0
$\begingroup$

What is the relationship between the frequencies of radiation absorbed by a material and those emitted? I know that Planck's law tells us the spectral radiance, but how does this map to those frequencies absorbed? For example, a black material at room temperature is absorbing all visible light and emitting infrared; what is the relationship between these two?

As an aside, what is the origin of the term "black body", since many black bodies aren't black!? I imagine that early realizations of black bodies indeed had coatings of black material, and this is where the term came from, an I correct?

$\endgroup$

2 Answers 2

0
$\begingroup$

By time reversibility, the tendencies to absorb and emit radiation of a given frequency (known as absorptivity and emissivity) are exactly equal. Thus, for example, a perfect mirror will reflect all radiation incident upon it, and it will never radiate away any of its internal energy in the form of photons. Einstein used the fact that a single atom in a thermal bath of radiation had to absorb and radiate in a time-reversal-invariant way, to derive his $A$ and $B$ coefficients, which give the rates of spontaneous and stimulated emission by the atom.

A "black body" is thus black in the sense that it is a perfect absorber; all radiation incident on its surface will be completely absorbed. Then the time reversal argument means that the surface will also produce the maximal amount of thermal radiation at all frequencies.

$\endgroup$
0
$\begingroup$

A black body is defined as not reflecting any, therefore absorbing all radiation falling on it. At finite temp erature such a body emits perfect black body radiation. Needless to say that thus is an idealisation. Real bodies absorb certain frequencies and reflect others. This means that the ideal emission spectrum is multiplied with a frequency dependent factor, the emissivity. It is equal to the absorptivity. So indeed a real body can only approach black body behaviour in a limited frequency range.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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