# Why is it important that black bodies absorb all wavelengths?

Physics Asked on May 9, 2021

In every definition of a blackbody I read, it always mentions how a black body is something that emits and absorbs at every wavelength. I understand why it is important that it emits at every wavelength, since otherwise we wouldn’t get the blackbody spectrum.

However, why is it important that it absorbs every wavelength? Will this affect the output? Can’t I supply it energy by some other means (like giving it a power supply?). In the common hole in the box, we are not providing every wavelength but rather just giving heat directly and getting a black body output.

It seems that the absorption part of the definition has little use – why do we keep saying it then?

One of the two (emission and absorption) necessarily follows from the other one by Kirchhoff's law, which, in simple terms, states that emissivity and absorptivity for any given body are the same.

Saying it must absorb every wavelength does not mean we have to put black body radiation in. It just says that whatever wavelength you shine at it, the body will absorb it. And this is really just the definition of something being black. Of course, energy can also be put in by some other means. Saying it is black and absorbs at every wavelength only means it can absorb at any wavelength, not that it will only absorb black body radiation or even that radiation is the only way to input energy.

Concerning the terminology, it is indeed not necessary to mention both, since one follows from the other. Possibly it's just to make it even clearer.

Answered by noah on May 9, 2021

If a blackbody reflected the light that was incident upon it at some wavelength range or allowed that light to pass through it, then the intrinsic spectrum of light emerging from the body, if it were in thermal equilibrium, would be missing the specific wavelengths that could not be absorbed.

The definition of a blackbody is that it absorbs all radiation incident upon it and that it is in thermal equilibrium at some unique temperature. Both parts of this definition are necessary.

In practical terms, a hole absorbs everything that is incident upon it so long as there is a negligible chance of any photon bouncing around whatever is behind the hole and coming out again and this could form the basis of an experimental blackbody.

Consider the thought experiment of putting two cavities, A and B, together, such that radiation can pass through a small hole from one to the other. We can allow them to equilibriate so that they attain the same temperature, in which case the flux of energy from A to B is the same as from B to A.

Now an ideal cavity absorbs everything that enters into it. But suppose that were not the case for B and that light over some range of wavelength bounces around and eventually emerges from the hole.

We can do the same experiment. A is now an ideal blackbody, but B is not. We put them together but this time insert a filter between them which only allows through the range of wavelengths that B will not absorb. We allow them to equilibriate at some temperature. The flux of radiation from A to B must equal that from B to A over the restricted wavelength range.

But all the flux going from B to A must be that which was emitted by A in the first place, since none of it is absorbed by B. That means if we pull the cavities apart and examine the spectra emerging from them, even though they are at the same temperature, the spectrum from B would have a big hole in it where it didn't absorb light. i.e. not a blackbody spectrum.

Note that a blackbody does not have to get its energy content by being illuminated. It just has to be capable of absorbing all radiation incident upon it.

Answered by ProfRob on May 9, 2021

With a black body, absorption is the defining property:

a black body is a body that absorbs all radiation incident upon it

One can then prove, using thermodynamic arguments (associated with Kirchoff) that if such a body is in thermal equilibrium then the spectrum emitted is the same as that of any such body at the same temperature. What is notable about this argument is that it does not involve the shape or quality of the body, chemical composition, and things like that.

Answered by Andrew Steane on May 9, 2021