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## Lightsabers and Thermodynamics – Poll Below

Posted by Jeff Id on June 4, 2011

UPDATE:  Ids poll answers –

Question 1 – Would a minus 180 C block of ice the size of the moon and in lunar orbit warm or cool the Earth.

Ids Answer – It  would warm the Earth.   No matter what you assumed the definition of lunar orbit to mean for the majority of the time, the ice chunk would find itself blocking a view of the blackness of space at 3K or minus 270C.  In place of that minus 270C radiation which continually strikes the earth, a small section of the sky has a hotter, brighter minus 180C emission.  Therefore – the Earth Warms.

Question 2 – Would it warm the Sun.

Ids Answer – It  would warm the Sun for the same reasons as above.  The tiny little pinpoint in the distance would probably add as much as a distant star, but it would add a little and that is all it takes.

Now, if you get all caught up in how much and could it be measured etc. etc. and so on, you are probably lousy at story problems. The answers are actually quite simple.  It will be interesting to learn where I am wrong in the inevitable follow up comments :D.

———————————–

Sometimes I’m too short with people.  It comes from some of the pressures of the rest of life, I really should show more patience though, after all we’re all just trying to understand.

There are a group of people on who have published articles on global warming (on the internet only) who claim to have disproven the effect that CO2 has on surface air temperatures.  There are a variety of these sorts of things but one, which is most annoying, is the claim that backradiation from a colder source (air at higher altitude) to a warmer one (the ground) somehow violates the second law of thermodynamics.  This argument is false.  Now I could spend time with the equations and demonstrate that the net heatflow is always from warmer to colder, which it is, but I think those who misunderstand this point are missing the fundamental understanding of what temperature is and how energy can also flow from cold to hot.  Therefore, we only need words today.

This first section discusses the physical mechanisms of thermodynamic conduction.

Thermodynamic laws are based on a bulk material reactions to temperature.  When two objects come in contact with one another, one hot the other cold.  Heat energy will transfer from the hot object to the cold one – this is basically the simplest conceptual form of the second law of thermodynamics.   What would you say if I claim that on a microscale, some heat goes the other way?

Temperatures taken in bulk are a combination of the motion of individual atoms, mass of the atoms and the density of the atoms.  When different temperature objects of equal density and material type are considered, the difference in temperature is in the physical velocity (vibration, rotation or linear) of the molecules.  If the hotter object is more dense than the other, it is possible for the individual atoms of each mass to have the same physical kinetic energy, but since the hot object is denser, it has a higher probability of transferring its kinetic energy at any particular moment in time.  Note though, that this second situation does not preclude the occasional transfer from the cooler object to the hotter one.  This is because not every atom in the material is vibrating/spinning/or translating by exactly the same amount.  In such collisions,  an individual atom can be said to be cooler or warmer but cool and warm are truly bulk concepts rather than microscopic.  This whole situation can get far more complicated so I’ll leave it there, but the point is on average, heat ALWAYS will flow from hot to cold at a pre-determined rate.   The bulk rate is based on the probability of energy transfer on an individual atom from a micro-scale analysis.

Before moving on to radiation transfer, imagine a container of Nitrogen gas  (N2) at room temperature.  Two balls on a spring.  Each molecule can spin like a dumbell, have linear velocity like a car or spring back and forth.  If you instantly compress the bulk volume of gas to half the volume, the temperature rises instantly.   What happened to the velocity of the individual molecules?

The answer is nothing.  All you did was push them closer together so that the temperature increase you measure is simply an  increase in probability of collision and energy transfer to the outside walls of your imaginary container.  The atoms are closer together so more of them hit the wall each second.  Eventually the energy will be transferred out of the container until room temperature is met and the actual molecules of Nitrogen in your container will individually have slowed until the probability of energy transfer into the container (on a microscale) is exactly equal to the probability of energy transfer out.

This concept deleted above is incorrect. Serioso kept saying so in the thread below and after some review, I have to agree that he is right.

So when we were discussing the temperature of Venus on the previous two threads, and I made the comment that it was a bad comparison because of pressure, a side discussion started with ScienceofDoom assuming that I believed the high pressure caused the high temp.  This is also a common fallacy on the internet which is prone to raising my blood pressure. Unfortunately (or rather fortunately), I was staring at a pile of biofuel in the woods while comforting a cold can of ale against the blazing IR heat and didn’t see any of the discussion all last weekend.  It might be interesting though to explain from a conceptual level the reason that the huge pressure on Venus doesn’t create ANY of the temperature gradient.

Take a column of gas in a perfectly insulated and very long tube (hundreds of miles).  Place the tube in an earth-like gravity field such that the bottom of the tube achieves 90 atmospheres, the top is 0.1 atmospheres.  The tube contains only gas and no precipitates form and no convection occurs. No heat is added or removed from the tube and the gas is inserted at a single temp until the tube is full.  In the beginning, the gas in the tube base will be far hotter than that of the rarefied gas at the top due to compression which causes the molecules to be close together at the base.  By the second law of thermodynamics, temperature inside the tube will equalize over time until it is exactly the same from top to bottom.  This is incorrect as Paul Lindsay points out below — the correct answer is that with height, velocity is lost in a gravitational field so the top will be cooler — see the correct answer in the comments.

Heat always flows from hot to cold – buckle up it’s the 2nd law!!  After stabilization, the base will be at a higher pressure but the temperature will have balanced throughout the tube.  For the second law of thermo to hold, the lower density gas molecules need to transfer energy to higher density molecules at an equal rate per unit time — on a bulk scale only — for stable temps.

This means that the low density molecules MUST have a higher kinetic velocity!  Interesting, no?  If they didn’t, the temperatures wouldn’t be balanced yet, and heat energy would still be flowing.

Again, the velocity (kinetic energy) can be spin, vibration or linear motion, but the probability of energy transfer throughout the entire tube length must balance for the thermodynamics, which drive so much of our lives, to be functional. Therefore, the temperature gradient (lapse rate) on Earth and Venus must be caused by something other than the pressure of the gas.  PV = mRT is absolutely NOT the direct cause of lapse rate, those who tell you it is, are wrong.  Those who tell you pressure has nothing to do with lapse rate are also wrong, but that is due to convection and that is another story.

On the previous thread, someone else linked an article which claimed that global warming is false because the principle of back radiation violates the second law of thermodynamics. ‘ Back radiation’ is what occurs in climate science when cooler air at a higher altitude spontaneously emits a photon in a random direction which happens to go downward.  It goes upward with equal probability but that wouldn’t be ‘back’ radiation.  This is exactly equivalent to shining a light in a random direction.  Now there really isn’t such a  thing as an expert in electromagnetic radiation as the stuff isn’t fully understood, but lets say I’ve spent a good deal of time with it and am reasonably educated for a human so bear with me.

Electromagnetic waves are governed by the principle of superposition.  Where two lasers intersect, the beam fields add together and continue propagating as though they had never met. It is incredibly interesting in that if you re-create the field at any point before during or after the intersection, you exactly recreate the pattern of the original beams.  Holography is based on this principle. It doesn’t matter if the beams are different frequencies, or traveling in opposite directions, they have no effect on each other outside of the local interference patterns created.

Like ducks in a lake, the waves add up and keep propagating in their own directions independently of each other.

From Wikimedia

Electromagnetic radiation is emitted by all objects having a temperature.  It is a spontaneous occurrence which again, when taken in bulk can be approximated by an equation.  Planck’s blackbody equation approximates both the spectrum and the energy of the emission of various materials at specific temperatures. If you’ve made it this far, you probably aren’t one of the most technical readers here (unless you’re trying to catch me in an error) so I’ll put the wiki link in.  There are simpler forms, but I’m lazy by nature so the reader is left to find simpler links.  Planck’s law is as correct as Einstein and Newton, it governs the color of your incandescent light bulb as well as the sun and even the burner on your electric stove.  The color of the output of a visibly hot object is mostly based on temperature.  Ever heard the phrase – red hot, or white hot?   That is what Planck’s law defines.  I am familiar with it so when I first studied global warming, the basic theory was hit you in the head obvious correct within minutes.  Actually, it was so obvious I don’t remember considering anything else.  Another point is that all of the technical AGW bloggers and climate scientists I know of, agree on these points.

Putting together the principle of superposition and the basic physics of Planck radiation, it becomes apparent that colder objects also emit electromagnetic energy and this energy cannot be stopped (or even altered) by other electromagnetic energy.  If the lightsabers of StarWars were actually light, they would simply pass through each other unaffected.  That would make a lousy sword fight though!

According to Planck,  things above absolute zero (Kelvin) always emit this radiation.  Ice cubes, the dark side of the moon, light bulb filaments, glass, liquid nitrogen, and even the background blackness of space has a temperature of 3 Kelvin.  That means that staring at an ice cube with a temperature of 273 K (0C) will warm your eyes when compared to the blackness of space. The ice emits IR which strikes you and warms you.   You can’t stop it with electromagnetic energy, even if you are the sun.

Now some claim this violates the second law of thermodynamics, but it does not.  While the ice is heating my eye, my eye is at 98.6 (or higher if I’m mad at the internet)  and emits more energy/unit area back at the ice!!  The net direction of energy flow is still always from hot to cold, the god of physics is happy and thus my Id’s universe is complete.

So when people send you links claiming that backradiation from the atmosphere to ground is a violation of the second law, perhaps the question back to them should be – have you seen my lightsaber?

————

1. ### Carricksaid

Jeff, what is really interesting to me is how many people will try and make thermodynamics arguments that clearly have no real understanding of what that theory even says.

Your ice cube analogy, made me think of this way to stay warm:

Inupiat Eskimo Igloo

2. ### Jeff Idsaid

Carrick,

When I first read it I thought it’s only a convection blocker, but then I remembered what it is like under a really open sky in the cold. Great example.

3. ### Oliver K. Manuelsaid

Thank you, thank you, Jeff, for all your efforts to bring sanity to climatology.

A comment on Professor Curry’s blog summarizes my opinion of today’s climatology:

On Extreme Weather:

“Extreme weather will remain a mystery until the scientific community “gets real” . . . Fortunately for us, a blanket of waste products has accumulated around the unstable pulsar (at the core of the Sun), almost – but not completely – insulating Earth from the violent spasms of the pulsar.”

With kind regards,
Oliver K. Manuel

PS – World leaders and government scientists will naturally be reluctant to admit that they have as much control of Earth’s climate and weather as a flea has over the direction of the elephant on which it rides.

4. ### Dereksaid

Amazing, simply amazing. AND, saddening.
Not a mention of latent heat, nor changes of state, particularly of water.
To say you missed the ball park would be an obvious, gross, and stupid understatement.
Such a stupid omission on your part, is patently and publicly an admission, and confirmation that you are in “denial”.

It might be worth me also mentioning / reminding you, that.
All object emit IR at a peak frequency BECAUSE of the temperature they are at.
This does NOT necessarily mean, or imply that, IR is the main, or even a significant contributor to, the objects temperature.
IR emitted by an object is merely the RESULT of the objects temperature.
IR emitted is the objects “temperature signal”.

The objects temperature, could be, and probably is (particularly within earth’s climate system), because of,
completely different heat / energy transfer reasons.
“Reasons” other than IR..

Conduction and convection of sensible heat,
AND,
convection (and downward transport) of latent heat,
for two massive, dominant, and ignored examples / “reasons”.
Two examples that are also so commonly, and incorrectly, treated as one,
when they are two, separate, different, completely different sized, “things”.

BUT, so the present IR obsession goes, it’s IR, ignore everything else.
Especially the latent heat of change of state, endothermic, and exothermic, heat / energy transfers thereof.

5. ### Dereksaid

” So when people send you links claiming that backradiation from the atmosphere to ground is a violation of the second law,
perhaps the question back to them should be – have you seen my lightsaber?”

BTW – What is the “range” of your light saber.

“range” = mean free path length of a photon – ie, Nasif Nahle.
Attachment to Post 6 in this thread.

6. ### Jeff Idsaid

Derek,

You are misunderstanding the post, again. Read it carefully and don’t fear it from preconceived notions because it makes no claims which are untrue. I don’t think it contradicts the mean free path of Nasif’s work at all, but without checking, Nasif has the longest average mean free path I recall reading. I wonder what frequency he used – but I don’t care enough to check.

7. ### Joel Heinrichsaid

Jeff:

“Electromagnetic radiation is emitted by all objects having a temperature.”

Except of course when it isn’t. What kind of EM-radiation is N2 radiating at 300K?

“It is a spontaneous occurrence which again, when taken in bulk can be approximated by an equation. Planck’s blackbody equation approximates both the spectrum and the energy of the emission of various materials at specific temperatures.”

Ok. could you show how you derive the spectra of CO2 and H2O out of the same equation for the same temperature?

“Planck’s law is as correct as Einstein and Newton, it governs the color of your incandescent light bulb as well as the sun and even the burner on your electric stove.”

Sure, but only if you are talking about blackbodies. I at least learned in undergrad studies that you cannot use Planck’s law for gases just like that.

“Another point is that all of the technical AGW bloggers and climate scientists I know of, agree on these points.”

And if they all jumped from a cliff…

“According to Planck, things above absolute zero (Kelvin) always emit this radiation. … The ice emits IR which strikes you and warms you.”

Only if this “things, objects of stuff” are some kind of blackbody. This IR radiation wouldn’t warm you if you were made of N2 or O2 or He or…

As for the first vote: I imagined this opaque chunk of ice standing just between Earth and Sun, so, yeah it would cool the Earth. 🙂

8. ### Jeff Idsaid

Except of course when it isn’t. What kind of EM-radiation is N2 radiating at 300K?

