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 $m \: \idv{\vb{u}_L}{t} = q \vb{u}_L \cross \vb{B}$, 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}

We know what $\vb{x}_L$ is, so we can write out $(\vb{x}_L \cdot \nabla) \vb{B}$ for $\vb{B} = (B_x, B_y, B_z)$:

\begin{aligned} (\vb{x}_L \cdot \nabla) \vb{B} = \frac{u_L}{\omega_c} \begin{pmatrix} \displaystyle \sin(\omega_c t) \pdv{B_x}{x} + \cos(\omega_c t) \pdv{B_x}{y} \\ \displaystyle \sin(\omega_c t) \pdv{B_y}{x} + \cos(\omega_c t) \pdv{B_y}{y} \\ \displaystyle \sin(\omega_c t) \pdv{B_z}{x} + \cos(\omega_c t) \pdv{B_z}{y} \end{pmatrix} \end{aligned}

Integrating $\sin$ and $\cos$ over their period yields zero, so the average vanishes:

\begin{aligned} \Expval{ (\vb{x}_L \cdot \nabla) \vb{B} } = 0 \end{aligned}

Moving on, we write out $\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$, suppressing the arguments of $\sin$ and $\cos$:

\begin{aligned} \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} &= \frac{u_L^2}{\omega_c} \begin{pmatrix} \cos \\ - \sin \\ 0 \end{pmatrix} \cross \begin{pmatrix} \displaystyle \pdv{B_x}{x} \sin + \pdv{B_x}{y} \cos \\ \displaystyle \pdv{B_y}{x} \sin + \pdv{B_y}{y} \cos \\ \displaystyle \pdv{B_z}{x} \sin + \pdv{B_z}{y} \cos \end{pmatrix} \\ &= \frac{u_L^2}{\omega_c} \begin{pmatrix} \displaystyle - \pdv{B_z}{x} \sin^2 - \pdv{B_z}{y} \sin \cos \\ \displaystyle - \pdv{B_z}{x} \sin \cos - \pdv{B_z}{y} \cos^2 \\ \displaystyle \pdv{B_y}{x} \sin \cos + \pdv{B_y}{y} \cos^2 \displaystyle + \pdv{B_x}{x} \sin^2 + \pdv{B_x}{y} \sin \cos \end{pmatrix} \end{aligned}

Integrating products of $\sin$ and $\cos$ over their period gives us the following:

\begin{aligned} \Expval{\cos^2} = \Expval{\sin^2} = \frac{1}{2} \qquad \quad \Expval{\sin \cos} = 0 \end{aligned}

Inserting this tells us that the average of $\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$ is given by:

\begin{aligned} \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} } &= \frac{u_L^2}{2 \omega_c} \begin{pmatrix} \displaystyle - \pdv{B_z}{x} \\ \displaystyle - \pdv{B_z}{y} \\ \displaystyle \pdv{B_y}{y} + \pdv{B_x}{x} \end{pmatrix} \end{aligned}

We use Maxwell’s equation $\nabla \cdot \vb{B} = 0$ to rewrite the $z$-component, and follow the convention that $\vb{B}$ points mostly in the $z$-direction, such that $B \equiv |\vb{B}| \approx B_z$:

\begin{aligned} \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} } &= - \frac{u_L^2}{2 \omega_c} \begin{pmatrix} \displaystyle \pdv{B_z}{x} \\ \displaystyle \pdv{B_z}{y} \\ \displaystyle \pdv{B_z}{z} \end{pmatrix} \approx - \frac{u_L^2}{2 \omega_c} \begin{pmatrix} \displaystyle \pdv{B}{x} \\ \displaystyle \pdv{B}{y} \\ \displaystyle \pdv{B}{z} \end{pmatrix} = - \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}

Assuming that $\vu{b}$ does not explicitly depend on time, i.e. $\ipdv{\vu{b}}{t} = 0$, we can rewrite the derivative using the chain rule:

\begin{aligned} \dv{\vu{b}}{t} = \pdv{\vu{b}}{s} \dv{s}{t} = u_{gc\parallel} \dv{\vu{b}}{s} \end{aligned}

Where $\dd{s}$ is the arc length of the magnetic field line, which is equal to the radius $R_c$ times the infinitesimal subtended angle $\dd{\theta}$:

\begin{aligned} \dd{s} = R_c \dd{\theta} \end{aligned}

Meanwhile, across this arc, $\vu{b}$ rotates by $\dd{\theta}$, such that the tip travels a distance $|\dd{\vu{b}}|$:

\begin{aligned} |\!\dd{\vu{b}}\!| = |\vu{b}| \dd{\theta} = \dd{\theta} \end{aligned}

Furthermore, the direction $\dd{\vu{b}}$ is always opposite to $\vu{R}_c$, which is defined as the unit vector from the center of curvature to the base of $\vu{b}$:

\begin{aligned} \dd{\vu{b}} = - \vu{R}_c \dd{\theta} \end{aligned}

Combining these expressions for $\dd{s}$ and $\dd{\vu{b}}$, we find the following derivative:

\begin{aligned} \dv{\vu{b}}{s} = - \frac{\vu{R}_c \dd{\theta}}{R_c \dd{\theta}} = - \frac{\vu{R}_c}{R_c} = - \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 $\idv{}{\vb{v}p}{t}$ 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.