback, and all you can say is you don’t know.
Likewise I have shown you that they made a huge mistake in assuming the Earth’s surface acts like a blackbody, and you also either don’t understand the ramifications, or just “don’t know.”
I gave you Johnson’s proof that backradiation cannot warm the surface. Do you know this, or is this also something you just “don’t know” whether it’s right or wrong. It seems others here are very convinced he’s wrong.
Does anyone else think for themselves, or am I the only one?
Looks like a duck, smells like a duck, etc..
Oh I can type in lasers into Google OK, thanks, and have done so and read many sites. But I can’t find what you say is there to be found, so I don’t believe you. Manufacturers have told me directly that their CO2 lasers do not cut steel.
Neither you nor anyone else on at least seven forums has ever produced evidence to refute Nahle’s September 2011 experiment in which he claims he proved that backradiation was not slowing the rate of cooling of the surface at night.
I don;’t believe anyone anywhere has ever proved that it is doing so with actual temperature measurements.
15 second search. Note the invisible beam of the CO2 laser cutting steel.
Doug, I know bloggers write things which are mean but you really should consider a shrink.
Doug,
I actually have a sheet of stainless steel in my office building both cut and scribed by the same CO2 laser. I have seen the machine which cut it work on steel. I stupidly agreed that we should give the company money to do it.
I believe we are done now.
Regarding lasers, the CO2 lasers I read about are 9 to 11 microns, peaking at 10.6 microns. This is a far cry from 15 microns.
The laser beam is however very different from spontaneously emitted thermal radiation, being far more focussed and powerful. Nevertheless, it won’t pass through glass or water.
I agree that at a power of 200W to 400W it can cut sheet metal. It penetrates about 2mm, but deeper cuts can be made by keyholing in which a metal gas is used to fill a prepared hole.
But, the process is very different from that when normal thermal radiation hits a surface. It is more like x-rays in that it causes ionization (removal of electrons) and this is how the thermal energy is generated – a process which is different from “normal” blackbody absorption.
In a nutshell, you are focussing far more energy into a very small space than would ever be absorbed naturally. The laser beam has an intensity which would never be found in the natural world coming from a black or grey body. Reflection is high (see below*) but cannot be high enough to stop significant energy input.
So the consideration of lasers is a red herring. Let’s just focus on what Johnson was taliking about – natural balckbody radiation.
“Both CO2 and Nd:YAG lasers operate in the infrared region of the electromagnetic radiation spectrum, invisible to the human eye. The Nd:YAG provides its primary light output in the near-infrared, at a wavelength of 1.06 microns. This wavelength is absorbed quite well by conductive materials, with a typical reflectance of about 20 to 30 percent for most metals. The near-infrared radiation permits the use of standard optics to achieve focused spot sizes as small as .001″ in diameter.
On the other hand, the far infrared (10.6 micron) output wavelength of the CO2 laser has an initial reflectance of about 80 percent to 90 percent for most metals and requires special optics to focus the beam to a minimum spot size of .003″ to .004″ diam. However, whereas Nd:YAG lasers might produce power outputs up to 500 watts, CO2 systems can easily supply 10,000 watts and greater.
As a result of these broad differences, the two laser types are usually employed for different applications. The powerful CO2 lasers overcome the high reflectance by keyholing, wherein the absorption approaches blackbody. The reflectivity of the metal is only important until the keyhole weld begins. Once the material’s surface at the point of focus approaches its melting point, the reflectivity drops within microseconds.”
#578 You appear to assume there will be “some” warming by CO2.
How then will that happen?
(a) Backradiation violating Johnson’s computations and transferring thermal energy from cold to hot?
(b) Warm air falling out of the sky and heating the surface to an even warmer temperature
(c) ?????
No chops Doug.
I give, you win.
Laser beams are stimulated (induced) emission. There is a threshold above which such induced emission occurs, and it is rare in nature.
It happens when photons are received at a higher rate that the “natural” resonating frequency of the molecule.The molecule can’t handle it and two identical photons are emitted.
The process I described above is in fact just this. The material is forced into a state of induced emission. Extra photons are generated and converted to thermal energy as a result.
You can read about stimulated emission on Wiki and elsewhere of course.
So the laser beam is “unnatural” and produces unnatural effects, being above the threshold for induced (rather than spontaneous) emission.
It is produced by stimulated emission and causes stimulated emission in the material it strikes.
#593 KK: Recoil on emission of radiation is really just a measure of energy (ie the difference between two electron states.)
I know that the energy which is then in motion in the radiation is sometimes referred to as momentum, but there is no mass involved, so it is a spurious inconsistent concept.
There is no necessity for that momentum to be captured (absorbed) by any particular molecule as in the physical case of an object hitting some other object.
#594 KK: What point of any relevance to backradiation are you trying to make re 16 micron lasers?
My post #589 applies to any laser – they all use stimulated (induced) emission and create stimulated emission in their targets, thus explaining the warming thereof in a totally difference process to that in spontaneous blackbody emission and absorption.
Dougie Pooh,
why are you asking about BACKRADIATION, a term popularized by Climate Scientists pushing an agenda, when all there is is electromagnetic radiation?
In answer to a question about how molecules “detect” the frequency of incident radiation I wrote ..
The “detection” takes place through resonance and near resonance usually over a narrow band of frequencies for any particular element at a given temperature. If the incoming frequency is too low there will be no effect because no resonance is created. If it is too high there will be “chaos” and thermal energy generated as a result, such as when the Sun’s light hits the surface. (For anyone not understanding “resonance” – A pendulum will “detect” if you are pushing it with the right natural frequency so that it keeps swinging.)
Kookie Kat
#598 Why do I need to try again? Didn’t you understand my previous explanation? At least read my post and do your best to understand.
#599 ditto Your examples are totally and utterly irrelevant and you apparently don’t understand why.
#600 It was quicker than typing “spontaneously emitted radiation from the atmosphere” which I’m sure you knew I meant.
How about answering the question.
PS The general idea on forums like this is to read what others say before showing everyone that you haven’t.. If there’s something you don’y understand in my #589 you’ve blown your chance to ask for a more detailed explanation ’cause I’m going to bed. Do some google searches yourself – then try thinking for yourself.
Regarding the above discussion of Johnson’s work, some may be aware that Judith Curry tried to rubbish him.
Judith Curry displays a distinct lack of understanding of Prof Claes Johnson’s “papers.” She says, in effect, that he ignores radiation in the atmosphere, whereas there is plenty of mention of such right there in the first half page. Then she quotes a conclusion without the slightest discussion of the lead-up to such, as if to imply it is sheer stupidity simply because it is contrary to the IPCC hypothesis.
The guts of what Johnson actually proves computationally (though I doubt she would follow the mathematics) is that spontaneous radiation coming from the atmosphere is not converted to thermal energy when it meets a significantly warmer surface.
I have done experiments which confirm for me the truth of this. So too has Prof Nasif Nahle in September last year when he showed the atmosphere cools faster than the surface at night. My 50 years of physics tells me it is right. Never before the days of the IPCC et al was there anything in physics textbooks claiming that thermal energy can in this way be transferred from a cool body to a significantly warmer body. It is not Johnson overturning physics, it is the likes of those misleading the IPCC who are doing that. No one has produced an experiment showing any backradiation actually warming anything. They can’t, because it doesn’t, if it even exists to anywhere near the extent claimed.
Yes Johnson does not see a need to treat EM radiation as having a (mass-less) particle nature. This was a desperate invention of Planck because he could not explain the UV catastrophe. Neither could Einstein, but he was uncomfortable about these “particles” until his dying days. Johnson has brilliantly derived a computational explanation of the UV catastrophe without resorting to a “particle” nature of radiation (just using a wave nature) and, in my view, yes he does deserve a Nobel Prize for advancing physics in this way and solving a problem that baffled even Einstein. It is absolute garbage to imply that Johnson has overturned 100 years of physics theory. Neither backradiation nor any particle nature of EM radiation ever reached the status of being theory.
Doug,
When your theory is proven wrong by observation, only you and paleoclimatology change the observation.
hide the decline!
Kunhkat:
How about one that is old enough it couldn’t possibly be related to AGW propaganda?
How about Reville and Seuss, 1957
First page, second column. I trust you can spot it without me having to circle the text for you, but if not…
Whether you like it or not, the quantity is exactly definable and calculable, and has practical uses.
(I think the term actually dates back to the early 1900s with Arrhenius, but if you are that interested, you can do your own leg work.)
Douglas Cotton #519
Everthing I have ever read indicates that your small hot marble will lose heat to the surrounding shell and will finish up at the same temperature as the shell. Clearly your much larger outer shell may be emitting more total power than the smaller ball, even though the temperature is lower. What you are neglecting is something called the view factor. This relates the fraction of the energy that leaves one body to the amount that reaches the second. This, of course applies in reverse to the fraction of energy emitted by the second body received by the first.
This link: http://mit.edu/16.unified/www/FALL/thermodynamics/notes/node137.html
will show you how this works in general but also deals specifically with concentric spheres. It is shown that the heat flow to or from the shell to the inner sphere is independent of the area of the outer sphere and only depends on the area of the inner sphere and the difference in the W/m2 emitted by the two surfaces. Naturally, when the two surfaces are at the same temperature they will emit the same W/m2 and there will be no heat flow between them.
I must say I was not very convinced by your reply to Steveta_uk at #565. It seems very airy-fairy compared to the lucid and simple explanation given in the above link.
kuhnkat:
Wrong take home message I believe.
The classic theory is easier to calculate, it does not make it more correct because of that, just more convenient to use. It’s been proven that the quantum theory REDUCES to the classical theory under the right conditions, so there are no results that the classical theory can compute that cannot in practice be computed with the quantum mechanic version of the theory. The proof is simple:
1) Make the appropriate assumptions so that the classical theory will apply.
2) Derive the classic theory from the quantum mechanical one.
3) Use the classic theory you’ve derived from the quantum theory to perform your calculations.
If you want to understand radiative physics, eventually you have to break it down to individual photons being exchanged between molecules. That’s the real physical level of the system. And what I am claiming is that+rationality leads inevitably to the picture in which the GHG effect is real.
As I mentioned above many of the objections of classically trained people with the quantum mechanically view and the coimpletely aptly named BACK-RADIATION (whether kunhkat likes a word is not on my list of criteria for whether it is appropriate to use), in which among other things they errantly claim that back-radiaiton violates the second law because it involves the exchange of heat energy from a colder object to a warmer object.
[Clausius’s Principle, a formulation of the 2nd law that is correct in some circumstances but possibly not all, actually states that the NET EXCHANGE of heat energy is from the warmer to the cooler object, without external work being done on the system. That and only that. This particular principle only holds macroscopically and does not apply to the individual exchange of photons.]
I noticed my Tellus and Suess link doesn’t work so Here’s another link. Just click on the PDF to see the full article, it’s not behind a pay wall.
@!!! Third try
Here’s the raw URL too.
http://onlinelibrary.wiley.com/doi/10.1111/j.2153-3490.1957.tb01849.x/abstract
So here is a smattering of articles using the concept of back-radiation not related to the discussion of the GHG effect.
Here’s a 1953 instrumentation article. .
Since kuhncat likes antenna design, here’s a year 2000 article that uses the term., again not related to atmospheric radiation.
A paper on boundary layer meteorology from 1968.
BTW, another accepted term in atmospheric physics is “down-welling radiation”. Down-welling refers of course to a particular orientation of the radiation path (vertical), so back-radiation is actually the more general term.
By the way for the bunnies who like to bury their head in sand so they don’t have to look at back-radiation from CO2:
Mar’s temperature profiles measured by the mini-TES (thermal emission spectrometer) on the Mars Rovers, reconstructed using CO2 emissions. Note that the temperature of the mini-TES is the same as the temperature of the rover, yet it’s efficiently receiving photons that are being emitted by much cooler air. More complete paper here.
There’s a variation on this instrument sold commercially that uses the 59 GHz O2 line.
And a device that uses the 183 GHz H2O line
Note that this device can detect photons emitted at the top of the troposphere (which is often what is meant by TOA).
These down-welling photons that are being measured are examples of back-radiation, which is an important distinction, because the device is on the ground and would have a bit of trouble measuring up-welling photons. 😛
Is 210K “significantly cooler” that 275K?
The operation of these devices have been confirmed by comparing them against non-radiometric measurements (e.g., weather balloons). There is a report on this comparison here.
Our group uses the Vaisala radiosonde system (mentioned in the report) because we go to much higher altitudes in our measurements (typically 35 km, but we’d prefer 60km). The Radiometrics system is also d*mned expensive.
#611, Carrick, Don’t tell anyone that the ‘cutoff frequency’ is arbitrarily chosen, the thread will never end.
Finally a lay-man definition of back-radiation that may even satisfy kuhnkat.
Connect a line between you, the observer, and the object you are viewing. Put an arrow at the of the line pointing away from you towards the object.
This is the “forward” direction from your perspective as the viewer.
The flipped arrow (the one pointing back towards you) is the backwards pointing direction.
From your perspective as an observer, forward radiation is radiation moving away from you, backwards radiation is radiation moving towards you according to these two arrows we’ve conceptually drawn.
This is physics 101 stuff. It has nothing whatsoever to do with AGW propaganda, and the argument that it represents propaganda is well just _________________ (insert appropriate phrase here).
Forward and backwards are operationally useful, because you generally can’t measure radiation unless it’s coming backwards towards you.
(It would require measurement of the properties of the media that vary as the radiation passes through in order to observe that forward moving radiation, but even then you need a signal of some sort traveling backwards towards you the observer.)
Carrick, that definition is trivial and irrelevant. As deployed, the term specifically refers to downward radiation which results from excitation of gas which has received/intercepted upwards-welling EM. Perhaps just add a syllable: back re-radiation.
Carrick:
Au contraire. Unless the detector is directional, it would preferentially receive photons from the nearest emitter(s), which is the ground. Since most is lateral anyway, this would in fact be much enhanced.
Brian, I guess you missed the part of the discussion where kuhnkat was claiming the word back-radiation was a propagandistic term, so to point out that such a “trivial” definition exists, is hardly irrelevant, and explains why the word can be found in such widely disparate fields of study.
I agree the atmospheric phenomenon under discussion could be written as described.
Brian H:
Are we getting a bit snitty here for a reason?
Who said the observer has to be on the ground? Generally when we do these conceptual sort of explanations we start by assuming a hypothetical observer in free space.
Of course the radiometric profilers I described above are highly directional.
Carrick;
Detectors know not whence their photons originate. There’s no difference between photons from Alpha Centauri, the Sun, TOA, a molecule of CO2 1″ in front of the detector, or your hand. Frequency and direction is all you get to detect.
Brian H:
Not as single photons, but as an ensemble, yes they can, both for the atmosphere, and for remote infrared sources. Do you need an explanation for how a radiometric profiler works? And for that matter an infrared telescope?
About “back re-radiation”: it is not strictly required that the GHG (e.g.) DLR be an immediate “return” of absorbed photonic energy. There’s a “mass effect” due to the high probability of thermal transfer to surrounding gas, and subsequent transfer from said gas. Then we get into the subtleties of thermal agitation and BB radiation vs. signature emission/absorption, etc.
http://wattsupwiththat.com/2012/01/01/a-big-picture-look-at-earths-temperature/#comment-852404
You did.
I just think the “bookkeeping” associated with back re-radiation is rather more complex and contaminated than the 1:1 instantaneity generally assumed. Energy flows are not so neat and choosy. “Paths of least resistance” open and close continuously. It is interesting in this regard that Singer’s study of OLR responsiveness to temperature increase found virtually instant matching, without the postulated warming lags mandated by GHG theory.
And yes I did.
This particular comment was in reference to a radiometric profiler, which is highly directional.
Otherwise it wouldn’t work.
This is a more complete context:
Where’s the confusion?
I think GE Smith is in danger, if not actually having crossed the line, of mixing quantum and classical perspectives in the excerpt in #621.
For example the part about “But when two such molecules collide, the electron clouds repel each other, and in a head on collision, they will come to a complete stop”. This is a classical perspective on what is a quantum mechanical system. In the linked text, he’s also appealing to Maxwell’s equations to describe what is clearly a quantum mechanical system, but Maxwell’s equations only apply to an ensemble of molecules, not to individual ones (which are governed by quantum electrodynamics).
I’m not sure you can give the explanation he’s given, as applied to a single molecule. I’m not an expert on this, nor do I claim any particularly great insight at the moment. I do know the standard reference is de Groot, but that’s as far as I’ve personally gotten on linking the quantum system with the classical one.
There is interesting data just published on World Climate Report that cloud altitude levels have been decreasing for a decade. (See http://www.worldclimatereport.com/index.php/2012/02/08/a-significant-measure-of-negative-feedback-to-global-warming/ )
However, I simply cannot agree with the aurthor’s logic about clouds “The higher up they are, the cooler they are, and thus the less radiation they lose to space, which means the surface stays warmer.”
One could say the higher they are the sooner they will reflect incident radiation back to space before it warms some of the atmosphere below it.
Less radiation to space may well mean the atmosphere gets a little warmer, but then we do still have carbon dioxide to help with the radiation to space and thus cool the atmosphere.
But my main point is that there is only one way in which thermal energy from the atmosphere can get back to the surface. It does not “fall” out of the sky by downward convection. It does not get radiated back down because radiation from a cooler source cannot transfer thermal energy to the warmer surface. The only way it can transfer back to the surface is by being transported downwards in precipitation. Yes, cooler clouds mean cooler rain and more hail and snow.
The fact that the clouds are getting lower and warmer may well be a reason why the world is perhaps not cooling quite as much as the 60 year cycle would appear to cause. But then, there is still a slight rise in the ~1,000 year cycle of about half a degree C per century and that may still have one or two centuries before reaching its maximum.
And, by the way, there is absolutely no “positive forcing from an increase in greenhouse gases from human activities.” The author needs to catch up with the fact that, as Prof Claes Johnson has proved in Computational Blackbody Radiation, (summarised on my site) radiation from the atmosphere has a frequency which is below the cut-off frequency for absorption by the surface, so it is merely reflected or scattered without slowing the rate of cooling or increasing the rate of warming of the surface one iota.
Brian H:
What lagging are you referring to? We’re talking a radiative mechanism here, and that’s what the GHG theory relates to.
The impact of this mechanism on climate depends on how climate really responds to the forcing associated with increasing CO2 concentration, which I think most of us will agree, is much more messy than the relatively straightforward effect we’ve been going back and forth on in this thread.
Doug is a blackbody which cannot absorb any knowledge greater than his own.
#625 Carrick: GES’s explanation (on WUWT) ends up with a cogent reason for the observed fact that the spectrum of downward radiation is equivalent to that for BB radiation at a temperature as observed just above the measuring instrument !!! This observation absolutely quashes the GHE hypothesis because carbon dioxide is playing such a small role in downwelling radiation which is primarily from ordinary air molecules and due to collision processes, not quantum energy jumps.
I suggest GES is correct, and, as I have already posted on WUWT, I stand corrected (and am going to alter my site and book) on the issue of CO2 cooling O2 and N2. Clearly O2 and N2 can do their own cooling. But CO2 does absorb some incoming solar radiation which would otherwise have warmed the surface.
A new paragraph now on my site: http://earth-climate.com/CaseAgainst.html
Older physics textbooks say that oxygen and nitrogen molecules do not emit very much radiation at atmospheric temperatures. This appears to be incorrect, because radiation can be emitted whenever molecules collide. The observed spectrum of downward radiation is equivalent to that of a blackbody radiating at a temperature found just above the measuring instrument. Thus all the “backradiation” supposedly coming from carbon dioxide is in fact primarily from ordinary air molecules, mostly nitrogen and oxygen. It is hard to believe that the IPCC was not made aware of the observed spectral distribution, because this in itself demolishes the hypothesis of a greenhouse effect.
Dougie pooh states:
“Thus all the “backradiation” supposedly coming from carbon dioxide is in fact primarily from ordinary air molecules, mostly nitrogen and oxygen. It is hard to believe that the IPCC was not made aware of the observed spectral distribution, because this in itself demolishes the hypothesis of a greenhouse effect.”
I am not going to interact with Dougie myself. I have enough mental issues without taking on his.
For others, I believe this is easily shown to be bunk by looking at a down welling spectra at night clear sky. The couple I have seen show primarily radiation in the 15 and 4 micron range where CO2 and water vapor overlap. Is that incorrect??
Carrick,
got any spectra showing horizontal radiation??
#625: Carrick: No you miss the main point of GES’s post which you should read, along with several other posts on that WUWT thread. GES has made a significant contribution to the discussion, not that I believe radiation from the atmosphere has any warming effect anyway. But it is interesting to note that ordinary air molecules do radiate through collisions. And it is especially interesting to note that the IPCC must deliberately avoid talking about the observed full spectrum in DLWR.
It is OK to take a classical approach to the issue of nucleii colliding. The whole point GES is making is that such collisions do not have to involve quantum steps of energy in order to radiate. The mere fact that they collide (not necessarily directly head on of course, and mostly grazing collisions) means that the electron fields are subject to acceleration and thus emit a full spectrum (due to random angles and velocities) as is observed. The observed spectrum of downward radiation is not just WV, CO2, methane etc lines.
Everything wrong again. Who would have guessed.
Can I suggest you title your book ‘scienceland’.
You’ve given me a good idea “Greenhouse Land”
I wrote this elsewhere, but feel it warrants repeating here as an overall summary of Johnson’s result …
The IPCC energy diagrams tend to treat the energy being transmitted by EM radiation as if it were thermal energy itself. Not so, of course, and hence you can’t just take a difference between two beams of light or other radiation and talk about net radiation. For a start the beams are really at all difference angles and may or may not be polarised. They do not normally cancel out.
When considering temperatures, it is invalid to add or subtract any energy in radiation until it is converted to thermal energy. Then you can add that thermal energy to whatever is already there. But you cannot assume all radiation will be converted to thermal energy when it strikes something. It can be transmitted, reflected, diffracted, deflected or scattered. It will only be converted to thermal energy if its peak frequency is above the peak frequency being emitted by the surface it strikes. So it has to come from a warmer source. If this were not so, then indeed any radiation could heat anything: so just stand outside at night and enjoy a hot “shower” in all that radiation going up and down. The longer you stand there the more you would absorb.
Hence thermal energy is not carried along in both directions when there are opposing beams of radiation. Only radiated energy is carried along. Thermal energy merely appears to be transferred but in fact it simply reappears only in a colder surface when the radiated energy is converted to thermal energy. At no point was it in existence anywhere along the way, so in that sense it does not travel. It is a bit like your voice being broadcast on radio waves and only appearing under certain conditions in a radio receiver.
Laser emission is actually different, because it is stimulated emission. We do find that, for example, 10.6 micron carbon dioxide lasers can melt metal when cutting it. This could not be done with normal spontaneous emission from carbon dioxide at atmospheric temperatures. Why is it so? My best guess is that it is because the intensity is such that the photons arrive faster than the resonating frequencies of the metal, so it can’t re-emit fast enough and has to convert the surplus to thermal energy because of the “chaos” created. Effectively the metal is then also undergoing stimulated emission, but the extra photons continue inwards and must cause warming.
(continued)
Because the energy in radiation from a cooler atmosphere cannot be converted to thermal energy when it strikes a (significantly) warmer surface, you have no thermal energy to affect either the rate of cooling each evening or the rate of warming each sunny morning.
So any such radiation from the atmosphere cannot in any way affect the temperature of the surface, or indeed the warmer areas of the atmosphere below the cooler level from which it was emitted. OK, there may be some rare weather conditions that result in warmer air a little above the surface, but these situations would be insignificant and have been happening since the Earth formed.
It should be clear from the above that a radiative greenhouse effect is a physical impossibility in the atmosphere..
If you don’t accept this, then you need to set up or find some experiment which actually demonstrates the opposite and actually shows thermal energy appearing to transfer from a cooler body to a warmer one. There is no middle ground. Either it happens or it doesn’t. You could have metal plates isolated in a vacuum container or some similar set up. As far as I can determine, this has never been achieved, yet the IPCC are in effect saying it is happening al the time as their “backradiation” slows the rate of cooling of the surface, and must also increase the rate of any warming.
The IPCC propagates this garbage, so they should attempt to prove it empirically. Their faces will be the only things warming.
Peter – the link you provided http://www.elmhurst.edu/~chm/vchembook/globalwarmA5.html is a joke. It implies all energy leaving the surface is by radiation and all energy absorbed by CO2 stays in the atmosphere. It does not explain how that energy goes back to warm the surface and it is totally different from the IPCC explanation involving backradiation, and also contrary to Trenberth’s energy diagram. If this represents your view of the GHE you have a lot still to learn my friend.
I posted a copy of my reply to a certain Peter Ridley above on this forum because it shows an example of how the public has been so misled by simplified (but thus totally incorrect) versions of the assumed GH effect. Read the link if you want a laugh.
I came across one of the guys who seems to think “heat” dams up in the atmosphere. If it were to, then the “valley” at the troppause should show some sign of warming.
Take a look at what’s really going on up there around the tropopause above which the temperature then starts to rise with increasing altitude in the stratosphere, so there’s a nice little “valley” for all your extra “heat”.
Now check out what NASA satellites found up there at say 56,000 feet. Then compare 2003 with 2011 – both almost precisely the same around the -63 deg.C value. Here’s the link: http://discover.itsc.uah.edu/amsutemps/
Check all the records available (for every year since 2002) and see how little variation there has been. Then tell me what would happen on Earth if it warmed up to, let’s say, -60 deg.C.
Carrick #613 has finally provided the proof the Doug and Prof Johnson are correct!
As has been know for over a century, light is constrained to travel at the speed of light on a vacuum, and at close to this speed in the atmosphere. It is very well established that photons cannot move at any other speed than the speed of light, or ‘c’
So radiation from the warm surface travels upwards at the speed ‘c’.
But the “back radiation” concept requires that photons move at ‘-c’, which is clearly impossible.
QEBD
Steveta_UK says,
” But the “back radiation” concept requires that photons move at ‘-c’, which is clearly impossible. QEBD”
Snicker.
Carrick,
“How about one that is old enough it couldn’t possibly be related to AGW propaganda?
How about Reville and Seuss, 1957”
Your link gave me a forbidden. I found it here:
http://uscentrist.org/about/issues/environment/docs/Revelle-Suess1957.pdf/view
I would point out that it was used as “back radiation” and was not defined. Watsup?? They were referring to the hypothesis of Arrhenius and Chamberlin. Wouldn’t their papers have established this term if it actually had a scientific meaning and not sloppy shorthand that became a term of Propaganda??I note that even Dougie used it as a shortcut rather than DLR.
And no, I do not accept your assertion that Revelle and Seuss (Dr. Seuss?) had nothing Gorebull Warming in mind in their research. Notice they are referring to Callendar and his claim that nearly ALL fossil fuel based CO2 is still in the atmosphere. A rather poorly supported claim even now. Even then they were assuming a figure for volcanoes with no possible way of quantifying the actual volcanic contribution etc. etc. etc. Manyof the same memes being bandied around by the IPCC and fellow travelers.
kuhnkat,
Doug has shown he’s crazy. I haven’t figured out just what you object to about radiation energy moving from cold to warm.
Jeff Condon,
” I haven’t figured out just what you object to about radiation energy moving from cold to warm.”
Probably because I haven’t expressed one. I DID express an objection to what turned out to be a pyroelectric effect being used to prove the surface absorbs DLR that slows its cooling.
Carrick,
“BTW, another accepted term in atmospheric physics is “down-welling radiation”. Down-welling refers of course to a particular orientation of the radiation path (vertical), so back-radiation is actually the more general term.”
All of the links were paywalled.
You are missing my point probably cause I ain’t making it well. Backradiation has not been defined in a meaningful way that we can agree or disagree on whether it is an appropriate term. One guy uses it to mean radiation that is REFLECTED from the atmosphere. The next guy is saying it is radiation emitted from the GHG’s that originated from the surface. One guy says that radiation heats the surface, Next guy says it slows the cooling. Dougie says it heats if it is above some breakpoint in frequencies. Where was it defined so that we can use it and know we are using it correctly??
Why even use it since it has become so politically charged and misused??
DLR is a much better term. Down welling long wave radiation.
OK, we’ll stick to DLR if it makes you happy. It is certainly more precise in this context.
Jeff #641 and others
Of course radiated energy can move (in a radiated “beam”) from cool to hot, or anywhere. I never said it couldn’t. I said it cannot be converted back to thermal energy by a target that is warmer than the source of such (spontaneous) radiation. If you still don’t understand what I am saying, then perhaps something in my post copied below will help ….
Bart says:
February 10, 2012 at 8:53 am
“There is no doubt at all that the Earth’s surface is hotter than it would be without an atmosphere “
__________________________________________________________
You are obviously unaware of the results of mathematical calculations by Claes Johnson (well-publish Professor of Applied Mathematics) in his Computational Blackbody Radiation” which is linked and summarised on the ‘Radiation’ page of my site http://climate-change-theory.com .
There is no physical process whereby atmospheric impedance to heat dissipation can cause thermal energy to be transferred back to the surface, (or even the air we stand in) other than, to a negligible extent, contained in precipitation. As Johnson and I have stated many times, radiation cannot transfer thermal energy from a cooler source to a significantly warmer surface or atmospheric ;layer.
I am not ruling out any accumulation of thermal energy in the atmosphere, but I am saying that such will not affect climate. If there had been any such accumulation, then ibt would surely show up in the tropopause – the temperature “valley” where temperatures stop declining and start to rise with increasing altitude. Check NASA satellite data at 56,000 feet for every year since 2002 on their site http://discover.itsc.uah.edu/amsutemps/ and you will see that it has been consistently around -63 deg.C every year with less than a degree of variation and certainly no rising trend.
The atmosphere does more in the way of cooling the Earth’s surface (by insulating it from solar radiation) and even the carbon dioxide actually also insulates and thus cools the surface. You have only to consider that the Moon’s surface can go about 100 deg.C to realise this. (I know it also goes below -150 deg.C but that is partly to do with not having core heat like the Earth and also the long 13 day cooling off period in the Moon’s night.) http://www.universetoday.com/19623/temperature-of-the-moon/
The earth’s adiabatic lapse rate is primarily determined by the mass of air in the atmosphere and the relative humidity, both of which mankind has no control over. This rate sets a temperature trend line in the atmosphere. That trend is actually an extension of the trend from the core to the surface. If the atmospheric trend were to rise to a new equilibrium at the surface end for some reason, probably greater solar radiance due to orbital variations, or variations in cloud cover, then the whole temperature plot from the core would also have to rise at the surface end because thermal equilibrium will be maintained at the surface/atmosphere interface. It would take a huge amount of energy to fill the gap under the new sub-surface plot, wouldn’t it. So, for a start, we would be seeing net flow into the surface rather than out of it as is currently observed. I am not saying it could not happen, and it obviously does in long natural cyclic patterns, but I am saying it takes a long time.
So, I repeat that there is no mechanism whereby any warming of the atmosphere will then warm the surface. All surplus energy above the well-embedded temperature trend line will be simply radiated away. Only when the surface itself is warmed, by stronger solar insolation or even by variations in thermal energy generated under the surface, and only when a new equilibrium is established in the core to surface plot will be see a new equilibrium at a higher surface temperature. Processes like this appear to happen naturally in long term natural cycles beyond the control of man.
All the above has been expanded upon in my book Greenhouse Land</i. soon to be published.
We have seen here at WUWT that there has been a reduction in relative humidity this century which should indicate that there is no radiative feedback from water vapour having any positive effect. That in itself demonstrates what Prof Johnson is saying that any radiation from a cooler atmosphere cannot transfer thermal energy to the warmer surface or to any (significantly) warmer atmospheric layer. This does more than just negate any amplification of any carbon dioxide forcing: it also negates the possibility of any GHG forcing whatsoever, which is in accord with what Johnson has proved. If WV does not warm with "backradiation" neither will any GHG.
I have written in other posts explaining how thermal energy merely appears to transfer only from warm to cool. It does not actually itself travel at the speed of light. Only radiated energy does so. Radiated energy only gets converted back to thermal energy when it meets a target which responds appropriately and converts the radiated energy to thermal energy. For this to happen the peak frequency of that radiation has to be above the cut-off frequency for the target itself, as Johnson explains. If it is below cut-off, the radiation (if not already reflected) will be transmitted or scattered and no energy left behind in the target. The very fact that thermal energy only ever appears to go from warm to cold is fully explained by Johnson’s result and no other process. If it could go from cool to warm there would be no end to the process and infinite spontaneous adiabatic warming would be a theoretical possibility by just surrounding a small object with numerous cooler radiating bodies.
If and only if there is conversion to thermal energy will there be any effect on the rate of warming or cooling of the target. So climate cannot be affected by radiation from the atmosphere.
Now, finally, there is considerable doubt that the capture of photons by carbon dioxide will necessarily lead to sharing of that energy among O2 and N2 molecules. These have different quantum energy steps for a start. Collisions may appear to transfer KE, but they in themselves can generate low amounts of radiation due to acceleration of electrons. What happens appears to depend very much on temperature also. If there is warming then there will be a greater propensity to radiate anyway. So, either radiation happens before any warming, or any warming causes more radiation. All radiation from the atmosphere eventually gets to space because the surface and warmer molecules lower down will not accept it.
Kookie Kat @ 643
“For others, I believe this is easily shown to be bunk by looking at a down welling spectra at night clear sky. The couple I have seen show primarily radiation in the 15 and 4 micron range where CO2 and water vapor overlap. Is that incorrect?? “
Well what I have looked at only shows ranges of wavelengths which in total cover the full spectrum. So they just assume that anything in a certain range must have come from WV or a trace gas. The process I described adds small amounts at virtually all wavelengths and is thus mixed in with (and so adds to) what they assume is from WV or CO2 etc.
Kuhnkat,
Are you saying that the bolometer example was pyroelectric? Truly sorry. I don’t want to go back and read the whole thing and it seems I’ve missed your point.
Jeff Condon,
you started with the IR camera. You didn’t add the bolometer until later.
Neither a bolometer nor the atmosphere can violate physics. Whatever radiation meets the surface from the atmosphere, whatever the intensity or wavelength, if it came from a cooler source it cannot add thermal energy to the surface.
See section 3.9.3 regarding the Second Law of Thermodynamics applying to heat transfer by radiation.
Click to access 0707.1161v4.pdf
Prof Claes Johnson proves computationally how and why it applies. If he were incorrect there would be a violation of the law.
This paper also mentions on p.74 the role of the core heat in climate, something which I have written about long ago on the ‘Explanation’ page of http://climate-change-theory.com I am relieved to find a paper that mentions such.
Only Doug is allowed is allowed to do that, on paper anyway. Man proposes, nature disposes, and proofs are for math class, not empirical science.
The 2nd law doesn’t apply to individual photons. Of course Claes gets around that by pretending photons don’t exist. At least he understands that’s what you have to do in order to not have a GHG effect. Never mind that it raises a million more questions than it answers (and it doesn’t answer anything except certain people’s need to not have a GHG effect).
Oh yeah and it’s dreck apparently to point out inconveniences like your hero Claes is a total wing-bat.
Claes has some serious bat competitors.
I am tired of the crap Doug. Three sentences on how IR lasers cut steel. No more. A second option, three sentences on how IR cameras can see objects colder than themselves.
If I have the energy, the rest will be snipped.
Trust the Germans to do it well …. knocking the AGW conjecture for six … take your time to read it all as I have – from this physics journal ….
Click to access 0707.1161v4.pdf
“Unfortunately, there is no source in the literature, where the greenhouse effect is introduced in harmony with the scientific standards of theoretical physics.”
“As already emphasized in Section 2.1.5 the constant appearing in the T4 law is not a universal constant of physics. Furthermore, a gray radiator must be described by a temperature dependent (T) spoiling the T4 law. Rigorously speaking, for real objects Equation (70) is invalid. Therefore all crude approximations relying on T4 expressions need to be taken with great care. In fact, though popular in global climatology, they prove nothing!”
Dear me, Carrick. “The 2nd law doesn’t apply to individual photons.”
So a few billion photons can just get together, each breaking the Law so the sum total of their effect also breaks the Law and, bingo, we have heat transferred from cold to hot, just like the Second Law says it can’t be.
You all confuse energy movement in the form of radiation going in all directions with heat transfer. When will you ever learn? Some of you don’t even realise that you can’t even slow down the rate of cooling (such as the surface cooling at night) without adding thermal energy. How are you going to do that from a cold atmosphere? The thermal energy has to be there first. Radiation itself does not slow down other radiation, let alone diffusion, conduction, convection, evaporation and chemical processes, all of which in total transfer more energy from the surface to the atmosphere than radiation. And these processes are “flexible” so if radiation from the surface is less, more can exit by other means.
The Germans explain it quite well, so I’ll leave it with them in the above linked peer-reviewed published physics journal. Allow two hours to study it well and it will change your viewpoint for ever if you dare.
.
@654
1. A laser uses electric power to generate highly focussed stimulated emission.
2. When stimulated emission strikes a target it also generates stimulated emission internally in the target.
3. This internal emission, which never happens due to spontaneous emission in nature, causes heating, burning or melting.
I predict the Germans are sure to pull out soon and perhaps lead European countries with them. See the third paragraph here … http://joannenova.com.au/
It’s all starting to happens guys. People respect German engineering and science.
The AGW conjecture is nearing the end of its days – within four years I predict – hopefully before the $100,000,000,000 bills start arriving in 2020.
The Second Law of Thermodynamics will hold fast and not be violated by any backradiation.
My friend Bart has put his foot in it again, but this may be of interest to any who really want to learn why the GHE is a fraud …
Bart: You display a distinct lack of knowledge about radiation when you say “In the case where there are higher energy emitters which are significantly stimulated,.”
No spontaneous radiation from the atmosphere can have radiated energy high enough to deposit thermal energy in a warmer surface, as official explanations of the GHE speculate happens. There are several different official explanations I know, which just demonstrates the confusion, but the IPCC says backradiation leads to warming somehow.
Why are you wrong in the above statement? Because peak frequency is proportional to absolute temperature according to Wien’s Displacement Law and the energy of a photon is proportional to its frequency and thus to the temperature of the emitter. Below the mesopause the atmosphere is colder than the surface, so any photons it emits have lower energy than the photons being emitted by the surface. And if any heat were transferred it would be in breach of the Second Law. And if no heat is transferred, there is no greenhouse effect causing any warmer climate down here. It’s that simple.
It is not I who “has it all wrong.”
My belief and understanding are exactly in accord with the published physicists who wrote this paper and made this statement therein.
Click to access 0707.1161v4.pdf
“Unfortunately, there is no source in the literature, where the greenhouse effect is introduced in harmony with the scientific standards of theoretical physics.”.
So if you, Bart, disagree with anything in the 100+ pages therein, write and publish an official rebuttal, or try to find one – which I would love to see.
(continued) Bart you really put your foot in it here: “the net flux is still out, so there is no 2nd law violation” because the Second Law has nothing to do with net radiative flux. It has to do with heat transfer. The IPCC tries to say there is heat transfer to the surface (thus slowing the rate of cooling in the evening and increaing the rate of warming in the morning and the maximum temperature for the day, whilst all the time there is net flux out of the system. This does not make it OK to break the Second Law by transferring heat to the surface.
This thought experiment should help to understand why the Second Law also applies to radiation, which is what Johnson proved and found the mechanism which ensures that it does apply.
Imagine three metal plates in parallel planes at equal separation in a vacuum.
1. The plate on the left is 100 cm^2 and is at 28 deg.C
2. The plate in the middle is 1 cm^2 and is at 30 deg.C
3. The plate on the right is 10 cm^2 and is 32 deg.C
(a) Net total radiative flux is left to right because the difference in surface areas dominates in S-B calculations.
.
(b) Heat flow is from right to left because of the temperature differences.
(c) The middle plate is not warmed by the cooler plate on the left despite the large flux.
Doug writes “Dear me, Carrick. “The 2nd law doesn’t apply to individual photons.” … So a few billion photons can just get together, each breaking the Law so the sum total of their effect also breaks the Law and, bingo, we have heat transferred from cold to hot, just like the Second Law says it can’t be.”
Have you heard of probability? Seriously Doug. Its a bit embarassing reading some of your posts. Sometimes its better to simply shut up, listen and learn. If your posts are representative of your knowledge, you’ve blown any chance of people taking you seriously on anything that you may be right on simply becasue you’ve demonstrated being ignorant on the basics so cant be trusted with any subtle argument.
Despite what TTTM seems to imply, the Second Law of Thermodynamics does apply for radiation. (All that Claes Johnson did was to prove what the mechanism involved was.)
All of you will find valid physics totally in line with what has been on my website and in my posts in the linked peer-reviewed published paper by the German physicists. http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf
There really is no need for me to explain what is in there to you as you can read the 100+ pages yourselves if your are genuimely interested in seeking the truth of the matter. I have not been influenced by this 2009 paper (as I have only just found it) but it does indeed echo my thoughts exactly, even that regarding the “thermal inertia” of sub-surface energy.