Yes there are spectra which different molecules prefer, you get one point for that.

No points though for the sitting between the Earth and sun as you failed to mention LaGrange and that was not an assumption of the problem 😀

9. ### Joel Heinrichsaid

Jeff #8:

“No points though for the sitting between the Earth and sun as you failed to mention LaGrange and that was not an assumption of the problem”

The poll mentions the lunar orbit. So, “in lunar orbit” and “between Earth and Sun” is indeed a well defined place in space (while only the duration of its placement there isn’t). 😉

10. ### Seriososaid

I think the seventh paragraph is incorrect: The RMS velocity of an ideal gas depends ONLY on the temperature. Thus, if the temperature rises, so does the molecular velocity.

I didn’t try reading in depth the rest of your post, and won’t until this error is corrected. But bear in mind that gas in a gravitational field has both kinetic energy and gravitational potential energy, and the general tendency is for the gas to be isoenergetic with altitude.

11. ### Joel Heinrichsaid

Let me emphasis this again:

“This IR radiation wouldn’t warm you if you were made of N2 or O2 or Ar or…”

If a theory won’t work on 95-99% of “things” it is trying to explain, why exactly is it supposed to be a good theory?

“Backradiation” is supposed to slow radiation into space and thus “warm” the surface while there is no mention of conduction and convection, which would also slow down radiation into outer space and thus “warming” the surface. Is there anywhere even a ‘guesstimate’ of its contribution to this “warming”?

12. ### Jeff Idsaid

Serioso,

“I didn’t try reading in depth the rest of your post, and won’t until this error is corrected. ”

I don’t intend to correct an error that doesn’t exist.

13. ### Steve Fitzpatricksaid

Jeff,

Nice post. One of the common problems with “skeptical” arguments is the frequent disconnect from basic, well known physics. Things that are obviously incorrect make easy straw men for ‘warmists’ to knock down. Insisting that the physical fundamentals are correct is the best thing you can do to add credibility to the skeptical POV.

14. ### Steve Fitzpatricksaid

Jeff,

As a ‘peer reviewed’ author in climate science (O’Donnell et al), you do have all the credibility you need.

15. ### Jeff Idsaid

Joel,

The concept is very simple. Mean path length of most long wave radiation is quite short in the atmosphere. If N2 doesn’t catch it H20 does. With so many chemicals to chose from, it always gets caught and the Planck curve is a good approximation.

I don’t have a clue why people can’t accept the basic physics of our world. Yes the greens are out of their heads, yes they exaggerate for political purpose but yes the basic physics is also correct. If you don’t know of the ‘guesstimates’ for the basic warming than you need to read more rather than have it spoon fed from me. That is because the basic numbers should come very early in your study and it means you haven’t put the hours in. No insult intended.

There is nothing controversial in this post, that’s what makes it fun sometimes but if people are interested they should study themselves. The problem I have is people like Derek who study with a completely closed mind.

16. ### Jeff Idsaid

Thanks Steve. I also think getting the basics right are key to making the skeptic argument.

17. ### JTsaid

Jeff, just a comment on the use of the word “heat”. It comes in two forms, a noun: “the heat”, as in “how much of the heat in the ocean comes from the sun and how much from the earth’s core?”, and a verb: “to heat”, as in “incident solar radiation tends to heat the surface of the earth”. It is one thing to say that it is impossible for a cooler body TO heat (verb) a warmer body; and another thing to say THE heat (noun) in a cooler body can’t flow into a warmer body. The first, using the verb form, is a statement of the second law of thermodynamics. The second, noun form, is false, as you have been at pains to point out. Heating (verb) is a process by which a temperature is increased. Heat (noun), it seems to me, is an often uncertain reference to more than one kind of energy.

18. ### Bill Illissaid

The Venus temperature issue is difficult – mainly because the greenhouse gas argument does not work and anyone working on the science of it is committed to global warming theory so cannot contradict it in any form. Therefore, there is no way to approach the issue with back-up.

But we all know there is more to it. It is moot point for now until another physicist unafraid of being blackballed takes it on.

19. ### Geoff Sherringtonsaid

There are different dominant mechanisms of energy transfer between substances, which is an incidental reason for the historical distinction of energy regions into groups like X-ray, ultra violet, visible, IR, microwave etc.

Next, energy transfer can involve the nuclei of atoms also the electrons thereof, as well as the physical motion of such relative to a fixed point. Importantly and commonly, the entity to consider is often the molecule rather than the atom, with its set of atoms and electrons, a more complex case.

Energy transfer is often considered as heat transfer, to the neglect of other effects such as light production and X-ray fluorescence.

Some processes are best explained by classical theory, some by quantum theory.

Sooner or later we need to define a specific problem and then decide which of the above compartments, or combinations of them, best provides a best answer.

One of the main present “problems” is comparing energy flow at Venus to that at Earth. I think this is semantically interesting, but not overly productive in explaining Earth climate. A second problem seems to be “can back scattered IR radiation in the atmosphere warm the land or sea below?”

People, including me, are jumping all over the place in this series of blogs. Jeff, can we please have a complete outline of a dominant “problem” that can be worked on to keep the discussion tighter?

It would also be good if people used correct physical terminology and units, and used historical “Laws” that are relevant to the problem. I found myself quoting BTUs last night, which was a cut and paste convenience but one that should have been resisted.

Moving now from school teacher lecture mode to a specific case, I think there is a good deal of confusion about reversibility as shown by discussion of the 2nd Law of Thermodynamics. There has also been a brevity of discussion about the mechanisms of excitation of molecules such as CO2 and their subsequent transformation to lower energy states with the emission of photons. Such a reaction can repeat over and over (and does in a CO2 gas laser), but it can be initiated only if the external energy input is above a certain threshold. (It is interesting to read a good account of how a CO2 gas laser works.)

So I guess this is a polite request to see if the discussion can be made more specific and in tighter terms.

20. ### Geoff Sherringtonsaid

There is also a matter that has confused me for a long time. What is temperature, in relation to how it is measured in a gas?

A mercury thermometer is simple enough. Heat energy causes expansion of the mercury, whose meniscus moves up a graduated tube. But, this assumes that a representative subset of the gas has hit the mercury to raise its energy. What happens as we approach the near vacuum of outer space? The probability of a gas atom or molecule hitting that big jump of mercury becomes vanishingly small, so we must record a temperature that is non-representative.

However, if we are in the vicinity of Earth, the Sun will be streaming vast numbers of photons to the mercury, so the temperature taken in the shade will be different to that taken in sunlight. Which temperature is more valid for which purpose? How do we know what we are measuring when we are in more dense air, split in the traditional conduction, convection and radiation trio? Does a mercury thermometer respond to IR radiation at 30 micrometers wavelength?

A reference to a text book covering these aspects would be most appreciated. I last studied quantum theory formally about 1966.

21. ### MikeFsaid

Is BB radiation a significant factor in atmospheric heating?
1 square meter @300k emits 450W, but between 0-10**11Hz only 0.00004W
Are the photons above 10**11Hz effectively absorbed by the atmosphere?

22. ### AusieDansaid

I feel that you are all getting fixated by the wrong problem.

What we know (if we had enought education in Physics):
* CO2 is a greenhouse gas
* the implications of the second law of thermodynamics
* ditto Planks Law
* ya de da de da (but please stick with me because I will make a serious point

We also know:
* Humans have been emitting CO2 emissions at an increasing rate for many years
* the CO2 content of the atmosphere has been rising at a linear rate for some time also

BUT (and here come the punch line)
* The temperature of the lower toposphere is the same now as it was in 1980, although it has been somewhat higher in between (according to Dr. Spencer’s satelite readings).

What is needed is some theory that starts from the observable fact of the lack of any measurable rise in atmospheric temperature and goes on to explain that.

The present arguements just go around and around, but they are not about reality.
It is just the same as in economics (with which I am more familiar) which sets up models of pure competition and economically rational consumers and then wonders why they did not predict the Great Financial Crash.

Neither economics nor climatology attempt to come to grips with reality.
The starting points and the choices of key principles are wrong in both cases.

23. ### AusieDansaid

I feel that I have not adequately explained myself.
There are two important questions:
*What is the definition of a greenhouse gas?
*Is it realistic to describe the atmosphere / ocean / earth system as a greenhouse?

Then please refer back to Jeff’s description of a long column of gas and how, presumably sealed within, temperature equibilates throught its height.

Is this a valid description of the atmosphere/ocean/earth system?

24. ### Cedric Katesbysaid

There are a group of people on who have published articles on global warming (on the internet only) who claim to have disproven the effect that CO2 has on surface air temperatures. There are a variety of these sorts of things but one, which is most annoying, is the claim that backradiation from a colder source (air at higher altitude) to a warmer one (the ground) somehow violates the second law of thermodynamics.

Yep. It’s funny how that PRATT got recyled from the creationists.

It’s great that you have decided to smack such a stupid (and sadly famous) argument down.
Might want to go for the “32,000 scientists” PRATT next.
That’s an oldie but a goodie.

25. ### Paul Linsaysaid

Jeff Id “After stabilization, the base will be at a higher pressure but the temperature will have balanced throughout the tube”

I don’t think so. You’re leaving out gravity and the increase of potential energy with height. Since kinetic + potential energy is constant, the molecules at the top of the tube will have lower kinetic energy, hence lower temperature. In fact, you can approximate the lapse rate from just this as

-m*g/k

where m is the mass of the molecule, g is the acceleration due to gravity, and k is Boltzmann’s constant.

26. ### Jeff Idsaid

Paul,

I hate being wrong, but you are right. I’ll update the text above.

Where were you on that one Carrick??

27. ### TerryMNsaid

Jeff,

I think you may be conflating “cool less” with “heat”. Reminds me of calling a smaller budget increase (say 10% this year compared to 15% last year) a budget cut, as our teachers union and several other departments are wont to do, or speaking of ocean acidification, when it is becoming less alkaline… the sign is wrong in all 3 cases, I think. Put another way, both space and the moon are drawing heat, it just may be a titch less on the vector pointing at the moon.

JMO, YMMV.

28. ### TerryMNsaid

In the first sentence I should have said “warm” instead of “heat” to stay consistent with your post wording, sorry. Replacing that, I still stand by the point. Have a great rest of the weekend and enjoy our belated warmth! I finally finished getting my garden in today – record late date for me since I moved here in 1991…

29. ### Jeff Idsaid

Terry,

“I think you may be conflating “cool less” with “heat”.”

I’m not conflating, I am disaggregating the two. My point is that heat energy is actually being transferred to the Earth in greater quantity than without the ice chunk. The net vector is away, as it was before, but by superposition, non-zero electromagnetic energy is still coming to the Earth from the ice.

This is different from the claimed breaking of the second law of thermo due to an imagined universe where flow cannot occur from cold to hot. Electormagnetic energy can (and does) flow in that direction as well, just not as fast.

30. ### timetochooseagainsaid

Are we expected to neglect the gravitational effects such a mass would have on the Earth’s orbit and rotation? Because while that may be a factor in this specific scenario (not sure, really) I guess that would miss the point. Although honestly I can’t see the Earth’s orbit shifting in such a way as to alter the mean conditions, as I suppose it would alternate pulling the Earth to and from the sun. I also tend to think it’s orbit would not be stable at the same distance as the moon, especially if the moon is still there. And it would change the tides…ARGH, my brain hurts!

31. ### Seriososaid

Jeff,

Paul Lindsay (#25) makes the same point I made in comment #10. And, really, the RMS velocity of a gas depends solely on temperature (and molecular mass), not on pressure or density.

I have now read the remainder of your post and find it accurate. I would only add that the spectrum of back radiation is different from the spectrum of radiation from a solid or liquid (such as the earth’s surface). The latter is very much like black-body radiation, while back radiation has the same spectral structure as the absorption spectrum of the gases that are radiating (i.e., water vapor and carbon dioxide).

32. ### Bryansaid

The effect of insulation.

4 temperatures to consider
1. Cold moon.(-180c)
2. Deep space (-270C)
3. Sun 6000K
4. Earth 15C

If 1 is between 4 and 2 the part of the Earth facing 1 will readjust to a slightly higher temperature.
If 1 is between 4 and 3 the part of the Earth facing 1 will readjust to a slightly much lower temperature.
Insulation impedes the transfer of heat in either direction.

Some of the cruder IPCC advocates often quote the bare rock earth as an excuse for the mythical 33C “greenhouse effect.
This they say comes from “backradiation”from the greenhouse gases CO2 and CO2.
The readership of this site know that this is tiresome nonsense, easily refuted.

But why not “backconducton” or “backconvection” if you think it through you can make an equally good case for return of energy compared to the famous “what otherwise would be” scenario.

Backradiation is the radiative component of insulation nothing more nothing less.

33. ### Bryansaid

Post above should be greenhouse gases CO2 and H2O.

34. ### Jeff Idsaid

Serioso,

Paul Lindsay (#25) makes the same point I made in comment #10. And, really, the RMS velocity of a gas depends solely on temperature (and molecular mass), not on pressure or density.

No he didn’t, he pointed out the energy loss with altitude and was correct. My paragraph 7 remains correct.

A density change of the material most certainly does affect temperature. Conduction rate (temperature) is a combination of kinetic energy per particle and the likelihood of it being communicated across the boundary. These factors are most certainly affected by density of the particles in question. Pushing them quickly into a tighter volume does not affect velocity but rather affects the number of times per second the outer particles interact.

35. ### Carricksaid

Jeff, technically Serioso is correct at least on this point: Temperature is proportional to mean kinetic energy, density plays no role in that relationship.

