# Simultaneous Conduction and Radiation Energy Transfer

Written by Pierre R Latour PhD, PE, Chemical Process Control Engineer

Abstract: The rigorous model of simultaneous thermal and radiant energy transfer proves energy transfer by radiation can flow from a colder radiator to a warmer one, heating the warmer one further. It explains and quantifies how dissimilar walls in a room can have different steady-state temperatures.

Only the general laws of thermal and radiant energy transfer and the First Law of Thermodynamics, conservation of energy, are employed. [Editor’s note: since publication PSI has negatively reviewed this post*]

Introduction: Many claim radiant energy can only transfer from a hot radiator to a colder one. Otherwise the Second Law of Thermodynamics would be violated.

While this is true for thermal energy transfer by conduction or convection, it is not true for radiant energy transfer. How does a cold radiator transfer energy to warmer surroundings? It depends on absorptivity, emissivity and intensity differences at each radiating surface and on the presence of simultaneous conduction. We will show why temperatures of dissimilar room walls at steadystate are not be equal.

Confusion: Many err claiming radiant energy cannot transfer from the colder radiator to the warmer one, heating it further, because they incorrectly assume the driving force (at a distance) is temperature difference, which is true for conduction/convection through a matter field, while the driving forces for radiant energy transfer between radiators are intensity differences at radiator surfaces through a radiation field.

As proved by Martin Hertzberg15 . Radiant energy does not transfer due to a temperature difference at a distance. Temperature is a point property of matter proportional to the kinetic energy of its atoms and molecules. The transfer directions switch at T = Ts, when Qc = Qr.

Radiation direction is always opposite to conduction direction according to the First Law. So when surroundings are Ts, a room wall at T can be greater or less than Ts, depending on its radiating properties compared to radiating properties of surroundings. Rate in by conduction = rate out by radiation = a nonzero constant at different temperatures, unless Kirchhoff’s Law applies.

When it does the temperatures are equal and no energy transfers between the radiator and surroundings either way. Atmosphere Since atmospheric temperature decreases with altitude, why doesn’t energy transfer up by conduction, equalizing T above?

Because thermal energy indicated by T, is kinetic energy of molecular motion and it must decrease as potential energy increases with altitude in Earth’s gravitational field to maintain fixed total energy of each m3 of gas.

Another energy mechanism is involved. Some popular explanations of Green House Gas Theory say radiant energy transfers from cold atmospheric CO2 down to warmer surface, which absorbs it, warming it further, i. e. global warming. Actually surface partly radiates directly to space because atmosphere has some transmissivity and the rest is absorbed by the atmosphere, including trace 400 ppm CO2, and then reemitted to space.

A recent paper2 derived rigorous equations for the coupled atmosphere and surface temperatures. Only system properties are needed, no empiricism. When one warming and three cooling mechanisms are included in the whole system, the net effect of CO2 on temperature is small and likely < 0.

The remaining question to quantify the effect of CO2 changes on temperatures is: how much does CO2 affect the atmosphere’s radiating properties: absorptivity and emissivity? Assuming2 a 1% increase in the atmosphere’s radiating properties, perhaps due to increased CO2, surface temperature change is – 0.76C and atmosphere change is -0.39C.

The net effect is slight cooling. Applying (3) we see Qr > 0 for transfer from surface up to and absorbed by atmosphere, even when Ts > Ta. That is because there is no conduction involved, as proved above. The presence of radiating CO2 increases atmosphere emissivity, a resistance to radiant energy transfer. CO2 is not an energy blocker or trapper; it is an absorber and transmitter.

No radiant energy transfers from cold atmosphere with CO2 down to warm surface, warming the surface. A shiny white car has greater reflectivity than rough black car. So it has lower absorptivity and emissivity. Since white absorbs less radiant energy than black, it emits less, causing it to be cooler.

Conclusion. According to the First Law of Thermodynamics, radiant energy transfers from a cold to a warmer radiator in the presence of energy transfer by conduction the other way. The temperature difference between radiators at steady-state depends on their radiating properties: absorptivity and emissivity. This paper is not a mere theory because it is based on well-known laws of physics and mathematics, confirmed by observation.

So the Hertzberg general rate law disproves the notion radiant energy transfer only flows from the hot radiator to the cold one. That is only true if one radiator is sufficiently hotter than the other or both radiators obey Kirchhoff’s Law, emissivity = absorptivity. That is not easy to guarantee. The Earth’s atmosphere has several energy transfer mechanisms within it and hence does not obey Kirchhoff’s Law.

Read the full paper with all equations at:

*This post was peer-reviewed by PSI and the equations used refuted. Read more at:

Heat Flow Cold to Hot when both Conduction & Radiation Occurring?

• ### Jeff Greenwell

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In one of his latest articles entitled “Carbon Dioxide: Utterly Useless In ‘Trapping Heat’ Or ‘Delaying Cooling’”, John wrote the following excerpt, something I agree with and is applicable to this discussion:

Backradiation does no work since work is a mechanical process like friction or like an adiabatic change, and so, backradiation would therefore have to function as heat if it were to cause the warmer surface to become warmer still; backradiation can not function as heat since heat can only flow from hot to cold, and cannot reverse and flow from cold to hot. Heat flow is irreversible and thermodynamics textbooks go on at great length about that.