So it’s all there – take it or leave it.
Doug,
TTTM and Carrick are correct.
The second law is a probability based phenomenon. The fact that you don’t realize that doesn’t surprise any of us anymore. Stop the sanctimonious lecturing and open your ears. If I get around to it later, I will snip your copied posts from other threads above as you were told.
I have written this summaryso I may as well post it here in conclusion …pardon a bit of repetition …
It seems some need to learn some basic facts of physics, rather than what they have been fed by climatologists…
It is physically impossible to slow the rate of anything cooling without adding thermal energy, and iIt is physically impossible to add thermal energy by conduction or spontaneous radiation from a cooler source.
The language I speak is that of physics, not that of the IPCC.
Go and read 100+ pages of the peer-reviewed published paper by those two German physicists which I linked above, but link again http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf
There is absolutely nothing in that physics paper which contradicts anything I have written on my website http://climate-change-theory.com or in recent posts, even though I have just found the paper yesterday.
The Second Law of Thermodynamics does apply for radiation and you need to remember that radiation is not itself thermal energy. It is only converted to thermal energy when it strikes a target which is cooler than its source – which is why heat only transfers from hot to cold and not from a cold atmosphere to a warmer surface. The target detects the temperature of the emitter by way of its peak frequency which is proportional to its absolute temperature. Radiation which is neither reflected nor converted to thermal energy is merely scattered and thus has no more effect than that which is reflected.
So there is no heat transfer from the cooler atmosphere (as per Second Law) and without any heat transfer from the atmsophere, there can be no slowing of the cooling rate of the surface.
I will not be replying to anyone who does not show evidence of reading the above physics paper at least in part. I am happy to explain any part of it, but will be unavailable for the next 24 hours.
$664 Garbage Jeff – with billions*billions of molecules in the atmosphere and surface don’t you think the probabilities even out?
Or put it this way, the probability of heat going from cold to hot when the temperature difference is significant (as I’ve always said it needs to be) is absolutely infinitesimal.
Don’t you think I am aware of why it is probability based – because the two distributions can overlap if temperatures are close. Really, you have a somewhat misguided idea of what I have learnt in thousands of hours spent on all this and far more in my physics studies over 50 years since gaining a high position in Physics in the State in my final year of school.
Anyway, it’s after 12.30am here in Sydney, so I’m off the air.
Strictly speaking I should have said it is impossible to slow the rate of cooling of anything in a vacuum without adding thermal energy. In practice, however, when the Earth is cooling at night the atmosphere at the surface is cooler and cooling faster than the surface, so no significant insulation is happening. Even more so, when calculating average rates of cooling over a 24 hour cycle it should be obvious that the rate of warming would also have to increase by whatever radiation process was slowing the cooling at night. It is perhaps easier to understand that this rate of warming requires addition of thermal energy, but it does also require addition of thermal energy to slow the rate of cooling once there is any significant difference in temperature at the interface.
Junior high physics is not a ‘high’ position.
Doug:
That’s pretty much it. Interactions between individual photons and atoms and molecules obeys time reversal invariance., large scale macroscopic ensembles of photons and atoms and molecules don’t
Entropy, which is the basis for Clausius’s Principle, isn’t even a microscopic quantity, it’s a description of a configuration of a large number of molecules (10^6 will do). When you say that you’ve transferred a quantity of heat energy delta Q between two bodies, you’ve also transferred an entropy dQ/T.
It’s pretty telling of your education level that you think that something that is only defined macroscopically should apply to microscopic particle interactions. Or that you don’t know it can be derived from the interactions of microscopic particles, themselves obeying time reversal invariance.
You might start here. Plenty of technical references at the end see the brief discussion of statistical thermodynamics too.
The first thing you will need to understand if Clausius’s Principle (that the net heat energy exchange between two bodies is from the hotter to the cooler in the absence of external work) is equivalent to the statement the “entropy of a closed system is either constant or increases with time.”
The second part you need to work out for yourself is how entropy relates to microscopic phenomenon via the number of possible configurations in the system.
Jars with lids can’t take on new water, so the third thing you will need to do is open your mind so you can learn something new.
You’re mixing concepts now. “Rate of cooling” refers to power, a rate of change, “thermal energy” refers to accumulated energy of the ensemble of particles.
Suppose you have a heat source, an electric space heater for sake of argument which is consuming an electric power P, and you measure the temperature beside it (put the thermometer in an aspirated radiometric shield so it reads a “true” mean temperature). Let’s say you measure a temperature “T1”.
Next, let’s put an insulating shell around the heater, give it a little while to reach a steady state, and measure the new temperature, T2. Assume in all of this that the thermal energy per unit time (thermal power) of the heater is constant.
For fun, let’s repeat it by adding a second insulating shell around the first. So now we’re double insulated. Yada yada … we measure temperature T3.
When you look at the three temperatures, you’ll find
T1 < T2 < T3
All the while the heater is drawing the same electric power P.
How is this possible?
And how is that different than a theory in which the atmosphere a GHG effect acts as an insulator (so the temperature gets raised by a mean 35°C on the surface) and an atmosphere without a GHG effect?
Conceptually they are the same thing. If I get a chance later I might give a short physics based model to demonstrate it.
#670, Whether there is a bolometer or not doesn’t matter. The photons are still being absorbed. I was aware of the sensor though. The bolometer was supposed to be the comeback if people still didn’t understand.
Carrick
People who believe in that 35 (or 33) degree C figure as you said you did (I quote you “raised by a mean 35°C”) have not got off first base in regard to understanding the fallacies in the IPCC argument and why you cannot apply SBL to the surface.
Go read why on either my site http://climate-change-theory.com or in the peer-reviewed published paper Falsification of the Atmospheric CO2 Greenhouse Effects Within the Frame of Physics
Click to access 0707.1161v4.pdf
I am not going to discuss your misrepresentative “examples” which you obviously think explain the atmosphere as if there were no pressure gradient – as in your insulated heater.
As I have said, and say again to everyone, the above paper sets out numerous fallacies in the arguments pertaining to the GH conjecture using real physics. I am happy to discuss any section in the paper, but nothing else, for the paper says what I have been saying almost to the letter, simply because physics is consistent when it is properly understood.
PS Stop trying to teach your grandmother to suck eggs. Of course I know about entropy and other things you mention, and I also know where you have gone wrong in your analogies and why they are not relevant. I have been talking about the physics of the atmosphere – you have not.
“Global climatologists claim that the Earth’s natural greenhouse effect keeps the Earth 33 C warmer than it would be without the trace gases in the atmosphere. About 80 percent of this warming is attributed to water vapor and 20 percent to the 0.03 volume percent CO2. If such an extreme effect existed, it would show up even in a laboratory experiment involving concentrated CO2 as a thermal conductivity anomaly. It would manifest itself as a new kind of `superinsulation’ violating the conventional heat conduction equation. However, for CO2 such anomalous heat transport properties never have been observed.
“Therefore, in this paper, the popular greenhouse ideas entertained by the global climatology community are reconsidered within the limits of theoretical and experimental physics.”
A machine which transfers heat from a low temperature reservoir to a high temperature reservoir without external work applied, cannot exist – even if it is radiatively coupled to an environment, to which it is radiatively balanced. A modern climate model is supposed to be such a variant of a perpetuum mobile of the second kind.
[Rahmstorf wrote:] ‘ ‘Some “sceptics” state that the greenhouse effect cannot work since (according to the second law of thermodynamics) no radiative energy can be transferred from a colder body (the atmosphere) to a warmer one (the surface). However, the second law is not violated by the greenhouse effect, of course, since, during the radiative exchange, in both directions the net energy flows from the warmth to the cold.’
“Rahmstorf’s reference to the second law of thermodynamics is plainly wrong. The second law is a statement about heat, not about energy. Furthermore the author introduces an obscure notion of \net energy flow”. The relevant quantity is the \net heat flow”, which, ofcourse, is the sum of the upward and the downward heat flow within a fixed system, here the atmospheric system. It is inadmissible to apply the second law for the upward and downward heat separately …”
#671 Jeff Of course photons are absorbed – the metal gets very hot doesn’t it?
I gave you the three sentences you asked for (repeating the concepts already explained in my earlier posts about the differences between stimulated and spontaneous emission) but after all that you still don’t understand. I’m sure you could google “stimulated emission.” I’m not going to do your work for you.
Well, it took 5 minutes at the most …
“What Einstein had realized is that light shined on an atom which is in an excited state can induce the atom to make a downward transition (emitting a photon) if the incoming light’s frequency matches the atomic transition energy. The incoming photon is a boson, and for this reason it stimulates the emission of a second photon in the same state, inducing an atomic transition. (Otherwise the “spontaneous emission” would happen randomly.)
Thus, in stimulated emission we have an example of “quantum causality.”
This process combined with reflection can yield many photons in the same state: coherent light. Stimulated emission underlies the laser. “
“Unfortunately, there is no source in the literature, where the greenhouse effect is introduced in harmony with the scientific standards of theoretical physics.”
This is why Carrick is wrong about 35 C …
“As already emphasized in Section 2.1.5 the constant appearing in the T4 law is not a universal constant of physics. Furthermore, a gray radiator must be described by a temperature dependent (T) spoiling the T4 law. Rigorously speaking, for real objects Equation (70) is invalid. Therefore all crude approximations relying on T4 expressions need to be taken with great care. In fact, though popular in global climatology, they prove nothing!”
The subject of this thread is about atmospheric mass – which, under the influence of gravity, causes pressure – which causes an adiabatic lapse rate which is a function of the acceleration due to gravity – which means an “automatic” temperature gradient must exist in the atmosphere with or without WV, CO2 et al.
So it has to be (much) warmer at the surface than at TOA with or without you know what.
When will you people wake up to what basic physics teaches?
PS Whatever you deduce using SBL can only be applied to the whole Earth-plus-atmosphere system as seen from space as that is the only system around here that emulates a blackbody. After all quite a bit of the radiation comes from the atmosphere, not the surface.
Hence, if you like a temperature approximately -18 C then it will be perhaps 10 or 12 Km up in the atmosphere as a weighted mean for the whole system.
Hence, with the automatic lapse rate (not requiring WV, CO2 etc) the surface will be warmer than the mean and the TOA will be colder.
Now that really is the last free lesson in physics.
From now on I will only discuss any quotes from the above two physicists in the paper quoted – iff you show genuine desire to learn by understanding the physics.
Oh yes, I forgot, carbon dioxide does indeed act as an insulator. Half the Sun’s spectrum is in the infra-red band, so it absorbs incoming solar radiation and sends some back to space, thus having the same effect as increasing the albedo.
So it insulates the Earth’s surface from some of the Solar Insolation
But nowhere near as much as the whole atmosphere insulates us from the heat of the SUn. If we had no atmosphere the temperature could go over 100 deg.C as it does on the Moon nearly every lunar day.
PPPS Sorry, couldn’t resist posing this one ….
And what about each sunny morning? The Sun is warming the surface, so is the radiation from the cooler atmosphere also adding extra thermal energy to the surface in breach of the Second Law?
Does the Second Law only work in the mornings perhaps?
A simple proof why Johnson is right and Jeff is wrong (in three sentences)
1. Suppose you start warming something on your kitchen stove.
2. The “something” also receives spontaneous radiation from cooler surfaces in your kitchen.
3. But the Second Law says no heat will be transferred from those surfaces to the “something”, so the photons from those cooler surfaces are not converted to extra thermal energy entering the warming “something” – just as Johnson predicted, because their frequency is too low.
Here is a simple proof in 10 easy steps why the Greenhouse Effect is a physical impossibility.
(1) The IPCC claim that radiation from a cooler atmosphere slows the rate of cooling of the (warmer) surface, thus leading to a greenhouse effect.
(2) The “rate of cooling” is a 24 hour worldwide mean, so wherever the Sun is warming the surface (any sunny morning) the rate of warming would have to be increased by whatever process is slowing the rate of cooling.
(3) Thus extra thermal energy must be added to the surface by such radiation in order to increase the warming rate in the morning and slow the mean rate of cooling calculated from both day and night rates.
(4) Now the Second Law of Thermodynamics relates to heat transfer which is not the same as energy transfer. Radiated energy can be two-way, but heat transfer between two points is always one way and it is invalid to split such heat transfer into two opposite components and try to apply the Second Law to each. Physics doesn’t work that way.
(5) Hence, the surface cannot warm faster in the mornings due to such an imaginary heat transfer, because that would be clearly breaking the Second Law no matter what. Nor can it slow the rate of cooling because of (4). And in general you would expect the same process to happen whether the surface is warming or cooling.
(6) So, those photons from the cooler atmosphere are not being converted to thermal energy in the warmer surface, as Prof Claes Johnson proved in Computational Blackbody Radiation.
(7) Hence the effect of the photons being either reflected or scattered is that there is no impact on the surface at all.
(8) It is also clear that there is no significant transfer by diffusion or conduction from the atmosphere to the surface because the surface absorbs more solar insolation than the lower atmosphere, and we observe that the atmosphere is generally cooler and even cools faster at night than the surface.
(9) So it really does not matter even if extra thermal energy is trapped higher up in the atmosphere because it does not affect what we call climate, and any such energy cannot make its way back to the surface, except possibly an insignificant additional amount in precipitation.
(10) Hence there is no valid physical way in which backradiation or absorption by carbon dioxide in the atmosphere will cause a significant atmospheric greenhouse effect.
If I haven’t convinced you, read this paper Falsification of the Atmospheric CO2 Greenhouse Effects Within the Frame of Physics http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf
Doug,
Your ignorance of physics is surprising considering the degree you claim. I don’t even believe you at this point. I just went through the whole thread and have discovered a pattern which indicates you haven’t even considered the real world examples we have provided. At the same time you have repeatedly declared victory. The pattern is denial of existence, recognition of existence, denial of function, wild explanation for function.
So many flatly unrealistic statements in one post that it truly hurts to read. Here are a few errors:
292 – sensor does not need to absorb radiation to detect it. —
#350 351 388 393 400 – insane denial that the bolometer examples even exist.
441 – cameras exist but designers are wrong!
#450 – complete misunderstanding of second law.
#458 – Crazy physics statement. “You don’t seem to believe that the receiving body can first detect the frequency (hence temperature of the emitting body) then “decide” whether to scatter the radiation or converts its energy to thermal energy.” Leading to a laser example
#466 – laser light is ‘totally different’
#470 – denial of the function of an infrared laser.
#496 – continued denial of IR laser
552 – more denial of IR laser existence
566,567 – IR lasers don’t cut
560 – denial of bolometers again
570,574 – denial of laser. Mixed up energy statement.
579 – faked inability to google for co2 laser
583 – denial that links exist when searched. Odd claim that seems to say manufacturers agree with you.
586 – admission that IR cutting lasers exist. Lengthly statement that they are absorbed by materials differently than anything else.
634 – mixed up statement about conduction process. No clue about 2nd law.
657 – crazy statements about laser energy being different
656 – denial of probabilistic foundation of second law
666 – denial of probabilistic foundation of second law
676 crazy laser stuff
You really should be listening rather than writing. We can’t even approach the discussion of absorption with you because your basic knowledge is so effed up. AND, excepting the denial of existence, you still haven’t even made up a fake explanation for the bolometric camera yet.
No chops.
More stuff that doesn’t work. Uncooled IR sensors
http://www.omega.com/guides/infrared.html
os423-ls -60C
os-dt8855w -50C
os1600 -45c
os425-ls
on and on. Literally dozens of examples of stuff you can buy today which violates Doug’s limits.
I didn’t correct one of Doug’s “free lessons in physics”, where he stated :
He’s wrong about the ensemble of photons being exchanged from the cold to the hot object and the hot to the cold object violating the second law. It doesn’t.
There are a larger number of more energetic photons emitted from the warmer object
It’s easy to see his confusion though, if he thinks it applies to the interactions between individual photons and molecules, why he would think that violated the 2nd law. Reviewing his comments, this seems to be his confusion throughout: He is trying to apply classical theory to quantum mechanical particles.
I wonder if this is Claes’s real problem too? There is an infrared catastrophe that is very similar to what he described above, that is resolved by taking into account all of the relevant corrections to quantum electro-dynamics. I don’t believe it’s exactly the same one, but I can imagine trying to mix quantum and classic concepts could lead to disastrous results. (Which is what is wrong with GE Smith claiming that individual molecules obey the classical Maxwell law).
In the real world, microscopic quantum mechanics knows nothing about bulk classical effects like entropy, which itself is a concept that owes its origin to the ensemble of a large number of particles.
This is easily enough demonstrated. If you have a box divided into two partitions with N molecules in it. Suppose all of the molecules started on one side of the partition at t=0 and the system is allowed to evolve. What is the probability after we’ve given the system a chance to return to equilibrium for all of the particles to return to the same side?
It’s 1/N of course, if it is a boson gas and much much smaller if the particles can interact with each other. If you make N very large at all, the process becomes irreversible.
Fun little animation here showing the concept.
The point is that individual photons don’t care squat about Clausius’s principle or any other classical concept grounded in large N configuration spaces.
Doug did get one thing right by accident. He quoted and even highlighted the quote:
It is inadmissible to apply the second law for the upward and downward heat separately
First thing you’d said right in weeks Doug.
So if the hot object emits N1 thermal photons of average energy E1 that are absorbed by the cold object, and the cold object emits N2 thermal photons of average energy E2 that are absorbed by the warm object, then the net thermal energy exchanged is:
Delta E = (N2 E2 – N1 E1)
But we know N2 > N1 and E2 > E1, so we know Delta E > 0, and Clausius’ s Principle is not violated by this process. See the lead-in comment of Doug’s above where he was applying the the second law for the upward and downward heat separately.
(BTW, you may wan to stop linking to people like Claes who don’t understand quantum mechanics and are trying to apply classical principles to molecular physics, governed by quantum mechanics, and requiring the full glory of quantum electrodynamics to get right.)
Sorry got my indexes reversed at the start. No preview button, blame the computer, the maid service, something. Hrumph.
I take this opportunity to revise and extend my remarks, this time with indices corrected.
Sorry for the confusion there. The net exchange of energy is from the warm object to the colder one.
Grr N2 > N1.
And for the record, Doug, I’m not going to respond to links to external sites where you claim you’ve explained something better. Copy the appropriate material to this thread is you think it explains it, or paraphrase it.
Secondly, I’m not going to respond to your constant appeal to authority in quoting Claes “there are no photons and no quantum mechanics” Johnson. You label yourself a fruit bat by endorsing his dreck, and so does anybody else gullible enough to swallow it. I do you a favor by neglecting to comment on such arguments, though in any case appeal to authority is a form of logical fallacy
The quote “There is no physical process whereby atmospheric impedance to heat dissipation can cause thermal energy to be transferred back to the surface, (or even the air we stand in) other than, to a negligible extent, contained in precipitation. As Johnson and I have stated many times, radiation cannot transfer thermal energy from a cooler source to a significantly warmer surface or atmospheric layer.” would be correct if the term thermal energy implied Heat Transfer or net energy transfer (which is what I think they meant). If it is meant to imply individual photon transfer and absorption is possible from cold to hot, but not net level, it is correct. If they meant no photons radiated down from cold to hot are absorbed, they are wrong. We need to clarify the term thermal energy in the use they stated. I personally think they are correct in their use, but have not decided on the rest of their analysis.
.The plain fact is that you are all stumped by my “10 steps.”
This has now been posted on 17 forums and no one can raise a valid objection any more than any of you can.
There are now several physicists saying the same thing, that the concept of backradiation having any effect on surface temperatures is a breach of the Second Law of Thermodynamics.
At the very least about a third of the backradiation hits the surface when it is warming due to solar insolation. So it is perfectly clear that no additional heat can be transferred then from the cold atmosphere to the surface. So what happens to those photons? They are scattered as I (and Claes Johnson and Nasif Nahle) have been saying all along.
Finally, the emissivity of the atmosphere is only about 20% and it is much colder than the surface, yet NASA diagrams show 117% up from the surface and 100% backradiation – tiotally impossible by SBL when you multiply by the emissivity
PS I don’t know how many times I have to say it, the Second Law of Thermodynamics is about heat transfer which is a totally different thing from net photon transfer or net raditive transfer or anything to do with the amount of radiation in any direction. If some of the radiation has no effect (eg reflected, transmitted or scattered) there is no corresponding heat transfer. Only the energy contained in that radiation from a warmer source will be converted back into thermal energy when it strikes an object which is both able to absorb it and cooler than it. That is how and why the 2nd Law works. That is correct physics, which no professor of physics should deny.
Now in addition to Claes Johnson, Nasif Nahle and myself saying it, you have two German physicists writing in a peer-reviewed published paper a complete demolition of the greenhouse conjecture based on this and many other cogent points. I bet you don’t even dare to read that 100 page paper linked at the end of my ten points. Keep on believing that probability theory or statistical mechanics or excessive backradiation reverses empirically observed physics if you so choose. Don’t get outside and do your own experiments like Nahle and I have, now with wide necked vacuum flasks filled with sand. You few people are a drop in the bucket compared with the numbers I will reach and the parliamentarians I will influence. Bye.
#687 PS What a joke for Carrick to imply that Claes Johnson might be confused because there is a UV catastrophe. You won’t appreciate the joke until you read his paper, but your statement proves you haven’t understood a word of his paper in which he discusses the UVC and provides a mathematical proof thereof using wave theory. In so doing, Claes provides a solution for the very problem that led to Planck guessing in desperation that light must also have a particle nature – a fact which Einstein was never satisfied with until the end of his days – a guess which Johnson has now proved was not necessary. Particles have mass. Things with wavelengths are waves.
#687 Carrick still misses the point. He says “So if the hot object emits N1 thermal photons of average energy E1 that are absorbed by the cold object, and the cold object emits N2 thermal photons of average energy E2 that are absorbed by the warm object, then …
The words in bold are incorrect. and this shows me that Carrick is simply denying what Johnson has proved without pointing out any error in his calculations.
For a start, the second transfer is breaking the Second Law if it did transfer heat from cold to hot.
Finally, suppose there were 99 such cold objects (maybe 18 deg.C, say) and only one warm one (22 deg.C say) all in a circle and all radiating at one in the middle which is 20 deg.C.
Claes and I and the Second Law and those other physicists all say that heat transfers from that one at 22 C to the one at 20C. and all those at 18C In addition, the one at 20C will transfer some heat to each of the ones at 18C. Heat only travels from hot to cold. That is the Second Law. :Otherwise lots of things would get heated by all the TV and radio transmissions and so such transmissions would not travel the distances that they do as their energy would be used up.
The words in bold are incorrect.
Don’t bold them then!
Doug, it would help if you could read, and then reason, in that order (otherwise GIGO).
I believe I was commenting on the infrared catastrophe not the ultraviolet one in reference to Claes. It’s completely unrelated to the ultraviolet one. I suspect he’s managed to discover another one that you get when you mix classical and quantum concepts. The infrared catastrophe is
Time for you to go google, spout some nonsense and pretend you know what you’re talking about.
Go ahead, get cracking.
And if Johnson were as brilliant and has revolutionized science as Doug thinks is the case, at least it would be a short commute for him to collect his Nobel Prize, but I’m afraid the only Prize he qualifies for is a Boobie Prize, the sort you get when you only half-way know what you’re talking about, and the only people nominating you are half-wits.
Carrick,
“…but I’m afraid the only Prize he qualifies for is a Boobie Prize, the sort you get when you only half-way know what you’re talking about, and the only people nominating you are half-wits.”
Are you claiming that Claes should get a Nobel PEACE Prize??
Doug: “For a start, the second transfer is breaking the Second Law if it did transfer heat from cold to hot.”
Nope. Only if there is net transfer of heat.
You really are a complete idiot.
#699 There would be a net transfer of heat when the surface is warming on a sunny morning. And net inward radiation. This would be additional hea from the backradiationt.
So you are wrong.
I’ll copy today’s post on WUWT as it explains more and compares radio waves. Maybe you need to think about why radio waves are not absorbed by the surface.
I have posted my “10 steps” post http://wattsupwiththat.com/2012/02/11/open-thread-weekend-7/#comment-890609 on many other forums (both for and against AGW) with a genuine desire to see if anyone can raise a valid counter argument. The following may help explain and reinforce what I have written there.
Firstly, the actual amount of backradiation must be grossly overstated because the emissivity of the atmosphere is only about 20% I understand and it is colder than the surface and emitting in all directions into a full sphere rather than a hemisphere. So how could NASA’s energy diagram possibly be right in showing similar values for emission from the surface as from the atmosphere. I postulate that the instruments calculate the radiative flux from the temperature assuming emissivity is unity and emission into a hemisphere.
But, whatever the amount of backradiation, it cannot transfer thermal energy from the colder atmosphere to the warmer surface as this would imply a heat transfer from cold to hot, which is against the Second Law of Thermodynamics. When the surface is warming on a sunny morning, for example, and net radiative flux is into the surface, how can additional heat be transferred from the colder atmosphere to the warmer surface against the Second Law. It can’t and all such radiation is reflected or (mostly) scattered and thus leaves no energy behind.
It follows that, since backradiation cannot add thermal energy to a warmer surface, then it cannot increase the rate of warming in the morning or slow the rate of cooling in the evening.
Backradiation is after all low energy radiation spontaneously emitted from a cold source. The peak frequency of the strongly attenuated spectrum is proportional to the absolute temperature (Wien’s Displacement Law) and such frequencies are usually lower than those in the radiation from the surface. Surface molecules “recognise” this and reject the low energy radiation which does not have enough energy to be converted to thermal energy. (See Prof Claes Johnson Computational Blackbody Radiation) http://www.csc.kth.se/~cgjoh/blackbodyslayer.pdf
We see examples of this in radio broadcasts where we know the radiation is even lower frequency than that in backradiation. It is of course artificially generated, but its frequency corresponds to much lower temperatures than normally experienced on earth or in the troposphere. For this very reason it is scattered by the surface and by the atmosphere and is not converted to thermal energy because it is “colder” even than the backradiation. If this were not the case, then it would have been quickly quenched as all its energy would have been used up warming whatever it struck. So it would not travel the distances that we know it does. Basically the same happens to backradiation and it just keeps getting scattered off molecules in both the atmosphere and the surface until it happens by chance to escape to space.
So the (latest version of) the atmospheric greenhouse conjecture is (like the first version) a physical impossibility.
😛
I would like to explain PV=nRT with respect to posting in this thread
Pressure placed on fringe theories (P) multiplied by volume of the fringe message as measured by bolding (V) can be equated to the number of posts (n) multiplied by redirected references (R) and logical science (T)
Indeed TTTM – Fringe theories (which are not in standard text books) have featured prominently in backradiation conjectures.
I quote from a well-known physicist ….
The original hypothesis of the greenhouse effect, which assumed that the Earth was warmed up because
of ―trapped‖ longwave thermal energy, was debunked through experimentation and observation [2], and
from measurements[3]. The assumed ―greenhouse gases‖, with exception of water vapor, are incapable of
―accumulating‖ longwave radiative energy as it had been assumed.
A consequence of the scientific demonstration is that the ―greenhouse effect‖ by ―trapped‖ longwave
radiative energy is imaginary[2], proponents of the nonexistent ―greenhouse effect‖ resort to another
explanation you will not find in any serious scientific literature; they invent a process of warming a warmer
surface by a cooler atmosphere through backradiation derived from thermal energy which has been
impossibly accumulated by a cooler atmosphere that in the real world does not reach such average of
thermal energy content.
AGW proponents and skeptics who argue about this hypothesis declare backradiation emitted from the
atmosphere warms the warmer surface, as if the atmosphere was a duplicator of thermal energy. As a
consequence, the second law of thermodynamics which determines the specific flow of thermal radiation
from warmer to cooler is dismissed[13
Funny that George E Smith also says CO2 radiation can’t warm the surface.
http://wattsupwiththat.com/2012/02/11/open-thread-weekend-7/#comment-891100
Doug,
George E. Smith agrees with us!!
Read carefully – a stupid thing for me to write after 700 comments.
“The key point is if the surface is hotter (it may not be), then the surface is also radiating, and likely more than the CO2, so there won’t be a net flow of EM radiation from the CO2 to the surface either, but a flow from the CO2 to the surface is a sure thing; but less than comes backup from the surface.”
He agrees with us, bolometers, radiotelescopes, microwaves, IR sensors, CO2 lasers, god etc….
In Doug’s world, radio doesn’t even work.
George is in the same boat as Claes, he’s applying classical physics to quantum mechanical systems and getting complete nonsense as a result. Poor fellows, their brains are stuck in the 19th century.
I have Brian H to point that out to me, that this is where the lot of them are getting confused: at the interface between classical and quantum theory.
As it turns out you can’t even use Schrödinger’s equation to model the interaction of a molecule with a photon, you have to use the full glory of quantum field theory (quantum electrodynamics), and even then it’s a d*mned tough nut to crack. (The infrared catastrophe I mentioned relates to that.)
Doug—you on the other hand are a loon. I think Claes and George both understand what “net transfer of heat” means, and would know better than to argue that if
dQ = Q2 – Q1 > 0
that we would have a violation of the 2nd law just because Q1 > 0, relating to the heat exchanged from the colder body to the warm one. Physics 001 (junior high physics) conceptual misunderstanding there on Douglas’ part. And he’s stuck on it, won’t give it up, Doug reminds me of Scrat chasing around that nut that he won’t give up. And of course, we can enumerate many of Doug’s conceptual misunderstandings, which aren’t limited to that Physics 001 problem above.
At least the other two are getting stuck in more substantial places, like applying Maxwell’s laws and classical dynamics to molecules in George’s case, and pretending you can just make photons and quantum mechanics go away in Claes’ case, and that this would explain more than the questions it opened up (in fact, it would open up vastly more questions than it would answer—which of course is the mark of a really bad theory).
In George’s case, well maybe not complete nonsense as Jeff points out. 😉
I’m not sure I fully follow what George’s saying here (if the surface has a temperature about 0K it’s radiating regardless), but anyway “flow from the CO2 to the surface is a sure thing” is right (I also would try and avoid the word “flow” but that’s me trying to impose 21st century notions on 19th century physics again).
No Carrick – George does not appear to understand what Johnson says at this stage. I have just tried to explain it to him as it happens with a new post on WUWT …
Thermal energy is not radiated energy and it does not travel along with radiation. Radiation can go in all directions, but “heat” does not actually travel itself or in any way cancel out due to opposing radiation.
When two bodies at different temperature radiate towards each other they are of course converting thermal or other energy into radiation. Assuming the bodies are not transparent to each other’s radiation, that radiation will be either converted to thermal energy or scattered without leaving any energy behind.
Thus “heat” only appears to travel, rather like sound being broadcast on radio waves. You see the temperature increase and say heat has travelled, but in fact it has just been “produced” on the spot by conversion of radiated energy.
This only happens if the peak frequency of the received radiation is (significantly) greater than the peak frequency of the target. As fhese peak frequencies are proportional to the absolute temperatures (by Wien’s Displacement Law) we only observe conversion to thermal energy when the emitter was warmer than the target.
The important thing is that it does not matter what radiation was being sent the other way. If you could block the radiation from the cooler body with a filter that still allowed the other radiation to pas through you would still see the same amount of warming.
This is the correct reason why radiation from a cooler atmosphere cannot be converted to thrmal energy in a warmer surface.
#708 No, radio waves in your world will add thermal energy to the surface every time they are supposed to bounce off the surface – just like all that backradiation warms the surface or slows its rate of cooling. After all, as you say, how can the surface tell the difference between one photon and another. Maybe we have AGW after all from all the new radio and TV stations.
So radio waves in your world will lose their energy when they get absorbed by the surface and get re-emitted as thermal IR radiation. .
So it is in your world that radio doesn’t even work.
Those who believe that thermal energy “moves” in the direction of net radiation should consider what happens when a blackbody faces another body, say, 10 degrees warmer but with much lower emissivity out in space, say. Suppose S-B calculations are such that the warmer body emits less than the cooler one because the warmer one’s flux is reduced by the lower emissivity. Hence the direction of net radiative flux is from the cooler one to the warmer one.
(1) What happens to the extra radiation that the warmer body does not absorb because of its low absorptivity?
(2) Does the warmer body convert to thermal energy any of the remaining radiation from the cooler body? If not all, what happens to that radiation?
(3) Is the Second Law of Thermodynamics obeyed or does the warmer body get warmer still?
(4) Does heat transfer from the cooler one to the warmer one, or vice versa?
Ric Werme at http://wattsupwiththat.com/2012/02/11/open-thread-weekend-7/#comment-891194
________________________________
Just in case you think lasers or microwave ovens or radar prove me wrong, the reason lies in other unique mechanisms to do with such things as stimulated emission (rather than spontaneous emission) and resonance in just certain molecules (mainly water molecules in the food) caused by microwaves which don’t heat everything as you know. So these are very different from so-called “blackbody” (spontaneous) radiation such as occurs naturally.
Anyone not knowing what stimulated (or induced) emission is can look up Wikipedia. Basically a laser beam is produced by stimulated emission and it will cause stimulated emission in a target such as sheet metal. Stimulated radiation arrives at such an intensity that additional radiation is generated in the same direction as the incident beam, and so more radiation penetrates deeper, and such is more than the material can scatter to regions outside its boundaries, so it converts this unnatural surplus to thermal energy. (Whilst it might not fully penetrate the sheet, a cut can be made using holes that are first cut and filled with special materials.) This is not usually a natural process having been generated artificially.
Doug,
You cannot explain the operation of bolometers, you completely deny their existence and function – not very scientific Doug. You cannot explain the pyrometers measuring to -60C. You have denied the existence of the CO2 laser – insanely – until you had to make up a fake reason for operation. I’m certain from your past behavior that you haven’t even realized that materials being hit by the laser have to warm up before they release their electrons. The metal starts out at room temperature – or even cryogenic temps. Like every comment you have written here, your ad-hoc arm waiving has no basis in physics.
We are done here Doug. Move along.
kuhnkat, lol. Sorry but he’s not a hippie or a hippie lover, so no not even that.
Jeff, I haven’t heard an explanation from him for how a radiometric profiler works either. they seem to function pretty well for something that’s impossible, according to Doug and Claes.
(I love the Mars profiler because it’s specifically using CO2 thermal emissions to generate the vertical temperature profile. As you know Mars atmosphere is mostly CO2.)
If you prefer Earth, The commercially available TP/WVP-3000 can generate profiles up to 10 km, which as I mentioned above is nominally TOA for CO2 emissions. So you can detect photons from the height needed for the GHG effect.
I get your point about the laser. Brian H did correctly state that individual photons don’t have many pieces of information, they have direction, energy and left or right circular polarization (he forgot that, no biggie), and that’s it. There’s no label to tell the receiving object that this photon came from black body versus any other mechanism.
And of course, Doug’s other big argument is based on a junior high school level misunderstanding of the 2nd Law of Thermodynamics. That word “net” doesn’t mean anything to Mr. Let Me Teach you Physics.
And finally, he doesn’t’ know that it only applies to macroscopic processes but to individual photon exchanges.
It’s impossible for Doug to admit that he’s this wrong, and possibly beyond his limited ability to comprehend.
I also understand (without having to torture myself) where Claes jumped the tracks. So yeah, I’d say it’s time to pack this one up and move on.
I note that no one has attempted o answer the four questions in #714.
Nor has anyone had any valid counter argument on any of the other forums on which I posted these questions.
I’ll come back in 24 hours to check.
Doug,
The questions are aphysical and thus cannot be answered.
Here’s one for you:
What color is seven?
A point was made on another forum about LED light having an apparent source temperature of ~2000 C. (This is another example of artificially generated radiation.) I replied …
You seem to think that light from your LED could not make something hot. You think that because you know its intensity is low. True. And in real life it doesn’t warm anything noticeably, because whatever small amount of thermal energy is created escapes just about as quickly. But under certain conditions with enough insulation and enough intensity LED light could warm something.
A good example is a solarium where UV light is generated artificially. They are closing down solariums because they kill people with sunburn that leads to skin cancer. That artificially generated UV light from a relatively cool machine is just as “effective” in burning skin as UV light from the far hotter Sun.
Jeff
Maybe you can answer these questions.
(1) Why is a microbolometer limited to “seeing” wavelengths only up to 12 or 14 microns (according to its specifications) when IR thermometers (the ones detecting frequency) can measure temperatures corresponding to much longer wavelengths?
(2) What mechanism limits the microbolometer if it is not the point at which the two spectra (which are moving apart as the temperature difference increases) no longer overlap?
My answer:
While ever there is an overlap of the spectrum for the temperature of the sensor and the temperature of the source of radiation then some signal can be detected due to some warming.
However, for wavelengths above 12 or 14 microns the spectra no longer overlap. This point obviously depends on the temperature of the sensor, so this may explain why different manufacturers specify different upper limits. You will note there is also a lower limit around 7.5 or 8 microns for the same reason when the spectra move apart in the opposite direction.
#719. Which of the four question(s) are “aphysical” and why?
The whole point in asking is that you are being asked to explain why you think such.
What exactly would happen in your view in this thought experiment – which is meant to make you think.
You have climbed out of your spacecraft on the shady side out of the sunlight. You have two metal plates with very different emissivity (as explained above) each preheated to different temperatures at least 10 degrees C apart. Each plate has a temperature sensing device. They are placed in parallel planes reasonably close together. There is much more radiation going from the cooler one than the hotter one due to the difference in emissivity. OK. so far?
Does any thermal energy transfer from the cooler one to the hotter one, such as is supposed to happen for radiation from a cooler atmosphere to a warmer surface?
——–
Doug,
I don’t think you should ask any more crazy questions until you answer real world observations. It really isn’t fair to the rest of us to take up our time trying to explain physics to you when you aren’t addressing the answers or questions.
[self snip]
[self snip]
Doug,
You are unqualified to discuss these matters.
Agreed Jeff. I don’t see any point in continuing this discussion with somebody who is not only unqualified, he’s unwilling to admit he’s unqualified, to be discussing this.
I don’t think he’s fooling himself, hopefully not even himself, at this point.
Leonard was the one who started the climagegate 1.0 thread also. He’s dangerous for a guest post.
This (copied) post really is all I need to say. If you guys wish to argue with Johnson, he has a blog on which you can post your objections to his mathematics, or whatever.
I think, as a well-published Professor of Applied Mathematics, Claes is “well qualified” to debate the issue with any on this forum.
__________________________________________
Prof Claes Johnson has proved in Computational Blackbody Radiation* that energy in radiation only gets converted to thermal energy if the peak frequency of the radiation from the source is above the peak frequency of the radiation from the target.
This essentially provides a mechanism which explains why the Second Law of Thermodynamics also applies for radiative heat transfer, as it does for heat transferred by conduction.
It is not the net radiative flux (or even its direction) which determines whether (and in which direction) thermal energy is transferred. For example, if the emissivity of two bodies is very different, there can be more radiative flux from the cooler one. But all that flux will be scattered by the warmer one and not converted to thermal energy. Only the flux from the warmer one (no matter how weak) will be converted to thermal energy in the cooler one. This “ensures” that the Second Law is valid in all cases because it depends on peak frequency which is proportional to absolute temperature – see http://en.wikipedia.org/wiki/Wien's_displacement_law
* http://climate-change-theory.com/RadiationAbsorption.html
[self snip again]
Why would I want to argue with Johnson? He didn’t write crap on an open science blog.
How typical of those who have no answersto uncomfortable questions or scientific findings.
(1) Find any weak excuse not to reply, or
(2) raise an issue of “qualifications” as if such are the be all and end all of knowledge, understanding and intelligent thought processes.
I doubt whether my physics degree course taught me 10% of the knowledge I have since gained from private study.
[snip]
Just gobsmacked.
You are all wrong because you have not addressed the situation in which the emissivity of one body is very different from another.
You cannot explain the Second Law if the hotter body emits less than the cooler body because the hotter one has lower emissivity, now can you? Net radiation in that case goes from cold to hot.
You need to read Prof Claes Johnson’s “Computational Blackbody Radiation” and/or my ‘radiation’ page at http://climate-change-theory.com
The temperature information is contained in the value of the peak frequency of the radiation. So a warmer surface can, and does, merely scatter radiation from a cooler atmosphere without converting it to thermal energy.
Hence the greenhouse conjecture is a physical impossibility.
In more detail …
The frequency distribution for a cooler emitter (the atmosphere) is fully contained within the frequency distribution of an emitter at a higher temperature (the surface), and it is shifted towards lower frequencies because the peak frequency is proportinal to the (absolute) temperature.
Hence any radiation from the cooler emitter can potentially resonate with the warmer emitter and be re-emitted immediately without any conversion to thermal energy. This is what Prof Claes Johnson has proved does happen and it is the only plausible explanation for the veracity of the Second Law of Thermodynamics applying to radiation.
In contrast, the frequency distribution of the warmer emitter does extend into higher frequencies than that of the cooler emitter. The area between the curves in this region gives a measure of the probability of conversion of such radiation to thermal energy. Thus this probability is nil until there is a temperature difference, and then increases as the temperature difference increases.