I agree when you have a complex coupled system like climate, you have to include density effects, but especially the gradient in density.

In acoustic-gravity waves, density shows directly in the overall scaling of the amplitude of the wave. Pretty much everything else involves the quantity rho'(z)/rho(z). E.g., see the the Brunt-Väisälä frequency.

36. ### Jeff Idsaid

Carrick and Serioso,

I’ve been reading the equations again while you left your comment and I am now of the opinion that Serioso is right and my paragraph 7 is wrong. Blogging stinks when you don’t put enough time into it! 😀

I’ll have to revise it later though.

Actually, I think that while it is conceptually accurate the way it is written, the instantaneous compression results in more collisions inside the gas which would then increase its average velocity. Like the higher density, low speed condition can’t exist for long. It is a little interesting but I’m working now and won’t be around the computer until tomorrow.

37. ### Carricksaid

Jeff:

Where were you on that one Carrick??

Sleeping on the job. 😉

It’s rare in practice to ever see the dry adiabatic lapse rate in nature (nor is the temperature gradient governed by purely adiabatic processes.) While, we’re being technical, it’s the 1st law that governs the lapse rate not the 2nd.

Real world, in a laboratory, the temperature is more complex than just being determined by the dry adiabatic lapse rate, as the tube itself (and the room, with its assorted heat generating equipment, and the building with its central heating an cooling system) and the temperature external to the building) have a much larger influence on the temperature gradient of the air within the tube than gravity does.

I know this from personal experience. One of the oddest things I’ve discovered is that the room convection is worse on a cold night in a heated room than it is in the accompanying daytime (and its the least in the middle of a hot day, when the roof is warmer than the A/C controlled rooms below it). In the outside atmosphere, this is reversed: The air is generally more stable at night, with a temperature inversion set up that suppresses large-scale convection (and less stable in the day time). On a cold night, in a room, you end up with a large negative temperature gradient that strongly drives room convection. (Convection…air motion in general…has a strong effect on my laboratory based measurements, so it’s something I have to control for.)

Anyway, without specifying your experimental setup, it would be pretty much impossible to predict the temperature gradient imposed on the air inside of the tube. WIthout active temperature control or very substantial insulation, the temperature in the tube will generally track the average room temperature gradient in the vicinity of the tube.

38. ### Carricksaid

Jeff:

Blogging stinks when you don’t put enough time into it!

Oh, it’s OK to be wrong occasionally… that’s the only way you learn anything new. (There are places where it is very bad to be wrong, being wrong in an internet post isn’t generally one of them.)

39. ### Carricksaid

Err… I got this reversed: “On a cold night, in a room, you end up with a large positive temperature gradient that strongly drives room convection”

It’s why houses are drafty on a cold winter night.

40. ### suricatsaid

Jeff, analogies can be confusing if not well thought through. Instead of analogising about ‘hot radiation’ and ‘cool radiation’ between ‘bodies of varying temperature’. Would it not be better to coin a more realistic name than “back radiation” for “radiative insulation”?

I think the term ‘insulation’ helps to explain the physical process better from outset. It would also bring ‘macroscopic’ and ‘microscopic’ understandings closer.

Best regards, Ray Dart.

41. ### Anonymoussaid

Jeff,

I was a little shocked to see a new post! Everything you say is correct (noticing the one minor issue was corrected), but significance of “back radiation” versus net is just academic exercise. It is more easily explained by “the sky has a temperature”. The impact of a trace gas on temperature more easily explained by comparing 280 PPM CO2 (some what well mixed) with H2O at a similar concentration in the upper troposphere and how changing the concentration of either changes the temperature of the sky at that point.

Maybe if you get bored with not blogging again, you could place your imaginary cube of ice in the tropopause and have a look around. 🙂

42. ### TimTheToolMansaid

Jeff Id writes : “If you instantly compress the bulk volume of gas to half the volume, the temperature rises instantly. What happened to the velocity of the individual molecules?

The answer is nothing. All you did was push them closer together so that the temperature increase you measure is simply an increase in probability of collision and energy transfer to the outside walls of your imaginary container.”

I think your answer violates the second law of thermodynamics.

Firsly imagine what happens if the compression is quick rather than instantaneous. In this case the velocity of the wall will speed up each molecule that collides with it during the compression.

But you said instantaneous…so what does that mean? Well if you have a single molecule in a box then all you need to do is wait until the molucule is in the middle, apply your instant compression and voila, you’re right. With two molecules you need to wait a bit longer until they’re in the right positions…but hey it still works.

But those examples aren’t what you’re meaning are they. What happens when there are 10^23 molecules? In order for them to all be in the “middle of the box” so you can simultaneously compress without collision, they need to have moved into positions that are disallowed by the second law…

Hence I think compressing a gas necessarily increases the velocity of the molecules within.

43. ### TimTheToolMansaid

Now I’m going to contradict myself.

Imagine a box full of gas. Assumption : Not every part of the walls of box is being collided at every moment. Hence it would be possible to opportunitically compress the walls a tiny bit at a time whenever that bit isn’t being collided upon. little dimples if you like.

This should be possible until the container has reached half its original size and wont have imparted any energy on the molecules.

Will the temperature have changed? What are the implications of this?

44. ### gallopingcamelsaid

Cedric Katesby @24,

I am probably the only one here who understands your fixation with PRATT.

Perhaps you could explain the philosophy of the “Open Parachute”.

I can guarantee you will be treated with more courtesy than you deserve.

45. ### gallopingcamelsaid

Jeff Id,

Where you differ from your opponents is that you admit that you may be wrong.

Can you imagine Michael Mann admitting that Steve McIntyre did a better job on Mann’s own data?

Mann is much too small to admit error and that is why he should should not be taken seriously by anyone.

46. ### gallopingcamelsaid

TimTheToolMan,

I recommend that you look up “Maxwell’s Demon”.

47. ### TimTheToolMansaid

Camel writes : “I recommend that you look up “Maxwell’s Demon”.”

Maxwell’s Demon is subtly different. In that case high energy molecules are being separated from low energy ones thus creating a temperature gradient where none should exist according to the second law.

In this case there is no such separation. There is volume decrease, pressure increase (more molecules hitting the same piece of wall per unit time) but possibly no temperature increase.

To be honest I’m not sure what it means apart from some vague feeling that information is a fundamental and very real property of matter.

48. ### amabosaid

If the second law seriously prohibited heat to travel from a less hot object to a more hot one, wouldn’t my keyboard then feel sort of like a block of ice when I type on it?

49. ### Phillip Bratbysaid

Jeff, I’m afraid I have to disagree with you or your terminology. Sure, the radiation from the ice will affect the energy flow from the earth, but not so as to warm it.

For just the angle subtended by the ice (i), the energy radiated by the earth (e) will change from a factor proportional to Te^4 – Ts^4 to Te^4 – Ti^4 (according to the way I always use to do heat transfer calculations, where s is space). The rate of change of the earth’s temperature will thus change accordingly. If it is daytime then the earth’s surface facing the sun will warm up more quickly if the energy received from the sun is greater than that radiated away. If it is night time, then the earth will cool more slowly. But in neither case will the ice warm the earth. It is the sun’s energy that warms the earth, the ice just affects the way the earth loses energy by radiation. It is no different from putting a blanket round a hot water bottle. The blanket won’t cause the hot water bottle to warm, it will act to slow its rate of cooling. The ice is acting as a radiative insulator and is no different from an insulator reducing conductive cooling or a barrier reducing convective cooling.

50. ### Jeff Idsaid

Phillip,

Can you tell me at which temperature you consider a body ceases being at least a cool sun and turns into a blanket?

51. ### Phillip Bratbysaid

Jeff,

I’m not sure what your question means. If a sun stops generating energy, i.e. fusion ceases, then once it is cooler than earth, it becomes a blanket to earth and slows earth’s cooling to space (assuming the earth’s internal energy generation continues to keep the earth warmer than the sun).

52. ### Jeff Idsaid

Phillip,

So I assume that in your view it stops emitting to Earth when it becomes cooler than earth but might still be warmer than a more distant earth and still be a sun?

53. ### Dereksaid

Jeff Id wrote in Post 15

” There is nothing controversial in this post, that’s what makes it fun sometimes but if people are interested they should study themselves. The problem I have is people like Derek who study with a completely closed mind. ”

Oh the irony… LOL.
Oh, the later corrections, and more yet to come.

Aussie Dan wrote in Post 22
” Neither economics nor climatology attempt to come to grips with reality.
The starting points and the choices of key principles are wrong in both cases. ”
A brilliantly simple and correct comment if I may say.

Jeff Id you just do not get it do you.
Your example is complete baloney.
IR maybe “dominant” in space (think large vacuum flask), but within earth’s climate system it is THE minor player.
IR is an objects temperature signal, and not much else really.
Your floating in space ice cube will not effect earth’s surface temperature, except
for where it (partially) shields the earth’s surface from solar input
(so it will lower surface temps slightly by blocking some incoming solar input and thereby reducing received solar input W/m2 at the surface).
AND, you are confusing two different situations, in space IR is “dominant”, because there is nothing to conduct or convect with.
But within the earth’s climate system conduction and convection of sensible heat, and convection of latent heat will completely overwhelm / make irrelevant, any IR from your orbiting ice cube (which any IR from is not absorbed positively anyway).

I have said before, and it is worth repeating now,
“skeptics” should be questioning the principles of AGW,
not merely quibbling the numbers of AGW.

Jeff Id, the greenhouse effect “theory” is a politically convenient, imaginary hobgoblin,
the reasons why this is so are as described by H L Mencken in this quote.
The whole aim of practical politics is to keep the populace alarmed
(and hence clamorous to be led to safety)
by menacing it with an endless series of hobgoblins, all of them imaginary.

H. L. Mencken.

So, Jeff when you wrote in Post 15
but yes the basic physics is also correct.
Who has a closed mind????
Your stating that, in your view, the basic physics is correct for an imaginary, non-existent, unproven, effect…

BTW – Solids, liquids, and gases have different IR receiving and emission properties,
this does seem to be ignored / confused / or unknown to and by many, Jeff Id included.
AND,
with relation to mean free path length of an emitted photon Jeff, CO2 emitted IR (at a specific frequency – because it is a gas) is NOT absorbed by Nitrogen (80% of the atmosphere), but it is absorbed by water vapour. Unfortunately such emitted IR would have to travel a long, long way before being absorbed by CO2 (which is what Nasif Nahle’s point actually is (that is not how you have portrayed it in this series of comments). But you have done that with me before now as well, misrepresenting my points, into your preferred and incorrect / unphysical views.
So, the vast majority of CO2 emitted IR is absorbed by water vapour in just a few (maybe occasionally up to tens) of meters maximum. But, more usually, such CO2 emitted IR is, within a couple, or tens of millimeters absorbed by water vapour.

I would of thought it obvious that,
if you increase the amount of gases within a mixture that have a radiative ability,
all you do is increase the mixtures ability to redistribute heat MORE QUICKLY.
Within the pressure gradient of a planet’s atmosphere this would mean that the atmosphere would be able to cool itself quicker.
BUT, within earth’s open to space atmosphere system the latent heat transport by water vastly dominants, or rather swamps into irrelevance such small changes of IR ability, particularly in the lower atmosphere.

54. ### Jeff Idsaid

Derek,

I do make mistakes, but I also admit them and learn. You (and Michael Mann) write garbage and never learn. I wonder if you will ever gain the ability to try.

55. ### Phillip Bratbysaid

Jeff,

I don’t know why you should “assume that in your view it stops emitting to Earth when it becomes cooler than earth but might still be warmer than a more distant earth and still be a sun”. I never said or implied that. No of course it does not stop emitting to earth (and in all directions).

56. ### Jeff Idsaid

Phillip,

I’m not assuming, I’m trying to understand your view. I think I do now but don’t understand the disagreement with my post.

57. ### Jeff Idsaid

Phillip,

You wrote.

“But in neither case will the ice warm the earth.”

So do you disagree that the earth would be warmer in its presence than without and if not, does that warming come from the additional emissions from the Ice body to Earth?

58. ### Phillip Bratbysaid

Jeff,

No, I didn’t imply it wouldn’t be warmer. I said the ice was slowing the cooling. The sun makes it warm, the ice impedes the cooling. Thus the earth would be warmer than it would be in the absence of the ice, but it is not the ice that makes it warmer, it is the reduced cooling that ice causes.

59. ### Jeff Idsaid

Phillip,

It is an inconsistent way to think of it IMHO, but it works fine. I don’t think it is accurate to say the Ice doesn’t warm the surface because it does transfer heat energy to Earth. You have obviously gotten past the point that back radiation is not impossible, which is the point of the post, but I don’t understand the reason for shifting thoughts on energy transfer when one body gets cooler than the other. A cold sun. It is just a net flow vector change.

60. ### Phillip Bratbysaid

I never was at the back-radiation not possible. Only that I never calculated such a thing, only net energy transfer as the T^4 dictates.

61. ### Dereksaid

Post 54 LOL.

Time will tell who is writing garbage, and not learning, nor improving their understanding.

I am certainly not into simply believing what supposedly “higher authorities” tell me,
can you honestly say the same Jeff.

“back radiation” comes down to is all radiation absorbed positively.
I am surprised no one wonders why man has not done the often quoted two bodies at different temps. in a vacuum experiment for real.
“We” have been to space often enough, surely….You’d have thought NASA knows.

