# 75 Adiabatic Invariants for a Charged Particle Moving in a Slowly Varying Magnetic Field

Michael Fowler UVa

Jackson 12.5

## Guiding Center Picture

As we’ve seen in the previous lectures, it turns
out that, over a wide range of particle energies and field strengths, the motion
of a charged particle in a slowly varying magnetic field is well-described as moving
in tight small circles (sometimes called Larmor circles) around a *guiding center*, which itself moves
slowly. And, by “slowly” we mean the
guiding center moves only a small fraction of the small circle radius during
each revolution, so the particle’s path is almost closed, the motion almost
repeats. For a constant magnetic field, the guiding center moves along the
field direction. We showed in the last lecture that the guiding center will
also drift sideways if the field strength varies or if the field lines are
curved.

In this
lecture, we’ll consider the center moving along a field line with negligible
curvature, *but* moving into a region
where the field *strength* is changing
significantly. This is different from the previous cases: here the size of the Larmor circle can change
in an essential way.

## Adiabatic Invariants: A Clue from the Old Quantum Theory

To think about how the Larmor circle itself might change on drifting into a region of different magnetic field strength, it’s instructive to consider a bit of ancient quantum history, where we’ll find a very similar situation. The first semi-successful model of an atom was Bohr’s hydrogen model, with electrons in circular orbits around the proton, but arbitrarily (it seemed) restricted in angular momentum$\u2014$which could only be $nh/2\pi ,$ with $n$ an integer and $h$ Planck’s constant (already known at the time from black body radiation). This recipe led to all the energy levels spectroscopically observed$\u2014$a fantastic triumph.

But still, what was the justification? Why these radii, and what about elliptical orbits? After all, the force was inverse square$\u2014$just like gravity, which certainly has elliptic orbits.

A rationale for the allowed radii was first
suggested by de Broglie, who conjectured
that the electron had a *wavelike*
nature, with wavelength depending on momentum using the *same* formula already familiar for photons, $p=h/\lambda .$ The allowed circular orbits were then those
that contained a *whole number of
wavelengths* $\lambda :$ so the wave fit smoothly and uninterrupted
around the path. This idea worked!

It’s often shown in textbooks as a real standing
wave fitting around the circle, but here we’ll take the *angular momentum* eigenstates, *traveling
*wavefunctions, proportional to ${e}^{ipx/\hslash}$ with $x$ measuring distance along the path. In the *n*^{th}
state, this complex-valued wave function $\psi \left(x\right)$ circles the complex $\psi $ plane origin *n* times as $x$ goes once around the real space path. ( The
real, *standing* wave representation
usually shown just takes the sum and difference of these traveling wave angular
momentum eigenstates.)

But wait, what about the elliptical orbits? Why can we ignore them? Sommerfeld solved *that* mystery by computing the general
elliptical orbit, that is, solving to find the momentum variable $p(x)$ at all points $x$ along the orbit, and then requiring a whole
number of$\u2014$now
variable$\u2014$wavelengths
to fit smoothly around the orbit,

$$\left(1/\hslash \right){\displaystyle \u2233p\left(x\right)dx=2\pi n}.$$

Remarkably, this gives the identical *same* set of energies as the simple Bohr
circular orbits! So, they were there all
along. This energy degeneracy for different shaped orbits is unique to the
inverse square force$\u2014$we got
lucky, or rather Bohr did.

Picture now in this semiclassical spirit the
general elliptical orbit in space, and the traveling complex wavefunction $\psi \left(x\right)$ around this orbit. As $x$ goes one time around the orbit in ordinary
space, the wavefunction $\psi \left(x\right)$ is a point circling the origin in its complex $\psi $ plane, completing exactly *n *turns,
because as $x$ returns to the orbital starting point, $\psi \left(x\right)$ must go to the original position and slope, it
must be a smooth wave function all the way around.

