Maxwell's equations in vacuum (and in MKS units) are \begin{eqnarray} \grad\cdot\EE &=& \frac{1}{\epsilon_0} \,\rho \label{Gauss}\\ \end{eqnarray}\begin{eqnarray} \grad\cdot\BB &=& 0 \\ \end{eqnarray}\begin{eqnarray} \grad\times\EE &=& -\Partial{\BB}{t} \label{Faraday}\\ \end{eqnarray}\begin{eqnarray} \grad\times\BB &=& \mu_0\,\JJ + \mu_0\epsilon_0 \Partial{\EE}{t} \label{Ampere} \end{eqnarray} where $\rho$ is the charge density, $\JJ$ is the current density, and the constants $\mu_0$ and $\epsilon_0$ satisfy (3) of §11.1. Equation ($\ref{Gauss}$) is just (the differential form of) Gauss' Law, ($\ref{Faraday}$) is Faraday's equation, and ($\ref{Ampere}$) is Ampère's Law corrected for the case of a time-dependent electric field. We also have the charge conservation equation \begin{equation} \grad\cdot\JJ = -\Partial{\rho}{t} \label{charge} \end{equation} and the Lorentz force law \begin{equation} \FF = q (\EE + \vv\times\BB) \end{equation}

The middle two of Maxwell equations are automatically solved by introducing the scalar potential $\Phi$ and the vector potential $\AA$ and defining \begin{eqnarray} \BB &=& \grad\times\AA \\ \EE &=& -\Partial{\AA}{t} - \grad\Phi \end{eqnarray}

What form do Maxwell's equations take if we rewrite them in tensor language? % Consider the following derivatives of $F^{\mu\nu}$: \begin{equation} \Partial{F^{\mu\nu}}{x^\nu} = \Partial{F^{\mu0}}{t} + \Partial{F^{\mu1}}{x} + \Partial{F^{\mu2}}{y} + \Partial{F^{\mu3}}{z} \end{equation} This corresponds to four different expressions, one for each value of $\mu$. For $\mu=0$, we get \begin{equation} 0 + \cm\Partial{E^x}{x} + \cm\Partial{E^y}{y} + \cm\Partial{E^z}{z} = \cm\,\grad\cdot\EE = \frac{\rho}{\cc\epsilon_0} = \cc\mu_0\rho \end{equation} where Gauss' Law was used to get the next-to-last equality. Similarly, for $\mu=1$ we have \begin{equation} -\cmsq\Partial{E^x}{t} + 0 + \Partial{B^z}{y} - \Partial{B^y}{z} \end{equation} and combining this with the expressions for $\mu=2$ and $\mu=3$ yields the left-hand-side of \begin{equation} -\cmsq\Partial{\EE}{t}+\nabla\times\BB = \mu_0\JJ \end{equation} where the right-hand-side follows from Ampère's Law. Combining these equations, and introducing the 4-current density \begin{equation} J^\mu = \pmatrix{\cc\rho\cr \JJ} \end{equation} leads to \begin{equation} \Partial{F^{\mu\nu}}{x^\nu} = \mu_0 J^\mu \label{tmax1} \end{equation} which is equivalent to the two Maxwell equations with a physical source, namely Gauss' Law and Ampère's Law.

Furthermore, taking the (4-dimensional!) divergence of the 4-current density leads to \begin{equation} \mu_0 \Partial{J^\mu}{x^\mu} = \Partial{}{x^\mu} \Partial{F^{\mu\nu}}{x^\nu} = 0 \end{equation} since there is an implicit double sum over both $\mu$ and $\nu$, and the derivatives commute but $F^{\mu\nu}$ is antisymmetric. (Check this by interchanging the order of summation.) Working out the components of this equation, we have \begin{equation} \cm\Partial{J^0}{t} + \Partial{J^1}{x} + \Partial{J^2}{y} + \Partial{J^3}{z} = 0 \end{equation} which is just the charge conservation equation ($\ref{charge}$).

What about the remaining equations? Introduce the dual tensor $G^{\mu\nu}$ obtained from $F^{\mu\nu}$ by replacing $\cm\EE$ by $\BB$ and $\BB$ by $-\cm\EE$, resulting in \begin{equation} G^{\mu\nu} = \pmatrix{ 0& B^x& B^y& B^z\cr \noalign{\smallskip} -B^x& 0& -\cm\,{E^z}& \cm\,{E^y}\cr \noalign{\smallskip} -B^y& \cm\,{E^z}& 0& -\cm\,{E^x}\cr \noalign{\smallskip} -B^z& -\cm\,{E^y}& \cm\,{E^x}& 0\cr } \end{equation} Then the four equations \begin{equation} \Partial{G^{\mu\nu}}{x^\nu} = 0 \label{tmax2} \end{equation} correspond to \begin{eqnarray} \grad\cdot\BB &=& 0 \\ -\cm\Partial{\BB}{t} - \cm\,\grad\times\EE &=& 0 \end{eqnarray} which are precisely the two remaining Maxwell equations.

Thus, Maxwell's equations in tensor form are just ($\ref{tmax1}$) and ($\ref{tmax2}$) — two tensor equations rather than four vector equations.

Some further properties of these tensors are \begin{eqnarray} \frac12 F_{\mu\nu} F^{\mu\nu} &=& -\cmsq\,{|\EE|}^2 + |\BB|^2 = - \frac12 G_{\mu\nu} G_{\mu\nu} \label{EM1}\\ \end{eqnarray}\begin{eqnarray} \frac14 G_{\mu\nu} F^{\mu\nu} &=& -\cm\>{\EE\cdot\BB} \label{EM2} \end{eqnarray} where care must be taken with the signs of the components of the covariant tensors $F_{\mu\nu}$ and $G_{\mu\nu}$. You may recognize these expressions as corresponding to important scalar invariants of the electromagnetic field.

Finally, it is possible to solve the sourcefree Maxwell equations by introducing a 4-potential \begin{equation} A^\mu = \pmatrix{\cm\,{\Phi}\cr \noalign{\smallskip} \AA} \end{equation} and defining \begin{equation} F^{\mu\nu} = \Partial{A^\nu}{x_\mu} - \Partial{A^\mu}{x_\nu} \end{equation} where again care must be taken with the signs of the components with “downstairs” indices. Furthermore, the Lorentz force law can be rewritten in the form \begin{equation} m \Partial{u^\mu}{\tau} = q u_\nu F^{\mu\nu} \end{equation} Note the appearance of the proper time $\tau$ in this equation. Just as in the previous chapter, this is because differentiation with respect to $\tau$ pulls through a Lorentz transformation, which makes this a valid tensor equation, valid in any inertial frame.