Retarded potentials

Equations (2.94) have the same form as the inhomogeneous wave equation (2.103), so we can immediately write the solutions to these equations as

$$\phi({\bfm r}, t) = \frac{1}{4\pi\,\epsilon_0} \int \frac{[\rho({\bfm r}')]}{\vert{\bfm r}- {\bfm r}'\vert}\,dV',$$ (178)

$${\bfm A}({\bfm r}, t) = \frac{\mu_0}{4\pi} \int \frac{[{\bfm j}({\bfm r}')]}{\vert{\bfm r}- {\bfm r}'\vert}\,dV'.$$ (179)

Moreover, we can be sure that these solutions are unique, subject to the reasonable proviso that infinity is an absorber of radiation but not an emitter. This is a crucially important point. Whenever the above solutions are presented in physics textbooks there is a tacit assumption that they are unique. After all, if they were not unique why should we choose to study them instead of one of the other possible solutions? The uniqueness of the above solutions has a physical interpretation. It is clear from Eqs. (2.141) that in the absence of any charges and currents there are no electromagnetic fields. In other words, if we observe an electromagnetic field we can be certain that if we were to trace it backward in time we would eventually discover that it was emitted by a charge or a current. In proving that the solutions of Maxwell's equations are unique, and then finding a solution in which all waves are emitted by sources, we have effectively ruled out the possibility that the vacuum can be ``unstable'' to the production of electromagnetic waves without the need for any sources.

Equations (2.141) can be combined to form the solution of the 4-vector wave equation (2.96),

$${\mit \Phi}^\mu = \frac{1}{4\pi\,\epsilon_0\,c}\int \frac{[J^\mu]} { r}\,dV.$$ (180)

Here, the components of the 4-potential are evaluated at some event $P$ in space-time, $r$ is the distance of the volume element $dV$ from $P$, and the square brackets indicate that the 4-current is to be evaluated at the retarded time; i.e., at a time $r/c$ before $P$.

But, does the right-hand side of Eq. (2.142) really transform as a contravariant 4-vector? This is not a trivial question since volume integrals in 3-space are not, in general, Lorentz invariant due to the length contraction effect. However, the integral in Eq. (2.142) is not a straightforward volume integral because the integrand is evaluated at the retarded time. In fact, the integral is best regarded as an integral over events in space-time. The events which enter the integral are those which intersect a spherical light wave launched from the event $P$ and evolved backwards in time. In other words, the events occur before the event $P$ and have zero interval with respect to $P$. It is clear that observers in all inertial frames will, at least, agree on which events are to be included in the integral, since both the interval between events and the absolute order in which events occur are invariant under a general Lorentz transformation.

We shall now demonstrate that all observers obtain the same value of $dV/r$ for each elementary contribution to the integral. Suppose that $S$ and $S'$ are two inertial frames in the standard configuration. Let unprimed and primed symbols denote corresponding quantities in $S$ and $S'$, respectively. Let us assign coordinates $(0,0,0,0)$ to $P$ and $(x,y,z,ct)$ to the retarded event $Q$ for which $r$ and $dV$ are evaluated. Using the standard Lorentz transformation (2.19), the fact that the interval between events $P$ and $Q$ is zero, and the fact that both $t$ and $t'$ are negative, we obtain

$$r' = -ct' =- c\gamma\left(t - \frac{vx}{c^2}\right),$$ (181)

where $v$ is the relative velocity between frames $S'$ and $S$, $\gamma$ is the Lorentz factor, and $r=\sqrt{x^2+y^2+z^2}$, etc. It follows that

$$r' = r\gamma\left(-\frac{ct}{r} + \frac{vx}{cr}\right) = r\gamma \left(1 + \frac{v}{c} \cos\theta\right),$$ (182)

where $\theta$ is the angle (in 3-space) subtended between the line $PQ$ and the $x$-axis.

We now know the transformation for $r$. What about the transformation for $dV$? We might be tempted to set $dV' = \gamma\, dV$, according to the usual length contraction rule. However, this is wrong. The contraction by a factor $\gamma$ only applies if the whole of the volume is measured at the same time, which is not the case in the present problem. Now, the dimensions of $dV$ along the $y-$ and $z-$ axes are the same in both $S$ and $S'$, according to Eqs. (2.19). For the $x$-dimension these equations give $dx' = \gamma(dx-v\,dt)$. The extremities of $dx$ are measured at times differing by $dt$, where⁵

$$dt = -\frac{dr}{c} = -\frac{dx}{c}\,\cos\theta.$$ (183)

Thus,

$$dx' = \left(1+\frac{v}{c} \,\cos\theta\right)\gamma\,dx,$$ (184)

giving

$$dV' = \left(1+\frac{v}{c} \,\cos\theta\right)\gamma\,dV.$$ (185)

It follows from Eqs. (2.144) and (2.147) that $dV'/r' = dV/r$. This result will clearly remain valid even when $S$ and $S'$ are not in the standard configuration.

Thus, $dV/r$ is an invariant and, therefore, $[J^\mu]\,dV/r$ is a contravariant 4-vector. For linear transformations, such as a general Lorentz transformation, the result of adding 4-tensors evaluated at different 4-points is itself a 4-tensor. It follows that the right-hand side of Eq. (2.142) is a contravariant 4-vector. Thus, this 4-vector equation can be properly regarded as the solution to the 4-vector wave equation (2.96).