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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.

Adrien-Marie Deschamps
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    Related: https://physics.stackexchange.com/q/670465/59023 – honeste_vivere Oct 08 '21 at 13:43
  • I took a look there. I have a question, is it correct to say that the Van Allen's belts are only possible because of the adiabatic invariance of the magnetic moment? Or because of all the adiabatic invariants? – Adrien-Marie Deschamps Oct 09 '21 at 00:15
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    @AdrienMarieDeschamps - The radiation belts are quasi-stable because of these adiabatic invariants, yes. Would there still be energetic particles if these did not hold? Probably yes, but transiently so not so consistently present... – honeste_vivere Oct 09 '21 at 18:38
  • What does it mean for something to be quasi-stable? – Adrien-Marie Deschamps Oct 10 '21 at 21:17
  • @AdrienMarieDeschamps - It means they can exist for periods of time longer than the relevant gyration/bounce/drift periods discussed in the answer I linked to above or they can be disrupted on shorter time scales. Some parts last longer than others and the lifetimes are dependent upon the energy of the particles and where they are located. For instance, the protons in the inner belt have lifetimes on the order of years but electrons in the outer belt typically don't last longer than a week. – honeste_vivere Oct 11 '21 at 12:44

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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 $$0=\frac{d}{dt}\frac{1/2 m v_\perp^2}{B_\parallel}=\frac{d}{dt}\mu.$$

Zach Johnson
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