We see that virtually all solar radiation reaching the surface (UV, visible and IR) will be converted to thermal energy because there is virtually no overlap between the spectra of incident solar radiation and that of emitted radiation from the surface.
Hence the surface absorbs solar radiation, converting it to thermal energy which can subsequently exit the surface layer (usually at night, but even months later in local winter) by diffusion, conduction, evaporation, radiation and chemical processes, followed by convection once the energy is into the air. There is subsequently negligible probability of any of that energy returning to the surface or remaining in the surface layer of the atmosphere for longer than it would otherwise have done just because of carbon dioxide, methane or similar trace gases.
Furthermore, the emissivity of the atmosphere is only about 20% and the radiation towards Earth is half that. Then the temperature of the atmosphere is colder, so even less radiation.
So “backradiation” can only be less than ~10% of upwelling radiation from the surface.
Nothing at all like those energy diagrams which fly in the face of SBL.
Elementary calculations you cannot fault, now can you?
How about addressing these issues ..
(1) The direction of net radiative energy flow can be the opposite of the direction of heat transfer. if you have a warmer object (say 310 K) with low emissivity (say 0.2) and a cooler object (say 300 K) with much higher emissivity (say 0.9) then net radiative energy flow is from the cooler to the warmer object. Yet the Second Law says heat transfer is from hot to cold.
(2) Any warming of a warmer surface by radiation from a cooler atmosphere violates the Second Law of Thermodynamics. Consider the situation when the surface is being warmed by the Sun at 11am somewhere. Its temperature is rising and net radiative energy flow is into the surface. How could additional thermal energy transfer from the cooler atmosphere to make the surface warm at a faster rate?
Clearly radiation from a cooler atmosphere cannot add thermal energy to a warmer surface. The surface molecules scatter radiation which has a peak frequency lower than the peak frequency of their own emission, and so no radiative energy from that radiation is converted to thermal energy. (This was proved in Johnson’s <Computational Blackbody Radiation.), So the atmospheric radiative greenhouse effect is a physical impossibility.
“How about addressing these issues .. ”
You haven’t grocked any of our points in the last 700 comments. Why would we keep trying to teach you?
Oh yeah. In #736, your conclusions are also completely wrong.
The conclusions are not wrong and you cannot give valid reasons for your statement. Even in (1) there is net radiation going into the warmer body, yet absolutely no increase in temperature because it is cooling.
Johnson is right.
I don’t require you to address it.
But you will learn if you think about it.
Humility or Hubris Doug,
It ain’t a tough call.
Seems like you’re the last man standing, albeit with words rather than logic.
Is it an omen that my middle name is Jeff ?
Bye
(I’ll just be on WUWT and Judith Curry’s site if anyone wants to take up the discussion with me.
The stampede will begin any time now …
LOL
Oh. add to that, Tallbloke’s Talkshop site. I just wrote this there …
[snip – as promised Doug]
Discussion implies two way conversation Doug. Not one where a person asks a question which goes on being ignored for days and cannot be reasonably addressed.
As a kindness, I suggest that you do the ACTUAL math of the emissivity posts here before putting that in your book.
I agree about discussion, but I only have time to do so on one main forum, namely WUWT in last Friday’s Open thread there.
So, just quickly this post I’ve just done on WUWT may help you all to understand better, but please only reply there
[snip, this is a technical blog doug. We cover details that cannot be covered or discussed at the center of the internet WUWT, or on a moderated forum. You have shown no respect for people here, I see no reason to continue allowing your nonsense no-content posts here until you address the issues pointed out. If you like, I could ask Anthony to do the same.]
Doug says “Fact: If there is any backradiation it is not even enough to melt frost (which is in the shade) in 8 hours, yet it is supposed to be a quarter as powerful as the Sun at noon” and asks why.
Because it cools as it sublimes and gradually diminishes.
Regarding the IR camera, from wikipedia: “A photodiode is a p-n junction or PIN structure. When a photon of sufficient energy strikes the diode, it excites an electron, thereby creating a free electron (and a positively charged electron hole)”
Are you saying that (1) the energy of the photon is directly related to the temperature of the source and (2) sufficient energy to free an electron is related to the temperature of the receiver? Or does the scattering and/or reflection not apply to this case?
Doug asks for someone to address this: “(1) The direction of net radiative energy flow can be the opposite of the direction of heat transfer. if you have a warmer object (say 310 K) with low emissivity (say 0.2) and a cooler object (say 300 K) with much higher emissivity (say 0.9) then net radiative energy flow is from the cooler to the warmer object. Yet the Second Law says heat transfer is from hot to cold.”
The higher emissivity object has equally high absorptivity so it will warm more than the lower absorptivity object. I hope that’s right, if not someone will correct me.
746.
Eric (skeptic) said
Are you saying that (1) the energy of the photon is directly related to the temperature of the source
________________________
Yes. Standard physics says it.
747.
Eric (skeptic) said
February 15, 2012 at 8:48 pm
The higher emissivity object has equally high absorptivity so it will warm more than the lower absorptivity object. I hope that’s right, if not someone will correct me.
_____________________________________________
Yes if the Sun is warming each simultaneously under identical conditions.
Eric,
“The higher emissivity object has equally high absorptivity so it will warm more than the lower absorptivity object. I hope that’s right, if not someone will correct me.”
This is not accurate. Planck’s emission spectrum also contains a magnitude component so the higher temperature object emits more inside its primary emission/absorption wavelengths than the cooler one. You need to do a little math to understand this problem. Doug is a very bad resource and has mashed yet another known physics point.
You all should read WUWT lead article today …
My post …
Suffice to say, the global warming hoax was the golden goose for everyone who received literally billions in public and private funding.”
Never was a truer word spoken. Any atmospheric “greenhouse effect” amounts to a complete violation of the laws of physics and is a travesty of such.
First they realised they had to give up on the concept of warm air acting as a blanket, so then they came up with the conjecture that second-hand “backradiation” would slow the rate of cooling of the surface.
Well, to slow the mean rate of cooling you also have to slow the rate of warming every sunny morning. After all, the same process (whatever it could possibly be) has to add thermal energy both morning and evening.
Now think about it. The Second Law of Thermodynamics has to apply between any two specific points wherever there is matter in the universe. And it applies for radiation just as much as it does for conduction. Can you imagine a long metal rod extending out of a patch of ground into the colder air with a heat flow going from the cooler air back into the warmer surface which is getting warmer and warmer due to the Sun at, say, 11am somewhere? This crazy reverse heat flow dreamed up by you-know-who is the conjecture upon which $ ??? billions are being betted.
It just isn’t physics..
Radiation from a cooler source always includes all the frequencies which a warmer surface is able to emit. So all its waves merely resonate and in effect get scattered back out again without any of its energy being converted to thermal energy – no energy is left behind. But radiation from a warmer source has additional frequencies above those of a cooler target, so these higher frequencies cannot resonate and their energy ends up being converted to thermal energy, as seen when the Sun warms something.
Hence, when analysing what would happen when the atmosphere and surface interact by radiation all the radiation from the cooler atmosphere should be disregarded. Only radiation from hot to cold is converted to thermal energy and this is how and why the Second Law of Thermodynamics works for radiation.
16. Leonard Weinstein said
January 23, 2012 at 12:10 am
When you lower the energy of some “hot” locations, you move that exact amount of energy to “cooler” locations. Using the T^4 law for the redistributed energy fluxes give a LOWER average surface temperature (but equal total energy flux). Break it into 2 parts and try it with energy and then average the area temperatures to see.
Indeed Leonard you do move energy from hot to cold, not the other way.
[you know what doug, snip. I really hate snipping but I hate stupidity and ignorance even more.
I think I’m getting the hang of this guys. It’s like firing people, your first dozen are tough, after that — meh..]
[you are in moderation. If you answer questions asked of you, I will let it through. If you post random false information misrepresented as science, you are on your own.]
Doug,
I can’t believe you would tell Dr. Weinstein to ‘read textbooks’ when you don’t know who he is and can’t understand his words. like telling Carrick to read your stupid links that you don’t even understand.
Get a clue man I’m tired of your spam.
Jeff, I don’t think I was clear. I think the answer is “it’s both” The warmer object has higher magnitude for each wavelength as you said. along with a shift to shorter wavelengths. In Doug’s scenario the high emissivity object is cooler (as he defined the scenario) and the low emissivity object is warmer. He then claims that standard physics would have more net photons flowing from the cooler, high emissivity object to the warmer, low emissivity object which would violate the Second Law.
But Doug did not consider that his hypothetical, cooler, high emissivity object must have high absorptivity and therefore absorbs more photons than the warmer, low emissivity (low absorptivity) object. So along with the higher power radiation from the warmer object, the warmer object warms the cooler object.
Let me put it another way, the emissivity is the fraction of the theoretical maximum black body radiation. If a somewhat cooler object has a much higher emissivity (say, 0.9) and a somewhat warmer object has a much lower emissivity (say 0.1) at the start of the scenario, then the somewhat cooler object would radiate more power per unit area than the somewhat warmer object. The “somewhat” needs to refer to a rather small difference because the radiative power is proportional to the 4th power of temperature.
But now we have to consider that the cooler, higher emissivity object (0.9) also must have much higher absorptivity (also 0.9 in this scenario vs 0.1), so the Second Law is not violated.
Since Doug is allowed to answer questions, I would like to ask him if cloudy nights are warmer than a clear nights (all other things being equal). Then if that is the case, is the warmth at the surface due to infrared radiation from the clouds? If not, then what causes the extra warmth?
Eric,
I don’t moderate this blog. Really.
Doug is the third person to make the list in close to 50K comments. The others were off the list in very short order.
He can answer you all he wants but now I have the hassle of approving a comment.
There are many, many situations Doug did not consider. If you wish to engage him, tis your IQ points. If he answers your questions, I’ll take him back off the spam filter and let er rip. When he stops answering you and starts posting insanely, back on again.
[snip ]
754.
Eric (skeptic) said
February 16, 2012 at 8:24 pm
The calculation of emissivity and absorptivity has either already taken into account the fact that radiation from a cooler object is not converted to thermal energy or it’s been calculated from solar radiation only.
You all need to stop assuming things that don’t happen. Just because a photon strikes a target does not mean it warms it. If it is not reflected it will resonate (if from a cooler source) be scattered and not be converted to thermal energy.
When you can show me an experiment where backradiation warms a flask of sand, or slows its cooling relative to a shielded flask, then we can continue this converstaion.
755.
Eric (skeptic) said
February 16, 2012 at 8:27 pm
Since Doug is allowed to answer questions, I would like to ask him if cloudy nights are warmer than a clear nights (all other things being equal).
________________________________
If there has been equal cloud cover during the day and equal daytime temperatures and if the relative humidity is the same (unlikely) there will be no difference in temperature at night.
747.
Eric (skeptic) said
February 15, 2012 at 8:48 pm
The higher emissivity object has equally high absorptivity so it will warm more than the lower absorptivity object. I hope that’s right, if not someone will correct me.
_____________________________________________
I’ll try again. Given that we are talking about the figures in my example with only 10 degrees difference, but a large difference in absorptivity, then Jeff’s #750 (the point in which is of course well known) does not alter the conclusion. Next hot day try walking on a black road in bare feet, then stand on the white concrete in the gutter. It’s cooler, right? Maybe you knew this since you were a kid.
So, yes if the Sun is warming each simultaneously under identical conditions you are correct in saying the one with (much higher) absorptivity would warm more than the other.
Now I’m not here to discuss basic physics questions any more. As Jeff does not seem to want the truth about backradiation spelled out any more, you can all read it in #751.
Jeff – you know I put screen captures of deleted posts on my website don’t you as proof of SkS and SoD policy of deleting posts they can’t answer. This will be featured in my book also. If you want to join them with another page on my site, go ahead …I haven’t started yet, but I may.
[snip – ok, sounds good. I’m certain that your posts here will get more reads than your book]
738.
Jeff Condon said
February 14, 2012 at 7:50 pm
Oh yeah. In #736, your conclusions are also completely wrong.
_____________________________________________________
If your bath tub is filling as fast as it can with the hot tap turned on fully it will indeed fill faster if you also turn the cold tap on.
If the Earth’s surface is filling with thermal energy (ie it is warming) as fast as it can on a sunny morning with the Sun shining fully it will indeed fill (warm) faster if you also radiate extra thermal energy from a colder atmosphere if and only if you violate the Second Law of Thermodynamics.
[doug, you are unqualified to discuss physics]
.
Those who think I am “writing loosely” or perhaps haven’t learnt any physics in my 50 years studying and teaching such might like to give their own explanation …
(1) of the point I often make about why the -18 deg.C is assumed to apply to the surface – which radiates nothing at all like a blackbody. Why doesn’t the -18 deg.C apply to the whole Earth+atmosphere system? Then we get -18 deg.C somewhere up in the atmosphere as a mean temperature. Hence, with a lapse rate (determined by pressure) we will of course have warmer temperatures at the surface and cooler at TOA – with or without water vapour or carbon dioxide etc. How on Earth can we say these things raised the surface 33 deg.C when the lapse rate is responsible?
(2) Let me hear your thinking on this. Remember the Second Law must apply from any point to any other point. It is not “averaged” over the whole Earth over 24 hours for example.
If your bath tub is filling as fast as it can with the hot tap turned on fully it will indeed fill faster if you also turn the cold tap on.
If the Earth’s surface is filling with thermal energy (ie it is warming) as fast as it can on a sunny morning with the Sun shining fully it will indeed fill (warm) faster if you also radiate extra thermal energy from a colder atmosphere if and only if you violate the Second Law of Thermodynamics.
Jeff (somewhere above) spoke of emission at higher temperatures. An object emits both at higher intensity and higher frequencies when its temperature rises. I guess Jeff knows about Wien’s displacement Law, so he must have forgotten about the fact that the peak frequency is proportional to the absolute temperature. See the first plot here …
http://scienceworld.wolfram.com/physics/WiensDisplacementLaw.html
[Doug, misrepresenting my comments will get you nowhere – Jeff]
Eric (skeptic) said
February 16, 2012 at 8:24 pm
You are still assuming that warming occurs because photons strike an object, regardless of the energy of the photons – ie the temperature of the emitting body. Also, photons in one direction don’t cancel out others in different directions. After all, the radiation has a different wavelength for a start if the temperatures are different. By what process could two such radiation “beams” affect each other?
Think of a triangle. A point on the Sun at “A” radiates to the surface at point “B” and (in the morning), point B is getting warmer.
Now think of a point C in a colder atmosphere radiating to point B. There can be no transfer of heat along this line without breaking the Second Law. Take the Sun out of it (at night) Again, no additional thermal energy (necessary to slow the rate of cooling) can travel along CB, so the cooler atmosphere cannot warm the surface more or slow its rate of cooling..
Now I have raised this point before and no one has responded with any valid answer.
Why does the “explanation” of the greenhouse conjecture apply Stefan-Boltzmann calculations to the Earth’s surface when …
(a) it would be logical to apply it to the whole earth-plus-atmosphere system?
(b) the Earth’s surface does not radiate anything like a true blackbody, as it sheds heat by several means other than radiation
Also, when the -18 deg.C should really be a mean for the Earth+atmosphere it will be found at a point somewhere perhaps 10Km to 12Km altitude.
So, in this case, the lapse rate (determined by gravity and relative humidity) would fully explain why the temperature plot is warmer than the mean at the surface and colder at TOA.
So, how can that 33 deg.C figure be blamed on water vapour, carbon dioxide etc.?
755.
Eric (skeptic) said
February 16, 2012 at 8:27 pm
Since Doug is allowed to answer questions, I would like to ask him if cloudy nights are warmer than a clear nights (all other things being equal).
_______________________________________
If the initial temperatures before the clouds rolled in were the same, and pressure and relative humidity are the same and there is no precipitation from the clouds, then there is no valid physical reason that I know of which could make such nights either cooler or warmer just because of the clouds. What valid physical reason did you have in mind involving transfer of heat from warmer to cooler regions as per the Second Law of Thermodynamics? What published empirical evidence do you have?
One wouldn’t wish to have to add another page of deleted posts for this site, like those for SkS and SoD …
[snip – keep all the records you like]
Just ore question …
It is a fact that about half of the Sun’s incident radiation is in the infra-red spectrum.
Carbon dioxide absorbs some of this incident IR radiation and sends at least half back to space – OK so far?
So this will prevent some radiation warming the Earth’s surface (just like the whole atmosphere does- so that the surface does not warm to over 100 deg.C like the Moon’s does in the Moon daytime. – Still OK so far?
So why don’t you deduct this cooling effect of carbon dioxide from the assumed warming effect which is only about 13% as much if we use IPCC calculations – or 0% by mine, since mine are restricted by the Second Law of Thermodynamics.
Finally, do you guys ever wonder why over 30,000 scientists have now signed a statement showing they disagree with the AGW conjecture? Do you ever wonder why we see an article such as on WUWT today in which one such scientist says …
Suffice to say, the “climate science” served up by the Intergovernmental Panel on Climate Change has been a pack of lies from the day it first convened. Its “science” was based on computer models rigged by co-conspirators that include Michael Mann of Penn State University and Phil Jones of the University of East Anglia.
The original leak of their emails in November 2009 instantly revealed the extent of their efforts to spread the hoax and to suppress any expression of doubt regarding it. A second release in 2011 confirmed what anyone paying any attention already knew.
The “warmists”, a name applied to global warming hoaxers, launched into a paroxysm of denial that has not stopped to this day. Their respective universities have since engaged in every possible way to hide the documentation they claimed supported their claims. Suffice to say, the global warming hoax was the golden goose for everyone who received literally billions in public and private funding.
One more question.
[snip, no more questions doug, you don’t answer mine and you don’t listen to answers.]
738.
Jeff Condon said
February 14, 2012 at 7:50 pm
Oh yeah. In #736, your conclusions are also completely wrong.
_____________________________________________________
If your bath tub is filling as fast as it can with the hot tap turned on fully it will indeed fill faster if you also turn the cold tap on.
If the Earth’s surface is filling with thermal energy (ie it is warming) as fast as it can on a sunny morning with the Sun shining fully it will indeed fill (warm) faster if you also radiate extra thermal energy from a colder atmosphere if and only if you violate the Second Law of Thermodynamics.
[reply: you have proven yourself completely unqualified to discuss thermodynamics.]
DJC – YOUR tub may fill faster with both taps open, but mine does not. The rest is, um, gibberish.
[snip]
.
JWR said
January 24, 2012 at 2:28 am
Inspired by Claes Johnson and in support of Doug Cotton, I compared one-stream and two-stream implementations of a single slab and of multi-layer models of the atmosphere. I found that two-stream models of heat flow, although giving athe same temperature distribution as the one-stream model, give in a symptomatic way spurious absorption.
Two-stream descriptions of heat flow and thereby back-radiation, as is used by IPCC authors, should be avoided.
______________________________________________________________________
You are of course correct JWR, and this points to the main fallacy in the greenhouse conjecture.
Only radiation from hot to cold has any effect because it contains frequencies (in the upper extremes of its spectrum) which are above those that can resonate with the target when the target is cooler. The energy in radiation with these frequencies is thus retained and must be converted to thermal energy.
Radiation from a cooler source always contains frequencies which can resonate with a warmer target and thus be scattered without any energy left behind to be converted to thermal energy.
This is easily seen from the first plot here http://scienceworld.wolfram.com/physics/WiensDisplacementLaw.html
As you can envisage from this plot, as the temperatures approach each other the overlap decreases and so the rate of heat transfer decreases until it ceases when the temperatures match.
I know that you may get similar results making calculations with two-way radiation, but situations can be hypothesised which would lead to invalid results.
Consider my funnel experiment concentrating radiation from, say, a large but cooler surface of 5 sq.m onto a smaller but slightly warmer surface of 0.5 sq.m. Even when temperatures become equal you would then have 10 times as much radiation in one direction, or a net of 9 times – all without warming because, if it did warm, the Second Law would be broken.
Thus only the passage of radiation from hot to cold is relevant and it fully explains all that happens in regard to heat flow and temperature changes. More importantly, it explains how and why the Second Law is valid for radiation.
You simply cannot refute this example – equal temperatures and yet net radiation in one direction. Why no further warming? Well, any further warming would certainly be creating energy.
It is little wonder that Claes Johnson was able to prove this computationally.
DJC (Doug), cloudy nights are indeed warmer than clear nights (without precip or wind, humidity the same and temp starting out the same). Here around DC there are three major stations with both cloud and temperature observations (DCA, IAD, BWI) and they show that a cloud pocket affecting one station will keep it warmer. There is a very large network of schools and other locations with thermometers (weatherbug) and they generally match the observations at nearby major stations although they do not measure clouds.
I know this pretty well since I am struggling to quantify a local heating effect at DCA due to gravel (more precisely: an increase in gravel area in recent years). But while comparing radiational cooling at DCA to other sites, I always have to check for clouds because that is the overwhelming factor in nighttime cooling.
Here’s an engineering evaluation of the effect of clouds: http://www.ba-pirc.org/pubs/nightcool/ that is empirical with effective cooling calculations from theory. They say “The impact of cloud on radiation is complex, but is estimated by Clark et al. (1981, 1982) to vary so that the net radiation from a cloudy sky can be roughly approximated by percent of cloud cover (Clark and Blanplied, 1979)”
The bottom line is that the effectiveness of their passive radiative roof cooling system is diminished with percentage of cloud cover.
One more question.
you don’t answer mine and you don’t listen to answers
______________________
Precisely which question(s) have I not attempted to answer to the best of my ability?
I “listen” to your answers but I don’t have to choose to believe them. That’s my prerogative.
These are totally inappropriate reasons you give, probably because the truth is you really can’t answer (or contradict my conclusions) using correct physics.
Only one of us can be correct, my friend.
]
[REPLY: Your answers have been contradicted in every single case. In every single case you have changed direction the moment you recognize the problem. How does the IR laser heat doug. Do you recognize that it doesn’t create sponteneous emission in a struck object at low power levels? Can you even begin to grasp that your arm waiving answer didn’t brush against solid ground?
I don’t really need to ask those questions, your obtuseness has already answered. ]
755.
Eric (skeptic) said
February 16, 2012 at 8:27 pm
Since Doug is allowed to answer questions, I would like to ask him if cloudy nights are warmer than a clear nights (all other things being equal).
________________________________________
I am not aware of any published peer-reviewed study that says whether (under identical conditions otherwise) cloudy nights have significantly different temperatures from clear nights. Are you?
754.
Eric (skeptic) said
February 16, 2012 at 8:24 pm
____________________________________________
Well, let me make the same point in a slightly different way. Suppose you have a funnel (reflective on the inside) and a metal plate size 10 square units seals one end and a similar plate only 1 sq. unit seals the other end, both plates being the same temperature and emissivity, say, 0.9. Then obviously more radiation hits the smaller plate having been focused from the larger plate. (You can boil water using sunlight with such funnels.) So, we have net radiation towards the small plate, but the Second Law of Thermodynamics says there can be no transfer of thermal energy to the smaller plate. This means there can be no slowing of the cooling rate thereof.. This would seem to compare with backradiation striking a surface which is not cooler than the source of the radiation. Any comments on this would be welcome.
753.
Jeff Condon said
February 16, 2012 at 8:05 pm
I can’t believe you would tell Dr. Weinstein to ‘read textbooks’ when you don’t know who he is and can’t understand his words. like telling Carrick to read your stupid links that you don’t even understand.
Get a clue man I’m tired of your spam
_________________________________________________________
I don’t agree with the claim he made that radiation from carbon dioxide can affect the rate of cooling of the surface. Basic physics regarding the Second Law of Thermodynamics appears to be lacking in his analysis of the situation, because such radiation cannot possibly add thermal energy each sunny morning when net radiation (mostly from the Sun) is into the surface. The Second Law says you cannot then transfer further thermal energy from some colder point in the atmosphere to this surface which is getting hotter by the minute. If you had a long metal bar from the surface to a point in the atmosphere, would thermal energy travel along the bar by conduction from the cold atmosphere to the warm surface? No. He and all of you do not appear to know that radiated heat transfer obeys the same Second Law. This assumption is built into all those models. That’s why I said he needs to read (again) basic physics textbooks.
There is a serious problem in Dr Leonard Weinstein’s thinking and analysis, as expressed in his article and subsequent post, so I expect him to answer my objection in the interests of humanity whom he appears to be deceiving by promulgating such pseudo-physics.
Doug Cotton
776.
Eric (skeptic) said
February 17, 2012 at 6:09 am
there are three major stations
hardly a worldwide study
I always have to check for clouds because that is the overwhelming factor in nighttime cooling.
I don’t suppose you might have had pre-conceived ideas? Anecdotal anyway, don’t you think?
the net radiation from a cloudy sky can be roughly approximated by percent of cloud cover (Clark and Blanplied, 1979)”
Does all the rest of the sky radiate nothing then? Doesn’t seem to agree with IPCC estimates on a clear night. No proof of actual warming caused by the radiation. No proof that the instrument doesn’t just appear to show more radiation because the clouds are there at a certain temperature. The instrument does not measure radiation directly – probably just temperature, so radiation is deduced using S-B in an inappropriate and incorrect manner.
The bottom line is that the effectiveness of their passive radiative roof cooling system is diminished with percentage of cloud cover.
Your biased assumption. There is no direct relationship between downward radiation and temperature.
752.
Doug Cotton said
February 16, 2012 at 6:02 pm
16. Leonard Weinstein said
January 23, 2012 at 12:10 am
I really hate snipping but I hate stupidity and ignorance even more
__________________________________________________________
The fact that I have a belief based on physics and yours is based on a hoax by Mann et al and the IPCC themselves, (rather than the Second Law of Thermodynamics) does not necessarily make me more stupid and ignorant than yourself.
[REPLY: It is your inability to follow logic and reason that makes you more ignorant. Were you to follow what was written and asked, we could have discussed the blindingly obvious flaws in your “work”. Since you cannot follow reason, ignore the most difficult questions, and continually spam the thread with victory celebrations, you are impossible to communicate with.
Thus, as you dig in your heels deeper, your ignorance continues and even grows.]
If you are communicating with Leonard Weinstein, who obviously ignores my posts, you might wish to advise him that the pseudo-physics that he and several others promulgate will be exposed in my self published and self funded book.
Don’t underestimate my marketing on such a book, nor the numbers that will be distributed free to media and influencial people worldwide.
Eric said elsewhere …:
clear sky conditions allows night sky radiation to reach its maximum potential
_________________________________________
You have a strange idea as to what controls the power of radiation leaving the surface or the roof of the house in this example. All that matters is the emissivity and the temperature difference between the top of the surface (or the roof) and the first millimetre of the air. Molecules in the surface don’t “know” whether the sky is clear or not.
After diffusion, conduction and evaporation have brought the temperatures very close to each other, you can then get an idea of how much might be radiated by using the Steffan-Boltzmann equation, provided you use it correctly and deduct a term based on the temperature of the air. So the closer the two temperatures (perhaps 1 to 2 degrees different) the less is the radiation.
As the relative humidity reduces the lapse rate, and as calmer conditions reduce temperature differences that can be caused by wind, you will get less radiation leaving the surface. So what? This is weather, not climate which is averaged over long periods. But I repeat, it has nothing to do with any radiation from the clouds which makes its way back to the surface where it is merely scattered and has no effect. How many times do I have to explain this to you?
My answer (below) to this person also answers some questions on this thread here …
Comment from: Martin MasonApril 11th, 2011 at 1:31 pm
Can anybody help me with a question on radiation? I instinctively believe that a cold body can’t transfer heat to a warmer body but it can radiate towards the warmer body. If the radiated wave back from GHGs in the atmosphere can’t be reabsorbed and re-emitted by the surface, what does it do?
It resonates with the target molecule and is effectively re-emitted rather like being reflected at the speed of light. None of its energy is converted to thermal energy. I prefer to use the term “scattered” in order to avoid implying that it is either reflected (in the true sense of the word) or absorbed – which most people assume means it does some warming.
Now, when and why does it resonate? Well, the frequency distribution of a blackbody has a peak which is proportional to absolute temperature. Study carefully the first plot here http://scienceworld.wolfram.com/physics/WiensDisplacementLaw.html and note that the plot for a warmer temperature always envelopes that for a cooler temperature. Hence radiation from a cooler source can only have frequencies which can resonate with those of a warmer body. So all such radiation never leaves thermal energy behind. In contrast, radiation from a warmer source will always have some frequencies (at the right) which cannot resonate with a cooler target. It is the energy in radiation with these frequencies which has to be retained and is thus converted to thermal energy. This is actually necessary for the Second Law of Thermodynamics to apply.
Hence spontaneous radiation from a cooler atmosphere cannot add thermal energy to a warmer surface. Since it cannot add thermal energy it cannot either increase the rate of warming of the surface in the morning or slow the rate of cooling on the evening.
Herein lies the collapse of the atmospheric radiative greenhouse conjecture.
Source: http://jennifermarohasy.com/2011/03/total-emissivity-of-the-earth-and-atmospheric-carbon-dioxide/?cp=5
Time to wake up Jeff, my boy.
This is your final warning that, if you continue to refuse to respond to posts such as this last one with a proper physics argument, rather than just deleting them or snipping them and thus making them meaningless (as SkS did often) then your site will be listed along with SkS and SoD on both my website and in my book. Don’t underestimate the effect of such!
You may delete this postt if you wish, but screen captures have been retained of many others.
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782. Jeff said: It is your inability to follow logic and reason that makes you more ignorant.
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I can not only “follow logic and reason” but I can also discern illogical arguments which are based on potential violation of the Second Law of Thermodynamics, for example.
If you are so proficient in following logic, then you would be able to pinpoint any errors in Johnson’s or Nahle’s or my arguments.
If you were able to do so, or even attempted such perhaps in error, then we could indeed have a meaningful conversation on a forum such as this.
So how about starting with my response to Martin Mason above.
(1) The IPCC claims that backradiation is transferring thermal energy from a cooler atmosphere to the (warmer) surface 24 hours a day.
(2) The Second Law of Thermodynamics has to apply between any two points (with matter) anywhere at any time. (You can’t just consider net transfer over a 24 hour period.)
(3) So it’s OK for radiation from a point on the Sun to transfer thermal energy to a point on the surface and make it get hotter and hotter any sunny morning. There will be net radiation into the surface at that time.
(4) But it’s not OK for the colder atmosphere to cause further additional warming (when the surface is already warming up in the morning) by sending additional thermal energy to the warmer surface. This second path of energy transfer would violate the Second Law.
OK – the challenge is to find any fault in my logic.
776.
Eric (skeptic) said
February 17, 2012 at 6:09 am
The bottom line is that the effectiveness of their passive radiative roof cooling system is diminished with percentage of cloud cover.
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And my “bottom line” is that radiation from neither water vapour nor carbon dioxide, methane, etc can be the cause of your bottom line – for the reasons set out in my posts. So if you are interested in finding the real reasons you can of course look elsewhere for such. I am not particularly interested in specific cases like this which are functions of weather conditions, and do not establish anything to do with climate I start and end with the physics of climate.
#788
1 – yes
2 -yes
3-yes
4-no
there.
#790
Your answer to Q.4 is wrong, Jeffrey. Go back to your physics textbooks.
The radiation beam from the atmosphere to the surface is quite independent of whether or not the Sun is shining or not. In fact, any radiation from any point to another point anywhere where there is matter in the universe is quite independent of any other radiation. There are vastly different frequencies anyway in this case, and all the frequencies in the atmospheric radiation are contained within the spectrum of emission from the warmer surface as explained in #785. So all the cooler radiation merely resonates in the warmer surface (as explained by Johnson) and is scattered without adding any thermal energy to the surface.
Why would two beams both entering the surface have any cancelling effect on each other?
The IPCC assumes they are both warming the surface while the Sun is shining.
The IPCC clearly assumesi the radiation from the atmosphere transfers thermal energy to the surface, so that there is a heat transfer from cold to hot.
This assumption breaks the Second Law of Thermodynamics because it is claiming heat transfers from cooler to warmer.
Hence the assumpton is wrong.
Hence there is no radiative greenhouse effect due to any radiation from the cooler atmosphere.
My post #785 explains the process.
There is absolutely no way in which the Second Law of Thermodynamics can work in all cases unless the radiation from cold to hot does not transfer thermal energy, and only radiation from hot to cold does so. My smple funnel experiment with net radiation from the larger plate to the smaller one at the same temperature cannot result in the smaller plate getting warmer, despite the surplus of radiation striking it compared with what it emits. QED.
Looks like you’re three quarters towards believing, Jeff. Just try a little more. I know it’s hard for you and will be very embarrassing to admit, but you are doing a great dis-service to the public running a site like this which influences others into believing a false conjecture.
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Don’t get me wrong. I really do want to help you all to understand how the IPCC has misled us all, myself included in the early stages – as any archives of my earth-climate.com site from early last year will demonstrate.
Perhaps my reply to the post below on WUWT will help you understand. I’m sorry, but this is quickler for me than re-typing something similar …
William M. Connolley says:
February 18, 2012 at 9:09 am
DC > Hence radiation from a cooler source can only have frequencies which can resonate with those of a warmer body.
This is wrong (or, perhaps more accurately, Not Even Wrong). Whether radiation is absorbed or not has nothing to do with the temperature of the body-that-might-absorb it.
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No it is you who is mistaken William. Go back and read in any physics textbook why the Second Law of Thermodynamics would be violated if any radiation “beam” from any cooler point transferred thermal energy to any warmer point at any time.
Radiation can travel both ways, but heat can only “travel” one way from hot to cold, no matter which way the most radiation is travelling..
Consider a reflective funnel which concentrates the radiation from a larger metal plate onto a small one. Assume initial temperatures, plate material and thickness are all the same and emissivity and absorptivity are 1.0 for each plate. We have more radiation from the large plate (because all radiation from the larger area is focussed down onto the smaller plate) and so we have net radiation into the smaller plate.
Is that radiation adsobed and converted to thermal energy even though absorptivity is 1.0? No,
Because if it were then there would be a transfer of thermal energy which would start to warm the smaller plate, so very soon there would be transfer of thermal energy from cooler to warmer which is physically impossible.
In the same way, when the IPCC assumes that, even when the Sun is warming the surface in the morning, extra radiation from the cooler atmosphere adds thermal energy to the warming (and warmer) surface they have assumed something which is physically impossible.
Hence an atmospheric radiative greenhouse effect is also a physical impossibility. It seems no one can prove me wrong on this, and no one has yet done so on any of several sites.
I want valid reasoning based on physics, WIlliam, not waffle about being wrong.
Readers may wish to note that William Connolley does not appear to have even three years of tertiary physics to his name, but I may be wrong. If he does, then it seems he has forgotten it all – and also where the documentation for his “PhD” was filed..
http://www.conservapedia.com/William_M._Connolley#Purported_qualifications
Anyway he’s done a comprehensive job changing many Wikipedia entries to fall in line with IPCC bluffing.
“This assumption breaks the Second Law of Thermodynamics because it is claiming heat transfers from cooler to warmer.”
Bull!
Work the math Douglas. Dig up those old (and obviously dusty) physics texts you claim to have read and work the math. I’m not doing it for you because you don’t understand even the simplest of words. If we threw equations at it, you probably would deny those exist in the same manner you have about lasers, bolometers, pyrometers and radiometers.
Why waste my life talking to a wall?
Doug Cotton said “And my “bottom line” is that radiation from neither water vapour nor carbon dioxide, methane, etc can be the cause of your bottom line – for the reasons set out in my posts. So if you are interested in finding the real reasons you can of course look elsewhere for such. I am not particularly interested in specific cases like this which are functions of weather conditions, and do not establish anything to do with climate I start and end with the physics of climate.”
Although that’s your bottom line for greenhouse gases, you have yet to provide an explanation of why clouds can reduce nighttime cooling and add it to your theory (and book). From this morning’s discussion from my local weather service office “temps have stayed fairly mild as clouds have inhibited radiational cooling.” (text translated from all-caps to lower case).
Another piece of evidence that I would like you to explain is my IR thermometer. Normally I use it for my wood stove for which its mediocre accuracy (1-2F) and emissivity assumptions are not an issue. To the best of my understanding the device measures incoming IR, makes an assumption about the emissivity of what it is pointing at and translates the IR measurement to a temperature.
Yesterday I pointed it at the sky in the late afternoon and got a reading of -30F. It was completely clear with low relative humidity. About 9PM I pointed it at the sky again and got a -10F reading due to incoming humidity and clouds. I just pointed it up at the sky again with lots of clouds and got a +17F reading, but this morning’s humidity is lower than yesterday afternoon’s due to dry air advection overnight. The primary difference today is the clouds. But there are other explanations for the current reading such as scattering of solar IR (the clouds are currently not that thick).
Nonetheless the reading supports the theory that my local weather service uses which is that the clouds are inhibiting radiational cooling. My device measures a higher apparent temperature and the higher that temperature is, the less radiational cooling we get.
If you choose to ignore the effect of clouds on radiational cooling and not provide an alternative explanation in your book, your book will be trivial to debunk.
Doug Cotton,
Since we are both in the skeptics camp, but for different reasons, I hate to rain on your parade. However, your ignorance on the nature of radiation transfer and repeated claims can’t go unanswered. I have repeatably stated that absorbed back-radiation is not the CAUSE of the so called atmospheric green-house effect. It is actually a result of it. However, the greenhouse gases and or clouds are necessary for, and do cause an increase in ground level (average) temperature compared to the lack of both in an atmosphere. The process is to raise the location of outgoing radiation to space, and the adiabatic lapse rate does the rest.
You continually repeat that radiation from a cooler source cannot be absorbed by a warmer source. In fact the only thing that effects matter that radiates or absorbs radiation is terms called the absorptivity and emissivity coefficients, which are measured and known for most materials. These are wavelength dependent, but water and much of the ground are nearly like black bodies in that they will absorb and radiate most wavelengths in the range of present interest. The absorbing gases of the atmosphere of interest here (CO2, Methane, and Water vapor) all absorb and radiate in known wavelength regions, mainly related to their vibration modes of excitation. These wavelength ranges include much of the thermal wavelength range of the radiating Earth (but not direct sunlight), so much of this Earth to gas radiation is absorbed in a modest distance. However, due to convective mixing, vaporization and condensation of water, and re-radiation in the gas, the energy is transported to an altitude (which actually varies for different wavelengths) to eventually be radiated to space. The average of the outgoing altitude increases if the “greenhouse gas” concentration and all feedbacks raises the final outgoing location. That is all there is to the “greenhouse gas” effect. The only issue is how much is the altitude raised, and I contend (bases on actual historical and present trends) that the effect is small, and is vastly dominated by natural effects (Sun, ocean currents, etc.), rather than human caused effects.
Your problem seems to stem from confusing radiation transfer with heat transfer. The ground does absorb the back radiation from the atmosphere, but it radiated more out than it absorbs in, and thus the HEAT transfer from radiation cannot heat the ground. It is only the difference of the two fluxes that matter for heat transfer, and the 2nd law only refers to the heat transfer. Radiation transfer is unlike heat transfer in liquids, solids, or even like conduction heat transfer in gases, and this is your source of confusion.
Addendum to 795,
Last paragraph should have:
and thus, on the average, the HEAT transfer from back radiation cannot heat the ground.
Eric – As you know I have answered your point on the Open Thread at WUWT which is the only place I will be posting after this, purely because I don’t have the time – not because of inability to answer or disinterest. Please understand I have a couple of businesses to run to support a wife and 3 children under my roof, and a book to complete. See http://wattsupwiththat.com/2012/02/19/open-thread-weekend-8/#comment-897295
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796.
Leonard Weinstein said
February 19, 2012 at 12:58 pm
The ground does absorb the back radiation from the atmosphere, but it radiated more out than it absorbs in, and thus the HEAT transfer from radiation cannot heat the ground.
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Leonard this is where you go wrong. You simply cannot explain how the Second Law of Thermodynamics operates for radiation if you try to calculate the effects of two-way radiation and take the “net” difference.
I have given an example with my thought experiment with a funnel which focuses radiation from a much larger plate at one end onto a smaller plate at the other end, each at the same temperature and with the same absorptivity. You have net radiation from large to small (agreed?) but the Second Law says you cannot have transfer of thermal energy, ie heat transfer. You can only get the computations to agree with reality if you don’t count any radiation from cold to hot and you only count all radiation from hot to cold. Whatever the temperatures, this funnel will not perform in accord with calculations which you do with net radiation calculated as a difference between the two-way radiative fluxes.
I keep talking about what happens when the Sun is already warming the surface every sunny morning. There is net radiation into the surface because it is warming. So how can extra radiation transfer thermal energy from the colder atmosphere so as to make the surface warm even faster? Obviously that would violate the Second Law. I don’t care if you don’t agree about what happens that evening when it is cooling. Some people can’t understand that thermal energy has to be added in order to slow a rate of cooling, like turning a tap on when the bath plug is out. But you should all be able to understand that the Second Law would be violated in the morning, and that is a part of the IPCC’s model calculations, now isn’t it? You cannot just take a 24 average “net” flow of energy and say all is OK. The Second Law has to apply between any two points at any particular time.
