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Retarded potentials

We already know the solutions to Eqs. (1439) and (1440). They take the form (see Sect. 4.9)
$\displaystyle \phi({\bf r},t)$ $\textstyle =$ $\displaystyle \frac{1}{4\pi \epsilon_0} \int
\frac{[\rho({\bf r}')]}{\vert{\bf r}- {\bf r}'\vert} dV',$ (1451)
$\displaystyle {\bf A}({\bf r}, t)$ $\textstyle =$ $\displaystyle \frac{\mu_0}{4\pi} \int
\frac{[{\bf j}({\bf r}')]}{\vert{\bf r}- {\bf r}'\vert} dV'.$ (1452)

The above equations can be combined to form the solution of the 4-vector wave equation (1442),
{\mit \Phi}^\mu = \frac{1}{4\pi \epsilon_0 c}\int
{ r} dV.
\end{displaymath} (1453)

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. (1453) 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. (1453) 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 $(x,y,z,c t)$ to the retarded event $Q$ for which $r$ and $dV$ are evaluated. Using the standard Lorentz transformation, (1346)-(1349), 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' = -c t' =- c \gamma \left(t - \frac{v x}{c^2}\right),
\end{displaymath} (1454)

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{c t}{r} + \frac{v x}{c r}\right)
= r \gamma \left(1 + \frac{v}{c} \cos\theta\right),
\end{displaymath} (1455)

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 incorrect. 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. (1346)-(1349). For the $x$-dimension these equations give $dx' = \gamma (dx-v dt)$. The extremities of $dx$ are measured at times differing by $dt$, where

dt = -\frac{dr}{c} = -\frac{dx}{c} \cos\theta.
\end{displaymath} (1456)

dx' = \gamma\left(1+\frac{v}{c}  \cos\theta\right) dx,
\end{displaymath} (1457)

dV' = \gamma \left(1+\frac{v}{c}  \cos\theta\right) dV.
\end{displaymath} (1458)

It follows from Eqs. (1455) and (1458) 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. (1453) is indeed a contravariant 4-vector. Thus, this 4-vector equation can be properly regarded as the solution to the 4-vector wave equation (1442).

next up previous
Next: Tensors and pseudo-tensors Up: Relativity and electromagnetism Previous: Gauge invariance
Richard Fitzpatrick 2006-02-02