Is there a thermodynamic heuristic argument on why a redshifted blackbody spectrum is a blackbody at a new temperature?

Without calculating it, it isn't obvious to me that if you take Planck's Law for the spectral radiance as a function of temperature of a black body and shift all the frequencies by the same factor, you will get a curve that is also a blackbody curve, but at lower temperature.

But because the blackbody curve is entropy maximizing, it seems there might be some sort of thermodynamic argument for this, e.g. a thought experiment which shows that if the redshifted spectrum were not a blackbody spectrum, it would be possible to build a perpetual motion machine with some sort of blackbody oscillating on a spring that sees different $z$-shifts at different times.

Is there such an argument, or something roughly similar?

• I ran into a similar difficulty here: physics.stackexchange.com/questions/343034/… Jul 8 '17 at 4:45
• the wiki has the calculation. en.wikipedia.org/wiki/… , and the arguments . Jul 8 '17 at 5:37
• @pentane See page 4, just below equation 3 astro.umd.edu/~miller/teaching/astr422/lecture04.pdf Sep 1 '17 at 6:25
• @pentane actually $B\left(\frac{\lambda}n,nT\right)=n^5B(\lambda,T).$ Sep 1 '17 at 7:04
• @pentane trial&error with the help of Wolfram Mathematica's Simplify function applied to the ratio of $B$s. Sep 11 '17 at 5:18

This is a neat fact! I think the first time it's usually encountered is in a cosmology course, where the expansion of the universe keeps the CMB temperature well-defined. Here's an argument why.

Consider adiabatic expansion of a photon gas at temperature $T$ from the standpoint of kinetic theory. In this point of view, each photon is a particle rapidly bouncing back and forth. Unlike for a classical ideal gas, every photon has exactly the same speed, so every photon must lose the same fraction of its energy. (This is because each one collides with the walls an equal number of times, picking up the same relativistic redshift factor every time.)

Thus adiabatic expansion causes the frequency shift you're talking about. Now we just have to show that adiabatic expansion also keeps the temperature well-defined. But this is immediate by thermodynamics: we can already run a Carnot cycle with a photon gas. If, after each adiabat, the gas were not in thermal equilibrium, we could do additional work by using this temperature difference, contradicting the Second Law.

• Nice argument! . Sep 4 '17 at 12:02

What does it mean that blackbody curve is entropy maximizing? It means that a black 1000W radiator radiates cooler, more entropic radiation, than a 1000W pink radiator.

When a black object cools by radiating, the radiation has some entropy. If we paint said body blue, does the entropy of the radiation change? No. If it changed, then it would be possible to build a perpetual motion machine of second kind, as we would be able to choose the entropy of the radiation produced when we cool objects.

If we paint a black object blue, it radiates energy at slower rate. And that is important. The energy packets (photons) contain more energy, but the energy packets are more spread out. And that's why entropy of radiation does not depend on the color of the radiating object.

If redshifted spectrum were not a blackbody spectrum, then we could change the color of an object by changing the speed of the object. And this way of changing the color of an object would not involve any change of cooling rate of the object. Thermodynamics requires that the cooling rate changes when the color changes, so redshifted spectrum of a black body must be a blackbody spectrum, so that there will be no color changes without cooling rate changes.