# Relativistic maxwell-boltzmann-distribution

1. Sep 22, 2012

### magicfountain

In thermodynamics (ignoring relativistic effects) you can use the maxwell-boltzmann-distribution to find the average speed of the gas particles.
$v^2=\frac{8kT}{\pi m}$

But there are high Temperatures that would have average speeds > c.
Are there distributions that describe gases with an average speed of 0.5 relativistically?

2. Sep 22, 2012

### Staff: Mentor

Well, you have to replace the nonrelativistic formula with the relativistic one.

The average squared velocity will go towards c^2, and all particles are always slower than c for every temperature.

3. Sep 22, 2012

### bcrowell

Staff Emeritus
...which to me is not very illuminating, since it's given without any derivation and written in terms of a goofy choice of variable.

Can't you just take the partition function and put in the relativistic expression for the energy? I.e.:

$$e^{-\beta\sqrt{m^2+p^2}}$$

This is in units with c=1, and beta is the inverse temperature.

4. Sep 22, 2012

### Bill_K

Yes, I think so. And I see the Maxwell–Juttner expression does consist of this Boltzmann factor, plus a function of T in front. To get the factor in front you have to normalize the distribution to N particles per volume, which involves integrating over the Boltzmann factor. Nonrelativistically the integral leads to (m/2πkT)3/2. But here we have to integrate over the relativistic Boltzmann factor, and that's where the K2(T) Bessel function comes from.

5. Sep 23, 2012

### magicfountain

@bcrowell
@Bill_K
that helped a lot. i guessed that i had to do lagrange multipliers with relativistic expressions, but i was too lazy to really think about it. thanks for reminding me that it actually just leads to the partition function and you have to plug in the rel. terms there.