Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 4 / vi

Antisymmetry gives $-F^{\rho \mu }{\partial }_{\rho }{A}^{\nu }=F^{\mu \rho }{\partial }_{\rho }{A}^{\nu }$. Combining this with the canonical stress-energy tensor removes the potential in favour of the electromagnetic field tensor:This is the electromagnetic stress-energy tensor in the $(+,-,-,-)$ convention. The product $F^{\mu \rho }{F}^{\nu }{}_{\rho }$ is symmetric in $\mu ,\nu$, since the contracted metric is symmetric. Its expression only in $F$ makes it gauge-invariant. These two properties hold without using the equations of motion.

For conservation, the source-free Maxwell equations are ${\partial }_{\rho }{F}^{\rho \mu }=0$. Hence the given correction equalson shell. This is an antisymmetric superpotential improvement: $B^{\rho \mu \nu }=-F^{\rho \mu }{A}^{\nu }$ is antisymmetric in $\rho ,\mu$, so ${\partial }_{\mu }{\partial }_{\rho }{B}^{\rho \mu \nu }=0$. Conservation of the canonical stress-energy tensor therefore implies ${\partial }_{\mu }{\Theta }^{\mu \nu }=0$. Equivalently, direct differentiation gives ${\partial }_{\mu }{\Theta }^{\mu \nu }=-F^{\nu }{}_{\rho }{\partial }_{\mu }{F}^{\mu \rho }$ after the homogeneous Maxwell equations cancel the field-strength derivative terms.

There are four conserved currents, one for each fixed $\nu$, with charges $P^{\nu }=\int d^{3}x{\Theta }^{0\nu }$. The Belinfante-Rosenfeld stress-energy tensor differs from the canonical tensor only by boundary contributions to these charges when the fields decay suitably.

Finally, in four spacetime dimensions,Thus all requested properties are