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My initial guess is yes there is a blackbody spectrum of photons inside a solid.

The process that emits thermal photons won't "know" that it is deep inside a solid and not near the surface, so the emission of thermal photons would be the same. They don't get very far.

So I'd like to ask: Is there a blackbody spectrum of photons inside a solid?

Let's assume that "inside" means in the center of a body with a temperature $T$ and that the body is at least several absorption lengths in size for photons near $k_BT$. Let's also assume the self-absorption is not so overwhelmingly strong that it significantly alters the emission process itself.

I believe that the answer will be the same for dielectrics, semiconductors and metals, but if the answer differs for these please mention it.

uhoh
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    Also would be worth thinking about the reflectivity, some photons are evanescent not because of absorption but reflection. – KF Gauss May 13 '19 at 00:00
  • You understand that most solids are not black bodies, yes? – DanielSank May 13 '19 at 05:52
  • @DanielSank yes indeed, which is why I've constrained the problem the way I have in terms of the medium being strongly self-absorbing. If you'd like to add an answer that addresses separately the shape of the spectrum from it's integrated power and mention emissivity, please do so. – uhoh May 13 '19 at 06:14

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In thermal equilibrium there are definitely photons inside a solid. They're bosons, so they will be described by the Bose-Einstein distribution. But they won't necessarily have a "blackbody spectrum", because that idea is specific to light in vacuum. Specifically, photons in vacuum obey the Bose-Einstein distribution with a trivial dispersion relation $\omega = c |\mathbf{k}|$, while in a general solid you might have something much more complex.

You probably won't see this commonly mentioned in solid state textbooks because most solids aren't blackbodies for light, i.e. light will pass directly out (so the absorbance is far below $1$, and you don't get the full thermal population in equilibrium) or electromagnetic waves of visible frequency are evanescent inside (so there aren't corresponding real $\omega$ at all). But the idea of a thermal spectrum of photons within a medium is definitely used in physics. For example, when people talk about radiative transfer within stars, they explicitly consider the thermal photons.

knzhou
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    "While this is an easy question to speculate upon, radiative transfer within various solids spanning a range of opacities is not a simple problem, but is one that must have been considered before, so I think this can have a definitive answer that draws from authoritative sources." +1 and thank you, but I've added the bounty to try to find something rigorous. – uhoh Feb 17 '24 at 03:32
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    @uhoh I know you wanted that, but I don't have any on hand -- figured I'd contribute a partial answer. – knzhou Feb 17 '24 at 03:33
  • and it certainly gives me more insight and a better way to think about the problem :-) A wise person once said "You can't always get what you want. But if you try sometimes, well, you might find... you get what you need." – uhoh Feb 17 '24 at 03:34
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    @uhoh Maybe for condensed matter references you can look into polaritons -- those are what photons become when they mix with various in-medium excitations. – knzhou Feb 17 '24 at 03:35
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In steady state condition (same temperature across the volume) situation is indeed slightly different. There is a continuous black-body emission spectra in both cases, but I cannot tell if they are absolutely identical.

Photons that are emitted deep inside - indeed don't know they are deep, and follow the same rules.

Main difference is that atom in the middle of the solid has vibrating lattice all around it, and hence see more field disturbances all around. Atom at the edge of the material - does not feel vibrating lattice from once side. Still spectra is continuous in both cases.

If material has some transparency windows - emission from deep levels will partly reach the surface and will change overall emission spectra.

BarsMonster
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    I'm pretty sure this is the beginning of the right answer, but I wish it linked to some published discussion or relied on some fundamental physics. If you ever take a break from overclocking wrist watches and photographing ducks, it would be great if you could go a little deeper with this answer. Thanks! ;-) – uhoh May 19 '19 at 13:25