Take particular note of “[back]radiation does no work since work is a mechanical process”, and therefore, Hertzberg’s equation is of no value to Dr. Latour’s hypothesis. For all of the mathematical reasons that Joseph Postma presents, Hertzberg, in this scenario, is in error.

Respectfully,
Jeff

• ### John O'Sullivan

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yes, indeed. And I would agree the following short sentence from Postma speaks volumes to better communicate the fatal flaw in supposed CO2 ‘ back radiation’ heating/delayed cooling:
“there’s no mechanism by which something slower can catch up to something faster.”

• ### Jeff Greenwell

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Hi John,

Joseph has encouraged me to write an article on that very subject, one that I am hoping I can craft to the standards of publication at PSI. It is all based upon my “bicycle wheel” analogy. I have expanded upon this analogy in some detail. I am beginning work on illustrations. It is pertinent to the conversation as it exhibits the behavior of all atoms, molecules, compounds and radiation in our universe. It is a very simple concept and one that I believe is irrefutable, also rendering any such “greenhouse effect” impossible within this universe.

I shall send to you when I think I have it complete. Looking forward to feedback and suggestion from you and Joseph.

Sincerely,
Jeff

• ### Joseph E Postma

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Latour’s argument hinges on his equation (3), which itself is derived in Hertzberg’s paper:

Hertzberg derives his equation on pg.4 of his paper not based on any sort of fundamental mathematical analysis, as can be seen for radiant heat transfer equations in the textbook I’ve linked to in other comments, but merely upon a verbal argument which states:

MH: “The warmer earth then absorbs a flux of radiation emitted from the colder atmosphere”

Thus, the cart is put before the horse and Latour’s resulting conclusion is based on Hertzberg’s verbal claim that a warmer surface absorbs radiation from a colder surface, which goes to say that a warmer object can be heated by a cooler object. Hertzberg’s is a verbal argument, created because it sounds like it makes sense; however, it is NOT derived on a formal mathematical proof such as one finds in the textbook linked to above.

Notwithstanding that error, Hertzberg himself actually writes:

“Since according to Kirchoff’s law a(c) = e(c), and a(h) = e(h), the net transfer of the
radiant energy flux between the two surfaces is:

I (net) = a(c) e(h) σ [T(h)4 – T(c)4] = a(h) e(c) σ [T(h)4 – T(c)4]

and, as always, is from the hot surface to the cold surface, as required by the Second Law of Thermodynamics.”

Look at Hertzberg’s I(net) equation: since a & e are positive, then the sign of heat flow only ever depends upon the difference in temperature terms, and these always guarantee that heat flow is hot to cold.

In fact, Latour’s equation 3) is:

“Qr = 5.67 * (α εs Ts4- αs ε T4)”

but that equation doesn’t actually appear at all in Hertzberg’s paper even though Latour has referenced it as originating from there.(!)

The way that Hertzberg *actually* writes his equation may actually be OK, but Latour’s equation 3 & 5 (I(net) = 5.67 [a(c) e(h) T(h)4- a(h) e(c) T(c)4] ) do not actually originate from Hertzberg. I think that Latour is mis-claiming his case.

All we really need to do is to go back to the mathematical fundamentals of thermodynamics: microstates and irreversibility of entropy, etc. Heat flow is basically work, and remember I’ve noted that there are only two ways to increase an object’s temperature: via heat, and work. Well, heat is really just another subtler form of work: Higher frequencies of light and faster vibrations of molecules can do work that lower frequencies and slower vibrations can’t. There is no mechanism by which lower frequencies and slower vibrations can induce higher ones to oscillate and vibrate higher still – it’s actually Newtonian physics at that level. If the lower frequencies and slower vibrations try to do work on the higher ones…they actually just end up having work done on them by the higher ones.

Latour should retract his paper and this post. And should also read the textbook linked, and others like it, and learn the mathematical fundamentals.

• ### Joseph E Postma

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And so I’ve been put into a position of criticizing Hertzberg, where Latour claims support from Hertzberg, but the reference that Latour claims does not actually exist in the Hertzberg paper and at this moment, actually looking at Hertzberg’s equation in his paper, his equation does seem to give sensible results, notwisthstanding some poor usage of language which I criticized above.

The way that Hertzberg writes his equations I would agree with in that they would never show heat flowing from cold to hot. Latour has created a new equation and claimed that it comes from Hertzberg, but it does not directly. That being said, I still think there might be a problem with Hertzberg’s equation given that it doesn’t have the form found in the radiant heat transfer text, with emissivities appearing on the bottom.

