Solving Maxwell’s equations; the potentials and the wave equation
Now we would like to do something mathematical; we want to write
Maxwell’s equations in a simpler form. You may consider that we are
complicating them, but if you will be patient a little bit, they will
suddenly come out simpler. Although by this time you are thoroughly
used to each of the Maxwell equations, there are many pieces that must
all be put together. That’s what we want to do. We begin with $\FLPdiv{\FLPB}=0$—the simplest of the equations. We
know that it implies that $\FLPB$ is the curl of something. So, if we
write
\begin{equation}
\label{Eq:II:18:16}
\FLPB=\FLPcurl{\FLPA},
\end{equation}

we have already solved one of Maxwell’s equations. (Incidentally, you
appreciate that it remains true that another vector $\FLPA'$ would be
just as good if $\FLPA'=\FLPA+\FLPgrad{\psi}$—where $\psi$ is any
scalar field—because the curl of $\FLPgrad{\psi}$ is zero,
and $\FLPB$ is still the same. We have talked about that before.) We take next the Faraday law, $\FLPcurl{\FLPE}=-\ddpl{\FLPB}{t}$,
because it doesn’t involve any currents or charges. If we write
$\FLPB$ as $\FLPcurl{\FLPA}$ and differentiate with respect to $t$, we
can write Faraday’s law in the form
\begin{equation*}
\FLPcurl{\FLPE}=-\ddp{}{t}\,\FLPcurl{\FLPA}.
\end{equation*}

Since we can differentiate either with respect to time or to space
first, we can also write this equation as
\begin{equation}
\label{Eq:II:18:17}
\FLPcurl{\biggl(\FLPE+\ddp{\FLPA}{t}\biggr)}=\FLPzero.
\end{equation}

We see that $\FLPE+\ddpl{\FLPA}{t}$ is a vector whose curl is equal to
zero. Therefore that vector is the gradient of something. When we
worked on electrostatics, we had $\FLPcurl{\FLPE}=\FLPzero$, and then
we decided that $\FLPE$ itself was the gradient of something. We took
it to be the gradient of $-\phi$ (the minus for technical
convenience). We do the same thing for $\FLPE+\ddpl{\FLPA}{t}$; we set
\begin{equation}
\label{Eq:II:18:18}
\FLPE+\ddp{\FLPA}{t}=-\FLPgrad{\phi}.
\end{equation}

We use the same symbol $\phi$ so that, in the electrostatic case where
nothing changes with time and the $\ddpl{\FLPA}{t}$ term disappears,
$\FLPE$ will be our old $-\FLPgrad{\phi}$. So Faraday’s equation can
be put in the form
\begin{equation}
\label{Eq:II:18:19}
\FLPE=-\FLPgrad{\phi}-\ddp{\FLPA}{t}.
\end{equation}

 We have solved two of Maxwell’s equations already, and we have found
that to describe the electromagnetic fields $\FLPE$ and $\FLPB$, we
need four potential functions: a scalar potential $\phi$ and a vector
potential $\FLPA$, which is, of course, three functions. Now that $\FLPA$ determines part of $\FLPE$, as well as $\FLPB$, what
happens when we change $\FLPA$ to $\FLPA'=\FLPA+\FLPgrad{\psi}$? In
general, $\FLPE$ would change if we didn’t take some special
precaution. We can, however, still allow $\FLPA$ to be changed in this
way without affecting the fields $\FLPE$ and $\FLPB$—that is,
without changing the physics—if we always change $\FLPA$ and $\phi$
together by the rules
\begin{equation}
\label{Eq:II:18:20}
\FLPA'=\FLPA+\FLPgrad{\psi},\quad
\phi'=\phi-\ddp{\psi}{t}.
\end{equation}

Then neither $\FLPB$ nor $\FLPE$, obtained from Eq. (18.19),
is changed. Previously, we chose to make $\FLPdiv{\FLPA}=0$, to make the equations
of statics somewhat simpler. We are not going to do that now; we are
going to make a different choice. But we’ll wait a bit before saying
what the choice is, because later it will be clear why the
choice is made. Now we return to the two remaining Maxwell equations which will give us
relations between the potentials and the sources $\rho$ and $\FLPj$. Once we can
determine $\FLPA$ and $\phi$ from the currents and charges, we can always get
$\FLPE$ and $\FLPB$ from Eqs. (18.16)
and (18.19), so we will have another form of Maxwell’s equations. We begin by substituting Eq. (18.19) into
$\FLPdiv{\FLPE}=\rho/\epsO$; we get
\begin{equation*}
\FLPdiv{\biggl(-\FLPgrad{\phi}-\ddp{\FLPA}{t}\biggr)}=\frac{\rho}{\epsO},
\end{equation*}

which we can write also as
\begin{equation}
\label{Eq:II:18:21}
-\nabla^2\phi-\ddp{}{t}\,\FLPdiv{\FLPA}=\frac{\rho}{\epsO}.
\end{equation}

