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Discussion > Simon Abingdon/Jonathan Jones/Radiative transfer

@Simon Abingdon

Let's try running through this very slowly from first principles. Start with disc 1. This is assumed to have one side perfectly insulated and non radiating, and the other side a perfect black body (an ideal radiator) which is exposed to space (which for convenience we will take as having a temperature of 0K, corresponding to truly empty space, rather than the 2.7K of local space, which is filled with the cosmic microwave background; the difference is negligible so let's stick to the simple case). Disc 1 also has a heater, but we'll start with that turned off.

Being a black body disc 1 will radiate (from its black side) according to the Stefan-Boltzmann law, so that the radiated power is given by


where A is the area, T the temperature (in K of course) and sigma, the Stefan-Boltzmann constant, is approximately 5.6704*10^-8. Any radiation will carry energy away from the disc, and it will of necessity cool down, as this radiation is simply lost to empty space and there is (as yet) no source of in-coming energy. How fast it cools down will depend on it's heat capacity: if the heat capacity is large then the temperature will fall slowly and vice versa. After a "long" time the temperature of disc 1 will fall asymptotically to zero, and the disc will effectively stop radiating (it no longer has any energy to lose!).

Now we turn on the heater on disc 1, with an assumed power of 100W/m^2. As energy is added to disc 1 its temperature will rise. (How fast it will rise will once again depend on the heat capacity.) As it warms up it will start to radiate, and as it gets hotter it will radiate faster, by the Stefan-Boltzmann law above. Eventually it will reach a temperature at which the energy loss by radiation is precisely 100W/m^2. we can easily work out what temperature this will be by solving


to get T=204.9K. So our calculation says that the temperature of disc 1 will rise asymptotically to 204.9K, with the time taken to rise to this temperature depending on the heat capacity of the disc (short time for a low heat capacity, long time for a high heat capacity).

(Note that in the calculation above we have assumed that 1m^2 of disc has a radiating area of 1m^2 because it only radiates from one side. This aside may seem unimportant now but will become important later. If the disc could radiate from both sides the final temperature would be only 172.3K; try running the calculation to see whether you agree!).

Does this make sense? If so we'll start to move onto disc 2, but I want to get the basics agreed first.

Apr 10, 2012 at 1:52 PM | Registered CommenterJonathan Jones

@Jonathan Jones

Yes I'm with you so far (although I am a little puzzled that in your second paragraph you assume the disc initially has a temperature greater than 0K which then dissipates asymptotically to 0K. You say "Any radiation will carry energy away from the disc, and it will of necessity cool down, as this radiation is simply lost to empty space and there is (as yet) no source of in-coming energy"). No matter, I'm ready for disc2.

Please continue.

@Simon Abingdon

Apologies for the confusion. With discussions of this kind there is a certain arbitrariness about exactly where you start the discussion. Logically it might make more sense to assume that we start with disc 1 at 0K, but that might raise the question of how it got there. So the initial discussion simply notes that if the disc doesn't start at 0K then we can always reach this state by simply waiting to allow the disc to asymptotically cool down. Then, when we switch on the heater, the disc asymptotically rises to T1.

In general it doesn't matter how disc 1 starts: it could even have started (for some unspecified reason) at an initial temperature higher than T1. All that matters is that if we switch on the heater and wait then the disc will eventually end up at T1. How quickly it gets there depends on the heat capacity of the disc. In a traditional "idealised" treatment we assume that the disc has negligible heat capacity, so the disc adopts temperature T1 almost instantly.

Now let's consider disc 2: this is identical to disc 1 except that it is a perfect black body on both surfaces. It starts off a long way away (sideways) from disc 1 so that the radiative interaction is negligible. If we wait long enough then it will have asymptotically cooled to 0K and we can take that as the starting state.

Now we come to the exciting bit. We slide disc 2 sideways so that it is directly in front of disc 1, and try to work out what happens.

Disc 1 is at equilibrium and so is radiating 100W/m2 from the side facing disc 2. Since this has a black body surface (a perfect absorber) facing disc 1 all this radiation will be absorbed. And so disc 1 starts to heat up. (As usual how fast it heats up will depend on its heat capacity, but this just determines the time scale on which the following events happen, not fundamentally what happens.) Because the disc is (by assumption) a perfect conductor the whole disc will be at the same temperature; in particular the far side will heat up.

The far side will now radiate (it is above 0K) losing energy to space. But because the near side has the same surface properties and the same temperature it will radiate in exactly the same way. Soi disc 2 is losing heat from boith surfaces, but because it is (initially) still quite cold it is not losing very much heat, so it continues to warm up.