Eventually you will see that Claes Johnson, a well-published Professor of Applied Mathematics does in fact know how to do the relevant computations. It was he who explained the resonating process (which is not absorption that leads to conversion to thermal energy) as you can read in his Computational Blackbody Radiation and I am not going to plagiarise his excellent work. All I have done is explain when the resonating takes place and why it must be related to the overlap and non-overlap of the frequency plots for the source and target. Only when the source is hotter will it have non-overlapping higher frequencies which cannot resonate and which thus have to be converted to thermal energy, as happens with solar insolation.
Thermal energy does not transfer with radiation – only radiated energy which does not know what it is going to strike until it does. If (like solar radiation) it includes visible light then that portion will probably appear as light and be refected as some colour. When they are not reflected, the invisible UV and the IR can be converted to thermal energy (as can that component of visible light which is not reflected) when there is the potential to be absorbed. But that potential (the absorptivity) is a measured empirical value which itself varies with temperatures of both the source and target. The reason it varies is because radiation from a cooler source never gets converted to thermal energy. In the empirical measurements, the radiation which is scattered may or may not get counted in the emissivity calculations and I suspect this possibility leads to errors. After all, it does not come back at the same angle as incident radiation, so they may not consider it part of the reflected component. But it had no effect on the temperature of the tearget. So what I am saying is, don’t hang your hat on such measured absorptivity (as WIlliam Connolly does) because you would need to know more about how it was calculated.
So “heat” only appears to transfer by the following process: at the source some thermal energy is converted to radiation, so the temperature of the source drops. When that radiation hits a cooler target (which can absorb it) then it will be converted back to thermal energy, so the temperature of the target rises and we say there is “heat” transfer. But, if the radiation hits a warmer target the Second Law says it cannot be converted to thermal energy because the Law says there cannot be heat transfer.
* * * * * * * * *
Finally, don’t forget the backradiation from carbon dioxide when it absorbs IR from incident Solar radiation and sends it back to space, thus cooling. This has been the most avoided point in all my posts everywhere.
PS Leonard
If the warming in the morning were increased by the addition of extra thermal energy (which had been converted from radiation from a colder atmosphere) then that extra thermal energy can remain in the surface for hours or even days or months.
When it does exit (usually by that evening) it may do so by other processes such as evaporation, conduction or diffusion followed by convection. So there has indeed been a warming effect and it is not just cancelled out by subsequent radiation that evening.
And so there are two scenarios …
(a) If you say the warming does happen in the morning then it is a clear cut violation of the Second Law at that time.
(b) If you agree with me that warming by the atmosphere does not happen on a sunny morning, then you are in effect agreeing that the whole conjecture that backradiation can do anything at all is wrong, and thus the whole concept of the radiative greenhouse effect is wrong, as are the models..
Don’t sit on the fence, Sir! Take a stand one way or the other!
Doug,
In the early morning, with clouds, and at very high latitudes, and for some other special cases,the ground is sometimes cooler than the air, and back radiation and conduction will heat the ground. If the ground is warmer, both will have the heating going upwards. However, on the average, the ground is warmer, and heat transfer is up. Your arguments are wrong. You clearly do not have an advanced Physics or Optics background, so quit using your arguments which are not based on science, and which are simply wrong.
“You clearly do not have an advanced Physics or Optics background,”
He also doesn’t realize how clear that fact is.
I’ve been trying to follow Doug’s arguments for some weeks, and others of a similar nature, which have been very common on Tallbloke’s site since the N&Z paper was discussed.
Once things I’ve noticed is that there seems to be a belief that SB can only be used to determine radiation from a black body in a vacuum, and that surface radiation must be greatly reduced due to the energy lost by conduction and convection, leaving far less radiant energy available.
But the surface is close to a black body in behaviour, and simply point your IR thermometer at the ground, and clearly it shows that the radiation is there.
So the only way to balance the energy flows is to assume that the is indeed a very large energy flow back to the surface from the atmosphere.
I.e. back radiation is REQUIRED by the 2nd law of thermo-dynamics.
800.
Leonard Weinstein said
February 20, 2012 at 10:13 am
Doug,
In the early morning, with clouds, and at very high latitudes, and for some other special cases,the ground is sometimes cooler than the air, and back radiation and conduction will heat the ground. If the ground is warmer, both will have the heating going upwards.
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I have always been specific in saying that radiation from a cooler atmosphere cannot add thermal energy to a warmer surface, for to do so would be a violation of the Second Law of Thermodynamics.
Even where usual weather events cause the air near the ground to be slightly warmer than the surface, most backradiation would probably come from higher in the atmosphere where the temperature was cooler. On a worldwide basis the amount of radiation from a source warmer than the ground represents an insignificant percentage of the total.
I’m sorry, Leonard, but your subsequent statement which implies that radiation will warm the surface when it is already warmer than the surface is totally incorrect and a total violation of the Second Law of Thermodynamics between the source of the radiation and the point on the surface which is supposedly warmed by such at that moment in time.
You have absolutely no idea of how much study I have put into this, thank you. It is you who needs to apply some pretty basic physics in an intelligent manner to the particular situation.
How on Earth can you imagine that thermal energy is transferred from a cooler source to a warmer target thus having “the heating going upwards?” – Really, Leonard!
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Corrections; Even where unusual …
I’m sorry, Leonard, but your subsequent statement which implies that radiation will warm the surface when it is already warmer than the atmosphere
802.
steveta_uk said
February 20, 2012 at 10:54 am
I
(1) But the surface is close to a black body in behaviour, and simply point your IR thermometer at the ground, and clearly it shows that the radiation is there.
(2) So the only way to balance the energy flows is to assume that the is indeed a very large energy flow back to the surface from the atmosphere.
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If you use a conventional IR thermometer of the original kind it measures frequency and determines temperature from that using Wiens; Displacement Law. Temperature tells you nothing about the radiative flux unless you know the emissivity and whether or not there are other heat losses. The temperature difference between ground and the first millimietre of the air might only be 2 degrees, say. So proper application of SBL requires that you subtract the term for radiation from a 2 degree cooler layer of air. This gives very little net radiation out of the surface and, if you like to think of it that way, you could say it has already allowed for backradiation.
I don’t care how much radiation goes from the atmosphere to the surface anyway. If the source point in the atmosphere was cooler than the point it struck on the surface that radiation will be scattered and have absolutely no effect on climate because the Second Law of Thermodynamics clearly states that there can be no heat flow from cold to hot anywhere, anytime between any two points where there is matter in the universe.
Leonard and others,
Iin summary the Second Law of Thermodynamics clearly states that there can be no heat flow from cold to hot at any time between any two points where there is matter in the universe. You have claimed that the Law can be violated when you say “If the ground is warmer, both will have the heating going upwards. However, on the average, the ground is warmer, and heat transfer is up.” The only thing that is warming the ground in the morning (when the surface layer of the atmosphere is cooler) is the radiation from the Sun because it comes from a hotter source.
A well-published Professor of Applied Mathematics has produced a computational proof. There is absolutely no empirical experiment showing backradiation warming anything on the ground more than an identical item shielded from such backradiation. My experiments show identical rates of cooling, whether shielded or not. I have explained a funnel experiment which focuses more radiation from a larger plate onto a smaller one at the same temperature, noting that the Second Law prevents warming of the smaller plate even though there is net radiation in that direction. Finally I have explained why Claes Johnson’s concept of resonating will apply except where the upper frequencies from a warmer body do not overlap the frequency plot of the cooler body. I have further explained that this mechanism is the only conceivable method by which the Second Law can apply for radiation as it does for conduction.
Al lthe above has been posted on at least half a dozen popular websites and no one has come up with any valid explanation based on physics which would refute any point.
In contrast, all the IPCC explanations of a radiative greenhouse effect can be easily demonstrated to violate the Second Law – as I have explained in the examples above.
PS Here’s an outline of how to do your own experiment to show that backradiation has no effect on temperatures of the ground.
http://wattsupwiththat.com/2012/02/11/open-thread-weekend-7/#comment-897444
Do you, Leonard, Jeff, or anyone, seriously believe that, when you turn on an electric radiator you could actually make it warm faster by holding a mirror beside it and reflecting its own radiation back onto itself? Try it if you do and time how long it takes to get to its maximum with or without the mirror. When it reaches its maximum, can you then make it hotter still with the mirror?
Backradiation would be even less effective than a mirror because it usually has lower frequencies than those originally emitted by the surface. It is ludicrous to imagine energy can be created this way. The Second Law of Thermodynamics would be so obviously violated.
Yet the IPCC says backradiation should be about a quarter as powerful as the Sun at noon. Well I know the Sun can heat sand on the beach until it blisters my feet, but when I tested the effect of backradiation on sand there was not even a tenth of a degree difference between it and the shielded sand in an identical wide necked vacuum flask.
Leonard, what exactly do you mean when you say both will have the heating going upwards.
Your “both” clearly refers to (1) backradiation and (2) what you call conduction (“diffusion” might be preferably) between the surface and the atmosphere.
Well, if “going upwards” just means the temperature of the surface is going upwards (increasing) it is already doing so due to solar insolation and my reply above is applicable because the Second Law would be violated.
If, on the other hand, you mean thermal energy is being transferred in an upward direction (ie from the surface to the atmosphere) then how does “backradiation” going the other way into the surface have anything to do with it, after all, you did say “both.” ???
Yoiu don’t really have to tell me that the surface will warm the cooler atmosphere by evaporation, chemical processes and diffusion (followed by convection) but why are you including backradiation which is normally thought of as coming from the atmosphere to the surface?
And, by the way, heat doesn’t “go” anywhere – energy gets radiated and then sometimes converted (back) to thermal energy. The disappearance of thermal energy at one point and reappearance of thermal energy at another point is called “heating.” The reappearance of the thermal energy happens if and only if the target hit by the radiation is cooler than the source which emitted the radiation spontaneously. Thus we have herein the process by which the Second Law of Thermodynamics operates for radiation.
“Not based on science” Leonard? So you don’t think the energy in the backradiation has to be first converted to thermal energy in order to increase the rate of warming of the surface or decrease the rate of cooling in the evening. Really? You really don’t think that, do you Leonard? Your backradiation somehow just comes into the surface and then all just “goes up” perhaps meaning out of the surface. Did it leave any energy behind? How does the radiative greenhouse effect work then?
You just think it all somehow goes into the surface and all comes out (or “up”) in some form which could be your “conduction” after all – but wait, it’s not thermal energy yet, or is it?
Really, Leonard, could you please talk in the proper language of physics and describe what you really think is actually happening to the energy in the backradiation every step of the way – both morning and evening, say. No more two liners please – hardly the type of description which would appear in a good physics textbook which I suppose you have read somewhere. Nobody pulls wool over my eyes, Sir.
To everyone;
Does the energy in radiation from a cooler layer of the atmosphere get converted to thermal energy when it strikes a warmer point on the surface which is already being warmed by the Sun at, say, 11am somewhere?
If it does, then this means there is a heat transfer from that cooler layer to that warmer point on the surface at that time, thus violating the Second Law of Thermodynamics.
If it doesn’t (as I say) then the Second Law is not violated and there is absolutely no radiative Greenhouse effect because there is no way in which such radiation can affect the temperature of the surface unless it is converted to thermal energy.
This really is fairly elementary physics well covered in upper levels of undergraduate courses throughout the world.
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“If it does, then this means there is a heat transfer from that cooler layer to that warmer point on the surface at that time, thus violating the Second Law of Thermodynamics.”
No Doug, it means ‘energy/power transfer’ from cold to warm. Entirely different thing due to the ancient terminology of thermo and NOT in violation of the probabilistic second law of thermo. Parse the difference and you will be on the same page.
“This really is fairly elementary physics well covered in upper levels of undergraduate courses throughout the world.”
Take one.
Jeff Condon said
February 21, 2012 at 12:08 am
“If it does, then this means there is a heat transfer from that cooler layer to that warmer point on the surface at that time, thus violating the Second Law of Thermodynamics.”
No Doug, it means ‘energy/power transfer’ from cold to warm.
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No Jeff, I clearly was speaking of a transfer of thermal energy. What sort of energy do you think you were talking about? What type of energy did the radiation leave behind? Does not the greenhouse effect have something to do with temperature, and thus thermal energy?
I’m really tired of your obvious lack of understanding of basic heat transfer theory as it pertains to radiation.
It is quite simple really Jeff:
(1) Thermal energy in Object A diminishes as it radiates this energy loss in the form of radiative flux.
(2) Object A gets cooler, so we say it has “lost heat”
(3) The radiation travels until it hits something
(4) If it hits a warmer Object B it merely gets deflected without leaving any energy behind. So now it’s heading somewhere else
(5) If it then hits a cooler Object C it has some higher frequencies which cannot resonate with C and so the energy in those higher frequencies has to stay in C and that energy has to be converted to thermal energy.
(6) So the thermal energy in Object C is (algebraically) increased.
(7) If Object C is losing thermal energy faster than the radiation is increasing it, then Object C will cool more slowly. Otherwise it will warm faster.
(8) So we say there has been a transfer of thermal energy from A to C (but not from A to B.)
(9) This transfer of thermal energy is also called “heat transfer”, or referred to as a “heating effect.”
Heat transfer happens if and only if thermal energy is transferred.
Do you get it yet Jeff? There is no energy transfer of any type between A and B.
We can’t have “energy / power transfer from cold to warm.” That’s against the 2nd Law.
PS If I have my furniture moved on a truck from House A and the truck drives by House B and then unloads the furniture at House C, I suppose you could say there was a furniture movement from A to B, but I moved house from A to C and that’s where all my furniture was transferred. if you follow the analogy. No furniture (thermal energy) was left at B and so there was no effect on the amount of furniture (the temperature) of B. The only transfer of furniture (heat) was from A to C. So all radiation from a cooler atmosphere can no effect on the temperature of the warmer surface, hence there is no radiative greenhouse effect. A is the atmosphere, B is the warmer surface, C is some cooler object in space that it may eventually hit, or a cooler region in the atmosphere above where it was originally radiated
This page http://mit.edu/16.unified/www/FALL/thermodynamics/notes/node133.html explains how reflected radiation is broken into two components …
(1) The specular (mirror-like) reflection where angle of incidence = angle of reflection.
(2) The “diffuse” reflection which is deflected at any random angle..
Here the second component is what I prefer to call deflected (or scattered) radiation, as the process is very different from specular reflection. But whatever you call it, it obviously does exist.
This is the process I have been talking about all along. As far as energy is concerned, as I have always said, it is the same as reflection and thus has no effect whatsoever on the temperature of the target, in this case the surface.
This diffuse reflection is what happens when the target is warmer than the source. If such radiation were absorbed and converted to thermal energy there would be a violation of the Second Law of Thermodynamics. That is why you need to know the temperatures of both source and target before you can know the overall absorptivity and emissivity. These factors will be affected by this diffuse reflection, which cuts in when the target starts to get warmer than the source.
Thus, all radiation from a cooler atmosphere undergoes diffuse reflection when it strikes a warmer surface. This is why an atmospheric radiative greenhouse effect is a physical impossibility.
(continued from my last post)
I am of course aware that “diffuse” reflection is usually applied to reflection of light which appears to be diffused because of a rough surface. On a microscopic scale, the rough surface may well comprise many small smooth surfaces which simply produce specular reflection at various angles.
So, strictly speaking, the “rejection” of low frequency radiation (from a cooler source) which meets a warmer surface and then resonates and is scattered is yet another different process. But it helps to think of it as diffuse reflection because the end result is the same. My main point is that it can have no effect on the surface temperature because it is as good as reflected.
When IR from the atmosphere strikes a rough Earth surface, you would not be able to distinguish between this scattered radiation and diffuse reflection. In general you will measure inflated values of emissivity as a result. I would suggest that true absorptivity should be expressed as a function of both source and target temperatures. Then its measure should reflect the proportion of the radiation from a source at that temperature which is actually absorbed and converted to thermal energy. In the case of a target temperature greater than a source temperature the absorptivity would thus always be zero.
So this is where models go wrong because they use mean absorptivity measurements which disregard the temperatures of source and target and probably include a lot of scattered radiation anyway. Thus the models end up assuming thermal energy is transferred to the warmer atmosphere simply because they do not rate absorptivity for the relevant temperatures as being zero. Thus they assume violation of the Second Law of Thermodynamics and are of course wrong as a result. There can be no radiative Greenhouse effect.
812.
Jeff Condon said
February 21, 2012 at 12:08 am
“If it does, then this means there is a heat transfer from that cooler layer to that warmer point on the surface at that time, thus violating the Second Law of Thermodynamics.”
No Doug, it means ‘energy/power transfer’
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No Jeff. Energy travels with the radiation and, if that radiation meets a warmer surface, it gets returned to the atmosphere instantly as if reflected. Thus there is no “transfer” of energy to the surface because no energy is left in the surface as a result. The energy will eventually be transferred to cooler air (probably above the original source) or to some cooler object in space – one of these years.
“No Jeff. Energy travels with the radiation and, if that radiation meets a warmer surface, it gets returned to the atmosphere instantly as if reflected. ”
You are ignoring an entire field of evidence on this matter (quantum mechanics included), in exchange for an arm waiving non-physical explanation. You are also ignoring all of the sensors and examples we have shown you that PROVE you are wrong in your hypothesis. A scientist looks at evidence which supports/rejects his theory. Your theory has been rejected by basic observation of “working instruments”.
Only in your mind is the second law violated. You have failed to understand lasers, bolometers, pyrometers, radios, radiometers, and even denied their existence where convenient. You have demonstrated no understanding of the second law of thermodynamics yet you continue to lecture me and others as though you have been ordained as some vessel of knowledge.
You have lost this one Doug, move on.
(continued)
In calm conditions on a clear day we have the Sun (and only the Sun) warming the surface. Its high frequency (high energy) radiation right across its spectrum can be converted to thermal energy (if not already reflected) and this energy flows by conduction deeper into the land surfaces. The radiation itself already penetrates a fair way into the oceans before it is converted to thermal energy in the depths. Meanwhile thermal energy is also escaping the surface more or less as fast as conduction in the land and convection in the oceans will allow it to get back to the surface.
In the morning the rate of warming exceeds the rate of cooling, and vice versa in the evening. In summer the longer hours of daylight may trap some energy that cannot escape before the next morning. This trapped energy may build up as the middle of summer approaches, but escape by the next winter as daylight hours reduce and there is more time for cooling at night.
Now, looking at the cooling process, at least half (maybe 70%) of the thermal energy escapes to the atmosphere by evaporation, chemical processes and diffusion, which involves molecular collisions between the surface and the adjoining air, as in conduction in solids. The remaining energy will be radiated.
However, experiments in such conditions show that the lower atmosphere is always cooler than the surface, and cools faster than the surface at night. Radiation can never transfer heat from a cooler source to a warmer target and neither can diffusion. So these processes can never make the atmosphere warmer than the surface. The Sun also usually warms the surface faster than the lower atmosphere, so it is only very unusual weather conditions which might leave the surface cooler than the adjoining air. The air which is warmed in the morning will rise by convection.
The radiation from the surface may escape to space, but most will be absorbed by some molecules in the atmosphere. These molecules are likely to be warmed and may share some of the thermal energy with other molecules, or simply radiate it in all directions in small bursts.
The radiation which is emitted by the cooler atmosphere will have frequencies which are generally lower than the original radiation from the surface. If some of this radiation gets to the warmer surface it cannot be converted to thermal energy. Instead it is simply scattered by the surface without leaving any energy behind. Its energy cannot be converted back to thermal energy until it collides with something cooler than the original layer of the atmosphere from which it was emitted. Such cooler air will usually be higher up. It may even escape to space and only warm some cooler object in space maybe years later.
So each time any radiation goes back to the surface it will have absolutely no effect on the surface temperature, but will instead make it further towards space on the next trip up, if indeed it doesn’t escape altogether. Clearly there can be no Greenhouse effect.
Footnore: In situations when the relative humidity is high, the moist adiabatic lapse rate is lower than the dry one, so such humidity (as well as clouds) can slow the rate of cooling of the atmosphere, but this can never lead to any thermal energy going back into the surface, so the rate of cooling of the surface need not be slower. In a sense, thermal energy is falling over a smaller temperature step, but it still falls over at the same rate. The air we stand in may well feel warmer partly because there is less evaporation off our skin. In any event, these are just weather conditions which average out and do not relate to or affect climate.
No Jeff. I have read the experiments and the so-called evidence probably every bit as much as yourself, having spent thousands of hours on all this in the last two years. I can see the faults which you perhaps can’t, the main one being that the effective absorptivity should take into account the temperature of the emitting source (or the frequency of the radiation) and compare that with the temperature of the target. If the target is warmer, the effectivie absorptivity has to be zero or otherwise the Second Law is violated in, for example, my funnel experiment. It only takes this one thought experiment to prove them all wrong about the greenhouse effect. So you have …
(1) You can’t explain my funnel experiment if you think radiation has a warming effect in each direction, even when it meets a warmer target.
(2) You, like the IPCC, have also overlooked, or never discussed the cooling effect due to carbon dioxide sending backradiation from IR solar radiation straight back to space.
So you still have two problems on your plate which no one on this forum or others has ever been able to successfully debunk.
Doug said “The radiation which is emitted by the cooler atmosphere will have frequencies which are generally lower than the original radiation from the surface. If some of this radiation gets to the warmer surface it cannot be converted to thermal energy.”
“Generally” or “always”? I thought in your frequency and vibration theory, that the source frequencies had to always be higher to warm the target object. Are you saying there is a possibility of some, limited amount of conversion?
Also, I have a scenario that I would like your comment on. Suppose there are two walls, one that was warmed by the sun, one that was not. It is now night, and the two walls are at temperatures 80F and 50F. Your body is 98F. Would you feel any difference standing about five feet away from each wall? How would your body, which is warmer than either wall, be able to detect the difference?
Doug,
You don’t know when to quit. If you place a mirror next to a heater, it will raise the heater temperature. If the heater is turned off, the presence of the mirror will make it cool off slower. The mirror needs to cover a large exiting angle to have a large effect, but even a small mirror will have some effect. It is not that the returned energy heats the heater, but it slows the escape rate. It acts like an insulator. Why do you think a thermos bottle holds heat (or cold) longer? I cannot for the life of me understand that you don’t believe that. Your comments can be taken as a typical skeptics position, and give the CAGW side some good ammunition on skeptics ignorance. Please learn some optical technology before commenting.
821. Eric (skeptic) said February 21, 2012 at 8:00 am
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Generally – yes – but I have mentioned the rare possibility of temperature inversion – that was all I meant as an exception.
Your body will not be warmed by the radiation from either wall – you should have been able to work that out for yourself. If the air was originally cooler than the warmer wall but warmer than the cooler wall, you may be able to detect a difference in air temperature due mostly to diffusion of thermal energy from the warmer wall to the air and from the air to the cooler wall.
822.
Leonard Weinstein said
February 21, 2012 at 10:13 am
If you place a mirror next to a heater, it will raise the heater temperature. If the heater is turned off, the presence of the mirror will make it cool off slower.
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No it doesn’t do either. You haven’t tried it. Show me a published experiment proving your result, or publish one yourself.
You are mistaken and your answer would require a violation of the Second Law of Thermodynamics.
If that is the level of understanding which you have of heat transfer then it is understandable that you are among the diminishing number of “scientists” who have been bluffed by the radiative greenhouse conjecture.
I shall leave you to your beliefs. Good luck.
When someone responds intelligently to my two points in #820 then we may be able to continue this discussion in the spirit of genuine interst and willingness to learn.
Why do you think a thermos bottle holds heat (or cold) longer?
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The reflective surfaces slow the rate of absorption into the walls. When thermal energy is conducted through the inner wall then the vacuum eliminates conduction and convection in the space between the walls, and low emissivity surfaces minimise radiation between the walls. So the overall rate of transfer of thermal energy is reduced significantly.
This has nothing to do with processes in the atmosphere because there is more than just radiation involved for a start. But the flask does demonstrate that backradiation reflected from the internal walls back into the coffee does not warm the coffee.
The atmosphere as a whole slows the rate of warming during sunlight hours (a cooling effect) because it absorbs and reflects some incoming solar radiation which would otherwise have warmed the surface more, as happens on the Moon which gets up to over 100 deg.C in its daytime. Even carbon dioxide in the atmosphere helps to cool by absorbing some of the solar IR radiation and sending some of it to space as backradiation.
However, the atmosphere is generally cooler than the surface, even at the interface. It also usually cools faster than the surface. The rate of heat loss from the surface is mostly restricted by the surface’s internal rate of conduction in the land and convection in the oceans as heat from the morning flows back out at night – see also #819 above. The atmosphere has not kept the surface in Antarctica anywhere near as warm as at the Equator, for example. But, broadly, yes the atmosphere has played a part in establishing a very long-term equilibrium temperature at the surface interface since the formation of the Earth. It does so because it has a natural mean adiabatic lapse rate, but this is not affected by carbon dioxide.
Backradiation cannot have any effect on temperature because it cannot transfer thermal energy. The rate of cooling could only be slowed if thermal energy were added – that’s simple arithmetic. Backradiation itself does not put the brakes on – as should at least be obvious for evaporation and diffusion, even if not obvious (though true) for outflowing radiation. In any event, what is not radiated can still exit by evaporation, chemical processes and diffusion followed by convection.
I have confirmed for myself with my own experiment (which you don’t have to believe, I know) that backradiation does not slow the rate of cooling of the surface at night. As far as climate is concerned, the mean adiabatic lapse rate is dependent mostly on the acceleration due to gravity and the relative humidity. If the climate is warming the relative humidity drops so that the lapse rate is higher (ie steeper gradient) which means that, as the temperature plot swivels around the 255 K mean (which is up in the atmosphere somewhere) the surface end of the plot will rise. But then the sub-surface plot from the core also has to rise by the same amount to achieve long term equilibrium, and that could take hundreds of thousands of years. Hence we have pretty stable climate which even over 1,000 year cycles has a long-term trend which rarely varies more than about 2 deg.C from its mean.
Guys, you can stop worrying about endless warming going up forever with carbon dioxide levels. It can’t happen.
The whole Earth system (including atmosphere) has to emit very close to the flux it receives from the Sun. So there will be some temperature – let’s say 255K – which is a mean and is somewhere up in the atmosphere.
The natural adiabatic lapse rate is determined, not by carbon dioxide, but by the acceleration due to gravity, the mass of the atmosphere and, to some extent, relative humidity which mostly averages out. The lapse rate sets the gradient of the atmospheric temperature plot which has to swivel about the 255K mean, so the surface end is warmer and the TOA colder. The drop in temperatures between the surface and the tropopause has been very close to constant in all the years of records shown on the NASA site since the end of 2002.
My point is that the mean surface temperature is dictated by these two values – the 255K (or whatever the exact figure is) and the lapse rate. Carbon dioxide cannot affect either, so neither are under mankind’s control.
Both the surface and the atmosphere will simply shed energy faster if they get a little warmer, thus tending back to the mean.
Don’t try to tell me there is a long term TOA net radiative flux difference. The net radiative flux varies between about 99.5% and 100.5% of incoming radiation. This is just random noise or short-term cycles. Longer natural cycles may have to do with variations in the effective power from solar radiation (affecting that 255K figure) and maybe the thermal energy generated under the surface. Small variations in the latter over many thousands of revolutions of the Earth could have a cumulative effect. The very fact that the terrestrial heat flow is low means that the massive quantity of thermal energy from the surface down to the core stays fairly much the same and brings about a stabilising effect as I have explained on the ‘Explanation’ page of my website. . This also is an additional comfort, so relax!
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Leonard says: “It is not that the returned energy heats the heater, but it slows the escape rate.” and he implies the same thing happens when radiation from a cooler atmosphere strikes the surface and somehow also “slows the escape rate.”
He has ruled out any conversion of the energy in the radiation into thermal energy because he says “It is not that the returned energy heats the heater.” Well I could have understood how thermal energy could be added to thermal energy. But I cannot for the life of me imagine how it could do it without converting to thermal energy.
Well maybe he or someone would like to described the exact physical processes involved, for I can see no way in which …
(a) Backradiation can affect the rate of diffusion of thermal energy (by molecular collision) from the surface to the atmosphere
(b) Backradiation can affect the rate of evaporation of water surfaces
(c) Backradiation (coming from all different angles and with lower frequencies) can somehow “react” with emitted radiation (having higher frequencies) and reduce the power of that radiation perhaps by reducing its frequency or by somehow blocking some of it.
Over to you, Leonard, to explain. But perhaps if you read and understand all my posts you will see it’s a pointless exercise because it simply doesn’t happen.
Doug, consider a single wall that is warmer than your body (say 120F). Presumably you would detect that warmth as it radiates. We can easily exclude conduction or advection by air by keeping the air at a constant temperature (e.g. still air, but unenclosed space so air warmed by the wall would escape upwards). As the wall cools to your body temperature, about 98F, you would notice less warmth. As we discussed before there are points at which the wall is slightly above body temp, right at body temp and slightly below body temp. Nature does not allow discontinuities in effects so in your theory there has to a fading of our body’s absorption of IR from the wall to zero as the wall cools below our body temperature. Is that drop in effect linear? Does it hit zero just below the temperature match point? How far below?
Suppose there is nothing but the one wall, now 80F and open space in all other directions. Suppose air temperatures are 50F along with all other objects in all other directions. Are you quite sure that the side of your body facing the wall would feel no different than the side of your body exposed to the 50F temperatures? If you believe that, I suggest trying it out. It should not be difficult to find similar conditions shortly after sunset with an isolated sun-warmed wall.
Eric: No the rate of heat transfer does not decline linearly as temperatures approach each other, because it is related to the area between two plots that are moving together. Consider the first diagram on this* page and imagine the 300K plot approaching the 250K plot. The area between the plots decreases, but not linearly.
* http://scienceworld.wolfram.com/physics/WiensDisplacementLaw.html
You should not do physics using human “feel” for temperatures. It should be easy enough to see if actual objects get warmed or not using thermometers, as I did with vacuum flasks filled with sand, one shielded from most of the backradiation and the other not.
The very outside layers of your skin have nerves which can of course detect if something is colder than body temperature because energy flows from your body to the ice or whatever. Your skin will stay a bit colder than normal for a while when you let go of the ice, but a little later you may not be aware that your hands are still a bit cold – until perhaps a woman tells you!
So radiation from the warmer wall may indeed warm the air in contact with your skin, which your skin will then sense is warmer than air on the other side, even though the air is not as warm as your body temperature. In other words, your skin does not have to be warmed by radiation for you to know if something is hot or cold. It’s all to do with nerves and conduction or diffusion processes, rather than radiation..
Eric (continued)
Although body temperaure under the tongue is about 37 deg.C, when I measured the temperature of the skin on my clasped hands just now it was 35.2 deg.C (with room temperature 25.9C). So obviously there is a small loss between my blood and the outer layer of skin. I guess this explains why my spa feels hotter than me when it’s 36 C and guessing temperatures based on feel may not be too accurate..
Doug, what was the temperature of your shield?
#831 Eric: I would not be able to measure it with the meat thermometer I used because the spike would be mostly exposed to the air. It was low emissivity anyway (being silver) so I can’t see any point in knowing its temperature which would be very similar to the air. Anyway, I plan to start again using glass with a shield on top. (I have been delayed by wet weather here in Sydney). Then I’ll also take more readings during the night like Prof Nahle did in his Sept 2011 experiment. BTW, there’s a lot of interesting info in the early pages here about IR measurements etc http://principia-scientific.org/publications/New_Concise_Experiment_on_Backradiation.pdf
Doug the shield was silvered fabric? It seems to me that a lot of object heat could be reflected back the object even with the silvered fabric held at an angle because the surface is too irregular. A very flat sheet of metal foil should work a lot better. The test is to use an IR thermometer, measure the effective temperature of the shield at various angles, like 90 and 45. If there is no difference, then it is not an effective shield because it is producing more emitted radiation than reflected radiation.
OTOH a perfect reflector (and suitable shield) would show the temperature of whatever object is 90 degrees in front of it (adding in a little extra for a warm hand holding the IR thermometer).
I just wrote this reply to a comment on another website, and I feel it may help people here …he wrote:
The obvious effect of this is that the radiative cooling of the hotter body will be slowed down by the presence of the cooler body, because some of the energy is being returned. This is completely different than there being a NET transfer of heat from the cooler body to the hotter body, which I agree is not possible.
No it’s not completely different. While the energy is still in the radiation it is not equivalent to thermal energy because it has not yet been converted. It cannot affect the temperature of the target unless and until it is converted to thermal energy. Only thermal energy can be added to other thermal energy with a resulting temperature change. You can only slow a rate of cooling by adding thermal energy. Hence the original thermal energy from the cooler body would have to end up being thermal energy in the warmer one before having any effect on temperature. Hence the 2nd Law would be violated.
My funnel experiment focuses more radiation from a large object onto a small one at the same temperature and made of the same metal material. What happens? Think about it. The only way the 2nd Law can apply is if you always disregard the radiation from cold to hot and only consider the radiation from hot to cold.
Any radiation from cold to hot merely resonates – as it can because the hotter body can always itself radiate at all the frequencies in the cooler body emission. It is as good as if diffuse reflection had happened – no energy is left behind and there is no effect on the temperature of the hotter body.
“My funnel experiment focuses more radiation from a large object onto a small one at the same temperature and made of the same metal material. What happens? Think about it. The only way the 2nd Law can apply is if you always disregard the radiation from cold to hot and only consider the radiation from hot to cold.”
Doug, I hate to break it to you but the focusing the total energy from a large solid angle to a small one violates optical etendue. Try it! – let me know how it goes.
Jeff –
I am not claiming optical entendue is reduced at all. Optical entendue is a function of the area of the source and the solid angle subtended at the source by the target, or more precisely, by the “pupil” of the target which could be the first lens in an optical system such as a camera’s zoom lens system. Entendue has units of area.
The funnel concentrates energy and more energy reaches the target than would do so without the funnel. That is my only point here, and I stand by such. Optical entendue is not reduced in the process – but energy is concentrated. These are quite different things.
The frequency does not change, so the target can still recognise the temperature of the source. So indeed the target will not get warmer than the source. It does not say that you can’t focus the total energy flow, as you in fact do with a funnel cooker concentrating sunlight onto food, or a magnifying glass focusing sunlight to burn paper.
The atmospheric radiative greenhouse effect is fallacious because it is based on false assumptions about absorptivity.
When scientists try to explain why the Second Law of Thermodynamics is valid for radiation (as well as conduction) they consider two parallel plates at different temperatures radiating at each other. In such a case it is easy to say, “Net radiation is clearly from hot to cold, so heat transfers from hot to cold, so all is in agreement with the Second Law which states that’s what should happen.”
They are, in effect, calculating the difference between the radiation in each direction as if that in each direction were having an effect in transferring some amount of thermal energy between the plates.
So, in this particular case, the net radiative flux also happens to be in the same direction as the net transfer of thermal energy.
But what happens if you can somehow filter out the radiation from the hotter body, so only the radiation from the cooler body reaches the hotter one? Or, what happens if you concentrate more radiation from the cooler body onto the hotter one using a concentrator such as a reflective funnel – like the funnels used to cook food with sunlight? Then, if you do your calculations based on two-way radiation concepts, you can get the wrong answer, now can’t you?
So the Second Law of Thermodynamics for radiation can only be valid in all circumstances if your calculations only include the component of radiation from hot to cold. This invalidates the atmospheric radiative greenhouse conjecture which assumes “backradiation” can further warm a warmer surface, even when it is getting hotter every sunny morning.
The way it must work is in fact the way Prof Claes Johnson has explained in his Computational Blackbody Radiation.. Radiation from the cooler body merely resonates and is rejected (or scattered) by the hotter one without its radiated energy being converted to thermal energy.
So, you cannot always apply mean absorptivity factors which were originally calculated empirically from radiation from a hotter source. The effective absorptivity when radiation from a cooler source strikes a warmer target must be zero.
Sorry = typo; etendue see http://en.wikipedia.org/wiki/Etendue
“The funnel concentrates energy and more energy reaches the target than would do so without the funnel. ”
Good luck with that.
#838 Yes – people cook food with such funnels using sunlight.- so I don’t need any “good luck” to know that theydo effectively concentrate radiation. See http://solarcooking.org/plans/funnel.htm
I guess this is just your way of acknowleging that there is obviously no valid way of refuting my post.
So we’ll leave it there, shall we, noting that there is no greenhouse effect.
You can always debate any details with me on Open threads at WUWT, but I will be posting less now that I have handled a wide range of objections and gained a better understanding of how people like yourself were originally misled by the AGW hoax, as indeed I was myself in the early stages until a year or two back.
My experience on forums will be the basis of a Q&A appendix in my book, but I won’t quote any names or exact comments.
Cheers
Doug
Email me if you wish at enquiries@douglascotton.com
“#838 Yes – people cook food with such funnels using sunlight.- so I don’t need any “good luck” to know that theydo effectively concentrate radiation”
Doug,
The sun is hot, you can fry ants with it. Try and concentrate a cold plate to heat a hot one. When you achieve your goal, send me an email because we will make billions.
Jeff
You still miss my point. I am just demonstrating that radiation can be concentrated with a funnel, regardless of its frequency (and so regardless of the temperature of its source).
Hence the small plate will receive more radiation at the small end of the funnel than the large plate receives. Surely that has been clear all along.
Now tell me what happens to the extra net radiation going towards the small plate, because it is not allowed to warm it by the Second Law.
I suggest it is not converted to thermal energy (as we know by the Law) because the absorptivity must go to zero when temperatures are equal.
This* indicates that absorptivity is actually determined, not by any warming effect, but by measuring reflected radiation in the visible spectrum.
I am saying that absorptivity cuts out (ie goes to zero) when the source becomes cooler that the target, which of course is not the case by a long shot when making these measurements using much higher frequency radiation than that contained in all spontaneous radiation from the atmosphere.
Hence, for proponents of the radiative greenhouse conjecture to use such measures of absorptivity of light (which are close to unity) and thus to assume the surface absorbs “backradiation” is a complete abuse of physics.
* http://naca.central.cranfield.ac.uk/reports/arc/cp/0601.pdf
Everyone
I have updated and expanded my Radiation page now and it seems to me you should find answers and detailed explanations there for any further queries you may have …
http://www.climate-change-theory.com/RadiationAbsorption.html
“Hence the small plate will receive more radiation at the small end of the funnel than the large plate receives. Surely that has been clear all along. ”
I keep telling you it won’t work.
If you can concentrate the energy from a ‘big’ solid angle plate onto a small one and heat it we will be able to make billions because then we CAN violate the second law and make energy pumps out of two equal temp objects. Again, we find that you don’t have a clue what you are talking about Doug.
Doug, I have a question about your radiation page. There you say “Note that the frequencies for cooler temperatures are a sub-set of those for any higher temperature. When radiation from a cooler source strikes a warmer target all its frequencies will resonate and not be absorbed.”
Your first sentence points out that all of the photons from a cooler source lie within the frequency range of a warmer source. In your diagram your show the frequency distribution of photons from various temperature sources. A warmer source can emit a lower frequency photon according to the distribution, but with a lower probability.
What happens when a lower frequency photon from the warmer source strikes a cooler target? Will it get absorbed? It seems to me that if the only information that the photon has is its frequency, that frequency will “resonate and not be absorbed” by the cooler target. If not, then the photon must have some other characteristic than frequency that is involved in its interaction with the target.
#843. Answer one question: Is there more radiation in total coming out of the larger plate than the smaller plate because its area is bigger? (Like, does the Earth radiate more to space than the Moon because it’s bigger?)
If you answer No, then you don’t understand the question.
If you answer Yes then you acknowledge my point that more radiation does not necessarily mean heat transfers in the direction of the greater radiation.
#844 Sensible questions, Eric.
The plot for the warmer temperature is always above that for the cooler one. The correct application of S-B Law comes out with a calculation which is equivalent to the area under the warm curve less the area under the cooler curve. So the radiative flux from warm to cool is represented by the area between the curves, in agreement with S-B calculations. So some radiation to the left of both peaks will exceed the maximum amount that can resonate. It is represented by the space between the curves.
Both curves are strongly attenuated on the left for lower frequencies. For any frequency there is a surplus of radiation from the warm one over the cool one. In general, the area between the curves is greater to the right of the peak in the cooler curve, so that is where the energy comes from for the vast majority of the warming. There is some warming to the left of that peak, but it is completely dominated by the warming on the right. But as the temperatures approach each other, there is no warming at all in the limit where the two plots coincide. . So what is on the left of the peak does not mean that any warming happens before the source is warmer.
