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Use the Minkowski metric with signature $(-,+,+,+)$. The source point is the intersection of the worldline with the observation event's past light cone: $R^{\mu }{R}_{\mu }=0$ and $R^0>0$. In components this is $c(t-t_r)=|x-y(t_r)|$ with $t_r<t$. For a timelike worldline the retarded solution is unique in the usual complete-source setting.

Writing $R=|x-y(t_r)|$, $n=(x-y)/R$ and $u^{\mu }=\gamma (c,v)$ gives $R\cdot u=-\gamma cR(1-n\cdot v/c)$. Thus the Lienard-Wiechert potentials are $\varphi =q{\mu }_0{c}^2/[4\pi R(1-n\cdot v/c)]$ and $A=q{\mu }_{0}v/[4\pi R(1-n\cdot v/c)]$. At leading order in $1/r$ and first order in $v/c$, set $R\simeq r,n\simeq \hat{r}$ and expand the denominator:

Now use $t_{-}=t-r/c$ and let $a=\dot{v}(t_{-})$. Derivatives of $1/r$ and $\hat{r}$ give near-field terms of order $r^{-2}$; the radiation terms come from ${\partial }_t{t}_{-}=1$ and $\nabla t_{-}=-\hat{r}/c$. The electric field and magnetic field are thereforeThey are transverse to the observation direction. The Poynting vector isIntegrating over a sphere uses $\int {\sin }^2\theta d\Omega =8\pi /3$, yielding the nonrelativistic Larmor formula $P=q^2{\mu }_0|a{|}^2/(6\pi c)$. For simple harmonic motion, $|a{|}^2=A^2{\omega }^4{\cos }^2(\omega t_{-})$, whose period average is $A^2{\omega }^4/2$. Multiplication by the period $2\pi /\omega$ givesThis is the leading radiation energy in the nonrelativistic and far-field approximation.

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

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.