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As I understand it, an ideal blackbody absorbs (and subsequently starts emitting) all incoming radiation. In typical setups like determining a planet's temperature given its albedo and distance from a star, we use the thermal equilibrium condition $P_{in} = P_{out}$ for the planet and use Stefan-Boltzmann's Law to obtain the planet's temperature (using approximations like emissivity, albedo, and so on).

Now, suppose we have an ideal spherical blackbody (A) inside a larger ideal blackbody in the shape of a spherical shell (B). This entire setup is in deep space. Say A has some source of power like fusion going on inside similar to a star. Here, the power emitted by A, $P_A$, is equal to the power supplied by the fusion reaction. Also, suppose that initially $P_A >> P_B$ (we can neglect any radiation by B initially; suppose no fusion is going on inside it). Now, as the outgoing $P_A$ is incident on B, after a while (when thermal equilibrium is reached), B starts emitting radiation with power $P_A$ (here, B emits $\frac{P_A}{2}$ outwards into deep space, and $\frac{P_A}{2}$ is emitted towards A). So, this $\frac{P_A}{2}$ is incident on A, which after a while starts emitting at $P_A + \frac{P_A}{2} = \frac{3}{2}P_A$. Then, this $\frac{3}{2}P_A$ will be incident on B, causing its emitting power to eventually raise to $\frac{3}{2}P_A$. From here, half of the power, i.e. $\frac{3}{4}P_A$ will again heat up A further. This in turn will heat up B, which will again heat up A. And so on and so forth.

Apparently, therefore, I am unable to see how the two interacting blackbodies will ever reach equilibrium. This is quite puzzling, and I think I might have some loopholes in the fundamental understanding of blackbodies in the first place. Any help is appreciated.

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    $\begingroup$ Are you just not aware that your confusion is simply solved by $1+r+r^2+r^3+\cdots=\frac1{1-r}$ with your $r=\frac12$? $\endgroup$
    – naturallyInconsistent
    Commented Jul 5 at 9:54
  • $\begingroup$ @naturallyInconsistent Well, I did think in this line initially; but r is actually 3/2 in this case, not 1/2. That leads to the series diverging, meaning the blackbodies get arbitrarily hot. To see why r is 3/2, try listing down the power of the inner blackbody at different stages: P, P + P/2, [(P + P/2) + 1/2 * (P + P/2)], .... This gives x_n = x_(n-1) + 1/2 * x_(n-1) = 3/2* x_(n-1). That is to say, half of the power is ADDED to the already existing power for the inner blackbody, making r equal to 3/2. $\endgroup$
    – Faiyaz
    Commented Jul 5 at 10:19
  • $\begingroup$ No, you simply made a mistake. It is [(P+P/2)+1/2(P/2)] +... $\endgroup$
    – naturallyInconsistent
    Commented Jul 5 at 10:26
  • $\begingroup$ @naturallyInconsistent Yes, I get my mistake now. Thanks for the dynamic approach to the solution! $\endgroup$
    – Faiyaz
    Commented Jul 5 at 15:18

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Here, the power emitted by $A$, $P_A$, is equal to the power supplied by the fusion reaction.

Here is the mistake. In equilibrium the power emitted by $A$ is given by $P_{in}=P_{out}$, and the $P_{in}$ term includes both the power supplied by the fusion reaction, $P_F$ as well as the power received from $B$. So the correct expression for $A$ is $$P_F+\frac{1}{2}P_B=P_A$$ and the correct expression for $B$ is $$P_A=P_B$$ This is two linear equations in two unknowns and is easily solved to obtain $$P_A=P_B=2P_F$$ There is no runaway solution in this case, although other cases certainly can produce runaway solutions

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    $\begingroup$ Thank you so much! $\endgroup$
    – Faiyaz
    Commented Jul 5 at 15:01
  • $\begingroup$ +1 for this clear explanation. One might take away from the result that the mechanisms of heat transfer and heat generation are acting in conjunction to somehow "leverage" or "amplify" $P_F$ by a factor of two when considering the intermediate heat transfer in each direction. I wonder if you could comment on this interesting "leveraging" or "amplification," if that's a reasonable interpretation. (Of course no work can be extracted from the heat transfer in either direction between two objects at the same temperature.) $\endgroup$
    – Chemomechanics
    Commented Jul 7 at 20:14
  • $\begingroup$ @Chemomechanics Thanks for your kind words. I don’t know if this is particularly insightful, but this is the way clothing works. We need to keep our body at a certain temperature. Clothing allows some of the thermal energy we lose to come back to us. This reduces how much energy we need to expend to heat our bodies. $\endgroup$
    – Dale
    Commented Jul 7 at 20:55

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