Guiding Center Motion

Consider the motion of a charged particle of mass $m$ and charge $e$ in the limit in which the electromagnetic fields experienced by the particle do not vary much in a gyroperiod, so that

$$\rho\,\vert\nabla{\bf B}\vert \ll B,$$ (2.21)

$$\frac{1}{{\mit\Omega}}\frac{\partial B}{\partial t} \ll B.$$ (2.22)

The electric force is assumed to be comparable to the magnetic force. To keep track of the order of the various quantities, we introduce the parameter $\epsilon$ as a book-keeping device, and make the substitution $\rho\rightarrow \epsilon\,\rho$, as well as $({\bf E},\, {\bf B},\, {\mit\Omega}) \rightarrow \epsilon^{-1}({\bf E},\, {\bf B}, \,{\mit\Omega})$. The parameter $\epsilon$ is set to unity in the final answer.

In order to make use of the technique described in the previous section, we write the dynamical equations in the first-order differential form,

$$\frac{d{\bf r}}{dt} = {\bf v},$$ (2.23)

$$\frac{d{\bf v}}{dt} = \frac{e}{\epsilon\,m} \,({\bf E} + {\bf v}\times {\bf B}),$$ (2.24)

and seek a change of variables (Hazeltine and Waelbroeck 2004),

$${\bf r} = {\bf R}(t) + \epsilon\, \mbox{\boldmath$\rho$} ({\bf R}, {\bf U}, t, \gamma),$$ (2.25)

$${\bf v} = {\bf U}(t) + {\bf u} ({\bf R}, {\bf U}, t, \gamma),$$ (2.26)

such that the new guiding center variables ${\bf R}$ and ${\bf U}$ are free of oscillations along the particle trajectory. Here, $\gamma$ is a new independent variable describing the phase of the gyrating particle. The functions $\rho$ and ${\bf u}$ represent the gyration radius and velocity, respectively. We require periodicity of these functions with respect to their last argument, with period $2\pi$, and with vanishing mean, so that

$$\langle \mbox{\boldmath$\rho$} \rangle = \langle {\bf u} \rangle = {\bf0}.$$ (2.27)

Here, the angular brackets refer to the average over a period in $\gamma$.

The equation of motion is used to determine the coefficients in the following expansion of $\rho$ and ${\bf u}$ (Hazeltine and Waelbroeck 2004):

$$\mbox{\boldmath$\rho$} = \mbox{\boldmath$\rho$} _0({\bf R}, {\bf U}, t, \gamma) + \epsilon\, \mbox{\boldmath$\rho$} _1({\bf R}, {\bf U}, t, \gamma) + \cdots,$$ (2.28)

$${\bf u} = {\bf u}_0({\bf R}, {\bf U}, t, \gamma) + \epsilon\,{\bf u}_1({\bf R}, {\bf U}, t, \gamma) + \cdots.$$ (2.29)

The dynamical equation for the gyrophase is likewise expanded, assuming that $d\gamma/dt \simeq {\mit\Omega} = {\cal O}(\epsilon^{-1})$,

$$\frac{d\gamma}{dt} = \epsilon^{-1}\,\omega_{-1}({\bf R}, {\bf U}, t) + \omega_0({\bf R}, {\bf U}, t) + \cdots.$$ (2.30)

In the following, we suppress the subscripts on all quantities except the guiding center velocity ${\bf U}$, because this is the only quantity for which the first-order corrections are calculated.

To each order in $\epsilon$, the evolution of the guiding center position, ${\bf R}$, and velocity, ${\bf U}$, are determined by the solubility conditions for the equations of motion, Equations (2.23) and (2.24), when expanded to that order. The oscillating components of the equations of motion determine the evolution of the gyrophase. The velocity equation, Equation (2.23), is linear. It follows that, to all orders in $\epsilon$, its solubility condition is simply

$$\frac{d{\bf R}}{dt} = {\bf U}.$$ (2.31)

To lowest order [that is, ${\cal O}(\epsilon^{-1})$], the momentum equation, Equation (2.24), yields

$$\omega\,\frac{\partial{\bf u}}{\partial\gamma} - {\mit\Omega}\,{\bf u}\times{\bf b} = \frac{e}{m}\left({\bf E} + {\bf U}_0\times{\bf B}\right).$$ (2.32)

