Let's say you have the ability to shine some light into a perfectly round sphere and the sphere's interior surface was perfectly smooth and reflective and there was no way for the light to escape.
If you could observe the inside of the sphere, what would you observe? A glow? And would temperature affect the outcome?
Seems silly, it's just something I've always thought about but never spent enough time (until now) to actually find an answer.
Answer
OK, the inside of the sphere is perfectly-reflecting, and there's an ideal optical diode to let light in but keep it inside. As you keep the light turned on, the photon density in the sphere goes up and up, of course. It "looks" brighter and brighter, but you don't see that because the light can't escape. After turning the light off, it stays bright, the photons just keep bouncing around. If you "stick your head in" to look, you see a bright uniform glow that quickly dies away because your head and eyes are absorbing all the photons.
But do the photons bounce around forever? No!! Even a perfectly-reflective sphere will still interact with the light, because of radiation pressure. Each time a photon bounces off a wall, the wall gets kicked backwards, gaining energy at the expense of the photon (on average). Light can't produce a smooth force, only a series of kicks with shot noise statistics, because one photon hits the wall at a time. These kicks eventually heat up the walls, and cool down the photons. (From the photon's point of view, the photon frequency is going down because of Doppler-shifts during reflection off the moving walls.) Eventually everything equilibrates to a uniform temperature, hotter than the sphere started out. I don't know how long that would take. [In any realistic circumstance this radiation pressure effect can be ignored, because it is much less important than the "reflection is not 100% perfect" effect.]
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