1) All positively. (Hotter body cools slower)
2) “Only above the peak frequency” – ie, Claes Johnson approach. (I am not sure what Claes says happens rate wise)
3) Relatively. (Hotter body cools faster)

No one actually knows though (and NASA ain’t saying).
In the end no one disagrees about the end result, both objects will be the same temperature,
nor does anyone really disagree about the net flow,
it is the how (and amount of time) that “we” do not know.
But, GH “theory” says IR must be all positively absorbed……

62. ### Carricksaid

Speaking of LOL.

Derek actually doesn’t think we’ve ever done radiation heat transfer measurements before, two body or otherwise.

Ignorance thy name is clueless.

63. ### Jeff Idsaid

Derek,

These things you splatter across your keyboard are well known, yet you seem to know none of them.

You need to read more, and not the crazy stuff.

64. ### Dereksaid

I come here for the crazy stuff?

Carrick said Post 62
” Derek actually doesn’t think we’ve ever done radiation heat transfer measurements before, two body or otherwise. ”

Brilliant news Carrick, please give an example of an actual experiment done, not a thought experiment.
Obviously two bodies in a vacuum.

I do vaguely remember Cement a friends comment of an actual example, but that did not show all positively absorbed,
in fact it showed relatively absorbed.

65. ### Jeff Idsaid

Derek,

Ever heard of a toaster? Really, a toaster. Or an incandescent light? Coleman tent heater?

Oh, you want a vacuum? Every satellite in orbit? The doors of the space shuttle, the cooling systems of the international space station. The two bodies in those systems are deep space and radiative coolers.

Then there are lunar temp calculations.

Imagine science being so stupid as to not test Planck’s law.

66. ### gyptis444said

Is there an astrophysicist in the house?
I thought all these issues were covered in stellar evolution.

67. ### Geoff Sherringtonsaid

The arguments still lack definitional exactitude.

Some of the argument here can be thought of in terms of what I wrote on the previous blog:

“This is part of the problem of assuming that a gas under pressure – returning to the Earth atmosphere case – is hot. No, it is hot only if the system is moving fast enough. Think of a bicycle pump. The pump heats the air by compression, it goes into the tire and it is hot. When pumping stops, it cools again to ambient, although it is still under pressure. However, if the tube has a hole and you have to keep pumping to maintain the tire pressure, the tire will get hot and stay hot so long as you pump. This is a type of analogy to the height in the atmosphere that forms a discontinuity because of inadequate convective/conductive transfer above it. Above that height, radiative transfer must dominate. (This does not mean that it might not dominate at lower altitudes in some circumstances).”

You have to bring dynamics of energy flow into the equation, not just the fact of energy flow.

Imagine the tire with a tiny leak, then with a large one. One will lead to a quasi equilibrium at a high temperature in the tire, the other can lead to fast transit and trivial change in the tire temp.

68. ### Dereksaid

Jeff Id Post 65.

” Imagine science being so stupid as to not test Planck’s law. ”

Imagine people being so stupid as to not realise it has not been done.
I asked for an experiment, any experiment, just a single solitary example of an experiment where a hotter and a cooler body in a vacuum are compared, on their own and together.
This has not been done to my knowledge.
The only examples are thought experiments, and they all fail.

” Then there are lunar temp calculations. ”

Which only go to show the regolith retains heat, as NASA admitted.
GH “theory” can not cope with that at all.
You just do not see this at all do you.
May I suggest that people download and re-read Alan Siddons pdf titled, Greenhouse_Effect_on_the_moon pdf,
attached to Post 3 on this thread at GWS,

In the above article Jeff Id writes,
” heat ALWAYS will flow from hot to cold at a pre-determined rate. The bulk rate is based on the probability of energy transfer on an individual atom from a micro-scale analysis. ”

FOR A SOLID. IR is emitted by solids because the “stuff” is constrained and knocks into other “stuff” in the solid.
Gases ARE different. They can move around, and tend to emit and receive at certain specific frequencies.
So, when you are talking about two solids (an orbiting ice cube and earth), “separated” by a mixed, open to space, and pressure gradient gases (atmosphere) “things” are NOT ALL SOLID.

” the fundamental understanding of what temperature is and how energy can also flow from cold to hot. ”

Temperature is, when measured with a thermometer, the number of “things” banging into the thermometer at the energy (vibration) level they are at.
At lower pressures the same thermometer would show a lower temperature, because there would be less “things” banging into it, even if they were at the same energy level as the higher pressure reading.

but one, which is most annoying, is the claim that backradiation from a colder source (air at higher altitude) to a warmer one (the ground) somehow violates the second law of thermodynamics. This argument is false. Now I could spend time with the equations and demonstrate that the net heatflow is always from warmer to colder, which it is,

Do you realise you have just admitted there is no GH effect.
GH “theory” says the same again as the solar input is returned to, and further warms the earth’s surface.
(ie, Yale and Harvard teach GH “theory” as, 239W/m2 from sun + 239W/m2 “back radiation” = 478W/m2 at the earth’s surface,
or rather -18C + -18C = +30C.)
You have just stated it can not.
(I would also question the amount of “back radiation” amounting to supposedly 239W/m2 ie, range and size of your lightsaber, AND
239W/m2 received from the sun, which is only that low because of P/4…)

It appears to me that you have confused the heat flow is always from warmer to colder, because of “all radiation is positively absorbed”.
So, either your arguing the 2nd Law of thermo is incorrect, or that all radiation is not positively absorbed.
I’d be interested to know which you are arguing please Jeff, AND,
do you admit you have just helped destroy GH “theory” as well.

NB – Have you removed some words from the piece Jeff?
I seem to remember you stating atmospheric nitrogen absorbs IR,
which as a very stable 0=0 gas it can not do much, if at all.
CO2 being a 0-0-0 gas is a radiatively able gas, and hence redistributes heat within the atmosphere,
aiding, and increasing the atmospheres ability to cool itself.

69. ### Carricksaid

Derek:

Brilliant news Carrick, please give an example of an actual experiment done, not a thought experiment.
Obviously two bodies in a vacuum.

There are whole textbooks full of descriptions of experiments. I’ll echo Jeff and suggest you spend a little time at the library, in the radiative physics section.

Seriously, Derek, we arrived at the current theory empirically (including Planck’s law). We didn’t just make them up. In fact, thermal radiative physics was instrumental in the discovery of the quantum nature of matter. I suppose you didn’t know that either.

We even have devices that work based on our understanding of radiation physics, see for example the radiometer.
The theory is well enough developed, it gets routinely applied (with highly accurate results) in a diverse set of fields

But no, I’m not doing your homework for you. If you have access to a technical library, I’m willing to suggest titles for you. But it’s not my job to teach unteachable people.

70. ### Dereksaid

Post 69 – Errr, thanks I think. Not so brilliant news then. No such experiment has been done.

BTW – Quanta were “invented” because of (as a workable “solution” to) the ultraviolet light catastrophe, and
as Einstein said (please excuse me, I’ll paraphrase him) way back in 1953,
“To the best of my knowledge there is not a man alive who knows or understands what a quanta is”
That still holds true today I think.
I certainly do not know, nor will I pretend that I understand or know what a quanta is.

71. ### Jeff Idsaid

Derek,

I hid nothing from my post, or any other on tAV.

You need to listen and read now. Your typing of nonsense is not teaching you anything. As before, I gave you your examples expecting from experience that you wouldn’t listen. You didn’t.

It is a very stupid person who can’t ever be wrong, even in the face of being wrong. I strongly suggest do some reading! Try the science section instead of Derek’s own imagination section.

You don’t have one clue what you are talking about. Loud, ignorant and deaf are an unappealing combination.

72. ### Dereksaid

Jeff do you ever, actually reason? Simply blind belief is also not appealing.
As for your ad homs and insults towards me, so what.

I admit when I am wrong when presented with FACTS, that PROVE such. I do change my mind when the facts change, but I am careful in assessing what are to my mind are actually “facts”. If facts change I have to be sure they have changed, AND as portrayed. Something I hope we all do.
You have already in this thread set a president for not having to admit to being wrong first time, or that you need persuading (quite correctly) of such.
ie, “Serioso kept saying so in the thread below and after some review, I have to agree that he is right.”
So,
I asked for an experiment to test a specific prediction of “is all radiation absorbed positively”, that has been done,
not irrelevant and different “examples” with no figures / data.

” It is a very stupid person who can’t ever be wrong, even in the face of being wrong. I strongly suggest do some reading! ”
Have you re-read Nasif’s paper yet?

Maybe you would be better addressing some of the points I raise, rather than avoiding doing so by throwing unwarranted, unjustified, and indeed, patently incorrect “stuff” at me.
I don’t mind though, it is your blog, and I am merely visiting, raising and pointing out a few things.
Time IS telling, as I said it would…

I like thermodynamics, it is so much more REAL than the present “radiation obsession”,
which is a delusion many including yourself are obviously suffering from.
It is figuratively and literally clouding your judgment from and of reality.

73. ### Jeff Idsaid

Derek,

I don’t recall you ever admitting your error.

https://noconsensus.wordpress.com/2010/12/19/fixing-the-basic-agw-calculations/

I have presented real world examples above to the same effect.

74. ### Jorgesaid

Derek,

You might like to see how the Planck result was obtained. It seems to me that it was a serious attempt to make theory fit the facts.

http://www.webpages.uidaho.edu/~crepeau/ht2009-88060.pdf

75. ### Dereksaid

Jeff wrote,

” I don’t recall you ever admitting your error. ”

P/4 is the error, it is unphysical.
Maths is determining (mangling) the physics, when maths should only describe the physics,
when applied to a given situation in a physically correct manner.

My “error” as such, with the “shape issue” piece you refer to is, the piece quantified a problem, before I could explain it properly.
I had hoped people here being “skeptics” would of grasped what was attempted to be described, and helped out with trying to explain it better.
For that I apologise, but the basic issue I tried to describe does exist within GH “theory” and the explanations / teachings of it,
so in that respect it is not an error as such, nor was I wrong.

Thanks Jorge, I’ll follow that up.

76. ### Carricksaid

Derek, experiments have been done. Tons of them, and the experimental body of measurements already fully overlaps almost any experiment you could imagine. But if you aren’t happy with the state of the theory, I’d suggest doing your own measurements. Maybe you could learn something new. I’m sure even an exact experiment of the type you demand that we dig up for you (because apparently you’re too lazy to do your own homework) wouldn’t convince you, unless you did it yourself.

As Jorge points out, the entire field started out as an empirical endeavor. It is interesting that as the experimental measurements piled up, we were forced to adopt a quantum mechanical picture for describing thermal radiation…this helped lead to an unprecedented revolution in science.

Radiative heat transfer is at the point, where it probably gets applied hundreds of times a day, if there were any glitches with the theory, it would have been noticed long ago.

A simple example that comes to mind for “back radiation” measurements are radiometric atmospheric profilers, such as the one described here.. These devices are placed on the ground and scan segments of the sky, starting from near horizontal to near vertical. They use the measurements from the series of segments to reconstruct temperature and humidity. These devices have been multiply calibrated against other ways of obtaining profiles, such as radiosondes and tall meteorological towers (such as the 1000-ft Erie, CO tower).

Fundamentally, the theory works exceedingly well… so much so, it has left the demesne of experimental science, and is now just another tool in the arsenal of applied scientists and engineers. The real limits currently aren’t in the radiative transfer part of the theory, but in modeling e.g. the effect of pressure and temperature on the emission spectra observed on the ground. (Pressure gives rise to “pressure broadening”, temperature to “Doppler widening”).

77. ### Dereksaid

Carrick wrote,

” Derek, experiments have been done. Tons of them, ”

Yet, you give not one example.

” But if you aren’t happy with the state of the theory, I’d suggest doing your own measurements. Maybe you could learn something new.
I’m sure even an exact experiment of the type you demand that we dig up for you (because apparently you’re too lazy to do your own homework) ”

Next time I’m in space with nowt to do, I promise to do such an experiment.

” Fundamentally, the theory works exceedingly well… ”

Which theory are you referring to?
Please tell me you mean and / or include the greenhouse effect “theory”.

78. ### Chris Colosesaid

Note Derek’s very similar debating style

79. ### Carricksaid

I gave you examples of working devices, Jorge gave you a paper that summarizes a number of experiments, and for the rest, I told you to do your own homework.

As to the theory, I think we were talking about the theory of radiative heat transfer, weren’t we?

It’s also telling that you aren’t clever enough to realize you can do these experiments on the Earth.

80. ### Jeff Idsaid

Derek,

Your equations are flawed. I pointed out the errors quite clearly as did others. They are flatly, 100% wrong. No ifs ands or buts, just flat wrong. The math is wrong and yet you still claim it wasn’t.

Fix your equations and the result becomes standard fare. Of course you have to listen in order to hear.

81. ### Dereksaid

” Your equations are flawed. ”

My equations ?? I used the standard equations, not mine. I just added up different “shapes”,
to show how they could be combined to produce a mathematically correct,
but physically imaginary “33 degrees effect” .

P/4 is unphysical. This is the error, and it is repeated within the radiative transfer described by GH “theory”.
GH “theory” as presently taught by the “consensus” AND accepted by figure quibbling only “mainstream skeptics”.

I will hopefully post later (permission is being sort) why GH “theory” as presently taught fails by it’s own logic of radiative transfer.
The one or two I have explained it to already, see it straight away.
I think everyone will.

Chris Colose – Grow up.

82. ### Cedric Katesbysaid

Note Derek’s very similar debating style…

It’s uncanny. Even the key words are the same.

GH “theory” as presently taught by the “consensus”

Creationists always get set off by the word “theory” and “consensus”.
Beware of the green dragon, folks!

