Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 35D / Solution

The retarded potential represents causal propagation: each source element contributes according to its charge density at the earlier time needed for a signal to reach the observer, with inverse-distance weighting. The printed formula uses units with the speed of light equal to one.

Insert a Dirac delta function in time to writeFor the moving point charge, spatial integration leaves $q/(4\pi {ϵ}_0)$ times $\int R(\tau {)}^{-1}\delta (t-\tau -R(\tau ))d\tau$, with $R(\tau )=|x-x_0(\tau )|$. The retarded time ${\tau }_r$ solves $t-{\tau }_r=R({\tau }_r)$. Since $R^{\prime }(\tau )=-v(\tau )\cdot n(\tau )$, the delta-function Jacobian is $1-v\cdot n$, positive for subluminal motion. Therefore the Liénard–Wiechert potential iswhere $R=x-x_0({\tau }_r)$ and $v={\dot{x}}_0({\tau }_r)$. Replacing charge density by the current density $qv{\delta }^{(3)}(x-x_0)$ gives $\boxed{A=v\varphi }$ in these $c=1$ units. Restoring SI units gives denominator $R-v\cdot R/c$ and $A=v\varphi /c^2$, with all source quantities still evaluated at retarded time.