Physics Asked on March 14, 2021
We have the Bohr-van Leeuwen theorem which tells that magnetism cannot be explained classically. The proof is simple; it turns out that classically the partition function is independent of the magnetic vector potential and therefore, the free energy is independent of the magnetic field which proves that there cannot be magnetization.
But we have a classical Langevin theory of paramagnetism where the partition function is not independent of $B$, and thus we get a magnetization. So how can we say that magnetism cannot be explained classically? Doesn’t the Bohr van Leeuwen theorem fail here?
One way to answer your question is that, citing J. H. Van Vleck,
when Langevin assumed that the magnetic moment of the atom or molecule had a fixed value $mu$, he was quantizing the system without realizing it.
If you do not assume the existence of a permanent magnetic moment, but try to derive it from the motion of electrons inside the atoms, then this is doomed to fail (classically), precisely because of the Bohr-van Leeuwen theorem. In other words, Langevin theory is not a classical theory, but a kind of semi-classical theory.
This is discussed, for instance, in this recent review paper.
Correct answer by Yvan Velenik on March 14, 2021
The Bohr-von-Leeuwen theorem applies to the absence of classical diamagnetism i.e to the magnetic effects on the classical theory of electron motion. On the other hand paramagnetism arsises from the field-induced alignment of the electrons instrinsic magnetic moment, an effect ignored in Bohr-von-Leeuwen. The electron's intrinsic magnetic moment is a quantum effect accounted for by the Dirac equation.
Answered by mike stone on March 14, 2021
Yes, the van Leeuwen-Bohr theorem isn't omnipotent. It assumes a specific scenario using classical physics, where charged particles without a magnetic moment in a finite region of space have Boltzmannian distribution of velocities, i.e. for every direction the opposite one is equally probable. Even on the boundary of the region, which is necessary for the theorem to hold.
If the particles themselves carry magnetic moment or if they move in a preferred direction, theorem's assumptions are not satisfied and the result does not hold.
Even if the particles do not have magnetic moment, they can still create magnetic moment by their motion. Consider weakly interacting charged particles moving in circles in external magnetic field. The motion is such that the set of particles behaves as diamagnetic body. This is possible because the particles' velocities aren't Boltzmannian; on the boundary of the region where the particles are, there is a preferred direction of motion consistent with the circling motion in the magnetic field.
From EM theory we know we can attribute to magnetic bodies a surface magnetization current. One way (Ampere's theory) to explain this current microscopically is that charged particles in the layer of space containing the body surface have a preferred direction in which they move. So the model of magnetized body can be classical but not assume Boltzmannian distribution for all particles. Not every system has to be Boltzmannian.
Answered by Ján Lalinský on March 14, 2021
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP