the Air Vent

Because the world needs another opinion

Fun Stuff

Posted by Jeff Id on January 21, 2012

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:

  1.  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.
  2. 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.

1,065 Responses to “Fun Stuff”

  1. Jeff Condon said

    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.

  2. Brian H said

    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.

  3. curious said

    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.

  4. Jeff Id said

    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.

  5. Leonard Weinstein said

    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.

  6. Leonard Weinstein said

    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.

  7. Jeff Condon said

    “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.

  8. Leonard Weinstein said

    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.

    • Brian H said

      Re: Leonard Weinstein (Jan 22 16:20),

      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

      But that’s exactly the “lag” whose energy buildup determines the temperature increase. Power x Time = Energy.

  9. Genghis said

    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.

  10. Leonard Weinstein said

    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.

  11. Leonard Weinstein said

    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.

  12. Jeff Condon said

    “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.

  13. Genghis said

    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.

  14. Robert Austin said

    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.

  15. kuhnkat said

    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??

  16. Leonard Weinstein said

    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.

  17. Leonard Weinstein said

    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.

  18. Genghis said

    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.

  19. Leonard Weinstein said

    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.

  20. steveta_uk said

    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.

    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…

  21. curious said

    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.

  22. JWR said

    No comment yet, I just want be informed on the various posts

  23. Leonard Weinstein said

    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.

  24. curious said

    23 – equals “I don’t have one”. Ok.

  25. Leonard Weinstein said

    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.

  26. Leonard Weinstein said

    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.

  27. curious said

    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.

  28. Leonard Weinstein said

    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.

  29. Carrick said

    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.

  30. AusieDan said

    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.

  31. curious said

    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

  32. curious said

    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.

  33. Leonard Weinstein said

    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.

  34. kuhnkat said

    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.

  35. AusieDan said

    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.

  36. 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.

    • JWR said

      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.

      http://www.tech-know.eu/uploads/IR-absorption.pdf

  37. AusieDan said

    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.

  38. Carrick said

    Curious:

    I looked at the sites LW quotes and I didn’t find a complete and continuous treatment on any of them.

    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.)

  39. Carrick said

    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.

  40. Duster said

    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.

  41. curious said

    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.

  42. curious said

    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.

  43. curious said

    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/

  44. Alan D McIntire said

    “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.

  45. steve fitzpatrick said

    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?

  46. Leonard Weinstein said

    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.

  47. Carrick said

    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.

  48. curious said

    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.

  49. curious said

    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! :-)

  50. Alan D McIntire said

    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

  51. Coldish said

    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!

  52. Coldish said

    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)!”

  53. Leonard Weinstein said

    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.

  54. gallopingcamel said

    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):

    http://maths.ucd.ie/met/msc/PhysMet/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?

  55. Leonard Weinstein said

    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).

    • kuhnkat said

      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

  56. Alan D McIntire said

    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.

  57. Leonard Weinstein said

    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.

  58. Leonard Weinstein said

    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.

  59. 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.

  60. 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.

  61. Brian H said

    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.

  62. 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.

  63. Coldish said

    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?

  64. JWR said

    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.

    http://www.tech-know.eu/uploads/IR-absorption.pdf

  65. Brian H said

    JWR said
    January 25, 2012 at 3:48 am | Reply w/ Link

    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.

    http://www.tech-know.eu/uploads/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! ;)

  66. curious said

    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.”

    ??

  67. Coldish said

    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?

  68. Leonard Weinstein said

    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.

  69. Jeff Condon said

    “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.

  70. Jeff Condon said

    “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.

  71. Jeff Condon said

    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.

  72. curious said

    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.

  73. Alan D McIntire said

    “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.

  74. curious said

    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.”

  75. JWR said

    @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.

    http://www.tech-know.eu/uploads/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.

    http://principia-scientific.org/publications/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.

  76. steve fitzpatrick said

    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.

  77. Leonard Weinstein said

    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.

  78. 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.

  79. 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.

  80. 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.

  81. Jeff Condon said

    “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.

  82. Jeff Condon said

    Doug,

    See 82.

  83. curious said

    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.

  84. 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.

  85. 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.

  86. 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.

  87. Duster said

    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.

  88. JWR said

    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.

