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If I put a perfect mirror(i.e. reflects with no attenuation) next to a blackbody radiator its spectra should be the same as the blackbody radiator.

Looking only at the spectra - is there any difference between a blackbody radiator and a perfect mirror?

For instance, suppose the mirror is accelerating as in the dynamic Casimir effect - does the spectra change in the same way as the blackbody radiator?

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This paper addresses your question, I believe, but I do not have access to full text from here. "Radiation from perfect mirrors starting from rest and the black body spectrum" – Mark Beadles Jan 19 '12 at 0:31

1 Answer 1

A mirror at rest (or moving at constant velocity) emits no thermal radiation whatsoever.

Detailed balance, i.e. 2nd law of thermodynamics, requires a relationship between absorbing incoming radiation and turning it into heat, versus emitting thermal radiation. A perfect mirror at rest does not absorb any incoming radiation, therefore it is a "whitebody", not a blackbody, and emits no thermal radiation.

By special relativity, a mirror moving at constant velocity should not emit radiation either.

An accelerating mirror, on the other hand, does emit radiation, at least according to the paper linked by @Mark Beadles. Don't ask me why, or what spectrum, I don't know!

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According to the paper it is black body radiation. But I don't know what temperature, or why - sadly I don't have time to read it more thoroughly and get a better idea. – Nathaniel Jan 19 '12 at 14:52

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