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 B\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 F\vb{F} with Newton’s second law:

F=mdudt=q(E+u×B)\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 δE\delta\vb{E} and δB\delta\vb{B}:

EE+δE(x,t)BB+δB(x,t)\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 u\vb{u} can be split into the guiding center’s motion ugc\vb{u}_{gc} and the known Larmor gyration uL\vb{u}_L around the guiding center, such that u=ugc+uL\vb{u} = \vb{u}_{gc} + \vb{u}_L. Inserting:

mddt(ugc+uL)=q(E+δE+(ugc+uL)×(B+δB))\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 mduL/dt=quL×Bm \: \idv{\vb{u}_L}{t} = q \vb{u}_L \cross \vb{B}, which we subtract from the total to get:

mdugcdt=q(E+δE+ugc×(B+δB)+uL×δB)\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 f\Expval{f} of a function ff over a single gyroperiod, where ωc\omega_c is the cyclotron frequency:

f02π/ωcf(t)dt\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 E\vb{E} and B\vb{B} are both uniform, such that δB=0\delta\vb{B} = 0 and δE=0\delta\vb{E} = 0:

mdugcdt=q(E+ugc×B)\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 b^B/B\vu{b} \equiv \vb{B} / |\vb{B}| makes all components perpendicular to B\vb{B} vanish, including the cross product, leaving only the (scalar) parallel components ugcu_{gc\parallel} and EE_\parallel:

mdugcdt=qmE\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 ugc\vb{u}_{gc\perp} is found by subtracting ugcu_{gc\parallel}’s equation from the original:

mdugcdt=q(E+ugc×B)qEb^=q(E+ugc×B)\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 ugc\vb{u}_{gc\perp} explicitly excludes gyration. If we try to split ugc\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 ugc\vb{u}_{gc\perp} does not depend on time. Therefore:

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

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

0=B×(E+ugc×B)=B×E+ugcB2\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 ugc\vb{u}_{gc\perp} is constant. The guiding center drifts sideways at this speed, hence it is called a drift velocity vE\vb{v}_E. Curiously, vE\vb{v}_E is independent of qq:

vE=E×BB2\begin{aligned} \boxed{ \vb{v}_E = \frac{\vb{E} \cross \vb{B}}{B^2} } \end{aligned}

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

vF=F×BqB2\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 B\vb{B} is non-uniform, but E\vb{E} is still uniform:

mdugcdt=q(E+ugc×(B+δB)+uL×δB)\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 rLr_L is small compared to the variation of B\vb{B}, we set δB\delta\vb{B} to the first-order term of a Taylor expansion of B\vb{B} around xgc\vb{x}_{gc}, that is, δB=(xL)B\delta\vb{B} = (\vb{x}_L \cdot \nabla) \vb{B}. We thus have:

mdugcdt=q(E+ugc×B+ugc×(xL)B+uL×(xL)B)\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:

mdugcdt=q(E+ugc×B+ugc×(xL)B+uL×(xL)B)\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 ugc=ugc\Expval{\vb{u}_{gc}} = \vb{u}_{gc}. The two averaged expressions turn out to be:

(xL)B=0uL×(xL)BuL22ωcB\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 xL\vb{x}_L is, so we can write out (xL)B(\vb{x}_L \cdot \nabla) \vb{B} for B=(Bx,By,Bz)\vb{B} = (B_x, B_y, B_z):

(xL)B=uLωc(sin(ωct)Bxx+cos(ωct)Bxysin(ωct)Byx+cos(ωct)Byysin(ωct)Bzx+cos(ωct)Bzy)\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\sin and cos\cos over their period yields zero, so the average vanishes:

