The R. W. Wood Experiment

Guest Post by Willis Eschenbach

Pushed by a commenter on another thread, I thought I’d discuss the R. W. Wood experiment, done in 1909. Many people hold that this experiment shows that CO2 absorption and/or back-radiation doesn’t exist, or at least that the poorly named “greenhouse effect” is trivially small. I say it doesn’t show anything at all. Let me show you the manifold problems with the experiment.

To start with, let me give a curious example of the greenhouse effect, that of the Steel Greenhouse. Imagine a planet in the vacuum of space. A residue of nuclear material reacting in the core warms it to where it is radiating at say 235 watts per square metre (W/m2). Figure 1 shows the situation.

steel greenhouse 1Figure 1. Planet in outer space, heated from the interior. Drawing show equilibrium situation

This planet is at equilibrium. The natural reactor in the core of the planet is generating power that at the planet’s surface amounts to 235 W/m2. It is radiating the same amount, so it is neither warming nor cooling.

Now, imagine that without changing anything else, we put a steel shell around the planet. Figure 2 shows that situation, with one side of the shell temporarily removed so we can look inside.

steel greenhouse 2Figure 2. As in Figure 1, but with a solid steel shell surrounding the planet. Near side of the shell temporarily removed to view interior. Vertical distance of the shell from the surface is greatly exaggerated for clarity—in reality the shell and the shell have nearly the same surface area. (A shell 6 miles (10 km) above the Earth has an exterior area only 0.3% larger than the Earth’s surface area.)

[UPDATE: Misunderstandings revealed in the comments demonstrated that I  lacked clarity. To expand, let me note that because the difference in exterior surface area of the shell and the surface is only 0.3%,  I am making the simplifying assumption that they are equal. This clarifies the situation greatly. Yes, it introduces a whopping error of 0.3% in the calculations, which people have jumped all over in the comments as if it meant something … really, folks, 0.3%? If you like, you can do the calculations in total watts, which comes to the same answer. I am also making the simplifying assumption that both the planet and shell are “blackbodies”, meaning they absorb all of the infrared that hits them.]

Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2. At that point the system is back in equilibrium. The planet is receiving 235 W/m2 from the interior, plus 235 W/m2 from the shell, and it is radiating the total amount, 470 W/m2. The shell is receiving 470 W/m2 from the planet, and it is radiating the same amount, half inwards back to the planet and half outwards to outer space. Note also that despite the fact that the planetary surface ends up much warmer (radiating 470 W/m2), energy is conserved. The same 235 W/m2 of power is emitted to space as in Figure 1.

And that is all that there is to the poorly named greenhouse effect. It does not require CO2 or an atmosphere, it can be built out of steel. It depends entirely on the fact that a shell has two sides and a solid body only has one side.

Now, this magical system works because there is a vacuum between the planet and the shell. As a result, the planet and the shell can take up very different temperatures. If they could not do so, if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell, then the two would always be at the same temperature, and that temperature would be such that the system radiated at 235 W/m2. There would be no differential heating of the surface, and there would be no greenhouse effect.

Another way to lower the efficiency of the system is to introduce an atmosphere. Each watt of power lost by atmospheric convection of heat from the surface to the shell reduces the radiation temperature of the surface by the same amount. If the atmosphere can conduct the surface temperature effectively enough to the shell, the surface ends up only slightly warmer than the shell.

Let me summarize. In order for the greenhouse effect to function, the shell has to be thermally isolated from the surface so that the temperatures of the two can differ substantially. If the atmosphere or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two.

Now, remember that I started out to discuss the R. W. Wood experiment. Here is the report of that experiment, from the author. I have highlighted the experimental setup.

Note on the Theory of the Greenhouse

By Professor R. W. Wood (Communicated by the Author)

THERE appears to be a widespread belief that the comparatively high temperature produced within a closed space covered with glass, and exposed to solar radiation, results from a transformation of wave-length, that is, that the heat waves from the sun, which are able to penetrate the glass, fall upon the walls of the enclosure and raise its temperature: the heat energy is re-emitted by the walls in the form of much longer waves, which are unable to penetrate the glass, the greenhouse acting as a radiation trap.

