# Why does a blackbody cavity with a hole emit black body radiation?

Why could we use nearly every material in a cavity with a hole to absorb/ emit every wavelength?

If the material doesn't have the exact energy gap, g, for example, it couldn't absorb a photon with frequency $$v_1 = \frac{g}{h}$$. How does reflecting the photon back and forth in the cavity help with absorbing it?

For the same reason, how could there be near a perfect black body radiation at any temperature at the hole? The only material which would have perfect black body radiation would be one whose energy gaps would be continuously (infinitely dense), so it's capable to emitting photons of every frequency (including frequencies which are typically only possible due to bremsstrahlung)...

So how could our cavity emit that photon with frequency $$v_1$$ (however it managed to absorb it in the first place??)?

• "why we could use nearly every material in a cavity" As noted below, this isn't exactly true. It may help to cite your source for this factoid. Chances are it's just plain wrong, or you've inadvertently taken it out of context (i.e., the smaller the hole is in your blackbody sphere, the less that it matters if the material inside doesn't have emissivity exactly equal to 1). Oct 15, 2023 at 17:16

I don't understand, why we could use nearly every material in a cavity with a hole to absorb/ emit every wavelength.

In practice, if you wanted to make a cavity that produces radiation close to blackbody radiation, all materials are not equal. The original measurements of equilibrium radiation (Lummer and Kurlbaum, and Lummer and Pringsheim) used heated metallic cavities with complicated inner surface covered with well absorbing "black" materials (metal oxides) to produce and contain the radiation.

The fact one could use any material is true in theory; if thermodynamic equilibrium is somehow present in the cavity, then details of the material inside don't matter, because spectrum of equilibrium radiation depends only on temperature, not on absorption/emission properties of the material.

However, one can ask whether a state close to thermodynamic equilibrium will establish itself spontaneously in cavity with any material. This requires some things: 1) that the cavity is well insulated from leaking radiation, either due to it being well reflecting, or due to it being made of thick, badly conducting material, 2) that the material inside interacts at least a little with radiation of any frequency.

In practice, no cavity is perfectly insulating and all materials are transparent to radiation of high-enough frequency, so perfect equilibrium at all frequencies can't be established.

If the material don't have the exact energy gap g for example, it couldn't absorb a photon

Material needs not have "exact energy gap". When radiation of frequency $$f$$ interacts with quantum system that has no available transitions of energy $$hf$$, the system will still be able to change states and get some energy from the radiation. The idea of gaps and transitions is a simplified picture, an approximation, giving the major part of material response. But interaction and energy exchange is not limited to cases where exact matches of $$hf$$ and energy levels occur.

If we had a perfectly insulating cavity, and the matter inside the cavity (or the material that the cavity is made of) had a very low absorptivity at some frequency (low ratio of absorbed and incoming intensity of radiation), establishing equilibrium radiation at that frequency will take a long time. The lower the absorptivity, the longer it takes. In the limit where the matter does not absorb at all at some frequency, equilibrium will never establish itself at that frequency.

Real bodies have non-zero absorptivity at all frequencies higher than zero; as far as I know, there are no perfectly transparent or perfectly reflecting bodies. So after a long enough time, in a well insulating cavity, radiation could get close to the equilibrium state.

If the material inside the cavity is completely incapable of absorbing at some frequency, then it cannot emit radiation at that frequency, and the radiation emerging from the hole would not be blackbody radiation.

The idea behind having a big cavity with a small hole is that it increases the probability of absorption by the number of times a photon bounces around inside the cavity before it might escape. Then, even if there is the tiniest probability of absorption (for example because of natural or thermal broadening), it can be greatly multiplied if the hole occupies a very small solid angle in the cavity. But, if the absorption cross-section for a single interaction with the walls were exactly zero, then it doesn't matter what you multiply it by.

The answer is a bit tautological. Except for the materials that are incapable of absorbing some range of frequencies, all the others are suitable to build a black body.

Notable exceptions are for example insulating materials like quartz that doesn't absorb photons in the range of the visible light. And all the others whose band structure allows some range of EM frequencies simply pass through, without being absorbed.