A.37 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, [41, 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

$\parbox{400pt}{\hspace{11pt}\hfill$\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

$$\fbox{$\displaystyle \frac{1}{c^2} \frac{\partial^2 \skew3... ...c\jmath}{\partial t} - \frac{1}{\epsilon_0} \nabla \rho $} %$$ (A.235)

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

$$\fbox{$\displaystyle \frac{1}{c^2} \frac{\partial^2 \skew2... ...} = \frac{1}{\epsilon_0 c^2} \nabla \times \vec\jmath\; $} %$$ (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 $\skew3\vec A_0$.

$\parbox{400pt}{\hspace{11pt}\hfill$\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

$\parbox{400pt}{\hspace{11pt}\hfill$\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

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

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

$\parbox{400pt}{\hspace{11pt}\hfill$\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,

$$\frac{1}{c^2} \frac{\partial^2 \chi}{\partial t^2} - \nabla... ...A_0 - \frac{1}{c^2} \frac{\partial \varphi_0}{\partial t} = 0$$

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),

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

From (4), af­ter us­ing (8),

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

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

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

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