Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
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!

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.