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Does relativistic statistical mechanics average over particles that exceed the speed of light? This article seems to imply as such:

It is interesting to note that the classical and quantum statistics of particles do not explicitly limit the speed of particles. The probability of gas particles and free electrons in their statistical models allows large velocities with a probability approaching zero as the velocity goes to infinity.

Could one strengthen this statement as follows?

Even relativistic statistical mechanics (e.g., statistical mechanics which accounts for general relativity) technically allows particles to surpass the speed of light, albeit with vanishing probability.

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  • $\begingroup$ Assuming the gases have mass, I would argue velocities with a probability approaching zero as the velocity goes to the speed of light. $\endgroup$
    – Cinaed Simson
    Commented Dec 7, 2019 at 8:01
  • $\begingroup$ I feel like the minkowski metric and the Lorentz transform should come into play in the relativistic statistical mechanics case to bound the velocity so you would have probability approaching zero but also a limit in the velocity $\endgroup$
    – Dabed
    Commented Dec 8, 2019 at 20:44
  • $\begingroup$ Your question is extremely unclear. First off, I don't see how you got that implication from the quote, which is talking about nonrelativistic statistical mechanics. Second off, saying that a theory "allows something to happen, albeit with vanishing probability" is physically meaningless. $\endgroup$
    – knzhou
    Commented Dec 9, 2019 at 4:19
  • $\begingroup$ Finally, in actual relativistic statistical mechanics, you never have particles going faster than light, as a look at any standard textbook will confirm. To me it looks like essentially every sentence of your question is either meanginless or the exact opposite of what is true. $\endgroup$
    – knzhou
    Commented Dec 9, 2019 at 4:20
  • $\begingroup$ @knzhou: Second off, saying that a theory "allows something to happen, albeit with vanishing probability" is physically meaningless Not at all. This phrase has a perfectly clear physical meaning in the sense that solutions with “something property” have measure zero on the space of all solutions. While the question may have problems this sentence isn't one of them. $\endgroup$
    – A.V.S.
    Commented Dec 11, 2019 at 14:05

3 Answers 3

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No, you cannot write that particles are allowed to surpass the speed of light but with a vanishing probability. The fundamental reason is that statistical mechanics for classical particles is constructed in phase space $(\vec r,\vec p)$ Therefore, in special relativity, the partition function of an ideal gas reads $${\cal Z}=z^N, \quad z=\int e^{-\beta\sqrt{p^2c^2+m^2c^4}}{d^3\vec rd^3\vec p\over h_0^3}$$ The position $\vec r$ is integrated over the volume and the momentum $\vec p$ over $\mathbb{R}^3$. However, $$\vec p={m\vec v\over \sqrt{1-v^2/c^2}}$$ so the speed $v$ remains always smaller than the speed of light.

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The Maxwell-Jüttner distribution describes the distribution in a relativistic gas:

$$ f(\gamma) = \frac{\gamma^2\beta}{\theta K_2(1/\theta)}e^{-\gamma/\theta}$$

with

$$ \theta = \frac{kT}{mc^2}$$

The Boltzmann factor is defined in terms of energy, not velocity, and $E = \gamma m$, so the probability of any unbounded energy is not a problem.

Note that it is classical: it ignores quantum mechanics.

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I think that to obtain statistical mechanics, one should start from Wick rotation of QFT (for flat spaces it works, it is enough). Then, after Wick rotation, statistical mechanics reproduces all properties of initial QFT theory. I mean that dispersion law for massive bosons will be $\sqrt{p^2+m^2}$ which is relativistic. It is not an answer, just draft.

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