 Solution

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.