Are there entropy changes associated with the transmission of energy from the sun to the earth? Does radiation differ from other modes of heat transfer with respect to the entropy changes?

Are the earth and sun in thermal equilibrium?

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Entropy is the Lack of order or predictability; gradual decline into disorder. As every attempt becomes less orderly over time entropy is associated with any system although in smaller timescales it is harder to measure accurately. So yes there is increasing entropy in the sun earth thermal system and no it is not a "perfect" thermal balance. –  Argus Jun 28 '12 at 3:14
If the earth and the sun were in thermal equilibrium, we would all be vaporized right now. –  user2963 Jun 28 '12 at 4:03
To make this question better I would ask (and would be interested in knowing) what an explicit calculable relation is between entropy increase and radiation (using formulas)... –  Chris Gerig Jun 28 '12 at 7:38
And actually a quick google search (which you should've done) shows such a calculation for blackbody radiation (on physicsforums.com) –  Chris Gerig Jun 28 '12 at 7:42

The Entropy article in Wikipedia includes a paragraph on the statistical mechanics definition of entropy. It is useful for such questions as you pose.

Specifically, entropy is a logarithmic measure of the density of states:

$$S = - k_B\sum_i P_i \ln P_i$$

where $k_B$ is the Boltzmann constant, equal to 1.38065×10$^{−23}$ J K$^{−1}$. The summation is over all the possible microstates of the system, and $P_i$ is the probability that the system is in the ith microstate. For most practical purposes, this can be taken as the fundamental definition of entropy since all other formulas for S can be mathematically derived from it, but not vice versa. (In some rare and recondite situations, a generalization of this formula may be needed to account for quantum coherence effects, but in any situation where a classical notion of probability makes sense, the above equation accurately describes the entropy of the system.)

In what has been called the fundamental assumption of statistical thermodynamics or the fundamental postulate in statistical mechanics, the occupation of any microstate is assumed to be equally probable (i.e. $P_i=1/\Omega$ since $\Omega$ is the number of microstates); this assumption is usually justified for an isolated system in equilibrium.[23] Then the previous equation reduces to:

$$S = k_B \ln \Omega$$ This means that any changes in the number of microstates increases entropy. Thus radiation does so, each new photon out of the sun increasing its number of microstates.

Thermal equilibrium means that the temperature is the same in the systems in thermal equilibrium and as @Zephyr commented above the earth would have the temperature of the sun, if we were in thermal equilibrium with the sun.

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You did mention the nonequilibrium entropy, but you didn't say that the nonequilibrium entropy is constant during the expansion of the light from the sun. Entropy doesn't increase until the photon hits the Earth, and then the Earth's microstates increase somewhat, but get reduced by the entropy of the radiation leaving the Earth, so that the Earth ends up in a steady state regarding microstates. –  Ron Maimon Jun 30 '12 at 12:46
@Ron Maimon Well, I did not go into what happens during the time the photons travel. In principle now and then they will have a Compton scattering with the intermediate dust and thus increase the total entropy too.And I would count radiation leaving the earth as extra microstates to be counted in the total sum as well as the one from the absorbed photons, so we differ there. –  anna v Jun 30 '12 at 13:21
p.s. also the sun is turbulent and radiation is not parallel. There will be photon photon scattering through higher order diagrams too, increasing the microstates –  anna v Jun 30 '12 at 15:57
The radiation is not parallel in any case, no blackbody radiation ever is. When leaving the sun, it doesn't matter what the surface turbulence is, because the radiation starts out thermal and in equilibrium with the matter. The two-photon scattering is negligible, as is dust. There's really no change in entropy as the photons go to the Earth. This is why you see the sun at a spot in the sky, and not as thermal radiation diffuse all around you--- you can extract information about the sun from the radiation, which is no longer thermal once it leaves the sun, but, well, coming from the sun. –  Ron Maimon Jul 1 '12 at 7:16
@RonMaimon Well, I am surprised that you think things are negligible. In my book, every interaction vertex counts in increasing the microstates, which is the simplest way to think about entropy changes, imo. I have never studied the information framework. In my accounting, entropy changes as I outlined above. The couplings may be small and tiny, but there are zillions of photons coming from the sun. I can compromise with "small" change in entropy, but not with "no" change. –  anna v Jul 1 '12 at 7:35

The Earth and the sun are not in thermal equilibrium, otherwise they would be at the same temperature. The quick answer is that there is no entropy change associated with light propagating freely through space, but this is a surprising thing. Freely propagating light just carries heat from emitter to absorber, and the heat is carrying an entropy which is 4/3 the energy density divided by the temperature.