Religious Right on Dangers of Environmentalism

83. ### Richard T. Fowlersaid

To all,

Without, hopefully, being seen to endorse either side in the backradiation debate that has raged above (since I freely admit I have not done sufficient homework to take any strong position) …

I would like to pose a question. As far as I am concerned, there is no particular answer to this question that I expect or require, and I’m not trying to lead somewhere specific with the question. I’m just throwin’ it out there to see where _it_ leads (if anywhere).

The following quote is found in the Crepeau paper from U. of Idaho that was linked to above:

———-
Although Stefan was an outstanding experimentalist, neither he nor any of his students performed (or at least published) radiation experiments themselves. It is not known why. For this reason, perhaps, Stefan was not completely confident in his model. He wrote that his analysis had a, “hypothetical nature and reasoned support for [it] was impossible, so long as measurements are not made of radiation to surroundings at absolute zero, or at least a very low temperature.”(translation from Dougal [11]) Interestingly, Stefan never computed a value of his proportionality constant, A, but from a straightforward analysis from Stefan’s paper, it can be easily determined to be 5.056 x 10-8 W/m2K4, which is about 11% lower than the currently accepted value of what we now know as the Stefan-Boltzmann constant.
———-

My question is this. It is apparent that the author believes that a difference of “11% lower” is a pretty close fit for the prevailing theory, or viewpoint, OR WHAT-EVER you want to call it (I really don’t care what “it” is called at this point, because that just detracts from discussion about its relative merit or lack thereof). I myself would have thought that 11% sounds pretty far away, or a very bad fit — in other words, so large that it actually sounds to me like evidence _against_ the prevailing theory. Am I definitely wrong about that … and why, or why not?

Thank you.

RTF

84. ### Dereksaid

Regarding P/4 –

There is no more common error than to assume that,
because prolonged and accurate mathematical calculations have been made,
the application of the result to some fact of nature is absolutely certain.

The problems of P/4 are at least 3 major areas.

1) P/4 is unphysical when applied to earth, with it’s one sun. It is incorrectly applied maths “determining”, and mangling to a meaningless mush, the actual physics of reality. If P/4 were true then we would have no day and night, we would have a constant 1/4 solar power haze. As this is not the case, then P/4 must be in earths case a time period of 24 hours, or multiples thereof (ie, full revolutions).

2) P/4 introduces a time period to the W/m2 unit used of 24 hours, or multiples thereof. A W/m2 is a timeless power unit, so dividing P by 4 in earths case results in a unit that is not a W/m2, because it has to be over a certain time period. In earths case this is obviously 24 hours, or one full revolution of the planet.
When looking at K&T type budgets it is immediately apparent that all the quoted W/m2 figures are calculated using different time scales / units, so all the units compared are actually different, and do not represent the physical reality of what they purport to represent. It is no wonder such budgets therefore lead to such daft, but seemingly “justified” conclusions. “justified” by numbers expressed in units of W/m2 that are not actually timeless W/m2 figures.

P/4 is one example, but just think about the number of thunderstorms that happen annually on earth. If these were all added up, then divided by the number of seconds in a year, and spread over the whole planet, how is that relevant / useful. But that is what W/m2 does when used, incorrectly, as it is.

If the W/m2 is not a timeless power figure, then it is not a W/m2.
If the unit used includes a time element, then it is an energy flow at that (averaged) power by the time element.
Which is still pretty meaningless because the average has mangled the physics.

3) P/4 introduces a principle of the GH (failed) hypothesis that is not applied throughout the “hypothesis” at present, when it is applied throughout the hypothesis, the hypothesis fails, by it’s own logic.
This is something I have come to realise, and have been able to illustrate because of some email exchanges I have had with Alan Siddons, and others recently. I am still seeking Alan’s permission to “run with this” as it is mainly his work / research I have built off and from. The resulting piece will be too long for a comment on a thread, and so will probably form a blog post at http://slayingtheskydragon.com/ in the near future.

To Richard T Fowler – Sorry to comment over your post, but I had to add this “update” to my post 81 above.
However, knowing and respecting your understanding of the W/m2 unit, I would greatly appreciate your thoughts on point 2 in particular.

85. ### RBsaid

Derek:
Regarding (2), you don’t understand the concept of ‘average’.

RB

86. ### RBsaid

In other words, what does integral(P*dt)/integral(dt) represent to you?

87. ### Jeff Idsaid

RB- It has been explained in great painstaking detail to no avail.

88. ### Richard T. Fowlersaid

Derek,

I’m glad to see you are understanding a little more than you were about watts and W/m^2. Thank you for the acknowledgment.

Regarding your point #2, I regret I cannot agree with you. If, as you say,

“When looking at K&T type budgets it is immediately apparent that all the quoted W/m2 figures are calculated using different time scales / units”

then I think you might have a point, though you’d still be a long way from overturning quantum physics and relativity. 🙂

Unfortunately, your premise is not at all apparent to me, let alone “immediately” apparent. Admittedly I haven’t looked at K&T type budgets in several months, so perhaps there’s something there that I’m not fully remembering.

I have to say that the P/4 issue is something that we’ve talked about before, and I find that Claes Johnson disagrees with you insofar as he clearly shows in one of his papers (the one I had cited) that night is being accounted for. It might not be with the granularity, or resolution, that we all would like. But clearly the overall watts-in are not half or 1/4 of what they should be.

Even if the defenders of Einstein and Planck are all wet with their arguments (which I’m not saying they are or they aren’t, I think both sides of this debate have unacknowledged weaknesses), they are 100% correct about this P/4 thing. The Claes Johnson paper is your proof. It proves, among other things, that 1) the current models are not using a flat disk, and it proves that 2) they are trying to approximate rotation by varying the solar input over the course of a day. The things Johnson attacks are much less serious than that. If either problem 1 or problem 2 that I mention two sentences above were happening, Johnson would have been all over that in his paper.

Regards,

RTF

p.s. I would also like to Jeff Id (thank you, sir), for having given me, Derek, and others (including Derek’s critics) another opportunity to discuss these matters on your blog; the more so since I had found Derek’s forum to be “somewhat” error-prone and thus very difficult to post on. Thanks, Jeff, and thanks for your unparalleled commitment to freedom of expression. That is simply priceless in this day and age. RTF

89. ### Dereksaid

Thank you for replying RTF, particularly,

Derek – “When looking at K&T type budgets it is immediately apparent that all the quoted W/m2 figures are calculated using different time scales / units”

RTF “then I think you might have a point, ”

Just to be clear, my point is that this mangles the physics, and maths CAN NOT be allowed to do this. But maths is allowed to by the GH hypothesis, AGW, and K&T type “global energy budgets”.

I also would like to add a 4th point to my post 84.

4) Hans Schreuder has recently stated another, most obvious, and probably most important “error” (ie the reason it is used) of P/4, (which explains why it is SO CENTRAL an issue.
I will paraphrase him here, and yes, I have his permission to do so.

Hans Schreuder –
How else, other than P/4, for the solar input received, do you get the earth’s surface temperature low enough to REQUIRE a GH “effect” ?????
Answer – You can not.

90. ### Richard T. Fowlersaid

Derek,

Just to reiterate, I don’t think you have a point.

Considering your quote from Hans Schreuder, I’m not familiar with him or his work, but it he wrote what you’re quoting, I don’t think he has a point, either.

I also think that if you had something that impeaches Claes Johnson’s clear belief that P/4 is correct AS AN AVERAGE, something that would show that his arithmetic skill is as horrible as you are claiming it is … you surely would have shared it by now. Instead, you only quote this Hans Schreuder who, in his quote, is not doing much more than re-asserting your previous suggestion that it is quite obvious.

If it’s so obvious, then, does that mean we’re all lying?

Am I lying?

Is Claes Johnson lying, by refusing to agree with you?

Or … is it that the truth of the hypothesis is just not quite as obvious as you have suggested?

RTF

91. ### Dereksaid

RTF wrote,
” then I think you might have a point, ” Post 88
then RTF writes,
” Derek, Just to reiterate, I don’t think you have a point. ” Post 90.

RTF wrote,
” I also think that if you had something that impeaches Claes Johnson’s clear belief that P/4 is correct AS AN AVERAGE, ”

The solar input at night is…..The P/4 average does not apply in the way it is applied, otherwise we would not have day followed by night.
Maths IS determining the physics (and divorcing them from reality), not simply and merely DESCRIBING the physics.

Furthermore, it is not that long ago that SoD (as RTF gleefully reported at the GWS forum) implied according to RTF that, I was lying because, SoD had (supposedly) found fault with Claes figures, again according to RTF.
RTF implied I was required to defend Claes, otherwise I was a liar.

I will let the readers decide for themselves what they think of RTF, I have judged him by his own actions.

Jeff and others, – re post 87 ” It has been explained in great painstaking detail to no avail. ”
I added up the figures for different shapes, to show how a mathematically correct, but in reality, imaginary 33 degrees Celsius effect could be, and in some instances is calculated,
over a range of power figures. Jeff effectively became “blinded by his own science” in explaining “averages” that are actually mathematical artifacts that are not, and bear no relation to, the actual physics.
Apparently that is myself not understanding what an “average” is, nor when it should, or could be applied to a given situation.

Maths and physics are seemingly in climate science as difficult to mix as religion and politics, but not many appear to have noticed.

I repeat for RTF’s and Jeff’s benefit this quote.
There is no more common error than to assume that,
because prolonged and accurate mathematical calculations have been made,
the application of the result to some fact of nature is absolutely certain.

I’d put it like this.
When maths determines different physics to those observed in reality,
the results of the maths are guaranteed to be DUFF IN REALITY.

92. ### Jeff Idsaid

Derek,

If the energy strikes half a sphere, you can integrate it spherically or skip the angle of incidence and determine the total energy striking the surface with a disk. The answer is identical to the googleplex decimal places. The emissions in both cases must be a sphere for surface temp calculations, cause that is what is emitting.

That is your error. Fix the math and we can talk.

93. ### Richard T. Fowlersaid

“RTF implied I was required to defend Claes, otherwise I was a liar.”

I did not imply as Derek states. I brought to his attention the words of ScienceofDoom posted on the Air Vent. I _reported_ the words of ScienceofDoom. I had no glee in doing this. I was concerned for Derek. My belief was (and still is) that Derek did not even know about the error in the Johnson paper that SoD cited. (That, by the way, is a different Johnson paper than the one I mentioned before.)

I brought SoD’s comment to Derek’s attention for the primary reason that, if I was right that he didn’t know about the error, he could correct the record. Instead he seems to have concluded that I was engaged in some kind of effort to smear him, or was celebrating for some reason. Neither is the case. I was seeking to learn the truth. I still am, about many things; though I don’t really care about that error much. It was clearly a typo, and doesn’t seem to have affected Johnson’s findings (or Derek’s, or anyone else’s, for that matter.)

As for the alleged discrepancy about ‘having a point’, there is no discrepancy. Derek, in my opinion, only has a point if it is “immediately apparent that all the quoted W/m2 figures are calculated using different time scales / units”. Since I stated that this is not at all apparent to me, I clearly was stating that Derek does not have a point.

I’m nearly done talking about this subject. It’s relatively simple, and Jeff has summed it up well enough.

My final comments on this will be as follows:

Claes Johnson’s work proves that even if none of us knows exactly how the models work, they cannot possibly be understating incoming radiation by even 1/4, let alone 3/4. Seeing as Johnson has a very elaborate hypothesis for how there can be no GH effect and how the models can falsely show that there is, there is no need to continue debating Derek’s hypothesis. Even if there _is_ no GH effect, Derek’s hypothesis, in its current incarnation, is still wrong. There is no need for Derek to defend Johnson, and there is no way that both Johnson’s and Derek’s explanations can both be true. They are mutually exclusive.

Derek is his own worst enemy because if there _is_ no GH effect, then he is actually inhibiting the ability of other people to see that fact.

End of my comments about Derek’s hypothesis.

RTF

94. ### Dereksaid

If you two were my “firing squad” I’d survive.

Jeff,
” If the energy strikes half a sphere, you can integrate it spherically ” – Maths determining and mangling the physics. That is your error. Fix the math and we can talk. Clue – only over 24 hours in earth’s case.

RTF –
” Claes Johnson’s work proves that even if none of us knows exactly how the models work, they cannot possibly be understating incoming radiation by even 1/4, let alone 3/4. ” – I DID NOT SAY THAT THEY DID, but K&T does dilute and spread incoming radiation unphysically by using P/4 (as does the GH failed hypothesis) .
I have been trying to find out if the models also dilute and spread incoming radiation unphysically by using P/4 http://www.globalwarmingskeptics.info/forums/thread-1039-page-2.html to no avail.

” Seeing as Johnson has a very elaborate hypothesis for how there can be no GH effect and how the models can falsely show that there is, there is no need to continue debating Derek’s hypothesis. ” – A new one to me this, there are two hypothesises, but we only have to talk about one.
Yet again I repeat, I have not put forward a hypothesis, I have simply added up, or subtracted the figures for different “shapes”, ie, a disc, a hemisphere, and a globe,

and shown mathematically one can produce a 33 degrees effect, that is a result of the maths, not a result of the physics.

Sophistry a go-go, in both cases.

95. ### Richard T. Fowlersaid

Derek on 12 June:

——-
I DID NOT SAY THAT THEY DID, but K&T does dilute and spread incoming radiation unphysically by using P/4 (as does the GH failed hypothesis) .
I have been trying to find out if the models also dilute and spread incoming radiation unphysically by using P/4
——-

I almost cannot believe I’m doing this, but I refuse to let anyone get away with such a blatantly false statement. I am sure that even Derek’s colleagues know that the above statement is false. I respond here only because I cannot suffer the idea that some newcomer to this debate might not know the facts.