I have always felt some doubt as to whether this action played any very large part in the elevation of temperature. It appeared much more probable that the part played by the glass was the prevention of the escape of the warm air heated by the ground within the enclosure. If we open the doors of a greenhouse on a cold and windy day, the trapping of radiation appears to lose much of its efficacy. As a matter of fact I am of the opinion that a greenhouse made of a glass transparent to waves of every possible length would show a temperature nearly, if not quite, as high as that observed in a glass house. The transparent screen allows the solar radiation to warm the ground, and the ground in turn warms the air, but only the limited amount within the enclosure. In the “open,” the ground is continually brought into contact with cold air by convection currents.

To test the matter I constructed two enclosures of dead black cardboard, one covered with a glass plate, the other with a plate of rock-salt of equal thickness. The bulb of a thermometer was inserted in each enclosure and the whole packed in cotton, with the exception of the transparent plates which were exposed. When exposed to sunlight the temperature rose gradually to 65 oC., the enclosure covered with the salt plate keeping a little ahead of the other, owing to the fact that it transmitted the longer waves from the sun, which were stopped by the glass. In order to eliminate this action the sunlight was first passed through a glass plate.

There was now scarcely a difference of one degree between the temperatures of the two enclosures. The maximum temperature reached was about 55 oC. From what we know about the distribution of energy in the spectrum of the radiation emitted by a body at 55 o, it is clear that the rock-salt plate is capable of transmitting practically all of it, while the glass plate stops it entirely. This shows us that the loss of temperature of the ground by radiation is very small in comparison to the loss by convection, in other words that we gain very little from the circumstance that the radiation is trapped.

Is it therefore necessary to pay attention to trapped radiation in deducing the temperature of a planet as affected by its atmosphere? The solar rays penetrate the atmosphere, warm the ground which in turn warms the atmosphere by contact and by convection currents. The heat received is thus stored up in the atmosphere, remaining there on account of the very low radiating power of a gas. It seems to me very doubtful if the atmosphere is warmed to any great extent by absorbing the radiation from the ground, even under the most favourable conditions.

I do not pretend to have gone very deeply into the matter, and publish this note merely to draw attention to the fact that trapped radiation appears to play but a very small part in the actual cases with which we are familiar.

Here would be my interpretation of his experimental setup:

r w wood experiment 2Figure 3. Cross section of the R. W. Wood experiment. The two cardboard boxes are painted black. One is covered with glass, which absorbs and re-emits infrared. The other is covered with rock salt, which is transparent to infrared. They are packed in cotton wool. Thermometers not shown.

Bearing in mind the discussion of the steel greenhouse above, I leave it as an exercise for the interested reader to work out why this is not a valid test of infrared back-radiation on a planetary scale … please consider the presence of the air in the boxes, the efficiency of the convective heat transfer through that air from the box to the cover plates, the vertical temperature profile of that air, the transfer of power from the “surface” to the “shell” through the walls of the box, and the relative temperatures of the air, the box, and the transparent cover.

Seems to me like with a few small changes it could indeed be a valid test, however.

Best regards,

w.

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davidmhoffer
February 9, 2013 6:50 am

richardscourtney;
Re Eltor
>>>>>>>>>>>>>>..
Richard, we may fail to grasp the stature of Mr Eltor. According to him:
I have my bona fides, that precious lambskin: after completion of four years’ BLISTERING work in 3-1/2, in the radiative transfer principles of everything mankind ever witnessed. Not that he ever built. That he ever witnessed.
When one is in the presence of a person who in 3.5 years studied all the radiative transfer principles that mankind has ever witnessed, refuted all the thermodynamics texts (which nonetheless remain in use to this day) while earning a degree in electronics engineering one is the presence of someone very special.

MikeB
February 9, 2013 7:07 am

Micro,
The SB equation has no T1-T2 term. It is simply
P = σ T4
Where P is the power in watts emitted per square metre of blackbody surface, sigma (σ) is the Stefan-Boltzmann constant and T is the absolute temperature in kelvin. The value of the constant σ is 5.67 * 10-8 (this is an easy constant to remember because the significant digits to remember go in sequence, 5,6,7,8 ). The power P is variously called the ‘emissive power’, ‘radiant power’, ‘radiant exitance’ or ‘radiant emittance’
For non-blackbodies the power is simply factored down by the emissivity.
You don’t

have to account for radiation being exchanged between both objects

.
The amount of energy radiated by an object depends only on its own temperature and emissivity. It is independent of where the object is or what is or the temperatures of things surrounding it.

dikranmarsupial
February 9, 2013 7:17 am

Will, no, they are not equivalent. The system will reach thermal equilibrium whether or not the vacuum is maintained.
The nuclear core generates energy at a constant rate, the outer shell radiates heat at a rate proportional to the fourth power of its temperature. Thus as the temperature of the outer shell increases, there will come a point where it radiates as much energy as the core generates and thermal equilibrium is reached. This is true whether the energy is transmitted from the core to the shell by radiation, conduction or convection, so whether the internal vacuum is maintained is irrelevant.
Willis is correct, there is no paradox, there is no double counting of energy.