What happens to the radiation leaving the near side of disc 2? Well it is heading towards disc 1, which is a black body (perfect absorber) and so it must be absorbed. This acts as an additional source of incoming energy on disc 1 and so it heats up, getting hotter than T1! Where does the energy needed to heat up disc 1 come from? That's easy: 100W/m^2 is still entering the whole disc 1/disc 2 system but now less energy is leaving the whole system (from the far surface of disc 2) and so energy is building up in the system, warming up disc 1.

Disc 1 heats up and so radiates more, and so loses more energy, all of which is absorbed by disc 2, and then half of which is radiated away by the far side of disc 2, while half is radiated back towards disc 1 by the near side of disc 2. This all sounds like it is going to get horribly complicated! But actually it is very easy to work out the point at which the whole system will restabilise. We have 100W/m^2 entering the system and at the new equilibrium point there must be 100W/m^2 leaving the system. It can only leave from the far side of disc 2 (by radiation into space) and this will occur when disc 2 is heated up to T1, so that it is radiating 100W/m^2 from its far side. But by symmetry disc 2 must also be radiating 100W/m^2 from its near side, and this must be readsorbed by disc 1.

So at the new equilibrium point disc 1 has 100W/m^2 coming in from the heater and 100W/m^2 coming in from disc 2, for a total input of 200W/m^2. Since it's at equilibrium it must be losing 200W/m^2 by radiation, and so by the Stefan-Boltzmann law its temperature must be T2=243.7K. This 200W/m^2 is absorbed by disc 2, and 100W/m^2 of it is reradiated back towards disc 1, while the remaining 100W/m^2 is lost by radiation to space. Since disc 2 is double sided it is losing 100W/m^2, and so its temperature is stable at T1.

Overall we have 100W/m^2 coming in and 100W/m^2 going out once equilibrium has been established. We also have an additional 100W/m^2 travelling each way between the discs; as Jorge says you could chose to nett this off, but if you do that it's not clear why disc 1 warms up. But when you consider all the individual radiative processes it should make sense.

How are we doing?

Apr 10, 2012 at 1:54 PM | Registered CommenterJonathan Jones

(butting in, slightly, but good topic)....

So in "layman's speak"...

Disc 1 is being heated not only by the heater inside, but also by the heat 'bouncing back' ( i.e. being absorbed and re-radiated) from Disc 2?

And the only thing stopping Disc 1 from getting hotter and hotter is the fact that Disc 2 stops becoming hotter when it reaches the point that it sends out the same energy as is coming into it?

It's the radiative gap that causes this temperature difference? If the discs were suddenly fused together, then the temperature of the new fused thicker disc would return to the same as Disc 1 on its own?

Isn't this just the 'first principles' description of a temperature gradient in a solid?

Apr 10, 2012 at 2:33 PM | Unregistered CommenterTheBigYinJames

As a lurker on this discussion at Unthreaded, and a only very occasional visitor to the Discussion threads, I'd just like to say that I am very much enjoying/appreciating all your work here. I'd especially like it you would also give a thought to real as opposed to thought experiments as well. There seems to have been a remarkable dearth of these despite this area being so important for policy that fabulous sums of money have been given to it in recent decades.

Apr 10, 2012 at 3:00 PM | Registered CommenterJohn Shade

@TheBigYinJames, @John Shade,

As the title implies I set this up as a "private" thread - so while you are very welcome to listen in, and indeed to butt in, I feel under no obligation to answer any questions!

@TBYJ Yes, yes, yes as long as everything is perfectly conducting, not really.

@JS Thanks, but this thread is sticking ruthlessly to naive theory.

Apr 10, 2012 at 4:25 PM | Registered CommenterJonathan Jones

Yes, I'm more or less happy. I know the answer is correct (the physics is long-established) and it feels intuitively correct. But I still have some uneasiness at apparently getting something for nothing.

Ahh, but are you actually getting anything for nothing? Note that there is a difference between raising the temperature of disc 1 from T1 to T2 and maintaining the temperature of disc 1 at T2. Once the discs are at equilibrium then disc 1 is radiating more energy than before, but it is getting precisely the same amount back, so there is no need for any additional external energy source.

What does need extra energy is the initial raising of the temperature of disc 1 from T1 to T2, which will require an additional amount of energy depending on the heat capacity of the disc. And we'll need some more energy to heat up disc 2 from 0 to T1. Where does this come from? Well, note that during the equilibration phase disc 2 is below T1, and so is radiating less than 100W/m^2, so some of the energy from the heater is trapped in the system. Once enough energy has been trapped to reach the new equilibrium temperatures disc 2 is radiating away 100W/m^2 and so no more energy will be trapped.

