Maxwell stress tensor

Remarks

For $C^1$ solutions of Maxwell's equations with sources $\rho$ and $J$, a computation gives $$\sum_{j} \partial_j \sigma_{ij} = f_i + \varepsilon_0 \mu_0 \, \partial_t S_i,$$ where $f = \rho E + J \times B$ is the Lorentz force density and $S$ is the Poynting vector. By the divergence theorem, the total force on the charges in a bounded region $V$ is $$F_i = \int_{\partial V} \sum_j \sigma_{ij} n_j \, dA - \varepsilon_0 \mu_0 \frac{d}{dt} \int_V S_i \, dV.$$ For static fields the last term is zero, and the force depends only on the fields on $\partial V$.

Example: for a uniform field $E = (0, 0, E_0)$ and $B = 0$, the tensor is $\sigma = \frac{\varepsilon_0 E_0^2}{2} \operatorname{diag}(-1, -1, 1)$. So the field pulls with tension $\varepsilon_0 E_0^2 / 2$ along the field lines and pushes with pressure $\varepsilon_0 E_0^2 / 2$ across them; this tension gives the attraction between the plates of a capacitor.

Non-example: the matrix $\varepsilon_0 E_i E_j$ alone is not the stress tensor; the trace terms with $\delta_{ij}$ are needed for the force identity above.

Some books use the opposite sign, so that the tensor is the momentum flux density $-\sigma$.