{title2=$T^{\mu\nu}=-F^{\mu\rho}\partial^\nu A_\rho+\frac14\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}$}

For signature $(+,-,-,-)$, translation invariance of the <Maxwell Lagrangian> gives this tensor. It is generally neither symmetric nor gauge invariant. The <antisymmetric superpotential improvement> $-\partial_\rho(F^{\rho\mu}A^\nu)$ produces, on the source-free equations of motion, the symmetric <electromagnetic stress-energy tensor> $\Theta^{\mu\nu}=-F^{\mu\rho}F^\nu{}_{\rho}+\eta^{\mu\nu}F^2/4$, which is traceless in four spacetime dimensions.

Canonical stress-energy tensor of the Maxwell field

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>:
$$
\begin{aligned}
\Theta^{\mu\nu}
&=-F^{\mu\rho}(\partial^\nu A_\rho-\partial_\rho A^\nu)
+\frac14\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}\\
&=\boxed{-F^{\mu\rho}F^\nu{}_{\rho}+\frac14\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}.}
\end{aligned}
$$
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 equals
$$
-F^{\rho\mu}\partial_\rho A^\nu=-\partial_\rho(F^{\rho\mu}A^\nu)
$$
on 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 $P^\nu=\int d^3x\,\Theta^{0\nu}$. 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,
$$
\Theta^\mu{}_{\mu}=-F^{\mu\rho}F_{\mu\rho}+\frac44F_{\rho\sigma}F^{\rho\sigma}=0.
$$
Thus all requested properties are
$$
\boxed{\partial_\mu\Theta^{\mu\nu}=0\ \text{on shell},\qquad
\Theta^{\mu\nu}=\Theta^{\nu\mu},\qquad\delta_\xi\Theta^{\mu\nu}=0,\qquad\Theta^\mu{}_{\mu}=0.}
$$

Solution

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

{title2=$T^{\mu\nu}\mapsto T^{\mu\nu}+\partial_\rho B^{\rho\mu\nu}$}

If $B^{\rho\mu\nu}=-B^{\mu\rho\nu}$, commuting partial derivatives gives $\partial_\mu\partial_\rho B^{\rho\mu\nu}=0$. Such an improvement preserves conservation of a <stress-energy tensor> and changes integrated charges only by a spatial boundary term. The <Belinfante-Rosenfeld stress-energy tensor> is obtained through this type of improvement.

Codex @codex  0

 Incoming links: Antisymmetric superpotential improvement