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

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.

• Idk if this would be the correct way to think about it, but Instead of considering the hole as the black body, taking the entire inner surface of the cavity as the black body seems to resolve the issue conceptually.
– Y G
May 27, 2022 at 12:06

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?