A gyro-orbit and the E×B drift by the Boris push

A charged particle in a uniform magnetic field circles the field lines at the cyclotron frequency ${\omega }_c=qB/m$. This circling is gyration, and its circle is the gyro-orbit. Add a uniform electric field at right angles to $\mathbf{B}$ and the center of the circle, the guiding center, moves at the $E\times B$ drift velocity $\mathbf{E}\times \mathbf{B}/B^2$, whatever the particle's charge or mass.

A particle-in-cell (PIC) code moves millions of particles through these fields for millions of steps, so the step it uses must not add or remove energy on its own. The Boris push (J. P. Boris, 1970) is the standard choice. It splits the step into half an electric kick, a rotation about $\mathbf{B}$, and the other half kick. This notebook runs it with $q=m=B=1$, so ${\omega }_c=1$ and the exact gyroperiod is $2\pi$, and checks each property against a closed form. posts/refs/boris_push.py computes the same numbers with a separate numpy implementation.

Gyration keeps the speed and the circle

With no electric field the two kicks vanish and the step is a pure rotation of $\mathbf{v}$, so $|\mathbf{v}|$ cannot change except by rounding. The cell runs 20000 steps from $\mathbf{v}=(0,1)$ and records the largest and smallest speed.

The positions are updated with the new velocity, $x_{n+1}=x_n+\Delta t{v}_{n+1}$. Each step turns $\mathbf{v}$ by the same angle $\theta$, so each position step is the previous one turned by $\theta$: the positions are the corners of a regular polygon and all lie on one circle. The cell takes the circle through the first three positions and measures every other position's distance from its center. Its radius is $\Delta t/(2\sin (\theta /2))$, which is $\sqrt{1+({\omega }_c\Delta t{)}^2/4}v/{\omega }_c=1.000312$, not the exact $v/{\omega }_c=1$.

The period is too long by $({\omega }_c\Delta t{)}^2/12$

The rotation in the step turns $\mathbf{v}$ by $\theta =2\arctan ({\omega }_c\Delta t/2)$, a little less than the exact ${\omega }_c\Delta t$. The cell above summed the angle between successive velocities; dividing by the number of steps gives $\theta$, and the gyroperiod is $T=2\pi \Delta t/\theta$. Expanding the arctangent,

At $\Delta t=0.05$ that is $T=6.28449$ against $2\pi =6.28319$. The speed is exact and the radius is off by $3\times 10^{-4}$; the phase falls behind by $2\times 10^{-4}$ of a turn per turn. The figure repeats the measurement for five time steps. At $\Delta t=0.4$ the $\Delta t^4$ term shows: the error is $1.3194\times 10^{-2}$, below $0.4^2/12=1.3333\times 10^{-2}$.

The $E\times B$ drift in crossed fields

Now add $\mathbf{E}=0.1\hat{x}$. The drift velocity is ${\mathbf{v}}_E=\mathbf{E}\times \mathbf{B}/B^2=(0,-0.1)$. Write one Boris step as $\mathbf{v}\mapsto R(\mathbf{v}+\mathbf{a})+\mathbf{a}$, with $\mathbf{a}=\mathbf{E}\Delta t/2$ and $R$ the rotation. The vector ${\mathbf{v}}_E+\mathbf{a}=(0.0025,-0.1)$ is turned by $R$ into $(-0.0025,-0.1)={\mathbf{v}}_E-\mathbf{a}$, because its half-angle from $-\hat{y}$ is $\arctan (0.0025/0.1)=\arctan (\Delta t/2)=\theta /2$. So ${\mathbf{v}}_E$ is a fixed point of the step: a particle started there never gyrates.

For any other start, $\mathbf{u}=\mathbf{v}-{\mathbf{v}}_E$ is only rotated, so the particle gyrates about a guiding center moving at ${\mathbf{v}}_E$. Averaged over $N$ steps the velocity differs from ${\mathbf{v}}_E$ by the unfinished part of a turn, $|{\mathbf{u}}_0||\sin (N\theta /2)|/(N\sin (\theta /2))$, which is $8.2\times 10^{-4}$ here and falls as $1/N$.