The conventional books on thermodynamics that I know, do not talk about the thermodynamics of relativistic systems. But in Cosmology, the thermodynamic concepts are often applied to the whole universe.

My question is about the concept of thermodynamic equilibrium for a relativistic system consisting of electrons, positrons and photons. In this system, there are processes of the type $$e^++e^-\leftrightarrow \gamma+\gamma.$$ However, for chemical equilibrium, the forward reaction must proceed at the same rate as the backward reaction. Now, if the temperature (of the environment) falls below a certain value, photons will not have enough energy required for pair production. In particular, if the temperature is such that the average photon energy falls below the rest energy $2m_ec^2$ the backward reaction will stop.

  1. Does it mean that the system fails to be in equilibrium below that critical temperature?

  2. Since the forward reaction continues to occur, shouldn't the system ultimately equilibrate with only photons (that obey the blackbody distribution?).


1) Thermodynamics for relativistic systems is pretty straightforward, and discussed in many text books (Landau, Greiner, $\ldots$). The only difference is that in relativistic systems the total number of particles is not conserved (and the associated chemical potential is zero), only the total charge (electric, baryon, $\ldots$) is conserved.

2) The fact that certain equilibration reactions become very slow as $T\to 0$ has nothing to do with relativity, it happens in non-relativistic systems as well. This just means that equilibration takes a long time.

3) In the particular case of electrons and photons there is no issue, because equilibration can take place via Bremsstrahlung $e^-+e^-\to e^-+e^-+\gamma$.

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  • $\begingroup$ About point 3: by Bremsstrahlung or electron-photon Compton scattering? @Thomas $\endgroup$ – SRS Mar 23 '18 at 20:46
  • $\begingroup$ Both ............... $\endgroup$ – Thomas Mar 24 '18 at 2:40

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