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Use the sign $s=\pm 1$ to label a point charge $sq$ at $s\mathbf{p}/(2q)$, with velocity $s{\mathbf{p}}^{\prime }/(2q)$. All quantities on the right below are evaluated at $t_0=t-r/c$. To first order,The denominator in each Liénard–Wiechert potential therefore expands assoIn the sum of the two electric potentials, the $q/r$ terms cancel and the finite terms add. This gives the retarded potentials of a point electric dipole:For the magnetic vector potential, the signed electric charge and signed velocity multiply to give the same sign for both contributions:Since ${\mu }_0{ϵ}_0{c}^2=1$,The ${\mathbf{p}}^{\prime }$ term in the electric potential arises from the moving-source denominator and remains finite in the point-electric dipole limit; discarding it would fail to give the full retarded electric field.

First retain the question's units $c=1$. Insert the retarded time Dirac delta:For a point charge, the spatial delta sets $x^{\prime }=x_0(t^{\prime })$. Put $R(t^{\prime })=|x-x_0(t^{\prime })|$, $\hat{R}=(x-x_0)/R$ and $v={\dot{x}}_0(t^{\prime })$. The retarded root $t_{\ast }$ satisfies $t-t_{\ast }=R(t_{\ast })$. The derivative of the delta's argument is $-1+\hat{R}\cdot v$, so for a subluminal trajectory its absolute value is $1-\hat{R}\cdot v$. Performing the time integral givesAll quantities in the denominator are evaluated at the retarded time. The retarded vector potential uses current $j=qv{\delta }^{(3)}(x^{\prime }-x_0)$ and the same Jacobian determinant, givingRestoring physical $c$, the root is $t-t_{\ast }=R_{\ast }/c$, both denominators become $R_{\ast }-v_{\ast }\cdot {\mathbf{R}}_{\ast }/c$, and ${\mu }_0=1/({ϵ}_0{c}^2)$. Thus $\mathbf{A}=v_{\ast }\varphi /c^2$. This is the Liénard–Wiechert potential; the Jacobian determinant, rather than merely substituting retarded position into a static Coulomb formula, accounts for the velocity factor.

The retarded potential represents causal propagation: each source element contributes according to its charge density at the earlier time needed for a signal to reach the observer, with inverse-distance weighting. The printed formula uses units with the speed of light equal to one.

Insert a Dirac delta function in time to writeFor the moving point charge, spatial integration leaves $q/(4\pi {ϵ}_0)$ times $\int R(\tau {)}^{-1}\delta (t-\tau -R(\tau ))d\tau$, with $R(\tau )=|x-x_0(\tau )|$. The retarded time ${\tau }_r$ solves $t-{\tau }_r=R({\tau }_r)$. Since $R^{\prime }(\tau )=-v(\tau )\cdot n(\tau )$, the delta-function Jacobian is $1-v\cdot n$, positive for subluminal motion. Therefore the Liénard–Wiechert potential iswhere $R=x-x_0({\tau }_r)$ and $v={\dot{x}}_0({\tau }_r)$. Replacing charge density by the current density $qv{\delta }^{(3)}(x-x_0)$ gives $\boxed{A=v\varphi }$ in these $c=1$ units. Restoring SI units gives denominator $R-v\cdot R/c$ and $A=v\varphi /c^2$, with all source quantities still evaluated at retarded time.

The consistent Liénard–Wiechert potential obeys $\mathbf{A}=\mathbf{u}\phi /c^2$. To first order in $\mathbf{\beta }=\mathbf{u}/c$, its large-distance form isTerms varying as $1/r^2$ give near fields and are negligible for the radiated power at large $r$. Keep the first-order velocity term in $\phi$: although small, its retarded-time derivative contributes a leading acceleration field. To the required order, $\partial t_r/\partial t\simeq 1$ and $\nabla t_r\simeq -\mathbf{n}/c$. Holding the observation direction fixed at leading order givesThus the outgoing electric field is transverse:Here $\mathbf{a}=d\mathbf{u}/d{t}_r$. The magnetic field follows from $\nabla \times \mathbf{A}$, or from the outgoing plane electromagnetic wave relation:The minus sign is essential. Both fields fall as $1/r$, are mutually orthogonal and obey $|\mathbf{E}|=c|\mathbf{B}|$. The corresponding Liénard-Wiechert field also contains nonradiative terms, omitted in this leading far-field calculation.

