The electromagnetic field is a 2-form $F$. Maxwell's equations imply that $F$ is closed, that is that \begin{equation} dF = 0 \end{equation} which leads to the assumption that \begin{equation} F = dA \end{equation} where $A$ is the 4-potential. In special relativity, we have \begin{equation} A = -\Phi\,dt + \AA\cdot d\rr \end{equation} where $\Phi$ is the electric potential, also called the scalar potential, and the 3-vector $\AA$ is the magnetic potential, also called the vector potential. The electric and magnetic fields are then \begin{align} \EE &= -\grad\Phi \\ \BB &= \grad\times\AA \end{align} both of which are 3-vectors, that is, have no component in the $t$ direction. In the language of differential forms, we have \begin{equation} F = E \wedge dt + {*}B \end{equation} where of course $E=\EE\cdot d\rr$, $B=\BB\cdot d\rr$, and where $*$ denotes the 3-dimensional Hodge dual operator.

The electric field of a point charge should take the form \begin{equation} \EE = \frac{q}{r^2} \>\rhat \end{equation} where in our geometric units we set $4\pi\epsilon_0=1$, so that, like mass, charge has the dimensions of length. We therefore expect the electromagnetic field of a point charge to take the form 1) \begin{equation} F = \frac12 F_{ij} \,\sigma^i\wedge\sigma^j = \frac{q}{r^2} \>\sigma^r\wedge\sigma^t \end{equation}

The components of the stress-energy tensor for an electromagnetic field can be shown to be \begin{equation} 4\pi T^i{}_j = F^{im}F_{jm} - \frac14 \delta^i{}_j F^{mn}F_{mn} \end{equation} which reduces to the diagonal matrix \begin{equation} 4\pi \left(T^i{}_j\right) = \frac12 \begin{pmatrix} -q/r^2 & 0 & 0 & 0 \\ 0 & -q/r^2 & 0 & 0 \\ 0 & 0 & q/r^2 & 0 \\ 0 & 0 & 0 & q/r^2 \end{pmatrix} \end{equation} for the electromagnetic field of a point charge.