Retarded Potentials: Fields That Remember

The potentials obey a wave equation

Recall the potentials: \mathbf{E}=-\nabla\phi-\partial\mathbf{A}/\partial t and \mathbf{B}=\nabla\times\mathbf{A}, with the scalar potential \phi and the vector potential \mathbf{A}. They are not unique: a gauge transformation \phi\to\phi-\partial\chi/\partial t, \mathbf{A}\to\mathbf{A}+\nabla\chi leaves \mathbf{E} and \mathbf{B} untouched. We exploit that freedom to simplify the equations.

Choose the Lorenz gauge \nabla\cdot\mathbf{A}+\tfrac{1}{c^2}\partial\phi/\partial t=0. With that condition, Maxwell's equations decouple into two identical inhomogeneous wave equations, one for \phi and one for each component of \mathbf{A}, driven by the charge and current.

\Box\phi = -\frac{\rho}{\varepsilon_0}, \qquad \Box\mathbf{A} = -\mu_0\mathbf{J}, \qquad \Box \equiv \nabla^{2}-\frac{1}{c^{2}}\frac{\partial^{2}}{\partial t^{2}}

In the Lorenz gauge each potential obeys a wave equation sourced by charge or current; \Box is the d'Alembertian.

The retarded solution

Strip the time derivative and the equation \nabla^2\phi=-\rho/\varepsilon_0 is just electrostatics, solved by the familiar Coulomb integral. Restore the time derivative and the solution has the same shape — but the source must be evaluated at the retarded time t_r=t-|\mathbf{r}-\mathbf{r}'|/c. The potential here-and-now is built from what each bit of source was doing exactly one light-travel-time earlier.

\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|}\,d^3r', \qquad t_r = t-\frac{|\mathbf{r}-\mathbf{r}'|}{c}

The retarded potential; \mathbf{A} has the identical form with \mu_0\mathbf{J} in place of \rho/\varepsilon_0.

From a point charge: Liénard-Wiechert

For a single moving point charge you cannot simply drop a delta function into the integral — because the charge moves while its signal is being emitted, a subtle geometric factor appears. The result is the Liénard-Wiechert potentials, carrying the tell-tale factor (1-\mathbf{n}\cdot\boldsymbol{\beta}), with \mathbf{n} the unit vector toward the field point and \boldsymbol{\beta}=\mathbf{v}/c, all evaluated at the retarded time.

\phi = \frac{q}{4\pi\varepsilon_0}\,\frac{1}{(1-\mathbf{n}\cdot\boldsymbol{\beta})\,R}\bigg|_{\text{ret}}, \qquad \mathbf{A} = \frac{\mathbf{v}}{c^{2}}\,\phi

The Liénard-Wiechert potentials of a point charge; R is the retarded distance and the vector potential simply tracks the velocity.

The fields of a moving charge

Differentiate the potentials — carefully, because t_r itself depends on position — and the electric field splits cleanly into two pieces. A velocity field \propto 1/R^2 carries the Coulomb field along and contains no acceleration; and an acceleration field \propto 1/R, linear in \mathbf{a}, transverse to \mathbf{n} — this is the radiation.

\mathbf{E} = \frac{q}{4\pi\varepsilon_0}\frac{R}{(\mathbf{R}\cdot\mathbf{u})^{3}}\Big[\underbrace{(c^{2}-v^{2})\mathbf{u}}_{\text{velocity field } \sim 1/R^{2}} + \underbrace{\mathbf{R}\times(\mathbf{u}\times\mathbf{a})}_{\text{radiation field } \sim 1/R}\Big], \quad \mathbf{u}=c\hat{\mathbf{n}}-\mathbf{v}

The exact field of an arbitrarily moving charge (Griffiths' form): \mathbf{R} is the retarded separation vector, R=|\mathbf{R}|, \hat{\mathbf{n}}=\mathbf{R}/R, and \mathbf{B}=\tfrac1c\hat{\mathbf{n}}\times\mathbf{E}.

Only the acceleration term survives to infinity, and it is transverse to \hat{\mathbf{n}} with E_{\text{rad}}=cB_{\text{rad}} and \mathbf{E}\times\mathbf{B} pointing outward — precisely the $1/r$ radiation field the kink picture sketched in guide 1, now derived exactly. In the radiation zone, far beyond a wavelength, only this piece matters.