# Two interacting blackbodies (one inside another) - when will thermal equilibrium be attained? [closed]

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.

• Are you just not aware that your confusion is simply solved by $1+r+r^2+r^3+\cdots=\frac1{1-r}$ with your $r=\frac12$? Commented Jul 5 at 9:54
• @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. Commented Jul 5 at 10:19
• No, you simply made a mistake. It is [(P+P/2)+1/2(P/2)] +... Commented Jul 5 at 10:26
• @naturallyInconsistent Yes, I get my mistake now. Thanks for the dynamic approach to the solution! Commented Jul 5 at 15:18

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
• +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.) Commented Jul 7 at 20:14