  89. Mark F said

    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?

  90. Jeff Condon said

    “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.

  91. Jeff Condon said

    Mark F,

    Yup. That is exactly right.

  92. Jeff Condon said

    #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.

  93. JWR said

    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.

  94. JWR said

    96 typing errors
    It has nothing to do with statistics. Do not make your case more ridiculous.

  95. 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….

  96. 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

  97. steve fitzpatrick said

    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.

  98. 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.

  99. Carrick said

    JWR:

    Again, you are a bad loser.

    ’tis better to be a bad loser, yet win—than lose, yet not know that you lost and really suck at physics to boot.

  100. steve fitzpatrick said

    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.

  101. 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.

  102. steve fitzpatrick said

    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.

  103. 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.

  104. 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.

  105. 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.
    That is unfair and I apologize. Give your link please.

  106. Carrick said

    Doug:

    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.

    LOL.

    You don’t have an open mind, nor that you know anything that you could teach us.

  107. steve fitzpatrick said

    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.

  108. 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.

  109. Carrick said

    Steve:

    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!

    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.

    • JWR said

      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 !.

  110. Carrick said

    Doug:

    a degree in Physics, but I have engaged in ongoing study and tertiary tutoring of Maths and Physics

    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.

  111. “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?

  112. Carrick said

    Jeff:

    How about a link?

    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.

  113. 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.

  114. Carrick said

    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!!!

  115. 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?

    • kuhnkat said

      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

  116. gallopingcamel said

    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.

  117. Terry said

    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.

  118. steveta_uk said

    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.

  119. curious said

    123 :-)

  120. 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.

  121. curious said

    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.

  122. 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.

  123. curious said

    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.

  124. Jeff Condon said

    #120

    “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

    Oops. Don’t forget that the moon is blocking a moon sized chunk of 4 degree Kelvin outer space. Want to try again?

    • kuhnkat said

      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

  125. 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.

  126. 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.

  127. 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.

  128. curious said

    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.”

  129. 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)

  130. curious said

    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

    http://iopscience.iop.org/1367-2630/14/1/013054/pdf/1367-2630_14_1_013054.pdf

  131. steveta_uk said

    DC, from your linked page:

    But let me pose a simple, well known situation in which backradiation is created by the silver reflective lining on the inside of a vacuum flask filled with coffee at, say, 97 deg.C. Yes, the reflection of radiation back to the source “inside” the coffee will indeed slow down the rate of cooling, as will the insulation created by a near vacuum between the walls. But the backradiation will not raise that 97 degree temperature at all.

    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.

  132. Carrick said

    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.

  133. Carrick said

    Doug:

    where I’ve been censored because they have no answers and cannot prove Claes wrong

    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.

  134. Carrick said

    Steveta_uk:

    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.

    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.

  135. curious said

    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

    *****

  136. curious said

    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.

  137. Jeff Condon said

    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.

    • kuhnkat said

      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.

  138. Carrick said

    Curious the book is Bohren and Albrecht, Atmospheric Thermodynamics.

    Regarding diurnal effects, how about this?

  139. Leonard Weinstein said

    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.

  140. JWR said

    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.

    http://principia-scientific.org/publications/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

  141. Jeff Condon said

    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?

  142. Jeff Condon said

    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.

  143. kuhnkat said

    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.

  144. Jeff Condon said

    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 object sensor the energy is striking.

    It is back radiation from a cooler source 100% confirmed for all to see.

  145. JWR said

    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

  146. JWR said

    Look into the paper eq (1)

  147. kuhnkat said

    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??

  148. Anonymous said

    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.

  149. Jeff Condon said

    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?

    • Anonymous said

      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)

  150. 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.

  151. Mark F said

    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?

  152. Jeff Condon said

    Doug,

    How does the camera work?

  153. Jeff Condon said

    #160 – You are aware that you have linked to an article proving my point right?

  154. 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/

  155. Jeff Condon said

    “You need to follow posts on WUWT as they discuss serious stuff, not fun stuff.”

    What is the difference?

  156. Carrick said

    Doug:

    You had better read what Prof Claes Johnson has proven, because you clearly have no understanding thereof

    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.

    Never in the history of mankind have more people been bluffed in more ways by so few.

