Invariance of Magnetic Moment

Let us now demonstrate that the magnetic moment, $\mu=m\, u_\perp^2 /(2\,B)$, is indeed a constant of the motion, at least to lowest order. The scalar product of the equation of motion, Equation (2.24), with the velocity ${\bf v}$ yields

$$\frac{m}{2} \frac{d v^2}{dt} = e\,{\bf v}\cdot{\bf E}.$$ (2.60)

This equation governs the evolution of the particle energy during its motion. Let us make the substitution ${\bf v} = {\bf U} + {\bf u}$, as before, and then average the preceding equation over gyrophase. To lowest order, we obtain

$$\frac{m}{2} \frac{d}{dt} (u_\perp^{2} + U_0^{2}) = e\,U_{0\,\parallel} \,E_\parallel + e\, {\bf U}_1\cdot {\bf E} + e\,\langle {\bf u}\cdot ( \mbox{\boldmath$\rho$} \cdot\nabla)\,{\bf E}\rangle.$$ (2.61)

Here, use has been made of the result

$$\frac{d}{dt} \langle f\rangle = \left\langle \frac{df}{dt}\right\rangle,$$ (2.62)

which is valid for any $f$. The final term on the right-hand side of Equation (2.61) can be written

$$e\,{\mit\Omega} \,\langle ( \mbox{\boldmath$\rho$} \times{\bf b})\cdot ( \mbox{\boldmath$\rho$} \cdot\nabla) \,{\bf E}\rangle = - \mu\,{\bf b}\cdot\nabla\times{\bf E} = \mbox{\boldmath$\mu$} \cdot \frac{\partial {\bf B}}{\partial t}= \mu\,\frac{\partial B}{\partial t},$$ (2.63)

where use has been made of Equation (2.44). Thus, Equation (2.61) reduces to

$$\frac{d K}{dt} = e\,{\bf U}\cdot{\bf E} + \mbox{\boldmath$\mu$} \cdot \frac{\partial {\bf B}}{\partial t} = e\,{\bf U}\cdot{\bf E} + \mu\, \frac{\partial B}{\partial t}.$$ (2.64)

Here, ${\bf U}$ is the guiding center velocity, evaluated to first order, and

$$K = \frac{m}{2} (U_{0\,\parallel}^{2} + {\bf v}_E^{2} + u_\perp^{2})$$ (2.65)

is the lowest order kinetic energy of the particle. Evidently, the kinetic energy can change in one of two different ways. First, by motion of the guiding center along the direction of the electric field, and, second, by acceleration of the gyration due to the electromotive force generated around the Larmor orbit by a changing magnetic field.

Equation (2.64) yields

$$m\,U_{0\,\parallel}\,\frac{dU_{0\,\parallel}}{dt}+ m\,{\bf v}_E\c... ...\parallel}\,E_\parallel +e\,{\bf U}_1\cdot{\bf E} + \mu\,\frac{\partial B}{dt}.$$ (2.66)

It follows from Equation (2.50) that

$$-m\,U_{0\,\parallel}\,{\bf b}\cdot \frac{d{\bf v}_E}{dt} + m\,{\b... ... + B\,\frac{d\mu}{dt} + \mu\,{\bf v}_E\cdot\nabla B = e\,{\bf U}_1\cdot{\bf E},$$ (2.67)

where use has been made of $dB/dt=\partial B/\partial t + U_{0\,\parallel}\,{\bf b}\cdot\nabla B + {\bf v}_E\cdot\nabla B$. However, ${\bf U}_1={\bf U}_{1\,\perp}$. Moreover, according to Equations (2.4), (2.35), and (2.51),

$$e\,{\bf U}_1\cdot{\bf E} = m\,{\bf v}_E\cdot\left[\frac{d}{dt}\,(U_{0\,\parallel}\,{\bf b} + {\bf v}_E) + \frac{\mu}{m}\,\nabla B\right].$$ (2.68)

Hence, Equation (2.67) reduces to

$$B\,\frac{d\mu}{dt} = m\,\frac{d}{dt}\left(U_{0\,\parallel}\,{\bf v}_E\cdot{\bf b}\right) = 0,$$ (2.69)

which implies that

$$\frac{d\mu}{dt}= \frac{d}{dt}\!\left(\frac{ m\,u_\perp^{2}}{2\,B}\right) = 0.$$ (2.70)

Thus, to lowest order, the magnetic moment, $\mu$, is a constant of the motion. Kruskal has shown that $m\,u_\perp^{2}/(2\,B)$ is, in fact, the lowest order approximation to a quantity that is a constant of the motion to all orders in the perturbation expansion (Kruskal 1962). Such a quantity is termed an adiabatic invariant.