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, [39, 20.35-45].

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

$$\begin{array} {@{\hspace{.7in}}c@{\hspace{.7in}}c@{\hspace{.6in}... ...{\epsilon_0} +\frac{\partial \skew3\vec{\cal E}}{\partial t} & (4) \end{array}$$

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

$\parbox{404pt}{\hspace{15pt}\hfill$\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

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

Similarly, for the magnetic field take the curl of (4) and use (2) and (3) to get

$$\fbox{$\displaystyle \frac{1}{c^2} \frac{\partial^2 \ske... ...= \frac{1}{\epsilon_0 c^2} \nabla \times \vec\jmath\; $} %$$ (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

$$\sin\Big(k(y-ct)+\varphi\Big)$$

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 $\skew3\vec A_0$.

$\parbox{404pt}{\hspace{15pt}\hfill$\displaystyle \skew2\vec{\cal B}= \nabla \times \skew3\vec A_0 $\hfill(6a)}$

Next define

$$\skew3\vec{\cal E}_\varphi \equiv \skew3\vec{\cal E}+ \frac{\partial \skew3\vec A_0}{\partial t}$$

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

$\parbox{404pt}{\hspace{15pt}\hfill$\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

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

where $\chi$ is any arbitrary function of $x$, $y$, $z$, and $t$, then still

$\parbox{404pt}{\hspace{15pt}\hfill$\displaystyle \skew2\vec{\cal B}= \nabla \times \skew3\vec A $\hfill(6)}$

since the curl of a gradient is always zero, and

$\parbox{404pt}{\hspace{15pt}\hfill$\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

$\parbox{404pt}{\hspace{15pt}\hfill$\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,

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

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),

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

From (4), after using (8),

$$\fbox{$\displaystyle \frac{1}{c^2} \frac{\partial^2 \ske... ...^2 \skew3\vec A = \frac{\vec\jmath}{\epsilon_0 c^2} $} %$$ (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

$$- \nabla^2 \chi = \nabla\cdot\skew3\vec A_0$$

Then the governing equations become

$$- \nabla^2 \varphi = \frac{\rho}{\epsilon_0}$$

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

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