Categories: Electromagnetism, Physics, Plasma physics.

# Guiding center theory

When discussing the Lorentz force, we introduced the concept of gyration: a particle in a uniform magnetic field $$\vb{B}$$ gyrates in a circular orbit around a guiding center. Here, we will generalize this result to more complicated situations, for example involving electric fields.

The particle’s equation of motion combines the Lorentz force $$\vb{F}$$ with Newton’s second law:

\begin{aligned} \vb{F} = m \dv{\vb{u}}{t} = q \big( \vb{E} + \vb{u} \cross \vb{B} \big) \end{aligned}

We now allow the fields vary slowly in time and space. We thus add deviations $$\delta\vb{E}$$ and $$\delta\vb{B}$$:

\begin{aligned} \vb{E} \to \vb{E} + \delta\vb{E}(\vb{x}, t) \qquad \quad \vb{B} \to \vb{B} + \delta\vb{B}(\vb{x}, t) \end{aligned}

Meanwhile, the velocity $$\vb{u}$$ can be split into the guiding center’s motion $$\vb{u}_{gc}$$ and the known Larmor gyration $$\vb{u}_L$$ around the guiding center, such that $$\vb{u} = \vb{u}_{gc} + \vb{u}_L$$. Inserting:

\begin{aligned} m \dv{t} \big( \vb{u}_{gc} + \vb{u}_L \big) = q \big( \vb{E} + \delta\vb{E} + (\vb{u}_{gc} + \vb{u}_L) \cross (\vb{B} + \delta\vb{B}) \big) \end{aligned}

We already know that $$m \: \dv*{\vb{u}_L}{t} = q \vb{u}_L \cross \vb{B}$$, which we subtract from the total to get:

\begin{aligned} m \dv{\vb{u}_{gc}}{t} = q \big( \vb{E} + \delta\vb{E} + \vb{u}_{gc} \cross (\vb{B} + \delta\vb{B}) + \vb{u}_L \cross \delta\vb{B} \big) \end{aligned}

This will be our starting point. Before proceeding, we also define the average of $$\expval{f}$$ of a function $$f$$ over a single gyroperiod, where $$\omega_c$$ is the cyclotron frequency:

\begin{aligned} \expval{f} \equiv \int_0^{2 \pi / \omega_c} f(t) \dd{t} \end{aligned}

Assuming that gyration is much faster than the guiding center’s motion, we can use this average to approximately remove the finer dynamics, and focus only on the guiding center.

## Uniform electric and magnetic field

Consider the case where $$\vb{E}$$ and $$\vb{B}$$ are both uniform, such that $$\delta\vb{B} = 0$$ and $$\delta\vb{E} = 0$$:

\begin{aligned} m \dv{\vb{u}_{gc}}{t} = q \big( \vb{E} + \vb{u}_{gc} \cross \vb{B} \big) \end{aligned}

Dotting this with the unit vector $$\vu{b} \equiv \vb{B} / |\vb{B}|$$ makes all components perpendicular to $$\vb{B}$$ vanish, including the cross product, leaving only the (scalar) parallel components $$u_{gc\parallel}$$ and $$E_\parallel$$:

\begin{aligned} m \dv{u_{gc\parallel}}{t} = \frac{q}{m} E_{\parallel} \end{aligned}

This simply describes a constant acceleration, and is easy to integrate. Next, the equation for $$\vb{u}_{gc\perp}$$ is found by subtracting $$u_{gc\parallel}$$’s equation from the original:

\begin{aligned} m \dv{\vb{u}_{gc\perp}}{t} = q (\vb{E} + \vb{u}_{gc} \cross \vb{B}) - q E_\parallel \vu{b} = q (\vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B}) \end{aligned}

Keep in mind that $$\vb{u}_{gc\perp}$$ explicitly excludes gyration. If we try to split $$\vb{u}_{gc\perp}$$ into a constant and a time-dependent part, and choose the most convenient constant, we notice that the only way to exclude gyration is to demand that $$\vb{u}_{gc\perp}$$ does not depend on time. Therefore:

\begin{aligned} 0 = \vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B} \end{aligned}

To find $$\vb{u}_{gc\perp}$$, we take the cross product with $$\vb{B}$$, and use the fact that $$\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$$:

\begin{aligned} 0 = \vb{B} \cross (\vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B}) = \vb{B} \cross \vb{E} + \vb{u}_{gc\perp} B^2 \end{aligned}

