In several papers I see something equivalent to the following expression for the entropy of radiation given by an astronomical object such as the Sun (assuming the object can be approximated as a black body):
$$
s = \frac{4}{3}\frac{u}{T},
$$
where $u$ is the total flux of radiation energy through a spherical shell surrounding the object, $s$ is the entropy flux through the same imaginary surface, and $T$ is the black-body temperature of the object (and hence also of the radiation). It is also generally stated that no entropy is produced during the process of emitting radiation.
In a much smaller number of papers I see a formula that corresponds to just $s=u/T$, which is what I would expect. So the short version of my question is, which of these is correct for an astronomical body? But please read the rest, so you can understand why I'm so confused about it.
I understand the derivation of the 4/3 formula by considering a photon gas enclosed in a piston (see below), but in the context of a continuously emitting body like the Sun it doesn't seem to make sense. The issue is that if, over a given time period, the body loses an amount of heat $Q$ then this must balance the increase in energy of the radiation field $U$, i.e. $U=Q$. The body loses entropy at a rate $Q/T$, and if radiation is really a reversible process then this should equal the gain in entropy of the radiation, which should therefore be $U/T$. But according to the above formula it's actually $4U/3T$, meaning that the total entropy increases by $Q/3T$.
The 4/3 formula above was derived by Planck (in his book "the Theory of Heat Radiation", of which I've read the relevant chapter), who considered a photon gas in a sealed cylinder of finite volume. At one end of the cylinder is a black body and at the other is a piston. The radiation comes into equilibrium with the black body and exerts a pressure on the piston. If one reversibly (i.e. slowly) allows the piston to move then this causes some heat to be lost from the black body. In this case it turns out that $U=3Q/4$, with the discrepancy being due to the fact that the radiation field loses energy when it does work on the piston. The entropy balance then requires that the entropy of the photon gas increases by $4U/3T$, as above.
The point is, I can't see how the radiation being emitted by the Sun can be seen as doing work on anything. At first I thought it might be doing work on the outgoing radiation field. So let's draw an imaginary shell around the Sun again, but this time let the shell be expanding at the speed of light. Perhaps the radiation inside the shell is doing work on the radiation outside it, and that's what's "pushing" it away from the Sun? But it seems to me that for anything inside the shell to have an effect on anything outside it, some kind of influence would have to travel faster than light, so I don't think that can be right.
In any case it's well known that for a normal gas (made of matter), expanding against a piston is quite different from just expanding into a vacuum. In the former case the temperature and internal energy decrease, because the molecules lose energy in pushing the piston, whereas in the latter case they both remain constant. I haven't found any source that addresses the question of why this would be different for a photon gas.
So it seems like the emission of radiation from a body like the Sun into space is quite different from the emission of radiation into a sealed piston, and I'm puzzled as to how the same formula can apply. Below I've listed some possible resolutions to this which seem plausible to me. I haven't been able to find any sources that directly address this issue, or state any of these positions.
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The emission of radiation into space is an irreversible process after all. $U=Q$, and the total entropy increases by $Q/3T$. (But then, what happens when radiation is absorbed by a body at a similar temperature? Surely the entropy doesn't decrease.)
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There is some weird sense in which the outgoing radiation can be thought of as doing work on something, so $U\ne Q$. (If so, why does nobody ever explain this subtle and important point?)
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$Q=U$, but the entropy of radiation emitted into space is actually different from the entropy of a photon gas in a sealed cylinder, such that its entropy is given by $U/T$, not $4U/3T$. (This one actually seems the most reasonable to me. The radiation in the closed cylinder has rays travelling in all directions, whereas the radiation coming off the Sun only has rays travelling in directions away from its surface, so it seems reasonable that they would have different entropies. This would mean that the 4/3 formula doesn't apply to the radiation emitted by astronomical bodies after all – but if that's the case then it's an extremely widespread mistake.)
Any insights into which, if any, of these is correct would be greatly appreciated – as would any references to sources that directly address the relationship between Planck's photon-gas piston and the emission of radiation into empty space.
Best Answer
Ok, so here is my answer to my own question:
Of the three options I presented in the question, the answer is 1: the emission of radiation into space is actually an irreversible process. At first I couldn't see how this could be the case, because the transfer of energy from the hot body to the outgoing radiation field doesn't involve a change in temperature, so it seemed like it should be reversible. What I had failed to consider is that the body interacts not only with the outgoing radiation field, but also with the incoming one (i.e. the cosmic microwave background in the case of a star.)
A useful metaphor is a heat exchanger, in which a pipe enters carrying cold water (temperature $T_C$), which is placed in contact with a body at a higher temperature $T_H$ until it equilibriates. Another pipe carries the warm water out.
Although the outgoing fluid is at the same temperature as the solid body, it is clear that this is an irreversible process. The entropy is produced not in the transporting away of warm water at $T_H$ but in the heating of water from $T_C$ to $T_H$.
A star may be seen as an analogous kind of "cosmic heat exchanger". In this case we have to imagine a small volume of space containing a small amount of cosmic microwave background energy at $3\:\mathrm{K}$ coming into contact with the star and being heated up to $6000\:\mathrm{K}$, and then moving away again.
These volumes of space should be thought of as moving at the speed of light. As with the heat exchanger, the transporting of the $6000\:\mathrm{K}$ radiation away from the star is a reversible process, but the "heating up of space" from $3\:\mathrm{K}$ to $6000\:\mathrm{K}$ is irreversible and produces entropy.
In my question I asked what would happen if this radiation were absorbed by a colder body, since it seems as if the entropy can decrease. For example, if $U$ Joules of thermal radiation at $4000\:\mathrm{K}$ (with entropy $\frac{4}{3}\cdot\frac{U}{4000} = \frac{U}{3000}\:\mathrm{JK^{-1}}$) were absorbed by a body at $3500\:\mathrm{K}$ then the body's entropy would increase by only $\frac{U}{3500}\:\mathrm{JK^{-1}}$ and it seems like the total entropy must have decreased. But if the second body can absorb all the radiation then it must be a black body, and hence it must emit black body radiation of its own according to the Stefan-Boltzmann law at a rate $A\sigma T^4$ (with A its surface area and $\sigma$ the Stefan-Boltzmann constant). It turns out that when you take the entropy of this outgoing radiation into account, the total entropy production is always positive. (Unless the absorbing/emitting body is at the same temperature as the radiation field, in which case it is zero, as we should expect in the case of thermal equilibrium.) This is like a heat exchanger operating the other way, with warm water coming into contact with a cold body, and cold water running out. If you forgot to take into account the entropy of the cold water it might seem as if the total entropy was decreasing.
Finally I should say why my argument about the outgoing radiation not doing work against a piston doesn't work. Imagine the following thought experiment. First we reversibly fill a piston with thermal radiation as described in the question. Then we make a small hole in the cylinder and let that radiation escape into space. This escaping radiation is exactly the same as black body radiation. The entropy of the radiation in the cylinder cannot decrease as it goes through the hole, and hence its entropy flux must be at least $\frac{4}{3}\frac{u}{T}$.