This is one equation relating $\phi$ and $\FLPA$ to the sources. Our final equation will be the most complicated. We start by rewriting
the fourth Maxwell equation as
\begin{equation*}
c^2\FLPcurl{\FLPB}-\ddp{\FLPE}{t}=\frac{\FLPj}{\epsO},
\end{equation*}

and then substitute for $\FLPB$ and $\FLPE$ in terms of the potentials, using
Eqs. (18.16) and (18.19):
\begin{equation*}
c^2\FLPcurl{(\FLPcurl{\FLPA})}-\ddp{}{t}\,
\biggl(-\FLPgrad{\phi}-\ddp{\FLPA}{t}\biggr)=
\frac{\FLPj}{\epsO}.
\end{equation*}

The first term can be rewritten using the algebraic identity:
$\FLPcurl{(\FLPcurl{\FLPA})}=$ $\FLPgrad{(\FLPdiv{\FLPA})}-\nabla^2\FLPA$;
we get

\begin{equation}
\label{Eq:II:18:22}
-c^2\nabla^2\FLPA+c^2\FLPgrad{(\FLPdiv{\FLPA})}+
\ddp{}{t}\,\FLPgrad{\phi}+\frac{\partial^2\FLPA}{\partial t^2}=
\frac{\FLPj}{\epsO}.
\end{equation}


\begin{align}
-c^2\nabla^2\FLPA&+c^2\FLPgrad{(\FLPdiv{\FLPA})}\notag\\[.5ex]
\label{Eq:II:18:22}
&+\ddp{}{t}\,\FLPgrad{\phi}+\frac{\partial^2\FLPA}{\partial t^2}=
\frac{\FLPj}{\epsO}.
\end{align}



It’s not very simple! Fortunately, we can now make use of our freedom to choose arbitrarily
the divergence of $\FLPA$. What we are going to do is to use our
choice to fix things so that the equations for $\FLPA$ and for $\phi$
are separated but have the same form. We can do this by
taking2
\begin{equation}
\label{Eq:II:18:23}
\FLPdiv{\FLPA}=-\frac{1}{c^2}\,\ddp{\phi}{t}.
\end{equation}

When we do that, the two middle terms in $\FLPA$ and $\phi$ in
Eq. (18.22) cancel, and that equation becomes much simpler:
\begin{equation}
\label{Eq:II:18:24}
\nabla^2\FLPA-\frac{1}{c^2}\,\frac{\partial^2\FLPA}{\partial t^2}=
-\frac{\FLPj}{\epsO c^2}.
\end{equation}

And our equation for $\phi$—Eq. (18.21)—takes on the
same form:
\begin{equation}
\label{Eq:II:18:25}
\nabla^2\phi-\frac{1}{c^2}\,\frac{\partial^2\phi}{\partial t^2}=
-\frac{\rho}{\epsO}.
\end{equation}
 What a beautiful set of equations! They are beautiful, first, because
they are nicely separated—with the charge density, goes $\phi$; with
the current, goes $\FLPA$. Furthermore, although the left side looks a
little funny—a Laplacian together with a $\partial^2/\partial t^2$—when we
unfold it we see
\begin{equation}
\label{Eq:II:18:26}
\frac{\partial^2\phi}{\partial x^2}+
\frac{\partial^2\phi}{\partial y^2}+
\frac{\partial^2\phi}{\partial z^2}-
\frac{1}{c^2}\,\frac{\partial^2\phi}{\partial t^2}=
-\frac{\rho}{\epsO}.
\end{equation}

It has a nice symmetry in $x$, $y$, $z$, $t$—the $-1/c^2$ is
necessary because, of course, time and space are different;
they have different units. Maxwell’s equations have led us to a new kind of equation for the
potentials $\phi$ and $\FLPA$ but to the same mathematical form for
all four functions $\phi$, $A_x$, $A_y$, and $A_z$. Once we learn how
to solve these equations, we can get $\FLPB$ and $\FLPE$ from
$\FLPcurl{\FLPA}$ and $-\FLPgrad{\phi}-\ddpl{\FLPA}{t}$. We have
another form of the electromagnetic laws exactly equivalent to
Maxwell’s equations, and in many situations they are much simpler to
handle. We have, in fact, already solved an equation much like
Eq. (18.26). When we studied sound in
Chapter 47 of Vol. I, we had an equation of the form
\begin{equation*}
\frac{\partial^2\phi}{\partial x^2}=
\frac{1}{c^2}\,\frac{\partial^2\phi}{\partial t^2},
\end{equation*}

and we saw that it described the propagation of waves in the
$x$-direction at the speed $c$. Equation (18.26) is the
corresponding wave equation for three dimensions. So in regions where
there are no longer any charges and currents, the solution of these
equations is not that $\phi$ and $\FLPA$ are zero. (Although
that is indeed one possible solution.)  There are solutions in which
there is some set of $\phi$ and $\FLPA$ which are changing in time but
always moving out at the speed $c$. The fields travel onward through
free space, as in our example at the beginning of the chapter. With Maxwell’s
new term in Eq. IV, we have been able to write the
field equations in terms of $\FLPA$ and $\phi$ in a form that is
simple and that makes immediately apparent that there are
electromagnetic waves. For many practical purposes, it will still be
convenient to use the original equations in terms of $\FLPE$
and $\FLPB$. But they are on the other side of the mountain we have
already climbed. Now we are ready to cross over to the other side of
the peak. Things will look different—we are ready for some new and
beautiful views.