Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 36C / Solution

Use $c=1$ and metric signature $(-,+,+,+)$, consistent with the supplied rest potential. In the Lorenz gauge, the Maxwell equations reduce to wave equations for the four-potential. A point worldline has current $j^a(x)=q\int u^a(s){\delta }^{(4)}(x-y(s))ds$, where $u^a=d{y}^a/ds$. Convolution with the retarded wave Green function $\theta (R^0)\delta (R^2)/(2\pi )$ gives the stated integral: its support is the future light cone of each source point and it excludes incoming advanced signals.

For uniform motion $y^a(s)=(\gamma s,0,0,\gamma vs)$, $u^a=\gamma (1,0,0,v)$. Let $R=x-y(s)$. The delta-function argument has derivative $d(R^2)/ds=-2R\cdot u$. Only the retarded root survives the step function, soPut $Z=z-vt$, $r_{\perp }^2=x^2+y^2$, and $\Delta =t-t_{\mathrm{ret}}>0$. The light-cone condition is ${\Delta }^2=r_{\perp }^2+(Z+v\Delta {)}^2$, whenceTherefore, away from the worldline,As an independent check, in the rest frame $z^{\prime }=\gamma (z-vt)$ and $r^{\prime }=\sqrt{x^2+y^2+{\gamma }^2{Z}^2}$. Its potential is $A^{\prime a}=({\mu }_{0}q/(4\pi r^{\prime }),0,0,0)$. A Lorentz transformation gives $A^0=\gamma A^{\prime 0}$, $A^z=\gamma v{A}^{\prime 0}$, exactly the expression above because $r^{\prime }=\gamma \sqrt{Z^2+(1-v^2)r_{\perp }^2}$. Restoring SI units would require restoring the factors of $c$ in both the four-potential convention and the supplied electrostatic normalization.