Let $z^{\mu }(\tau )$ be the point electric charge's world line, with proper time $\tau$ and four-velocity $U^{\mu }=d{z}^{\mu }/d\tau =\gamma (c,\mathbf{u})$. For the field event $x^{\mu }$, choose the retarded source event: $R^{\mu }=x^{\mu }-z^{\mu }({\tau }_{\mathrm{ret}})$ is future-directed and null. Write $R^{\mu }=(R,R\mathbf{n})$, where $R=|\mathbf{x}-\mathbf{z}(t_{\mathrm{ret}})|=c(t-t_{\mathrm{ret}})$. In the instantaneous rest inertial frame of that source event,Consequently the covariant expression matching this Coulomb potential isThe denominator is a Lorentz scalar, so the right-hand side is a four-vector with the required instantaneous-rest components. This describes the retarded potential in Lorenz gauge, with no added free electromagnetic wave. Lorentz covariance alone does not establish the boundary conditions or exclude additional terms depending on four-acceleration. The physical input is the retarded point-source solution of Maxwell equations: the potential at the retarded event depends on the instantaneous four-current, while its derivatives generate the acceleration field. Equivalently, evaluating the retarded source integral gives a Jacobian $1/(1-\mathbf{n}\cdot \mathbf{\beta })$ from the light-cone constraint; this fixes that input for arbitrary smooth subluminal motion.^[ref]

In any inertial frame, put $\mathbf{\beta }=\mathbf{u}/c$ and $\kappa =1-\mathbf{n}\cdot \mathbf{\beta }$. Then $-R\cdot U=\gamma cR\kappa$, and the Liénard–Wiechert potential isDifferentiate using $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$, including the position and time dependence of the retarded time. For example, $\partial t_{\mathrm{ret}}/\partial t=1/\kappa$ and $\nabla t_{\mathrm{ret}}=-\mathbf{n}/(c\kappa )$. The resulting Liénard-Wiechert field isHere the dot denotes a derivative with respect to source coordinate time. The first term is the velocity or non-radiative field: it falls as $R^{-2}$ and exists even without acceleration. The second is the acceleration or radiation field, falling as $R^{-1}$ and transverse to $\mathbf{n}$. Its Poynting flux falls as $R^{-2}$, so its outward power integrated over a large sphere stays finite as its radius tends to infinity. The first term's squared-field flux instead vanishes in that limit. This large-distance distinction defines radiation; a moving near field can carry local energy without constituting radiation escaping to infinity.

The past light cone of a spacetime event $O$ is the past-directed part of its light cone. In Minkowski spacetime, its spacetime events satisfy $|\Delta \mathbf{x}|=-c\Delta t$ with $\Delta t<0$. They are the possible emission events of light rays arriving at $O$. The interior also contains events that can send slower signals to $O$. The retarded potential of a moving point electric charge, given by the Liénard–Wiechert potential, uses the emission event where the source's world line intersects this boundary.

An ideal electric dipole at the origin with smooth electric dipole moment $\mathbf{p}(t)$ has Lorenz gauge electric potential and magnetic vector potential, for $r>0$,For a short proof, place electric charges $sq$ at $s\mathbf{p}/(2q)$, $s=\pm 1$, and let $q\to \infty$ with $\mathbf{p}$ fixed. The corresponding retarded times are $t_0+s\mathbf{n}\cdot \mathbf{p}(t_0)/(2qc)+O(q^{-2})$. Each Liénard–Wiechert potential has denominator $r-s(\mathbf{n}\cdot \mathbf{p}+r\mathbf{n}\cdot {\mathbf{p}}^{\prime }/c)/(2q)+O(q^{-2})$. The opposite $q/r$ terms cancel, leaving the displayed electric potential. The products of signed electric charge and velocity add, giving the displayed magnetic vector potential. Bounded time derivatives on the relevant interval justify the Taylor expansions.