Understanding blackbody radiation

Question

I am studying blackbody radiation and modelling a cavity as a blackbody. However, I am encountering a number of confusions in this description:

1. Many textbooks mention that the cavity consists of metallic walls that act as perfect reflectors. Light entering through the hole, after multiple reflections, get absorbed. However, if the cavity walls are perfectly reflecting, how is the radiation getting absorbed by the walls?
2. For studying the radiation emitted by the blackbody, the textbooks model the radiation inside (emitted by walls due to heating) as standing waves. But how are standing waves being emitted by the cavity? Because standing waves are supposed to be confined inside.

Can anyone please explain in simple terms?

Answer 1

Perfectly reflecting cavity is not a blackbody. However, if inside the cavity there is some piece of absorbing matter, then a small hole in the cavity will behave as black body. The piece of matter can be arbitrarily small, and some sources neglect to mention it.

Perfectly reflecting cavity is a superfluous concept in derivations of the Planck spectral function. All that is needed is the Fourier decomposition of EM field, which can be done even for imaginary cuboid in space, no special walls are needed.

The reflecting walls are carried on in expositions on this topic probably because that is one sure way, theoretically, to keep radiation inside the cavity long enough so that it can come to thermodynamic equilibrium with matter inside. But it is not needed for the derivation itself; what is needed is the assumption that equilibrium statistical physics applies.

In practice, blackbody radiation was prepared and studied in cavities that are metallic, but are covered with black oxides on its internal faces. The black oxides are there to absorb great part of incoming radiation and thus make the radiation interact with matter and facilitate establishing of thermodynamic equilibrium between radiation and matter. The usual derivation of the Planck function does not take into account this absorbing layer, because it is not needed for the derivation.

Textbooks sometimes mention standing waves, but don't imagine radiation inside to actually be a standing wave; it is random fluctuating radiation, which can be also thought of as superposition of travelling waves. "Standing waves" is here just to describe the Fourier decomposition of the EM field inside in case the walls are perfectly reflecting; then electric field vanishes on the walls, and one can re-express the Fourier sum over travelling waves as sum over standing waves.

Answer 2

Here is a more correct and hopefully easier way to visualize the process.

What is wrong here is the basic idea that the walls of the cavity are perfect reflectors of EM radiation. They are not. They are instead perfect radiators of EM radiation, and absorbers of EM radiation. If the walls are a little hotter than the radiation inside the cavity, they will radiate into the cavity. If the walls are a little cooler than the radiation inside the cavity, they will absorb radiation from the cavity. A dynamic equilibrium is then achieved, where the rates of radiation and absorption balance out and the walls achieve the same temperature as that of the radiation.

Now, the idea that there will be "standing waves" building up inside the cavity is a bit of a kludge because there is no obligation for the cavity to be resonant in order for the radiation inside to come into equilibrium with the walls of the cavity. Furthermore, any atom in the wall surface is free to emit a photon of EM radiation in any random radial direction it wants, which means that photon will bounce off an adjacent portion of the wall at any possible distance within the cavity that geometry dictates, and not at one single distance (analogous to the length of a resonant organ pipe, for example).

Now since the radiating wall atoms are not rattling around at one single frequency, the wavelengths of the EM waves they radiate will vary from one atom to the next and from one vibration to the next which means there will be a spectral spread of photon frequencies/wavelengths occupying the cavity, where the cavity dimensions have no effect on that spectrum. If the cavity dimensions did affect that spectrum, then every cavity of different size would equilibrate at a different temperature which we know isn't the case for blackbody radiation.

Finally, note that the hole in the cavity just furnishes a convenient way to peek inside the cavity and deduce what its interior temperature is, by measuring the spectrum of radiation leaking out of the hole.