Derek on 19 December:

——-
Present computer climate models are modelled on a surface that is a perfectly flat disc of an area equal to the earth’s surface area.
This IS “disc world”, which is constantly under a one quarter solar actual strength “haze”.. There is no night in “disc world”.
The “physics” used in the computer models are “supplied” by another computer software program known as MODTRAN, which is “held” by the American navy.
The reason why the computer “disc world” is perfectly flat is that when “they” try to model hills, mountains, etc,the models produce supersonic winds, not seen in reality. There can be little faith in such “physics” but that is for a laterdate.
Here I will (only…) continue to try to explain the “issues” with “SHAPE” that is presently modelled in the “disc world” computer climate models and portrayed in the K&T hemisphere type energy budgets at present.
I will also have to refer to “lit hemisphere only world” that are the K&T global energy budgets.

[. . .]

These ARE the basic building blocks of the presently dominant (false) paradigm that is AGW pseudo science.
The reason the K&T (hemisphere) type plots and (disc world) computer models have to be BOTH discussed, is that they are one and the same thing. Two “halves” that when added together become the unproven hypothesis that is AGW, or rather the present false paradigm that is modern “climate science”.
Quite simply, when “shape” is mathematically incorrectly applied, then a “33 degrees” effect is created, that is actually a “quirk” of the differences between the shapes and the maths used.
This is why the “33 degrees effect” is not a physical effect that can be observed in reality.

There are three main problem areas with the present computer modelling “disc world” and the K&T (hemisphere) global energy budgets approach to planet earth’s climate system.
1) The complete ignoring of planet earth’s permanently unlit side(hence the “disc world” and “evenly lit sphere” approaches) This avoids “explaining” why the earth is no where near as cold as it should be (according to AGW and black body) on it’s dark side. The computer models and the K&T type plots BOTH ignore night.

2) The confused and commonly unrealised use and inappropriate mixing of “disc world”, hemisphere, and sphere average temperatures, effects, and influences. This causes differences to be omitted, and / or included, almost all contrary to the known laws of physics. This also further “allows” the use of a mathematical “quirk” to “show” a “33 degrees greenhouse effect” that is actually just the differences due to the difference in shapes calculated.

[. . .]

However, AGW unproven hypothesis times 4“disc world” does not even begin to try to explain the real reason why the sun is diluted over the whole globe by AGW unproven hypothesis. The reason for that is surely the real question to be answered.

[. . .]

The least number of dimensions “disc world” could possibly use because the sphere is only half lit, is 3 dimensions (length and breadth, and top and bottom), but AGW unproven hypothesis steadfastly adheres against reality to a 2 dimensional approach., length and breadth ONLY.
This is plainly unphysical, disconnected from reality in any meaningful way, and not a little ridiculous [. . . .]

AGW unproven hypothesis has ignored the problem completely of explaining night time temperatures upon earth that are simply too warm, and daytime temperatures that are simply too cool, appearing to be forlornly hoping the questions will go away.. They will not just “go away”, when the reasons to ask them are so obviously depicted by the “hemisphere” K&T type global energy budgets themselves, and modelled by the “disc world” computer climate models.

[. . .]

How one works out, for example, a global sphere temperature from a “disc world” haze, as AGW does, is actually beyond me, it really is not worth bothering with, except to show where AGW went wrong.

[. . .]

When you change the number P is divided by, you must also change the number to calculate the maximum temperature with. In the case of a sphere you can not divide by four with only one source of P, so the only reasonable interpretation is that by dividing by four you are calculating for a disc [. . . .]

[End of quotes from Derek’s presentation.]
——-

Further Derek commentary on 19 December – 4 January:

——-
The use of the “quirk” explains [. . .], the difference between disc computer models, and hemisphere K&T budgets [. . . .]

[. . .]

What does dividing P actually do.?
In the case of comparing it to earth, or a black body receiving energy from a single source, dividing P moves the object away from the source.
This must be the case as that is the only way the W/m2 received by the black body could be reduced, with a single and constant source of P.
If we agree that earth is nominally 93.5 million miles from the sun, then dividing P by two doubles the distance between the, and if P is divided by 4, then it quadruples the distance between them (187 million miles, and 374 million miles). [Not exactly by those amounts, but this is just to show Derek’s statement that radiation is reduced in a disk model. –RTF]

[. . .]

But as I do not see the point in considering a disc 187 million miles from the sun, or a disc 374 million miles from the sun, when earth is only 93.5 million miles from the sun, I suggest we are all in error, and it is not a small error.

[. . .]

This “hemisphere model” approach would be quite versatile, and it would seem to me to be capable of far more than merely a “snap shot” view, it could easily be doubled into a “sphere”, and / or “rotated” for example, and incorporate many things not presently included in modeling, such as surface heating, retention and varying later release….Oh, and night of course…..

[. . .]

For a (saucer like 4 x diameter of earth) “disc” of appropriate surface area A, (374 million miles from the sun) 1366/4 times A = Volume X

[. . .]

I would further inquire of Eli, please explain how a single source of P can illuminate a whole globe, as far as I understand this is unphysical, so the “globe” must be a saucer like disc to maintain a constant W/m2 input over it’s whole surface. The “globe” model is NOT a globe at all (unless it can be shown a single source of P can illuminate a whole globe…), it is a “disc” that is 4 times further away than it should be to the source of P.

[End of Derek quotes from 19 Dec. – 4 Jan.]

I am also confident that if I look on Derek’s forum, I will find a veritable cornucopia of quotes along similar lines, provided they will not have been deleted. But I think the point is clear enough. For Derek to be telling the truth as he remembers it when he says, “I DID NOT SAY THAT THEY DID”, he must have forgotten all of the above statements posted by him between 19 December and 4 January.

I have a pretty thick skin, and gave serious thought to not responding. The statement, “I DID NOT SAY THAT THEY DID”, defies explanation. Any reasonable person who has been following or participating in this already knows that such a claim not only has been made by Derek, but is the very centerpiece of his explanation. When he says “P/4”, what does he mean, if not his oft-quoted disc climate model, which he has stated unequivocally is the only type of climate model that can produce the appearance of a GH effect?

I repeat again Derek’s quote from higher up in this post of mine:

——-
“The reason the K&T (hemisphere) type plots and (disc world) computer models have to be BOTH discussed, is that they are one and the same thing. Two “halves” that when added together become the unproven hypothesis that is AGW, or rather the present false paradigm that is modern “climate science”.
——-

Having based my analysis on this and similar statements from Derek, I am now accused of “sophistry a go-go” because I actually, for some ‘strange’ reason, believed that Derek had stated that climate models “[understate] incoming radiation by [as much as] 3/4.”

Derek’s latest response doesn’t even rise to the level of sophistry.

I can’t fathom his reason for the statement “I DID NOT SAY THAT THEY DID”. What a sad conclusion to this whole debacle.

RTF

96. ### Dereksaid

RTF – I have come to realise that I can not simply state the climate models and the K&T budgets do exactly the same thing. I have looked into this, and found that I can say K&T does use P/4.
What the climate models actually do, has eluded me. I have been checking what I said, and updating / changing my mind when I find proof of what I say, or can not find proof of what I say.

An example – http://www.globalwarmingskeptics.info/forums/thread-1039-page-2.html Post 21. I wrote, and later added.
I think sceptics are loosing the “shape” argument with modeling BECAUSE they insist on a “flat disc”.
Yes, the disc’s surface is smooth (because MODTRAN can not handle the physics), and yes the disc is an area equal to the earth’s entire surface area, BUT,
the disc is not flat, it is curved.
Later edit – I retract the above statement, as it is now obvious to me that
to date (April 2011) the question of what “shape” is actually modeled still can not be answered.

So, I can say that the “shape” issue and P/4 that unphysically dilutes and spreads incoming solar radiation power DOES apply to K&T type plots, AND, the failed GH effect hypothesis,
but I can not say for definite at present that this also applies to climate models, although I strongly suspect that it does.

So goes researching an issue, not from a static view point.

97. ### Richard T. Fowlersaid

I am glad Derek has admitted to having lied so that we can put at least that behind us. I am not glad that he lied in the first place, nor that he refused to apologize.

But so goes life. I find most liars are like that.

Also, speaking as a skeptic, I have never countenanced, and would never in the age of teraflop computing, countenance a disc climate model. I highly doubt any other skeptic would today, either.

The models are not disks today. Johnson’s paper proves it. No ifs, ands, or buts about it. Unless you think Johnson’s a liar, which I don’t.

RTF

98. ### Dereksaid

Congratulations RTF, the best destroy the man not the argument approach I have seen in a long, long time.
I take it as a compliment such was worth so much effort on your part.

RTF wrote – ” speaking as a skeptic ” Yeah, yeah, pull the other one, it has got bells on.

99. ### Richard T. Fowlersaid

Once I realized the argument -ahem hypothesis- was invalid, I had no choice but to try to get it altered, and failing that, to try to knock it down, or “destroy” it if you insist.

Man is a sacred creation and one must never seek to destroy him.

If you would just admit your uncertainties promptly when they become apparent to you, you would be fine. You did this to yourself. Anyway, it can be recovered from. Please don’t take it so hard.

Regarding skepticism, I still wonder if there’s a greenhouse effect! And I’ve taken a lot of hits for that position. There appear to be at least a small number of people here who really don’t even feel like even talking to me because of that. I would call myself a skeptic’s skeptic. An AGW skeptic. A GW skeptic (for which I’ve become somewhat infamous over at Climate Audit, though you wouldn’t know it because I get “zambonied” so much). A GH skeptic, thanks in part to you. A negative-impacts skeptic. A government-solutions skeptic. And on, and on.

I’m sorry you don’t agree, but I am what I am. I just happened to believe that you weren’t doing skepticism any favors — including GH skepticism. I’m sorry if you don’t agree with that either, but I’m taking a break from this starting right now, so that’ll have to be a topic for some other day.

Climate Thermodynamics by Claes Johnson — read it! Try to impeach it if you can. If no one can, then … well, then we’ve all learnt something, haven’t we? And if someone can, well, that’s all right too, because there’s still plenty more to be skeptical about!

RTF

100. ### Dereksaid

Genuinely – Thanks to Jeff Id and Richard T Fowler, you have been a great help along the way for me.
Hence this blog and yourselves get honorable mentions.

P/4 – Why it is THE issue that destroys GH and AGW.

Short excerpt –
” One world famous main stream climate “skeptic” recently said –
Derek What is P/4? And do try to imagine a world without politics. Would ‘it’ be better or worse? ” ”

If only I could imagine a world without politics….
There would be no GH “theory” in that world, it is a politically motivated “issue”, not a science “theory”.

101. ### Jeff Idsaid

Derek,

It seems that you don’t mind the incoming energy per meter squared being calculated by a disk

a = pi r^2

But why don’t you agree with the outgoing energy per meter squared being calculated by a sphere?

A = 4 pi r^2.

The calculation also works with incident light on a hemisphere but you have to integrate by the cosine of the angle of incidence which from -pi to pi becomes a factor of 2 and actually breaks down from the hemisphere – exactly (letter for letter) to the disk equation.

If you want emitted energy per meter squared on a sphere of nearly equal night/day temperature (you have to think Kelvin for radiative emission), it becomes exceedingly obvious that you need to use the surface area of the sphere.

Therefore you have made a basic conceptual error that should be corrected so that others aren’t confused.

102. ### Dereksaid

P can be calculated as a “disc” for a point calculation, that I have no problem with.

Jeff wrote, ” But why don’t you agree with the outgoing energy per meter squared being calculated by a sphere? ”

Because of the time, and therefore physical processes ignored, from solar input received to radiation lost to space.
AND, it assumes a balance, that there is plenty of proof for, does not exist. Particularly on shorter times scales.

Yes, P/2 is a hemisphere average, that I do not disagree with.

Jeff wrote, ” If you want emitted energy per meter squared on a sphere of nearly equal night/day temperature ”

OK, here is a way to think about, or rather question to ask, I do not think many have done so far.
If we view earth as an object from space it is supposed to be an object (on average) at -18C. I do not disagree with this.
However, where are you viewing earth from?

If we are say, 30 million miles above earth and orbiting the earth,
what temperature is the object (hemisphere actually) earth, that you would view, because it will not be (on average) -18C from all positions of the orbit.
Directly between the sun and earth (12 O’clock), looking at earth you would see the lit hemisphere only, what temp would that be, on average?
At 90 degrees to the above viewing point (3 or 9 O’clock) you would see a half lit hemisphere that would be -18C, on average.
A further 90 degrees on (6 O’clock) you would see the dark side of the planet only, what temperature would that be, on average?

Have I, with that question illustrated what P/4 unphysically mangles, using divorced from the actual reality maths?

103. ### Dereksaid

Slight correction.

Yes, P/2 is a lit hemisphere average, that I do not disagree with.

My apologies.

104. ### Anonymoussaid

Sorry, but what a load of nonsense! A -180C moon warming a 15C earth! Nonsense. That does not “prove” the GHE at all. Cold cannot heat warm, that is basec Thermodynamics and there can be NO exceptions, and I will explain why on a molecular/subatomic scale. Iif the moon were at -180C then it would be cooler than it is now and would reflect less radiation back to the earth, therefore a cooler moon would cool the earth. However, a moon at -180C could not warm the earth as it is colder than the earth. I will now explain why.

