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Suppose you were inside a thick spherical shell of inner radius $R$, which was a perfect black body at some temperature T. What would be the power a sphere of radius $r$ would absorb located inside the shell?

I assume the location does not matter, given the spherical symmetry and using similar arguments to the gravitational force inside a shell. If the location does matter, let's place the sphere at the origin. I am tempted to think that you could solve this problem with the Stefan-Boltzmann law, thus I believe all the information necessary is provided.

I am almost lead to believe that the power absorbed is $P = \sigma_{SB} T^4 A_{abs}$, where $A_{abs}$ is the surface area of the absorbing sphere; but I would like to hear thoughts on how to approach this problem.

How I arrived at this result seems rather non-intuitive since the black body flux per unit area of the emitting shell is $\sigma_{SB} T^4$. So then $P = \sigma_{SB} T^4 A_{abs}$ would be the power emitted by the sphere if it was in thermal equilibrium with the shell. Which I suppose if you waited long enough and the shell was always held at a fixed T, the sphere would enter thermal equilibrium. Thus since the flux from the shell never changed, but it is clearly $P = \sigma_{SB} T^4 A_{abs}$ in thermal equilibrium, then this must have been the power absorbed the entire time.

While I am slightly satisfied with the logic, I am hoping for a more intuitive derivation, that doesn't invoke thermalization but rather talks directly about the flux from the shell at the location of the absorbing sphere.

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Thermal output radiation from the small sphere is given by the Stefan-Boltzman law,

$Power = A σ T^4 = σ 4π r^2 T^4$

but of course the NET radiated power is more interesting; output minus input

$Net\space{power} = σ 4π r^2 (T_{inner} ^4 - T_{outer}^4)$

The important thing to remember, is that the area of that small sphere is part of the scale of its thermal radiation, but the enclosing sphere area is not (because the enclosing sphere radiates to itself as well as to the small sphere, but the small sphere faces only the larger one).

For the thermal equilibrium to occur at equal temperature, there can be no different emitting area and absorbing area, by the zeroth law of thermodynamics: when two items are in contact, heat flows from hotter to cooler.

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If the smaller sphere behaves as a black body, it absorbs all radiation of the large sphere that falls on it and your argument is valid; energy absorbed per unit time by the smaller sphere is independent of its temperature and depends only on its surface area and temperature of the large sphere. So we can calculate this energy for any situation and the result is general. The calculation is easy when the system is in equilibrium and so is the preferred way to do it. The result is as you wrote: the small sphere absorbs energy per unit time

$$ \sigma\pi r^2 T^4. $$

The same answer should be obtained by summing the radiation powers that come from the large sphere elements and hit the small sphere.

However in reality, although one may construct a sphere to function as a container for equilibrium radiation, there is no material for the small sphere so that it would behave as black body, as all materials reflect part of the radiation and the power absorbed is therefore less than the above formula suggests.

For a real body, the amount of energy absorbed depends on frequency of the radiation and temperature of the reflecting body. So for a real object, the above reasoning is not really applicable. The power absorbed when the body is cold will be different than the power absorbed when it is in equilibrium with the large sphere.

The right way to find the answer to your question would be to do an experiment. For example, find or make a good isolating cavity, set up equilibrium radiation, put a ball inside and measure how its temperature and other measurable properties vary in time.

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  • $\begingroup$ Some of the radiation from the outer sphere misses the smaller inner sphere, going past it to hit & be absorbed by the larger school itself. When you do the geometry, you find this effect is the right size to make the equilibrium work right. $\endgroup$ – Bob Jacobsen Dec 6 at 5:09
  • $\begingroup$ You are right, but I didn't mean to say otherwise. I reworded my answer. $\endgroup$ – Ján Lalinský Dec 6 at 23:55

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