Eric. (continued)
Remember we are talking about emission from an (approximate) blackbody, not just one molecule and one photon. So the statistical probabilities come into play and basically the temperatures of the small part of the atmosphere we are talking about and the small part of the surface are in fact the temperatures that correspond to the peak frequencies as per Wien’s Displacement Law. So when we talk about a cool or warm body it is assumed it is emitting a continuous distribution of frequencies as shown. (Integrating over all frequencies gives SBL.) So, yes, there will always be these sort of things when you look at any one frequency, but no one frequency determines the temperature – only the full distribution. What you measure with a thermometer corresponds to the peak frequency.
When the peak frequency increases for the smaller one (as it gets warmer than the other) it will start to transfer heat to the larger one, even if it is sending less radiation to the larger one than it is receiving from it, due to a concentrator like the funnel. Maybe you can explain this to Jeff better than I can.
I really have to leave this now – all my answers are already on the Radiation page of my site.
#843 <
If you can concentrate the energy from a ‘big’ solid angle plate onto a small one and heat it we will be able to make billions
Nonsense – it’s being done with funnel cookers. http://solarcooking.org/plans/funnel.htm
But when you can use a funnel arrangement through the roof of a house to a small plate in the ceiling and make that plate hotter with backradiation when the sun is not shining (thus warming the house for nothing at night) then you can make billions. It will warm the house in sunlight, but not at night with backradiation because the latter is impossible because the 2nd Law says so.
Why is it so, Jeff?
The only way you can answer is with Johnson’s “Law.”
Finally, the IPCC models assume backradiation is about a quarter as powerful as the Sun at noon at the Equator.
So if you did use a funnel with an opening four times the size of a small plate you might think you could heat the plate by backradiation at night just as quickly as you could in sunlight without the filter. The total radiative flux would be over 1200 W/m^2 – and yet it would not warm the plate even 0.1 deg C, because it it did someone would be making a lot of money warming houses at night with free energy from the backradiation from the cold atmosphere. So what happens to the radiative flux? Answer: Absorptivity = 0 when the source is cooler.
. .
I just read the whole webpage on the funnel cooker. What do you think happens at night?
It certainly doesn’t slow the rate of cooling due to concentrated backradiation. It acts as a refrigerator and cools well below ambient temperature.
Hi Doug,
I looked at the funnel cooker page. The funnel cooker reflects sunlight onto the item at the bottom of the funnel. It seems to me that short wavelengths will reflect more than long wavelengths especially off of that aluminum foil.
Does foil like this http://inhomeprojects.com/reflective-foil-insulation/ really reflect IR or does it absorb and re-emit it?
Questions I asked myself: If we built a large enough funnel with this foil would we reflect enough backradiation to heat something at the bottom? (even though the sky is colder than the object at the bottom). Would it heat the object faster on a cloudy night (since according to what I said above, clouds reflect IR which causes the surface to cool more slowly)
My answers: The sun is a point source and can be reflected, refracted and thus concentrated. Backradiation is a diffuse source so it cannot be concentrated. I don’t know a way to perform this experiment.
#851 Eric. The longer wavelengths should not stop reflection – at worst it would be diffuse reflection rather than spectral. Aluminium foil will always have low absorptivity. The funnel will concentrate backradiation. Most of what goes in the larger opening at the top will reflect down to the smaller end. Try it – no matter how much you multiply backradiation it can never have any effect on surface temperatures because to do so would violate the Second Law. The only effect you will get at night is the refrigerator effect as was observed, but that is rather complex to explain here.
Douglas,
You don’t know the first thing about the second law.
Your idea of how a ‘magic’ funnel would concentrate light from the sky proves you don’t know a thing about optics.
Your idea that a CO2 laser doesn’t cut steel proves that you don’t know a thing about light.
Your idea that bolometers can’t measure below their own temps proves you don’t know a thing about radiation.
Your failure to even recognize that the commercially available pyrometers which measure to temperatures lower than your challenge way back in the thread above, proves the same thing again.
Your misunderstanding of Carrick’s radiometer – ditto.
Microwaves – ditto
Your replies unequivocally demonstrate that you hadn’t even considered these devices exist prior to this thread. How about a walkie talkie Doug? Fancy thing emits very long wavelength light that is ABSORBED into the antenna of another comparatively VERY hot device.
Get a clue – Please!!
Thats it. The only thing which Doug understands is that he is right. I’m tired of this thread and am considering deleting the whole thing.
Stupid hurts my brain.
not only does it hurt the brain, it hurts the wallet. 855 comments is many times over your average byte count per page load.
Here is another chart of the breakdown of solar insolation. http://climate-change-theory.com/insolation.jpg
Note the reference to most of the infra-red being absorbed by the atmosphere.
Clearly that in the visible spectrum is by no means “most” of the energy. For a start, the UV, X-rays etc have much higher energy than light as you all must know. So the atmosphere has a significant cooling effect during daylight hours, and water vapour has a net negative feedback partly because of this absorption and also due to reflection off clouds.
Seeing that backradiation does not affect climate in any way, there is no way WV could have a positive feedback as assumed by IPCC, thus amplifying CO2 effects they claim.
There simply cannot be an atmospheric radiative greenhouse effect without violating the Second Law of Thermodynamics, as I have proved on my website, and also because every ray of radiation has to be treated as a separate process.
There is no physical meaning associated with, and no physical entity corresponding to “net” radiation. Radiation rays do not combine like, for example, force vectors.
A warm body will not absorb any radiation from a cooler source, no matter how much of such radiation is sent in its direction, as shown in my funnel experiment. And all such radiation has no effect on the normal spontaneous outgoing radiation, let alone the heat loss by evaporation and diffusion followed by convection.
Leonard
I note that you are on Roy Spencer’s site disagreeing with Claes Johnson and myself.
I’ll try just once more to show you why we are right about the resonance and the application of the Second Law …
Maybe this solar cooker funnel experiment will help you and others understand better ….
You simply cannot explain what happens to the extra radiation in my funnel experiment from the much larger (though slightly cooler) plate to the smaller one. Clearly far more radiation gets concentrated onto the small plate, yet heat flow must be from hot to cold, ie opposite to net flow.
Net radiative flow has no physical entity and is a meaningless concept. Yes there is two-way radiation, but heat only flows one way because only the radiation from hot to cold has any effect.
The effect that radiation from the hotter source has is that the excess over that which can resonate with the cooler one is converted to thermal energy. Radiation which fits under the curve of the cooler one can resonate either way (hot to cold or cold to hot) because those frequencies are common to hot and cold bodies. Such radiation is scattered and the effect is the same as diffuse reflection. The warmer body can scatter any amount of such energy without its own outward radiation being affected and without receiving any thermal energy from the cooler one.
What happens when the Sun is warming the surface in the morning? The net flow of radiation is into the surface, right? So how could the IPCC models be right in saying extra thermal energy (also from radiation in the same direction) flows from cool atmosphere to warm surface against the Second Law.
The Second Law must apply to every individual “transaction” or radiated beam between any two points. You cannot just say all will be fixed up that evening when net flow is finally outwards. Besides, the energy might come back out by diffusion or evaporation rather than radiation.
There are no two ways about it. Only radiation from hot to cold can transfer thermal energy. That from cold to hot does nothing. The amount transferred is represented by the extra frequencies / extra radiation in the area between the curves because these frequencies (coming from the hot body) are only in its distribution and thus cannot resonate with the cooler body. (The result is the same as SBL calculations in normal situations, but SBL does not give the right answer for a funnel.) In contrast, all the frequencies in the cooler body’s radiation can resonate with the warmer one.
For more detail see the ‘Radiation’ page my website http://climate-change-theory.com and a paper (which I have oompleted) will be available in due course – to be advised.
Jeff – Claes paper is about blackbody radiation – not artificial radiation or battery boosted signals such as I found out are used in some IR thermometers etc.
Your idea of how a ‘magic’ funnel would concentrate light from the sky proves you don’t know a thing about optics
Well, seeing is believing: http://solarcooking.wikia.com/wiki/Solar_Cookers_World_Network_(Home)
These “magic funnels” are being used and are great for developing countries.
There you go again Doug putting tags on photon’s butts. What is this mysterious photonic property you have invented now – Stink? So photons have wavelength, polarization and stink now right?
‘Tis the stink of the photon which prevents absorption.
Rejected to the sky as an unwanted portion.
In search of an object with a similar smell.
Where the photon will stop, only Doug can tell.
hehe
Focusing the sun is just fine. When light comes from a limited solid angle you can increase its energy transfer by increasing the solid angle (apparent size) to another object. Of course the radiation transfer is bi-directional – unless you don’t believe in the reversibility of light. I suppose that since we have challenged the existence of basic instruments, we can challenge pretty well anything. Try to focus the sky on your house and make it hotter than it actually is.
You are a quack who is apparently unable to realize you have completely fabricated an easily falsified universe just so you can say global warming isn’t real. You suffer the same disease as Michael Mann, a loser I also have little patience for.
Jeff – please action my email to you
Clivere,
If you mean remove the email address from your other comment. I did. I’m not sure about your phrasing.
The Second Law of Thermodynamics s so obviously violated by the assumptions in the IPCC energy diagrams that I find it hard to believe that anyone with a drop of physics in their blood could not see this.
Clearly, any sunny morning there is already a net flow of thermal energy into the surface from the Sun. We know this because the temperature is increasing. Yet the models assume that more thermal energy can be added to the Sun’s heating effect by radiation from a colder atmosphere. So cold is warming up hot even more. Crazy!
But the IPCC, Trenberth et al try to say it’s all OK because the other side of the Earth is radiating outwards in the cool of the night.
Where in any physics textbook does it say the Second Law does not have to apply in any local area?
Jeff – radiation has frequency. You cannot eclipse this fact by saying it’s all just a string of identical mass-less photons. Why do x-rays have a different effect to a torch beam? They are all just identical photons, aren’t they?
Jeff – if you build a simple reflective funnel, which is bigger at one end and you seal it off with two metal plates at each end having different sizes, then the radiation from the inside of each plate is fully trapped within the sealed off enclosure. So where’s it going to go? Fair chance it will reach the other plate, don’t you think? Just like the Sun’s rays entering the large opening at the top of the solar cooker mostly get reflected down onto the juicy food.
Jeff – Ypu are quoting physics about light which is acting in the absence of focusing mirrors. That’s where you go wrong in your thinking. (see #864)
Stop all your waffle about me being wrong, and start thinking about the physics of the real world. How the hell could radiation beams at different angles and frequencies interact with each other to form some physical new radiation beam which is a net sum of the two? Then you want to get this net sum by combining beams on the opposite side of the Earth – see #862.
The bluff of AGW is over, Jeff. Physicists throughout the world are turning their attention to what has been claimed by climatologists (who perhaps did a bit of undergraduate physics and remembered a few formulae, but never really understood when to apply them) and physicists are saying this is not the physics of the real world.
They are saying that absorptivity varies a lot with the temperatures of the source and target. Experiments prove this.
The only way the Second Law can work in nature is if absorptivity reduces to zero when the temperature of the source is no longer than that of the target,
Yes I have written a paper on this (submitted on Saturday) and am awaiting to see if it gets published. If not I will “publish” it on my website and include it in my book which I’m holding back while awaiting their response. But I have to keep it confidential at this stage, including the results of my refined experiment, But I’ve given you all enough of a hint as to its content.
.
Correction: The only way the Second Law can work in nature is if absorptivity reduces to zero when the temperature of the source is no longer warmer than that of the target,
“They are all just identical photons, aren’t they?”
No, they have frequency and polarization. Are you admitting they don’t have magic Doug stink?
——–
Jeff – if you build a simple reflective funnel, which is bigger at one end and you seal it off with two metal plates at each end having different sizes, then the radiation from the inside of each plate is fully trapped within the sealed off enclosure. So where’s it going to go? Fair chance it will reach the other plate, don’t you think?
===
Finally a good question Doug.
The radiation from the physically larger source can only be concentrated by increasing its apparent solid angle. Make the source appear larger and you increase its influence. The effect is bi-directional.
Wanna play light? Of all things, I’ve got that one.
To make it really simple. If you concentrate light too far, it turns around in the concentrator and shoots back out the top.
Jeff, radiation can’t get out the top of my funnel because the metal plate on top is sealed over the whole opening, as is the small plate at the bottom. I’m sure I described it clearly in an earlier post. You may assume the walls are made of thick polished aluminium and the whole device has four sides coming off the edges of the square metal plates but tapering inwards at about 60 and 120 degree angles to the large and small plates respectively. No radiation escapes anywhere.
[reply: Placing the plate at the top of the “funnel” doesn’t stop etendue physics.]
Jeff – you say the effect is bidirectional – but you forget the reflection and concentration. All radiation from the whole hemisphere on each internal side of each plate is trapped. So the total from each plate is a function of its area.
[Reply: nope]
Here’s today’s post going on several forums …
[snip – not this one]
Snip? Just because you have no counter argument I guess, Jeff?
Well you and others can watch for replies on Spencer, Curry, WUWT and tallbloke . . . [snip again – keep it up and you may actually exceed me as the most snipped person here. You are allowed to address comments but I’m really tired of constant references to other blogs when they make no point.]
#869 Etendue physics does not counter my argument. You need a better understanding of etendue physics, Jeff my boy.
[reply: Doug, you are not qualified to talk to me about physics. You have not answered one single critique satisfactorily.
Etendue is based on math and it directly refutes your funnel. Again, we find you have not one clue about that which you speak.
We have beaten your fake universe to death from every angle and you have yet to notice any of it.]
1.
Jeff Condon said
January 21, 2012 at 5:26 pm
————————————————————————-
Jeff, thank you for addressing this. In other discussions I stated this same assertion/question several times, and I used the term, “back conduction”, as in an entirely non GHG atmosphere the atmosphere would then have to back conduct the specific heat to the surface in order to release it.
Now I have my own little physics law, called, well what else, David’s law, (-; which goes like this;
“At its most basic only two things can effect the energy content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system.
This concept appears very critical to understanding earth’s energy budget. The larger any systems heat capacity, the more energy it can lose or gain in any persistent change in input. A systems heat capacity is also a function of time. The longer any given input can stay within a system, the greater its heat capacity.
It is apparent and not contoversial that if one increases the density of an atmosphere, through either an increase in the atmospheres gas, or an increase in gravity, then the specific heat capacity per M2 will increase, and, while the energy per molecue may not change, the energy per M2 will, over time, increase to a higher eqalibrium. It is also logical that in a case where there are no GHGs in the atmosphere, this energy and heat can only escape via back conducting to the surface, and the surface will at the same time still be recieving the same TSI (input) it always did, plus back conducted specific heat.
If all this has been addresed please point me to the posts.
Thanks.
This is typical of the type of mistake which is made when trying to explain the (radiative) greenhouse conjecture. On his website Dr RoySpencer wrote this very strange comment when trying to “prove ” that the Second Law is not violated.
But the same objections could be made against many systems which create very high temperatures. You can pump energy into a system at a certain rate, and insulate the system so that it cannot lose heat easily and thus increase temperatures to very high levels.
And he seemed to imply this “heat furnace” concept applies in the atmosphere. Let me quote Wikipedia (Second Law .. ) The second law declares the impossibility of machines that generate usable energy from the abundant internal energy of nature by processes called perpetual motion of the second kind.
Any “pumping” up of temperatures in the atmosphere would have to raise the temperature up there to more than the surface temperature at the time before any spontaneous radiation from the atmosphere would warm the surface.
It cannot happen, Roy, and it doesn’t.
And before you come back at me with discussion of “net” radiation, tell me what physical entity you think net radiation actually corresponds to. Are two rays on opposite sides of the World (day and night) going to have a combined effect? Hardly! Nor would they even if only a metre apart in parallel with each other.
The only way any effect of radiation in one direction can be altered by radiation in the other direction is via thermal energy addition. This means the energy in each ray has to be converted to thermal energy first. And remember, the energy might exit the surface by evaporation or some other non-radiative process.
Each ray has to be considered as a separate process. So any conversion to thermal energy involving radiation from a cooler atmosphere to a warmer surface violates the Second Law. Other rays in the opposite direction cannot justify the violation.
The reason it does not happen is because the absorptivity of the surface does in fact reduce to zero for radiation coming from a cooler source. Such radiation merely resonates or is “rejected” in some way, just as if it underwent diffuse reflection. Only radiation from hot to cold counts when it comes to anything to do with temperatures of the target.
My funnel experiment proves that this must be the case and that the warmer surface can in fact handle any amount of such radiation without it affecting its own rate of emission or its temperature.
(I give notice that I have submitted a paper on this and do not wish to reveal the explanatory mechanism detailed in that paper at this point for obvious reasons.)
#873 The reason etendue considerations are inapplicable to the funnel are that, if the large plate is the image and the small plate is the aperture, then the reflective funnel is external to the “system” being discussed, since that system starts at the small plate where it also finishes in this case. Normal considerations involving etendue math do not have the focusing device
I should not have to say that a car headlight will brighten a subject at a distance far better than the globe without the parabolic reflector focusing the beam..
Jeff You say the smaller plate (which is warmer) “sees” a bigger angle. But that is not going to make it radiate more than the maximum “allowed” by SBL, that being far less than the maximum for the large plate. So you are confusing things with red herrings once again, trying to give the impression you know what you are talking about better than I do. No one pulls wool over my eyes.
When I know that Claes and I are onto the truth of the matter, there will always be an explanation for any apparent contra-argument.
When two plates face each other they both radiate at the maximum allowed by SBL, contrary to popular belief. You only think they radiate less because there is less heat transfer when temperatures are close. But the real reason for that is that each plate rejects an amount of radiation equivalent to the maximum emitted by the cooler plate, because such radiation resonates mutually in each one and leaves no energy in either target plate. It gets scattered and goes on to warm something else that is cooler, perhaps in space.
Only the warmer plate transfers energy (equal to the area between the Planck curves) to the cooler body. All other radiation from either plate goes elsewhere to eventually warm a cooler object (or objects or air above) elsewhere.
Have you ever considered that proper application of S-B Law would result in no radiation from a cooler atmosphere to a warmer surface, because you would have to subtract a larger value from a smaller one and get a negative result.
But I come to the rescue and say that there will in fact be downward radiation, because after all you would expect radiation to go in all directions within a full spherical angle.
I say about half of the full amount of SBL radiation will head towards the Earth’s surface – but it won’t have any effect and will be rejected and not warm anything that is warmer than its source was. So, after coming back up from the surface, it can only warm cooler air (usually higher up) or get through to warm some object in space one of these years.
.
Markus Fitzhenry is saying the same as myself about radiation from cooler sources having no thermodynamic effect on a warmer target. See http://judithcurry.com/2012/02/22/teaching-the-controversy-2/#comment-176757
“#873 The reason etendue considerations are inapplicable to the funnel are that, if the large plate is the image and the small plate is the aperture, then the reflective funnel is external to the “system” being discussed, since that system starts at the small plate where it also finishes in this case. Normal considerations involving etendue math do not have the focusing device”
I can’t even follow you Doug. It has been almost 24 hours since I last designed a “funnel” though and it is true that I have designed less than 500 such ‘funnels’ in my lifetime so I might be a little rusty in my non-imaging optics. Don’t worry about me though, I’m going to be taking a refresher course non-imaging concentrator design in about an hour (yes it is Sunday). I will let you know if I succeeded again in about 6 weeks.
Doug, sorry I was away from the thread. The sun being a point source results in funnel reflection into one location and cooking food at that location. Radiation is a diffuse source cannot do that.
Also the shield that you describe in your experiment (424) with an angle of 45 degrees is partly reflecting the radiation from whatever is at a 90 degree angle from the object on the ground to that object on the ground. It is of course an imperfect reflector, but then that means it is also directly emitting its own BB radiation onto the object on the ground.
Jeff, would you have any thoughts to enlighten my assertions/questions in my comment #874?
Concerning the 2nd law discussion could not anyone do an experiment. Set four rooms to the same T and atmosphere. Heat four 2×2 metal plates to the same T and hang in the center of the room. Heat four other plates, one to the ambient room T, each of the other to a higher T on a considtent gradient, but each one cooler then the first four plates. Hang those, one in each room 10” from the hotter plates. Masure how fast they cool. I am betting it is not the same.
#880 & #881 All the radiation from the whole hemisphere which the large plate radiates into is trapped by the sealed enclosure and such radiation will nearly all get reflected down to the small plate, even if several reflections are required. Likewise all the radiation from the small plate’s hemisphere will get reflected back to the large plate. It’s quite simple really. Jeff is just getting confused with etendue issues that usually relate to lens systems as in cameras and telescopes. I’ll prepare a diagram for my book soon, but am delaying the book pending possible publication of the paper I completed on Saturday. If not accepted, I’ll include the paper as an Appendix in the book, and also on my website. The paper explains it all much better, I feel, and it’s pretty similar to what Markus Fitzhenry is also saying as I discovered only yesterday when I first came across the above linked post of his.
#881 Regarding the shield, my latest experiments with the vacuum flasks use a sheet of plate glass at an angle of about 15 degrees with the original screen above the glass and a small gap between. The glass should absorb IR without reflecting much..
I have to go now as I have an appointment at Macquarie University this morning (Monday.).
“nearly all get reflected down to the small plate, even if several reflections are required.”
Keep reading doug, because if you can get it all down there to that little bitty plate, you are right and I need to close our company down. You will be personally responsible for laying off a hundred people.
Of course, if you can do it, we will make literally trillions as I know how to get rid of old fashion power supplies.
#882 Heat transfer rates between radiating plates have been frequently measured and there are mathematical formulations for such which we all accept based on Stefan-Boltmann’s Law. None of this is in dispute.
I am just saying the actual physical processes involved have to be different from whaat has been supposed, and such processes do not involve each lot of radiation in each direction either interacting or both transferring heat. The radiation each way is “full blast” as per the full area under the Planck curve for each. But the only thermal energy transfer is from hot to cold represented by the area between the curves, that being the only extra radiation from the hot body that does not merely resonate and have no thermal effect.
David,
“Jeff, would you have any thoughts to enlighten my assertions/questions in my comment #874?”
Yes.
“Jeff, thank you for addressing this. In other discussions I stated this same assertion/question several times, and I used the term, “back conduction”, as in an entirely non GHG atmosphere the atmosphere would then have to back conduct the specific heat to the surface in order to release it.”
Your terminology is too mixed up to reply to. You seem to have an idea what you are saying but I cannot parse it.
” to back conduct the specific heat”
No idea what this means.
“A systems heat capacity is also a function of time.”
No it isn’t. The problem here is that we don’t have a common language. You have obviously not been formally trained in physics or engineering because you use the terms incorrectly. We might sit across a table for several hours until I understood your point but until we use a common language, we cannot communicate.
The idea that somehow the basics of AGW are mixed up is my least favorite subject. I don’t enjoy the conversation much because it consists of misconception and random theory based on whichever individual you are talking to.
Hi Jeff. Please understand that I am more asking questions then making an assertion. Also, as Mr Cotton has taken over this thread I understand you being weary.
Quickly and simply then. I used the term, “back conduction”, as in an entirely non GHG atmosphere the atmosphere, which would heat up via conduction from the surface would then have to back conduct the specific heat to the surface in order to release it. (A Simple assertion, I do not know if it is true.)
“A systems heat capacity is also a function of time.”
By this I mean simply that the large the heat capacity, think the oceans, assume they are at equalbrium, now change the input, let us say an increase in SWR reaching the surface. Because there heat capacity is large, it will take longer for the oceans to reach a new equalbrium T. The same change in the atmosphere may take only one day to reach equalibrium. The oceans, having far greater heat capacity due to the fact that the residence time of the SWR entering the oceans is far longer then in the atmosphere. So in this sense residence time of energy in a defined system, in this case the oceans verse the atmosphere, is what determines the heat capacity.
At the end of this long thread I understand Jeff’s patience capcity to be full. (-;
David,
I don’t mean to be impatient. Your ‘back conduction’ is exactly as real as ‘back radiation’ although ‘back’ does not need to be added to either of the terms. The net flow direction is the result of probability but the energy transfer happens in both directions, so your intuition on the matter is correct.
Heat capacity is a measure of how fast something changes temperature with a specific addition of energy – Joules. It is usually expressed in terms of a standard mass – a mole or Kg or something. The ocean’s ‘heat capacity’ is therefore a measure of how many joules it would take to raise the temp x degrees.
“The oceans, having far greater heat capacity due to the fact that the residence time of the SWR entering the oceans is far longer then in the atmosphere. ”
The residence time of SWR is not related to heat capacity. You seem to have a different idea of the definition so I truly cannot understand your last paragraph.
http://en.wikipedia.org/wiki/Heat_capacity
In the real world there is usually a small step down in temperature between the surface and the first few mm of the air in calm conditions. It is highly improbable that much thermal energy goes back from the atmosphere to the surface by any means such as diffusion or radiation.
Jeff – with all your funnel-stuff, maybe you’d like to explain why, with all the radiation from the atmosphere at night, a solar funnel acts like a refrigerator and cools more than 10 deg.C even though the radiation is supposed to be about a quarter as strong as the Sun at noon..
#884: The radiation from the large plate in the closed funnel will also be scattered by the large plate itself if it comes back onto itself. That’s why most of it will also strike the small plate.
The message I have been giving you all along about radiation from the cooler atmosphere having no thermal effect on the warmer surface is now getting out on WUWT and the semi-skeptics are being challenged for their half agreeing with the IPCC. http://wattsupwiththat.com/2012/02/26/the-skeptics-case/#comment-905670 and my comment following his.is this …
If you look at the NASA plots for all available sea surface data from 2003 you will see a very regular annual pattern with a maximum in March, a dip in June, slight rise till August and a minimum around the end of November. Select sea surface and tick all years except 2012: http://discover.itsc.uah.edu/amsutemps/
This indicates very tight control and very little random noise. It also indicates no rise over that period, but my point is that the lack of rise cannot be blamed on random noise. The plain fact is that carbon dioxide is having absolutely no effect – not just a little effect as semi-skeptics like the author would have you believe.
It is time for true sceptics to debunk statements like http://discover.itsc.uah.edu/amsutemps/ altogether and not to acknowledge that radiation from the atmosphere can have any thermal effect on the warmer surface. It can’t because to do so would be a breach of the Second Law of Thermodynamics. You cannot “excuse” it all by saying “net” radiation is out of the surface. It isn’t on a warm sunny morning. And in any event, “net radiation” is a totally meaningless expression with absolutely no physical entity matching its description.
Radiation goes “full blast” with all the power allowed in the area under the Planck curve – in each direction. But only the surplus in the higher frequencies from the warmer surface has any thermal effect, namely warming the atmosphere. All the radiation from the cooler atmosphere, and a matching amount from the warmer atmosphere merely resonates without transferring thermal energy.
So any radiative greenhouse effect is a physical impossibility as several, including myself, have been pointing out for a while now.
It’s time for the semi-skeptics to become full skeptics with a unified, proven message.
The IPCC models make use of absorptivity measurements for the Earth’s surface which were measured using visible light. But they apply them to far-IR radiation from the atmosphere, even though it is well known that absorptivity reduces very significantly for much lower temperature radiation. This is obviously important when determining the assumed warming effect of radiation from the atmosphere – which, by the way, is assumed to help the Sun with its warming every sunny morning – all quite against the Second Law of Thermodynamics which they think it isn’t because somewhere on the other side of the Earth at night some radiation is turning it all into totally unphysical “net” radiation which cannot be a physical entity. But, never mind, I diverge.
The question is Can someone link me to any empirical measurement of absorptivity by the surface of radiation in the IR bands emitted by the atmosphere?
You’d kinda think the IPCC would have got this part sorted out before spending all that money on the models. So show me where they did – anybody!
888.
Jeff condon said
February 26, 2012 at 8:42 pm
David,
……………….The oceans, having far greater heat capacity due to the fact that the residence time of the SWR entering the oceans is far longer then in the atmosphere. ”
The residence time of SWR is not related to heat capacity. You seem to have a different idea of the definition so I truly cannot understand your last paragraph.”
———————————————————–
Thanks for the time Jeff. First of all, you are correct concerning my unprofessional terminology and I rarely post here, as being a layman, I learn more then I asssert, and my assertions are usually for feedback, which you have kindly given. I used the term “back conductiton” as a imitation of the term “back radiation” to make a point of how a gas world with ZERO GHGs, could only cool via conduction, and it appears possible that the process of conduction is slower then the process of radiation.
Concerning the resdidence time of energy within a given system, and relating that to heat capacity, well, I think I can assure you that heat capacity has an element of time in it. Your own answer gave the relationship of heat capacity to time. “Heat capacity is a measure of how FAST something changes temperature with a specific addition of energy – Joules…”
This is percisely why “David’s law works; ““At its most basic only two things can effect the energy content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system. Imagine an ocean suddenly five degrees cooler in the top 100 meters. Now if ALL the SWR which enters that ocean today, all leaves today, then it will still be 5 C cooler the next day. However, if some of todays energy has a longer residence time, then tomorrow that energy will still be there, and with todays added energy the oceans will no longer be 5C cooler. In the oceans the residence time of SWR can be years, decades, even longer.
I hope this traffic analogy works…
1. On a highway if ten cars per hour enter the highway, and the cars are on the road for ten hours before exiting, there will be 100 cars on the road and as long as these factors remain the same the system is in balance. If you change the INPUT to eleven cars per hour, then over a ten hour period the system will increase from 100 cars to 110 cars before a balance is restored and no further increase occurs. The same effect as the increase in INPUT achieves can be realized by either slowing the cars down 10% or by lengthening the road 10%. In either case you have increased the energy in the system by ten percent by either increasing the residence time or the input.
2. Now lets us take the case of a very slow or long road with the same input. Ten cars per hour input, 1000 hours on the road, now you have ten thousand cars on the road. Now lets us increase the input to eleven cars per hour just as we did on the road with a ten hour residence time. Over a 1,000 hour period we have the same 10% increase in cars (energy) How ever, due to the greater capacity on that road, the cars (energy) have increased 100 times relative to the 10 hour road with a 10% increase in input. (1,000 car increase verses a 10 car increase.) Any change in the input or the residence time his 1,000 hour road will have a 100 times greater effect then on the 10 hour road if the input change endures for 1,000 hours. The ocean of course is the 1000 hour road, the atmosphere is the 10 hour road.
I hope a reread of my post 874 will not appear more cogent to your post #1.
I hope a reread of my post 874 will now appear more cogent to your post #1.
Ugh, there changed “not” to “now’. Sorry, but, but I am tired after spending 30 minutes trying to convince Myrrh not to bomb thread everything with his ideas on the 2nd law, regardelss of the veracity of his assertions.
Look, I will agree that you do have to take account of the fact that the large plate scatters “backradiation” – just as the Earth’s surface does. I was always imagining it this way, but strictly speaking that is probably assuming the result of what I’m trying to prove, hence is circular.
So let’s not talk about funnels anymore.
All you need to consider is what is happening each sumny morning. The surface is warming at, say, 11am. So there is net solar radiation into the surface. Now the IPCC wants you to believe that additional radiation from the cooler atmosphere is supposedly transferring thermal energy from the atmosphere to the surface, thus cooling the atmosphere and warming the (already warmer) surface. This process (which must be considered stand alone) clearly violates the Second Law.
I don’t think I got an answer to my question about the funnel. The sun is a point source, the sun’s rays reflect off the funnel onto the object at the bottom. Radiation from the atmosphere is a diffuse source and even if it reflects, there will be no focal point.
David,
“Concerning the resdidence time of energy within a given system, and relating that to heat capacity, well, I think I can assure you that heat capacity has an element of time in it. Your own answer gave the relationship of heat capacity to time. “Heat capacity is a measure of how FAST something changes temperature with a specific addition of energy – Joules…”
Heat capacity is NOT related to time. If you add 1joule per second or per week to an insulated system you get the same temp increase according to its “heat capacity”. If you are interested in being more of a physicist, look at the units of heat capacity, particularly for seconds, and it should become more clear. Be careful though as sometimes units are compounds of other units and may divide out. In formal training, you spend a lot of time with units of a number. I don’t mind explaining, just try and realize that I’ve spent my life working with these units and am not a layman, you aren’t going to convince me to change all of physics because of my use of the word ‘fast’. Perhaps it is more accurate to say – heat capacity is a measure of how much the temperature of a material changes with the addition of energy.
“David’s law” is approaching very close to standard GHG theory now. The residence time of energy flowing through the system does determine temperature. Energy is measured in Joules (and other units) and power or energy flow is in Joules/second which equals Watts. I would suggest you replace “heat capacity” in your argument with energy content.
The car analogy doesn’t work as well as Carrick’s sink example because, intuitively, the pressure at the drain and thus the outflow increases with water depth. Tighten the drain a little, and the water level deepens until the drain is again balanced. In standard science, more delta temperature between two objects and you get more flow.
In a GHG the drain is upward, the only release of energy is by radiation to deep space and adding more gas increases the probability of a photon being captured again before release. This increases the probability of conduction as well but the conduction process doesn’t release energy to space and is therefore balanced. It can be ignored- almost. So you add GHG and you will increase residence time in the lower atmosphere and surface temps absolutely rise – guaranteed.
What isn’t guaranteed are things like convection feedback, H20 feedback etc. If Doug came here and stomped around saying he had proven the effect is 0.001C/doubling of CO2 (very small), I would be skeptical but really cannot disagree. I have read no proof that this isn’t the case. This is where the real discussion gets interesting. Leonard had a comment recently on another thread that nobody has replied to. I don’t really agree with it but it requires a mathematical treatment to demonstrate one way or the other what is correct.
Don’t get me wrong, there are many reasons to be skeptical about AGW. I just wish that everyone knew that they are based on actual physics rather than some odd theory of radiation which contradicts everything we have learned. The media portrays skeptics in the light of the Doug’s of the world because he makes us look like an easy target. Unfortunately for climate science, other professionals can read my/OUR words quite easily and immediately get the point.
#897 Eric, yes you’re right and the discussion here has led me to remove reference to the funnel in a revision of my paper which I worked on until midnight, last night (Monday) and resubmitted. It was not vital for my argument.
The reason radiation from a cooler object slows down the radiated heat transfer to itself from a warmer body is not because there is a compensating transfer of thermal energy back to the warmer body, because such would violate the Second Law. Rather it is because a standing wave is established which is represented by all the area under the Planck curve for the cooler body. This area represents the frequencies and intensities that are common to both the warm and cool objects.
The atmosphere (with over 50 gases and water vapour) does not radiate everything that a true blackbody would, but water vapour does help fill the area under that curve. So there is a standing wave, but its total power is not as much as a true blackbody. This is why some radiation escapes directly through the atmospheric window.
The standing wave has no thermal effect because none of its energy is ever converted to thermal energy. It just sends information back to the warmer body and a part of the warmer body’s radiation goes into the standing wave. The energy radiated by the warmer body which is represented by the area between the curves does get converted to thermal energy because it cannot resonate and thus contribute to the standing wave. The calculations of course agree with accepted physics, but the mechanism is not a two-way transfer of heat, as many appear to have supposed.
But there is no build up of the effect of carbon dioxide due to multiple repetitions of the capturing and re-emitting process envisaged in the IPCC energy diagrams and models. Each carbon dioxide molecule can only play a single role in a very limited sub-section of the total standing wave. Its contribution per molecule would be no more than a molecule of water, and so its total overall effect is comparable with its relative proportion to WV and other emitters in the atmosphere – insignificant.
Furthermore, there must be a compensating effect for reduced radiation by way of additional evaporation, diffusion etc because the very stable temperatures not far underground will be reflected in the close thermal equilibrium at the surface / atmosphere interface.
Jeff and others
I really must sign off now from this forum. I do want you to know that there have indeed been some helpful comments (eg re the funnel) which have helped me see where I was wrong on that particular issue. It wasn’t crucial to my argument, however.
Now that I have read about the standing waves I am convinced that this is the real reality. Basically I have revised my paper accordingly and devoted a whole page on my website to the content of the above post which will be my last main post I hope.
You either accept it or you don’t, but it makes good sense to me.
I will answer genuine questions that arise perhaps from insufficient description in the above post, but I am not going to argue with those who just say it’s wrong.
It would be interesting to measure the relative cooling rates of a near blackbody in the presence of a container of carbon dioxide and then another container with 80% N2 and 20% O2.
Doug,
There ain’t going to be a lot of standing in multi-frequency waves passing in opposite directions with “different polarizations” .
It might sound good to you, but it ain’t.
The real answer is that the second law is a bulk law. Energy goes both ways in conduction, radiation and even convection – strange eh? Is there no air pressure below an updraft?
You have shifted again to an easily dismissed position in order to come to the same conclusion. At least you recognized the funnel problem.
I used to do holographic interferometery as a means of measurement. Published in that too. Our holography used a standard method of recording a standing wave pattern in a film thickness. Superposition of light is also interesting to me but it doesn’t do what you seem to suspect. Hell, I’m not sure it does what anyone ‘expects’, but it does do it. Light travelling through glass – also cool. Reflection, interesting but like transmission you have to grock math.
Observation over theory, except in matters of faith.
If there wasn’t a standing wave between two particular molecules – one on the surface and one in the atmosphere, then, let’s say a Rock A is cooling. It is sending all its potential radiation under its Planck curve. But what causes the atmosphere to send thermal energy back to that particular Rock A rather than Rock B? Rock A could keep cooling to absolute zero and Rock B get far hotter. Also that part of the atmosphere would get too cold.
There can be no control of the rate of cooling by some sort of compounding of thermal energy, because that requires conversion of the cold to hot radiation back into thermal energy. That particular “transaction” breaks the Law. No, you can’t have “bulk” or net results – when one side of the Earth is warming in the Sun, there is no balancing act with radiation on the other side of the world, for example.
Without the tight one-on-one correlation between source and target (right down to individual molecules in fact) which a standing wave creates, you cannot “get the effect” by compounding either radiation itself or thermal energy which you think is generated by conversion of radiated energy travelling each way.
I have indeed thought this through very carefully, and a standing wave (which others are talking abouit and which fits with Claes Johnson’s resonance concept) is the only feasible reality. (I presume you read the post on the standing waves by Markus.)
“Thus, at the molecular level, radiation and conduction are two sides of the same coin.
“Furthermore, because that part of the radiated energy intercepted by the warmer body is standing wave communicating information between the emitter/absorber states, on both bodies, it can do no thermodynamic work.
“When the cooler body is at absolute zero, the exchange energy is zero. When the temperatures are equal, it is the same as the radiation emitted by either body.
“However, it can still do no thermodynamic work and it can only be detected by blocking the energy from the warmer body to the colder body.”
(written by Markus Fitzhenry) – first link in #892
)
Doug,
I can’t even read your post well enough to disagree.
I’ll try again.
There is no way that radiation at different angles and frequencies could interact and cancel out or compound in any way. Only thermal energy could be compounded.
You have to decide between two possible ways in which a cool plate could slow the cooling of a warmer plate by radiation.
(a) By adding thermal energy to the warmer plate, thus reducing the rate of loss of thermal energy
(b) Some other way not involving thermal energy
Now (a) violates the second law because every “transaction” has to stand alone. For example, the surface might be mostly losing heat by diffusion and evaporation, so radiation out might be less than radiation in. In any event, different points in the surface and atmosphere would be involved. So these are all different transactions and each must obey the Law because it has no way of “knowing” what’s going on elsewhere.
So (b) is the only way. And (b) must involve radiation which is not converted to thermal energy as I have been saying all along.
Claes Johnson said the same too. And he suggested it resonates.
Resonating can lead to standing waves because we are talking about frequencies which are common to both the surface and the atmosphere.
So standing waves seem to be the only method.
Note also that the amount of energy transferred is the area between the Planck curves. As expected, this has a limit of zero as the temperatures approach each other. The radiation from the hot one has to stop at that point (disregarding any other cooler targets) so how does it know to stop? Well in fact it was always split in two – some in the standing wave, and some causing the heat transfer, so just the latter approaches zero and the standing wave keeps going.
.
“There is no way that radiation at different angles and frequencies could interact and cancel out or compound in any way. ”
Now we don’t believe in Maxwell either?
Just what is a hologram then?
The most important thing to come out of the standing wave concept is this: The models assume CO2 creates about as much backradiation as all the WV because CO2 supposedly works harder and captures and re-emits photons faster than WV.
Well that may be true regarding any warming of the atmosphere by the radiation between the curves which is the only radiation from the surface that can warm anything. But the radiation in the standing wave has no thermal effect. Furthermore, there is just one standing wave for each match up between a molecule on the surface and one in the atmosphere. So carbon dioxide cannot create more backradiation per molecule than WV can. So WV dominates the generation of backradiation purely because there are many more molecules of it.
The assumed interaction of two-way radiation that can happen physically is in a standing wave. Any other compounding can only be in the effects of radiation – more heat and/or more light. The radiation beams (at different angles and frequencies) do not combine first and then get converted to heat or light. There is no such physical entity as “net radiation” or “bulk radiation” as all such rays remain separate entities, even if the human eye cannot distinguish them. The energy in the radiation has to be converted to heat or light before the compounding can take place – and this means violation of the second law if the radiation was from cool to warm.
