Spherical waves from a point source
In Chapter 18 we found that Maxwell’s equations could be solved by letting \begin{equation} \label{Eq:II:21:2} \FLPE=-\FLPgrad{\phi}-\ddp{\FLPA}{t} \end{equation} and \begin{equation} \label{Eq:II:21:3} \FLPB=\FLPcurl{\FLPA}, \end{equation} where $\phi$ and $\FLPA$ must then be solutions of the equations \begin{equation} \label{Eq:II:21:4} \nabla^2\phi-\frac{1}{c^2}\,\frac{\partial^2\phi}{\partial t^2}= -\frac{\rho}{\epsO} \end{equation} and \begin{equation} \label{Eq:II:21:5} \nabla^2\FLPA-\frac{1}{c^2}\,\frac{\partial^2\FLPA}{\partial t^2}= -\frac{\FLPj}{\epsO c^2}, \end{equation} and must also satisfy the condition that \begin{equation} \label{Eq:II:21:6} \FLPdiv{\FLPA}=-\frac{1}{c^2}\,\ddp{\phi}{t}. \end{equation}  Now we will find the solution of Eqs. (21.4) and (21.5). To do that we have to find the solution $\psi$, of the equation \begin{equation} \label{Eq:II:21:7} \nabla^2\psi-\frac{1}{c^2}\,\frac{\partial^2\psi}{\partial t^2}= -s, \end{equation} where $s$, which we call the source, is known. Of course, $s$ corresponds to $\rho/\epsO$ and $\psi$ to $\phi$ for Eq. (21.4), or $s$ is $j_x/\epsO c^2$ if $\psi$ is $A_x$, etc., but we want to solve Eq. (21.7) as a mathematical problem no matter what $\psi$ and $s$ are physically.  In places where $\rho$ and $\FLPj$ are zero—in what we have called “free” space—the potentials $\phi$ and $\FLPA$, and the fields $\FLPE$ and $\FLPB$, all satisfy the three-dimensional wave equation without sources, whose mathematical form is \begin{equation} \label{Eq:II:21:8} \nabla^2\psi-\frac{1}{c^2}\,\frac{\partial^2\psi}{\partial t^2}= 0. \end{equation} In Chapter 20 we saw that solutions of this equation can represent waves of various kinds: plane waves in the $x$-direction, $\psi=f(t-x/c)$; plane waves in the $y$- or $z$-direction, or in any other direction; or spherical waves of the form \begin{equation} \label{Eq:II:21:9} \psi(x,y,z,t)=\frac{f(t-r/c)}{r}. \end{equation} (The solutions can be written in still other ways, for example cylindrical waves that spread out from an axis.)  We also remarked that, physically, Eq. (21.9) does not represent a wave in free space—that there must be charges at the origin to get the outgoing wave started. In other words, Eq. (21.9) is a solution of Eq. (21.8) everywhere except right near $r=0$, where it must be a solution of the complete equation (21.7), including some sources. Let’s see how that works. What kind of a source $s$ in Eq. (21.7) would give rise to a wave like Eq. (21.9)?  Suppose we have the spherical wave of Eq. (21.9) and look at what is happening for very small $r$. Then the retardation $-r/c$ in $f(t-r/c)$ can be neglected—provided $f$ is a smooth function—and $\psi$ becomes \begin{equation} \label{Eq:II:21:10} \psi=\frac{f(t)}{r}\quad(r\to0). \end{equation} So $\psi$ is just like a Coulomb field for a charge at the origin that varies with time. That is, if we had a little lump of charge, limited to a very small region near the origin, with a density $\rho$, we know that \begin{equation*} \phi=\frac{Q/4\pi\epsO}{r}, \end{equation*} where $Q=\int\rho\,dV$. Now we know that such a $\phi$ satisfies the equation \begin{equation*} \nabla^2\phi=-\frac{\rho}{\epsO}. \end{equation*}  Following the same mathematics, we would say that the $\psi$ of Eq. (21.10) satisfies \begin{equation} \label{Eq:II:21:11} \nabla^2\psi=-s\quad(r\to0), \end{equation} where $s$ is related to $f$ by \begin{equation*} f=\frac{S}{4\pi}, \end{equation*} with \begin{equation*} S=\int s\,dV. \end{equation*} The only difference is that in the general case, $s$, and therefore $S$, can be a function of time.  Now the important thing is that if $\psi$ satisfies Eq. (21.11) for small $r$, it also satisfies Eq. (21.7). As we go very close to the origin, the $1/r$ dependence of $\psi$ causes the space derivatives to become very large. But the time derivatives keep their same values. [They are just the time derivatives of $f(t)$.] So as $r$ goes to zero, the term $\partial^2\psi/\partial t^2$ in Eq. (21.7) can be neglected in comparison with $\nabla^2\psi$, and Eq. (21.7) becomes equivalent to Eq. (21.11).  To summarize, then, if the source function $s(t)$ of Eq. (21.7) is localized at the origin and has the total strength \begin{equation} \label{Eq:II:21:12} S(t)=\int s(t)\,dV, \end{equation} the solution of Eq. (21.7) is \begin{equation} \label{Eq:II:21:13} \psi(x,y,z,t)=\frac{1}{4\pi}\,\frac{S(t-r/c)}{r}. \end{equation} The only effect of the term $\partial^2\psi/\partial t^2$ in Eq. (21.7) is to introduce the retardation $(t-r/c)$ in the Coulomb-like potential. 