The answer to the second question is that this is slightly different from other physical systems. If you allow a box of gas to expand into empty space, it's entropy usually increases during the expansion due to particle collisions. This goes on until the gas is dilute enough to turn into free particles, and them the entropy is constant (in the appropriate definition). The free particles are no longer in thermal equilibrium, and as the gas expands it carries a large amount of information about how it started--- the fast particles run ahead of the slow ones, revealing the velocity distribution.

### Expansion of gas

I will explain the classical analog first--- a classical noninteracting gas expanding into vacuum. The blackbody light is analogous to a box of very dilute gas. Imagine some non-interacting particles that have reached thermal equilibrium with the walls. The particles have a Maxwell distribution on the velocities, and they fill up the box randomly, so that the probability distribution for each particle's position and velocity is the same, and given by:

$$\rho(x,v) \propto e^{-v^2\over 2mkT} B(x)$$

Where B(x) is 1 inside the box, and zero outside the box. The joint probability distribution for all the particles is the product

$$\rho(x_1, v_1) \rho(x_2,v_2) .... \rho(x_N,v_N)$$

This is the Boltzmann distribution for the particles of the gas, it is $e^{-\beta H}$ where H is the energy for the free particles as a function on phase space and $\beta={1\over kT}$ is the reciprocal of the temperature measured in energy units.

Now the walls vanish. During this collisionless expansion, the gas goes completely out of thermal equilibrium. The total information in the probability distribution stays constant, but the total physical volume that the particles occupy grows enormously, so that the positions and velocities very quickly become highly correlated (you can even use this system to experimentally measure the Maxwell distribution).

The entropy of a classical probability distribution $\rho$ can be defined as the Shannon entropy:

$$S = - \int \rho \log\rho$$

If you write $\rho$ as the canonical ensemble probabiity distribution $\rho = {1\over Z} e^{-\beta H}$, this formula reproduces the thermodynamic entropy, so that it is a proper generalization. But it extends the concept to any probabilistic description of any classical system. The Shannon entropy is given meaning by the noiseless coding theorem (see this: Proof of $S=-\sum p\ln p$? ), it is the minimum number of bits on average (bits if you take log base 2, usually physicists take natural log, so its the number of "nats") that you need in order to specify in order to pin down the full state of the system (the position and velcoities of all the atoms) given a random pick from the probability distribution. You need infinitely many bits to describe a classical state to infinite precision, and this is reflected in the fact that the entropy is only defined up to an additive constant classically (see linked answer).

When the particles are freely expanding, this classical entropy is constant. The probability distribution at time t becomes

$$\rho_t(x,v) = \rho_0(x-vt,v) = e^{v^2\over 2mkT} B(x-vt)$$

Since this is just a shear transformation (a v-dependent shift of x) of the original distribution, the entropy integral is constant.

### Entropy of sunlight

The same is true for light from the sun. The light, when it leaves the surface of the sun, is (almost exactly) in thermal equilibrium, but it propagates freely outside the sun towards the Earth, and the part of the light that reaches Earth is not in thermal equilibrium at all --- it is all coming from one direction, namely the sun. The total entropy of all the light we receive is the same as the entropy this light had on the surface of the sun. The entropy is now the quantum information, not the classical information:

$$S= - tr(\rho\log\rho)$$

And it is calculated by a sum over the discrete quantum states, just as the classical entropy is calculated by an integral over all phase space.