#907 Stop introducing red herrings like holograms.
Radiated beams cannot be componded as can, for example, two or more forces.
Shine a torch through a beam of sunshine and watch the sunbeam bend the torch beam so they both become one slightly bigger beam in a slightly different direction to the sunbeam? Garbage!
“#907 Stop introducing red herrings like holograms.”
As though I have introduced a single one!
Holograms are interference patterns created by off axis radiation.
“There is no way that radiation at different angles and frequencies could interact and cancel out or compound in any way.”
Radiation at different angles and frequencies interacts all the time. Tis the basis of the hologram Doug. It is also the basis of basic electromagnetic theory.
The hologram is produced (recorded) and displayed using laser light and it is important to ensure that the reference beam and the object beam started out with the same frequency, amplitude and phase. So in fact a beam is split and then caused to spread out. The fringes then contain the information that is required and is stored in the film.
This is nothing at all like “radiation at different angles and “frequencies” and I could have added different phases – as found in the surface and atmosphere system that we are discussing.
Radiation with different frequencies from A to B does not get together with radiation from C to D and form a combined ray from X to Y, as if the rays were forces that can be compounded by vector addition – ie the diagonal of a parallelogram.
If the bank tells you you can no longer make charges on your credit card, they do not mean you just have to ensure the balance decreases each month. They mean no debits at all made by you (just them .) So it is with the Second Law.
Only standing waves ensure there is no net transfer of energy either way. They are of course like two opposing waves with the same frequency and exactly the same path between Molecule A in the atmosphere and Molecule B in the surface. So, yes, the do interact and become a standing wave that can “keep going” for ever if nothing changes.
This is why you need to shield upwelling radiation in order to detect “backradiation.” But some other radiation starts in the atmosphere because it uses energy that originally came from phase change, convection or absorption of solar insolation. This still leads to standing waves usually at even lower frequencies.
I keep asking, how can the energy in radiation from the atmosphere enter the surface at one point and eventually exit at another point by evaportation or diffusion? It must be converted to thermal energy which would then be transferred by the surface to the exit point. Thermal energy is the common currency.
Does the “heat” say “It’s OK for me to be generated from the energy in the (cooler) radiation because I plan to exit the surface again some time in the future. I have a six month visa to hang around till winter. I just can’t do any work while I’m here.”
No. So this conversion to thermal energy would violate the Second Law.
Dang it Doug, it is like negotiating a religion with you. Drives me nuts.
“The hologram is produced (recorded) and displayed using laser light and it is important to ensure that the reference beam and the object beam started out with the same frequency, amplitude and phase. So in fact a beam is split and then caused to spread out. The fringes then contain the information that is required and is stored in the film.”
Not phase. Not amplitude. Frequency yes but not always. Fringes is a strange term for the microscopic interference pattern in my experience but not unacceptable.
“This is nothing at all like “radiation at different angles and “frequencies” and I could have added different phases – as found in the surface and atmosphere system that we are discussing”
Well since the radiation IS coming from different angles and phases, is it the frequency that you object to? Red herrings my ass.
I can’t even parse 912 excepting a few points. That is one of the more dyslexic comments I’ve read here.
Not sure why you’re having trouble parsing the grammar in #912. Anyway, try reading it carefully and I think you’ll get the meaning. Otherwise, just read what Markus said about the standing waves that don’t transfer any heat.
I stand by my statements on the hologram, which are supported by at least one website from which I quote …
You need the right light source to see a hologram because it records the light’s ,phase and amplitude like a code. Rather than recording a simple pattern of reflected light from a scene, it records the interference between the reference beam and the object beam. It does this as a pattern of tiny interference fringes.
Would you like to open a thread for “negotiating a religion” or similar? Only trouble is, Christianity is not negotiable because that guy who helped out with the Creation told us about how He came down from Heaven and how only those who believed in Him would get back up there with Him – rather than suffer His Father’s eternal punishment for not believing in Himself.
I note that BobC agrees with Markus too now (LOL) …
Author: BobC
Comment:
REPLY: I will join when my brain stops working. Do “FULL” skeptics have a signup sheet or do we just repeat phrases like believers?
P.S. It’s all in one short summary page now ..
http://climate-change-theory.com/RadiationAbsorption.html
[REPLY: Didn’t I ask you to stop this kind of thing Dr. Spam? ]
So, if we can compound any two radiation rays or beams as Jeff says,
Q.1 What do we get when we compound solar radiation coming in at 20 degrees from the horizontal in the late afternoon with radiation from directly overhead in the atmosphere coming from a source at, say, -30 deg.C. Assume the surface temperature is 15 deg.C.
Q.2 Which components (the Sun or both) warm the surface by transferring thermal energy?
Q.3 Does the combined (compound) ray or beam strike the ground at 20 degrees, 90 degrees or somewhere between?
Q.4 Does the compound ray or beam warm the surface more than the Sun’s radiation would have on its own.
Q.5 Does a part of the atmosphere cool because it warmed the surface, if it did?
Q.6 Did CO2 absorb some of the incoming IR radiation from the Sun?
Q.7 If it did, would about half of it have been sent back to space? Or would all of it have set up a standing wave with the Sun?
Don’t forget to let me know if you’d like to “negotiate a religion” – maybe some of the 190 university students ‘I gave copies of the New Testament to yesterday may be able to help you out.
http://SavedByTheLamb.com
There was a nice critique by Dikran Marsupial at SkepSci of Postma’s thought experiment in
http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf (page 6) where Postma states that ” it is absolutely fundamentally impossible for a blackbody to further warm itself up by its own radiation.” In his example a mirror reflects radiation back to a blackbody which is in thermal equilibrium with a light bulb. Postma says the blackbody would be the same temperature whether or not there was a mirror reflecting back radiation.
Dikran says:
“To remain at the same temperature, it would have to be radiating energy at the same rate that it is absorbed (Kirchoff’s law). If you increase the amount absorbed using the mirror, the amount emitted must increase as well. However the Stefan-Boltzman law says that the rate at which a blackbody radiates energy is proportional to the fourth power of its temperature, so it can’t increase emissions without an increase in temperature. Thus Postma is wrong on the fundamental application of the laws of thermodynamics.”
( http://skepticalscience.com/argument.php?p=1&t=51&&a=509#75803 )
The applicability to this discussion is that if the mirror is instead a greenhouse gas and the blackbody is the earth’s surface heated by the sun, the GHG will return approximately 50% of the IR that it absorbs back to the surface. Although the GHG is cooler than the surface, its presence will raise the temperature of the surface
Carrick and others,
The Sun is the source of energy to heat the Earth’s surface. For a given average absorbed solar energy, and assuming the average Earth (including atmosphere, ground, and oceans) temperature is constant (no storage), the heat transfer from Earth to space is CONSTANT. This means the net heat transfer (energy flux rate) from the surface up by radiation to space, radiation to absorbing gases, back radiation, conduction, convection, etc., is not dependent on the presence of greenhouse gases. This is so even in the presence of greenhouse gases results in a much warmer surface and atmosphere than without. This means that the larger the greenhouse effect, the more the convection has to transfer heat up. i.e., the smaller the net radiation heat transfer. In the extreme case of near total absorption of outgoing radiation near the surface, the altitude of outgoing radiation to space is raised, so the surface gets hotter. However, for that case radiation heat transfer goes to near zero. Thus for the case of the largest back radiation, the heat transfer is almost all due to conduction and convection up to a high enough altitude to radiate to space. Since the atmosphere is in LTE, the gas absorption and eventual radiation is just a sea of radiation doing nothing heat transfer wise, and saying the back radiation is heating the surface is just wrong. It is effectively just radiation insulation, but the convection prevents that from being any actual insulation effect on heat transfer. Back radiation is a result of, not cause of the greenhouse effect.
Leonard,
That’s the one. I may make it a headpost in a couple of days. My opinion is similar to your own except that I think the causality can be disputed to the same result.
We know energy from backradiation is added to the surface. Insulation does the same thing. We also know that heat flows the other way. LTE is a relative term though and the delay in flow of energy to space caused by the absorption-emission time increase can be integrated in a calculus like manner to a non-zero sum. Net radiative energy flows from ground to space. I therefore don’t agree that LTE means the radiation is doing nothing, just not very much for each dV.
#918 Standing waves do not count as either absorption or emission as far as S-B is concerned. That’s why Johnson emphasises that they only resonate and are not absorbed. So what you point out strengthens the argument for standing waves.
Discuss standing waves with Markus from now on please.
#919 If radiation from the surface varied then diffusion and evaporation compensate. They have to because of the reasons on my Explanation page.
#920 Jeff saidWe know energy from backradiation is added to the surface. Totally wrong and against the 2nd Law. Keep believing your pseudo-physics. I can see you’re non-negotiable.
Cheers
Doug,
I work in standard physics, sorry I’m not buying into your alternate reality. Every physical example we gave you of working instruments contradicts what you have now written. Your claims of second law violation again show that you cannot read or understand basic thermodynamics. You would fail a first year college course on the matter. You still haven’t even admitted that the bolometers or pyrometers violate your own universal conditions. Hell, every radio receiver on the planet does too but lets not go there.
At least we taught you a little about lasers and optics. Only god can teach you physics though I’m afraid.
920 Jeff – if it fits in with your post I’d be grateful if you can add some context re: diurnal and geographical location impacts. Re: the geographical impacts the type of thing I’m thinking is not only spherical geometry but the actual physical nature of the planet at the point the radiation arrives. For example, looking at the globe between the tropics shows a pretty high percentage of the surface is ocean. If you can also tie in comment re: lapse rate impacts on temp with altitude that would be appreciated. Apologies to hit you with a “shopping list” when I’ve not had time to bottom out the helpful links Carrick gave up thread – nor have I had time to follow recent discussions at Tallblokes where lapse rates have been discussed and possibly bottomed out. FWIW I saw a comment from Leonard at SoD on the potential temperatures thread which could be read that he is currently considering these issues.
Doug, the standing waves between the object and mirror correspond to vibrations of molecules matching (probabilistically) the frequencies in the spectrum of radiation of the object. That vibration in the molecules of the object is heat. There is no other way to resonate that I am aware of. More standing waves means more heat.
A physical analogy is if I put a loudspeaker on an open field and send out a few relatively low frequencies (which make it easier to set up standing waves in this case). Then I place sheets of solid material (that reflect sound waves) at distances from the speaker that are multiples of the frequencies coming from the speaker. In between the reflectors there are standing waves. Once I have enough reflectors I will destroy the speaker. How does the speaker get destroyed?
.
#924 More standing waves means more heat. No it doesn’t, because a standing wave is symmetrical at each end, so it would have to create heat at each end, which is of course impossible.
All the examples that you and Jeff cite are for machines (lasers, amplifiers, etc) of some type which involve energy input by electricity.
How about just keeping to the natural environment – like the surface and the atmosphere?
Jeff is still wrong. Everything has to be considered on the macro scale. That’s when the 2nd Law is meant to operate, and does wherever there is matter in the Universe.
Jeff said we know energy goes from the (cooler) atmosphere to the (warmer) surface. It can’t. And it’s no use saying net radiation and heat flow is out of the surface, because “net” involves other “transactions” at different times and locations. And it’s no use just telling someone they wouldn’t pass an exam, rather than presenting a cogent physics argument.
You too, Jeff, can always go and argue with Markus Fitzhenry who also talks about the standing waves which are the backradiation (and equivalent component of the upwelling radiation) both of which transfer no energy.
Nothing but a standing wave (which conveys information, but not energy) will slow the rate of radiation. But even then, it never adds thermal energy, not even momentarily. Only a standing wave can block other radiation of exactly the same frequency and intensity
A carbon dioxide molecule has no more effect than a water molecule when forming standing waves, so CO2 has less than 1% the effect of water vapour on climate, even if that is anything at all. It probably isn’t, because evaporation and diffusion will compensate. The massive amount of thermal energy held under the surface due to the slow rate of terrestrial heat loss, provides long term climate stability.
I have no intention of arguing a different person on your points. like even Claes Johnson, you will likely differ. Instead I would like you to explain this crazy point:
“All the examples that you and Jeff cite are for machines (lasers, amplifiers, etc) of some type which involve energy input by electricity. ”
How does the input affect the output photon such that it no longer needs to follow standard physics?
#926
Electricity can be used to generate radiation, such as broadcast radio waves. laser beams etc. The frequency distribution of such artificial radiation may not fit a Planck curve. The machine that generates the radiation does not have to be hot or cold, ie at an equivalent temperature to a natural blackbody or surface which would emit those frequencies. A laserbeam (having been produces by stimulated emission) can have totally different effects due in part to the fact that there are pairs of identical photons.
In some ways most artificial radiation can be similar to natural spontaneous radiation with the same frequency. Hence broadcast radio waves are also unlikely to warm the surface. After all, if they did, they would be quickly attenuated and not likely to reach across a whole continent as they often do. I suggest this simple observation proves my point anyway.
However you choose to imagine your mass-less photons, the radiation still has a frequency and the mode of the frequency distribution indicates the temperature of the source. The standing wave corresponds to the area under the Planck curve for the cooler boody (all of which is also under the Planck curve for the warmer one) and the energy transferred from hot to cold corresponds to the area between the curves – thus giving the same result as S-B Law calculations.
It’s all on the two pages in the Radiation section of my site and in more detail in my paper which is now being reviewed.
OK – some excerpts …but no more till it is published or released on my own website ….
But how can our adjacent plate seemingly slow the rate of cooling of the warmer plate, even though, as we have seen, it cannot actually transfer thermal energy to it? The answer appears to be that a standing wave is set up which effectively transmits information about the temperature of the cooler body to the warmer one. This standing wave has two-way energy transfer represented by the area under the Planck curve for the cooler body. When this wave resonates with the warmer source, it “blocks out” identical potential radiation from the warmer body, leaving only that energy represented by the extra area under the Planck curve for the warmer body. So only the radiation represented by this extra area causes any thermal energy transfer.
3. Only thermal energy itself can be compounded, not radiation
So we now have a mechanism whereby nature ensures the Second Law always applies, even when the Sun is already warming the surface every sunny morning.
We saw in Section 2 that radiation does not appear to compound in any way when it meets other radiation with different direction and wavelength. But it can have an effect on heat transfer rate when it forms a standing wave between a source and target. So it can affect the rate of heat loss by other radiation.
However, the real world is more complex and, if the heat loss by radiation between the surface and the atmosphere is reduced there is a possibility that extra heat loss will occur by the other processes such as evaporation, diffusion, convection etc.
********************************************
So, in straight forward cases the calculations are effectively working with the area between the two curves. Can any physical significance be placed upon this area between the curves if we are going to use just that area and consider only the radiation from the warmer source, disregarding the radiation from the cooler target?
It is indeed necessary to place a physical significance on such an area between the curves, if and only if it represents all the radiation from the warmer body which is actually absorbed by the cooler one. It must do just that, because that is the area which approaches zero when the temperatures approach each other. This area has a corresponding actual heat transfer, whereas the total areas under each curve do not.
Such an hypothesis requires the assumption that the portion of radiation from the warmer body which is represented by the area under the curve for the smaller one is all “rejected” by some physical process, and thus not converted to thermal energy. The end result is that it produces standing waves corresponding to that part of the incident radiation. This explains why absorptivity reduces to zero when the temperatures are equal and there is no remaining area between the curves. It also explains why there is no absorption when the source is cooler than the target.
**********************************************
[excerpt from Conclusions]
Radiation in the form of standing waves is totally different from incident solar radiation for which the frequency distribution does not overlap that of the surface. Solar radiation does not set up standing waves with the atmosphere or the surface because the frequencies are so different that they cannot resonate. In contrast, radiation from the atmosphere can only set up standing waves and none of it is converted to thermal energy in the surface. The only thing it can do is slow the radiative component of heat loss, but this may be rectified by other heat loss processes in order to maintain long-term equilibrium with sub-surface temperatures.
So Jeff. what is it that “we still know ..” about backradiation which is comprised of standing waves which do not transfer energy either way between their ends?
And yes, Jeff, I suppose if I sat an exam which was marked by a young examiner who had been brainwashed with AGW conjectures in his own education (something I escaped) then, when I defiantly answered with the truth (revealed by God if you like – perhaps you’re right) I guess I would be marked down due to his misconceptions. So might Einstein.
Doug, when you say a standing wave conveys information, specifically in what form? The idea that a standing wave can block other radiation of the same frequency (and intensity!) is at odds with other media. The standing wave in my example destroys the loudspeaker by imparting energy to it. The electricity that generated the original audio is not relevant, only the difference between the presence of standing waves and not having standing waves (the case without passive reflectors).
My quick review of your conclusion is that using words like “totally” should be avoided. Better to spell out the differences since there are also similarities.
It believe that “slowing the radiative component of heat loss” is one of the ways to better understand the phenomena of clouds (my example) and others given above.
#924 Eric – You’d better read about standing waves: http://en.wikipedia.org/wiki/Standing_waves and note there is no energy transfer. It’s a bit like what happens to the current in the cable in your house when you turn a power point off. How does current coming in from the street “know” to stop as soon as the power point down the other end of the cable is turned off? I suggest a standing wave “blocks” new current that wants to come down the line – just as a standing wave between hot and cold plates blocks some (but not all) of the energy wanting to be radiated from the hot one.
#931 Yes perhaps I could delete “totally” but I am not getting into discussing any system with an electric input anywhere. I choose to stick to what I have specialised in with my study, namely atmospheric physics. I’m the first to admit I’m not a specialist in optics, for example. But optics itself is a very specialised branch of radiation theory I would suggest, and not normally needed for atmospheric radiation studies. No physicist is a specialist in every branch of physics.
The “information” is basically its frequency spectrum. Consider radiation from a small “parcel” of air containing WV and over 50 gases. Within that radiation will be numerous separate frequencies making up the overall shape of a Planck distribution, but with some fine gaps. The Earth’s surface would have a more complete spectrum because of the greater number of elements. So, anything the atmosphere can dish up, the surface can match. So lots of separate resonance events take place, each forming a standing wave at that frequency until the limit of that frequency is reached. So, for all the area under the small Planck distribution for the cooler atmosphere there will be trillions of trillions (or whatever) of standing waves. These prevent the warmer body emitting in these frequencies up to the maximum for each frequency as dictated by the temperature of the cooler body. So in this way the information about the cooler body is sent to the warmer body (the Earth’s surface). via the standing wave.
So, in effect, we take out the radiation represented by the area under the curve for both the warm surface and the cooler atmosphere. All that just cancels out as standing waves not transferring thermal energy. Note I was careful to say “thermal.” You may well argue that an equal amount of radiated energy somehow goes each way, but that clearly could have no effect. The main point is that none of the radiated energy gets converted to thermal energy – ie no heat is transferred.
So, the only energy that is transferred is that in the warmer body which is represented by the area between the curves – ie the extra area in the Planck distribution for the warmer surface. ..
S-B Law calculations have always subtracted what they thought was radiation from the cooler body in the opposite direction from the total from the warmer body. Mathematically the result is the same as the area between the curves. But its physical significance is very different. there are not two large (but slightly different) amounts of heat transfer in opposite directions. Instead there is only heat flow in one direction and that is only the difference. You’ll remember Claes Johnson talked about two-way radiation but one way heat transfer.
Anyway, it won’t be long before I know if my paper will be published as initial interest has been shown and the draft knocked into shape. That will explain the basics, though this post goes into even more detail, so keep it as an addendum to the main paper which I’ll put on my website if it is not published somewhere in the next two or three months.
“is no energy transfer”
No NET energy transfer – if the sources are equal. Of course I’ve been waiting for doggie to realize what happens when one source has more output than the other. Ya know, when the emission curves are different or what happens when the polarization of the EM field has a different vector.
Wha???
Don’t hold your breath though, Doug is too old for math.
As you say Doug, the frequencies are discrete, there is a standing wave for each frequency. The presence of a particular standing wave for each frequency requires some tiny amount of atomic vibration in both ends of the wave at that frequency. Does the amount of vibration on both ends increase with the presence of the standing wave?
Does the presence of a standing wave for a particular frequency slow the radiative component of heat loss for that frequency by keeping a larger number of atoms in our black body vibrating at that particular frequency (over time)?
I really don’t have time to keep pointing you both to references like Wikipedia for which I provided a link in #932 and which you don’t seem to have read. I’ll spoon feed you with a copy and paste, where they are talking about standing waves which must of course have the same frequency …
it can arise in a stationary medium as a result of interference between two waves traveling in opposite directions. In the second case, for waves of equal amplitude traveling in opposing directions, there is on average no net propagation of energy.
In addition to the standing (or stationary) wave there is additional one way radiation from hot to cold and its frequencies are represented by the area between the Planck curves, which is the same as SBL effectively calculates by subtracting the area under the smaller (cooler) curve from the area under the larger (warmer) curve.
What is it that you don’t understand? Obviously it’s all different with electrically generated radiation such as lasers which don’t form standing waves because natural bodies can’t normally generate stimulated laser emission spontaneously, now can they?
Eric, you need to understand that when there is “vibration” between energy levels the energy required to excite = energy emitted on relaxation in the case of resonating standing waves.
So how could any extra energy appear from nowhere and get converted to thermal energy? A whole new and different process is required for that conversion. You seem to keep imagining physical vibration causing friction or something. It’s not like that. Energy cannot be created in the process of resonance associated with standing waves.
All radiation from the cooler atmosphere to the warmer surface comprises standing waves transferring no net energy either way. Only the additional “top portion” of the radiation from the warmer surface is separate radiation which does cause heat transfer from warm to cool.
I warned you at the outset that Claes Johnson’s Computational Blackbody Radiation is ground breaking physics extending the work of Einstein and Planck. You are not going to find it in textbooks, but that doesn’t mean it’s wrong. There is far more to it than just imaging a lot of identical mass-less photon particles crashing into surfaces and transferring thermal energy.
Any textbook which tells you that radiation between two plates transfers the full SBL amount in each direction is wrong, because there simply cannot be any transfer of thermal energy along a different path from cold to hot as it violates the Second Law. “Net” radiation has no corresponding physical entity and is thus meaningless.
Only standing waves have an identical path and can thus interfere with each other if they have equal frequency and amplitude, as explained in Wikipedia.
The Second Law applies to every individual path between two particular points. Standing waves may be considered as two opposing waves, but they do of course have the same path, and that makes all the difference. It’s up to you whether you want to take an interest in these new developments in physics or stick to your old beliefs so you can feel good running web forums like this and practising what you preach by driving around in little cars with solar panels or whatever.
“feel good web forums like this practising what you preach by driving around in little cars with solar panels or whatever.”
You really don’t know this blog do you?
Let me help you “feel good” Doug.
“I really don’t have time to keep pointing you both to references like Wikipedia for which I provided a link in #932 and which you don’t seem to have read. I’ll spoon feed you with a copy and paste, where they are talking about standing waves which must of course have the same frequency …
it can arise in a stationary medium as a result of interference between two waves traveling in opposite directions. In the second case, for waves of equal amplitude traveling in opposing directions, there is on average no net propagation of energy.
”
All of this is correct.
No argument at all.
What you are ignoring is that as soon as one of the two bodies is slightly warmer, the balance is upset and ‘net’ transfer begins as the difference of area between the curves. Your hero Claes johnson is proven wrong by this statement you endorse. From your own quote -“represented by the area between the Planck curves, ” The slightly warmer body emits almost the exact same frequency range yet the amplitude at each frequency goes up. Are you giving up on your silly ‘cutoff frequency’ now? It certainly cannot make sense that both a cutoff frequency and the area under the curve are true. In fact, were you sharp enough to notice this as I did when you first proposed the cutoff, you wouold have realized that ALL of the net energy transfer rates are changed between bodies by an arbitrary cutoff. It would take minimal math to prove this concept wrong but we can’t begin discussing math at your level.
If the cooler body planck curve overlaps a much warmer body planck curve, is there something different about those photons which prevents them from being absorbed? Doug’s photonic stink property again?
I assure you that this situation where much a warmer body output overlaps the curve of the cooler one at a higher amplitude. You have just admitted two slightly different temp bodies absorb when amplitude of one exceeds the other. Now you have the problem of defining when the difference in planck area doesn’t apply.
uh oh!!
I have a lot more uh oh’s for your new whacko theory as well. Of course they are based on silly things like observation and science. Sorry about that.
Then we can begin to discuss the concept of NET vs actual.
Eric is right Doug.. Read it carefully.
Doug, when you say “Eric, you need to understand that when there is “vibration” between energy levels the energy required to excite = energy emitted on relaxation in the case of resonating standing waves.”
This is my understanding; Conversion of thermal energy (random continuous motions of atoms and particles in a body) to radiation is a probabilistic process of conversion of a quantum of that thermal energy into a photon. The presence of a standing wave stops the process of thermal energy loss for that (atom or similarly small area) because another photon or quantum of energy always replaces the one just lost. Without the standing wave, the photon is lost and not replaced. Hence the removal of a mirror or another blackbody with the same temperature results in more cooling or the addition of a mirror or nearby blackbody will result in less cooling.
This link seemed to be a reasonable explanation:
http://Galileo.phys.Virginia.EDU/classes/252/black_body_radiation.html
Eric,
:”The presence of a standing wave stops the process of thermal energy loss for that:
Consider first that for a ‘standing wave’ to exist the em fields of the resonant electrons need identical polarization. A random occurrence with low probability. Then you have a surface atom with vibrationally energized electrons. Fodder for a neighbor atom bumping into it. Any bumping = thermal conversion = nope.
There is nothing magic about a standing wave, except that I think it what we noisy blogging baryon clumps are formed of. Shake the surface of anything, standing wave or not, and you make warm. End of story.
I was rather hoping you would come to that same point here because Doug was slowly stepping into this other “reality trap”.
Jeff, I agree that standing waves are a random occurrence and that they only last short periods of time, etc. But standing waves will occur in some quantity (that I can’t identify) between two blackbodies. I would also argue that more standing waves will occur as two blackbodies get closer in temperature. Those will have the effect that I indicated, but probably be greatly exceeded by waves between atoms of unmatched vibrations (although Doug will probably disagree with that).
That does not even consider wave interference, superposition, etc I think part of the interest in standing waves comes from the cavity and pinhole model of a blackbody. But my understanding is that is just a model and doesn’t apply to the external interactions we are talking about.
Eric,
Just keep in mind that anything which vibrates the electric field of an atom creates heat. There isn’t any choice in the matter – some pun intended. I was rather expecting you to corner Doug on this for several comments.
Claes explains “cut-off” with a proviso that the distributions are strongly attenuated. No, it’s not a “perfect” description I agree, but it is a reasonable way of explaining it to the public at large. When you see my paper you’ll see that my only reference to Claes is a two line footnote noting that he also spoke about resonance. That is the key point I take from him and (along with Markus Fitzhenry and some others) I consider the resonance must cause standing waves. Much radiation is not polarised, so I see no major problem there. Even if polarised the “reflected” component of the standing wave would pick up the polarisation.
If radiation meets a target and can neither resonate nor be absorbed, it probably continues elsewhere looking for a better match to marry up with and form a standing wave. I imagine molecules get engaged in a standing wave, so the next ray of radiation passes on if it is notfrom the area between the curves. Of course other events like a physical collision with another molecule may breakup the standing wave, like a divorce, so I guess its lifetime is similar to the time between those collisions that involve energy transfer, like divorces that cause money transfer.
You are saying what I’m saying about the extra area between the Planck curves. I know that represents the energy transferred.
But you argue that there are exceptions when you say If the cooler body planck curve overlaps a much warmer body planck curve, is there something different about those photons which prevents them from being absorbed? Doug’s photonic stink property again? Nothing is different – that’s why they resonate because each photon has the same frequency and thus the same energy. The warmer body just has a surplus of warmer (higher frequency) photons. I could just as easily refer to “radiation” rather than “photons” here. (The difficulty with the photon concept is that they apply a frequency to the photon, but how do you imagine its corresponding wavelength?) There are more of the lower frequency ones as well as some higher frequency ones which have no counterpart in the emission from the cooler body.
I will just repeat what I have been saying all along: Before any of the radiation from a cooler body can be converted to thermal energy in the target, there has already been established a standing wave resonating between the source and target in a mutual two-way handshake that is equal in all respects in each direction and which does not transfer any thermal energy from one to the other.
Take a closer look when the curve is plotted against frequency rather than wavelength – and you will see that the cooler body’s distribution is always fully contained within that for the warmer body. http://climate-change-theory.com/freqdist.jpg
When one body gets slightly warmer the mode (peak) of the frequency plot will always be to the right and greater than that for the cooler body because peak frequency is proportional to absolute temperature. That’s established by Wien’s Displacement Law..
Standing waves need not last long in terms of whole seconds – their life is probably far shorter. But I suggest new ones are formed if one is broken.
Any warm body can of course radiate towards several different bodies at different cooler temperatures. The radiation going to the coldest one will have a greater surplus with which to warm it. If we were just talking about metal plates suspended in a vacuum in a room they would of course all end up at the same temperature in an equilibrium state, and then there would be only standing waves. On a micro scale in the real world there could be very small differences in temperature which would then be corrected by heat transfer, and of course this happens all the time between things on the Earth’s surface. But the adiabatic lapse rate comes into play to establish a temperature gradient in the atmosphere which is related mostly to the acceleration. due to gravity. The temperature at the surface end of the atmospheric trend is, in the very long term (tens of thousands of years at least) tightly controlled by the massive amount of thermal energy in the core, mantle and inner crust which we know does not vary much because the terrestrial heat flow is so relatively low.
“Take a closer look when the curve is plotted against frequency rather than wavelength – and you will see that the cooler body’s distribution is always fully contained within that for the warmer body. http://climate-change-theory.com/freqdist.jpg”
Exactly!!
There is always an overlap. So how can a sensor EVER detect a cooler object in your VERY MUCH NEW resonant wave theory?
Answer: It cannot.
Solution: Another theory broken.
A million observations cannot prove something right, yet a single observation can prove a theory wrong.
Stick a fork in that one too.
What’s next?
If a sensor can detect a reasonable “section” of the right tail of the Planck distibution then computations could be applied to determine the mode (peak) of the distribution even if it cannot be detected directly.
Normally it would be possible to determine the peak frequency anyway if the sensor is just reacting to the electric or magnetic field in the radiation and does not depend upon being physically warmed itself. IR thermometers which are not microbolometers use this principle of determining temperature from peak frequency.
Wien’s Displacement Law also enables calculation of the maximum frequency and this is proportional to the peak frequency being about 2.898 times the peak. So some radiation from a body at 100 K can get up to the peak frequency emitted by a body at 289.8K.
Hence, it is not surprising that some radiation from a cold source can be detected and have an effect on the rate of heat loss by the sensor. But it is not warming the sensor. So then it is just a matter of calibrating the instrument using known temperatures of other sources.
You all need to understand that there is a huge difference between adding thermal energy to the surface from the atmosphere (ie the impossible) and just slowing the radiative component of the heat loss to the atmosphere which does happen purely because the atmosphere is not at absolute zero temperature. A metal plate would cool faster in space for example than anywhere on the surface or in the atmosphere.
But, when only the radiative component of heat loss is reduced (by an infinitesimal amount by extra carbon dioxide) then the heat loss by diffusion and evaporation will compensate by increasing. So there is no net effect on climate.
What this means is that carbon dioxide molecules cannot have any more effect than water vapour molecules. Each supports standing waves in the same way. So CO2 is not capturing photons at a much greater rate (as assumed) than WV – hence its effect is less than 0.5% that of water vapour, based on relative numbers of molecules in the atmosphere.
Doug said “You all need to understand that there is a huge difference between adding thermal energy to the surface from the atmosphere (ie the impossible) and just slowing the radiative component of the heat loss to the atmosphere which does happen purely because the atmosphere is not at absolute zero temperature.”
I understand that. It is also true that the slowing of that heat loss depends on the temperature of the atmosphere as viewed from the surface. The radiative component of heat loss is greatest during radiational cooling nights. I can point my IR thermometer at the sky and get -30F or less reading. Contrast to last night, my IR thermometer said +38F apparent temperature (of the sky) while a conventional thermometer said the air temperature was 45F.
Then Doug said “But, when only the radiative component of heat loss is reduced (by an infinitesimal amount by extra carbon dioxide) then the heat loss by diffusion and evaporation will compensate by increasing. So there is no net effect on climate.”
That may be true. Note my IR thermometer readings showing -35F the night before last (clear and dry), +38F last night in the light rain, -20F this morning (clear but very damp). Obviously the added effect of CO2 against that type of variation is trivial. The radiative cooling I experience is 99%+ determined by the clouds and water vapor in my location.
BTW, I should mention all those readings are at night (no sun). When I say “morning” I mean 4AM when I get up.
Eric
I do consider that +38F apparent temperature of the sky at night seemingly somewhat high. We know atmospheric temperatures average well below freezing point, so you must be reading warm air globules at very low altitudes. Does your IR thermometer have a setting for emissivity and, if so, what did you set it on and why? Did you try several different angles?
To me this seems to show just how variable and inaccurate IR thermometers can be when measuring atmospheric temperatures when you don’t know what it is detecting or what the emissivity should be. I’d also like to know if it is a microbolometer. See #944 where I trust I have explained satisfactorarily for Jeff how these things measure rates of heat loss rather than heat gain. In practice the variations are so small that you wouldn’t know what it is measuring, but physics says it must be the rate of heat loss. This rate of heat loss will vary and increase as the temperature of the surrounds decreases, all the way down to absolute zero in fact. That’s why a true blackbody has to be in space.
Doug, unfortunately my thermometer makes gross assumptions about emissivity (that the object being measured has relatively high emissivity) and it is not adjustable. The only thing I can say reliably is that there is a large difference between the clear sky (-35F) and the rainy sky (+38F with air temp of +45F). It is also possible that some rain hit the sensor, but subsequent readings were ok elsewhere against indoor and outdoor surfaces. Also you are right, they are not very accurate even when pointed at an object with high emissivity (+/- 1-2F). I am only using it for gross readings, not a scientific study.
I’m not sure how my model works, but I will find out.
Doug,
“If a sensor can detect a reasonable “section” of the right tail of the Planck distibution then computations could be applied to determine the mode (peak) of the distribution even if it cannot be detected directly. ”
No, no, no. you can’t have it both ways. Is it the ‘area under the curve’ or the tail on the right? Come on, Doug, you are contradicting your own crazy theory.
#951 Try reading the next bit of that post which completely rubbishes your concept that a bolometer proves my hypothesis wrong. Any “body” (ie small globule of air) at any temperature above absolute zero will slow the rate of radiative cooling of the Earth’s surface, as well as the sensors. But it will not warm either. Now continue with second para of #949 and then note the ridiculous fact in #950 that emissivity is assumed close to 1.0 when, for the atmosphere, it could well be less than 0.1.
The assumption about high emissivity by my IR “thermometer” means that the sky temperature reading will be biased low since the thermometer must do a crude estimate of temperature from measured IR. But the relative measurements should be acceptable (comparing various sky measurements). A higher apparent temperature with low clouds and rain makes sense.
DJ,
You do recognize that you have changed your theory several times since this thread started right? If the bolometer’s temperature changed exactly equal to the difference in the area under the Planck curves (as the rest of us expect it to), what happened to the insane cutoff frequency? Can we now say that is crap and move on?
I recognize no assumption by Eric, he just used the instrument.
#953 It’s not reading the radiation itself. It reads frequency and converts that to temperature using Wien’s Displacement Law. (See Johnson’s paper)
Then, the AGW guys use high emissivity to calculate that the radiation is higher than it is.
#954 No the rate of temperature change of the sensor doesn’t relate at all closely to the full area under the Planck curve unless it were floating in space, in the shadow of the Earth. If it receives radiation from any “body” that is above absolute zero then its cooling rate will relate to the area between the Planck curves for itself and that body. Surely you know this basic physics.
I have been quite consistent in telling youi that there can be no radiated transfer of thermal energy from any point A to any other point B if the temperature of B is greater than that of A. The cut-off is quite precise when the temperatures are equal. Likewise the area between the curves approaches zero as the temperatures approach. So clearly the area between the curves is more likely to represent the heat transfer rather than the area under the whole curve.
Wait now for my article Radiated Energy and the Second Law of Thermodynamics,
#954 – Look, I know you said the “difference in the areas under the Planck curves” and i know that’s the same as the area between. My point is that physically it does not involve two-way thermal energy “swaps” represented by the total areas under the curves. The only physical thermal energy transfer is represented by the area between the curves. The rest is a standing wave.
Becuase of this, carbon dioxide molecules can make no greater contribution than water vapour molecules. And in fact water vapour molecules can support far more standing waves than CO2 ones.
So that puits CO2 in its place.
#957,
Actually if you use the area between the curves, you come to what is en-bulk mathematically identical situation to the one we have in reality. The Johnson cutoff is crap. What’s more is that the two are mutually exclusive which from what I recall, was Johnsons point.
If you go with the ‘area under the curve’ you will put CO2 exactly where the IPCC does. Of course they overstate the problem a little.
I said; The only physical thermal energy transfer is represented by the area between the curves.
You said; if you use the area between the curves, you come to … the one we have in reality
Are we in agreement on the above, then?
Johnson covers himself by pointing out that the Planck curves are “strongly attenuated.” So he is working with the peak frequencies only and these are proportional to temperature. Hence, when they approach each other we get (in the limit) no heat transfer.
The main thing is that he correctly shows that, on a macro scale, radiation from cold to hot merely resonates and is not converted to thermal energy. I suggest that the “resonating” must be forming standing waves which can not transfer any energy from one end to the other. Thus pairs of points in the atmosphere and surface are temporarily linked by a standing wave. This wave is the information system that tells the warmer body to cool more slowly as the temperature difference reduces. The “cooling more slowly” cannot be the result of two lots of thermal energy transfer in roughly opposite directions (but not usually along precisely the same path) because the cold to hot violates the Second Law because it is an independent transaction. Radiation can only interfere with other radiation in a significant way when it is along the same path and has the same frequency and amplitude. That’s what my article is all about and it may be published online within a week or so following peer review.
.
Consider a metal plate enclosed on one side with a “perfect” insulator. It is dangling out of a satellite and collecting the full blast of the Sun on its uninsulated side. Say it is between the Sun and Earth and its plane is perpendicular to the line between Sun and Earth. Assume it is not affected by radiation from the satellite.
Say the Sun warms it to 330K at equilibrium. Now pull it into the shade of the satellite and turn it 90 degrees so it faces space and not the Earth or satellite. It will radiate virtually all the amount represented by all the area under its Planck curve, like a blackbody. So it “wants” to do this. You time how long it takes to cool to, say, 200K
Now repeat the experiment, but when it is in the shade of the satellite face it towards the Earth.
We know the radiation from the Earth has the same flux (close enough) to that from the Sun which it receives. So will it stay at 80 deg.C because of this? I suggest the IPCC’s energy diagram concepts might suggest that it would in fact do so, because they imply that the temperature would be a function of the number of photons received..
No. It will cool more slowly than before, but how does it “know” to do so? It still “wants” to cool faster. The slower cooling cannot be caused by a transfer of thermal energy from the cooler Earth system. The only other way is for radiation from the Earth to interfere with its own radiation, and this can only happen with a standing wave.
However, it will stop cooling when in equilibrium with the Earth.
But only some of the Earth’s radiation will have any effect. Its rate of cooling will be slowed down and gradually come to a halt at whatever is the weighted mean temperature of the Earth and atmosphere – perhaps around 255K.
So at equilibrium the Earth’s radiation is split into two “sections” – some which is below its mean temperature and thus has no effect because it forms standing waves with the plate at 255K, and some above that 255K “cut-off” which is maintaining the plate’s temperature also at 255K.
It was the UV, visible and maybe some near-IR in the Solar radiation that was able to heat it to 330K.
“I said; The only physical thermal energy transfer is represented by the area between the curves.
You said; if you use the area between the curves, you come to … the one we have in reality
Are we in agreement on the above, then?”
It depends on your personal definition of ‘thermal energy’. I can state that energy absolutely changes hands in both directions and the net works out to the area under the curve. If you can say that the differential area under the curve matches your understanding, we can agree on that and work on the rest.
Claes would NOT agree with this.
Doug, I have this thermometer: http://microtempusa.com/mtpro.php The web site tells me nothing about it. From other sites the construction of my device might be a thermopile in series with a reference resistor with the voltage between the two fed into an amplifier. A document http://www.diva-portal.org/smash/get/diva2:14337/FULLTEXT01 indicates that my cheaper design is not a bolometer, but does thermalize incoming photons which measures of flux magnitude in a particular frequency range which could be narrow or wide.
I’m guessing the that the current through the thermopile induces a fairly precise amount of heating and the incoming photons from the measured body cause the thermopile to cool more slowly. The amount of cooling changes the resistance, then they measure the voltage. Probably a lot more complicated, but the fundamental principle is variation of resistance by temperature.
