Adiabatic Invariants

Poincaré invariants are generally of little practical interest unless the curve $C$ closely corresponds to the trajectories of actual particles. For the motion of magnetized particles, it is evident from Equations (2.25), (2.38), and (2.40) that points having the same guiding center at a certain time will continue to have approximately the same guiding center at later times. An approximate Poincaré invariant may thus be obtained by choosing the curve $C$ to be a circle of points corresponding to a gyrophase period. In other words,

$${\cal I} \simeq I = \oint {\bf p}\cdot \frac{\partial{\bf q}}{\partial\gamma}\, d\gamma.$$ (2.75)

Here, $I$ is an adiabatic invariant.

To evaluate $I$ for a magnetized plasma recall that the canonical momentum for charged particles is (Jackson 1998)

$${\bf p} = m\,{\bf v} +e\,{\bf A},$$ (2.76)

where ${\bf A}$ is the vector potential. Let us express ${\bf A}$ in terms of its Taylor series about the guiding center position:

$${\bf A}({\bf r}) = {\bf A}({\bf R}) + ( \mbox{\boldmath$\rho$} \cdot\nabla)\,{\bf A}({\bf R}) + {\cal O}(\rho^2).$$ (2.77)

The element of length along the curve $C(t)$ is [see Equation (2.39)]

$$d{\bf r} = \frac{\partial\mbox{\boldmath$\rho$}}{\partial \gamma}\,d\gamma = \frac{{\bf u}}{{\mit\Omega}}\,\,d\gamma.$$ (2.78)

The adiabatic invariant is thus

$$I = \oint \frac{{\bf u}}{{\mit\Omega}} \cdot \left( m\,[{\bf U} +... ...oldmath$\rho$}\cdot \nabla)\,{\bf A}\right]\right)d\gamma + {\cal O}(\epsilon),$$ (2.79)

which reduces to

$$I = 2\pi\,m\,\frac{u_\perp^{2}}{{\mit\Omega}} + 2\pi\,\frac{e}{{\mit\Omega}}\, \langle {\bf u}\cdot( \mbox{\boldmath$\rho$} \cdot \nabla)\,{\bf A}\rangle + {\cal O}(\epsilon).$$ (2.80)

The final term on the right-hand side is written [see Equations (2.41) and (2.44)]

$$2\pi\,e\,\langle ( \mbox{\boldmath$\rho$} \times{\bf b})\cdot ( \mbox{\boldmath$\rho$} \cdot \nabla)\,{\bf A}\rangle = -2\pi\,e\,\frac{u_\perp^{2}}{2\,{... ..., {\bf b}\cdot \nabla\times{\bf A} = -\pi\,m\,\frac{u_\perp^{2}}{{\mit\Omega}}.$$ (2.81)

It follows that

$$I = 2\pi\, \frac{m}{e}\,\mu + {\cal O}(\epsilon).$$ (2.82)

Thus, to lowest order, the adiabatic invariant is proportional to the magnetic moment, $\mu$.