    WGAFF what you think? You don’t even believe photons exist as quantized particles.

  157. unkwoop said

    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?

  158. Jeff Condon said

    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.

  159. Terry said

    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.

  160. curious said

    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.

  161. Carrick said

    2nd try on the diurnal link.

  162. steve fitzpatrick said

    #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.

  163. curious said

    170 – thanks Carrick, works. Just had a quick read – looks good. Will read again when feeling sharper!

  164. Carrick said

    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. ;-)

  165. Carrick said

    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.

  166. Leonard Weinstein said

    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.

  167. beng said

    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”.

  168. http://wattsupwiththat.com/2012/01/29/making-things-matters/#comment-878941

  169. Jeff Condon said

    So now we’re going to leave links to comments rather than actual comments?

  170. kuhnkat said

    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.

  171. Jeff Condon said

    “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?

    • kuhnkat said

      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.

  172. Carrick said

    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.

  173. 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.

  174. Jeff Condon said

    Doug,

    So how does the camera work.

    • Carrick said

      Doug Cotton:

      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.

      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?

  175. Carrick said

    Latest science of doom blah blah blah vs equations:

    It is not surprising that the people most confused about basic physics are the ones who can’t write down an equation for their idea.

    [….]

    Over the last few days, as at many times over the past two years, people have arrived on this blog to explain how radiation from the atmosphere can’t affect the surface temperature because of blah blah blah. Where blah blah blah sounds like it might be some kind of physics but is never accompanied by an equation.

    Sound familiar?

  176. Carrick said

    Kuhnkat:

    can you give me an example of ir or visible absorption that knocks an electron out of place without being intensity dependent??

    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).

    i would much rather make sure there is absolutely no doubt in anyone’s mind.

    Fair enough. Fair enough.

    • kuhnkat said

      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?

  177. 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.

  178. 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?

  179. Jeff Condon said

    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.

    • kuhnkat said

      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?

  180. Leonard Weinstein said

    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.

  181. Leonard Weinstein said

    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.

  182. Carrick said

    kuhnkat:

    Any hope of a link to solving the 2 slot with photons?

    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).

  183. Carrick said

    Leonard:

    . 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.

    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.

  184. Carrick said

    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.)

  185. steve fitzpatrick said

    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.

  186. steve fitzpatrick said

    Sorry, that should have been: Leonard #193.

  187. steve fitzpatrick said

    Carrick,
    Great reference! I hope Leonard reads it.

  188. Carrick said

    Thanks, Steve. And that was an excellent exposition in #199. Hopefully people read and understand that too.

  189. Leonard Weinstein said

    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.

  190. Carrick said

    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.

  191. steve fitzpatrick said

    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.

  192. Jeff Condon said

    “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?

  193. 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.

  194. Jeff Condon said

    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.

  195. 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?

  196. 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

  197. Jeff Condon said

    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?

  198. Jeff Condon said

    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.

  199. 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.

  200. 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.

  201. Carrick said

    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.

  202. steve fitzpatrick said

    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.

  203. kuhnkat said

    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.

  204. Carrick said

    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).

  205. TimTheToolMan said

    From Pierrehumbert himself.

    http://geosci.uchicago.edu/~rtp1/papers/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.

  206. TimTheToolMan said

    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.

  207. 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.

  208. 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.

  209. 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

  210. TimTheToolMan said

    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.

  211. TimTheToolMan said

    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.

  212. Jeff Condon said

    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.

    • kuhnkat said

      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.

    • kuhnkat said

      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.

  213. 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.

  214. Jeff Condon said

    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!!

  215. Jorge said

    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.

  216. Leonard Weinstein said

    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.

  217. Carrick said

    Leonard:

    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

    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:

    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

    That is the statement that Steve and I both were really reacting to.

  218. Carrick said

    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:

    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.

    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.

  219. steve fitzpatrick said

    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.

  220. curious said

    230 – Leonard – please can I check what constitutes “Putting a layer of insulation” an object in a vacuum?

  221. Jeff Condon said

    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.

  222. steve fitzpatrick said

    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.

  223. Mark F said

    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.

  224. steve fitzpatrick said

    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.

  225. Jeff Condon said

    MarkF,

    Besides some conflict with Dougs theory, the IR sensor it is tuned to higher energy photons.