ThePhysicsGuy
February 9, 2013 7:33 am

Willis Eschenbach says:
February 6, 2013 at 5:01 pm
Here’s an example to consider. Mount a light bulb socket on a table. Insert a 60-watt bulb. Measure the temperature, we’ll say it’s 150°C. Try a 100-watt bulb, it’s hotter, say 200°C.
Now, mount another socket right next to the first one. Insert a 100-watt bulb in one and a 60-watt bulb in the other. What will the final temperatures of the bulbs be?
I say both of them will run hotter in that situation, because A warms B, and B warms A
You also stated that, “..of course it has been proven, many times over. It’s in every college thermo textbook”.. In 1850 Clausius stated, “Heat cannot of itself pass from a colder to a hotter body.” That statement (or similar version) of the Second Law of Thermodynamics has been in every college textbook since that time. Yet you indicate that the colder light bulb will warm the warmer light bulb. So, what will happen with your version of thermodynamics? The 150°C bulb will cause the 200°C. bulb to become warmer. But now the warmer 200°C+ bulb (B) will then warm bulb (A) even more, with the cycle going back and forth until both bulbs thankfully explode, before melting a hole to China.
Care to explain yourself?
REPLY: Seems like an easy experiment to perform to find out. I think I’ll actually do this – Anthony

dikranmarsupial
February 9, 2013 7:57 am

Anthony – light bulbs are not designed to radiate heat but light, and also have a rather low mass and are hollow, so it may be difficult to get reliable temperature measurements. I think this would be a useful experiment, but I would suggest electrically heated metal spheres instead, so the thermal inertia makes the equilibrium temperature easier to measure. I’d certainly be interested to see the results.

February 9, 2013 8:11 am

MikeB,
Here’s the equation
http://upload.wikimedia.org/math/9/c/a/9ca1177f3d75ec3bb4a98ab7ce668297.png
And it indeed has a T1 – T2 term.

February 9, 2013 8:29 am

Anthony, if the bulbs are say 6″ apart, measure between them, then 3″ on the outside of both, and with a say 60w and 100w bulbs you will get 3 different readings. Also note the filaments will both be similar temps, since they both output similar color spectrums (color temps).

tjfolkerts
February 9, 2013 8:35 am

ThePhysicsGuy asks ” … with the cycle going back and forth until both bulbs thankfully explode, before melting a hole to China.
Care to explain yourself?

Every true physics guy is quite familiar with convergent series.
http://en.wikipedia.org/wiki/Convergent_series

Greg House
February 9, 2013 8:42 am

MiCro says, February 9, 2013 at 5:28 am: “For those who argue there’s no back radiation, no “Net”, notice SB equations have a T1-T2 term, if you look up the definition of Net, it say one definition is a difference.”
============================================================
Stefan–Boltzmann law has nothing to do with your “T1-T2 term”. Your “T1-T2 term” has apparently no basis in real science, because it has apparently never been confirmed experimentally. The same goes for this “Net”, it is equivalent to that fictional “T1-T2 term”.

February 9, 2013 8:47 am

Anthony,
You might want to modify the experiment some since the two filaments will have roughly the same temp (but different visible light flux). Test with two 100w clear bulbs, with one on a dimmer circuit. Also align both filaments in parallel, and it might be interesting to do one test with one perpendicular to the other.

richardscourtney
February 9, 2013 8:59 am

davidmhoffer:
re your post addressed to me at February 9, 2013 at 6:50 am.
I admit to being bemused at this thread, and it is not only Eltor whose posts I fail to understand. The following listed history explains my puzzlement.
1.
At February 7, 2013 at 11:48 am I wrote a post to ‘wayne’ which I mistakenly thought would overcome any confusion about the difference between ‘heating’ and ‘reducing heat loss’.
2.
But that failed, and Greg House explained why at February 7, 2013 at 6:30 pm. He asserted that there is no physical demonstration of a cooler object making a warmer object hotter.
3.
So, at February 8, 2013 at 3:37 am, I refuted that by saying my microwave oven IS a cooler object that makes a warmer object hotter.
4.
Ryan responded (at February 8, 2013 at 7:28 am ) saying

That’s all very well Richard, but this is an energy conversion process, not a heat energy transfer in a closed system.The argument we are having is whether a cooler object can make a hotter source even hotter, or if this would break the First or Second Law of Thermodynamics.

