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Given a collection of photons obeying the Planck distribution at some temperature $T$, is it possible to compute the average electric field they produce and argue that a perfect blackbody radiation is completely unpolarized? I am looking for a mathematical understanding of the nature of polarization of perfect thermal radiation.

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    $\begingroup$ The radiated energy does not depend on polarization, so every polarization is represented there. $\endgroup$ Jun 29, 2018 at 17:06
  • $\begingroup$ Vaguely related: Slow thermal equilibrium. $\endgroup$ Jun 29, 2018 at 17:07
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    $\begingroup$ Equipartition? $\endgroup$
    – rob
    Oct 2, 2018 at 15:29
  • $\begingroup$ Hey SRS, what's the status on this post? Do we still need to improve the answers? $\endgroup$
    – DanielSank
    Oct 9, 2018 at 18:39
  • $\begingroup$ @DanielSank The existing answers are pretty good indeed. I'll take some more time before I accept one. $\endgroup$
    – SRS
    Oct 10, 2018 at 6:49

2 Answers 2

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Yes, you surely can compute things about the electric field of black body radiation.

As stated in the question, the electromagnetic field obeys the Planck law. The Planck law usually refers to the spectral radiance of a black body. To simplify things, we can instead refer to Bose-Einstein statistics, which tell us the mean occupation of each mode of the electromagnetic field. The Bose-Einstein law is $$n_i(E_i) = \frac{1}{\exp(E_i/k_b T) - 1}$$ where $E_i$ is the energy of the $i^\text{th}$ mode and $k_b$ is the Boltzmann constant. Noting that $n_i$ depends only on energy, and that two polarizations of the electromagnetic field (in vacuum) have the same energy, we can see that there's is no preferred occupation of any particular polarization.

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  • $\begingroup$ Can we actually compute the resulting electric field and show that the field does not oscillate in a fixed direction? @DanielSank $\endgroup$
    – SRS
    Jun 30, 2018 at 12:17
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    $\begingroup$ @SRS, the local electric field is 0, on average, at any point in the thermal radiation field. The squared electric field is non-nil on average and is proportional to the energy density. $\endgroup$
    – Cham
    Jun 30, 2018 at 12:50
  • $\begingroup$ @Someone Why do you say that the local electric field is zero? $\endgroup$
    – SRS
    Jun 30, 2018 at 12:51
  • $\begingroup$ Because it's a thermal field. It's constantly fluctuating. At equilibrium, it cannot have any fixed direction. $\endgroup$
    – Cham
    Jun 30, 2018 at 12:54
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    $\begingroup$ @SRS the electric field in a thermal state isn't a fixed quantity. We can, however, compute its statistical properties. As Someone already mentioned, the mean field at any particular point is zero. I suppose we could try to come up with a quantity like the "average polarization" and show that it too is zero. $\endgroup$
    – DanielSank
    Jun 30, 2018 at 16:07
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The simplest way to see this is by thermodynamics. The radiation an ideal blackbody can emit is completely determined by what it can absorb. Since blackbodies can absorb all polarizations of radiation, they must be able to emit all polarizations of radiation. Hence they emit a statistical mixture of all of them.

Of course, if the surface of the "blackbody" was a reflective polarization filter, which reflected horizontally polarized light, then it would not emit horizontally polarized light either.

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