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The moving point charge has distributional charge density and current densityLetThe given retarded fundamental solution identity immediately implies that the retarded potentialssolve $\square \varphi =-\rho /{ϵ}_0$ and $\square \mathbf{A}=-{\mu }_0\mathbf{J}$ in the sense of distributions. Integrating the spatial Dirac delta function first givesand the analogous integral for $\mathbf{A}$ has numerator ${\mu }_{0}q{\mathbf{v}}_0(t^{\prime })/(4\pi )$.

Assume the physical trajectory has subluminal speed and the retarded time exists. At its root $t_r$, with ${\mathbf{n}}_r=\mathbf{R}(t_r)/R(t_r)$,The positive factor $\kappa =1-{\mathbf{n}}_r\cdot {\mathbf{v}}_0/c$ also proves uniqueness of retarded time for a subluminal source. The Dirac delta function substitution formula gives $\delta (g(t^{\prime }))=\delta (t^{\prime }-t_r)/\kappa$, soThese are the Lienard-Wiechert potentials. The denominator includes the time-substitution factor, rather than just the retarded distance. Away from the trajectory the electric potential and magnetic vector potential are ordinary smooth functions; on the trajectory their wave equations have the indicated distributional sources.

For completeness, the charge continuity equation ${\partial }_{t^{\prime }}\rho +{\nabla }^{\prime }\cdot \mathbf{J}=0$ also verifies the Lorenz gauge. Using $\nabla K=-{\nabla }^{\prime }K$ and ${\partial }_{t}K=-{\partial }_{t^{\prime }}K$, integration by parts in the retarded integrals giveswith the retarded boundary prescription. Thus the construction also has the required Lorenz gauge condition.

With Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$, the relativistic momentum and four-momentum areThe Newtonian formula $a=F/m$ fails because $p$ is not $mv$: both the magnitude and direction of $v$ affect $dp/dt$. Sincethe relativistic force isTaking the dot product with $v$ gives $F\cdot v=m{\gamma }^{3}v\cdot a$, and substitution yields the inverse relationThis is the requested sum of a vector parallel to $F$ and one parallel to $v$.

In the constant electric field, $dp/dt=qE$ and the particle starts with $p(0)=0$, so $p=qEt$. The relativistic energy-momentum relation impliesThus the speed increases monotonically but remains subluminal, and $v$ tends to $c$ in the direction of $qE$ as $t\to \infty$.

PutBecause $dR/ds=-n\cdot v$, differentiation givesA massive charge has the subluminal speed $|v|<c$, so $F^{\prime }(s)>0$. Hence $F$ is strictly increasing, and $F(s)=t$ has at most one root. This is the uniqueness of retarded time for a subluminal source.

For $v=|\mathbf{v}|>0$, decompose the velocity as $\mathbf{u}={\mathbf{u}}_{\parallel }+{\mathbf{u}}_{\perp }$, parallel and perpendicular to $\mathbf{v}$. A rotation-free Lorentz boost gives the velocity-addition formulaThese two components are orthogonal, soThe last equality is Lagrange's identity for the cross product. For physical subluminal speeds, $D>0$, givingThe $\mathbf{v}=0$ case follows directly or by continuity. An independent form useful for checks is $1-|\mathbf{w}{|}^2/c^2=(1-u^2/c^2)(1-v^2/c^2)/D^2$, obtained from the transformed four-velocity normalization.

A timelike curve has a nonzero tangent vector with negative Minkowski norm in metric signature $(-,+,+,+)$ at every regular point. A physical massive observer's world line is also future-directed. Parametrizing it by coordinate time gives $-c^2+|\mathbf{v}{|}^2<0$, or $|\mathbf{v}|<c$: the local tangent condition is exactly the subluminal speed condition. It is stronger than checking only that two endpoints have timelike separation.