5.
I replied to that at February 8, 2013 at 8:14 am, saying

I agree that I described an “energy conversion process”: the energy of electromagnetic radiation is converted to heat in the contents of the microwave oven and is powered by a source of energy flow (i.e. mains electricity in the microwave oven).
Similarly, the temperature rise of Willis’ hypothetical planet’s surface is an “energy conversion process”: the energy of electromagnetic radiation is converted to heat in the planet’s surface and is powered by a source of energy flow (i.e. the nuclear reactions in the planet)..
I fail to understand what you mean by it is “not a heat energy transfer in a closed system”: in each case electromagnetic energy is emitted from one object and is converted to heat in a hotter object such that the temperature of the hotter object is raised.
And neither case violates the First or Second Law of Thermodynamics.
Your statements about mixing water are irrelevant. They say nothing about how electromagnetic radiation interacts with matter.

6.
At that point I had thought there was nothing left to say except to explain the issues of mechanism and mathematical description of the kind you have been providing.
7.
Clearly, I was wrong because subsequently a series of people have ignored the contents of my posts and have iterated the same points which I had explained are erroneous.
Those who have made those posts include Ryan, jae, mkelly, jae43 (who is perhaps jae), Steve Richards, gbaikie, Allen B. Eltor, Will, ThePhysicsGuy, Greg House, and (if I understand him) johnmarshall.
8.
None of them has came back to me about my explanations to refute them or to ask for clarifications. Instead, they have continued – with no reasoning and/or evidence – to assert the points I had refuted.
9.
I fail to understand why these people are refusing to consider the explanations which I took the trouble to provide. (The post provided by MikeB (at February 8, 2013 at 3:17 pm) suggests they have a “conceptual” problem, but that does not explain their refusal to consider the explanations I provided.
Hence, I am bemused by the thread.
Richard

tjfolkerts
February 9, 2013 9:17 am

MiCro says: “MikeB, Here’s the equation …”
Micro, that is not “the SB equation”, but rather a “radiative heat transfer equation”. The equation you referenced is specifically for the transfer of heat between two specific black bodies. It includes “F” which deals with the specific geometry involved.
The SB equation is given just a little further down on that same wikipedia page as P = σ A T^4 (or the more general case general case when ε is not equal to 1: P = ε σ A T^4).
http://en.wikipedia.org/wiki/Thermal_radiation#Radiative_power
** SB gives the power radiated out from each surface.
** The other equations gives the net power (power out – power in).

February 9, 2013 9:45 am

davidmhoffer says:
February 9, 2013 at 6:34 am
“Your statement that back radiation exists but no thermalisation occurs is a violation of the Law of Conservation of Energy, First Law of Thermodynamics and Stefan-Boltzmann Law and to be true would require that both the joule and watt as fundamental units of physics be re-defined.”
I rest my case, you are clue-less.
Ice cannot melt ice because it cannot thermalise its own or lower flux density radiation. If it could there would be no ice.
dikranmarsupial says:
February 9, 2013 at 7:17 am
No further comment.

ThePhysicsGuy
February 9, 2013 9:51 am

Alright Anthony! Take some photos for us!. To really put the theory to the test, in addition, consider surrounding the higher wattage bulb with 3 lower wattage bulbs, and then surround the whole experiment with tin foil to act as an IR reflector. (OK, maybe that is overkill)

Greg House
February 9, 2013 9:58 am

richardscourtney says, February 9, 2013 at 8:59 am: “…Greg House explained why at February 7, 2013 at 6:30 pm. He asserted that there is no physical demonstration of a cooler object making a warmer object hotter.
…So, at February 8, 2013 at 3:37 am, I refuted that by saying my microwave oven IS a cooler object that makes a warmer object hotter.”

===========================================================
Let me give your another example of equal scientific value: your warm finger makes your warm house colder! (by pressing a button on the air condition). Of course, it goes also like that: your cold finger makes your house warmer!

donald penman
February 9, 2013 10:13 am

What insulation is in contact with the planetary nuclear core ?How is this insulation affecting the temperature of the planetary nuclear core? If we put extra mass directly in contact with the nuclear core this would also increase the temperature of the core because heat is lost more slowly from the core.If instead of adding more mass to make a shell you took some of the mass close to the nuclear core and moved it out until it formed a shell as described would the nuclear core not cool down and the energy lost to space increase.

