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If you had a container that had perfect white walls on the inside and you somehow filled it with light, would the light always be reflecting inside the container because the color white does not absorb any light rays?

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  • $\begingroup$ Does 'white' reflect the entire electromagnetic spectrum? $\endgroup$ – Alfred Centauri Jun 24 '17 at 3:05
  • $\begingroup$ @AlfredCentauri The OP mentions 'light'. He might have only the visible electromagnetic spectrum in mind. $\endgroup$ – SRS Jun 24 '17 at 5:15
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Yes, a perfect reflector container could trap light. There are a couple of problems, of course, that make this impractical: there are no perfect reflectors, and there's no way to INSERT light (or observe it) if the container is closed.

A clean silver surface might reflect 99% of all light that strikes it, but a 1 liter container (circa 0.1m in size) implies that in a single second, there are three billion collisions of light with wall. That leaves about zero light in the container.

The only exception would be if your container were to rise in temperature and emit its own light. Odd as this sounds, if it's not a perfect reflector, a container WILL come into thermal equilibrium with its internal trapped light. The 'perfect reflector' hypothetical situation implies that no thermal contact, and no thermal equilibrium, occurs.

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There are two concepts:

  • First, if you refer to a perfectly reflecting surface (=a perfect mirror) then the light would always get reflected by assumption.
  • Second, if you refer to a system where the walls are in thermal equilibrium with its environment then the amount of emitted light is equal to amount of absorbs light. However, the "photons" are not reflected, however, I guess the color of the paint is not important. It just shifts the amount of absorbed and emitted light. See here.
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Because no surface is perfectly reflective eventually all the energy is absorbed. The energy is converted to thermal and radiates away in every direction.

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  • $\begingroup$ No surface that we know of is perfectly reflective. $\endgroup$ – Lambda Jun 25 '17 at 3:21
  • $\begingroup$ @Lambda you're right but at this point I would consider perfect reflection without energy loss the same as perpetual motion. $\endgroup$ – Bill Alsept Jun 25 '17 at 3:26

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