Invariance of the magnetic moment and Bohr-Van Leeuwen theorem In plasma physics, particularly in magnetic mirros a very well-known result is the invariance of the magnetic moment of the particles:
$\dfrac{d\mu}{dt}=0$.
I've been looking into some ways of proving this important result. As a naive glance, I thought about using Bohr–Van Leeuwen theorem: $\langle u\rangle = 0$, that is, the thermal average of the density of the magnetic moments is zero. But I'm struggling at how to relate this fact to the invariance of the magnetic moment, if that approach is possible.
If anyone can point some directions out to me, I'll be grateful.
 A: The Bohr-Van Leeuwen theorem I don't think is relevant here since it is a thermodynamic statement and really $\mu$ is conserved for individual particles .
The derivation I would do is specific to the case of converging magnetic fields, such as in a magnetic mirror.
Assume the particle is gyrating with velocity $v_L$ around some $B_\parallel$, additionally with some parallel velocity $v_\parallel$.
The $B$ does no work, so
$$\frac{1}{2}m\frac{d}{dt}(v_\perp^2 + v_\parallel^2)=0$$
Then we simplify $\frac{d}{dt}(v_\parallel^2)=2v_\parallel \frac{d}{dt}v_\parallel$.
The $B$ is divergence free, so roughly one must have a $B_\perp$ pointing inwards, $B_\perp=\frac{r}{2}\frac{d}{dz}B_\parallel$.
The Lorentz force averages to
$$F_\parallel=\frac{d}{dt}v_\parallel=-\frac{e v_\perp B_\perp}{m}$$
$$\frac{d}{dt}v_\parallel=-\frac{e v_\perp r}{2m}\frac{d}{dz}B_\parallel$$
Multiplying by $v_\parallel$ (the velocity along the z axis),
$$v_\parallel \frac{d}{dt}v_\parallel=-\frac{e v_\perp r}{2m}v_\parallel\frac{d}{dz}B_\parallel$$
Or
$$2v_\parallel \frac{d}{dt}v_\parallel=-\frac{e v_\perp r}{m}\frac{d}{dt}B_\parallel$$
We can insert this into our energy equation. Recall $r=v_\perp/\Omega$ where $\Omega$ is the gyration frequency $eB/m$. Simplifying gives
$$0=\frac{d}{dt}v_\perp^2 -\frac{ v^2_\perp }{ B_\parallel}\frac{d}{dt}B_\parallel,$$
$$0=B_\parallel\frac{d}{dt}\frac{v_\perp^2}{B_\parallel}. $$
Inserting constants gives
$$0=\frac{d}{dt}\frac{1/2 m v_\perp^2}{B_\parallel}=\frac{d}{dt}\mu.$$