MikeB
February 9, 2013 10:17 am

richardscourtney
I appreciate your problem. I gather, like me, you and David Hoffer are not a true believers in the Global Warming catastrophe, but, as I said earlier, we sometimes feel morally obliged to correct some of the more embarrasing misconceptions that would-be supporters of our case have, before they give all sceptics a bad name.
Coming from a scientific background, your initial expectation is that once you point someone in the right direction they might at least go and check it out. But that is wishful thinking. They carry on like those zombies who get knocked down but won’t admit to be dead. Makes for good films but very frustrating.
What I like to do is guess what they are going to come back with next (apart from just ignoring you). For instance when Micro talks about a non-existant T1-T2 term in the SB equation I know the mistake he has made and that he is going to come back quoting some other equation with aT1-T2 term in it. And that is exactly what he does. Thanks to tjfolkerts for pointing that out.
OK… maybe I have had too much wine after dinner.

February 9, 2013 10:22 am

I would call it the two body version of S-B, but if that’s not what it’s called, I’m fine with that.
Greg, it’s tested out in millions of things everyday, your protests are just plain wrong.

davidmhoffer
February 9, 2013 10:30 am

richardscourtney says:
February 9, 2013 at 8:59 am
>>>>>>>>>>
Richard,
Like you, I am bemused, but also frustrated. There is so much wrong with the CAGW meme that can be debunked with good solid science. How to get to that when the skeptic side is littered with people that cling to a belief system that can be debunked by the simplest of observational evidence? People who the warmists can point to in order to discredit skeptics? People who attempt to influence others to join them in their folly, forcing a tremendous amount of energy to be expended trying to set the record straight.
On setting the record straight, one wonders if we’re even being successful. The same people come back with the same tired arguments over and over again, debunked by the same simple examples over and over again, but what do the silent majority think? Are they influenced one way or the other? I don’t know.
But I am beginning to see a pattern. The fact is that the notion that cold things radiate energy to warm things is counter intuitive. But the notions that the earth is round and that the earth orbits the sun are also counter intuitive. Yet somehow most people get past those.
The examples are easy. Venus, despite being further from the sun than Mercury, is warmer than Mercury. Venus has an atmosphere, Mercury doesn’t. The Earth is warmer than the Moon, despite being (on average) the precise same distance from the sun. Earth has an atmosphere, the Moon doesn’t. Then we have every day experience to draw upon. Cloudy nights are warmer than clear nights. The clouds are much colder than earth surface. Deserts cool off rapidly at night due to lack of ghe effect form water vapour while jungles which have very high water vapour cool off at night very slowly. Are these simple observations so hard to understand?
But then out come the nay sayers with their version of science. Cold things can’t heat up warm things, because the Second Law of Thermodynamics says so. Point out that the 2nd Law refers to net energy flux and they’ll start quoting Clausius. Pointing out that the 2nd Law has been reworded several times since 1850 as our understanding of physics has improved, is meaningless to them. Pointing out that Clausius wording refers to heat transfer which is not the same as energy flux similarly falls on deaf ears. But it is at this point that I find the discussion gets truly bizarre.
At this point I sometimes link to ARM or a paper on the topic showing that we can measure downward LW in the middle of the night, with no moon, and get readings in the hundreds of w/m2. We know from instrumentation we have in orbit that downward LW above the atmosphere is on the order of 2.7 w/m2. So, at night, no moon, nothing to produce a couple hundred w/m2 of downward LW but the atmosphere itself, where does it come from?
That’s when the real lunacy begins. Oh, back radiation exists someone will shout, but it doesn’t “thermalize” anything. I find that those who scream the loudest about the 2nd Law become even more shrill (but unconvinced) when I point out that this would violate Conservation of Energy and the 1st Law. And Stefan-Boltzmann Law. And the very definitions of joules, watts, and temperature itself.
They start hollering that I am claiming that you can pressurize a tank with an inlet pressure that is lower. A difficult thing to debunk because it is true. It just has zero to do with how radiated energy works. In the meantime, through all this, millions of engineers around the world used the exact physics we’re trying to explain to design and build everything from the mundane kitchen stove to the environmental control systems for the space station. Tjfolkerts, mikeB, bart, MiCro and others are having a conversation amongst themselves that I for one have learned from. They are refining their collective knowledge and learning from each other rather than shouting belief systems at each other while ignoring valid points that someone else advances. That’s science in action and we need more of it. As for the nutters who want to selectively choose which laws of physics to use and which not…
Bemused? Yes.
Frustrated? Yes
Exasperated? Bingo.