And I will reiterate: All we really need to do is to go back to the mathematical fundamentals of thermodynamics: microstates and irreversibility of entropy, etc. Heat flow is basically work, and remember I’ve noted that there are only two ways to increase an object’s temperature: via heat, and work. Well, heat is really just another subtler form of work: Higher frequencies of light and faster vibrations of molecules can do work that lower frequencies and slower vibrations can’t. There is no mechanism by which lower frequencies and slower vibrations can induce higher ones to oscillate and vibrate higher still – it’s actually Newtonian physics at that level. If the lower frequencies and slower vibrations try to do work on the higher ones…they actually just end up having work done on them by the higher ones.

• ### Joseph E Postma

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Latour actually creates his new equation 3) out of Hertzberg’s equations 3) & 4), but dispenses with Kirchoff’s Law and thus creates a new one which Hertzberg never presented and did not discuss.

And on that point, the way that Latour combines Hertzberg’s 3) & 4) does not follow what we find for such scenarios as developed on mathematical fundamentals in the radiant heat transfer textbook.

The onus is on the climate alarmists, and Latour, etc., to show that heat can spontaneously flow from cold to hot without the assistance of outside work being supplied. It is not on us to disprove such claims since the entire body of thermodynamics mathematical fundamentals, and modern physics, already disproves it.

The scientific course of action Latour should take is to formally develop his new maths on first principles, and show where the existing one is wrong, and then not only that, but experimentally demonstrate it. The attempt to get heat to flow from cold to hot is what created modern thermodynamics with it being discovered, and mathematically demonstrated, to not be able to occur, just like there are possibilities and impossibilities in Euclid’s Elements. It (heat flow from cold to hot) would be industrially useful and it could be patented by someone to make themselves and their lineage infinitely wealthy, or given away for free to make everyone wealthy.

But is just doesn’t happen, because there’s no mechanism by which something slower can catch up to something faster.

• ### Jeff Greenwell

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… there’s no mechanism by which something slower can catch up to something faster.

And that my friends, is what you call “Priceless”

Oh, but Joseph, what about the “Flux Capacitor” ?? .. hmmmm??? … LOL!

Sincerely,
Jeff

• ### Joseph E Postma

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“there’s no mechanism by which something slower can catch up to something faster.”

That’s basically heat flow right there. The slower things, i.e. the lower frequencies and the slower vibrations, can’t catch up to and bump the faster things to make them faster.

The slower one can be made faster, but the faster one is never made faster.

• ### Jeff Greenwell

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Indeed sir (Joseph), you nail it with one very simple and irrefutable sentence. [mic drop]

• ### Pierre R Latour

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Hi Joe,
I reply to all three posts, avoiding the reiterate repetition in your second one.
You have not yet explained the unnatural prediction you found in your equation. Until you do we must agree it is wrong. Not good form to just sluff it off.

I have no problem with your Martin Hertzberg quote: “MH: “The warmer earth then absorbs a flux of radiation emitted from the colder atmosphere”. It is just that the rate is less that the surface’s emitting rate.

I agree “All we really need to do is to go back to the mathematical fundamentals of thermodynamics:” That is what I did.
My equation follows Hertzberg development terms directly by simple subtraction, a well-established 3rd grade mathematical fundamental. I give him the credit. I did not mis-claim him at all. Unfair charge. His subsequent equations are only valid if his Kirchhoff’s Law assumption is applies, which both Hertzberg and I clearly stated. But my work is general, I do not assume Kirchhoff. I did not “dispense with” Kirchhoff’s Law, I merely did not assume it. I suspect all your statements are only valid if Kirchhoff’s Law applies, which it rarely does in the real world. I go to nature first. No harm there.
I included his proof in Appendix which you have not yet disproven. You even said it sounds reasonable. If you claim my work is false because his equation is, it would help if you proved your claim against Hertzberg. I find his rate law in harmony with second law, and I explained why.

How do you explain thermal energy transfer by conduction from warmer to colder body and simultaneous radiant energy transfer from warmer to colder, both in the same direction, at steady state? That would violate the First Law. If energy flowed into a radiator in the same direction by both mechanisms, the colder would quickly drop to zero and steady state between them will never be reached. That is the unreal problem I have solved.

So how does a hot radiator at 500C with low emissivity of 0.1 and corresponding intensity of 2026 w/m^^2 transfer radiant energy to a colder radiator at 400C with high emissivity of 0.9 and corresponding intensity of 10,478 w/m^^2? It can’t. Energy flows from 400 to 500. How does the 500C know the temperature of the impinging radiator with 10,478 is > 500C?.Answer: it can’t. How does an absorber know the temperature of the emitter? It doesn’t.
Forget about conduction, place dissimilar radiators in vacuum. If they are at steady state, Q is zero and their temperatures remain different, with different emissivities and equal intensities. Obviously.

You often say low intensity radiation from a cold radiator cannot be absorbed by the warmer one because it would raise the warmer one temperature higher. But that is not so, output rate increases also. The first law says temperature rises, dT/dt > 0 when input rate > output rate. I explained that carefully in my paper. Suggest you reread it.

Further you claim: “there are only two ways to increase an object’s temperature: via heat, and work.” That is obviously incorrect; radiation, electricity, chemical and nuclear reactions can do it too.

