A particle of charge of $q$ moves along a trajectory $y^{a}(s)$ in spacetime where $s$ is the proper time on the particle's world-line.

Explain briefly why, in the gauge $\partial_{a} A^{a}=0$, the potential at the spacetime point $x$ is given by

$$A^{a}(x)=\frac{\mu_{0} q}{2 \pi} \int d s \frac{d y^{a}}{d s} \theta\left(x^{0}-y^{0}(s)\right) \delta\left(\left(x^{c}-y^{c}(s)\right)\left(x^{d}-y^{d}(s)\right) \eta_{c d}\right)$$

Evaluate this integral for a point charge moving relativistically along the $z$-axis, $x=y=0$, at constant velocity $v$ so that $z=v t .$

Check your result by starting from the potential of a point charge at rest

$$\begin{aligned} \mathbf{A} &=0 \\ \phi &=\frac{\mu_{0} q}{4 \pi\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}} \end{aligned}$$

and making an appropriate Lorentz transformation.