The potentials of a moving charge; the general solution of Liénard and
Wiechert
In the last section we made a simplification in calculating our integral for $\FLPA$ by considering only low velocities. But in doing so we missed an important point and also one where it is easy to go wrong. We will therefore take up now a calculation of the potentials for a point charge moving in any way whatever—even with a relativistic velocity. Once we have this result, we will have the complete electromagnetism of electric charges. Even Eq. (21.1) can then be derived by taking derivatives. The story will be complete. So bear with us.  Let’s try to calculate the scalar potential $\phi(1)$ at the point $(x_1,y_1,z_1)$ produced by a point charge, such as an electron, moving in any manner whatsoever. By a “point” charge we mean a very small ball of charge, shrunk down as small as you like, with a charge density $\rho(x,y,z)$. We can find $\phi$ from Eq. (21.15): \begin{equation} \label{Eq:II:21:28} \phi(1,t)=\frac{1}{4\pi\epsO} \int\frac{\rho(2,t-r_{12}/c)}{r_{12}}\,dV_2. \end{equation} The answer would seem to be—and almost everyone would, at first, think—that the integral of $\rho$ over such a “point” charge is just the total charge $q$, so that \begin{equation*} \phi(1,t)=\frac{1}{4\pi\epsO}\,\frac{q}{r_{12}'}\quad(\text{wrong}). \end{equation*} By $\FLPr_{12}'$ we mean the radius vector from the charge at point $(2)$ to point $(1)$ at the retarded time $(t-r_{12}/c)$. It is wrong.  The correct answer is \begin{equation} \label{Eq:II:21:29} \phi(1,t)=\frac{1}{4\pi\epsO}\,\frac{q}{r_{12}'}\cdot \frac{1}{1-v_{r'}/c}, \end{equation} where $v_{r'}$, is the component of the velocity of the charge parallel to $\FLPr_{12}'$—namely, toward point $(1)$. We will now show you why. To make the argument easier to follow, we will make the calculation first for a “point” charge which is in the form of a little cube of charge moving toward the point $(1)$ with the speed $v$, as shown in Fig. 21-5(a). Let the length of a side of the cube be $a$, which we take to be much, much less than $r_{12}$, the distance from the center of the charge to the point $(1)$.  Now to evaluate the integral of Eq. (21.28), we will return to basic principles; we will write it as the sum \begin{equation} \label{Eq:II:21:30} \sum_i\frac{\rho_i\,\Delta V_i}{r_i}, \end{equation} where $r_i$ is the distance from point $(1)$ to the $i$th volume element $\Delta V_i$ and $\rho_i$ is the charge density at $\Delta V_i$ at the time $t_i=t-r_i/c$. Since $r_i\gg a$, always, it will be convenient to take our $\Delta V_i$ in the form of thin, rectangular slices perpendicular to $\FLPr_{12}$, as shown in Fig. 21-5(b).  Suppose we start by taking the volume elements $\Delta V_i$ with some thickness $w$ much less than $a$. The individual elements will appear as shown in Fig. 21-6(a), where we have put in more than enough to cover the charge. But we have not shown the charge, and for a good reason. Where should we draw it? For each volume element $\Delta V_i$ we are to take $\rho$ at the time $t_i=(t-r_i/c)$, but since the charge is moving, it is in a different place for each volume element $\Delta V_i$!  Let’s say that we begin with the volume element labeled “1” in Fig. 21-6(a), chosen so that at the time $t_1=(t-r_1/c)$ the “back” edge of the charge occupies $\Delta V_1$, as shown in Fig. 21-6(b). Then when we evaluate $\rho_2\,\Delta V_2$, we must use the position of the charge at the slightly later time $t_2=(t-r_2/c)$, when the charge will be in the position shown in Fig. 21-6(c). And so on, for $\Delta V_3$, $\Delta V_4$, etc. Now we can evaluate the sum.  