Why do all black bodies at the same temperature exhibit the same emission spectrum?

Question

I’m reading Eisberg & Resnick Quantum Physics, and there’s a part in the introduction that’s confusing to me. They say, “independent of the details of their composition, it is found that all black bodies at the same temperature emit thermal radiation with the same spectrum. This general fact can be understood on the basis of classical arguments involving thermodynamic equilibrium.”

I understand that an idealized black body absorbs all incident radiation, and that to be in equilibrium with its surroundings it must also emit radiative energy equivalent to what it absorbs. What I don’t understand is why all black bodies at a given temperature must have exactly the same emission spectrum.

Why is temperature the only factor that can change the emission spectrum of a black body? Why are material composition and the spectrum of incident radiation irrelevant? More specifically, how can this be “understood on the basis of classical arguments involving thermodynamic equilibrium?“

In other words, how can you prove from classical arguments alone, not what the black body emission distribution function is, but that it must be the same for all black bodies at the same temperature?

Answer 1

Imagine you have a black body consisting of a highly absorbent cavitity with a small hole with which to let out some of the radiation within their interior. In this idealised situation, the only way heat can get in or out of the cavity is through the hole - it is otherwise isolated from its surroundings.

By definition, the blackbody (the hole in this case) absorbs all light that is incident upon it and it is in thermal equilibrium - i.e. the cavity temperature must not change with time.

Now let's introduce a second blackbody, constructed of some other material and perhaps of a different shape or size, and place it such that the the two apertures of the blackbodies are in contact, so that any radiation leaving BB1 will enter BB2 and vice versa. If you leave this system for sufficient time then they will achieve the same temperature, such that in equilibrium just as much power exits BB1 and enters BB2 as vice versa.

However, if you were to then insert a filter between the two apertures that only allowed a narrow range of wavelengths to pass through, then if the spectrum of radiation emitted by the two apertures were different, then one cavity would heat up and the other would cool down. But if that happened they would not be in thermal equilibrium and so they would not be blackbodies. Thus genuine "blackbody radiation" must have a universal spectrum that only depends on temperature, not on the material, shape or size of the blackbody.

Answer 2

Black body radiation was one of the first experimental observations that forced the theory quantization and finally the theory of quantum mechanics on the physics of the microcosm.

There are no classical arguments, just Planck's intuition in trying to fit observed spectra.

The amount of radiation emitted in a given frequency range should be proportional to the number of modes in that range. The best of classical physics suggested that all modes had an equal chance of being produced, and that the number of modes went up proportional to the square of the frequency.

But the predicted continual increase in radiated energy with frequency (dubbed the "ultraviolet catastrophe") did not happen. Nature knew better.

Read on in the link.

The functional form fitted the data (approximately) , it is an observational fact. The best fit to the bb curve is by the Cosmic Microwave Background radiation. The sun's fit is approximate. In general emissivity and absorptivity must be taken into account for real measurements.