= Solution
In the magnetic-dominated case the suitable <Lorentz boost> instead has
$$
\boxed{\mathbf w=\frac{\mathbf E\times\mathbf B}{B^2}
=\frac EB\hat{\mathbf z},\qquad
\mathbf E'=0,\qquad B'_y=B\sqrt{1-E^2/(c^2B^2)}.}
$$
The hypothesis gives $|w|<c$. The <electromagnetic field invariants> now have $E^2-c^2B^2<0$, allowing a purely magnetic <inertial frame>. The initial <velocity> is again $-w\hat{\mathbf z}$, perpendicular to $\mathbf B'$. Since a <magnetic field> does no <work>, the <electron>'s <relativistic energy> and <speed> in this <inertial frame> remain constant. Its trajectory is a circle in the $x'z'$ plane with <Lorentz factor> $\gamma_w$ and <cyclotron frequency> $\Omega=eB'/(\gamma_wm)$.
For $E,B>0$, set $\theta=\Omega t'$ and $\rho=w/\Omega$. A parametrization with the correct initial <velocity> and <Lorentz force> is
$$
x'=\rho(\cos\theta-1),\qquad z'=-\rho\sin\theta,\qquad
u'_x=-w\sin\theta,\quad u'_z=-w\cos\theta.
$$
Transforming the <world line> back to the laboratory gives
$$
\boxed{x=\rho(\cos\theta-1),\qquad
z=\gamma_w\rho(\theta-\sin\theta),\qquad
t=\frac{\gamma_w}{\Omega}(\theta-\beta_w^2\sin\theta).}
$$
The laboratory trajectory is a stretched <cycloid>, with cusps at $\theta=2\pi n$ where the <electron> is momentarily at rest. It first accelerates toward $-x$, then curves toward $+z$, returning to its original $x$ coordinate each cycle while advancing in $z$. Its mean drift is $\Delta z/\Delta t=w=E/B$, the <E-cross-B drift>. Its laboratory <kinetic energy> oscillates as the <electric field> does positive and then negative <work>; it does not grow indefinitely. The strict inequalities in the two branches exclude the null case $|E/B|=c$, where neither field can be removed by a finite <Lorentz boost>.
Back to article page