It is nonsense because a -180C moon is emitting radiation that cannot possibly warm a 15C (or -18C etc) temp earth because the matter of the earth is in higher energy state being at a warmer temperature and is ALREADY emitting radiation at this wavelength that corresponds to this -180C temperature and higher. Therefore the radiation from -180C moon cannot excite the atoms and molecules on the earth because the atoms, molecules, bonds etc are already at this state of excitation and higher. All of the lower energy states are filled that correspond with the radiation that will be emitted by a -180C body. Therefore the energy states of the bonds and electrons that correspond to -180C are already occupied- therefore the earths matter cannot be excited by this radiation and therefore the earth cannot be warmed by this radiation and the Lunar -180C radiation will merely be scattered back to the moon and to outer space. A -180C moon cannot therefore warm a 15C earth.

The same applies to the earths atmosphere and the earths surface and between layers in the earths atmosphere. Cold can never transfer heat to warm, not even by radiation. This is the simple basic physics (and actually chemistry!) that cannot be broken. Not by your -180C moon example, and not by alleged “greenhouse” gasses.

Thermodynamics is the mathematical expression of what happens (does not happen in this case) and my paragraph above explains WHY it cannot happen (cold heating warm) on the molecular and atomic scale with radiation and energy states of electrons and bond orbitals.

The result is that alleged GH gasses do not act like a greenhouse and merely scatter radiation and can only warm when heat is moving from a warm body to a cooler body, NEVER the other way around.

105. ### Anonymoussaid

A further thought. Day time temperatures of the lit side of the Moon reach well over 100C, Therefore the more of the sunlit moon that is visible from the earth, the more the earth will be warmed by the moon as it currently is.

106. ### Jeff Idsaid

#104, There is net flow from warm to cold, but the cold does project radiative electromagnetic energy to the warm. What happened to that energy?

#102, Derek, You are not paying attention. The equations are for energy received and lost to space. The explanation shows your error. Your follow up is on the nuance of the difference in temperature of the re-radiation which is a complete non-sequitur to the first problem and I refuse to address it as you have not grasped the first so furthering into more complex discussion is a bad idea.

107. ### Anonymoussaid

“#104, There is net flow from warm to cold, but the cold does project radiative electromagnetic energy to the warm. What happened to that energy?”

Yes, there is a flow of em radiation, but it is of too low energy (too long a wavelength) to warm the already hotter Earth. The energy from the -180C moon would be of far longer wavelength and the radiation itself would have little energy compared to the “hotter”, i.e. shorter wavelength radiation that the earth is already emitting. Wavelength of radiation emitted by a body is inversely proportional to its temperature:- http://en.wikipedia.org/wiki/Wien's_displacement_law. This is because as you heat a body all the lower energy states of that body are filled as the temperature increases- and the body emitts shorter and shorter wavelength radiation. Thus longer wavelength radiation from a colder body CANNOT excite a warmer body because the warmer body is already excited to this level and beyond.

Therefore, the radiation from a -180C moon is incapable of heating a warmer object- i.e. the average 15C earths surface. Therefore the greenhouse effect from back radiation is a myth. The back radiation is there but it is merely scattered as it is incapable of warming the already hotter excited surface of the earth. The “colder” radiation is merely scattered and is not converted into heat.

Heat always moves from hot to cold, therefore a -180C body cannot warm a 15C body.

108. ### Jeff Idsaid

“Yes, there is a flow of em radiation, but it is of too low energy (too long a wavelength) to warm the already hotter Earth.”

So what happens to the energy? Does it reflect off 100% or go away?

109. ### Anonymoussaid

Further to that, :- “Conversely all matter absorbs electromagnetic radiation to some degree. An object that absorbs all radiation falling on it, at all wavelengths, is called a black body. When a black body is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. Its emission is called blackbody radiation.” http://en.wikipedia.org/wiki/Black_body

There is a difference between radiation and heat- radiation is not necessarily heat, only if it can be converted into heat by a body that is of sufficiently low excited state.

It is possible that a black body may absorb lower energy state radiation and then immediately re-emmit that radiation if the black body’s energy excitation states that the radiation energy would fill are already filled. (be they vibrational, electron levels, bond stretches and waggles etc). This happens if the black body is warmer than the body that emitted the radiation. This is why cold cannot warm hot, even where electromagnetic radiation is concerned.

110. ### Anonymoussaid

“So what happens to the energy? Does it reflect off 100% or go away?”

The body cannot use this energy to make heat so yes, it is merely scattered in a random direction and goes away till it finds a body with a temperature of less than 180C in the case of em radiation from your cold moon, in which case the body below -180C will be warmed.

111. ### Jeff Idsaid

#110 So from your theory, a colder body can not radiatively affect a warmer body right?

112. ### Anonymoussaid

“#110 So from your theory, a colder body can not radiatively affect a warmer body right?”

Yes, sort of, it cannot radiatively heat a colder body, because the warmer body has all it’s lower quantum energy states filled. EM radiation may be passed from the cooler body to the warm but not converted into heat. The only exception could be if the cooler body was travelling very fast and the long wave radiation squashed up by the Doppler effect. Say close to the speed of light. So the long wave -180C radiation will be scattered by the warmer body. The cooler radiation may be absorbed into the warmer body, provided the warmer body is not opaque to that radiation and then scattered back out of the body.

BTW, when the sun rises, clearly all IR interacting gases , H2O, CO2 etc will be excited first by the INCOMING IR of those discrete wavelengths with which these IRIG gases can interact with. Hence the IR gaps in the sun’s spectrum as seen from the earth’s surface.

BTW, also a fire can heat you in the woods via EM radiation because the fire is hotter than you are.

All radiation is not equal. The shorter the wavelength the more energy is packed into the wave/photon, therefore the shorter the wavelength the “hotter” the radiation may be considered. LW IR radiation from CO2 and water vapour is of a longer wavelength and “cooler” than the surface of the earth when visible light is shining onto the earths surface.

I refuse to use the term “Greenhouse gas”, these gases do not act as a greenhouse, they merely scatter

113. ### Anonymoussaid

Sorry! errata! I mean “cannot radiatively heat a warmer body”

114. ### Jeff Idsaid

“because the warmer body has all it’s lower quantum energy states filled”

That is an interesting statement. I wonder how all of the low energy states remain filled if the warmer body emits at all?

Lets ignore that for a moment. If light from a cooler body simply scatters from a warmer body, how do bolometric cameras work when viewing cooler objects than themselves?

115. ### Anonymoussaid

I guess like all digital cameras by converting light to electricity. Again, not by converting light to heat!

116. ### Anonymoussaid

I guess like all digital cameras by converting light to electricity. Again, not by converting light to heat!
” Today, most bolometers use semiconductor or superconductor absorptive elements rather than metals. ”

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

117. ### Jeff Idsaid

But how would a radiometer of any sort view any object cooler than its own detector? By your claim, it would be impossible for the sensor to pick up the difference in two cooler objects right?

118. ### Anonymoussaid

Not true.

You can convert EM radiation into electricity. That does not mean the cooler body warms the hotter! Plus EM detection devices have a power supply, so work input is involved. This is different to the passive radiating of one body onto another.

The only way you can get around cold heating warm thermodynamics is when work input is involved- hence a refridgerator has a pump and relies on expansion and contraction of gasses. This is not a passive process. Energy is required from a power supply.

119. ### Anonymoussaid

No such work input is involved in the alleged “Greenhouse effect”.

120. ### Jeff Idsaid

If the EM radiation is absorbed by an electron, you are claiming that no heat is generated? I thought the photon was diffused from your previous statement and left the system entirely??

Here is a camera with an uncooled sensor.

http://cgi.ebay.com/ICI-7320-FLIR-Infrared-Thermal-Imaging-Camera-NEW-/230267306283

Here is the same camera imaging ice:

It uses a microbolometer: http://www.alibaba.com/product-free/217346823/ICI_7320_S_Series_Infrared_Thermal.html

Bolometer defined here:

http://www.evilprofessor.co.uk/bolometer-theory/

It is a heat absorbing element which physically changes temperature in response to IR striking the surface. By your theory, it would be impossible for the heat of the ice to change the warmer surface of the bolometer sensor.

121. ### Jeff Idsaid

You can see in the histogram at the side, the violet of the ice sits around 5C.

122. ### Anonymoussaid

These devices have a power supply and thus work input is involved in the detection process, it is not a simple case of a cool body warming a hotter one. They are not a valid example as you do not have one body radiating onto a passive body. Turn the power supply off and see if they still work!

“If the EM radiation is absorbed by an electron, you are claiming that no heat is generated? I thought the photon was diffused from your previous statement and left the system entirely??”

This is not a passive system as a power supply and charges are involved in the detection of the EM radiation. The ice block will not be heating the camera, the camera merely detects the EM radiation. I’m not an expert on cameras, if you know more than me then answer the question with superior knowledge!

As i said, with work input, i.e. a power supply in the case of a camera, the detection of cold EM radiation is possible, much in the same way as with work in put of energy, the cold contents of a fridge can warm the fridge radiating element.

No such work input or power supply exists between your -180c moon and the 15C Earth, or between IRIG’s and the earths surface. Your examples are not valid examples of passive heat transfer from cold to hot. The laws of thermodynamics are not broken here. I sense that you are trying to distract from my theory by bringing in invalid comparisons.

Let’s stay on the topic!

123. ### Anonymoussaid

Again, not valid comparisons as a power supply, i.e. WORK INPUT is involved, not valid examples. let’s get back to the -180C moon and 15C earth and the IRIG’s in the atmosphere and the earths surface!

I detect an attempt to de-rail my argument by bringing in irrelevant examples. See if your cameras still work with no power supply!

124. ### Anonymoussaid

“The ICI 7320 operates on 1 watt of power via USB connection. ” see, it needs a power supply!

AGAIN, a power supply is needed in your Bolometer! It shows current going in and this gives the presence of “free electrons”. However, be aware that the Bolometer theory in that article may itself be incorrect, especially if the proposers of this theory used the same logic as Greenhouse effect from back radiation advocates.

Further down in the article:- ” A dc bias current flows through the bolometer generating a voltage . Changes in the incoming radiation power give rise to changes in the resistance , and therefore in the output voltage ”

It is likely that changes in the resistance may be causing changes in temperature, not the incoming radiation itself! So we have work input, an electrical current. So it is not a passive warmer body receiving the EM radiation so therefore not a valid comparison between our -180C moon and thw 15C earth.

125. ### Jeff Idsaid

“Turn the power supply off and see if they still work!”

While the deactivated device won’t produce an image on a TV screen, a second bolimeter would be able to image the sensor of a deactivated bolimeter sensor and record the temp differences. That may be difficult to find an example of but your assertion that the voltage in the sensor provides the work to allow the sensor to heat is false.

Let’s move on then.

Stand 5 feet back from your freezer, open the door with a broom or something and your face will immediately feel cool from the radiation inside the freezer. Close the door, and it will feel instantly warmer. — Try it, it works.

Yet both the freezer and the freezer door are cooler than your face?

How does that work?

126. ### Anonymoussaid

” That may be difficult to find an example of but your assertion that the voltage in the sensor provides the work to allow the sensor to heat is false.”

Not true! There is work input in a Bolometer, i.e. a voltage applied across something. Without this work input the device will not work. Your article say pretty much this.

As for the face feeling colder when you open the freezer, you’re just getting silly now! Either the effect is 1) imaginary, 2) the result of a cold blast of air being moved to your face by the action of opening the freezer door. Your example if done by radiation (which I think it is not!) is the opposite of your -180C moon and 15C earth.

In the case of cold moon and warm earth, you claim that the cold earth can further warm the warm earth (which is wrong!) and yet in the fridge example you claim that radiation from the fridge now cools your face?? Radiation cannot cool by transmission from cold to hot, but it can from warm to cold.

BTW I just tried opening the Freezer door with the broom handle and apart from looking ridiculous, I did not notice anything happening till after a few seconds when the slight blast of air from opening the door moved with it some cold air from the freezer, I felt this on my arm and not my face. I expect radiation from my face started to warm the freezer so I shut the door before my food de-frosted. Again a freezer is not magical, work input is involved.

127. ### Anonymoussaid

I agree though, that if the surface of the Earth at 15C were made of the same material as a Bolometer and had a voltage passed through it that the photon from the -180C moon may cause warming by changing the resistance. But this is a special case and work input (the voltage) is required to affect this process.

The laws of thermodynamics are to heat what gravity is to matter. So matter always moves downhill, UNLESS you apply work to that matter, i.e. use a car to drive up the hill or apply energy to your muscles. An object with no power supply cannot get itself up the hill. So it is with heat and hot and cold objects.

128. ### Jorgesaid

Hi Anonymous,

I had a look at how these microbolometers seem to work.

It appears that they use a thin strip of vanadium oxide as the temperature detector rather than the traditional platinum. There are two circuit techniques that may allay your concerns. Firstly, a bridge circuit is used which only looks at the difference in temperature between one strip that is exposed to radiation and another strip which isn´t. This removes most of the problem with self heating due to the bridge supply current as both strips are heated almost equally.

Secondly the bridge is only powered for a short period, just long enough to take the reading for that pixel. This reduces any self heating to a minimum but also means that the temperature equilibrium of the particular strip mainly happens when there is no current flowing through it.

The conventional explanation seems quite adequate to me. When the strip receives a lot of radiation by being focused on a hot target it has a net gain in energy because it receives more than it is radiating and when it is aimed at a cold target it radiates more than it gains by radiation from the low temperature source. In both cases heat (net energy) flows from hot to cold. In reaching equilbrium, in both cases, the temperature of the strip will rise or fall altering the amount it radiates until it balances the amount of incoming radiation.

Other explanations are no doubt possible but they all seem to involve more complex behaviour on the part of the photons.

129. ### Anonymoussaid

Jorge, the point is the bridge STILL HAS TO BE POWERED for a photon to be detected!