Purists might point out that there's also a quantity of energy trapped in the circulating radiation, but this value (100W/m^2 times the transit time of photons between the two discs times two) will be small in any sensible case. Where does this come from? As before it's just subtracted from the total energy radiated away by disc 2.

However many discs there were, the outermost would still be transmitting the original 100W while successive discs working back towards disc1 would I think be stabilising at ever more extreme temperatures. (Please ignore the idea that in the limit that was the same as conducting the experiment in an insulated container).

That's right. You can add as many discs as you like and at the new equilibrium the outermost one will always be at T1, while discs further in will get successively hotter. The analogy with insulation is dangerous and might lead to unwarranted conclusions; that's why I'm not using any analogies in this discussion.

Apr 10, 2012 at 4:41 PM | Registered CommenterJonathan Jones

As a layman I am enjoying reading this too. Thank you.

Apr 10, 2012 at 4:49 PM | Unregistered CommenterJack Cowper

I feel under no obligation to answer any questions!


I covered all this as part of my foundation Physics module as part of my Electronic Materials course many moons ago, but had forgotten it in the intervening years, so nice to see a refresher.

Also, when I see people proposing that they've 'disproved' basic thermodynamics, it gets me worried that I can't pick holes in it myself. I'm even more skeptical of ideas that go against established science than I am about AGW.

Apr 11, 2012 at 10:53 AM | Unregistered CommenterTheBigYinJames

Apr 10, 2012 at 1:54 PM | Jonathan Jones

"Disc 1 is at equilibrium and so is radiating 100W/m2 from the side facing disc 2. Since this has a black body surface (a perfect absorber) facing disc 1 all this radiation will be absorbed. And so disc 1 starts to heat up. (As usual how fast it heats up will depend on its heat capacity, but this just determines the time scale on which the following events happen, not fundamentally what happens.)"

Jonathan, I think there may be a typo here. I think it is disc 2 that starts to heat up.


Apr 11, 2012 at 12:15 PM | Unregistered CommenterJorge

Well, as I now know, it does (to 204.9K) but so does disc1 (to 243.7K).

Apr 11, 2012 at 1:11 PM | Unregistered Commentersimon abingdon

Jorge, correct, that should read "And so disc 2 starts to heat up." Thanks for spotting this.

Simon Abingdon, yes. Have you tried calculating what happens with 3 discs? It's not too difficult to solve for the steady state, and the result is quite pretty.

Jack Cowper, welcome to the thread!

Apr 11, 2012 at 3:25 PM | Registered CommenterJonathan Jones

Disc3 will radiate 100W to outer space and therefore 100W back to disc2 which was radiating 100W before the introduction of disc3 so now has to radiate 200W towards disc3 and therefore 200W back to disc1 which now has to radiate 300W towards disc2 (its original 100W plus 200W back towards disc2).

So the radiation now emitted by each of the three discs is in the ratio of 3/2/1 which must equal T1^4/T2^4/T3^4. We know that T3 = 204.9K and all I now need is a suitable calculator to work out T1 and T2.

If this is right the calculator's all I now need. I'll look for one if everything's OK so far.

Apr 11, 2012 at 6:39 PM | Unregistered Commentersimon abingdon

I realised I didn't need a particularly sophisticated calculator: one which does square roots is enough since the fourth root is the square root of the square root.

I got T1 = 269.7, T2 = 243.7 (same as T1 was in the two disc set-up) and T3 (the outermost) 204.9 as usual. Is that right?

Interesting that it's getting harder to warm up T1 by adding more discs. I don't think it can be asymptotic though, can it?

Best regards, simon

Apr 11, 2012 at 7:20 PM | Unregistered Commentersimon abingdon

No of course it can't be asymptotic since (counting back from the outermost disc) the Watts are in the ratio 1/2/3/../n and n can increase without limit therefore so can T^4 and consequently T. (Sorry if I wasted even a bit of your time on a stupid question).

Apr 11, 2012 at 7:31 PM | Unregistered Commentersimon abingdon

Your argument as given strikes me as a bit dodgy, but it's basically right and the answer is certainly correct: if you have n discs and number them from the outside in then you find that Tn=T1*n^(1/4) with T1 being the temperature of a single isolated disc. As you say adding more discs gets steadily less effective: if you want to double the temperature then you need 16 discs!

Apr 11, 2012 at 7:42 PM | Registered CommenterJonathan Jones

So now we agree on the basics, we can return to the key question: what is @mdgnn on about?