But now suppose we add in some small slow change
to the potential, perhaps another atom moves *slowly* into the neighborhood. This is termed an “adiabatic”
change. What happens to the wavefunction
$\psi \left(x\right)$?
Obviously, it will change. But, visualizing $\psi \left(x\right)$ circling around the origin in the complex $\psi $ plane as $x$ traces the physical path, unless the
perturbing potential is strong enough to drive the wavefunction to zero at some
point, *the total number n of turns about the **$\psi $** origin for the complete path will not change. *

**Note: ** From a *quantum*
perspective, the photon energy necessary for $\psi \left(x\right)$ to move up from *n *to *n*+1 turns is close
to $h\nu ,$ with $\nu $ the inverse orbital time, so if external
changes are small on a time scale set by the orbital time, the quantum number
is very unlikely to change.

That is, for a particle in a periodic orbit, $\u2233p\left(x\right)dx$ does not change when parameters such as the
external field vary slowly. This is called an *adiabatic invariant*, adiabatic here meaning slow, smooth change.

In classical
mechanics, this adiabatic invariant is the *action
integral *defined as

$$J=\left(1/2\pi \right){\displaystyle \oint pdq}$$

around one cycle, $q$ being an angle parameter and $p$ the canonical momentum. This is fully discussed in my lecture.

*History*:
The connection to quantum mechanics was elucidated by Einstein and
Sommerfeld at the Solvay Conference in 1911: in fact, the main conference agenda was
attempting to reconcile classical and quantum physics.

The adiabatic invariant is the key to understanding how the Larmor orbit develops under smooth changes as it moves through the varying magnetic field.

## Adiabatic Invariance of Flux Through Orbit of Particle

### Magnetic Field and Wave Function Phase

We’ll consider field strengths, and particle momenta, to be such that the path is well-described (as in the preceding lecture) by a circling motion about a slowly moving guiding center, successive circles having centers much closer together than the circle radius.

We will continue in the semiclassical picture
described above, and *if we ignore the magnetic field*, we would take the
local wave function at a point on the ring $\sim {e}^{ipx/\hslash},$ with local kinetic momentum $p=\gamma mv,$ and going around the ring would yield a net phase
increase of $2n\pi .$ (?)

But this is wrong! Recall (from lecture 72) that to get the experimentally observed behavior of the classical electron in magnetic and electric fields we needed to add a term to the Lagrangian (in fact the simplest possible Lorentz invariant term linking the field to the particle)

$$S={\displaystyle \underset{a}{\overset{b}{\int}}Ldt}={\displaystyle \underset{a}{\overset{b}{\int}}\left(-m{c}^{2}d\tau -\frac{e}{c}{A}_{\mu}d{x}^{\mu}\right).}$$ so we see that adding the above term to the
action is adding $eA/\hslash c$ to the *phase
*of $\psi .$ With this addition, the waves correspond in
wavelength to the *canonical momentum*

$${P}_{i}=\frac{\partial L}{\partial {u}_{i}}=\gamma m{u}_{i}+\frac{e}{c}{A}_{i}.$$

In other words, we’ve given a hand-waving semiclassical
derivation for the adiabatic invariance of the classical action integral (with
the *canonical *momentum!)

$$J={\displaystyle \oint {P}_{i}d{q}_{i}}.$$

### Action Integral for Larmor Loop

The action corresponding to the circling motion ${\overrightarrow{v}}_{\perp}$ perpendicular to the field is

$$J={\displaystyle \oint {\overrightarrow{P}}_{\perp}\cdot d\overrightarrow{\ell}}={\displaystyle \oint \gamma m{\overrightarrow{v}}_{\perp}\cdot d\overrightarrow{\ell}}+\frac{e}{c}{\displaystyle \oint \overrightarrow{A}\cdot d\overrightarrow{\ell}}$$ In the adiabatic limit, we can take ${\overrightarrow{v}}_{\perp}\parallel d\overrightarrow{\ell}$ so (with circle radius $a,\text{\hspace{0.33em}}{v}_{\perp}=a{\omega}_{B},\text{\hspace{0.33em}}\text{\hspace{0.33em}}d\ell =ad\theta ,$ ${\omega}_{B}$ the circling frequency in field $B.$ )

