Potential due to a moving charge

Suppose that a particle carrying a charge $e$ moves with uniform velocity ${\bf u}$ through a frame $S$. Let us evaluate the vector potential, ${\bf A}$, and the scalar potential, $\phi$, due to this charge at a given event $P$ in $S$.

Let us choose coordinates in $S$ so that $P=(0,0,0,0)$ and ${\bf u} = (u,0,0)$. Let $S'$ be that frame in the standard configuration with respect to $S$ in which the charge is (permanently) at rest at (say) the point $(x', y', z')$. In $S'$, the potential at $P$ is the usual potential due to a stationary charge,

$${\bf A}' = {\bf0},$$ (1518)

$$\phi' = \frac{e}{4\pi \epsilon_0 r'},$$ (1519)

where $r'=\sqrt{x^{'2}+y^{'2}+z^{'2}}$. Let us now transform these equations directly into the frame $S$. Since $A^\mu= (c {\bf A}, \phi)$ is a contravariant 4-vector, its components transform according to the standard rules (1397)-(1400). Thus,

$$c A_1 = \gamma\left(c A_1' + \frac{u}{c} \phi'\right)= \frac{\gamma u e}{4\pi \epsilon_0 c r'},$$ (1520)

$$c A_2 = c A_2' = 0,$$ (1521)

$$c A_3 = c A_3' = 0,$$ (1522)

$$\phi = \gamma\left(\phi' + \frac{u }{c} c A_1'\right)= \frac{\gamma e} {4\pi \epsilon_0 r'},$$ (1523)

since $\beta=-u/c$ in this case. It remains to express the quantity $r'$ in terms of quantities measured in $S$. The most physically meaningful way of doing this is to express $r'$ in terms of retarded values in $S$. Consider the retarded event at the charge for which, by definition, $r'= -c t'$ and $r=-c t$. Using the standard Lorentz transformation, (1346)-(1349), we find that

$$r'=-c t' =-c \gamma (t-u x/c^2) = r \gamma (1+u_r/c),$$ (1524)

where $u_r = u x/r = {\bf r}\!\cdot\!{\bf u}/r$ denotes the radial velocity of the change in $S$. We can now rewrite Eqs. (1520)-(1523) in the form

$${\bf A} = \frac{\mu_0 e}{4\pi} \frac{[{\bf u}]}{[r+ {\bf r}\!\cdot\!{\bf u}/c]},$$ (1525)

$$\phi = \frac{e}{4\pi \epsilon_0} \frac{1}{[r+ {\bf r}\!\cdot\!{\bf u}/c]},$$ (1526)

where the square brackets, as usual, indicate that the enclosed quantities must be retarded. For a uniformly moving charge, the retardation of ${\bf u}$ is, of course, superfluous. However, since

$${\mit\Phi}^\mu = \frac{1}{4\pi \epsilon_0 c}\int \frac{[J^\mu]}{r} dV,$$ (1527)

it is clear that the potentials depend only on the (retarded) velocity of the charge, and not on its acceleration. Consequently, the expressions (1525) and (1526) give the correct potentials for an arbitrarily moving charge. They are known as the Liénard-Wiechert potentials.