Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 37B (Electrodynamics)

Dot the vacuum Ampère-Maxwell equation with $\mathbf{E}$ and Faraday's law with $\mathbf{B}/{\mu }_0$. The vector identity
$\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot \nabla \times \mathbf{E}-\mathbf{E}\cdot \nabla \times \mathbf{B}$
gives the Poynting theoremThusis the Poynting vector, the electromagnetic energy-flux density.

Differentiate $\mathbf{g}={ϵ}_0\mathbf{E}\times \mathbf{B}$, use the vacuum Maxwell equations, and use $\nabla \cdot \mathbf{E}=\nabla \cdot \mathbf{B}=0$. Componentwise vector identities then givewhereThis is the negative of the conventional Maxwell stress tensor; with the sign used here, its divergence is the outward momentum flux.

With $x^0=ct$, so that ${\partial }_0=c^{-1}{\partial }_t$, the $\nu =0$ component of
${\partial }_{\mu }{T}^{\mu \nu }=0$ isand the $\nu =j$ component isThese are exactly the energy and momentum conservation laws from parts (a) and (b), now combined into conservation of the electromagnetic stress-energy tensor.

Let $J_k=\int {ϵ}_{k\ell j}{x}_{\ell }{g}_j{d}^{3}x$. Local momentum conservation and integration by parts, with the boundary term vanishing for a localized field, givebecause ${\sigma }_{ij}$ is symmetric. Hence the total electromagnetic angular momentum is independent of time.

For $T^{\mu \nu }$ to be symmetric in every inertial frame, its mixed components must satisfyFor arbitrary fields this requires