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What is the correct formulation of the relation between the speed and temperature in the relativistic kinetic gas? I have found one mentioned in problem 3.24 of the third edition of Pathria's Statistical Mechanics, and seen another based upon the relativistic mass relation in one of the blogs in physics stack exchange. I would like to hear more views on the topic.

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The answer depends. Are the particles relativistic because they're massless? If that's the case, then speed and temperature are, of course, uncorrelated and the only relevant question is whether the particles are Bose-Einstein or Fermi-Dirac distributed. Assuming the temperature is high enough that Maxwell statistics are a good approximation and that the particles are massive then the partition function gives the following phase space density for a single particle: $$ \rho(\mathbf{x},\mathbf{p}) = \frac{1}{Z h^3} \operatorname{e}^{-\frac{c\sqrt{p^2 + m^2 c^2}}{kT}}, $$ where $Z$ is a normalizing factor known as the relativistic partition function. The formula for $Z$, is:$$Z = 4\pi V \left(\frac{mc}{h}\right)^3 \frac{kT}{mc^2}\, K_2 \left(\frac{mc^2}{kT}\right),$$ with $K_2$ a modified Bessel function of the second kind.

The question is somewhat ambiguous in that speed can mean any number of things. Do you mean the mean speed of the particles? The root-mean-square (rms) speed (the usual definition)? For these the mean speed the formula is:$$\begin{align}\langle v\rangle & = \left\langle \frac{p c}{\sqrt{p^2 + (mc)^2}}\right\rangle \\ & = \frac{c}{K_2\left(\frac{mc^2}{kT}\right)} \int_0^\infty \frac{p^3 }{m^2 c kT \sqrt{p^2 + (mc)^2}} \operatorname{e}^{-\frac{c\sqrt{p^2 + m^2 c^2}}{kT}} \operatorname{d}p \\ & = \frac{c}{K_2\left(\frac{mc^2}{kT}\right)} \int_{mc^2/kT}^\infty \left(\left[\frac{u kT}{m c^2}\right]^2 - 1\right) \operatorname{e}^{-u} \operatorname{d} u \\ & = 2\frac{c}{K_2\left(\frac{mc^2}{kT}\right)} \left(1 + \frac{kT}{mc^2}\right) \frac{kT}{m c^2} \operatorname{e}^{-mc^2 / kT}.\end{align}$$

For rms speed we get:$$\begin{align} \langle v^2\rangle & = \left\langle \frac{p^2 c^2}{p^2 + (mc)^2}\right\rangle \\ & = \frac{c^2}{K_2\left(\frac{mc^2}{kT}\right)} \int_0^\infty \frac{p^4 }{m^2 c kT [p^2 + (mc)^2]} \operatorname{e}^{-\frac{c\sqrt{p^2 + m^2 c^2}}{kT}} \operatorname{d}p \\ &\hphantom{=}\mathrm{let\ } p= mc\sinh\phi,\ \mathrm{and\ } x=\frac{m c^2}{kT} \\ &= \frac{c^2 x}{K_2\left(x\right)} \int_0^\infty \frac{\sinh^4 \phi}{\cosh \phi} \operatorname{e}^{-x\cosh\phi} \operatorname{d}\phi \\ & = \frac{c^2 x}{K_2\left(x\right)} \left[\operatorname{Ki}_{-2}(x) - 2\operatorname{Ki}_{-1}(x) + \operatorname{Ki}_1(x)\right],\end{align}$$ where $\operatorname{Ki}_\alpha(x)$ is the Bickley function.

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Temperature is related to the kinetic energy of harmonic oscillators associated with the degrees of freedom in the medium. In a gas, there are degrees of freedom for translation (i.e. velocity) of the particles, and also rotation if the particles are not monatomic. Thermal energy is partitioned equally between translation (i.e. velocity) in the x, y, and z axes, rotation around one or two axes, etc.

Any formula for temperature as a function of speed is only using speed as a proxy for the kinetic energy which is balanced between all the oscillators. At relativistic speeds, you should use kinetic energy and momentum, not velocity.

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