Prove Van Leeuwen's Theorem: Diamagnetism Does Not Exist in Classical Physics

  • Thread starter Thread starter Henk
  • Start date Start date
  • Tags Tags
    Theorem
Henk
Messages
22
Reaction score
0
For a statistical mechanics course we have to prove Van Leeuwen's theorem: Diamagnetism does not exist in classical physics.

I know that in an external magnetic field H the Hamiltonian Ha goes from Ha(p1,p2,--------,pN ,q1,q2,-------qN) to Ha(p1-(e/c)A1, p2-(e/c)A2, ------- pN-(e/c)AN, q1,q2,------.qN)

I also know that the induced magnetization M = kT*d(log QN)/dH

So the problem is finding QN. I know how to calculate it for a perfect gas without the magnetic field but I can't seem to solve the integral when Ha changes.
 
Physics news on Phys.org
Try shifting integration variables.
 
Ah I found bij changing p - (e/c)A to p' and then integrating. Thanks a lot.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top