4
$\begingroup$

In plasma physics, particularly, in magnetic mirrors, we have a very well-known result, which is the invariance of the magnetic moment of the particles trapped in the mirror.

I've saw in some notes that, going further into the "invariance", we actually have three adiabatic invariant of charged particles, the first one being exactly the magnetic moment.

I've been not able to prove this fact. If anyone could give me some hints, I'd be grateful.

$\endgroup$

1 Answer 1

2
$\begingroup$

The following comes from Jackson [1999], pages 592-593.

Suppose we start with the action integral given by: $$ J_{i} = \oint \ dq_{i} \ p_{i} \tag{0} $$ where $p_{i}$ and $q_{i}$ are the usual generalized canonical coordinates and the integration is over a complete cycle of $q_{i}$. Then from Jackson [1999]:

...For a given mechanical system with specified initial conditions the action integrals $J_{i}$ are constants. If now the properties of the system are changed in some way (e.g., a change in spring constant or mass of some particle), the question arises as to how the action integrals change. It can be proved that, if the change in property is slow compared to the relevant periods of motion and is not related to the periods (such a change is called an adiabatic change), the action integrals are invariant.

First we need to define a few things. These are the gyroradius, $\rho_{cs}$, and cyclotron frequency, $\Omega_{cs}$, of species $s$ given by: $$ \begin{align} \rho_{cs} & = \frac{ \gamma \ m_{s} \ v_{\perp} }{ e \ B } \tag{1a} \\ \Omega_{cs} & = \frac{ e \ B }{ \gamma \ m_{s} } \tag{1b} \end{align} $$ where $e$ is the fundamental charge [C], $\gamma$ is the Lorentz factor [N/A], $B$ is the magnitude of the magnetic field [T], $m_{s}$ is the mass of species $s$ [kg], and $v_{\perp}$ is the speed orthogonal to the magnetic field vector [km/s]. Note that in Gaussian units there is a factor of $c$ (speed of light) in the numerator of $\rho_{cs}$ and denominator of $\Omega_{cs}$.

Now for a charged particle in a magnetic field, we know the canonical momentum (Gaussian units) is given by $\mathbf{P} = \mathbf{p} + \tfrac{ e }{ c } \mathbf{A}$, where $\mathbf{p}$ is the 3-vector momentum and $\mathbf{A}$ is the vector potential. Then we can write the action integral as: $$ \begin{align} J & = \oint \ d\mathbf{l} \cdot \mathbf{P} \tag{2a} \\ & = \oint \ d\mathbf{l} \cdot \left( \gamma \ m_{s} \ v_{\perp} \right) + \frac{ e }{ c } \oint \ d\mathbf{l} \cdot \mathbf{A} \tag{2b} \\ & = \oint \ d\phi \ \gamma \ m_{s} \ \Omega_{cs} \ \rho_{cs}^{2} + \frac{ e }{ c } \int_{S} \ dA \ \hat{n} \cdot \nabla \times \mathbf{A} \tag{2c} \\ & = 2 \pi \ \gamma \ m_{s} \ \Omega_{cs} \ \rho_{cs}^{2} + \frac{ e }{ c } \int_{S} \ dA \ \hat{n} \cdot \mathbf{B} \tag{2d} \end{align} $$ where we've used Stokes' theorem for the second integral and took advantage of the fact that $d\mathbf{l}$ and $v_{\perp}$ are parallel and that $d\mathbf{l} = \rho_{cs} d\hat{\phi}$ in the first integral. The line element $d\mathbf{l}$ orbits $\mathbf{B}$ in a counterclockwise sense, thus the outward unit normal, $\hat{n}$, is anti-parallel to $\mathbf{B}$. The second integral is then an integral over the area of a gyroradius, which reduces it to $\pi \rho_{cs}^{2}$. After a little manipulation one sees the second integral is one half the first integral and negative resulting in the action going to: $$ J = \pi \ \gamma \ m_{s} \ \Omega_{cs} \ \rho_{cs}^{2} = \frac{ e }{ c } \left( \pi \rho_{cs}^{2} B \right) \tag{3} $$

The quantity in ()'s in Equation 3 is the magnetic flux through the particles gyro orbit. So now we ask how this relates to adiabatic invariance. If we look at Equation 3 we can see that for slowly varying magnetic fields, $J$ must be constant. The reason we can say this is the following:

If B increases, then $\rho_{cs}$ decreases such that $\left( \pi \rho_{cs}^{2} B \right)$ remains constant.

This statement can be reduced to three so called adiabatic invariants of motion for charged particles in slowly varying fields given by: $$ \begin{align} \left( \pi \rho_{cs}^{2} B \right) & \equiv \text{magnetic flux conservation} \tag{4a} \\ \frac{ p_{\perp}^{2} }{ B } & \equiv \text{transverse momentum} \tag{4b} \\ \gamma \mu & \equiv \text{magnetic moment} \tag{4c} \\ \int_{a}^{b} \ ds \ p_{\parallel} & \equiv \text{parallel momentum} \tag{4d} \end{align} $$ where $\mu = \tfrac{ e \ \Omega_{cs} \ \rho_{cs}^{2} }{ 2 \ c }$ is the particle magnetic moment.

These are linked to particle motions in, for instance, Earth's magnetosphere.

  • Equation 4a relates to particle drifts in the azimuthal direction around the Earth. If B increases slowly compared to the drift-orbital period of the particles, the particles drift orbit radius must decrease accordingly. The inward motion of these particles is related to a phenomena called radial diffusion, important for radiation belt dynamics.
  • Equations 4b and 4c relate to the gyration of particles about the magnetic field. If the field varies slowly compared to the gyroperid of the particle, then an increase in B will result in a increase in the momentum transverse to $\mathbf{B}$. This is merely the gyration of charged particles about the magnetic field.
  • Equation 4d relates to what is called bounce motion of particles along $\mathbf{B}$. If B varies slowly compared to the time it takes to bounce between magnetic mirror points, then the action integral of Equation 4d is constant, e.g., the particles will bounce between the north and south magnetic poles of Earth's dipole magnetic field.

So we have three periodic motions (from fastest to slowest): gyration, bounce, and drift.

References

  • J.D. Jackson, Classical Electrodynamics, Third Edition, John Wiley & Sons, Inc., New York, NY, 1999.
$\endgroup$
1
  • 1
    $\begingroup$ I had not yet quite studied Lagrangian mechanics, after achieving the concepts of general and canonic coordinates , action integral and some others, this development became clear, and helped me a lot. Thank you. $\endgroup$ Commented Oct 18, 2021 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.