    • kuhnkat said

      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.

  226. steveta_uk said

    DC #227

    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.

    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?

  227. Leonard Weinstein said

    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.

  228. Jeff Condon said

    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.

  229. Leonard Weinstein said

    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.

  230. TimTheToolMan said

    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)

  231. 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.

  232. Jeff Condon said

    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.

    • kuhnkat said

      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.

  233. 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.

  234. steve fitzpatrick said

    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.

  235. Jeff Condon said

    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.

  236. Jeff Condon said

    And who said physics isn’t fun!

  237. 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.

  238. Jeff Condon said

    #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.

  239. 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 ?

  240. Jeff Condon said

    Jesus dude, C.A.M.E.R.A!!!

    HINT: There is a plate inside it!!

    Answer the question!!!

  241. 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.

  242. 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.

  243. Jeff Condon said

    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.

  244. 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

  245. 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.

  246. 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

  247. Mark F said

    Um, Doug? Stop digging. Now.

  248. Jeff Condon said

    #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?

  249. Jeff Condon said

    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.

  250. 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.

  251. 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

  252. 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°)

  253. 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.

  254. kuhnkat said

    Here is Wiki’s version of the IR Camera explanation:

    http://en.wikipedia.org/wiki/Thermographic_camera

  255. 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

  256. steveta_uk said

    #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?

  257. 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.

  258. 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)..

  259. Jeff Condon said

    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.

  260. 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.

  261. Jeff Condon said

    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.

    • kuhnkat said

      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??

  262. Jeff Condon said

    Kuhnkat,

    “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.

    I’m not sure the minimum energy threshold for the photoelectric effect.

    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.”

    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.

    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?

    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.

    • kuhnkat said

      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…

  263. steveta_uk said

    I’m amazed that Doug seemed to think I was supporting his argument. He completely ignored the heated plates section of my post (#274).

  264. Jeff Condon said

    Here is another link

    http://www.glolab.com/pirparts/infrared.html

  265. steve fitzpatrick said

    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.

  266. steveta_uk said

    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.

  267. Jeff Condon said

    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.

  268. 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

    .

  269. Jeff Condon said

    Claes Johnson couldn’t explain the camera either.

  270. 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

  271. 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.

  272. Jeff Condon said

    “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.

  273. TimTheToolMan said

    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.

  274. 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.

  275. 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.

  276. 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)

  277. 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.

    .

  278. Jeff Condon said

    “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?

  279. 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.

  280. Jeff Condon said

    hehe.

    What about -20 C?

    I read the “cutoff” apparently you don’t follow Johnsons math well.

  281. 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..

  282. Jeff Condon said

    “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?

  283. 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

  284. TimTheToolMan said

    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?

  285. 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.

  286. 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.

  287. correction: I did not say “no radiation goes from cold to hot”

  288. TimTheToolMan said

    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?

  289. TimTheToolMan said

    Doug writes “correction: I did not say “no radiation goes from cold to hot””

    Then precisely stated, what ARE you saying?

  290. Jeff Condon said

    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.

  291. 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?

  292. TimTheToolMan said

    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?

  293. 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?

  294. Tim – you won’t understand until you read Johnson or my site.

  295. Jeff Condon said

    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:

    http://144.206.159.178/FT/CONF/16414843/16414861.pdf

    How does 100 Kelvin sound?

    Good thing I didn’t write a book.

  296. TimTheToolMan said

    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?

  297. TimTheToolMan said

    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?

  298. Jeff Condon said

    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!

  299. Jeff Condon said

    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.

  300. TimTheToolMan said

    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.

  301. Jeff Condon said

    #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.

  302. steve fitzpatrick said

    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! ;-)

  303. TimTheToolMan said

    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…

  304. Tim

    I am not saying what you state I am.

    I am saying what is on my Radiation page.

  305. Jeff Condon said

    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.

  306. TimTheToolMan said

    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.

  307. 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.

  308. 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.

  309. TimTheToolMan said

    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” ?

  310. Jeff Condon said

    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.

  311. TimTheToolMan said

    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.

  312. 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.

  313. 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.

  314. Now how about discussing Hansen’s error – see my post #61 and Home page at http://climate-change-theroy.com

  315. Jeff Condon said

    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.