Rearranging this shows that $$\vb{u}_{gc\perp}$$ is constant. The guiding center drifts sideways at this speed, hence it is called a drift velocity $$\vb{v}_E$$. Curiously, $$\vb{v}_E$$ is independent of $$q$$:

\begin{aligned} \boxed{ \vb{v}_E = \frac{\vb{E} \cross \vb{B}}{B^2} } \end{aligned}

Drift is not specific to an electric field: $$\vb{E}$$ can be replaced by a general force $$\vb{F}/q$$ without issues. In that case, the resulting drift velocity $$\vb{v}_F$$ does depend on $$q$$:

\begin{aligned} \boxed{ \vb{v}_F = \frac{\vb{F} \cross \vb{B}}{q B^2} } \end{aligned}

## Non-uniform magnetic field

Next, consider a more general case, where $$\vb{B}$$ is non-uniform, but $$\vb{E}$$ is still uniform:

\begin{aligned} m \dv{\vb{u}_{gc}}{t} = q \big( \vb{E} + \vb{u}_{gc} \cross (\vb{B} + \delta\vb{B}) + \vb{u}_L \cross \delta\vb{B} \big) \end{aligned}

Assuming the gyroradius $$r_L$$ is small compared to the variation of $$\vb{B}$$, we set $$\delta\vb{B}$$ to the first-order term of a Taylor expansion of $$\vb{B}$$ around $$\vb{x}_{gc}$$, that is, $$\delta\vb{B} = (\vb{x}_L \cdot \nabla) \vb{B}$$. We thus have:

\begin{aligned} m \dv{\vb{u}_{gc}}{t} = q \big( \vb{E} + \vb{u}_{gc} \cross \vb{B} + \vb{u}_{gc} \cross (\vb{x}_L \cdot \nabla) \vb{B} + \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} \big) \end{aligned}

We approximate this by taking the average over a single gyration, as defined earlier:

\begin{aligned} m \dv{\vb{u}_{gc}}{t} = q \big( \vb{E} + \vb{u}_{gc} \cross \vb{B} + \vb{u}_{gc} \cross \expval{ (\vb{x}_L \cdot \nabla) \vb{B} } + \expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} } \big) \end{aligned}

Where we have used that $$\expval{\vb{u}_{gc}} = \vb{u}_{gc}$$. The two averaged expressions turn out to be:

\begin{aligned} \expval{ (\vb{x}_L \cdot \nabla) \vb{B} } = 0 \qquad \quad \expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} } \approx - \frac{u_L^2}{2 \omega_c} \nabla B \end{aligned}

With this, the guiding center’s equation of motion is reduced to the following:

\begin{aligned} m \dv{\vb{u}_{gc}}{t} = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg) \end{aligned}

Let us now split $$\vb{u}_{gc}$$ into components $$\vb{u}_{gc\perp}$$ and $$u_{gc\parallel} \vu{b}$$, which are respectively perpendicular and parallel to the magnetic unit vector $$\vu{b}$$, such that $$\vb{u}_{gc} = \vb{u}_{gc\perp} \!+\! u_{gc\parallel} \vu{b}$$. Consequently:

\begin{aligned} \dv{\vb{u}_{gc}}{t} = \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} + u_{gc\parallel} \dv{\vu{b}}{t} \end{aligned}

Inserting this into the guiding center’s equation of motion, we now have:

\begin{aligned} \dv{\vb{u}_{gc}}{t} = m \bigg( \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} + u_{gc\parallel} \dv{\vu{b}}{t} \bigg) = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg) \end{aligned}

The derivative of $$\vu{b}$$ can be rewritten as follows, where $$R_c$$ is the radius of the field’s curvature, and $$\vb{R}_c$$ is the corresponding vector from the center of curvature:

\begin{aligned} \dv{\vu{b}}{t} \approx - u_{gc\parallel} \frac{\vb{R}_c}{R_c^2} \end{aligned}

With this, we arrive at the following equation of motion for the guiding center:

\begin{aligned} m \bigg( \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} - u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} \bigg) = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg) \end{aligned}

Since both $$\vb{R}_c$$ and any cross product with $$\vb{B}$$ will always be perpendicular to $$\vb{B}$$, we can split this equation into perpendicular and parallel components like so:

\begin{aligned} m \dv{\vb{u}_{gc\perp}}{t} &= q \vb{E}_{\perp} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\perp} B + m u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} + q \vb{u}_{gc} \cross \vb{B} \\ m \dv{u_{gc\parallel}}{t} &= q E_{\parallel} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\parallel} B \end{aligned}

