The four-potential of a moving charge
We saw in the last chapter that the potential $A_\mu=(\phi,\FLPA)$ is a four-vector. The time component is the scalar potential $\phi$, and the three space components are the vector potential $\FLPA$. We also worked out the potentials of a particle moving with uniform speed on a straight line by using the Lorentz transformation. (We had already found them by another method in Chapter 21.) For a point charge whose position at the time $t$ is $(vt,0,0)$, the potentials at the point $(x,y,z)$ are \begin{equation} \begin{aligned} \phi&=\frac{1}{\overset{\phantom [}{4\pi\epsO\sqrt{1-v^2}}}\, \frac{q}{\biggl[ \dfrac{(x-vt)^2}{1-v^2}+y^2+z^2 \biggr]^{1/2}}\\[1.5ex] A_x&=\frac{1}{\overset{\phantom [}4\pi\epsO\sqrt{1-v^2}}\, \frac{qv}{\biggl[ \dfrac{(x-vt)^2}{1-v^2}+y^2+z^2 \biggr]^{1/2}}\\[1ex] A_y&=A_z=0. \end{aligned} \label{Eq:II:26:1} \end{equation}  Equations (26.1) give the potentials at $x$, $y$, and $z$ at the time $t$, for a charge whose “present” position (by which we mean the position at the time $t$) is at $x=vt$. Notice that the equations are in terms of $(x-vt)$, $y$, and $z$ which are the coordinates measured from the current position $P$ of the moving charge (see Fig. 26–1). The actual influence we know really travels at the speed $c$, so it is the behavior of the charge back at the retarded position $P'$ that really counts.1 The point $P'$ is at $x=vt'$ (where $t'=t-r'/c$ is the retarded time). But we said that the charge was moving with uniform velocity in a straight line, so naturally the behavior at $P'$ and the current position are directly related. In fact, if we make the added assumption that the potentials depend only upon the position and the velocity at the retarded moment, we have in equations (26.1) a complete formula for the potentials for a charge moving any way. It works this way. Suppose that you have a charge moving in some arbitrary fashion, say with the trajectory in Fig. 26–2, and you are trying to find the potentials at the point $(x,y,z)$. First, you find the retarded position $P'$ and the velocity $v'$ at that point. Then you imagine that the charge would keep on moving with this velocity during the delay time $(t'-t)$, so that it would then appear at an imaginary position $P_{\text{proj}}$, which we can call the “projected position,” and would arrive there with the velocity $v'$. (Of course, it doesn’t do that; its real position at $t$ is at $P$.) Then the potentials at $(x,y,z)$ are just what equations (26.1) would give for the imaginary charge at the projected position $P_{\text{proj}}$. What we are saying is that since the potentials depend only on what the charge is doing at the retarded time, the potentials will be the same whether the charge continued moving at a constant velocity or whether it changed its velocity after $t'$—that is, after the potentials that were going to appear at $(x,y,z)$ at the time $t$ were already determined.  You know, of course, that the moment that we have the formula for the potentials from a charge moving in any manner whatsoever, we have the complete electrodynamics; we can get the potentials of any charge distribution by superposition. Therefore we can summarize all the phenomena of electrodynamics either by writing Maxwell’s equations or by the following series of remarks. (Remember them in case you are ever on a desert island. From them, all can be reconstructed. You will, of course, know the Lorentz transformation; you will never forget that on a desert island or anywhere else.)  First, $A_\mu$ is a four-vector. Second, the Coulomb potential for a stationary charge is $q/4\pi\epsO r$. Third, the potentials produced by a charge moving in any way depend only upon the velocity and position at the retarded time. With those three facts we have everything. From the fact that $A_\mu$ is a four-vector, we transform the Coulomb potential, which we know, and get the potentials for a constant velocity. Then, by the last statement that potentials depend only upon the past velocity at the retarded time, we can use the projected position game to find them. It is not a particularly useful way of doing things, but it is interesting to show that the laws of physics can be put in so many different ways.  It is sometimes said, by people who are careless, that all of electrodynamics can be deduced solely from the Lorentz transformation and Coulomb’s law. Of course, that is completely false. First, we have to suppose that there is a scalar potential and a vector potential that together make a four-vector. That tells us how the potentials transform. Then why is it that the effects at the retarded time are the only things that count? Better yet, why is it that the potentials depend only on the position and the velocity and not, for instance, on the acceleration? The fields $\FLPE$ and $\FLPB$ do depend on the acceleration. If you try to make the same kind of an argument with respect to them, you would say that they depend only upon the position and velocity at the retarded time. But then the fields from an accelerating charge would be the same as the fields from a charge at the projected position—which is false. The fields depend not only on the position and the velocity along the path but also on the acceleration. So there are several additional tacit assumptions in this great statement that everything can be deduced from the Lorentz transformation. (Whenever you see a sweeping statement that a tremendous amount can come from a very small number of assumptions, you always find that it is false. There are usually a large number of implied assumptions that are far from obvious if you think about them sufficiently carefully.) 