My original theory, that he was just worrying about the netting off, is looking very dodgy: if that were true then he should come up with the same values for T1 and T2 in the two disc case, albeit with a different explanation of why this happens. But if I understood him correctly then he said that in the two disc case under the conditions I specified then both discs would be at temperature T1 (he actually said that the inner disc would be slightly hotter, but I think that was reflecting an imperfect thermal conductivity in the discs).

In essence he seems to be claiming that the correct way to treat radiative coupling is to consider the two discs as being in direct thermal contact. So I have been trying to think about why he might think that, and have come up with a few ideas. More soonish.

Apr 11, 2012 at 7:53 PM | Registered CommenterJonathan Jones

mdgnn is correct.

Consider your plate 1 at thermal equilibrium, with an energy input of 100 J/s and radiating to outer space.
Now consider two plates, 1 and 2, firmly held together with the same conditions as plate 1.

Now separate the plates slightly, do you think plate1 increases in temperature?

Prevost demonstrated that two bodies at the same temperature continually interchange e/m radiation.
with no change in temperature of either body. Essentially, both plates have exactly the same distribution of molecular kinetic energy (i.e. temperature) and even though they are physically separate, they are coupled electromagnetically with standing waves providing the coupling.

In your two plate example there is a source of energy (100J/s) and a sink (outer space). The space between the two plates acts like a matched, lossless transmission line between source and sink.

Apr 11, 2012 at 9:05 PM | Unregistered CommenterRonaldo

People who keep posting about Prévost would do well to be aware that his 1791 work Mémoire sur l'Equilibre du feu was based on a variant of the caloric theory of heat, and as such can hardly be taken as a statement about the interchange of electromagnetic radiation, a concept which arose almost a century later. Prévost's big idea as far as I understand it was to show that the concept of frigoric was unnecessary, as it could be replaced by the interchange of caloric.

Consider me unimpressed.

Apr 11, 2012 at 10:24 PM | Registered CommenterJonathan Jones

The man from 1791 says "mdgnn is correct". Consider me amused. Great thread, Jonathan, thank you.

Apr 11, 2012 at 10:37 PM | Unregistered CommenterRichard Drake

On a more positive note, I have already stated on the original thread that the traditional radiative treatment will break down once the two discs are "close enough", and that this should be the case once the separation is not large compared to the relevant wavelength (in this case about 10 micrometres). On this scale the concept of photons breaks down, and some sort of classical electromagnetic coupling treatment is likely to be more appropriate.

What makes @mdgnn and his supporters unusual (beyond their fondness for caloric) is the claim that these effects persist over arbitrarily large distances.

Apr 11, 2012 at 10:42 PM | Registered CommenterJonathan Jones

An experiment will confirm the predicted result beyond dispute.

Apr 11, 2012 at 10:53 PM | Unregistered Commentersimon abingdon


I should have thought a little longer before posting.
Your arguments are sound, please accept my apologies.


Apr 12, 2012 at 4:25 AM | Unregistered CommenterRonaldo

Simon, the experiment with thin plates would have to be done in a vacuum to avoid losses due to air convection. Also you would have to approximate the inifinite heat sink which disc 2 radiates to - any glass or metal wall within distance will affect the experiment.

I'm not sure the experiment needs to be done physically (again) - we're talking about a well established principle in thermodynamics. I know, take nothing on authority :) I just imagine that if it was wrong, we'd have planes falling out of the sky, exploding distilleries, etc. all based on it.

Apr 12, 2012 at 8:22 AM | Unregistered CommenterTheBigYinJames

The ancient Greeks loved these philosophical mind games, but trying to disprove classical physics without firm empirical evidence is a bit of a dead end really.

Apr 12, 2012 at 9:22 AM | Unregistered CommenterRKS

Ronaldo Apr 12, 2012 at 4:25 AM.

I don't have any arguments: I just did some manipulations of numbers using an internally-consistent algebra under the instruction of Prof Jones to get an answer. Whether that algebra and answer reflect the real world I can't know for sure without the results of experiments.

Prof Jones talks about the possibility of "netting off" (ie counting as zero) the "circulating energy". If you do that you haven't got enough information left to calculate the "right" answer (269.7 243.7 204.9) and maybe you're back wondering about mdgnn's equality (204.9 204.9 204.9).

Like Doubting Thomas I need a convincing experiment to be sure the "right" answer is indeed right.

TheBigYinJames Apr 12, 2012 at 8:22 AM.

Of course I had in mind a practically realizable equivalent of the thought experiment.

And while I'm really quite sure you must be right, would you like to convince me (and perhaps others) that "we'd have planes falling out of the sky, exploding distilleries, etc" if the temperature of disc1 was after all unaffected by the introduction of disc2?

Apr 12, 2012 at 9:32 AM | Unregistered Commentersimon abingdon