Can a perfect mirror melt from radiation?

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mezarashi
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Is it possible to melt a perfect mirror by radiative means? (reflects light of all frequencies with no absorption)

Which leads me to ask. What is the mechanism (at the atomic level), for the reflection of light? I've never had a satisfactory answer on the mechanism of reflection. Is it necessary for the atoms to firstly absorb the incoming photon, and then re-emit it? If so, what makes it make sure that the re-emitted light is identical to the incoming?
 

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Such a mirror cannot exist at finite temperature since it could not possibly come into thermal equilibrium with its environment, i.e. a temperature gradient would have to exist without the flow of heat. Thermal equilibrium requires both emission and absorption of radiation and they occur equally in order to equilibrate.

Reflection from the surface of a mirror (metallic, e.g.) occurs as a result of electrons responding to the applied electromagnetic field in such a way that they exclude electric fields within the material. Accelerated charges radiate so that, in effect, they cancel the applied fields by reradiating EM fields out of phase with the applied fields within the material. The reradiated field outside the material is what we call the reflected wave.
 
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mezarashi said:
Is it possible to melt a perfect mirror by radiative means? (reflects light of all frequencies with no absorption)
Which leads me to ask. What is the mechanism (at the atomic level), for the reflection of light? I've never had a satisfactory answer on the mechanism of reflection. Is it necessary for the atoms to firstly absorb the incoming photon, and then re-emit it? If so, what makes it make sure that the re-emitted light is identical to the incoming?

There have been several threads discussing such issue. You may look at one of them here...

https://www.physicsforums.com/showt...+reflection+transmission+conduction+electrons

Zz.
 
Sorry about that. The title of the thread was not immediately suggestive of the discussed contents. Thanks for the response and link.