Your Newtonian mechanics analogy does not disprove my conclusion. Lowering the intensity of one radiator does increase the energy flow to it from a higher intensity radiator. Second Law at work. I applaud learning from the fundamentals. I identified the ones I used. You put yourself in your position of criticizing Hertzberg when you said his equation “was wrong”. Don’t blame me.
While I invoked Second Law in my analysis for direction of flow, I do not believe Second Law demands radiant energy must flow from a higher temperature radiator to lower temperature one. My paper proves that. The reason it says so for conduction/convection is because temperature indicates the molecular kinetic energy intensity gradient through any matter field. Just like intensity indicates the energy gradient through the EMR energy field. Loke water, energy flows downhill.

You advise, “The scientific course of action Latour should take is to formally develop his new maths on first principles, and show where the existing one is wrong, and then not only that, but experimentally demonstrate it.” I did that and recommend you do the same.

You say “Latour should retract his paper and this post.” I have been reading your emails and invitations to your website since you wrote me on 7Sep11. I have disagreed with you from time to time, but I wouldn’t dream of attempting to shut you up, stifle debate, discredit physics and math or demand a retraction. I await one reason to justify my retraction. You failed to prove any error in my work or Hertzberg’s equation. You even said my logic was correct. I am skeptical you read and understood what I wrote. Calling me a climate alarmist is unfair, unscientific and incorrect.

• ### Jef Reynen

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@pierre latour
I had the intention to write you personally, but I understood that you prefer open discussions.
It is a pity to see that on the “slayers” site, people continue the use of the two stream formulation.
You are not the only one, also Joe Postma continues since years to use in his blogs (and for reasons which are not clear, in the PSI blog
the papers with the two way
formulation continue to remain on the first row!)
Claes Johnson in the Slayers book make it clear that the correct SB equation is to be written between a pair of surfaces. Of course, one of the elements of the pair can be at zero K, and in that case one can write the so-called Prevost flux
qPrevost=sigma*T^4.
But the heat transport between two surfaces, with the same emission coefficient should be written as
q(1 to 2) = sigma*(T1^4-T2^4) and not as
q(1to2) = Prevost1-Prevost2
= sigma*T1^4 -sigma*T2^4).

In case the emission coefficients of the two surfaces are equal, one argues that the result is the same.
When the emission coefficients of the two surfaces are not equal, the expression for the heat flow
q(1to2) = Prevost1 -Prevost2
is not correct.
This was already written by Christiaansen in 1888.
The reference to that paper and the application of it is given in a PROM paper on PSI:

This paper gives the derivation of the Christiaansen relation:
1/eps12=1/eps1+1/eps2-1.

• ### Joseph E Postma

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Latour’s argument hinges on his equation (3), which itself is derived in Hertzberg’s paper http://climaterealists.com/attachments/ftp/05-Hertzberg.pdf.

Hertzberg derives his equation on pg.4 of his paper not based on any sort of fundamental mathematical analysis, as can be seen for radiant heat transfer equations in this textbook https://climateofsophistry.files.wordpress.com/2015/01/heat_4e_chap13-radiation_ht_lecture-pdf.pdf, but merely upon a verbal argument which states:

MH: “The warmer earth then absorbs a flux of radiation emitted from the colder
atmosphere of”

Thus, the cart is put before the horse and Latour’s resulting conclusion is based on Hertzberg’s verbal claim that a warmer surface absorbs radiation from a colder surface, which goes to say that a warmer object can be heated by a cooler object. Hertzberg’s is a verbal argument, created because it sounds like it makes sense; however, it is NOT derived on a formal mathematical proof such as one finds in the textbook linked to above.

Notwithstanding that error, Hertzberg himself actually writes:

“Since according to Kirchoff’s law a(c) = e(c), and a(h) = e(h), the net transfer of the
radiant energy flux between the two surfaces is:

I (net) = a(c) e(h) σ [T(h)4 – T(c)4] = a(h) e(c) σ [T(h)4 – T(c)4]

and, as always, is from the hot surface to the cold surface, as required by the Second
Law of Thermodynamics.”

Look at Hertzberg’s I(net) equation: since a & e are positive, then the sign of heat flow only ever depends upon the difference in temperature terms, and these always guarantee that heat flow is hot to cold.

In fact, Latour’s equation 3) is:

“Qr = 5.67 * (α εs Ts4- αs ε T4)”

but that equation doesn’t actually appear at all in Hertzberg’s paper even though Latour has referenced it as originating from there.(!)

The way that Hertzberg *actually* writes his equation may actually be OK, but Latour’s equation 3 & 5 (I(net) = 5.67 [a(c) e(h) T(h)4- a(h) e(c) T(c)4] ) do not actually originate from Hertzberg. I think that Latour is mis-claiming his case.

All we really need to do is to go back to the mathematical fundamentals of thermodynamics: microstates and irreversibility of entropy, etc. Heat flow is basically work, and remember I’ve noted that there are only two ways to increase an object’s temperature: via heat, and work. Well, heat is really just another subtler form of work: Higher frequencies of light and faster vibrations of molecules can do work that lower frequencies and slower vibrations can’t. There is no mechanism by which lower frequencies and slower vibrations can induce higher ones to oscillate and vibrate higher still – it’s actually Newtonian physics at that level. If the lower frequencies and slower vibrations try to do work on the higher ones…they actually just end up having work done on them by the higher ones.

