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We know that in classical thermodynamics $$v_{rms} = \sqrt{\frac{3k_B T}{m}}$$ However we immediately see that this is wrong for high temperatures as there is no upper bound on velocity. How do I get the exact equation?

My approach-

We have, $E = \sqrt{m_o^2c^4 + p^2c^2}$

Now from thermal energy we have total energy to be(sum of rest energy and thermal energy) $E = m_o c^2 + \frac{3}{2}k_B T$

Thus, $$m_oc^2 + \frac{3}{2}k_B T = \sqrt{m_o^2c^4 + p^2c^2}$$

Here, $p = mv$ & $m = \frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}} $

Then we can solve for $v$ $ ( \sim v_{rms})$

I am not sure if this is right. Can someone correct me? Can you give me atleast the final result if not the entire drivation?

Indigo1729
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    Unfortunately the classical result that $KE = \frac{3}{2}k_BT$ depends on the equipartition theorem and the fact that classical kinetic energy is quadratic in velocity, so it will not generalise to a relativistic setting. Good try though. – By Symmetry Jun 06 '18 at 10:57
  • Additional trouble: At temperatures where $T \approx mc^2$ pair production will become relevant, while this might not change the velocity distribution, it makes predictions for other quantities based on the velocity distribution and original particle number wrong. – Sebastian Riese Jun 06 '18 at 13:25

3 Answers3

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The assumption that the thermal energy is $\frac{3}{2}k_bT$ is actually only valid at non-relativistic temperatures. In general we have to use the equipartition theorem to find the relation between temperature and energy:

\begin{equation} \left< x_m \frac{\partial E}{\partial x_n} \right> = \delta_{mn} k_BT, \end{equation} where $x$ can be a coordinate or conjugate momentum.

Just taking the one-dimensional case for simplicity, in the Newtonian regime, $E = \frac{mv^2}{2}$, so that $v=\sqrt{\frac{k_BT}{m}}$. But in the relativistic case, $E = \sqrt{p^2c^2 + m_0^2c^4}$. This means that

\begin{equation} \frac{c^2p^2}{\sqrt{p^2c^2 + m_0^2c^4}} = k_BT, \end{equation} so \begin{equation} p^2 = \frac{k_B^2T^2c^2 \pm \sqrt{k_B^4T^4c^4 + 4k_B^2T^2m_0^2c^8}}{2c^4}. \end{equation}

As $T \rightarrow \infty$ we get $p = k_BT/c$, or $E=k_BT$ (as the mass-term in the energy becomes negligible compared to the momentum). This is the well-known energy-equation for the ultra-relativistic gas. As $T \rightarrow 0$ we get $v = \sqrt{\frac{k_BT}{m_0}}$, the Newtonian result.

Chris Cundy
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What you are looking for is the mean of the Maxwell–Jüttner distribution, $$f(\gamma)=\frac{\gamma^2\beta}{\theta K_2(1/\theta)}e^{-\gamma/\theta}$$ where $\beta=v/c=\sqrt{1-1/\gamma^2}$, $\theta=k_BT/mc^2$ and $K_2$ is the Bessel function of the second kind.

So the expected $\gamma$ would be $$E[\gamma]=\frac{1}{\theta K_2(1/\theta)}\int_1^\infty \gamma^3 \left (\sqrt{1-1/\gamma^2} \right )e^{-\gamma/\theta}d\gamma$$ Unfortunately there doesn't seem to be any closed form formula for it. Here is a plot of my numeric results in Matlab: Mean gamma and beta in MJ distribution

Here is a derivation of anisotropic Maxwell–Jüttner distributions, which may or may not be useful.

Anders Sandberg
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If I am correct, the Maxwell–Jüttner distribution has

$$E[\gamma]=\frac{1}{\theta K_2(1/\theta)}\int_1^\infty \gamma^3 \left (\sqrt{1-1/\gamma^2} \right )e^{-\gamma/\theta}d\gamma=\frac{K_1\left(\frac{1}{\theta}\right)}{K_2\left(\frac{1}{\theta}\right)}+3\theta,$$

where $K_n$ is a modified Bessel function of the second kind.

atr
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Jean-Christophe Pain
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