Since the thickness of each $\Delta V_i$ is $w$, its volume is $wa^2$. Then each volume element that overlaps the charge distribution contains the amount of charge $wa^2\rho$, where $\rho$ is the density of charge within the cube—which we take to be uniform. When the distance from the charge to point $(1)$ is large, we will make a negligible error by setting all the $r_i$’s in the denominators equal to some average value, say the retarded position $r'$ of the center of the charge. Then the sum (21.30) is \begin{equation*} \sum_{i=1}^N\frac{\rho wa^2}{r'}, \end{equation*} where $\Delta V_N$ is the last $\Delta V_i$ that overlaps the charge distributions, as shown in Fig. 21-6(e). The sum is, clearly, \begin{equation*} N\,\frac{\rho wa^2}{r'}=\frac{\rho a^3}{r'}\biggl(\frac{Nw}{a}\biggr). \end{equation*} Now $\rho a^3$ is just the total charge $q$ and $Nw$ is the length $b$ shown in part (e) of the figure. So we have \begin{equation} \label{Eq:II:21:31} \phi=\frac{q}{4\pi\epsO r'}\biggl(\frac{b}{a}\biggr). \end{equation}  What is $b$? It is the length of the cube of charge increased by the distance moved by the charge between $t_1=(t-r_1/c)$ and $t_N=(t-r_N/c)$—which is the distance the charge moves in the time \begin{equation*} \Delta t=t_N-t_1=(r_1-r_N)/c=b/c. \end{equation*} Since the speed of the charge is $v$, the distance moved is $v\,\Delta t=vb/c$. But the length $b$ is this distance added to $a$: \begin{equation*} b=a+\frac{v}{c}\,b. \end{equation*} Solving for $b$, we get \begin{equation*} b=\frac{a}{1-(v/c)}. \end{equation*} Of course by $v$ we mean the velocity at the retarded time $t'=(t-r'/c)$, which we can indicate by writing $[1-v/c]_{\text{ret}}$, and Eq. (21.31) for the potential becomes \begin{equation*} \phi(1,t)=\frac{q}{4\pi\epsO r'}\,\frac{1}{[1-(v/c)]_{\text{ret}}}. \end{equation*} This result agrees with our assertion, Eq. (21.29). There is a correction term which comes about because the charge is moving as our integral “sweeps over the charge.”  When the charge is moving toward the point $(1)$, its contribution to the integral is increased by the ratio $b/a$. Therefore the correct integral is $q/r'$ multiplied by $b/a$, which is $1/[1-v/c]_{\text{ret}}$.  If the velocity of the charge is not directed toward the observation point $(1)$, you can see that what matters is the component of its velocity toward point $(1)$.  Calling this velocity  component $v_r$, the correction factor is $1/[1-v_r/c]_{\text{ret}}$. Also, the analysis we have made goes exactly the same way for a charge distribution of any shape—it doesn’t have to be a cube. Finally, since the “size” of the charge $q$ doesn’t enter into the final result, the same result holds when we let the charge shrink to any size—even to a point. The general result is that the scalar potential for a point charge moving with any velocity is \begin{equation} \label{Eq:II:21:32} \phi(1,t)=\frac{q}{4\pi\epsO r'[1-(v_r/c)]_{\text{ret}}}. \end{equation} This equation is often written in the equivalent form \begin{equation} \label{Eq:II:21:33} \phi(1,t)=\frac{q}{4\pi\epsO[r-(\FLPv\cdot\FLPr/c)]_{\text{ret}}}, \end{equation} where $\FLPr$ is the vector from the charge to the point $(1)$, where $\phi$ is being evaluated, and all the quantities in the bracket are to have their values at the retarded time $t'=t-r'/c$.  The same thing happens when we compute $\FLPA$ for a point charge, from Eq. (21.16). The current density is $\rho\FLPv$ and the integral over $\rho$ is the same as we found for $\phi$. The vector potential is \begin{equation} \label{Eq:II:21:34} \FLPA(1,t)=\frac{q\FLPv_{\text{ret}}} {4\pi\epsO c^2[r-(\FLPv\cdot\FLPr/c)]_{\text{ret}}}. \end{equation}  The potentials for a point charge were first deduced in this form by Liénard and Wiechert and are called the Liénard-Wiechert potentials.  To close the ring back to Eq. (21.1) it is only necessary to compute $\FLPE$ and $\FLPB$ from these potentials (using $\FLPB=\FLPcurl{\FLPA}$ and $\FLPE=-\FLPgrad{\phi}-\ddpl{\FLPA}{t}$). It is now only arithmetic. The arithmetic, however, is fairly involved, so we will not write out the details. Perhaps you will take our word for it that Eq. (21.1) is equivalent to the Liénard-Wiechert potentials we have derived.2