  316. 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.

  317. Jeff Condon said

    Another link:

    http://www.icsoconference2008.com/cd/pdf/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……..

  318. Jeff Condon said

    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.

  319. #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.

  320. Jeff Condon said

    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.

  321. Jeff Condon said

    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.

  322. Jeff Condon said

    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.

  323. 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

  324. Jeff Condon said

    “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.

  325. Jeff Condon said

    “With sensor temperature at TOA about 200K ”

    The Mercury sat sensor does not EVER go below -20C.

  326. Jeff Condon said

    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.

    • kuhnkat said

      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.

  327. Carrick said

    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.)

    • kuhnkat said

      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??

    • kuhnkat said

      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.

  328. 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.

  329. 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.

  330. Jeff Condon said

    “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?

  331. 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.

  332. #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.

  333. Leonard Weinstein said

    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.

  334. Jeff Condon said

    “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.

  335. Leonard Weinstein said

    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.

  336. 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.

  337. 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?

  338. Jeff Condon said

    “, as I did for you”

    Yeah right.

    What a joke.

  339. 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” ?

  340. 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.

  341. 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

  342. 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

    • kuhnkat said

      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.

  343. Jeff Condon said

    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.

  344. steveta_uk said

    “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?

  345. Leonard Weinstein said

    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.

  346. Carrick said

    Kuhnkat:

    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??

    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.

  347. steve fitzpatrick said

    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.

  348. Carrick said

    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 .

  349. Jeff Condon said

    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.

  350. steveta_uk said

    “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.

  351. steveta_uk said

    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.

  352. Jeff Condon said

    Steveta,

    Yup. It bothers me that he hasn’t figured that out. Your point in 369 pretty well stuck a fork in the thing.

  353. 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.

    • Carrick said

      Doug Cotton:

      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.

      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).

  354. 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”

  355. 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.

  356. Jeff Condon said

    “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.

  357. 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

  358. 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.

  359. Jeff Condon said

    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:

    http://www.icsoconference2008.com/cd/pdf/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.

  360. 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!

  361. 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.

  362. Carrick said

    Doug Cotton:

    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

    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.

  363. Jeff Condon said

    “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?

  364. Carrick said

    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.

  365. 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.

  366. 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.

  367. Anonymous said

    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.

  368. TimTheToolMan said

    Drat… that was me above rather than some anonymous poster who agrees with me ;-)

  369. #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

  370. 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?

  371. Carrick said

    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.

  372. 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?

  373. “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.

  374. 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.

  375. steveta_uk said

    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:

    We can also compare with a common teacher-class situation with an excited/high temperature teacher emitting information over a range of frequencies from low (simple stuff) to high (difficult stuff), which by the class is absorbed and re-emitted/repeated below a certain cut-off frequency, while the class is unable to emit/repeat frequencies above cut-off, which are instead used to increase the temperature or frustration/interest of the class. The temperature of the class can then never exceed the temperature of the teacher, because all coherent information originates from the teacher. The teacher and student connect in two-way communication with a one-way flow of coherent information.
    The net result is that a warm blackbody can heat a cold blackbody, but not the other way around. A teacher can teach a student but not the other way around.

    And we’re supposed to take Johnson seriously?

  376. TimTheToolMan said

    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.

  377. 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.

  378. 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?

  379. 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.

    .

  380. Jeff Condon said

    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.

  381. Carrick said

    Tim:

    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.

    Why wouldn’t they reradiate it? If they absorb a photon at that frequency, they certainly can reemit it (just isotropically).

  382. Carrick said

    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.

  383. Carrick said

    Left the link out for Inamdar and Ramanathan, sorry.

  384. TimTheToolMan said

    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.

    • kuhnkat said

      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?

  385. Jeff Condon said

    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.

  386. TimTheToolMan said

    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.

  387. Jeff Condon said

    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.

  388. TimTheToolMan said

    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.

  389. TimTheToolMan said

    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.

  390. Jeff Condon said

    “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.

  391. TimTheToolMan said

    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 :-)

  392. TimTheToolMan said

    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.

  393. Carrick said

    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.

  394. 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.

  395. 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.

  396. 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.

  397. Jeff Condon said

    self snip…

    What is your expected temperature for each model if correct?