The parallel part simply describes an acceleration. The perpendicular part is more interesting: we rewrite it as follows, defining an effective force $$\vb{F}_{\!\perp}$$:

\begin{aligned} m \dv{\vb{u}_{gc\perp}}{t} = \vb{F}_{\!\perp} + q \vb{u}_{gc} \cross \vb{B} \qquad \quad \vb{F}_{\!\perp} \equiv q \vb{E}_\perp + m u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\perp} B \end{aligned}

To solve this, we make a crude approximation now, and improve it later. We thus assume that $$\vb{u}_{gc\perp}$$ is constant in time, such that the equation reduces to:

\begin{aligned} 0 \approx \vb{F}_{\!\perp} + q \vb{u}_{gc} \cross \vb{B} = \vb{F}_{\!\perp} + q \vb{u}_{gc\perp} \cross \vb{B} \end{aligned}

This is analogous to the previous case of a uniform electric field, with $$q \vb{E}$$ replaced by $$\vb{F}_{\!\perp}$$, so it is also solved by crossing with $$\vb{B}$$ in front, yielding a drift:

\begin{aligned} \vb{u}_{gc\perp} \approx \vb{v}_F \equiv \frac{\vb{F}_{\!\perp} \cross \vb{B}}{q B^2} \end{aligned}

From the definition of $$\vb{F}_{\!\perp}$$, this total $$\vb{v}_F$$ can be split into three drifts: the previously seen electric field drift $$\vb{v}_E$$, the curvature drift $$\vb{v}_c$$, and the grad-$$\vb{B}$$ drift $$\vb{v}_{\nabla B}$$:

\begin{aligned} \boxed{ \vb{v}_c = \frac{m u_{gc\parallel}^2}{q} \frac{\vb{R}_c \cross \vb{B}}{R_c^2 B^2} } \qquad \quad \boxed{ \vb{v}_{\nabla B} = \frac{u_L^2}{2 \omega_c} \frac{\vb{B} \cross \nabla B}{B^2} } \end{aligned}

Such that $$\vb{v}_F = \vb{v}_E + \vb{v}_c + \vb{v}_{\nabla B}$$. We are still missing a correction, since we neglected the time dependence of $$\vb{u}_{gc\perp}$$ earlier. This correction is called $$\vb{v}_p$$, where $$\vb{u}_{gc\perp} \approx \vb{v}_F + \vb{v}_p$$. We revisit the perpendicular equation, which now reads:

\begin{aligned} m \dv{t} \big( \vb{v}_F + \vb{v}_p \big) = \vb{F}_{\!\perp} + q \big( \vb{v}_F + \vb{v}_p \big) \cross \vb{B} \end{aligned}

We assume that $$\vb{v}_F$$ varies much faster than $$\vb{v}_p$$, such that $$\dv*{\vb{v}_p}{t}$$ is negligible. In addition, from the derivation of $$\vb{v}_F$$, we know that $$\vb{F}_{\!\perp} + q \vb{v}_F \cross \vb{B} = 0$$, leaving only:

\begin{aligned} m \dv{\vb{v}_F}{t} = q \vb{v}_p \cross \vb{B} \end{aligned}

To isolate this for $$\vb{v}_p$$, we take the cross product with $$\vb{B}$$ in front, like earlier. We thus arrive at the following correction, known as the polarization drift $$\vb{v}_p$$:

\begin{aligned} \boxed{ \vb{v}_p = - \frac{m}{q B^2} \dv{\vb{v}_F}{t} \cross \vb{B} } \end{aligned}

In many cases $$\vb{v}_E$$ dominates $$\vb{v}_F$$, so in some literature $$\vb{v}_p$$ is approximated as follows:

\begin{aligned} \vb{v}_p \approx - \frac{m}{q B^2} \dv{\vb{v}_E}{t} \cross \vb{B} = - \frac{m}{q B^2} \Big( \dv{t} (\vb{E}_\perp \cross \vb{B}) \Big) \cross \vb{B} = - \frac{m}{q B^2} \dv{\vb{E}_\perp}{t} \end{aligned}

The polarization drift stands out from the others: it has the opposite sign, it is proportional to $$m$$, and it is often only temporary. Therefore, it is also called the inertia drift.

## References

1. F.F. Chen, Introduction to plasma physics and controlled fusion, 3rd edition, Springer.
2. M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.