Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 37B (Electrodynamics)

Assume $qE>0$ and putThe nonzero four-velocity equations arewith $u^0(0)=c$ and $u^x(0)=0$. HenceIntegrating from the origin gives the uniformly accelerated trajectory

The light ray has $x_L=-h+ct$. At interception, $x_L=x$, soA finite solution exists exactly whenThenand, writing $H=\alpha h/c$,At $h=h_c$ interception is approached only as $t\to \infty$, and for $h>h_c$ it never occurs. The limiting backward light ray is the Rindler horizon of the accelerated particle.

Again let $\alpha =qE/(mc)$. For $\mathbf{B}=(0,0,E/c)$, the Lorentz-force equations areThe sum $u^0+u^y$ is constant and initially equals $c$. ThereforeA further integration givesEliminating $\tau$ between $x$ and $y$ yieldsa semicubical parabola, with the physical branch selected by $y\le 0$ for the signs shown.