Latour should retract his paper and this post. And should also read the textbook linked, and others like it, and learn the mathematical fundamentals.

• ### Carl Brehmer

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“Heat flow is basically work, and remember I’ve noted that there are only two ways to increase an object’s temperature: via heat, and work. Well, heat is really just another subtler form of work:”

If one is going to strictly adhere to fundamental thermodynamics then one must acknowledge that “heat” is definitively not “work”. Yes, both are forms of thermal energy transfer but the First Law of TD separates them because they are distinctly different in nature.

∆U = ∆Q-∆W
where
U =internal energy
Q = heat
W = work

If “heat” and “work” were the same reality then the equation for the First Law of TD would not have them separated. “Heat” is the transfer of thermal energy due to a temperature differential, while “work” is the transfer of energy due to a mechanical action of some kind. Take for example riding you motorcycle without a face shield with the wind hitting your face. After a time your face will be wind burned even though the air through which you are traveling is cooler than your skin temperature. The air molecules are in mass hitting your face and are thus doing “work” on your skin and causing said wind burn.

• ### Pierre R Latour

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I appreciate the way you promptly confirmed the validity of my paper based on Martin Hertzberg’s radiant energy transfer law (5). Well done.

To be precise, Hertzberg combined two steps into one, combining his four emission and absorption terms, which he did not write, but I did
I(net) = a(c) e(h) σ T(h)4 – a(h) e(c) σ T(c)4 (5)
and then assuming Kirchhoff’s law for both radiators, a(c) = e(c) and a(h) = a(c), so he could factor them out to write the less general but common version
I(net) = a(c) e(h) [T(h)**4 – T(c)**4] (6)
So, any use of simplified (6) requires verification Kirchhoff’s law applies to the situation under study. Otherwise conclusions are suspect. This is why I worry that all your work with (6) is limited narrowly to Kirchhoff Law systems.

Hertzberg’s general (5) requires no verification; he used rigorous First Law physics. I did not use any slight-of-hand or deception for eqn (5); in fact Hertzberg should have included it to help readers follow his development. Just because he combined two simple algebraic steps does not mean (5) should not be attributed to him. I claim no credit for my simple one step addition.
I found your proposed alternate equation in your reference book pg 29, Table 13-3, eqn 13-38. https://climateofsophistry.files.wordpress.com/2015/01/heat_4e_chap13-radiation_ht_lecture-pdf.pdf
There was no derivation. If you look at pg 21 you will find an equation at the bottom for black bodies. Hertzberg’s more general eqn (3) reduces to it. There is one close to yours on pg 28 for enclosures, which does not apply to Hertzberg’s flat-plate situation. If you use his general equation for absorbing gases and black surroundings, it reduces to pg 42, eqn 13-58. I don’t see anything that falsifies Hertzberg’s equation. Most of your book article is about view factors. Your reference book came from my source, Hoyt C Hottel, “Radiant-Heat Transmission” Chapter, John H Perry, “Chemical Engineer’s Handbook”, 1950, which I bought in 1960.

Jef Reymen brought to my attention your equation was derived by Christiaansen in 1883 and provided its derivation. He assumed absorptivity is zero. It is for mirrors. Can be derived from general Hertzberg equation.

• ### Pierre R Latour

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This was a reply to Postma, not Brehmer.

• ### Jeff Greenwell

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In my own observations, for the past 55 years, I have observed every single material or compound having but one common property. They all cool over time .. that is, without additional work, all materials in this universe cool (entropy). If what Dr. Latour has hypothesized were true, then one would expect at least one example to the contrary, and yet, I have never seen one, nor have I ever heard of one hypothesized. Considering the broader range, or the “big picture” if you will, I cannot imagine our universe in successful existence if it behaved as Dr. Latour hypothesizes. Countless “observations” throughout our universe would tend to contradict this hypothesis by Dr. Latour.

Respectfully,
Jeff

• ### Pierre R Latour

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Jeff,

When my furnace goes on my house warms. When a star collapses to a nova, it heats up. When spring arrives my yard warms. Need any more counter examples?

I think radiant energy transfers at the rate I showed. I know of no observations that contradict my hypothesis.

I fail to see the relevance of your comment to my paper.

You just disagree with it. Not sure why.

• ### Jeff Greenwell

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Dr. Latour, everything you have describe takes “work” in one form or another. My “observation” and statement still stands.

