Gyrating proton near magnetic pole of the Earth

Question

Imagine a proton from space which approaches the magnetic pole in the Northern hemisphere of the Earth. The proton spirals around the magnetic field lines. The $\vec{B}$-field is stronger near the magnetic pole, where the 'converging' magnetic field lines are closer to each other and slightly curved. $\vec{B}$ has both a 'longitudinal' component $\vec{B}_{long}$ directed at the pole and a 'radial' component $\vec{B}_\bot$.

Let us ignore the gravitational force on the proton and assume that the proton doesn't interact with molecules in the upper atmosphere.

The lorentz force doesn't do work on the proton, so the kinetic energy of the proton doesn't change. The circular part of the motion of the proton around field lines is equivalent to an electric current loop with a magnetic (dipole) moment $\mu=W_\bot/B$ where $W_\bot$ is the kinetic energy of the proton associated with its circular motion around the field lines. As the proton approaches the pole, the lorentz force will reduce the 'forward' speed of the proton and make the circular speed higher.

If the spatial variation of $\vec{B}$ is slow (as seen by the moving proton), the magnetic moment can be shown to be adiabatic invariant. This is usually shown by expressing the lorentz force in cylindrical coordinates, or using a Hamiltonian approach with canonical variables. That's too difficult for my pre-universty audience.

Question a. Is there a simpler way to demonstrate that $\mu$ is (almost) constant along the proton's path towards the magnetic pole?

It seems to me that $|\mu|$ is equal to the angular momentum $|L|$ up to a constant factor. As the magnetic field gets stronger, the angular momentum of the proton will increase due to the increasing lorentz force.

Question b. How can $\mu$ be almost constant while $|L|$ increases?

Answer 1

Let's call the cyclotron (angular) frequency $\omega$, the cyclotron/Larmor radius $r$, the velocity perpindicular to the magnetic field $v_\perp$, the particle charge $q$ and mass $m$. The angular momentum is $L=mr^2\omega$. The magnetic moment is (lots of possible formulas) $$\mu=mr^2\omega^2/2|\mathbf{B}|=qr^2\omega/2=qv_\perp r/2=qv_\perp^2/2\omega$$ As you point out, angular momentum is proportional to magnetic moment. In case it wasn't clear to you, the approximation that $\mathbf{B}$ is slowly changing (with respect to the cyclotron frequency) is VERY good here - the magnetic moment and angular momentum are nearly perfectly conserved, except when collisions with other particles are involved. Even if the proton was moving a comparable fraction of the speed of light parallel to the magnetic field, the adiabatic parameter is small. $(c/r_\text{earth})(10\text{ }\mu\text{T=roughly the magnetic field magnitude})q_e/m_p=0.05$

Your assertion that $|L|$ is changing is not correct. You may have been confused by the fact that $v_\perp$ (the velocity perpendicular to the magnetic field) is increasing when $|\mathbf{B}|$ increases. But the cyclotron/Larmor radius decreases to compensate, conserving both angular momentum and $\mu$.

As for what's the easiest way to derive this (trying not to mention the conservation of action per cycle), I don't think I have the best answer - I wonder actually if there's a faster argument using this fact about angular momentum. Here's a derivation of this fact though:

EDIT: I just realized this is the "lorentz force in cylindrical coordinates" argument you explicitly didn't want to use... I'll leave it here because I typed it up, but the answer is no I don't know an easier way

Let's say the particle is following a $z$ oriented magnetic field line, and the field is increasing. I'll use cylindrical coordinates, and assert cylindrical symmetry. So I have: $$ \mathbf{B}(z,r=0,\theta)=(B_0+kz)\hat{z} $$ In order for my $\mathbf{B}$ field to satisfy the laplace equation, I need an inward pointing field at nonzero $r$ $$ \mathbf{B}(z,r,\theta)=(B_0+kz)\hat{z}-(kr/2)\hat{r} $$ This inward pointing field, coupled with $v_z$, gives me a force in the $\hat{\theta}$ direction, speeding up the particle. $$ F_\theta=qv_z(kr/2)\hat{\theta} $$ So the particle gains perpindicular velocity at a rate: $$ \frac{dv_\perp}{dt}=\frac{qv_zkr}{2m}=\frac{qv_zkv_\perp}{2m\omega}=\frac{v_zkv_\perp}{2|\mathbf{B}|}=\frac{v_\perp(d|\mathbf{B}|/dt)}{2|\mathbf{B}|} $$ $$ \frac{1}{v_\perp}\left(\frac{dv_\perp}{dt}\right)=\frac{1}{2|\mathbf{B}|}\left(\frac{d|\mathbf{B}|}{dt}\right) $$ $$ \frac{d\ln v_\perp}{dt}=\frac{1}{2}\frac{d\ln |\mathbf{B}|}{dt} $$ $$ \ln v_\perp=\frac{1}{2}\ln|\mathbf{B}| +C $$ $$ v_\perp=C\sqrt{|\mathbf{B}|} $$ $$ \frac{v_\perp^2}{|\mathbf{B}|}\propto\mu=\text{constant} $$