$$\begin{array}{c}J={\displaystyle \oint \gamma m{\omega}_{B}{a}^{2}d\theta}+\frac{e}{c}{\displaystyle \oint \overrightarrow{A}\cdot d\overrightarrow{\ell},}\\ =2\pi \gamma m{\omega}_{B}{a}^{2}+\frac{e}{c}{\displaystyle \underset{S}{\int}\overrightarrow{B}\cdot \widehat{\overrightarrow{n}}da.}\end{array}$$ At this point, it's crucial to get the signs right, because these terms partially cancel.

Let's suppose the magnetic field is pointing in the positive $z$-direction, and the circling motion is therefore in the $x,y$ plane. Take the particle to be positively charged, and think of the moment it crosses the positive $x$-axis, as it circles the origin. The force $(e/c)\overrightarrow{v}\times \overrightarrow{B}$ has to be towards the center of its circle, the origin, so it must be moving downwards, meaning circling clockwise. Since it's part of the same integral, the $\oint \overrightarrow{A}\cdot d\overrightarrow{\ell}$ must also be taken clockwise, so that second integral must give

$$\frac{e}{c}{\displaystyle \underset{S}{\int}\overrightarrow{B}\cdot \widehat{\overrightarrow{n}}da}=-\frac{e}{c}\pi {a}^{2}B.$$
Noting that ${\omega}_{B}=eB/\gamma mc,$ we find the first term in $J$ is twice the magnitude of the second, and
positive, so
$$J=\frac{e}{c}\left(B\pi {a}^{2}\right).$$
This means that as the guiding center moves from a
weak field to a strong field, the circle orbit will shrink to keep *the same total magnetic flux through the
circle*.

Equivalent invariants are ${p}_{\perp}^{2}/B$ and $\gamma \mu ,$ where $\mu =e{\omega}_{B}{a}^{2}/2$ is the magnetic moment of the current loop. (*Exercise*:
prove these!)

This invariant gives some insight into why a circling charged particle can be repelled by a magnetic pole, such as one of the Earth's magnetic poles. Jackson considers the case of a field mainly in the $z$-direction, but of varying strength:

Imagine a charged particle, circling in the $x,y$ plane with velocity ${\overrightarrow{v}}_{\perp},$ but also coming in from the left with a nonzero drift velocity ${v}_{\parallel}$ in the $z$ -direction.

Since the only force on the particle is from the static magnetic field, its kinetic energy cannot change, ${v}_{\perp}^{2}+{v}_{\parallel}^{2}={v}_{0}^{2}.$

From the adiabatic theorem, $$\frac{{v}_{\perp}^{2}}{B}=\frac{{v}_{\perp 0}^{2}}{{B}_{0}},$$ so $${v}_{\parallel}^{2}={v}_{0}^{2}-{v}_{\perp 0}^{2}\frac{B\left(z\right)}{{B}_{0}}.$$ This implies that motion in the $z$-direction is the same as a particle in a one-dimensional potential, evidently with a turning point if the field strength gets sufficiently strong to make the right-hand side zero.

A charged particle can be confined in a "magnetic bottle" having field lines coming together at both ends. This is the basic idea behind early attempts to confine a plasma at very high temperatures in hopes of gaining confined nuclear fusion. Unfortunately, the one-particle model ignores sideways drift, and the plasma interactions occurring at finite density, which render the confinement unstable.

However, there are excellent examples in nature of this kind of magnetic bottle, such as the van Allen radiation belts, toroidally shaped region surrounding the Earth, and filled with charged particles. These are shaped by the Earth's magnetic field (from Wikimedia Commons):