Respectfully,
Jeff

• ### Pierre R Latour

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Good points, Carl.
1. Like Euclid, I lack capability to verify my theorem by experiment. Therefore, I do not wish to test the validity by experiment myself. I leave that to expert experimentalists like you to disprove or confirm.
2. The wall properties you mention do affect the transient responses. I did not claim there may be other explanations for dissimilar steady-state temperatures. I simply assumed they were well insulated and should have said so. While thermal properties of radiators affect the numerical result in each case, they do not refute the general conclusion I discovered.
3. All I showed is the role of absorptivity and emissivity on steady-state temperatures and in some real situations, radiant energy can transfer from a cooler radiator with high emissivity to a warmer one with low emissivity. It nicely explains why rooms with dissimilar walls can have different temperatures at steady-state. I think my conclusion fits with reality. Welcome any data that refutes it, like dissimilar walls always have same steady-state temperatures. Without that, my comment is based on a very rigorous observation of nature. With all due respect.
4. I don’t think you found any physics, mathematics or logical error.
5. It is quite true scientists are biased toward their own hypotheses. Cannot be helped. It is a trait of human nature; not a problem because it is unsolvable. Just recognize and account for it. That is what scientists and critics do. Nature rules. I pray you do not classify me as biased like nutty catastrophic global warming promoters. If you do, please describe my bias against nature.
6. You state my hypothesis correctly. I clearly incorporated Second Law, which proves it will. It determines the direction of radiant energy transfer from higher intensity to lower intensity at surfaces. Since you disagree, it is incumbent upon you to support your claim by proving my result violates Second Law. Just stating cold radiators cannot transfer radiant energy to warmer ones won’t due. I proved they can.
7. If you don’t accept Hertzberg’s radiant energy transfer law, your dispute is with him, not me.

• ### Pierre R Latour

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In my item #2 change “be other” to “be no other”.
I apologize for my English mistake.

• ### Joseph E Postma

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PL: “Just stating cold radiators cannot transfer radiant energy to warmer ones won’t due. I proved they can.”

You proved that they can by using an equation from Hertzberg which is unique to him, and which does not come from radiant transfer theory and is not found in such textbooks. In other words, you proved something incorrect by using an incorrect equation. The logic follows: start with a wrong equation, end with a wrong conclusion.

I can see why the equation was derived in the way Hertzberg did it, but although your conclusions from his equation are certainly logically correct in that regards, the starting point is flawed. Your logic was excellent extrapolating things in the way that you did, but unfortunately, the premises, the original equation, were and was not correct.

So, the proof was incorrect, although it correctly followed the (incorrect) axioms set up for it at the beginning.

• ### Pierre R Latour

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Joe,
I am glad you agree with the obvious, my paper hinges on the correctness of Hertzberg’s equation. I said as much.
We both found your textbook equation is incorrect. Your next move is to prove Hertzberg is not correct and find the correct law. When I am convinced it is correct, I will consider redoing my paper. No guarantees.

• ### Pierre R Latour

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Your claim is not quite right.
Should be: while your proof is valid, your conclusions are incorrect because the Hertzberg rate law you used is incorrect.

• ### Joseph E Postma

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Yes that is a better way to state it: “while your proof is valid, your conclusions are incorrect because the Hertzberg rate law you used is incorrect”.

PL: “We both found your textbook equation is incorrect. Your next move is to prove Hertzberg is not correct and find the correct law. When I am convinced it is correct, I will consider redoing my paper. ”

The textbook equation is derived after thirty-some pages of development and derivation based on mathematical fundamentals. It is still your and Hertzberg onus to show where that derivation is wrong, and to prove your own new equation which is new to thermodynamics. After you think you have a valid proof based on mathematical fundamentals (which you do not have at this point), then the onus will still be on demonstrating it experimentally. I will await such developments, but for now I am still going with the textbook derivation based on mathematical fundamentals. I note that you know of Euclid, and such procedures, and how they are relevant to the physical world.

• ### Carl Brehmer

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“I don’t think you found any physics, mathematics or logical error.”

1) The physics error is that your hypothesis is out of sync with the Second Law of TD. According to your hypothesis two bodies of matter in close proximity that are the same temperature and are thermally isolated from their surroundings will not necessarily stay the same temperature if their emissivities/absorptivities are mismatched. One will lose thermal energy to the other, which will cool it and the other will gain thermal energy from the former, which will warm it. This by definition is a decrease in entropy.

2) The logical error is Argument from ignorance (in which ignorance represents “a lack of contrary evidence”). You said, “I know of no observations that contradict my hypothesis” and “I lack capability to verify my theorem by experiment. Therefore, I do not wish to test the validity by experiment myself.” You then assert that if I disagree with your hypothesis, which either hasn’t or cannot be tested, then the burden is on me to prove you wrong, because just the fact that your hypothesis is out of sync with the Second Law of TD is insufficient.

The argument from ignorance “asserts that a proposition is true because it has not yet been proved false (or vice versa).” Under this standard I could describe in detail a very sophisticated civilization of super beings that exists on an Earthlike planet some 100,000 light years away. When challenged I could say “I know of no observations that contradict my hypothesis” and “I lack capability to verify my theorem by experiment. Therefore, I do not wish to test the validity by experiment myself.” “If you disagree that a very sophisticated civilization of super beings exists on an Earthlike planet some 100,000 light years away then prove me wrong.”

Science then devolves into “everything is true that cannot be proven wrong”. This is akin to the religious argument that God exists because you cannot prove that He doesn’t.

