A.36 Maxwell’s wave equations

This note derives the wave equations satisfied by electromagnetic fields. The derivation will use standard formulae of vector analysis, as found in, for example, [40, 20.35-45].

The starting point is Maxwell’s equations for the electromagnetic field in vacuum:

$\displaystyle\begin{array}
{@{\hspace{.7in}}c@{\hspace{.7in}}c@{\hspace{.6in}...
...\epsilon_0} +\frac{\partial \skew3\vec{\cal E}}{\partial t} & (4)
\end{array}$
Here $\skew3\vec{\cal E}$ is the electric field, $\skew2\vec{\cal B}$ the magnetic field, $\rho$ the charge density, $\vec\jmath$ the current density, $c$ the constant speed of light, and $\epsilon_0$ is a constant called the permittivity of space. The charge and current densities are related by the continuity equation
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\frac{\partial \rho}{\partial t} + \nabla\cdot\vec\jmath = 0
$\hfill(5)}$

To get a wave equation for the electric field, take the curl, $\nabla\times$, of (3) and apply the standard vector identity (D.1), (1) and (4) to get

\begin{displaymath}
\fbox{$\displaystyle
\frac{1}{c^2} \frac{\partial^2 \ske...
...math}{\partial t}
- \frac{1}{\epsilon_0} \nabla \rho
$} %
\end{displaymath} (A.235)

Similarly, for the magnetic field take the curl of (4) and use (2) and (3) to get
\begin{displaymath}
\fbox{$\displaystyle
\frac{1}{c^2} \frac{\partial^2 \ske...
... = \frac{1}{\epsilon_0 c^2} \nabla \times \vec\jmath\;
$} %
\end{displaymath} (A.236)

These are uncoupled inhomogeneous wave equations for the components of $\skew3\vec{\cal E}$ and $\skew2\vec{\cal B}$, for given charge and current densities. According to the theory of partial differential equations, these equations imply that effects propagate no faster than the speed of light. You can also see the same thing pretty clearly from the fact that the homogeneous wave equation has solutions like

\begin{displaymath}
\sin\Big(k(y-ct)+\varphi\Big)
\end{displaymath}

which are waves that travel with speed $c$ in the $y$-​direction.

The wave equations for the potentials $\varphi$ and $\skew3\vec A$ are next. First note from (2) that the divergence of $\skew2\vec{\cal B}$ is zero. Then vector calculus says that it can be written as the curl of some vector. Call that vector $\skew3\vec A_0$.

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\skew2\vec{\cal B}= \nabla \times \skew3\vec A_0
$\hfill(6a)}$

Next define

\begin{displaymath}
\skew3\vec{\cal E}_\varphi \equiv \skew3\vec{\cal E}+ \frac{\partial \skew3\vec A_0}{\partial t}
\end{displaymath}

Plug this into (3) to show that the curl of $\skew3\vec{\cal E}_\varphi$ is zero. Then vector calculus says that it can be written as minus the gradient of a scalar. Call this scalar $\varphi_0$. Plug that into the expression above to get
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\skew3\vec{\cal E}= - \nabla \varphi_0 - \frac{\partial \skew3\vec A_0}{\partial t}
$\hfill(7a)}$

Next, note that if you define modified versions $\skew3\vec A$ and $\varphi$ of $\skew3\vec A_0$ and $\varphi_0$ by setting

\begin{displaymath}
\varphi = \varphi_0 - \frac{\partial\chi}{\partial t}
\qquad
\skew3\vec A= \skew3\vec A_0 + \nabla \chi
\end{displaymath}

where $\chi$ is any arbitrary function of $x$, $y$, $z$, and $t$, then still
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\skew2\vec{\cal B}= \nabla \times \skew3\vec A
$\hfill(6)}$
since the curl of a gradient is always zero, and
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\skew3\vec{\cal E}= - \nabla \varphi - \frac{\partial \skew3\vec A}{\partial t}
$\hfill(7)}$
because the two $\chi$ terms drop out against each other.

The fact that $\skew3\vec A_0,\varphi_0$ and $\skew3\vec A,\varphi$ produce the same physical fields is the famous gauge property of the electromagnetic field.

Now you can select $\chi$ so that

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\nabla \cdot \skew3\vec A+ \frac{1}{c^2} \frac{\partial \varphi}{\partial t} = 0
$\hfill(8)}$
That is known as the “Lorenz condition.” A corresponding gauge function is a “Lorenz gauge.”

To find the gauge function $\chi$ that produces this condition, plug the definitions for $\skew3\vec A$ and $\varphi$ in terms of $\skew3\vec A_0$ and $\varphi_0$ into the left hand side of the Lorentz condition. That produces, after a change of sign,

\begin{displaymath}
\frac{1}{c^2} \frac{\partial^2 \chi}{\partial t^2} - \nabl...
...0
- \frac{1}{c^2} \frac{\partial \varphi_0}{\partial t} = 0
\end{displaymath}

That is a second order inhomogeneous wave equation for $\chi$.

Now plug the expressions (6) and (7) for $\skew3\vec{\cal E}$ and $\skew2\vec{\cal B}$ in terms of $\skew3\vec A$ and $\varphi$ into the Maxwell’s equations. Equations (2) and (3) are satisfied automatically. From (2), after using (8),

\begin{displaymath}
\fbox{$\displaystyle
\frac{1}{c^2} \frac{\partial^2 \var...
...al t^2} - \nabla^2 \varphi
= \frac{\rho}{\epsilon_0}
$} %
\end{displaymath} (A.237)

From (4), after using (8),
\begin{displaymath}
\fbox{$\displaystyle
\frac{1}{c^2} \frac{\partial^2 \ske...
...a^2 \skew3\vec A
= \frac{\vec\jmath}{\epsilon_0 c^2}
$} %
\end{displaymath} (A.238)

You can still select the two initial conditions for $\chi$. The smart thing to do is select them so that $\varphi$ and its time derivative are zero at time zero. In that case, if there is no charge density, $\varphi$ will stay zero for all time. That is because its wave equation is then homogeneous. The Lorenz condition will then ensure that $\nabla\cdot\skew3\vec A$ is zero too.

Instead of the Lorenz condition, you could select $\chi$ to make $\nabla\cdot\skew3\vec A$ zero. That is called the “Coulomb gauge” or “transverse gauge” or “transverse gauge.” It requires that $\chi$ satisfies the Poisson equation

\begin{displaymath}
- \nabla^2 \chi = \nabla\cdot\skew3\vec A_0
\end{displaymath}

Then the governing equations become

\begin{displaymath}
- \nabla^2 \varphi = \frac{\rho}{\epsilon_0}
\end{displaymath}


\begin{displaymath}
\frac{1}{c^2} \frac{\partial^2 \skew3\vec A}{\partial t^2}...
...
- \frac{1}{c^2} \nabla \frac{\partial \varphi}{\partial t}
\end{displaymath}

Note that $\varphi$ now satisfies a purely spatial Poisson equation.