The Maxwell-Boltzmann velocity distribution in 3D space is
$$
f(v)dv = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 \exp\left(-\frac{m v^2}{2k_B T}\right)dv$$
It gives the probability for a single particle to have a speed in the intervall $[v,v+dv]$. But this probability is not zero for speeds $v > c $ in conflict with special relativity.
Is there a generalization of the Maxwell-Boltzmann velocity distribution which is valid also in the relativistic regime so that $f(v) = 0$ for $v>c$ ? And how can it be derived? Or can a single particle distribution simply not exist for relativistic speeds, because for high energies, we always have pair-production meaning the particle number is not conserved and a single particle distribution can not be defined in a consistent way?
Best Answer
The point of a Boltzmann distribution is that it maximizes entropy given a fixed energy. The concept applies to systems with other degrees of freedom besides translational kinetic energy. The general distribution, from Wikipedia is
Thus, the simple adjustment to the Maxwell-Boltzmann distribution you cited is to replace the Newtonian kinetic energy $\frac{mv^2}{2}$ with the relativistic kinetic energy $(\gamma - 1)mc^2$ everywhere it appears in the distribution.
Pair creation is a separate issue that I'll leave to someone else.