  398. 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

  399. Carrick said

    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).

  400. Jeff Condon said

    And after you do, then tell us why the cameras work.

  401. 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.

  402. 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

  403. Carrick said

    Doug:

    But if suspended they will receive radiation from the ground

    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.

  404. 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?

  405. 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.

  406. Carrick said

    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.

  407. steveta_uk said

    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.

  408. 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.

  409. 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.

    .

    • kuhnkat said

      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.

  410. 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.

  411. Jeff Condon said

    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.

  412. 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.

  413. Jeff Condon said

    “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?

  414. Carrick said

    Douglas:

    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.

    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.,)

  415. Carrick said

    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.

    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.

  416. steveta_uk said

    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.

  417. Carrick said

    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?

  418. 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.

  419. Jeff Condon said

    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.

  420. 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.”

  421. Anonymous said

    Douglas:

    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!

    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:

    As a point of fact, it’s quite well know in meteorology that the surface layer responds quite rapidly to temperature, and 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.

    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?

    He also makes statements like “the object under the barrier will be warmer ..” without a slither of evidence..

    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.”)

  422. Carrick said

    Sorry that was me.

  423. Jeff Condon said

    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?

  424. Jeff Condon said

    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.

  425. 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.

  426. Carrick said

    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. ;-)

  427. Carrick said

    Douglas:

    Well you gave no indication of the depth of what you refer to as the surface layer

    I shouldn’t have to for somebody with 40 years of science experience.

    It’s a well defined concept already.

    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,

    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.

    Meanwhile read the emails etc in the above link.

    Link whore alert.

  428. 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.

  429. 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?

  430. 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?”

  431. Jeff Condon said

    “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. :D

    Define significant difference, I will provide a new example of a sensor which absorbs radiation exceeding your difference.

  432. Jeff Condon said

    Carrick,

    I can’t find any AVHRR data which is current enough.

  433. 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 “

  434. Jeff Condon said

    “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.

  435. 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

    ,

  436. 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

  437. Jeff Condon said

    #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?

  438. 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.

  439. 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.

  440. 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.

  441. Carrick said

    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

    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.

    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):

    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!

    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.)

  442. 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/

  443. Jeff Condon said

    “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?

  444. Carrick said

    #472=distraction.

  445. Jeff Condon said

    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!

  446. Jeff Condon said

    #471 Was stuck in the spam trap due to links. Enumerated references between here and there are offset.

  447. Carrick said

    Thanks Jeff. Nonetheless, like a fine, well worth the wait. :-P

  448. Carrick said

    grrr.. “Like a fine wine, well worth the wait.”

  449. 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.”

  450. Jeff Condon said

    “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.

  451. Carrick said

    #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.

  452. Jeff Condon said

    #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.

  453. Jeff Condon said

    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!

    • kuhnkat said

      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.

  454. 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.

  455. Carrick said

    Doug:

    I really don’t have time to give free online tuition in basic tertiary physics here on this site.

    What a twat.

    You aren’t even competent to explain F=ma. You’d probably explain how it was really F=mv.

  456. steve fitzpatrick said

    So very much time wasted on one person’s misunderstandings.

  457. 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.

  458. 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)

  459. Carrick said

    More quackery from Doug. Obvious CAGW plant.

  460. Carrick said

    Speaking Douglas Cotton and quackery…

  461. 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.

  462. “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.

  463. 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.

  464. 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.

  465. 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

    • kuhnkat said

      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.

  466. 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

  467. steveta_uk said

    Doug’s attempted distraction makes me think ho doesn’t actually read the items he links to.

    Skeptic readers should not think that the book will fortify their existing skepticism of CO2 causing warming. The authors agree it does. but have major qualms about the assumed positive CO2-related feed-backs and believe the sun plays a far greater role in the whole scheme of things.

    I hate to ask, Jeff, but do you ever end up banning commenters?

  468. Jeff Condon said

    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.

  469. Jeff Condon said

    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.

  470. steveta_uk said

    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?

    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.

  471. Jeff Condon said

    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.

  472. steveta_uk said

    Jeff, did my reply (a minute ago) get lost, or do you filter anything with a link to Prof Johnson in it?

  473. Carrick said

    Jeff:

    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.