• ### John O'Sullivan

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Pierre,
Just read through the math. I can see why he thought he arrived at the conclusion he did, but the equation was wrong.

PL’s conclusion:

“I(net) = σ a(c) e(h) [T(h)4 – T(c)4] = σ a(h) e(c) [T(h)4 – T(c)4] (5)

So the Hertzberg general rate law disproves the notion radiant energy transfer only flows from the hot radiator to the cold one. That is only true if one radiator is sufficiently hotter than the other or both radiators obey Kirchhoff’s Law, emissivity = absorptivity. That is not easy to guarantee. The Earth’s atmosphere has several energy transfer mechanisms within it and hence does no obey Kirchhoff’s Law.”

The correct equation for heat flow when there are differing emissivities can be found on pg. 32 of this radiant heat transfer textbook. You can see that Pierre’s/Marty’s equation isn’t correct.

The equation should be (and I(net) should be Qdot):

Qdot = = σ[T(h)4 – T(c)4] / (1/e(h) + 1/e(c) – 1)

The emissivities do not enter at the top of the equation for each T term, but at the bottom and in their summed inverse, minus 1. The example for which this equation is derived (following on from earlier first-principles) is for a heat shield with *highly reflective* barriers, such barriers of course having high albedo (low absorptivity), and these do not actually enter the equation.

I can see why Hertzberg derived his equation in the way that he did, but it’s not correct.The point being, that heat still doesn’t flow from cold to hot, when using the *correct* equation.

• ### Pierre R Latour

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So Postma disagrees with Hertzberg. When Hertzberg accepts Postma, I will consider redoing my paper.
Postma equation does not include absorptivities of radiators, which seems unnatural. It may assume Kirchhoff’s Law, which I resist assuming.

• ### Joseph E Postma

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Well it’s an equation from a radiant heat transfer textbook (linked below), so it’s not mine. That is however Hertzberg’s equation as it does seem unique to him. I suggest reading the linked textbook…always the best place of all to go back to fundamentals. I side with textbook physics…as long as they’re not climate alarmist textbooks!

With emissivities entering as they do as given by the textbook, then you don’t get the reversal of heat flow from cold to hot. If you were to enter the absorptivities in the way Hertzberg does, then you would get:

Qdot (or improperly, “I(net)”) = σ[a(c)T(h)4 – a(h)T(c)4] / (1/e(h) + 1/e(c) – 1)

The emissivity term is always positive so it has no role in reversing the sign.

And so if the a(c), absorptivity of the cold, were sufficiently low, and the absorptivity of the hot, a(h) = 1, then that could make the equation reverse sign and hence say that heat flows from cold to hot. However, think about that: just because the cold object has low absorptivity, and the hot object high, shouldn’t mean that the cold object can heat the warmer one. And so the derivation in the textbook is correct, and Hertzberg’s equation is not correctly derived in the first place, not placing the emissivities where they should be.

I suggest reading through the examples and the derivations from the textbook. It looks strange the way the emissivities enter, but if you follow the math then it makes sense.

There is no need for you to wait for Hertzberg…I suggest both of you refer to the textbook on radiant heat transfer theory, and follow the way the radiant heat equation is derived – it’s not mine, and it is what disagrees with Hertzberg. I just followed the textbook:

• ### Pierre R Latour

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Joe,
Thank you for your alternate radiant energy transfer law from a textbook. I agree denominator > 0. I agree with your conclusion from it, if a(c) is low, sign of Q (I define Q to be a rate, not amount) becomes negative and flow reverses from T(c) to T(h). This confirms my Hertzberg conclusion using your different textbook equation.

I also agree with your conclusion from your equation that low absorptivity of cold radiator should not increase transfer from cold. Makes no physical sense. Hertzberg equation says the opposite. Can you see this is obviously where your equation does not represent reality? And so the derivation in the textbook must be incorrect. Can you post a pdf of it?

The only thing you can do now is abandon it and adopt Hertzberg.

BTW Hertzberg equation checks with Hoyt Hottel in Perry’s Chemical Engineers Handbook, 1950.

• ### Joseph E Postma

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PL: “I agree with your conclusion from it, if a(c) is low, sign of Q (I define Q to be a rate, not amount) becomes negative and flow reverses from T(c) to T(h). This confirms my Hertzberg conclusion using your different textbook equation.”

The point of that was to show that if one simply inserts terms and creates new equations arbitrarily, merely because it might seem like the right thing to do, you end with illogical results: just because the cold object has low absorptivity, and the hot object high, shouldn’t mean that the cold object can heat the warmer one. The correct equation is only the one from the textbook, linked here again:

PL: “I also agree with your conclusion from your equation that low absorptivity of cold radiator should not increase transfer from cold. Makes no physical sense. Hertzberg equation says the opposite. Can you see this is obviously where your equation does not represent reality? And so the derivation in the textbook must be incorrect. Can you post a pdf of it?”

If you agree that Hertzberg’s way of deriving equations leads to making no physical sense, then you should no longer agree with it. Hertzberg’s equation is simply wrong…and you can see that if you go through a correct derivation of the radiant transfer equation developed over 30 pages in that linked textbook.

