= Solution
For a translation with the sign used in the paper, the field variation at fixed coordinates is $\delta A_\rho=\beta\epsilon^\nu\partial_\nu A_\rho$, and $\delta\mathcal L=\partial_\mu(\beta\epsilon^\mu\mathcal L)$. Differentiating the <Maxwell Lagrangian> with respect to the field gradient gives
$$
\frac{\partial\mathcal L}{\partial(\partial_\mu A_\rho)}=-F^{\mu\rho}.
$$
Thus <Noether's theorem> yields $j^\mu=\epsilon_\nu T^{\mu\nu}$, where the <canonical stress-energy tensor of the Maxwell field> is
$$
\boxed{T^{\mu\nu}=-F^{\mu\rho}\partial^\nu A_\rho-\eta^{\mu\nu}\mathcal L
=-F^{\mu\rho}\partial^\nu A_\rho+\frac14\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}.}
$$
It is conserved on the source-free <Maxwell equations>, but it is not generally symmetric. Indeed,
$$
T^{\mu\nu}-T^{\nu\mu}
=-F^{\mu\rho}\partial^\nu A_\rho+F^{\nu\rho}\partial^\mu A_\rho
$$
need not vanish. For example, take $A_0=a x^1$, $A_2=b x^0$, and the other components zero, with nonzero constants $a,b$. This produces constant field strength and satisfies the source-free equations, but $T^{12}=0$ and $T^{21}=ab$.
Nor is it <gauge-invariant>: while $F$ and $\mathcal L$ are unchanged, a <gauge transformation> gives
$$
\boxed{\delta_\xi T^{\mu\nu}=-F^{\mu\rho}\partial^\nu\partial_\rho\xi,}
$$
which is generally nonzero. These shortcomings motivate the improvement in part (vi); they do not contradict conservation of the <canonical stress-energy tensor>.
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