Bohr - Van Leeuwen theorem and the velocity distribution

In summary, the theorem in statistical physics states that the thermodynamical quantities of a system of charged particles in a cubic container are not influenced by a nonuniform external magnetic field. This is because when calculating the partition function integral for the system, the magnetic field can be substituted out and the resulting distribution function does not depend on the field. However, the velocity distribution of the particles may be affected by the magnetic field, resulting in a non-Maxwellian distribution. This can be observed through changes in particle energy levels and trajectories, as well as by quantifying the deviation of the distribution from the Maxwell-Boltzmann's distribution.
  • #1
Jano L.
Gold Member
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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
[tex]
a \times a \times a
[/tex]
in the external static magnetic field.
The argument is that when calculating the partition function integral for the system of particles in the box
[tex]
Z =
\int{
e^{-\beta \left(
\sum_a \frac{(\boldsymbol{p}_a -\boldsymbol{A}_a)^2}{2m})
\right) } d\boldsymbol{r} d\boldsymbol{p}},
[/tex]

we can substitute all canonical momenta [tex]\boldsymbol{p}_a[/tex] back to kinetic momenta
[tex]\boldsymbol{\pi}_a = \boldsymbol{p}_a - q\boldsymbol{A}_a[/tex] 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 [tex]\boldsymbol{A}[/tex] 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?
 
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  • #2
If yes, how can this difference be observed?Yes, it is possible that the resulting velocity distribution is different than for the case without the magnetic field. The difference can be observed in various ways, such as the change in particle energy levels or in the average energy of particles. In an external magnetic field, the Lorentz force acts on the charged particles and alters their trajectories, resulting in a non-Maxwellian velocity distribution. This can be seen experimentally by measuring the temperature of the system or by observing the effect of the field on the motion of the particles. Additionally, the deviation of the velocity distribution from the Maxwell-Boltzmann's distribution can be quantified by calculating the deviation parameter, which is defined as the ratio of the number of particles in the highest energy level to the number of particles in the lowest energy level.
 

1. What is the Bohr-Van Leeuwen theorem?

The Bohr-Van Leeuwen theorem states that in classical physics, the entropy of a system of charged particles in thermal equilibrium cannot depend on the magnetic field. This means that the magnetic field does not affect the average motion of the particles and therefore does not contribute to the entropy of the system.

2. How does the Bohr-Van Leeuwen theorem relate to the velocity distribution?

The Bohr-Van Leeuwen theorem implies that the velocity distribution of charged particles in a magnetic field is the same as that of non-interacting particles in the absence of a magnetic field. This is because the magnetic field does not affect the average motion of the particles, so the velocity distribution remains unchanged.

3. Can the Bohr-Van Leeuwen theorem be applied to quantum systems?

No, the Bohr-Van Leeuwen theorem only applies to classical systems. In quantum mechanics, the particles have discrete energy levels and the magnetic field can affect the motion of the particles, so the theorem does not hold.

4. How does the Bohr-Van Leeuwen theorem relate to the third law of thermodynamics?

The third law of thermodynamics states that the entropy of a perfect crystal at absolute zero temperature is zero. The Bohr-Van Leeuwen theorem is consistent with this law, as at absolute zero temperature, the particles have no thermal motion and therefore the magnetic field has no effect on the entropy.

5. What are the implications of the Bohr-Van Leeuwen theorem for the study of magnetism?

The Bohr-Van Leeuwen theorem shows that the classical theories of magnetism, such as Langevin's theory, are not consistent with the laws of thermodynamics. This led to the development of quantum theories of magnetism, which take into account the discrete energy levels and the effects of the magnetic field on the motion of charged particles.

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