Why can the hole of a cavity-hole model be considered a radiating blackbody?

Question

It is easy to understand why the small hole of the cavity-hole model simulates an absorbing blackbody as almost all incoming radiation into the hole is captured. Why can we consider the hole an emitting black body when the cavity is heated evenly to some temperature? what defines a blackbody when we consider emittance?

I understand that by Kirchoff's radiation law, given some temperature $T$, any object that absorbs like a blackbody should always a have maximum monochromatic radiation exitrance $M_{\lambda_0}(T)$, which is just an equivalent definition of a black body. But in our case, how do we make sense of $M(\lambda,T)$ for the hole when it doesn't really "exist" and have a temperature? Why are we allowed to assume the temperature of the cavity as the temperature of the hole? Theoretically, how do we know that the hole would have $M_{\lambda_0}(T)$, apart from assuming it is a "normal surface" and should therefore abide by Kirchoff's radiation law?

Note that the issue about the hole's temperature is evaded when we see it as an absorbing black body since the monochromatic absorptance of a black body is always 1, regardless of temperature.

Thanks.

Answer 1

Why can we consider the hole an emitting black body when the cavity is heated evenly to some temperature? what defines a blackbody when we consider emittance?

Because the radiation inside is assumed to have characteristics of radiation in thermodynamic equilibrium at that temperature. The hole then emits in all directions just as a surface patch of a solid black body would.

This isn't guaranteed just because the radiation is coming off a cavity. In general, reflecting cavity does not have radiation in thermodynamic equilibrium inside. In order to have such radiation inside, the cavity has to have matter inside which allows for establishment of thermodynamic equilibrium between radiation and matter. Metal oxides, lamp black, soot, etc.

But in our case, how do we make sense of $M(\lambda,T)$ for the hole when it doesn't really "exist" and have a temperature? Why are we allowed to assume the temperature of the cavity as the temperature of the hole?

Because the temperature is that of the radiation inside, and the emission function characterizes this radiation, not the hole itself.