    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).

  474. Carrick said

    Jeff:

    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.

    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:

    The idea of photon particles as carriers of heat energy is primitive, confusing and has led to the unphysical concept of back radiation. Radiation as wave phenomenon makes much better sense.

    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.

  475. Jeff id said

    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.

  476. steveta_uk said

    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.

  477. Carrick said

    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.

  478. steve fitzpatrick said

    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….. ;-)

    • kuhnkat said

      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?

  479. Carrick said

    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).

  480. Carrick said

    My implied question there is “Would these same ‘skeptics’ be arguing against back radiation then?” were this the IPCC “front page story”?

    • kuhnkat said

      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.

  481. Jeff Condon said

    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?

    • kuhnkat said

      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.

  482. Jeff Condon said

    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.

  483. 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.

  484. 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.

    • kuhnkat said

      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>)

  485. Jeff Condon said

    “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.

    ?

  486. 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.

  487. Jeff Condon said

    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?

  488. Jeff Condon said

    Actually, I’m quite confused as to why a bolometer doesn’t prove any concept. Perhaps that would be a good place to start.

    • kuhnkat said

      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.

  489. 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.

  490. Carrick said

    kuhnkat:

    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

    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.

    why the term Backradiation?? It is no different from any other with the same attributes

    Because it is downward pointing and it is the term that accounts for the totality of the classic GHG effect.

    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?

    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.

    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.

    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.

    • kuhnkat said

      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.

  491. Jeff Condon said

    “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.

  492. Jeff Condon said

    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.

  493. Jeff Condon said

    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.

  494. Jeff Condon said

    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.

  495. Jeff Condon said

    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.

  496. Carrick said

    kuhhkat:

    This is part of the disinformation with the use of the term backradiation

    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.

    BackRadiation is a PROPAGANDA term

    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.

    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

    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.

    Sorry, I make up my own rules and games.

    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.

    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

    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.

    • kuhnkat said

      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.

    • kuhnkat said

      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.

  497. Carrick said

    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.”

  498. Jeff Condon said

    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.

  499. Carrick said

    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?

    • kuhnkat said

      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.

  500. Jeff Condon said

    “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.

  501. TimTheToolMan said

    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.

  502. Jeff Condon said

    Climate models are smashmatics at their best.

  503. Carrick said

    I would say climate models are more useful in helping us understand the past than predicting the future.

  504. Jeff Condon said

    Finally, something we can disagree on.

    Too many arbitrary DOF for use IMO. The Hadley cell section of CAM3 really poisoned them for me.

  505. Carrick said

    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.

  506. Jeff Condon said

    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. :D

    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.

  507. Carrick said

    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).

  508. Carrick said

    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.)

  509. Jeff Condon said

    “(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.

  510. 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.

    • kuhnkat said

      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??

  511. $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.

  512. #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.

  513. 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?

  514. #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.

  515. #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.

  516. #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.

  517. 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!

  518. 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.

  519. 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?

  520. 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.

  521. steveta_uk said

    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.

  522. Jeff Condon said

    “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.

  523. 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.

  524. Jeff Condon said

    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.

  525. Carrick said

    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.

  526. steveta_uk said

    #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?

  527. #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.

  528. 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.

  529. #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.

    • kuhnkat said

      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??

  530. #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.)

  531. $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.

  532. 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.

  533. Jeff Condon said

    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.

  534. [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.

  535. 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.

  536. #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?

  537. 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.”

  538. Jeff Condon said

      “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.

  539. 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.

  540. Jeff Condon said

    “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.

  541. #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

    • kuhnkat said

      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/

  542. Jeff Condon said

    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.

  543. #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?

  544. Jeff Condon said

    Looks like a duck, smells like a duck, etc..

  545. 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.

  546. Jeff Condon said

    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.

  547. Jeff Condon said

    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.

  548. 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.”

  549. #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) ?????

  550. Jeff Condon said

    No chops Doug.

    I give, you win.

  551. 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.

  552. #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.

  553. #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.

    • kuhnkat said

      Dougie Pooh,

      why are you asking about BACKRADIATION, a term popularized by Climate Scientists pushing an agenda, when all there is is electromagnetic radiation?

  554. 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.)

  555. 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.