The partition function can be calculated from the explicit formula

$$Z = \prod_k \sum_{n_k} e^{-\beta n_k |k|}$$

To simplify the formulas, I have set $\hbar=c=k_B=1$ (this is a relativistic thermal quantum system), I will restore them at the end by dimensional analysis. The free energy density in d-dimensions is

$$-\beta F = \log Z = {\sigma\over d} {V\over \beta^d}$$

And this determines everything else. The scaling law is determined by dimensional analysis--- it is by rescaling the k-integral to make it independent of $\beta$. The free energy is negative (this means that the entropy times the temperature is bigger than the energy). The constant $\sigma$ is defined with a factor of $1\over d$ to make the formula for the energy density nice:

$${\sigma\over d} ={ N_p S_{d-1} \zeta(d+1) \Gamma(d)\over (2\pi)^d }$$ $$\sigma ={ N_p S_{d-1} \zeta(d+1) \Gamma(d+1)\over (2\pi)^d }$$

Where $N_p$ is the number of polarizations, 2 for the photons. The energy density is

$$U = d |F| = \sigma T^{d+1} V$$

This is the only dimensionally correct extensive expression, since when $\hbar=c=k_B=1$ temperature/energy is an inverse time which is an inverse length. The entropy is

$$S = (d+1) |F| = {d+1\over d} U$$

This all follows from thermodynamics alone, since F is a power-law in temperature. In 3d, the entropy of the emitted radiation is ${4\over 3}$ the energy density.

The sun produces light at temperature T carrying entropy per unit volume which is $4 a T^3$. In a unit of time $\Delta t$, the pressure that an equilibrium radiation gas would exert on the sun (considered as a blackbody) is one third the energy density. This pressure is double the light that the sun (considered as a blackbody) would absorb in a unit of time, so it is the amount of light emitted by the blackbody sun.

The entropy of the outgoing light is 4/3 the emitted light power.

$$\dot{S} = {4\over 3} \sigma T_s^3 (4\pi R_s^2)$$

This is the Stephan Boltzmann law for emitted entropy---4/3 the emitted power. The constant $\sigma$ is the Stephan Boltzmann constant.

The fraction of this entropy that hits the Earth is given by the solid angle subtended by the Earth, so the entropy that hits the earth in time $\Delta T$ is

$$S_i = {4\over 3} \sigma T_s^3 (4\pi R_s^2 c \Delta t) \pi{R_e \over A}^2$$

Where $R_e$ is the radius of the Earth and A is the astronomical unit. In the same unit of time, a shell of thermal infrared light leaves the Earth, carrying entropy

$$S_o = {4\over 3} \sigma T_e^3 (4\pi R_e^2 c \Delta t)$$

So the total entropy produced by the Earth is the difference between $S_o$ and $S_e$. The ratio of the two is:

$$(\pi {R_s\over A})^2 ({T_s\over T_e})^3$$

The first factor is the radian solid angle subtended by the sun in the sky, and the second factor is the ratio of the entropy densities of the equilibrium radiation on the surface of the sun and on the surface of the Earth. The temperature ratio is 20:1, so the second factor is 8000, while the first factor is the solid angle of a quarter-degree circle, or $\pi({\pi\over 720})^2$, so the Earth is dumping about twice the entropy into space that it receives from the sun.

The total entropy per second we dump is a bound on the maximum information transformation the Earth can do-- the maximum number of bits you can zero out per second. This bit-rate (restoring hbars and c's and dropping order 1 factors) is

$$I ={ R_e^2 (kT_e)^3 \over c^2\hbar^3 \log(2) }$$

which is about $10^{38}$ bits per second. This was suprising to me, because it is not that much bigger than the total information in all the biomass on Earth.