• ### Jeff Greenwell

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When Hertzberg accepts Postma, I will consider redoing my paper.

I’m sorry to be so persnickety Dr. Latour, however, you stated previously

When I am convinced it is correct, I will consider redoing my paper.

If you want people to take your paper and conclusions seriously, then I would suggest you be more careful to be accurate, not just in your maths, but in your strategy as a whole. I believe Joseph has sufficiently proven that Hertzberg has it wrong, in particular for the case in which you are attempting to apply.

Respectfully,
Jeff

• ### Pierre R Latour

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Jeff,

My first comment is based on the assumption Postma is correct. Second comment is based on the assumption Postma is correct. They mean the same thing. Where is the inaccurateness of which you speak?

I am not yet convinced Postma proved Hertzberg wrong.

Postma has not provided derivation of his equation. I provided Hertzberg derivation of his equation, which seemed reasonable to me.
It should help Postma disprove Hertzberg, if possible.

I just posted argument that Postma disproved his own textbook equation.

• ### Jeff Greenwell

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Well, in the first example, you indicate “When I am convinced”, in the second case you indicate “When Hertzberg accepts Postma”

To be clear, I believe you are indicating that, when you, yourself, are convinced that Joseph has disproved Hertzberg for your application, you will consider revising your hypothesis.

Again, I’m sorry to be persnickety about this, but to me it is important to be specific here. It is you that Joseph needs to satisfy. I personally believe he has done so, but I am not the arbiter of this discussion, you are. I do have confidence Joseph will successfully do so however. He is quite a talented guy and I trust his skill.

Respectfully,
Jeff

• ### Jef Reynen

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@Pierre Latour
I had the intention to write you personally, but I understood that you prefer open discussions.
It is a pity to see that on the “slayers” site, people continue the use of the two stream formulation.
You are not the only one, also Joe Postma continues since years to use the two stream formulation in the figures in his blogs (and for reasons which are not clear, in the PSI blog in the paper with the two way formulation pictures which continues to remain on the first row, since january)
Claes Johnson in the Slayers book maked it clear that the correct SB equation is to be written between a pair of surfaces. Of course, one of the elements of the pair can be at zero K, and in that case one can write the so-called Prevost flux
qPrevost=sigma*(T^4- zeroK^4)= sigma*T^4.
The heat transport between two surfaces, with the same emission coefficient should be written as
q(1 to 2) = sigma*(T1^4-T2^4) and not as
q(1 to 2) = qPrevost1-qPrevost2 = sigma*T1^4 -sigma*T2^4).
In case the emission coefficients of the two surfaces are equal, one argues that the result is the same.
When the emission coefficients of the two surfaces are not equal, the expression for the heat flow q(1 to 2) = qPrevost1 -qPrevost2
is not correct.
This was already written by Christiaansen in 1883.
The reference to that paper and the application of it is given in a PROM paper on PSI:
This paper gives the derivation of the Christiaansen relation of 1883:
1/eps12=1/eps1+1/eps2-1
The heat flow between unequal surfaces becomes for T1> T2:
q(1 to 2) = eps12*sigma*(T1^4-T2^4)

BTW the evacuation from the heat from the surface of the planet is not that much by radiation (and of course even less by conduction) but mainly by convection.

• ### John O'Sullivan

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Thanks Carl. Pierre has said he is very happy to discuss this openly on the PSI comments section rather than privately by email. This demonstrates well our wider commitment to open and honest debate of this complex and contentious subject.

• ### Jeff Greenwell

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John, myself, and probably many others, are extremely grateful to you and PSI for maintaining exactly this policy. There are few other resources available to learn in such a manner as this.

With extreme gratitude to you and your colleagues at PSI,
Jeff

• ### Carl Brehmer

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You assert that you hypothesis is “confirmed by observation” yet the only empirical observation offered in the article is, “It explains and quantifies how dissimilar walls in a room can have different steady-state temperatures.”

With all due respect, that different walls within a room might have different steady-state temperatures is hardly a rigorous scientific experiment. How thick are the walls? What is the conductivity of the material that the wall is made out of? What is on the other side of the walls? Is one wall an inside wall and another an outside wall? What temperature is it outside of the house? Where in the room is the fireplace located?

All of these factor can create a steady-state temperature differential between the walls within a room.

One of the universal problems that we see in science is that scientists’ tend to be biased towards there own hypotheses. This not infrequently leads a skewed perception of reality. Those who believe vehemently in catastrophic man-made climate change, for example, are well know for their ability to blindly ignore the scientific data that is out of sync with what they believe the truth to be.

If you wish to truly test the validity of your hypothesis using the scientific method, construct a scientific experiment that has eliminated all of the the variables–a scientific experiment that can be replicated by other scientists.

Your real hypothesis is this: a plate of material suspended within a vacuum chamber whose walls have a unified temperature and a unified inner surface will not necessarily attain thermal equilibrium with the inner walls of the vacuum chamber over time. The second law of thermodynamics–the law of entropy–says that they will.