Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 36A (Electrodynamics)

The electromagnetic four-potential is the field generated by the four-current $J^{\mu }$: every source volume element contributes with the inverse-distance Green-function factor. Electromagnetic influence propagates at speed $c$, so causality requires the source to be evaluated where its worldline intersects the field event's past light cone. ThusThe source time is exactly the retarded time $t_{\mathrm{ret}}=t-|x-x^{\prime }|/c$, rather than the observation time $t$.

PutBecause $dR/ds=-n\cdot v$, differentiation givesA massive charge has the subluminal speed $|v|<c$, so $F^{\prime }(s)>0$. Hence $F$ is strictly increasing, and $F(s)=t$ has at most one root. This is the uniqueness of retarded time for a subluminal source.

For a point charge, the charge density and current density areInsert these into the retarded electromagnetic potential. The argument of the Dirac delta function depends on $x^{\prime }$ throughAt its unique zero $x^{\prime }=y(t_{\mathrm{ret}})$, the Jacobian determinant contributes the factorSince the source is subluminal, this quantity is positive. Writing $\mathbf{R}=x-y(t_{\mathrm{ret}})$, $R=|\mathbf{R}|$, and evaluating $\mathbf{v}$ at $t_{\mathrm{ret}}$ gives the Lienard-Wiechert potentialsThe identity ${\mu }_0{ϵ}_0{c}^2=1$ equivalently gives $\mathbf{A}=\mathbf{v}\varphi /c^2$.

Assume that $\mathbf{v}$ is constant. SetImplicit differentiation of $t=\tau +R(\tau )/c$ givesSince $d\mathbf{R}/d\tau =-\mathbf{v}$, direct differentiation of $D$ yieldsThereforeThe Lienard-Wiechert potentials for uniform motion have $\varphi =q/(4\pi {ϵ}_{0}D)$ and $\mathbf{A}=\mathbf{v}\varphi /c^2$. The preceding identity consequently impliesAs $\mathbf{v}$ is constant,Thus the explicit uniformly moving potentials satisfy the Lorenz gauge.