# Entropy of radiation emitted into space

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.

1. 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.)

2. 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?)

3. $Q=U$, 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.

• I've spent some time discussing this with someone who's more of an expert on these these things than I am, and he reckons the answer is my possibility 1 above: entropy is produced when a body radiates into space, even though the body and the emitted radiation have the same temperature. The answer to my question about what happens when that radiation is absorbed by another body at a similar temperature is that the other body will also be emitting radiation according to the Stefan-Boltzmann law at its own temperature, and it turns out that this always results in an increase of entropy overall. – Nathaniel Oct 5 '11 at 13:42

## 5 Answers

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 $\frac{4}{3}\frac{u}{T}$.

• I would like to make sure I understand the picture. In the water pipe, you need input water to conserve the overall quantity of matter. This is not the case for the emission of a blackbody, which generates the photons it emits. Also, if emitting an energy $Q$ generate some entropy, how could absorbing energy at the same time reduce the total entropy without somehow destroying some ? And could you detail the balance you consider with the body at 3500K ? – DSuchet Feb 14 '17 at 7:27
• @Penangol (I'm aware that your comment is a year old - I somehow missed it at the time): in the water pipe analogy, the analog of the fluid is empty space. This is counterintuitive, but you can make sense of it if you imagine a perfect adiabatic box containing a vacuum. Energy can be put into this vacuum (in the form of thermal radiation), just as energy can be put into water (in the form of heat). This is why I drew packets of "empty space" moving around - they're analogous to packets of water entering and leaving the heat exchanger. – Nathaniel Feb 15 '18 at 9:06
• For your second question I'm not sure I understand it properly, can you clarify (if you're still interested)? For your final point, I agree, my answer would be improved by putting in the full calculation --- perhaps I will do that one day. – Nathaniel Feb 15 '18 at 9:06

Let us consider a leaky container of black body radiation. The process of radiation leaking out has been reversed when the leaked radiation has been collected in a container that is the same size as the original container. The radiation can be directed towards the collecting container using mirrors. But all of the radiation can not be catched into the container, unless there is a Maxwell's demon kind of thing opening and closing some kind of door. (because photons arrive at different times)

It seems that all of the entropy of black body radiation is caused by photons having different creation times, because if there was no difference of creation times, the radiation process could be reversed.

• That's an interesting observation and gives a plausible physical reason why my option 1 would be correct, i.e. entropy does increase when a body emits radiation. (I'm now pretty sure that option 1 is correct - I've been meaning to post it as an answer for a while.) – Nathaniel Jan 2 '12 at 14:50
• Side note: your argument doesn't show that /all/ the entropy in the radiation comes from the different emission times - rather, it shows that spreading out the emission over time increases the entropy to a greater value than it would have if they were all emitted at once. This is because the radiation already has some entropy when it is sitting in the container prior to leaking out. – Nathaniel Jan 2 '12 at 15:00

This paper is crystal clear: http://www.csupomona.edu/~hsleff/PhotonGasAJP.pdf

Blackbody radiation do not carry only "heat", also "work", it could move a piston.

Q is not always equal to U (only in a irreversible receiver), in a a reversible receiver 1/3 U could be converted to work

First thermal radiation is emitted by matter as a result of its finite temperature. Matter has zero entropy when its temperature equals zero Kelvin. Increase in temperature results increase in entropy. If a solid substance putted in vacuum, it radiates resulting decrease in temperature and also decrease in entropy. Thermal radiation itself is entropy decreasing processes.

Second, the second law of the thermodynamics states that for closed system entropy is equal zero or increase for any processes.

All above are known subject. As you try to understand what about the entropy change of sun radiation. Many text including theory of heat radiation by Max Planck define the magnitude of the entropy S of black body radiation as S=4/3*AσT3 or S=4/3*aT3*V (page 65 eqn. 80 in theory of heat radiation by Planck).

Max Planck used special cylinder-piston arrangement that its base is black surface and the rest surfaces are perfect reflector. This arrangement has pdV work output. There is volume change with positive work while T is constant. Energy and entropy change is found for this arrangement.

Based on above entropy calculation and according to the second law of thermodynamics for a case like sun thermal radiation, entropy change is tried to show always positive. This may be mistake because second law does not dictate that the entropy of the system must be positive. It may be negative but total entropy including surrounding must be zero or positive. So sun is not closed system itself and not required to have positive entropy. Its entropy change may be negative and including its surrounding total change become positive. Many text and papers state that pure thermal radiation always cause positive entropy change against the entropy decreasing nature of the thermal radiation. Sun is not a closed system itself and not required to have positive entropy change.

If a black body surface has the magnitude of the entropy S= 4/3*aT3*V (as in the page 65 of the theory of heat radiation), next page sec. 67 explain that the increase in energy of radiation by 1/3(U’-U). This excess heat is added to do external work accompanying the increase in the volume of radiation.

Last discussion clarify the situation that without work, entropy of the thermal radiation must become S= AσT3. But Planck do not consider this and this may be assumed as mistake.

As a result I think third situation is the correct one.

Also I want to ask that is there any condition that the second law of thermodynamics has an exception?

According to me the answer of this question is yes. Detail is presented in
hsoylu.wordpress.com/2012/03

I will discuss briefly radiation heat transfer between two surfaces. For thermal equilibrium qnet=0 and surface temperature is given by below equation.

$$T1 / T2 = (A2* F 2→1 / A1* F 1→2 )1/4$$

T1 and T2 are surface temperature, A1 and A2 surface area, F1→2 and F2→1 view factors. As it is seen left side of equation is represents the physical properties and right side of equation represents the geometrical characteristics. Inside of the bracket known as the reciprocity equation and it gives one. It is obvious that surface temperature depends to geometric data during thermal equilibrium. It is assumed for ideal black surface emission is independent of the direction. But in reality and also according to the electromagnetic theory of emission it depends to the direction. Direction dependence is the geometric property of the emission than it changes the result of the bracket in the equation different from one and the surfaces temperature also changes. This means that if the surface temperatures are equal at the beginning, they will differ for thermal equilibrium at the end.

This temperature difference violates the second law of the thermodynamics. For thermal equilibrium, heat transferred to low temperature surface to high temperature surface.

• A couple of things: firstly, when I talk in the question about entropy being produced by the emission of radiation, I mean in the increase in entropy of entropy of the sun plus its radiation field. I know that the sun isn't a closed system, but the sun together with its outgoing radiation is, to a good degree of approximation. Secondly, regarding your exception to the second law, a real surface will indeed have some directionality to its emission - but it will also have the same kind of directionality in how it absorbs radiation. In equilibrium it will always emit as much radiation as it ... – Nathaniel Mar 19 '12 at 22:00
• ... absorbs, at every frequency and at every angle independently. (There's a name for this law but I can't remember it. But it's really just a simple consequence of the second law.) So at equilibrium the temperatures in your system will be equal. – Nathaniel Mar 19 '12 at 22:02

The process you've described isn't reversible: imagine playing the film backwards. There's a hot star, at the focus of a rather wierd field. This star absorbs all the incident radiation, but doesn't radiate itself. I forget the exact argument, but that isn't compatible with the laws of thermodynamics.

The emitted radiation is far from thermal equilibrium, and I expect that even defining its entropy is tricky.