path: root/source/know/concept/guiding-center-theory/index.md

---
title: "Guiding center theory"
sort_title: "Guiding center theory"
date: 2021-09-21
categories:
- Physics
- Electromagnetism
- Plasma physics
layout: "concept"
---

When discussing the [Lorentz force](/know/concept/lorentz-force/),
we introduced the concept of *gyration*:
a particle in a uniform [magnetic field](/know/concept/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](/know/concept/electric-field/).

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:

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


{% include proof/start.html id="proof-averages" -%}
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](/know/concept/maxwells-equations/) $$\nabla \cdot \vb{B} = 0$$
to rewrite the $$z$$-component,
and follow the convention that $$\vb{B}$$
points mostly in the $$z$$-direction,
such that $$B \equiv |\vb{B}| \approx B_z$$:

$$\begin{aligned}
    \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
    &= - \frac{u_L^2}{2 \omega_c}
    \begin{pmatrix}
        \displaystyle \pdv{B_z}{x} \\
        \displaystyle \pdv{B_z}{y} \\
        \displaystyle \pdv{B_z}{z}
    \end{pmatrix}
    \approx - \frac{u_L^2}{2 \omega_c}
    \begin{pmatrix}
        \displaystyle \pdv{B}{x} \\
        \displaystyle \pdv{B}{y} \\
        \displaystyle \pdv{B}{z}
    \end{pmatrix}
    = - \frac{u_L^2}{2 \omega_c} \nabla B
\end{aligned}$$
{% include proof/end.html id="proof-averages" %}


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](/know/concept/curvature/),
and $$\vb{R}_c$$ is the corresponding vector from the center of curvature:

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


{% include proof/start.html id="proof-curvature" -%}
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:

$$\begin{aligned}
    \dv{\vu{b}}{s}
    = - \frac{\vu{R}_c \dd{\theta}}{R_c \dd{\theta}}
    = - \frac{\vu{R}_c}{R_c}
    = - \frac{\vb{R}_c}{R_c^2}
\end{aligned}$$
{% include proof/end.html id="proof-curvature" %}


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.