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Write the electromagnetic field tensor as $F^{\mu \nu }={\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu }$. Substitution into the Covariant Maxwell equation with the minus-plus-plus-plus metric givesChoose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. Since ${\partial }_{\mu }{\partial }^{\mu }={\nabla }^2-c^{-2}{\partial }_t^2$, the electromagnetic four-potential obeysConvolution with the outgoing retarded Green function gives the retarded electromagnetic potential

For the point particle, ${\tau }_{\ast }(x)$ is its unique retarded proper time, characterized byThus $y({\tau }_{\ast })$ is where the particle's worldline intersects the past light cone of the observation event $x$; the displayed expression is the covariant form of the Lienard-Wiechert potentials.

On the $z$-axis put $D=\sqrt{R^2+z^2}$. The retarded time is $t_{\ast }=t-D/c$. For the circular orbit, the spatial velocity at that time isBecause the displacement from the retarded source point to the observation point is orthogonal to this velocity, $|R^{\nu }{\dot{y}}_{\nu }|=\gamma cD$, while ${\dot{y}}^{\mu }=\gamma (c,\mathbf{v})$. The factors of $\gamma$ cancel, leaving

The outgoing Green function of the electromagnetic wave equation has support on the past light cone. Convolution with that retarded Green function yields the retarded electromagnetic potentialwhere the source is evaluated at the retarded timeThis is the causal solution with no incoming radiation.

Let $R=|\mathbf{x}|$, $\hat{\mathbf{R}}=\mathbf{x}/R$, and $t_R=t-R/c$. To leading nonzero multipole expansion order, the loop is a time-dependent magnetic dipole withExpanding the retarded electromagnetic potential across the small loop givesThe $R^{-2}$ term is the near magnetic-dipole field and the $R^{-1}$ term is the magnetic dipole radiation field. Substituting the Fourier series for the current givesThe $n=0$ summand vanishes for the stated sine series.

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 $\mathbf{A}=\mathbf{v}\varphi /c^2$.