Why does blackbody radiation formula work for radiator?

Question

I'm just trying to understand this from a purely thermodynamical standpoint. I'm comfortable with the scenario of a blackbody cavity's interior being in thermal equilibrium with the photon gas contained, and they have identical temperature, and then the Planck distribution would apply to photon energy. However, why would such an argument extend to a radiator in space where the photon gas is obviously not in equilibrium? The photons shoot out from the surface, never crossing path with it ever again, and the photon gas' numerical density is inversely proportional to distance squared.

Answer 1

If I have well understood your question, it is equivalent to asking why radiation from a small hole in the cavity emits radiation distributed according to the internal blackbody distribution or why the distribution of velocities of particles coming out of a small hole in a box containing gas at equilibrium turns to be Maxwellian.

In this case, the answer is simply that if the dominant flux arriving at a receiver is directly coming from the hole, one is just measuring the internal distribution of energies.

This is the trivial part of the answer. A little less trivial is the analysis of apparently different situations like, for example, the case of the radiation coming from a star, where there is no small hole and the entire surface behaves like a radiator of equilibrium radiation. However, what really matters to be able to measure the equilibrium distribution is only the request that the overwhelming part of the radiation arriving at the receiver originates from the region where thermodynamic equilibrium is dominant.

This is true in the case of the stellar surface because of the opacity of the interior part of the star. The radiation we can observe is dominated by that coming from the base of the stellar photosphere, which can be assumed at a local thermodynamic equilibrium. Even if there are additional emitting regions at different physical conditions on the line of sight toward the observer, it remains true that the measured radiation flux is dominated by the part leaving the local equilibrium region at the base of the photosphere.

Of course, the example of the star can be extended to any other radiator at equivalent conditions.

Answer 2

I think the key here is that the sort of characteristic time over which the photon gas explores state space is very short compared to the characteristic time over which the photons escape. That is, the half-life of the photon gas is way, way longer than the timescale of fluctuations in the photon gas. As such, it's a good approximation to treat the photon gas as being in equilibrium.

Similarly, one can speak of a gas in a piston undergoing compression as being at a temperature. Strictly speaking, temperature is an equilibrium property and the gas is changing state, so it's not at equilibrium. But it's so close to equilibrium that this is a very good approximation and so you can use $PV = nRT$ to calculate the pressure the piston generates.

Answer 3

The outer surface of the radiator is doing the same thing as the inner surface of a cavity, namely absorbing and radiating at a temperature and being in accord with Planck's distribution.

The difference is really in the handling of reflections. The cavity is designed to absorb all reflections, but obviously in the radiator's case, reflections might be lost, so that the spectrum from the radiator would have to take this into account.

However, as long as the surroundings is not sending a lot of light onto the radiator to reflect, the main distribution from the radiator is still Planckian, and so that is fine. As long as you agree that the cavity emits Planck's distribution, then you should also agree that the radiator is mostly emitting some Planckian distribution, except affected by the fact that the coefficient of emissivity is not unity.