# How to combine thermal velocity concept and special relativity?

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?

• 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. Commented Jun 6, 2018 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. Commented Jun 6, 2018 at 13:25

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:

$$\left< x_m \frac{\partial E}{\partial x_n} \right> = \delta_{mn} k_BT,$$ 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

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

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.

• Also just for the sake of record could you keep the k's in Boltzmann constant lowercase. Commented Jun 6, 2018 at 13:56
• Valid for free particle only. Commented Jun 7, 2019 at 11:21

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:

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

• This distribution is what I would point out as well, but shouldn't you be finding $\mathbb E[\beta]$? Commented Jun 6, 2018 at 13:11
• @KyleKanos - Finding $E[\gamma]$ looked easier. Commented Jun 6, 2018 at 22:37
• Possibly, but you then need to somehow invert that to get $\beta$ since you're looking for $v_\text{therm}$, no? Commented Jun 6, 2018 at 22:46
• ...where $E$ means expectation value, not energy. Commented Jun 7, 2018 at 8:03

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.