“Secondly the bridge is only powered for a short period, just long enough to take the reading for that pixel. ”

so again, as I have said, work input in the form of a voltage is applied for the photon to be detected.

In the case of a passive body (i.e. one with no work input) then the photon that is of insufficient energy cannot be permanently absorbed as heat, it will be scattered.

Bolometric cameras therefore do not in any way provide evidence for a Greenhouse type effect that contravenes the 2nd law of thermodynamics.
Cold cannot heat warm, it is impossible! Like expecting water to flow uphill of it’s own accord- which is again impossible without some form of work input, in this case a pump.

130. ### Jeff Idsaid

Jorge,

It is amazing the gymnastics some will go through to maintain the false belief in how the second law of thermo works. You are exactly right, the individual elements are cooled or heated by the background they are focused on and the measurement energy is deliberately low enough to not disturb the temperature — or else the camera wouldn’t work.

And none of it violates the second law.

131. ### Jeff Idsaid

If you walk into a room at 70F and one wall of the room is -200F, you will be able to sense the direction of that wall all the times because of the more rapid loss of energy to that wall even though all the other walls are also at a temperature cooler than your own 98.6 degrees. Your body will physically transfer more energy and work to the cold wall -because despite the fact that all walls are cooler, the cold wall is emitting less back to you.

Where in Planck radiative law is that second body included to calculate the emission of another? It isn’t. The lack of emission is what creates the difference. Thu a cold chunk of moon sized ice, blocks the colder chunk of moon sized space and the Earth loses less energy.

132. ### Jorgesaid

Hi Anonymous,

It certainly is quite difficult to measure the resistance of anything without applying a voltage for as long as it takes to make the measurement.

From what you are saying the strip can only gain energy by the radiation from the cold target during the period when the bridge is energised and when it is not active the radiation is not absorbed but scattered instead. On the other hand the radiation from the hot target can be absorbed all the time. This would seem to lead to an asymmetry in the heat uptake or loss of the strip depending on whether the strip was hotter or colder than the target. I should say that I believe the time for a strip to reach an equilibrium temperature is much longer than the brief measurement period and is probably closer to the time between measurements.

As far as I know there is no such change in sensitivity between sensing hot or cold targets. In fact, if your idea was true the temperature of the strip would fall rapidly as it would be radiating all the time but would only receive energy from the cold target for brief intervals. The other possibility you may wish to put forward is that the cold radiation is in fact being absorbed in the period when the bridge is not energised but will subsequently scatter unless a brief energising pulse is applied within a specific time.

If such an effect was present I think someone would have noticed it by now. Perhaps we should ask the chip makers to increase the time between measurements to see if readout for cold targets suddenly become really really cold.

133. ### Anonymoussaid

The “sensing of walls temperature” idea is ridiculous. Would not work at all unless there is a slight breeze from the cold wall to your skin. That is how you would sense the cold wall. You would not be able to “feel” the cold wall just through cold radiation! Besides which the body is still losing heat to the cooler walls, regardless of their relative temperatures. Neither colder wall can or is capable of warming an already warm body!

“Jorge,

It is amazing the gymnastics some will go through to maintain the false belief in how the second law of thermo works. You are exactly right, the individual elements are cooled or heated by the background they are focused on and the measurement energy is deliberately low enough to not disturb the temperature — or else the camera wouldn’t work.

And none of it violates the second law.”

Jeff and Jorge, it amazes me the gymnastics some are going through to prop up the discredited Greenhouse effect theory and the ridiculous -180C moon warming a 15C earth by radiation example! Invoking irrelevant examples such as Bolocameras seems to me to be a debating tactic to try and divert the argument and confuse and distract your “opponent”. Sorry, but those tactics are not working. And that is because the example with which this thread was started is plainly wrong! And I am not the confused one here!

“From what you are saying the strip can only gain energy by the radiation from the cold target during the period when the bridge is energised and when it is not active the radiation is not absorbed but scattered instead. On the other hand the radiation from the hot target can be absorbed all the time. This would seem to lead to an asymmetry in the heat uptake or loss of the strip depending on whether the strip was hotter or colder than the target. I should say that I believe the time for a strip to reach an equilibrium temperature is much longer than the brief measurement period and is probably closer to the time between measurements.”

It is not necessary for the strip to reach equilibrium as it is merely converting a photon to an electrical signal, via the use of an applied voltage!

The asymmetry in heat uptake is the 2nd law of thermodynamics- and that is because HEAT ALWAYS MOVES FROM HOT TO COLD. NEVER from cold to hot. That is an asymmetry of heat uptake, as I keep repeatedly saying you cannot move heat from a cold object to a hot object without work input. I’m not an expert on Bolocameras but that is the only way I can see the camera working is through work input from the voltage applied to the detector.

A much better example is gravity: Like expecting water to flow uphill of its own accord- which is again impossible without some form of work input, in this case a pump.

You would not argue that water can flow uphill would you?? So why try and make out a case for heat moving from cold to hot. The fact that this is radiation is irrelevant. Maybe in a parallel universe with different laws of physics to ours but certainly not in our universe.

134. ### Jeff Idsaid

#133 the voltage measures the change in temperature of the bolometer element. The voltage is not present during the warming.

That’s all it does.

If you used a bolometer to image the sensor of a bolometer you would see the image focused on the sensor.

“HEAT ALWAYS MOVES FROM HOT TO COLD. NEVER from cold to hot. ” CAPS don’t prove anything. NET heat always flows from hot to cold, that does not in any way prevent energy flow in the opposite direction it only determines that the magnitude must be less.

135. ### Jeff Idsaid

The “sensing of walls temperature” idea is ridiculous. Would not work at all unless there is a slight breeze from the cold wall to your skin. That is how you would sense the cold wall.

You are wrong.

136. ### Jorgesaid

Anonymous

I would not consider myself to be any kind of expert on bolometers, micro or otherwise but it seems that Jeff and I do have a conventional explanation of how they function. Your explanation does seem to involve some novel physics to say the least.

Neither of us claim that heat (net energy) moves from cold to hot, whether it is in the bolometer, moon or earth to atmosphere examples. To me it is very simple.

Ein – Eout = Heat

Ein is the energy absorbed from incoming radiation and Eout is the energy the surface loses by radiation because it is not at absolute zero. It is the difference (Heat) that changes the temperature of the body. It is not necessary to enquire into the origins of the incoming photons and neither is the absorbtion of the incoming photon affected by the temperature of the receiving surface.

The situation with gases is more complex because it is not easy to identify a surface within the bulk but in principle a volume of gas will absorb some incoming radiation and emit some as well. The difference (Heat) determines whether the temperature will rise or fall after taking account of other heat flows from convection or possible phase changes.

Anyway, I have enjoyed the exchange of views even if no minds have been changed.

137. ### Anonymoussaid

Well, it would indeed be novel physics if radiation from a colder body could passively warm the detector of the already hotter Bolometer! As I said (which you don’t seem to accept), the Bolometer won’t work without an applied voltage. That means, in practice a flow of electrons. While I know about what happens when photons are absorbed by electrons within atomic and molecular orbits- i.e. the discrete energy states I was talking about- which require certain energy, i.e. certain wavelengths of photon are absorbed and raise the energy level of that electron, which may lead to kinetic heating of this molecule. A certain amount of energy is needed to effect this raising of electron orbital state. A photon from a cold body would have insufficient energy to raise electron orbitals, therefore a cold body cannot warm a hotter body by radiation WITHOUT work input.

In the case of the Bolometer, this is the applied voltage to the detector. I accept that with this applied voltage, we have “free” electrons (from reading your link) and while I’m not familiar with the interaction of “free” electrons and photons (more likely enhanced energy state electrons brought about by the applied voltage!) , I have come up with an idea- The applied voltage raises the energy state of electrons in the detector and this applied voltage may raise some electrons to such a level that the weak long wavelength photon from the colder body can have an affect, be absorbed, raise the energy level of the electron, change the voltage and thus the resistance and thus be detected. So you still need a helping hand from the voltage, so as I have said previously work input is involved (voltage across the detector element). So therefore the Bolometer camera is NOT a valid comparison to your -180C moon to the 15C earth.

This is not “novel physics” at all! This is the quantum interaction between photons and electrons and bonds- a tried and tested area of physics that borders on chemistry called Spectroscopy. Maybe “novel physics ” to you but I studied this at university and spectroscopy formed part of my previous science career.

Finally:-

“Jorge said
June 17, 2011 at 3:54 pm

-snip-

Neither of us claim that heat (net energy) moves from cold to hot, whether it is in the bolometer, moon or earth to atmosphere examples. To me it is very simple.”

On that I agree- heat does not move from cold to hot. But you are confusing yourself. EM energy is not necessarily heat! EM energy may only be converted into heat if it passes through a material with which the EM radiation can interact.

“Jeff Id said
June 17, 2011 at 2:50 pm

The “sensing of walls temperature” idea is ridiculous. Would not work at all unless there is a slight breeze from the cold wall to your skin. That is how you would sense the cold wall.

You are wrong.”

Sorry, Jeff but it is you that are wrong. I am right! I feel that the issue is you are not understanding or are not willing to understand my explanations.

138. ### Myrrhsaid

Visible, Light, of the AGWScience energy budget as claimed in KT97, cannot covert the heat the land and oceans of Earth.

Concurrently, the claim is that it is these “Solar” energies, Visible and shortwave UV and Near Ir either side, which do so, and that no Thermal IR plays any part in heating the land and oceans.

Nonsense.

The heat we feel from the Sun is Thermal IR. If we can feel it, it is certainly sure that the oceans and land can feel it too, and warm up from it just as we do.

139. ### Anonymoussaid

See Professor Nasif Nahle’s ground breaking experiment where he repeats Prof RW Wood’s experimental demolition of the Greenhouse effect from back reflected IR.

http://climaterealists.com/index.php?id=8073&page=1

there you go Jeff ID- experimental proof that back reflected IR does not cause warming. The greenhouse effect is disproved!

140. ### Jeff Idsaid

#139,

haha. Yup, that’s proof alright. Nonsense statements slammed together in such a way that the undereducated can’t parse them.

Sorry if that is too harsh but that is the reality.

141. ### Anonymoussaid

What harsh reality???? By nonsense statements, I presume you mean all your statements, Jeff about a cold body being able to heat a hotter body, by radiation, bolo camera or not, it does not matter, which is clearly impossible by the laws of thermodynamics and radiative physics and by quantum mechanics.

Face it Jeff, you are a luke-warmer and your precious theory of the “greenhouse effect” from “back radiation” has been disproved and is indeed impossible. Did you even read the excellent article about Professor Nahle’s work??? Did you understand how he has proven that back radiation cannot heat a warmer body from a cooler body.

And as for nonsense statements

Harsh reality indeed to you Mr ID.

“Jeff Id said
June 17, 2011 at 2:50 pm

The “sensing of walls temperature” idea is ridiculous. Would not work at all unless there is a slight breeze from the cold wall to your skin. That is how you would sense the cold wall.

You are wrong.”

No, Jeff, it is you who is wrong! Sensing of colder walls by radiation is pure nonsense. Think about it for a minute.

Jeff, I respectfully suggest you should get some education on the subjects of physics and chemistry, in particular thermodynamics and quantum mechanics.

142. ### Anonymoussaid

and another thing:-

“Jeff Id said
July 25, 2011 at 6:38 pm

#139,

haha. Yup, that’s proof alright. Nonsense statements slammed together in such a way that the undereducated can’t parse them.

Sorry if that is too harsh but that is the reality.”

One knows when one has won the argument and yet one’s opponent can’t admit it when one’s opponent resorts to name- calling.
Say what you like Jeff about “uneducated” but you know you have lost the argument.

143. ### Anonymoussaid

“Jeff Id said
June 17, 2011 at 2:23 pm

#133 the voltage measures the change in temperature of the bolometer element. The voltage is not present during the warming.

That’s all it does.

If you used a bolometer to image the sensor of a bolometer you would see the image focused on the sensor.

“HEAT ALWAYS MOVES FROM HOT TO COLD. NEVER from cold to hot. ” CAPS don’t prove anything. NET heat always flows from hot to cold, that does not in any way prevent energy flow in the opposite direction it only determines that the magnitude must be less.”

Yes, heat always flows from hot to cold. There are NO exceptions. You are confusing heat with radiation. They are not the same thing at all. IR radiation may warm a cooler body, but IR photons passing from a cooler body to a warmer one will not warm the warmer body. The IR radiation of those wavenumbers will merely be scattered from the warmer body back to the cooler body as the warmer body is already emitting that radiation wavenumber as it is already above that temperature. Heat has not passed from the cold body to the warmer body, net or otherwise CAPITALS or lower case. It matters not Jeff. It is merely IR that leaves the cooler body which passes to the hotter body and yet may not warm it.

Oh, and the voltage is necessary for the Bolometer to work. A Bolometer could not detect a cooler source than itself without an applied voltage across the detector.

Hence the cooler atmosphere cannot warm the hotter surface by back radiation, hence the greenhouse effect is a myth.

144. ### Anonymoussaid

” Nonsense statements slammed together in such a way that the undereducated can’t parse them.”

You just reviewed your article Jeff.

145. ### Carricksaid

Is this our noobie friend Douglas Cotton again?

Sounds like him:

“See Professor Nasif Nahle’s ground breaking experiment where he repeats Prof RW Wood’s experimental demolition of the Greenhouse effect from back reflected IR.”

146. ### Jeff Condonsaid

Gotta be Doug.

Don’t care though.

147. ### Jeff Condonsaid

It may not be Doug. Different IP.

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