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I was surprised at the result for the number of bits per second, because this is about the biomass bit-capacity of the Earth, so the limit is close to achieved. It seems that the Earth's plants are using up a sizable fraction of the solar photon entropy for computation, at least there isn't a factor of a million left over. On another note, when editing long posts, I have noticed that the stackexchange automatic tex-renderer makes extremely slow recently. Is this a change in software? –  Ron Maimon Jun 28 '12 at 8:45
Addressing your comment, where did you get your figure for the "biomass bit-capacity of the Earth", and what do you mean by that phrase? Energetically, only a small fraction of incoming light is used in photosynthesis, and only a tiny, tiny fraction of that is used for anything connected to "information processing" in the sense of DNA copying or transcription - so if your figures suggest that the entropy produced by the information processing is of a similar magnitude to the total entropy production then something has gone wrong somewhere! –  Nathaniel Jun 28 '12 at 9:08
This is a completely obfuscated way to explain a rather simple phenomenon. No one wants an essay :) –  Chris Gerig Jun 28 '12 at 9:20
I get a lower figure than you for the outgoing entropy. Earth is well approximated by a black body that emits $Q=238\,Wm^{-2}$ at $T=255\,K$, so the entropy production is $$A\frac{4}{3}\frac{Q}{T} \approx 6.3\times 10^{14}\,W$$ or about $7\times 10^{37}\,\text{bits}$. –  Nathaniel Jun 28 '12 at 9:25
That should be $7\times 10^{37}$ bits per second, obviously. –  Nathaniel Jun 28 '12 at 9:31

View this as an extended-comment. Hopefully answering the following questions will do some justice: Does radiation cause change in entropy? Is there an explicit relation between radiation and entropy?

There is no entropy change associated with photons expanding in vacuum (information is not lost); but it can change the entropy of your system. The entropy carried by blackbody radiation is given as follows. Photons in a volume $V$ at temperature $T$ induces an energy density $u=\frac{\sigma}{c} T^4$, and a small calculation gives $S_\text{rad}=\frac{16\sigma V}{3c}T^3$. If you now add a heat bath with temperature $T_r$, then, assuming the heat capacity of the cavity material is negligible, the overall change in entropy of the universe (after the system and heat bath reach thermal equilibrium) is $\triangle S=\frac{4\sigma V}{3c}T_r^3[1-4(\frac{T}{T_r})^3-3(\frac{T}{T_r})^4]$.

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What does it mean to say "There is no entropy change associated with photons expanding in vacuum (information is not lost); but it can change the entropy of your system." How does the entropy of a "system" change when there is no change in entropy from the expansion? There is no change in entropy during free propagation. There is a change from absorption and reemission at a new temperature. This is the main content of my answer. I might have gotten a wrong prefactor for the entropy, I didn't start with the energy radiated, but I calculated the entropy directly. I'll check the prefactor. –  Ron Maimon Jun 28 '12 at 15:08
i.e. we need not just view radiation as "photons expanding in vacuum" alone. –  Chris Gerig Jun 28 '12 at 16:07
How else can you view it? If it's not photons, there is no statistical equilbrium possible, and no blackbody radiation. The "photons expanding in vacuum" is not an equilibrium concept, and I calculated the entropy transfer in a non-equilibrium way that I realize now is not what books do. I think it makes more sense the way I do it (although I think my prefactor is different by a rational ratio, and I probably made an arithmetic mistake). The process becomes equilbrium if you superpose additional photons going the other way, and then you can use detailed balance arguments. –  Ron Maimon Jun 28 '12 at 19:24
These arguments are not made often in books, instead they say "using the formula dQ/T we find the entropy going into space by blackbody radiation..." without justifying why this formula applies (although it isn't hard to justify, the entropy of the outgoing photons themselves is a strange idea, since it is a gas expanding isentropically and becoming out of equilibrium). I really don't see any simpler way of thinking about it than what I wrote, please enlighten me if I am wrong. I also shortened it a bit, to remove internal and external redundancies, and I'll fix the prefactor after I redo it. –  Ron Maimon Jun 28 '12 at 19:26
No all I am saying is that if you simply consider the radiation as photons moving (and not interacting with anything), then it carries entropy but doesn't change. But usually we consider the radiation interacting with a system, in which case there is entropy change. –  Chris Gerig Jun 29 '12 at 4:14