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

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

uL×(xL)B=uL2ωc(cossin0)×(Bxxsin+BxycosByxsin+ByycosBzxsin+Bzycos)=uL2ωc(Bzxsin2BzysincosBzxsincosBzycos2Byxsincos+Byycos2+Bxxsin2+Bxysincos)\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\sin and cos\cos over their period gives us the following:

cos2=sin2=12sincos=0\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 uL×(xL)B\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} is given by:

uL×(xL)B=uL22ωc(BzxBzyByy+Bxx)\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 B=0\nabla \cdot \vb{B} = 0 to rewrite the zz-component, and follow the convention that B\vb{B} points mostly in the zz-direction, such that BBBzB \equiv |\vb{B}| \approx B_z:

uL×(xL)B=uL22ωc(BzxBzyBzz)uL22ωc(BxByBz)=uL22ωcB\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:

mdugcdt=q(E+ugc×BuL22ωcB)\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 ugc\vb{u}_{gc} into components ugc\vb{u}_{gc\perp} and ugcb^u_{gc\parallel} \vu{b}, which are respectively perpendicular and parallel to the magnetic unit vector b^\vu{b}, such that ugc=ugc ⁣+ ⁣ugcb^\vb{u}_{gc} = \vb{u}_{gc\perp} \!+\! u_{gc\parallel} \vu{b}. Consequently:

dugcdt=dugcdt+dugcdtb^+ugcdb^dt\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:

dugcdt=m(dugcdt+dugcdtb^+ugcdb^dt)=q(E+ugc×BuL22ωcB)\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 b^\vu{b} can be rewritten as follows, where RcR_c is the radius of the field’s curvature, and Rc\vb{R}_c is the corresponding vector from the center of curvature:

db^dtugcRcRc2\begin{aligned} \dv{\vu{b}}{t} \approx - u_{gc\parallel} \frac{\vb{R}_c}{R_c^2} \end{aligned}

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

db^dt=b^sdsdt=ugcdb^ds\begin{aligned} \dv{\vu{b}}{t} = \pdv{\vu{b}}{s} \dv{s}{t} = u_{gc\parallel} \dv{\vu{b}}{s} \end{aligned}

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

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

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

 ⁣db^ ⁣=b^dθ=dθ\begin{aligned} |\!\dd{\vu{b}}\!| = |\vu{b}| \dd{\theta} = \dd{\theta} \end{aligned}

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

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

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

db^ds=R^cdθRcdθ=R^cRc=RcRc2\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:

m(dugcdt+dugcdtb^ugc2RcRc)=q(E+ugc×BuL22ωcB)\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 Rc\vb{R}_c and any cross product with B\vb{B} will always be perpendicular to B\vb{B}, we can split this equation into perpendicular and parallel components like so:

mdugcdt=qEquL22ωc ⁣B+mugc2RcRc+qugc×Bmdugcdt=qEquL22ωc ⁣B\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 F ⁣\vb{F}_{\!\perp}:

mdugcdt=F ⁣+qugc×BF ⁣qE+mugc2RcRcquL22ωc ⁣B\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 ugc\vb{u}_{gc\perp} is constant in time, such that the equation reduces to:

0F ⁣+qugc×B=F ⁣+qugc×B\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 qEq \vb{E} replaced by F ⁣\vb{F}_{\!\perp}, so it is also solved by crossing with B\vb{B} in front, yielding a drift:

ugcvFF ⁣×BqB2\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 F ⁣\vb{F}_{\!\perp}, this total vF\vb{v}_F can be split into three drifts: the previously seen electric field drift vE\vb{v}_E, the curvature drift vc\vb{v}_c, and the grad-B\vb{B} drift vB\vb{v}_{\nabla B}:

vc=mugc2qRc×BRc2B2vB=uL22ωcB×BB2\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 vF=vE+vc+vB\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 ugc\vb{u}_{gc\perp} earlier. This correction is called vp\vb{v}_p, where ugcvF+vp\vb{u}_{gc\perp} \approx \vb{v}_F + \vb{v}_p. We revisit the perpendicular equation, which now reads:

mddt(vF+vp)=F ⁣+q(vF+vp)×B\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 vF\vb{v}_F varies much faster than vp\vb{v}_p, such that d/dvpt\idv{}{\vb{v}p}{t} is negligible. In addition, from the derivation of vF\vb{v}_F, we know that F ⁣+qvF×B=0\vb{F}_{\!\perp} + q \vb{v}_F \cross \vb{B} = 0, leaving only:

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

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

vp=mqB2dvFdt×B\begin{aligned} \boxed{ \vb{v}_p = - \frac{m}{q B^2} \dv{\vb{v}_F}{t} \cross \vb{B} } \end{aligned}

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

vpmqB2dvEdt×B=mqB2(ddt(E×B))×B=mqB2dEdt\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 mm, and it is often only temporary. Therefore, it is also called the inertia drift.


  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.