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}

Proof

Proof. 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}

Proof

Proof. 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

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M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.