# Bohr - Van Leeuwen theorem and the velocity distribution

1. May 10, 2010

### Jano L.

There is an interesting theorem in statistical physics which states that the thermodynamical quantities of the system of charged particles are not influenced by the magnetic field (even if it is nonuniform). Imagine we have many particles in a cubic container with dimensions
$$a \times a \times a$$
in the external static magnetic field.
The argument is that when calculating the partition function integral for the system of particles in the box
$$Z = \int{ e^{-\beta \left( \sum_a \frac{(\boldsymbol{p}_a -\boldsymbol{A}_a)^2}{2m}) \right) } d\boldsymbol{r} d\boldsymbol{p}},$$

we can substitute all canonical momenta $$\boldsymbol{p}_a$$ back to kinetic momenta
$$\boldsymbol{\pi}_a = \boldsymbol{p}_a - q\boldsymbol{A}_a$$ and integrate the kinetic factor out (boundaries for canonical and kinetic momenta being the same, -+infinity).
What rests is the distribution function in coordinates only, which does not depend on the magnetic field. Finally integrating over coordinates, we obtain a field independent partition function. And as all thermodynamic quantities are derived from the partition function, they are all insensitive to the field.

However, what happens to the velocity distribution of the particles? To calculate it, we take the partition function of just one particle. To obtain the distribution function for momentum (or velocity), we would like to integrate over the coordinates first. But that is a problem: the system is closed in a container and coordinates are bounded, which makes the integral with position dependent $$\boldsymbol{A}$$ very hard, even in homogeneous field. Despite we do not know how to express it in a simpler form, it seems that even after the integration is done, the magnetic field does not disappear from the expression so easily.
Is it possible that the resulting velocity distribution is different than for the case without the magnetic field - the Maxwell-Boltzmann's distribution?