Eric,
The first time I saw a pyrometer, I was amazed with it. The PhD I worked for at the time explained about emissivity corrections and that the number was not to be trusted on surfaces you didn’t know the correction factor for. I remember asking, is the difference it really that much?
This was way before I had figured out why of course.
Doug has gone back to school on this thread and he doesn’t know his prof’s yet. You have an odd style, but you are too often correct for it to be accidental.
#961 This definition will do me: : http://www.ifpaenergyconference.com/Thermal-Energy.html
As in the second paper linked in #962, it is appropriate to talk about the energy in radiation being “thermalised.” In general, thermal energy has to be contained in matter, not radiation. That’s why I used the title “Radiated Energy …” to avoid ambiguity. There has to be thermalisation in order to observe heat transfer. I say that does not happen with the rays of radiation from cold to hot. And each ray is an independent “transaction.” You cannot excuse an assumed violation of the Second Law on the grounds of what happens with another ray. The only energy that gets thermalised is that corresponding to the area between the curves and going from hot to cold.
#963 Some say the emissivity of the atmosphere is very low. This makes sense if only WV and trace elements have to do all the emitting. These molecules only make up ~2% so maybe emissivity could be as low as 0.02, but don’t quote that. In a globule of air the O2 and N2 will reduce the mean emissivity dramatically. We know the temperatures at various altitudes, so work out maximum S-B flux and then multiply by very low emissivity and you don’t get anything like 100% of mean incident solar radiation in the backradiation.
#962: Thanks for those links, Eric. Although the Introduction for the second one talks about thermalising the radiation, that may only be half true – ie it is true when the object’s temperature is higher than the sensor’s.
When the object is cooler than the sensor they wouldn’t know the difference if some sensors just cooled at different rates. it would be easy to assume the warmer ones had been warmed by thermalising the radiation, but I suggest that it is really a case of the rate of cooling of the sensor being slowed down more when the cooler object is warmer than other ones that are yet cooler. It’s all relative and an IR camera will also detect relative differences in temperatures in order to form an image.
Jeff, I’m a middle-aged EE, so I can’t be expected to remember all my physics. But I aced circuit theory and graded it and remember some of the possible mistakes. The measurement concept is pretty simple, a semiconductor produces a lower resistance with higher temperature because as the temperature rises there are more free electrons. From the curves shown 4.9 in the thesis it is clear from those that low temperature measurements on the flatter part of the curve will have much less accuracy than room and high temperature measurements on the steep parts of the curves. The varying current through the semiconductor can be measured as voltage across the precision resistor.
Doug, when the device is turned on, the current through the semiconductor warms it, and the circuitry obtains a reading. It then takes a series of readings and from the baseline reading it accurately calculates the change in temperature of the semiconductor. If the semiconductor warms faster than expected it is due to more photons being absorbed. Slower than expected warming means fewer photons were absorbed. The fact that I can get different readings from the clear sky (-35F last night) and cloudy sky (from -20F up to 38F with the rain and fog) means that the device is measuring different numbers of absorbed photons all of which originated from colder objects (there were no warmer objects above the sensor).
There are complications that I am aware of and not aware of. One problem with measuring the falling rain is that it might reflect ground radiation back at the device. But in this case the ground at about 45F was also colder than the device.
Eric
Although the semiconductor gets hotter more quickly it is not because photons are being absorbed from a cooler source. It is because, even though it is warming due to electrical input, it is simultaneously radiating in a cooling process which is not as fast as the warming. It is that rate of cooling which is affecting by the standing wave. So the net rate of warming is also affected. Would it take longer to boil a pot of water in the open air in Antarctica than it would in your 23 deg.C kitchen with the same equipment? Yes, because there is a faster cooling rate.
If one body is cooler than the other you can ignore its radiation to the warmer one because all that happens is that a standing wave is set up corresponding to the amount of radiation the cooler one could emit. You then deduct this amount which the cooler one could emit from that which the warmer one could emit. The difference represents that portion of the warmer one’s radiation which is absorbed and converted to thermal energy in the cooler one.
That’s what actually happens physically. I know you can get the same result by assuming radiation from the cooler one is absorbed in the warmer one, but it isn’t and the warmer one’s absorptivity for radiation from the cooler one is zero.
This is why radiation from a cooler atmosphere cannot transfer thermal energy to a warmer surface. It can only slow the rate of radiative cooling by forming standing waves which do not transfer any energy between the atmosphere and the surface. Other cooling processes like evaporation and diffusion followed by convection will increase to compensate. Carbon dioxide molecules can play no greater role than water vapour molecules. Hence their effect in regard to any greenhouse conjecture would be less than 4% of the effect of water vapour. Work out the consequences for yourself.
See my paper which I expect to be published within the next 4 days. It will also be translated into German.
#966 Eric – You could always collect some of the rain (with a funnel) and measure its temperature with a normal thermometer and compare with the IR reading. I’d say they would be fairly close.
BTW February was slightly cooler than January and was 0.12 deg.C below the mean since 1979
Spencer’s 13 month mean is almost back to the peak in 1988.
http://www.drroyspencer.com/
Doug, isn’t a standing wave due to the resonance between two equal and opposite waves from either end? With a wave from only one end, there will be no standing wave. It seems to me that Jeff is correct above, there are standing waves in some quantity but they are not of any relevance. When they occur, it is by happenstance that an identical frequency photon is emitted simultaneously on both ends. That will be more likely as the two bodies are closer in temperature, but still rather rare. Also there is no single “standing wave” between the objects, there is only a number of transient standing waves between atoms in the one object with atoms in the other object. Importantly there are many photon interactions that do not result in standing waves even when the objects are at the same temperature. That is rather easy to see in practice. In theory a “perfect” chunk of matter would be made up of identical atoms all at the same energy level and standing waves would be ubiquitous. In reality the the imperfections in matter will produce variations in energy levels and a photon impacting from the other end would not match the energy level and not resonate and not produce a standing wave.
I found a simple explanation of why the second law can appear to be violated for a single photon interaction, but still be satisfied for the bodies here: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC406177/pdf/plntphys00467-0011.pdf As the author says at the end, the individual photon interactions are limited by the conservation of energy and conservation of momentum.
#969 Who said the standing wave was from only one end? I quite agree that backradiation exists and is the second “half” of the standing wave, while the original radiation from the surface to the atmosphere is the first half.
When the temperatures are close standing waves predominate because very little energy is transferred (as per the area between the curves) only from the warmer to the cooler.
In the macro state, there can never be any spontaneous transfer of thermal energy from a cooler source to a warmer target, either by conduction or radiation. That’s what the Second Law dictates, because otherwise entropy would decrease. http://simple.wikipedia.org/wiki/Second_law_of_thermodynamics
You’d better wait till you can study my 10 page paper when it is published this week hopefully.
#969 Regarding your linked paper: Entropy will always increase in any spontaneous process, but entropy itself can only be observed in a macro (ie many molecule) state. It’s all about probabilities which result in more molecules relaxing to a lower state than getting excited to a higher state – so that (net) entropy increases (because it may be thought of as a mean) and there is spontaneous emission as an object cools..
Why do you suppose Harvard makes a point of saying the gas which is aborbing is cool and the blackbody hot? Experiments show that when the gas is hotter than the blackbody there are no absorption lines.
https://www.cfa.harvard.edu/~jbattat/a35/cont_abs_em.html
Just as Claes Johnson said.
And again …
“We usually see absorption spectra from regions in space where a cooler gas lies between us and a hotter source.”
from http://coolcosmos.ipac.caltech.edu/cosmic_classroom/ir_tutorial/spec.html
I don’t believe they and Harvard would each make a point of the fact that the gas is cooler unless they knew there is no absorption when the gas is warmer than the source. Hence ..
(1) The gas can “distinguish” between when the source is hotter or colder than itself. (Johnson’s “cut-off” frequency.)
(2) In the case of a gas the absorption stops when the source is no longer warmer than the gas.
(3) Whatever process causes the gas not to absorb could just as easily apply to liquids and solids
(4) Radiation emanating from a cooler part of the atmosphere and heading towards the surface would not be absorbed by a warmer layer of the atmosphere, even if it struck water vapour or carbon dioxide molecules identical to that which emitted it.
(5) Even if this only applied to gases, we thus know that radiated heat can only transfer to cooler areas of the atmosphere and so heat transfer is generally upwards except in the stratosphere.
There are two broad classes of radiation from the atmosphere. ..
(1) That which results from spontaneous emission after a molecule has ended up in a warmer (maybe excited) state due to radiative absorption or from molecular collision. Any of this radiation which heads to warmer targets (lower atmosphere or surface) will by the first half of a standing wave and the second half comes back to it from the warmer target it hit.
(2) That which emanates from the surface. This heads into the atmosphere and strikes cooler targets at various altitudes. Some of its energy thermalizes, and some just forms the first half of a standing wave which comes back to it. The cooler the target was, the more it will be warmed and the less will be the effect of the standing wave on the rate of cooling of the surface. So molecules closer to the surface have more effect on the rate of cooling, but each CO2 molecule is no more effective than each water vapour molecule.
[In contrast, the IPCC models assume CO2 sends far more photons back per molecule than water, so they claim the overall effect of only about 4% as much CO2 as WV is that each contributes similar amounts in total. But the concept of standing waves puts each molecule on an equal footing, so CO2 has nowhere near the effect of WV.
Doug, the point of my link in 969 is that the second law (just like entropy) applies to the body as a whole, not individual photon interactions.
Regarding your two links to emission and absorption spectra, suppose in their diagrams they show the spectrum of a black body at some temperature (continuous) versus the same emissions passing through a gas at the same temperature. The gas would absorb parts of that spectrum at its resonant frequencies but also emit the same magnitude in those same resonant frequencies. So the two spectral graphs, the BB emission direct and the BB emission passing through gas at the the same temperature as the BB, would be identical. Then there would be nothing to show.
My question for you is, given a spectrum of emission from a BB at some temperature T, what would that spectrum look like if it passed through a cloud of gas at the same temperature T as the BB. Would the two look identical (my answer)? If not identical, what would be the difference?
#975 When DeWitt Payne wrote about this (SoD Backradiation Part 3) he said the gas only started to absorb when the emitter started to get warmer than the gas. When temperatures are equal there is still no absorption. It seems all this is common knowledge among those working with spectroscopy, yet the IPCC appears ignorant and probably assumes backradiation warms the lower atmosphere as well. If gases can act this way, I suggest Claes is right in saying any blackbody can do so also.
Since then I read what Markus Fitzhenry said about standing waves, and I am convinced this is what must happen, such standing waves being between individual molecules in the gas and molecules in the emitter. These rays of standing waves will have all the frequencies that are common to the gas and emitter and the total flux at any frequency is limited by the Planck curve. In fact, they happen to some of the radiation from the warmer body, and always to all of the radiation from the cooler body. So, whatever the cooler body can emit goes into standing waves with the warmer body, thus leaving the difference above that to be absorbed – ie that corresponding to the area between the Planck curves.
This is why each CO2 molecule probably has less effect that each WV molecule in slowing down radiative cooling of the surface. Considering that there’s only about 4% CO2 v. WV, and if each CO2 has fewer frequencies resonating in standing waves, then its total effect could be way under 1% that of WV. And even if WV slows the rate of cooling, evaporation and diffusion will increase to compensate.
The Second Law applies to entropy, and thus to the transfer of thermal energy on a macro scale between any matter. The word “macro” here refers to anything from (I guess) a few hundred or thousand molecules upwards because with such numbers the probabilities “average out.” You could also think of it requiring a few hundred or thousand photons for the same reason. But we are talking about climate after all.
I see no point in discussion what individual molecules or photons do. The Second Law recognises that, due to the probabilities involved, in the limit entropy always increases. Whenever you measure temperatures you are taking an average of the contributions of many millions of molecules in the local region.
#975 The warmer gas does not absorb and then re-emit. It forms a standing wave which runs along the same path as the incident ray and can thus interfere with it before any absorption takes place.
This has to be the case. Consider a ray of backradiation penetrating a small distance into the ocean. If it were to get converted to thermal energy by absorption, the energy will not be re-emitted straight away via radiation from a few cm/inches under the surface of the ocean. Instead, the warmer water will rise to the surface by convection and at least some of it will evaporate and only some be radiated. There would have been a “completed (macro) transaction” in which that.radiation (from lots of molecules in a cooler atmosphere) warmed (a layer of) water below the surface of the ocean. This clearly would violate the 2nd Law.
Now do you see the difference between this and a standing wave? If there is conversion to thermal energy then there has to have been a violation of the Second Law.
Doug, at the photon and molecular level, the processes are somewhat unknown because the structure of matter is unknown. It is possible that the emission of a photon from an object always coincides with the emission of a photon from a distant object of the same temperature to form a standing wave in all cases. They work out the problems of causality using some equations (which I find hard to grasp). Otherwise I find it hard to imagine that the photon emitted from one object can always be matched by a photon emitted from the other, which is the basic condition for a standing wave.
In the case of backradiation heating the ocean, the ocean’s surface evaporates which extracts heat from the air above it and the cooling from this process is partly offset by the warming from the backradiation. Unless we dive down to the molecular level (which you and I both don’t want to do), the macro level suffices to show that the energy from backradiation causes the top of the ocean (including the surface layer and absorption layer below the surface) to cool more slowly than it would have otherwise.
Eric (and Jeff if you’re there),
You need to think more in terms of waves and vibrations which resonate. Think of kids making waves holding both ends of a rope. Their arms going up and down are like the electrons going up and down between energy states and in resonance. The frequencies coming in match the frequencies that the other body could have emitted. The vibrations at the ends synchronise because an initial condition must be that there is an integral number of wavelengths between them. That’s why I say they only last a very short time, but new waves will form. Many attempts at forming waves will fail I guess until the radiation continues on and finds a molecule it can dance with at the right distance to be an integral number of wavelengths..
You might well argue that the energy may not go back along the same path and that the process might be more like 100% scattered (diffuse) reflection. That actually makes no difference, provided you don’t assume any energy is transferred either way and that the rate of outgoing radiation from the warmer body is still reduced.
But Eric, you fell for the trap regarding the backradiation entering the oceans. The backradiation does not enter because it will be a standing wave ending at the surface where it started. There can be no warming of any water as that violates the Second Law, which should be obvious because of the time lag for the convection towards the surface after the assumed warming.
Doug, as many attempts to form standing waves fail, what happens to the energy transferred in those attempts (in both directions)? Also as the vibrations at each end synchronize won’t they just as easily desynchronize? For starters, that would be caused by random molecular motions on each end.
I see your point about the time lag except that the evaporative cooling is always ongoing elsewhere. A parcel of water heated by backradiation can appear to violate the second law while the second law is maintained for the ocean and atmosphere as a whole. IOW the surface and subsurface (and thus the lag) can be merged into one body for purposes of calculating energy and entropy flow.
A similar situation exists when a random bunch of photons (some nontrivial number) coming from a cooler object happen to warm a small portion of a warmer object. That will always be the case in real world objects with imperfections. As that apparent violation happens, there are other offsetting occurrences so that net energy transfer is from warmer to colder.
Tried the IR thermometer outside tonight with some cumulus clouds, but otherwise dry and chilly (31F air temperature). I got readings from down to -68F without clouds and from -40 to -10F with 1/4 to 1/2 clouds at roughly 9000 feet. It appears to have very poor accuracy at that type of measurement. I tried the fridge (34F) and the freezer (5-10F). I don’t see any other plausible (i.e. inexpensive) way this device works other than to absorb photons and measure a temperature change.
Eric
while the second law is maintained for the ocean and atmosphere as a whole
It can’t work this way. If you took this to an extreme you could say it balanced out on the other side of the World – or even somewhere in space. The Second Law always applies on a point to point basis in the macro scale (macro being as distinct from molecular level)
I’m sorry my paper has been delayed till later in the week, but at the very end I say ….
The only other choice is to assume the “rejected” radiation undergoes a process similar to reflection, but which (unlike reflection) does still block other radiation of the same frequency. Either concept debunks the anthropogenic global warming conjecture all together.
I am never emphatic about the standing wave concept, but whatever happens, the radiation is not converted to thermal energy. I still believe standing waves happen, but if one can’t form with the first molecule the radiation just continues till it finds one it can resonate with. That’s why some gets through the atmospheric window all the way to space. It must hit molecules en route.
No, you won’t get a small portion that you are able to observe getting warmed. The 2nd Law works probably even for just 1000 molecules, which you could never observe.
.
There is an Appendix to my paper with seven FAQ’s. Here’s a copy of the two that relate to devices etc.
Q.4 How can a microbolometer (IR thermometer) measure cooler temperatures?
The instrument has sensors which are warmed by electric current whilst they are simultaneously radiating energy towards the cooler object whose temperature is being measured. So there are two contributions to the warming of the sensors, one positive and one negative. (It is not a case of two positives, even though the manufacturers might say this.) As explained, the negative rate will vary in accord with the temperature of the object, and so the overall rate of warming will be reduced by an amount which depends on that temperature. Note that the original IR thermometers effectively measured the frequency of the radiation and deduced the temperature using Wien’s Displacement Law.
Q.5 Do holograms, lasers or microwave ovens disprove the hypothesis?
These all depend on artificially generated radiation which can have characteristics quite different from natural spontaneous emission with its Planck distribution of frequencies. There is indeed interference between two rays used in a hologram, but these started out as a single ray which was split. One half is reflected of a mirror and the other half off the subject, but both had the same frequency, phase and amplitude initially.
The radiation generated in a laser is stimulated emission which can have a very different effect on the target, basically because the target cannot handle a type of doubling up effect in the radiation, so the surplus that cannot resonate has to be converted to thermal energy.
Microwaves, like broadcast radio waves, are not absorbed much by a composite surface. However, in a certain frequency range water molecules, as well as some fats and sugars, do in fact absorb microwaves and convert their energy to thermal energy. Food is cooked because it contains water, but, as you know, many materials are not affected. So microwaves and other radio waves do not have much effect on blackbodies, and that is why broadcast radio waves can travel long distances without being absorbed by the atmosphere, or the surface. The equivalent temperature of a blackbody emitting radio wave frequencies would be colder than the atmosphere, so this is in keeping with the hypothesis.
Doug, thanks for the snippets from your paper. The issue of radiation absorption can probably be resolved in some of these cases through careful measurement and quantitative analysis. An instrument like mine is useless in that respect, I can get validate measurements relative to each other and I can validate some closed box measurements by comparing to air temperature (e.g. inside the fridge) but I still don’t know the emissivity assumptions and absorption-to-temperature conversion factor to sufficient accuracy (exacerbated by the nonlinearity of the resistance to temperature curve)
If I were to draw a box around some top section of the ocean I would have to quantitatively show that the energy flows, including back radiation, balance out. That is also difficult to measure in practice, but does not assume some undefined balancing out on the other side of the world or in space.
I think your hypothesis can and must be tested by measurement and quantitative analysis. It is not adequate to point to the alleged lack of global warming or similar ill-defined concepts.
Eric
Yes good comments. I was going to suggest using a meat thermometer (with a spike – they cost about $15) and comparing IR measurements in light sand, dark soil, grass and water surfaces. The problem is that emissivity of the atmosphere varies a lot and is generally very low. These IR thermometers have to be calibrated as I’ve now mentioned in a revised version of the paper. They are not really intended for atmospheric measurements.
The other main thing I have done (in a new revision of my paper) is remove all reference to “standing waves” and replace with a coined term “resonant scattering” which introduces the idea of resonating and then immediately re-emitting in a different direction. I never did find a reference to the standing waves, but I think it must have related to the usual “two parallel metal plates” experiment. I agree with you that it probably doesn’t work with back radiation. All we need is Claes Johnson’s result. However, when radiation from a warmer body strikes a cooler one its energy is split so that some goes into thermal energy (the area between the curves) and the rest into resonant scattering. So this latter part continues as “new” radiation in a different random direction until it strikes its next target, or escapes to space and eventually strikes something up there. Obviously there’s a probability of some going back to the surface, but that is just another warmer target which scatters it back into the atmosphere, just like diffuse reflection.
The main point is that none of any radiation from cold to hot adds extra thermal energy to the target, whilst radiation from hot to cold is split, with some adding thermal energy (corresponding to the area between the curves) and the rest immediately continuing as new radiation without adding any more thermal energy to the target.
I was up till 2:30am doing a re-write of the paper to remove the standing wave references and make other improvements to the explanation. My best example I think is that, if backradiation penetrated a little into a lake or ocean, and if it were converted to thermal energy below the surface then that is clearly a stand-alone completed “transaction” (with no immediate contra transaction) and so it clearly would violate the Second Law. Hence it doesn’t happen and the backradiation is scattered at the very top of the water surface.
One reviewer has suggested I write a bit more introductory stuff, which I will do today about climatologists and photon concepts etc, but they are happy with the rest. There might be a slight delay as the main coordinator will be unavailable Thursday to Sunday this week, but it’s important to get it right.
Doug, in a perfect world with perfect blackbodies I could see a possibility that only the energy between the two Plank curves will be absorbed and the rest won’t in some theoretical way (perhaps Feynman could explain but he is no longer with us). However the reality of imperfect blackbodies means that some will get absorbed and some reflected in some theoretical way and we would have to figure out how much.
In my interpretation of your theory, a “violation” of the Second Law is plausible at the individual photon level, coming from the cooler object, striking the warmer object and, due to a random molecular circumstance, being absorbed. Essentially any very small portion of the warmer material can be “cooler” for an instant and allow the photon to be absorbed.
The area between the curves is a net balance dictated by the Second Law. It is not a law unto itself.
#986 Eric
No – Imperfect BB’s always emit less than perfect.
All frequencies for the cooler object are always contained within those for the warmer one, so resonant scattering will occur without any thermal energy transfer.
See http://joannenova.com.au/2012/01/dr-david-evans-the-skeptics-case/#comment-1002913
I’ve just been notified of a slight delay with my paper (now about 13pp) as the coordinator is unavailable for a few days. However, they are very happy with it and expect to launch it next week.
#986 At” the individual photon level” the temperature of the room you are in could be over 500 deg.C if its frequency was three times the peak frequency of the distribution. But we base climate observations on temperature measurements made at the macro level. That’s where the Second Law must operate – the rest is just all about probabilities as I’m sure you understand.
O H Dahlsveen says on WUWT:
March 8, 2012 at 1:14 pm
I do not believe the theory of the Atmospheric Greenhouse Effect (AGHE) to be a “misnomer”. I think it is just plain wrong
_______________________________________
Basically you are correct. There can be no transfer of thermal energy from the cooler atmosphere to the warmer surface by any physical process, radiation or otherwise. However, we have to acknowledge that radiation from the atmosphere does slow the rate of radiative energy transfer from the surface to the atmosphere. This is why it can be warmer on moist nights. However, on balance, other processes, mostly evaporation and diffusion (conduction) will make up for any reduction in radiative flux, because of the stabilising effect of the massive store of thermal energy beneath the outer crust, which is not due to the very slow rate of terrestrial energy flow.
There is also a cooling effect due to water vapour and CO2 etc as these absorb downwelling IR radiation from the Sun and send upward backradiation to space.
The temperature gradient in the atmosphere is determined by the mass of the atmosphere and the acceleration due to gravity, both close enough to being constants. All the claims about 255K are based on the false assumption that the surface is anything like a blackbody. It’s not because it’s not insulated from losses by diffusion and evapoation. Less than half the energy exits by radiation. So, not only is that 33 degree figure based on a totally incorrect 255K figure, but it also ignores the fact that there is an adiabatic lapse rate that has nothing to do with backradiation.
This is a very brief summary of my peer-reviewed paper being published next week.
in 1951 the American Meteorological Society (AMS) had
already condemned the GHE to the trashcan of failed theories.
Perhaps it might be fair to say that Hansen helped uncrumble it from the trashcan.
And, by the way, show me an “explanation” of the GHE dated prior to the 1980’s which refers to backradiation.
in case you are not aware, the IPCC dropped the original Arrhenius version of trapped “heat” in the atmosphere and replaced it with the backradiation conjecture from the 1980’s. I just thought i’d mention that in case you didn’t know, because the two concepts are totally different mechanisms, both wrong anyway, but nevertheless, totally different.
Prof Claes Johnson has fully endorsed my 6,600 word Radiated Energy and the Second Law of Thermodynamics and it’s now ready for publication within about 30 hours at http://principia-scientific.org/ a site which I chose as being most in keeping with the principles I subscribe to.
“Principia Scientific International is a self-sustaining community of impartial scientists from around the world deliberating, debating and publishing cutting-edge thinking on a range of issues without a preconceived idea of outcomes.
“PSI has identified that there are currently two opposing methodologies at conflict:
“Traditional scientific method: borne of the Age of Enlightenment and which gave rise to the technological advances of the industrial revolution.
“Post-normalism:* pre-deterministic approach where policy and outcome dictate the kind of ‘science’ needed to justify it. The most culpable purveyors of this modern malaise are national governments, NGO’s and big corporations.
“PSI ASSOCIATES are steadfast in their support of the traditional scientific method … “
I often make the point that the use of relative surface areas for land v. sea (roughly 30:70) is not appropriate. The amounts of thermal energy are in the ratio of more like 6:89 I believe. Whatever ratio is estimated ought to be used to assess so called “global temperatures” because it is total thermal energy we are needing to measure. I know temperatures are not a direct measure, but weighting this way is at least better than 30:70. Clearly a square km of deep ocean has more thermal energy than an equivalent land area.
We nearly always seem to see slower rates of increase in the underlying 1,000 year cycle when we consider sea surface temperatures. Obviously any gradient should be calculated over a 60 year (perhaps 59.6 year) period so as to eliminate the effect of the 60 year cycle. When you do this, say from March 1952 to Feb 2012 I believe you will find a trend gradient of only about 0.05 to 0.06 C degrees per decade. Even if you then bring in land temperatures with about 6% weighting it obviously won’t make much difference.
I would further suggest that 0.06 C degrees per decade is more in keeping with the variations from the mean of the 1000 year cycle, giving us a total variation in 500 years between maxima like MWP and minima like LIA of 3 C degrees. At the most, I suggest that history shows no more than plus or minus 2 C degrees above or below the level trend of that 1000 year sinusoidal-like cycle. So, a mean rate of increase of 0.08 C / decade seems likely. The fact that the current rate is lower (and perhaps only 0.05 C / decade) indicates an approaching maximum probably within 50 to 200 years.
My point is, of course, that there is no evidence of any effect at all from CO2 if you agree with my estimate of 0.05 C deg / decade in the last 60 year period.
The reason carbon dioxide has no significant effect on the rate of radiative cooling of the surface is that it has a limited number of frequencies “standing up against” the full Planck spectrum radiated by the surface. Furthermore, it cannot affect the rate of cooling by evaporation and diffusion (followed by convection) and these rates will increase to compensate.
The rates of warming are shown in a plot at the foot of my Home page http://climate-change-theory.com and the physics is covered in my peer reviewed paper Radiated Energy and the Second Law of Thermodynamics being published within the next 12 hours at http://principia-scientific.org/
PS The paper is published now at http://principia-scientific.org/publications/psi_radiated_energy.pdf
Doug, thanks for the link to the paper. I will read it tomorrow or Thursday and comment.
I accept that, technically, “points” are dimensionless and I used the term colloquially in the Abstract for the paper, but I would also say that what I am talking about can be physically a very small volume of matter. I suggest, some such volume with only perhaps a million molecules would be quite sufficient for the Second Law of Thermodynamics to be applicable.
Wikipedia puts it this way: In classical thermodynamics, the second law is a basic postulate applicable to any system involving measurable heat transfer …
It is all based on probabilities, of course, but we just need radiation with the full range of frequencies indicated by the Planck curve in order for temperature information to be conveyed. In reality, molecules react to frequencies and it is all to do with frequency distributions, not any characteristic of an individual photon or molecule.
At the outset, let me be honest and say that I too do not necessarily agree with everything that every author in Slaying the Sky Dragon has written. There are some subtle contradictions in fact between authors. I do however, agree with Prof Claes Johnson’s general concept that radiation from a cooler blackbody merely resonates with molecules in a warmer blackbody, without any of its energy being converted to thermal energy. And Claes also read my paper prior to publication and commented that I was one of only a few who understood his Computational Blackbody Radiation and that he fully endorsed my paper. (I understand that Claes is not a member of the Slayers.) But I chose Principia Scientific International because they have a growing number of scientists joining their ranks who participate in “open review” of the papers they publish on their site – six in total now. PSI comprises many more scientists than the few authors of the book.
May I ask that people do in fact read the paper before commenting. You will find, for example, that I explain why lasers, microwave ovens and microbolometers do not disprove the hypothesis. Whilst I don’t mention it, I anticipate that there will be experiments published later this year using spectrometers to demonstrate that warm gases do not absorb emission from cooler sources.
It is not appropriate to assume, for example, that I am discussing thermal energy accumulating in the atmosphere somehow warming the whole Earth system. That concept, I understand, has been dismissed by the IPCC who now argue that it is all about “backradiation” slowing the rate of cooling of the surface. Thermal energy is not transferred from a cold atmosphere to a warmer surface (nor to warmer layers of the lower atmosphere where we live) by any physical process. Thus the slowing of the cooling process is not due to the addition of thermal energy to the surface. Rather, it is due to resonance of the radiation itself, which does not involve absorption in the usual sense involving conversion of radiated energy to thermal energy.
Radiation from a cooler source merely undergoes what I call “resonant scattering” when it strikes a warmer target. As I have said, there is no conversion of its radiated energy into thermal energy, which is quite a different thing. If the radiation from the cooler (macro) source is (close to) that of a blackbody it will have frequencies across the appropriate Planck curve. Most radiation from the atmosphere will not have all these frequencies, but it will (to some extent) oppose equivalent radiation from the warmer surface while it experiences resonant scattering by the surface.
When it is scattered by the surface, it becomes a part of the emission of the surface, but, because it already has its own energy, it does not need energy from the surface itself. Thus it slows the rate of cooling of the surface because it “uses up” some of the potential radiation frequencies which the surface would otherwise have used to dispose of its own energy.
However, carbon dioxide does not radiate like a blackbody, so its few spectral lines are relatively ineffective compared with even water vapour radiation, let alone a blackbody.
So water vapour is the major contributor, having probably at least 100 times the effect of carbon dioxide when you take into account its greater presence and greater effectiveness per molecule.
Even so, only the radiative cooling process is affected by radiation from the atmosphere, not all of which is actually “backradiation” as such, because it may have originated from energy carried up by convection.
Now, there are other processes, mostly evaporative cooling and diffusion (sometimes called conduction) which involves molecular collision between surface and atmospheric molecules.
These other processes are not affected by radiation from the atmosphere. Yet they probably account for more than half the thermal energy transfer between surface and atmosphere, and they will tend to compensate by increasing their rate if the radiation rate decreases.
There are reasons for this explained in the Appendix.
Now, some don’t realise just how much of the incident solar radiation is actually in the near infra-red. Some of this is absorbed by water vapour and, to a small extent, also by carbon dioxide. This SW-IR radiation has much more energy per photon than does the LW-IR radiation from the surface. Some will be absorbed and this helps explain why the thermosphere gets very hot, often well above 400 degrees K in fact. By sending backradiation to space a cooling effect results, which is almost certainly greater than any warming effect due to carbon dioxide.
Temperatures on the Moon (without an atmosphere) vary from about -153°C at night to +107°C during the lunar day. Over 40% of solar insolation is either reflected or absorbed by the Earth’s atmosphere, so our atmosphere keeps the surface cooler than the Moon’s in daylight hours, by reducing incident solar radiation. Then, both day and night, the atmosphere slows the rate at which solar radiation (which was absorbed by the surface) then exits back to into the atmosphere and to space.
What does not happen is any transfer of thermal energy from cooler regions of the atmosphere to warmer regions on the surface, for any such heat transfer would violate the Second Law of Thermodynamics. For example, radiation from the atmosphere does not penetrate even 1cm below the surface of warmer water and add thermal energy to that sub-surface water. If it did, such warmer water could then rise to the surface by convection and its thermal energy could then get back into the atmosphere by evaporative cooling. Hence we would have had a stand-alone process transferring thermal energy from a cooler atmosphere to a warmer sub-surface layer of the water and warming it even more. Such a process would violate the Second Law.
Over the course of 4 billion years an approximate equilibrium point has been reached at any particular location on the Earth’s surface. Even though the atmosphere is roughly similar at the South Pole, the equilibrium temperature is very different from that at the Equator, due to different mean solar radiative flux over the course of each year. This clearly indicates that the temperature is mostly determined by the Sun’s radiation, not so much the properties of the atmosphere.
In regard to experiments, some are being arranged. My own “backyard” experiments with sand and soil in wide necked vacuum flasks indicated no difference in cooling rates between the contents of the flask which was shielded from backradiation at night, and that which was not. Try it yourself using a digital meat thermometer and a sheet of plate glass with an additional shield on top of it, all about 20cm above one flask and at a 10 degree angle to the horizontal to allow convection.
I suggest the onus should have been upon the IPCC to produce evidence to the contrary with a similar obvious experiment. I suspect it has been tried and failed, thus never being published. Correct me if I’m wrong anyone, and link me to any experiment showing backradiation warms anything.
I am the first to agree that it can slow that component of the surface cooling which is by radiation. However, in the context of anthropogenic effects, the role of carbon dioxide is minuscule because of its limited radiation frequencies and the fact that it is only one molecule in over 2,500 other molecules. Because it also has a cooling effect radiating energy to space, it is highly improbable that it causes any net warming at all.
It would be appreciated if people would actually read the paper and this comment in full.
Even though many clearly believe what has been the “usual” explanation involving heat transfer in both directions, it should be apparent that Prof Claes Johnson and myself disagree with this and are putting forward a hypothesis that there is another mechanism that explains what actually happens and yet still gives the same quantitative result as does application of SBL.
I really do not need to hear again the “standard” explanation of photons supposedly transferring thermal energy to everything they collide with – and “not knowing” the temperature of the source. You will find all these matters are addressed in the paper.
But, as I politely asked above, either please read the paper and all of this comment before commenting, or otherwise consider refraining from joining the discussion herein. I believe the paper itself, (perhaps with the additional explanation in this comment which may help some to understand) covers all the objections anyone has thus far raised, both here and on other forums as well.
My paper will now be subjected to “open peer review” by dozens of members of Principia Scientific International (PSI) who will shortly receive an email from the organisation.
Click to access psi_radiated_energy.pdf
Also: http://tallbloke.wordpress.com/
Eric and others:
If you say “The transfer of energy from a cooler body to a hotter one is compensated by the transfer of energy from the hotter body to the cooler one.
____________________________
With respect. the whole point of my paper is to prove the above statement must be incorrect. If, for example, IR from a cooler atmosphere is converted to thermal energy, (say going from A to B) then there is absolutely no physical reason why there has to be a “compensating” or greater transfer at that time or between those points, ie back from B to A. (It might go, for example, from B to C in the surface, then from C to D in the atmosphere.)
Physics down through the ages has never said this could happen. Each movement of thermal energy has to stand alone and satisfy the Second Law of Thermodynamics in its own right. Movement of radiation does not necessarily mean movement of thermal energy between the same two points.
This is vital in consideration of atmospheric physics for the following reasons …
(1) It eliminates the possibility of any particular “parcel” of thermal energy going from a cold atmosphere to a warmer surface and then back out by evaporation or diffusion.
(2) It restricts the effect to one of slowing the rate of radiative cooling, but not the other rates of cooling by evaporation and diffusion followed by convection.
(3) Furthermore, when the physical process is understood (as is explained in the paper) then it is obvious that carbon dioxide can have far less effect per molecule because it emits far fewer separate frequencies (wavelengths.) All the carbon dioxide in the atmosphere must have less than 1% of the effect of all water vapour. So, just because it gets warmer on a humid night, does not mean carbon dioxide can create the same
effect.http://tallbloke.files.wordpress.com/2012/03/radiated_energy.pdfhttp://tallbloke.files.wordpress.com/2012/03/radiated_energy.pdf
PS My photo was taken at Glenelg Beach, Adelaide SA.
The following points are made in my peer-reviewed paper now published on at least four sites and linked from my site http://climate-change-theory.com
(1) Radiation from cooler parts of the atmosphere to warmer parts of the surface cannot transfer thermal energy, but can slow just the radiative component of surface cooling, which is less than half the cooling, probably about a third.
(2) When radiative cooling is slowed, the rates of evaporative cooling and diffusion (conduction) followed by convection will increase for reasons explained in my paper.
(3) The energy in each photon is proportional to the frequency of the associated radiation.
(4) Short wave (high frequency) infra-red radiation (making up about half of the total solar radiation) thus has far more energy per photon than does long wave (low frequency) infra-red radiation from the atmosphere, which is mostly well below freezing point.
(5) The effect which radiation from the atmosphere has on radiative cooling of the surface depends upon both the temperature of the region from which it originated and the density of frequencies in that radiation.
(6) Carbon dioxide radiates far fewer frequencies than water vapour, and each radiates fewer than a blackbody.
(7) Hence each carbon dioxide molecule has far less effect on the radiative rate of cooling than each water vapour molecule, of which there are usually about 20 to 50 times as many.
(8) So carbon dioxide is like a picket fence with most of its pickets missing, standing up against full blast radiation from the surface.
(9) Any warming effect of carbon dioxide is cancelled because of the reasons in (2) and, because of those in (4) there is a significant cooling effect as it sends back to space at least half of the high energy photons it captures from solar radiation.
(10) Hence carbon dioxide has a net cooling effect, but such is absolutely minimal compared with the effect of water vapour which also has radiative cooling effects, but possibly some warming effects also about which we can do nothing.
Doug,
Your paper would receive a worse review here than it did at Tallblokes site.
I tried to read it but gave up at the second page.
Here’s ours, let me know if you make it farther in: http://www.climateaudit.info/data/odonnell/4%2020101126%20Submission.pdf
Jeff – that’s why you don’t understand what it says.because the main points are on pages 3 to 12.
Study the graphic on this page http://www.theresilientearth.com/?q=content/why-i-am-global-warming-skeptic
There you will see just how much of the solar radiation is absorbed by water vapour compared with how much less terrestrial radiation is absorbed by it. Yet, according to the IPCC, water vapour is supposed to have a significant positive feedback, amplifying any effect of carbon by a factor of about 2.5.
You will also see the carbon dioxide absorption of Solar near IR (as on the graphic in my Section 6) at just over 2μm.
Note that carbon dioxide does not absorb much of the majority of the surface radiation in the 7.5 to 11μm range, only that which in the 16μm range which is equivalent to about -90 deg.C – colder than anywhere in the troposphere or the surface.
So you would only get significant radiation from carbon dioxide up near the mesopause at the top of the mesosphere (above the stratosphere) where temperatures just about get down to -100 deg.C. And what it absorbs up there would be minimal.
Furthermore, photons emitted at -90 deg.C (183 K) each have only about 13% of the energy of those at 2μm which are radiated by parts of the Sun that are about 1450 K.
So, when carbon dioxide does most of its capturing and emitting at about 16μm (and -90 deg.C – way above the stratosphere) each photon captured has only 13% of the energy of each of the photons it captured from the Sun, at least half of which it radiated back to space, thus preventing that energy reaching the surface, and so having a cooling effect. How much of the radiation from layers at -90 deg.C do you suppose gets back to the surface?
This information has always been readily available, and there is no excuse for the “cover up” of absorbed solar IR and the focus on terrestrial emission only.
Yes Jeff, I’ve always been interested in Antarctica since talking with my grandfather, Prof. Leo Cotton who went there with Shackleton in 1907.
Did you notice any emission down there in the 16 micron range that could be absorbed by carbon dioxide at about -90 deg.C up in the mesosphere?
http://adb.anu.edu.au/biography/cotton-leo-arthur-6325
PS Jeff
You’re very welcome to do a review of my paper. It might make you actually read what it actually says. You’ll get comments from me like those on my tallbloke thread – which absolutely no one has been able to rebut successfully.
Just be careful, won’t you to avoid misquoting me, misinterpretting what I’m saying in the paper, or introducing red herrings like micrometers which you will find are dismissed in the FAQ in the Appendix.
Let me put this way …
(1) Radiation sets out on its journey from an object with a frequency distribution represented by the appropriate Planck curve for its temperature.
(2) Not all the possible frequencies may be present if the object is not “composite” but is, perhaps, a single gas or several gases with limited emission bands. If this is the case, this radiation will have less effect on the radiative cooling rates of warmer composite bodies than would perfect blackbody radiation with the full range of frequencies that are allowed by the Planck curve
(3) When the radiation strikes a blackbody target, that target, if cooler, will only be able to match and resonate with a portion of the emitting source’s radiation. The extra source radiation, corresponding to the energy between the Planck curves will be converted to thermal energy, but the rest will undergo resonant scattering. So the radiation is split.
(4) If the target is warmer, all the radiation from the source undergoes resonant scattering.
(5) When radiation strikes a target, that portion that resonates (be it all or part) takes the place of radiation for which the target would have had to use its own energy. Hence the target, even if warmer, will cool more slowly, as is observed.
