Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 16D (Electromagnetism)

Let $x^{\mu }=(ct,\mathbf{x})$, let $\tau$ be proper time, and define the four-velocity and four-momentum byIf $F^{\mu \nu }$ is the electromagnetic field tensor, the covariant Lorentz force law isThe temporal component and three spatial components respectively giveHere $\mathcal{E}=\gamma m{c}^2$ is the relativistic energy and $\mathbf{p}=\gamma m\mathbf{v}$ is the relativistic momentum. In the nonrelativistic limit, $\gamma \to 1$, so the spatial equation becomesthe usual Lorentz-force law.

The temporal component found in part (a) gives directlyFor a constant electric field, this is the work done by the field along the particle trajectory:

With $\mathbf{B}=0$ and $\mathbf{E}=(0,0,E)$, the spatial Lorentz force equation givesThe relativistic energy-momentum relation therefore yieldsSince $\mathbf{v}=c^2\mathbf{p}/\mathcal{E}$,Integration from the initial position $z(0)=0$ gives

The $x$ component of the velocity isThusSolving this equation for $t$ givesSubstitution into the expression for $z(t)$ and the identity $1+{\sinh }^{2}u={\cosh }^{2}u$ produce the trajectoryThis is relativistic motion in a constant electric field with transverse momentum.

For $t\to +\infty$ with $qE>0$, an asymptotic expansion givesso the longitudinal velocity tends to the speed of light. More generally, its limiting sign is the sign of $qE$.

For small $t$, the taylor series of the square root givesIn the nonrelativistic limit ${\mathcal{E}}_0\simeq m{c}^2$, this becomesThe transverse motion is then uniform:Eliminating $t$ gives the nonrelativistic parabolic trajectory