Stress-energy tensor for a massive vector field I am trying to work out the symmetric stress-energy tensor for a free massive vector field and show that it is conserved. The Lagrangian density and resulting EoM are:
$$L=-\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta}+\frac{1}{2}m^2 A^{\beta}A_{\beta}$$
$$\partial_\mu F^{\mu\nu}+m^2 A^\nu=0$$
Where I am using the metric $\eta_{\mu\nu}=diag(+1,-1,-1,-1)$.
Evaluating the usual expression for the canonical stress energy tensor:
$$T_C^{\mu \nu}=\frac{\partial L}{\partial \left(\partial_{\mu} A_{\lambda}\right)}\eta^{\nu\beta}\partial_{\beta} A_{\lambda}-\eta^{\mu \nu}L$$
$$\frac{\partial L}{\partial \left(\partial_{\mu} A_{\lambda}\right)}=F^{\lambda\mu}$$
$$T_C^{\mu \nu}=F^{\lambda\mu}\eta^{\nu\beta}\partial_{\beta} A_{\lambda}+\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta}\eta^{\mu \nu}-\frac{1}{2}m^2 A^{\beta}A_{\beta}\eta^{\mu \nu}$$
According to what I can find on this site and elsewhere, the symmetric stress energy should be (adding the mass term to the form found here):
$$T^{\mu\nu}=\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta}\eta^{\mu \nu}+F^{\mu\lambda}\eta_{\lambda\kappa}F^{\kappa\nu}-\frac{1}{2}m^2 A^{\beta}A_{\beta}\eta^{\mu \nu}$$ 
When I calculate the difference between these I get:
$$T^{\mu\nu}-T_C^{\mu \nu}=S^{\mu\nu}=-F^{\beta\nu}\partial_\beta A^\mu$$
It should be that $\partial_\mu S^{\mu\nu}=0$ so that $T^{\mu\nu}$ is conserved just as $T_C^{\mu\nu}$ is. But I calculate:
$$\partial_\mu S^{\mu\nu}=-\left( \partial_\mu F^{\beta\nu}\right) \partial_\beta A^\mu-F^{\beta\nu}\partial_\beta \left( \partial_\mu A^\mu\right)$$
The second term is zero since $\partial_\mu A^\mu=0$ by the EoM. But I can't figure out how to show that the first term is zero, nor can I see that $\partial_\mu T^{\mu\nu}=0$ when I try to calculate it directly. Where am I going wrong?
 A: *

*The Lagrangian density for massive E&M is
$${\cal L}~=~\sqrt{|g|}L, \qquad L~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\mp \frac{1}{2}m^2 A^2, \tag{1}$$
for Minkowski sign convention $(\mp,\pm,\pm,\pm)$. The Euler-Lagrange (EL) equations read
$$ d_{\mu}F^{\mu\nu}~\stackrel{(1)}{\approx}~\pm m^2A^{\nu} .\tag{2}$$ 

*The mass term breaks gauge-invariance but not translational invariance, so that the canonical stress-energy-momentum (SEM) tensor
$$\mp T_C^{\mu}{}_{\nu}~:=~\frac{\partial L}{\partial(\partial_{\mu}A_{\lambda})}\partial_{\nu}A_{\lambda}-\delta^{\mu}_{\nu}L
~\stackrel{(1)}{=}~F^{\lambda\mu}\partial_{\nu}A_{\lambda}-\delta^{\mu}_{\nu}L\tag{3}$$
is conserved cf. Noether's first theorem

*The Hilbert/metric SEM tensor is
$$ \mp T_{\mu\nu}~:=~\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g^{\mu\nu}}
~\stackrel{(1)}{=}~- F_{\mu}{}^{\lambda}F_{\nu\lambda}\mp m^2A_{\mu}A_{\nu} - g_{\mu\nu}L.\tag{4}$$
In eq. (4) it is important that we think of $A$ as a co-vector/one-form rather than a vector.

*The difference becomes
$$S^{\mu}{}_{\nu}~:=~T^{\mu}{}_{\nu}-T_C^{\mu}{}_{\nu}
~\stackrel{(3(+(4)}{=}~\mp F^{\mu\lambda}d_{\lambda}A_{\nu}+ m^2A^{\mu}A_{\nu}
~\stackrel{(2)}{\approx}~\mp d_{\lambda}(F^{\mu\lambda}A_{\nu}),\tag{5}$$
so that
$$d_{\mu}S^{\mu}{}_{\nu}~\stackrel{(5)}{\approx}~\mp d_{\mu}d_{\lambda}(F^{\mu\lambda}A_{\nu})~=~0,\tag{6}$$
by (anti)symmetry of indices $\mu\leftrightarrow\lambda$. Hence the conservation law for the two SEM tensors (3) & (4) are equivalent, cf. OP's question.