The solubility condition (that is, the gyrophase average) is

$${\bf E} + {\bf U}_0\times {\bf B} = {\bf0}.$$ (2.33)

This immediately implies that

$$E_\parallel \equiv {\bf E}\cdot {\bf b} \sim \epsilon\,E.$$ (2.34)

In other words, the rapid acceleration caused by a large parallel electric field would invalidate the ordering assumptions used in this calculation. Solving for ${\bf U}_0$, we obtain

$${\bf U}_0 = U_{0\,\parallel} \,{\bf b} + {\bf v}_E,$$ (2.35)

where all quantities are evaluated at the guiding center position, ${\bf R}$. The perpendicular component of the velocity, ${\bf v}_E$, has the same form—namely, Equation (2.4)—as that obtained for uniform fields. The parallel velocity, $U_{0\,\parallel}$, is undetermined at this order.

The integral of the oscillating component of Equation (2.32) yields

$${\bf u} = {\bf c} + u_\perp \left[\sin \,({\mit\Omega}\,\gamma/\omega) \,{\bf e}_1 +\cos\,({\mit\Omega}\,\gamma/\omega)\,{\bf e}_2\right],$$ (2.36)

where ${\bf c}$ is a constant vector, and ${\bf e}_1$ and ${\bf e}_2$ are again mutually orthogonal unit vectors perpendicular to ${\bf b}$. All quantities in the previous equation are functions of ${\bf R}$, ${\bf U}$, and $t$. The periodicity constraint, combined with Equation (2.27), requires that $\omega={\mit\Omega}({\bf R}, t)$ and ${\bf c} = {\bf0}$. The gyration velocity is thus

$${\bf u} = u_\perp \left(\sin\gamma \,{\bf e}_1+ \cos\gamma\,{\bf e}_2\right),$$ (2.37)

and, from Equation (2.30), the gyrophase is given by

$$\gamma = \gamma_0 + {\mit\Omega}\,t,$$ (2.38)

where $\gamma_0$ is the initial gyrophase. The amplitude, $u_\perp$, of the gyration velocity is undetermined at this order.

The lowest order oscillating component of the velocity equation, Equation (2.23), yields

$${\mit\Omega}\,\frac{\partial\mbox{\boldmath$\rho$}}{\partial \gamma} = {\bf u}.$$ (2.39)

This is readily integrated to give

$$\mbox{\boldmath$\rho$} = \rho\,(-\cos\gamma\,{\bf e}_1+\sin\gamma\,{\bf e}_2),$$ (2.40)

where $\rho = u_\perp/{\mit\Omega}$. It follows that

$${\bf u} = {\mit\Omega}\, \mbox{\boldmath$\rho$} \times {\bf b}.$$ (2.41)

The gyrophase average of the first-order [that is, ${\cal O}(\epsilon^0)$] momentum equation, Equation (2.24), reduces to

$$\frac{d{\bf U}_0}{dt} = \frac{e}{m}\left[ E_\parallel\,{\bf b} + ... ...langle{\bf u}\times(\mbox{\boldmath$\rho$}\cdot\nabla) \,{\bf B}\rangle\right].$$ (2.42)

All quantities in the previous expression are functions of the guiding center position, ${\bf R}$, rather than the instantaneous particle position, ${\bf r}$. In order to evaluate the last term, we make the substitution ${\bf u} = {\mit\Omega}\,$   $\rho$$\times {\bf b}$, and calculate

$$\langle ( \mbox{\boldmath$\rho$} \times{\bf b})\times ( \mbox{\boldmath$\rho$} \cdot\nabla)\,{\bf B} \rangle = {\bf b}\,\langle \mbox{\boldmath$\rho$} \cdot ( \mbox{\boldmath$\rho$} \cdot\nabla)\,{\bf B} \rangle - \langle \mbox{\boldmath$\rho$} \,\,{\bf b}\cdot( \mbox{\boldmath$\rho$} \cdot\nabla)\,{\bf B} \rangle$$

$$= {\bf b}\,\langle \mbox{\boldmath$\rho$} \cdot ( \mbox{\boldmath$\rho$} \cdot\nabla)\,{\bf B} \rangle - \langle \mbox{\boldmath$\rho$} \,( \mbox{\boldmath$\rho$} \cdot\nabla B)\rangle.$$ (2.43)