Bryan
Reply to  davidmhoffer
February 9, 2013 10:47 am

David Hoffer says
“But then out come the nay sayers with their version of science. Cold things can’t heat up warm things, because the Second Law of Thermodynamics says so.”
David you do not have a clue about what the word ‘Heat’ means.
Heat is always capable of doing thermodynamic work.
Go find a textbook covering the Carnot Cycle.
This is the usual introduction to the second law.
Then you will not make such elementary mistakes.

Greg House
February 9, 2013 10:40 am

tjfolkerts says, February 9, 2013 at 9:17 am: “** SB gives the power radiated out from each surface.
** The other equations gives the net power (power out – power in).”

=============================================================
The problem is that this “other equation” about that “net” thing has apparently never been confirmed experimentally. Therefore it can not be considered to be a scientific fact.
On the other hand, of course, anyone has the right to produce any equation. For example, I like this one: “2 oranges + 3 apples = 5 tomatoes”, why not, but the difference is that I do not claim it to be a scientific fact.

Greg House
February 9, 2013 10:58 am

davidmhoffer says, February 9, 2013 at 10:30 am : “Cold things can’t heat up warm things, because the Second Law of Thermodynamics says so. Point out that the 2nd Law refers to net energy flux and they’ll start quoting Clausius. Pointing out that the 2nd Law has been reworded several times since 1850 as our understanding of physics has improved, is meaningless to them.”
==========================================================
The problem is that in linguistic terms your rewording (I mean the insertion of the word “net” into the 2nd Law) produces a formal contradiction to the 2nd Law. This is important to understand. The Clausius statement and your statement formally contradict each other. Both can not be true at the same time.
So, we have your statement versus Clausius statement. Two different things.
The Clausius statement means that something is not possible, but your statement declares it possible. Your statement is however not the 2nd Law, this is important to understand. If you change words you get formally another statement, however, it is still possible that those 2 different statements are physically equivalent. So, you need first either to prove the equivalence or just prove your statement scientifically, thus refuting the Clausius statement. I am looking forward to it.

davidmhoffer
February 9, 2013 11:03 am

….and as if on cue:
Bryan says:
February 9, 2013 at 10:47 am
David Hoffer says
“But then out come the nay sayers with their version of science. Cold things can’t heat up warm things, because the Second Law of Thermodynamics says so.”
David you do not have a clue about what the word ‘Heat’ means.
Heat is always capable of doing thermodynamic work.

dikranmarsupial
February 9, 2013 11:11 am

Greg House – The “net transfer” is a consequence of viewing the transfer of heat as being mediated by the radiation and absorption of IR photons. Unless you can come up with a mechanism by which a cold body can know which directions it can safely emit a photon without it hitting a warmer body, or for a body to choose not to absorb photons from a cooler body (impossible as there is no way to determine the temperature of the body that emits an individual photon), then an exchange of photons (and hence energy) is inevitable. However, the net transfer is still from warmer to colder body, so there is no violation of the second law of thermodynamics.

Greg House
February 9, 2013 11:12 am

davidmhoffer says, February 9, 2013 at 10:30 am: “That’s when the real lunacy begins. Oh, back radiation exists someone will shout, but it doesn’t “thermalize” anything. I find that those who scream the loudest about the 2nd Law become even more shrill”
==========================================================
The Wood experiment demonstrates how back radiation “thermalize” anything, see above.

PJF
February 9, 2013 11:18 am

Will wrote:
“The colder substance/object radiates IR with a lower flux density than the warmer object. The IR from the colder substance/object cannot be thermalised by the warmer substance/object and therefore is just scattered/re-emitted at light speed.”
The notion that matter cannot be thermalised by the infrared radiation from cooler objects seems to be the core concept for those who do not agree with the basis of Willis’s article
There are practical implications to this idea. One is that a vacuum flask would not work other than as a device to prevent heat loss by conduction and convection. So a vacuum flask would not work in space (or other vacuum) at all; the contents would lose heat by radiation at the same rate as if those contents were not in the flask.
This should be easy to demonstrate.

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