Appendix: Mathematical Details

### The Stress Tensor for a Point Charge

The electromagnetic field is a 2-form $F$. Maxwell's equations imply that $F$ is closed, that is that $$dF = 0$$ which leads to the assumption that $$F = dA$$ where $A$ is the 4-potential. In special relativity, we have $$A = -\Phi\,dt + \AA\cdot d\rr$$ 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 $$F = E \wedge dt + {*}B$$ 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 $$\EE = \frac{q}{r^2} \>\rhat$$ 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) $$F = \frac12 F_{ij} \,\sigma^i\wedge\sigma^j = \frac{q}{r^2} \>\sigma^r\wedge\sigma^t$$

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

1) If this expression reminds you of curvature 2-forms, that is no coincidence. Gauge field theory, including electromagnetism, can be formulated in the geometric language of fibre bundles; the corresponding notion of curvature in the electromagnetic case is precisely $F$.