The averages are specified by

$$\langle \mbox{\boldmath$\rho$} \mbox{\boldmath$\rho$} \rangle = \frac{u_\perp^{2}}{2\,{\mit\Omega}^{2}} \,({\bf I} - {\bf b}{\bf b}),$$ (2.44)

where ${\bf I}$ is the identity tensor. Thus, making use of ${\bf I}:\nabla{\bf B} = \nabla \cdot {\bf B} = 0$, it follows that

$$-e \,\langle {\bf u} \times ( \mbox{\boldmath$\rho$} \cdot\nabla )\,{\bf B}\rangle = \frac{m\,u_\perp^{2}}{2\,B} \,\nabla B.$$ (2.45)

This quantity is the secular component of the gyration induced fluctuations in the magnetic force acting on the particle.

The coefficient of $\nabla B$ in the previous equation,

$$\mu = \frac{m\,u_\perp^{2}}{2\,B},$$ (2.46)

plays a central role in the theory of magnetized particle motion. We can interpret this coefficient as a magnetic moment by drawing an analogy between a gyrating particle and a current loop. The (vector) magnetic moment of a plane current loop is simply

$$\mbox{\boldmath$\mu$} = I\,A\,{\bf n},$$ (2.47)

where $I$ is the current, $A$ the area of the loop, and ${\bf n}$ the unit normal to the surface of the loop. For a circular loop of radius $\rho = u_\perp/{\mit\Omega}$, lying in the plane perpendicular to ${\bf b}$, and carrying the current $e\,{\mit\Omega}/2\pi$, we find

$$\mbox{\boldmath$\mu$} = I\,\pi\,\rho^2\,{\bf b} = \frac{m\,u_\perp^{2}}{2\,B}\,{\bf b}.$$ (2.48)

We shall demonstrate, in Section 2.6, that the (scalar) magnetic moment, $\mu$, is a constant of the particle motion. Thus, the guiding center behaves exactly like a particle with a conserved magnetic moment $\mu$ that is always aligned with the magnetic field.

The first-order guiding center equation of motion, Equation (2.42), reduces to

$$m \,\frac{d{\bf U}_0}{dt} = e\,E_\parallel\,{\bf b} + e\,{\bf U}_1\times {\bf B} -\mu\,\nabla B.$$ (2.49)

The component of this equation along the magnetic field determines the evolution of the parallel guiding center velocity:

$$m\,\frac{dU_{0\,\parallel}}{dt} = e\,E_\parallel - \mbox{\boldmath$\mu$} \cdot\nabla B - m\,{\bf b}\cdot\frac{d{\bf v}_E}{dt}.$$ (2.50)

Here, use has been made of Equation (2.35), and ${\bf b}\cdot d{\bf b}/dt=0$. The component of Equation (2.49) perpendicular to the magnetic field determines the first-order perpendicular drift velocity:

$${\bf U}_{1\,\perp} = \frac{{\bf b}}{{\mit\Omega}} \times\left( \frac{d{\bf U}_0}{dt} + \frac{\mu}{m}\,\nabla B\right).$$ (2.51)

The first-order correction to the parallel velocity, the so-called first-order parallel drift velocity, is undetermined to this order. This is not a problem, because the first-order parallel drift is a small correction to a type of motion that already exists at zeroth order, whereas the first-order perpendicular drift is a completely new type of motion. In particular, the first-order perpendicular drift differs fundamentally from the ${\bf E}\times{\bf B}$ drift, because it is not the same for all species, and, therefore, cannot be eliminated by transforming to a new inertial frame. Thus, without loss of generality, we can absorb the first-order parallel drift into $U_{0\,\parallel}$, and write ${\bf U}_1={\bf U}_{1\,\perp}$.

We can now understand the motion of a charged particle as it moves through slowly varying electric and magnetic fields. The particle always gyrates around the magnetic field at the local gyrofrequency, ${\mit\Omega}=eB/m$. The local perpendicular gyration velocity, $u_\perp$, is determined by the requirement that the magnetic moment, $\mu=m\,u_\perp^{2}/(2\,B)$, be a constant of the motion. This, in turn, fixes the local gyroradius, $\rho = u_\perp/{\mit\Omega}$. The parallel velocity of the particle is determined by Equation (2.50). Finally, the perpendicular drift velocity is the sum of the ${\bf E}\times{\bf B}$ drift velocity, ${\bf v}_E$, and the first-order drift velocity, ${\bf U}_{1\,\perp}.$