# Why are good absorbers also good emitters?

Yeah, good emitter and good reflector are definitely not the same property. Maybe the best way to visualize it is just as the time inverse, under the mapping $$t\mapsto-t$$, of absorption is emission. This also gives something of a hint of why this relationship might hold, albeit I only know the derivation for the case of thermal radiation.

For thermal radiation, you bring two bodies of the same temperature into radiative contact, one of one material, one of the other. If either one emits more radiation than it absorbs, then it spontaneously cools and the other one spontaneously heats up, violating the second law of thermodynamics. Very simple argument. May require sealing the two in some idealized setup of mirrors or so, to make the proof work, but aside from that it appears to have immediate physical relevance.

A blackbody must be in thermal equilibrium. Therefore if it absorbs everything incident upon it, it must emit just as much radiation. In principle this should be true at all frequencies. If it is not, then the object is not a blackbody.

At a microscopic level this boils down to the principle of detailed balance, that says that for systems in equilibrium, every process must be in balance with its reverse process.

When it comes to the absorption and emission of light, well these are each other's respective reverse processes. If we put an object into equilibrium (enclosed in a heat bath say) then it must be true that the absorption and emission processes are in balance (as in a blackbody). However, the relationship we could derive in this situation between the Einstein coefficients that govern emission and absorption at a microscopic level are properties of the material itself, and are not influenced by the fact that it is in equilibrium. Therefore it is generally true that an object that is a good absorber is also a good emitter, regardless of whether it is in equilibrium or not.

Reflection is an entirely different matter. If an object reflects light, then that light is obviously not absorbed. By definition a reflective object cannot be a blackbody and is both a poor absorber and emitter.

According to Wikipedia:

Blackbody:
A black body or blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. The name "black body" is given because it absorbs radiation in all frequencies, not because it only absorbs: a black body can emit black-body radiation.

So, you can see that a black body does emit radiation. An ideal blackbody must not only absorb radiations of all frequency but it must also not reflect and transmit them. An ideal blackbody must have two important properties:

• The body must be an ideal emitter i.e., at every frequency it must emit radiation as much as any other body at that temperature.
• It must also emit energy isotropically i.e., uniformly in all directions.

To add to these points a blackbody first absorbs all the electromagnetic radiations it comes in contact with. And then it emits thermal radiation. To be more clear, a perfect blackbody is one which is a perfect absorber and a perfect emitter of all kind of radiation. So, you can consider our sun(as well as other stars) as an almost perfect black body.
Wait a minute then does sun absorb light?
Believe me or not, it does!
And, that's one of the craziest things that I've heard so far. You can find more details about that over here in Astroquizzical