(6) The target becomes a new source. It, like the original source, still radiates all it can under its Planck curve, but it is using some or all of the energy it received from the original source. Hence “new” radiation sets out which may be thought of as containing some or all of the energy of the original source, plus some of its own it it was warmer than the source.
(7) If you were to follow the passage of any particular parcel of energy, some of it would be “dropped off” at each target that was cooler than the last, but not targets which were warmer than the last.
(8) So the surface also just scatters “cooler” radiation from the atmosphere, and all radiation gets to space eventually. There it continues in the same way, striking targets perhaps every few years or centuries or whatever, until its energy gets down to the base level of about 2K to 3K I understand, as is observed in background space radiation. This means nothing much is colder than that out there.
The above process overcomes the problem of assuming two-way radiation is associated with two-way heat transfer, including transfer from cold to hot. The latter would violate the Second Law, and so cannot happen.
Hence the above postulate is a far more likely alternative.
Sorry about typo in #2004 – microbolometers (of course)
Reminds me, there’s an interestin post by on Spencer about microwaves …
Turnedoutnice says:
March 18, 2012 at 11:07 PM
KR: a microwave oven does not use randomly-emitted 5 K photons to heat food at 290 K.
It sets up a series of standing waves at whose nodes the energy amplitude and the resonant coupling is such as to oscillate water molecules sufficiently that the local heating heats the food.
As an exercise, I’ll ask you to calculate the equivalent arrival rate of focused 5 K photons. Imagine a silicon lens focusing IR!
“Just be careful, won’t you to avoid misquoting me, misinterpretting what I’m saying in the paper”
Have I done a lot of that?
Your argument, if it can still be called that, has been rebutted so many ways right in this thread it isn’t funny.
Jeff, you’re right. Check out the comments at tallbloke.
Many of them cover the same territory we already went over, including the mixing of quantum mechanical and classical views. This is the same problem that guy who studies numerical algorithms that Doug likes to much made (Clae Johnson).
They are so far out of their own water, their work doesn’t even smell fishy anymore —more like dead fish.
Everyone should note Turnedoutnice’s comment at http://www.drroyspencer.com/2012/03/global-warming-as-cargo-cult-science/#comment-39322
Roughly 60% of the thermal energy from the surface to the atmosphere does not transfer by radiation. Only about 40% is radiated back. So, if radiation from a cooler atmosphere really did transfer thermal energy to a warmer surface (and the Sceond law of Thermodynamics really could be thus violated) then only 40% of that thermal energy would then exit (a second time) by radiation. So you have much more radiation into the surface than out of it. You have expected water to run uphill by itself and fill a water tank at the top of the hill, just because you know the water is going to flow down pipes to people’s homes sometime in the future.
Thermal energy does not transfer both ways between hot and cold as Planck and Co originally claimed. Botzmann got the right result mathematically, but he did not guess the right physical process. I suggest that is what is in my paper.
Click to access jo120314.pdf
If you think you can better explain how radiation from a cooler source does in fact slow the radiative rate of cooling of a warmer body then please put forward your ideas. But any ideas which depend upon compounding thermal energy, some of which is assumed to result from the radiation from the cooler body to the warmer one, will not wear with me.
#1008 Well, Carrick or Jeff
See if one of you can be the first to explain in cogent physics just exactly what is the mechanism by which incident radiation slows the rate of radiative cooling of a body, a fact which is well known and acknowledged in the final paragraph of Section 5 in my paper which is written up here: http://www.webcommentary.com/docs/jo120314.pdf
PS You won’t get far if you try the two way heat transfer garbage of Planck and Boltzmann.
PS Hardly appropriate to make a point of the fact that I didn’t respond to a post on tallbloke while I was asleep..
Everyone
There is an informed discussion on Roy Spencer’s site about all this, with over 400 posts so far.
You might care to pick up here http://www.drroyspencer.com/2012/03/slaying-the-slayers-with-the-alabama-two-step/#comment-39517
Because I really don’t have time to focus on more than just two threads now, this one of Roy’s and my own thread on tallbloke will have to do..
On tallbloke we are moving on to discussing FAQ 1 in the paper.
Link to paper here http://www.drroyspencer.com/2012/03/slaying-the-slayers-with-the-alabama-two-step/#comment-39517
Link to tallbloke recent posts start here: http://tallbloke.wordpress.com/2012/03/13/doug-cotton-radiated-energy-and-the-second-law-of-thermodynamics/#comment-20365
Correction
Link to paper here http://www.webcommentary.com/docs/jo120314.pdf
Regarding lasers
Here is a response I received …
For the commenter who claims that the heating of a warm surface by a laser beam from a laser source that is “colder” in some way invalidates the Second Law, here is the correct response. For the laser beam to function there must be an inversion of energy state populations such that the upper level has a higher population of energy states than the lower level of the specie that is supplying the stimulated emission. That inversion must be maintained by an external energy source. For a system in thermal equilibrium the Maxwell-Boltzmann distribution makes that impossible. So, in a sense, the laser source is operating at a high, non-equilibrium temperature that is unattainable by the warm surface that it is heating. So, on the basis of its distribution of energy states, the laser is indeed hotter than the warm surface it is heating. Or looked at from another point of view, the agent maintaining the population inversion of energy states in the laser is an external agent like the compressor and working fluid in your refrigerator that allows heat transfer from the cold refrigerator to the warmer room outside.
(name witheld)
Doug,
The comment you pasted is correct, although the wording of the refrigerator example is a bit confusing. You are the commenter who believed the IR laser violated the second law of thermodynamics. You may recall the all-knowing thermodynamic genius who said – “Why the hell would anyone use such a low energy laser to cut steel”
One wonders if you even realize that you have cut and pasted a comment from classical thermodynamics which goes directly against your description of the magical “cutoff frequency” for the first 700 comments of this thread?!!
Your argument is now coming full circle to classical physics – and you don’t even seem to realize it.
All this from the narcissist who started by telling us he knew best, and he didn’t have time to teach us. The Air Vent should charge you a fee at this point, yet you are still stomping all over blogland claiming you are the true genius and all of thermodynamics is wrong. This is crazy behavior Doug. Do you really think you know everything and that nobody else can reach your understanding?
I did not “believe the IR laser violated SLoT.” When there are plenty of lasers to choose from, why not choose the one which is best for the job – CO2 for mild steel and sheeting, yes, but hardly a steel beam or even a thick sheet.
There are no royalties or payment of any kind for my paper, but please note the Acknowledgements. I appreciate your help in some specialty areas, but I wonder a little why you could not have just made a statement such as the above regarding lasers, rather than trying to claim that they disproved my theory, or that microbolometers do either. This is rebutted in the Appendix Q.4.
It is of course not hard to explain how a refrigerator works and does not violate the SLoT. http://home.howstuffworks.com/refrigerator.htm
I’m a bit busy now responding to posts about my paper which has been published on at least five sites.
Anyway, what I am saying is in the paper and no one has rebutted anything successfully yet to my knowledge.
I leave you with this quote from Roy Spencer’s site …
“Gordon Robertson says:
March 23, 2012 at 10:11 AM
Watchman…”Peer reviewed? Isn’t this “paper” in Principia Scientific? Are you suggesting this is a “peer reviewed” scientific journal?”
Where in the scientific method does it require a paper to be peer reviewed? Peer review began as a means of preventing laymen from flooding journals with inane science. Today it is a method of preventing people who don’t support the official consensus from being published. That is, peer review is now censorship, and top IPCC official are participating in it (Climategate).
I have just spent a couple of interesting hours reading through material at PSI that is soundly based in science. I would never get such information from peer reviewed mind control.
So when will I see you on the internal mailing list at Principia Scientific International, Jeff
I quote from http://www.epiloglaser.com/tl_yaginfo.htm
Technical Library: YAG vs. CO2 Lasers
Two of the most common types of lasers you hear about are YAG and CO2. While Epilog manufactures a CO2 laser system, we wanted to give you some information about the two different types of laser systems and which one may be right for your application. Both CO2 lasers and YAG lasers generate a very concentrated beam of light, but from there the lasers become very different in their uses and how they work. In this article we will first look at the different uses of the two laser types, and when each one is right for your application, and finally break down the benefits of each system.
What are the different uses for YAG and CO2 lasers?
YAG lasers and CO2 lasers react very differently on different materials because of the differing wavelengths of the laser beams. The wavelength of a YAG laser (1.064 microns) is exactly ten times smaller than the CO2 wavelength of 10.64 microns, which makes it ideally suited for absorption in most metals, but this small wavelength inhibits its ability to be absorbed by many other materials (wood, acrylic, plastics, fabrics, etc.)
A CO2 laser beam is not easily absorbed by metal, but can easily be absorbed by many organic materials such as wood, acrylic, rubber, etc, while it tends to reflect off of most metal surfaces. It’s the different wavelengths of the two beams that are mainly responsible for the different types of materials that they will react with. There are a number of other differences between the two lasers; thermal efficiency, heat transfer, minimum and maximum power output, etc. and these characteristics all have an affect on the materials that the beams react with.
PS I also appreciated your advice regarding getting peer-review and publishing a paper. I decided this was better than a self published book, and I really appreciated the help and efforts of the four reviewers over a three week period prior to the launch.
The main point is that CO2 has limited frequencies, especially at atmospheric temperatures, and, when you understand the concept of resonant scattering, you will realise such limited frequency density can have only a minuscule effect on the rate of radiative cooling of the surface which emits far more frequencies “against” the atmospheric radiation. I then explain why the other cooling processes (over 60% in total) will speed up to compensate, thus not changing the overall cooling rate at all.
The Abstract is reproduced on this site, with an article below and a link to the paper, which is of course also linked from my site.
http://climaterealists.com/index.php?id=9281
PSI is a growing organisation with another 17 new members announced recently – make me the 18th as I missed the count. We are working together on new papers for launch later this year.
Doug,
Your denial of the function of an IR laser is very well documented on this thread. Your denial of that denial D^2, is a unique trait.
I don’t want to waste my time with your paper. I did finally read your explanation of a laser (that you have pointed me to literally a dozen times), and as expected, it is ridiculous to an extreme. I suggest removal of section q5, as it makes very clear that you don’t have a clue.
The microbolometer section is also screwed up.
“The instrument’s sensors are warmed (using electric input) but while they are warming they are also radiating energy to the object whose temperature is to be measured.”
To keep thermal noise to a minimum, internal sensor warming is avoided to the maximum extent possible. The best ones are cryogenically cooled. You again have no clue AND in both cases you contradict yourself on the early half of this thread.
blah!
_____________________________
Yes Jeff. I can improve the description of the laser, because experts at PSI have helped me with that. And I have a better explanation of how a microwave oven heats water by causing friction between molecules as they flip through 180 degrees each half wavelength and collide with other water molecules, but there is no such friction in ice. Friction is not absorption.
So Jeff, if ice cubes from your frig are placed in a microwave bowl which itself does not warm, then the fairly high intensity, but low frequency radiation does not melt the ice, even though it boils an equivalent amount of water in a similar bowl at the same time.
But, radiation from an electric radiator of similar power will melt the ice, now won’t it, because its frequencies are higher, whereas the microwave oven’s frequencies were lower than those of the ice.
Almost enough to remind you of what Claes and I have been saying, now isn’t it Jeff?
______________________________
Oh, and Jeff, I forgot to mention that the sensor of a microbolometer does have a slower radiative cooling component the warmer the object being measured, and vice versa – as is well known in standard physics, and discussed in my paper, Section 5, last paragraph.
It doesn’t matter if the sensor is being warmed or not, though it is in fact warmed with battery power as I have correctly explained in the Appendix to my paper, because otherwise it would get too cold. It has to be kept at approximately the same temperature with thermal energy supplied by battery power providing a warming effect, whilst radiation and sensible heat loss do the cooling, just like the Earth’s surface.
#1019
Jeff said The best ones are cryogenically cooled.
http://www.iecinfrared.com/photovoltaic-versus-microbolometer-cameras.html
says
On the other hand, microbolometers do not have to be cooled in order to work. They can operate at or near room temperature. While it is useful to maintain a stable temperature on the microbolometer array, they are considered “uncooled” devices because they do not use cryogenic coolers
http://gs.flir.com/technology/Imaging_Systems/Uncooled_IR_Detectors/
Microbolometers are generally temperature stabilized by means of a thermo-electric (TE) device. The TE device heats or cools the detector in response to a particular voltage
Seeing that it is cooling by radiation, I suggest most of the TE action would be warming – just like I said in the paper which you thought I “screwed up.”
_______________________________________________________________________________________
So Jeff, why don’t microwaves melt ice when IR rays from an electric radiator of similar power do melt ice?
_______________________________________________________________________________________
Doug,
The camera example above is an ‘UNCOOLED’ camera.
Nice to see you are paying attention.
The two sections I read of the paper you wrote, are pure fantasy. I did expect that after this conversation. I’m sure many of us did.
You should be paying very careful attention to every critique on this thread. You have been wrong in every single instance of conflict above and each one could teach you something you are not understanding.
Jeff or anyone.
Why does a microwave oven not melt ice?
How could carbon dioxide radiation thus melt snow or ice anywhere on the planet, when similar more powerful LW microwave radiation does not get absorbed and converted to thermal energy in the ice cubes, just like Claes said would be the case?
.
Summary of key points
Firstly, any slowing of the radiative cooling due to carbon dioxide, methane, etc is absolutely minuscule. The Energy diagrams try to attribute all the backradiation to about 2% of the molecules in the atmosphere. For some strange reason they also assume more goes towards the surface than to space. These 2% of molecules at colder temperatures than the surface, are accused of radiating about as much as the maximum which S-B law permits the surface itself to emit. How can they? Their own Planck curves must limit them to only 2% of what the surface radiates if they were the same temperature, and closer to 1% when they are on average about 40 degrees cooler than the surface.
So the energy diagrams overstate the radiation from WV and trace gases by about 100 fold.
In fact, oxygen and nitrogen can cause radiation by acceleration of electrons in grazing collisions, rather than emission due to quantum energy steps which requires much greater temperatures. So it is apparent that the remaining radiation from the atmosphere comes mostly from oxygen and nitrogen molecules as they collide.
Furthermore, as a result of the limited frequencies in carbon dioxide emission, its effect on rates of radiative cooling is far less than that of a blackbody or a metal plate in the classic experiment with two metal plates radiating towards each other. This is explained more in my paper.
It is doubtful that carbon dioxide would slow the radiative cooling each afternoon and evening by more than, say, a minute. There is ample time at night for that minute to be “made up” with extra cooling by evaporation and sensible heat transfer, which could be faster and/or last longer until the surface gets back to the long term base temperature supported by the “thermal inertia” of sub-surface temperatures, as discussed in Appendix Q.3.
For more detail on how the process works, refer to a page I wrote last September here:
http://climate-change-theory.com/explanation.html
A footnote for PB I did “hunt” for a difference. I had a shielded and unshielded wide necked vacuum flask, and the shielded one cooled more slowly. The unshielded one cooled almost as fast as the air. You seem to be forgetting that the IPCC energy diagrams show more backradiation than they do incident solar radiation. Even allowing for day/night variations in solar insolation, there is very clearly no effect anything like what can be calculated from the energy diagrams if in fact the radiation were absorbed.
But if a microwave oven can’t melt ice, what effect can much weaker backradiation from carbon dioxide have on all the snow and ice covered areas of the globe, just for a start?
_________________________________
Perhaps the most fundamental mistake made by climatologists when they attempt to apply physics is to do what PB does, namely to pull out the equations of Stefan-Boltzmann and Kirchhoff and apply them in circumstances where they just don’t apply, and were never intended to apply.
There is a tendency to assume everything acts like a blackbody and all we have to do is adjust for emissivity and absorptivity, which themselves are assumed always equal (when in fact they are not) and constant, which they also are not for different temperatures of source and target.
For example, we know the surface absorbs SW solar radiation and yet emits LW radiation. Even the timing is not simultaneous, as more absorption takes place in full sunlight. But most importantly, the surface acts nothing like a blackbody just adjusted for emissivity. In fact, the sensible heat transfer (diffusion followed by convection) and evaporative cooling dominate radiation. These processes keep the air temperature which we measure (1.5 to 2 metres above the surface) just a little cooler than the surface, rarely more than 4 C degrees cooler except in unusual weather events. If you then apply S-B equations to determine radiation you should be deducting the “backradiation” in the process of doing so. This is the correct way to apply SBL. You don’t then double count the backradiation as further radiation from the clouds or whatever. It has already been taken into account in determining what is in fact a very small amount of radiation from the surface. In fact, there is more radiation, but it will be counted as radiation from the atmosphere after it is scattered by the first target it hits. So, again, we have to be careful not to double count.
But, as far a cooling the surface itself goes, radiation plays a relatively small part. Calculations show that the radiative loss would only exceed the other losses if the surface were about 100 deg.C. But it seems that the physics formulas for evaporative cooling and diffusion have been overlooked.
So you need to forget concepts like emission = absorption and everything gets radiated according to S-B equations and so on, and think about what’s really happening and whether there is any marginal effect due to extra carbon dioxide, being about 1 molecule in 2,500 other air molecules, trying to slow down the rate of radiative cooling of the whole surface in a David and Goliath battle which David is losing this time.
.
.
Microwaves not melting ice provide an example of what I deduce in the paper, namely that there must be another process apart from reflection, transmission and absorption resulting in thermalisation. There is at least some LW radiation entering the ice which is not reflected, not transmitted and not being thermalised. Indeed, I did read papers which referred to scattered radiation observed coming from ice. Whatever you call the process, “resonant scattering” or “pseudo scattering” it does exist and explains why radiation does not always have to be thermalised in a solid. The SLoT is obeyed because this process “selects” when it occurs by the temperature differences between source and target.
In practice, it is just a matter of corresponding frequencies resonating when they find their match. So you can visualise the two Planck curves like picket fences (with the shape of the curves) facing each other. Carbon dioxide only has a few pickets compared with the surface whose pickets are so close there are virtually no gaps. The “fence” for a cooler source will always be fully contained within that for a warmer one – see Section 3. This is the process whereby radiation from one body has an effect on the radiative cooling rate of another body.
The Earth’s surface, as it sheds energy to the atmosphere, uses up some of its radiating capacity handling the resonant scattering in which it is easier and faster to use the energy in the incident radiation (rather than its own supply of thermal energy) for the new radiation which exactly matches that component of the incident radiation which is also under its Planck curve. If the source was cooler, all such incident radiation will have a distribution that is under its own curve. Only if the source were warmer would there be a surplus (between the Planck curves) that does some warming. So heat transfer by radiation can only be from hot to cold (as we knew by SLoT) and this mechanism which is described in more detail in the paper appears to be what happens.
.
“Microwaves not melting ice provide an example of what I deduce in the paper,”
No they don’t.
“There is at least some LW radiation entering the ice which is not reflected, not transmitted and not being thermalised.” – false
“In practice, it is just a matter of corresponding frequencies resonating when they find their match.” – bull. Evidence please!
“So you can visualise the two Planck curves like picket fences (with the shape of the curves) facing each other. ”
What kind of nonsense are you onto now? What happened to ‘area under the curve’?
“If the source was cooler, all such incident radiation will have a distribution that is under its own curve. Only if the source were warmer would there be a surplus (between the Planck curves) that does some warming. So heat transfer by radiation can only be from hot to cold (as we knew by SLoT) and this mechanism which is described in more detail in the paper appears to be what happens.”
Standard stuff, FINALLY. Thank god.
Standard stuff, FINALLY
No it’s not. Read the paper if you don’t understand the significance in physical terms between
(1) heat transfer in two directions each represented by areas under their respective Planck curves, and
(2) heat transfer in one direction represented by the area between the two Planck curves.
Before you comment that there is no mathematical difference (which is true) you will find my response in the Appendix Q.7 but you will never fully understand unless and until you read the whole paper.
http://climate-change-theory.com
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No Virginia, microwaves and LW radiation from carbon dioxide do not melt ice anywhere on Earth, just like the Slayers have been saying all along.
Microwaves do not even excite anything in water, let alone ice. If you do the calculations they have far too little energy per photon.
In water they cause 180 degree flipping of water molecules in synchronisation with the wave motion of the radiation. As the molecules flip they create friction with other flipping molecules, and that is what causes generation of thermal energy, not the normal process which happens when solar radiation strikes water – totally different. The microwaves even have to be such that standing waves are created in order to achieve the flipping. But in ice there is no room for the flipping. And such is the case in nature when low energy LW radiation from a cooler source strikes a warmer surface – there is no generation of thermal energy (ie no heat transfer from cold to hot) which the IPCC energy diagrams wrongly imply happens
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All Jeff can do is quote the “old” physics which is wrong when it says there is two-way heat transfer between hot and cold objects. He is saying I am wrong because I am saying the old physics is wrong just in this regard. But he has no proof – as I shall now explain …
No one has ever detected such separate heat flow and there never will, because the cold to hot flow would violate the Second Law of Thermodynamics as it would be a separate, complete, stand alone process which, in practice, ought to be able to be distinguished, but never can be. For example, if it happened you could detect warming of a layer of water just below the surface of water where the energy could not escape back immediately. So, if it happened, you could detect it and you would be detecting a violation of the SLoT. But of course you can’t, because the SLoT is never violated.
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In case anyone has read Pete Ridley’s attempt to rebutt my microwave experiment, I responded …
Firstly, the flipping action (in synch with the wave motion) is illustrated here – see the next page also
http://www.colorado.edu/physics/2000/microwaves/water_rotates3.html
Pete has used plastic bowls which probably got warmed a bit themselves. I used special microwave bowls made of special material that does not get warmed at all in a microwave oven.
My whole point was to demonstrate that there was ample energy in the first minute (maybe 90 seconds if volumes are greater) to boil an equivalent volume of water from, say, 5 deg.C, but not melt the ice. I demonstrate that there is ample energy in the water to then melt the ice within the same time frame when I poured the boiling water onto the ice.
So clearly virtually all the energy went into the water and hardly any into the ice, even when they were there together at the same time and for the same length of time. There was ample energy to melt the ice and warm the water quite a few degrees and achieve the end result after mixing. But that did not happen.
Obviously, if you extend the period you will eventually warm the container (especially a normal plastic container such as Pete probably used) and the container will warm and melt the outer layer of the ice, and the water so formed will start to melt the ice. But the ice is melting by conduction from warmed water. If the microwaves were melting it there would be melting throughout the ice cubes, not just at the surface. This is why defrosting mode turns power on and off, in order to allow time for conduction from heated water to melt the ice in frozen food.
Even the fact that many materials do not warm much in microwave ovens shows that the heating is mainly achieved by radiation at just the right frequency to flip the H2O molecules through 180 degrees and heat them by friction between molecules. The molecules in ice are too close to flip, as the above linked website explains. This is not thermal absorption, because the frequencies are too low, just as they are for radiation from a cooler atmosphere which also will not transfer thermal energy to all the ice and snow covered land surfaces.
And this makes an absolute farce of the IPCC claim that the most warming on the whole globe will occur at the North Pole – even up to 6 or 8 degress by 2100, so they say.
So we know the radiation enters water, but there is no quantum excitation involved which brings about conversion of radiated energy to thermal energy in the same way that solar radiation warms water. We have demonstrated that the frequency makes a difference, and the fact that ice and other materials are not warmed, proves that it is as described, with heating being by friction only.
Sorry, Pete, but once again you are not right.
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The aim is to prove that there are things which don’t get heated by the LW radiation in microwave ovens. What happens to all those photons? IPCC assumes photons transfer thermal energy to solids and liquids (like the land surfaces and oceans) regardless of the frequency of the radiation. They don’t if they come from colder regions of the atmosphere, which is nearly always the case. Microwave ovens demonstrate that at least some LW radiation doesn’t warm some things.
The black metal plate in my microwave oven does not get warmed by the oven’s LW radiation, whereas the same metal plate does get warmed in front of an electric bar radiator also emitting LW radiation with intensity of the same order of magnitude. The difference is in the frequency.
And this is the whole point of what we are discussing. Radiation with lower peak frequency than that for the emission which the target is emitting, will not transfer thermal energy to such a warmer target.
The computational proof is in Ref [3] – which is why I don’t need to include computations in my paper. Furthermore, the quantification of the process I describe has the same mathematics as is well known and well documented for calculating heat transfer between hot and cold bodies, so I also don’t need to repeat those computations in my paper. A simple explanation that the difference in the areas under the Planck curves is obviously the area between those curves is sufficient, knowing that the smaller curve is (at all frequencies) below the larger curve for the warmer body.
As I have said several times, people will not understand my argument until they read and study the whole paper. It would be preferable if commenters would indicate that they have read the paper and then quote Section or FAQ numbers and paragraph numbers to which they are referring.
There are other topics in the Appendix, including the temperature analysis which shows carbon dioxide has had absolutely no effect on climate. Do I assume that, since no one has argued about that point (which I first raised days ago) that you all agree with it? That point alone is sufficient to overturn the AGW conjecture, I suggest, and it is empirical evidence in my paper supporting the theory. So why do some of you keep saying there is no empirical evidence in my paper, or no computations when I have explained why such are not required?
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“The aim is to prove that there are things which don’t get heated by the LW radiation in microwave ovens. What happens to all those photons?”
Photons are transmitted or absorbed according to the complex index of refraction of the material at the photon’s wavelength Doug. That’s it. Your incredibly naive experiment is so flat stupid that it can neither prove nor disprove any theory, no matter how simple or convoluted as your current mess is. Most high school students could point this out to you.
You have chased the technical people from the thread with this nonsense.
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Jeff and others
Virtually nothing at room temperature converts the low frequency radiation in a microwave oven to thermal energy by absorption which causes quantum energy changes.
It is not because the objects are transparent. The black metal plate certainly isn’t. Why would ice be transparent and water opaque?
The process involves friction as explained on this page and the next two ..
http://www.colorado.edu/physics/2000/microwaves/water_rotates3.html
Why should low frequency radiation from the cold atmosphere be all that different when it strikes a warmer surface?
What empirical ground do the IPCC have for their implicit assumption that all photons will warm the land and ocean surfaces?
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Doug,
We have done what we can to teach you.
Dumber than two boxes of rocks and a shovel.
Doug,
I spent two minutes and requested that your paper be redacted. I’m sorry but your work is crap and doesn’t even deserve a place on the Internet.
Jeff
You have not provided a single word of valid physics, or any appropriate linked peer-reviewed papers that cogently refute any point I have made in the paper.
I eagerly await a proper peer-reviewed rebuttal of my paper which I look forward to rubbishing, as I have been able to do for every single attempt to rebutt my argument on any of the five climate sites on which I have been discussing the content.
The lack of science on your site here has been very noticeable of late. Just look back over your responses to some of the new points I have been making lately in support of the content of the paper.
See if you think anyone has made a valid rebuttal that I could not show to be incorrect anywhere in the tallbloke thread dedicated to my paper.
Regarding microwave ovens, explain what’s wrong if anything on this page and the next two …
http://www.colorado.edu/physics/2000/microwaves/water_rotates3.html
You have not even recognised that the fact that microwave ovens warm water but not ice (directly) clearly indicates that the concept of all photons warming anything is wrong. After all, LW radiation from an electric radiator sure melts the ice.
Address an issue like this, Jeff, because assertive statements like yours above are like water off a duck’s back.
Doug,
Bull.
Bad news for you is that I’ve been invited to rebut the thing. I’m not sure if I will take the time still, but if I do, there will be plenty of valid physics to go around for you.
Your explanations for the bolometer and laser are complete fantasy. I wonder why you haven’t addressed that yet?
Garbage Jeff
I have provided an explanation of lasers from a top expert in the field. In short, the inversion makes them act like a far hotter source.
I have explained how microbolometers detect differences in their radiative cooling rates due to the presence of objects of different temperatures. What is it that you don’t understand, Jeff? I thought you knew basic physics like the use of S-B for two parallel plates. It’s mentioned in my paper and so well known that it hardly seemed necessary to reiterate the equations.
And nor do microwave ovens disprove the hypothesis, as they heat water by friction as they flip molecules through 180 degrees, in synch with the wave motion of the radiation.
Sorry, Jeff. I’m in good company now with plenty of internal emails circulating within about 36 like minded members of
Principia Scientific International – all of whom are dedicated to upholding established, traditional physics and refuting the “post physics” that is now politically and financially motived.
It’s pretty obvious you have some incentive to uphold the status quo, Jeff, but the continued propagation of the AGW hoax is going to cost millions of lives due to misdirected funds which could otherwise have served more humane causes. You ought to think about that and come down off your pedestal. Maybe you’ll be wiser by the time you’re my age.
Regarding the rebutting, the whole team at PSI will absolutely rubbish you Jeff – go ahead – I can’t wait.
I’ll hold back on my book until I can include a thrashing of your rebuttal. You have no idea of the financial resources I can then plough into self publishing and promoting the book throughout the world – and there for all to read will be the rubbishing of Jeff Condon.
The problem is Jeff, you just don’t take the trouble to understand what I say, and, knowing you, you won’t cross your T’s and dot your I’s.
PS You’d better be very careful regarding any personal defamation – I can afford, and use none but the best of lawyers.
Just to be clear, are you saying you will look for any possible method to sue me if I attempt to rebut your garbage work?
I will look for personal defamation like “Dumber than two boxes of rocks and a shovel” of which I have kept a screen capture, should you attempt to delete the evidence.
You win Doug. Rebutting your work is not worth the expense.
😉
Just a few excerpts from the PSI website …
PSI AND POLITICS: THE BASIC PRINCIPLES
PSI is divested of any political purposes, and shall not engage in political activities; and strides for the advancement of the traditional scientific method (as per the ideas of Karl Popper) and resolute opposition to ‘post-normalism’ in science.
RE-ASSERTING TRADITIONAL SCIENCE METHODS
Climate science was the most conspicuous victim of a hijack attempt of science for global political policy purposes and stands as a stark warning of the insidious power of post-normalism. As President Dwight D. Eisenhower foretold, the monolithic juggernaut that is government science (a key driver of post-normal rationale) is all too easily able to subvert open science debate and deny empirical evidence when it suits.
All around the world supporters of the traditional scientific method are expressing fears that unless principled people unite to stem its ascendancy; ‘post-normal’ science will rise again and again in various guises.
It is the science blogging community that has become the self-appointed watchdogs of science ethics and through them has emerged Principia Scientific International.
Thanks to the modern miracle of the Internet a grassroots movement exposed the myth of so-called runaway ‘greenhouse gas’ warming. Further landmark achievements are sure to follow as long as openness and transparency reigns.
OUR BROAD STRATEGY FOR CHANGE
PSI aims to restore the image of the vast majority of scientists by proactively applying viable strategies (e.g. creating educational publications, lobbying, blogging, public speaking, media/marketing and pursuing legal remedies as a last resort) by:
•lobbying policymakers to address perceived misconduct
•exposing bias and promoting transparency (via ‘PROM’ process)
•confronting poor professional practice
•correcting misinformation
PSI welcomes scientists and all concerned citizens who support the Scientific Method and wish to see corruption in science exposed and remedied. PSI commits itself to advancing the formal creation of a new concept: open online peer-to-peer review. We believe such real time transparent assessment of the latest scientific findings will become the new ‘gold standard’ of peer review to replace the secretive and readily corruptive ‘pal review’ system at the root of the malaise within modern science.
BTW Do you confirm that “Jeff Id” is the same person as “Jeff Condon” ?
Nope,
Jeff Id is a grumpy SOB, and Jeff Condon is his happy second personality.
Which one do you like better?
.
Why carbon dioxide does not affect climate
Firstly, let me clear up a common misconception about microwave ovens heating things with “cooler” radiation. When microwave ovens heat water there is no quantum energy excitation as is involved when a cooler object is warmed by a hotter source of radiation. All I am wishing to demonstrate is that this natural thermalisation process does not occur in nature when the spontaneous (natural) radiation comes from a cooler source. So I am dismissing microwave ovens as being any contra-proof because a totally different process is involved. The very fact that they do not warm most composite solids does in fact support the hypothesis. At the same time, it provides contra evidence against the IPCC assumption that all photons warm any solid or liquid regardless of relative temperatures.
But let’s now keep the discussion to what’s relevant in natural processes between the atmosphere and the surface. We are not going to see spontaneous emission of such low energy (“cool”) radiation (equivalent to < 1K) having any noticeable effect on the surface. Microwave ovens work because they deliver (artificially) far higher intensities of such radiation than would occur naturally.
I am very well aware of what the "post normal" science claims about a greenhouse effect, having studied their propaganda extensively. Absorbing of such radiation does not make a greenhouse effect. Those that absorb are the very ones emitting radiation which will cool the atmosphere as it takes all its energy to space, even if it is scattered off the surface of the Earth first.
Water vapour and carbon dioxide also absorb direct incident Solar radiation for which the photons have about five times the energy of those leaving the surface. When at least half of these incident photons are sent back to space I suggest we see a clear-cut cooling process. Indeed, all the absorbing and reflecting of solar insolation prevents nearly 30% of the Sun's "heat" reaching the surface in daylight hours.
Now, narrowing it down to carbon dioxide and perhaps methane, these gases have very little effect on the warmer surface. Their radiation does not actually transfer additional thermal energy to the surface. All it can do is slow down radiative cooling, which is just one component of the cooling of the surface. The radiation itself is not slowed, just the cooling because some of the energy needed comes from the incident radiation, not the energy in the surface.
So, the surface "looks" out the atmospheric window and sees space (at 2.7 K) to which it can radiate full blast. It also sees a few spectral lines of radiation from about 1% of the atmosphere which is water vapour. And it sees even fewer spectral lines from even fewer carbon dioxide and methane molecules. This radiation from the atmosphere comes with a variety of effective temperatures, so the warmer radiation is more effective than the cooler.
However, the subsurface is (at say 200 metres deep) at a very constant temperature that is unlikely to change significantly within a thousand year cycle. Even if it does, there is indication that it would do so in a 1000 year cyclic pattern, there being also longer and shorter natural cycles. This sub-surface temperature controls and stabilises climate, basically because of the massive "thermal inertia" of all the energy beneath the outer crust. The stabilisation process causes other heat loss processes to compensate for any slower radiation, if the latter were even significant.
Finally, a closer look at the temperature records does in fact display evidence of a 60 year cycle superimposed on the 1000 year cycle. The effect of the 60 year cycle can be removed in order to study the underlying 1000 year cycle. It is then seen (Appendix Q.1) that the rate of increase has been reducing over the last 100 years or so from about 0.06 C degrees / decade to about 0.05 C degrees / decade. This is to be expected because we can extrapolate the 1000 year cycle to a maximum in about 50 to 200 years' time at the most. Absolutely all climate change can be attributed to these natural cycles and there is no effect whatever that relates to anthropogenic causes.
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Seven comments in rapid fire mode? Are you worried about something?
Jeff, I have a concern for those in developing countries, both health wise and spiritually. A friend of mine who has just returned from Brazil (where a team of Gideons distributed 1,050,000 copies of the New Testament in two weeks) was describing to me only last night at our monthly meeting, just a glimpse of their living standards and lack of safety over there.
Lives are being lost needlessly, and you, Jeff are one of those helping to maintain the greatest hoax in the history of the World when you rubbish (with words, not science) those like TurnedOutNice, myself and other PSI members etc who are in fact revealing the errors in the AGW conjecture. Huge funds are being misdirected – funds which could have far greater effect in the field of health and medicine, funds which will be totally wasted trying to control climate.
Let me quote a couple of recent posts on http://DrRoySpencer.com …
Turnedoutnice says:
March 29, 2012 at 10:40 AM
RW: stop trying to fool me, yourself and the rest.
The claim that the Earth’s surface emits more energy than is claimed to impact it as SW energy is, on average and except for the occasional case of a temperature inversion, the kind of mistake I would expect would not be made by anyone with more than normal education to age 16.
161 W/m^2 comes in as SW. There is no way that the Earth’s surface can on average produce more than 160.1 W/m^2 [there’s the 0.9 ‘permanently absorbed’, a nice little fudge factor for further confidence tricks perhaps?]
Because this is well below what the S-B equation predicts for 16 °C average temperature it seems these poor saps have had a brain wave that there has to be ‘back radiation’ to make up the difference.
But surely they’ve heard of something called the heat transfer coefficient for combined conduction and convection? And of course, the radiation part of that can have an emissivity <1.
You can't have transfer of heat energy from a colder body to a warmer body. You may have net zero heat transfer where the air temperature is nearly the same as the Earth's temperature and high relative humidity so no evapo-transpiration.
To summarise, my assessment of the Energy Budget is that the claim of 396 W/m^2 from the Earth's surface is radiation, and the thermals and evapo-transpiration is on top of this. If so, it's the biggest mistake in scientific history.
And the proof that my suspicion is correct is very simple. Take away the 333 W/m*2 'back radiation' and the sums still add up. Therefore, 2.7 times more atmospheric warming is claimed than reality, 4.3 times greater IR warming.
Do your sums add up correctly this way as well? They don't.
Tell it to the marines; it's a fundamental failure to understand real physics and practical heat transfer.
Reply
Doug Cotton says:
March 29, 2012 at 7:20 PM
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RW
TurnedOutNice is, as usual, correct. Though I would say “It’s the biggest hoax in the history of the world.” Let me assure you, he knows his physics and, more importantly, he thinks about the situation.
Maybe this net energy diagram will help you understand where the “heat” transfers take place. It’s all one way traffic from the surface to space.
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Doug,
You are in luck. I tried but your ignorance was beyond assimilation!!
here is the beginning of my rebuttal which will never be finished.
Dear XXXXX,
Your letter convinced me to give it a shot. I tried but got 4
paragraphs into the paper and gave up. This is not a scientific
article, but rather is a manifesto of stupidity. There is no logic in
the thing. I was unable to identify a paragraph without an egregious
error from the beginning.
I’m sorry but I cannot waste my time with something that a high school
physics student would reject. My recommendation stands, but it is
your blog.
Here is the start of what I did, use it as you wish:
Review of “Radiated Energy and the Second Law of Thermodynamics”
This paper purports to be a landmark document presenting new theories
on thermodynamics and more specifically, radiation in the context of
atmospheric physics and anthropogenic global warming. To achieve
such a hurdle, one would expect a rigorous application of past
fundamental thermodynamic equations and the new theory mathematically
expressed, such that science can move forward under the author’s new
paradigm. On the fact of it, not only is standard theory not
represented, there is no equation presented to expand or even
contradict standard theory. In fact, not one single equation of
thermodynamics is presented in a paper which sports the conclusion:
“Thus radiation from a cooler atmosphere cannot transfer thermal
energy to a warmer surface.”
Can words be enough to reverse the science of an entire society with a
proven energy production that is based on bulk thermodynamic laws?
Possibly, but the standard is certainly quite high. What’s more is
that the new standard must result in the nearly the same equations of
today’s science in the regimes for which science has already been able
to physically verify. We all know that Einstein’s ‘theory’ reduces
to Newton in low differential velocity frames of reference.
Thermodynamic radiation theory is the foundation of literally millions
of functioning devices, from satellite radiators to a child’s two-way
radio. All are forms of thermodynamic transfer of electromagnetic
radiation.
Individual actions of subatomic particles are governed by laws only
partially known to human physics. However, these observed
interactions have very well defined probabilities. How often a photon
is absorbed by a solar cell is very well quantified and is based on
the material in question. It is worth noting, and should not be
lightly brushed aside, that the temperature of the electromagnetic
sensor is not generally included in the equations for the detection
abilities of the sensor. If sensors rejected radiation based on
temperature, why wouldn’t engineers include thermometers on the sides
of electromagnetic sensors? Surprisingly, this landmark paper
doesn’t even attempt to explore these obvious avenues.
Best regards,
Jeff
A better explanation regarding microwave ovens and lasers
[REPLY: Doug, you are thin skinned and have admitted a ridiculous intent to sue for anything you see as defamatory. You are a moron in my opinion and can’t hold a candle to the general audience here. You know, the one chased away by your ignorant discussion.
I don’t like people who sue for stupid reasons. It is bad for business.
Go away. – and go back to school because your comment was standard physics. – Jeff]
Well Jeff, according to Rog, my thread on his site has attracted several thousand views. [snip]
[REPLY: Good for you Doug. Several hundred people telling you that you are wrong and you won’t listen to any of them. That is about 1 day’s views here when I’m not blogging. Go away, and go to school.
The Air Vent has 3.5 million views, 46000 comments and you are the first to be banned from what used to be a nice science blog.
Be proud!]
[Snip.]
You are not welcome here.
Just for you Jeff – you don’t have to let this post stick, and of course you won’t. …
[snip, and I didn’t read it either. Go away Douglas. You wish to conduct science by lawyer, do it elsewhere.]
[SNIP, didn’t read ]
Omg this thread was worth a laugh.
Too bad that even people like Doug have their followers who are now copying his claims across the forums.
Omg this purely was worth a laugh. GJ Jeff.
Doug needs meds. Seriously.