In Misner, Thorne and Wheeler's Gravitation exercise 21.3, the authors suggest

For a more ambitious project, show that any stress-energy tensor derived from a field Lagrangian by the prescription of equation (21.33) will automatically satisfy the conservation law $T_{\alpha\beta}^{\hphantom{\alpha\beta};\beta}=0$.

Equation 21.33: $$T_{\alpha\beta}=-2\frac{\delta L_{field}}{\delta g^{\alpha\beta}}+g_{\alpha\beta}L_{field}.\tag{21.33}$$

I was able to show that for a scalar field Lagrangian of the form $L(\phi,g^{\mu\nu}\phi_{,\mu}\phi_{,\nu})$ the Euler-Lagrange equation together with the definition of the stress-energy tensor in equation (21.33) imply that $T_{\alpha\beta}^{\hphantom{\alpha\beta};\beta}=0$. The key point seemed to be the connection between derivatives with respect to $g^{\alpha\beta}$ and those with respect to the field and its derivatives given by the Euler-Lagrange equation. In the case of a vector field, however, the possible forms of terms in the Lagrangian seem to be infinite (it seems to always be possible to contract a higher tensor power of $\nabla A$ along combinations of "slots" that do not come from combinations of lower powers, unlike the scalar case).

Is this "ambitious project" well-defined and doable? If so, what is the right way to approach the vector field (or more general) case?

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    $\begingroup$ Does this answer your question? When is stress-energy tensor defined as variation of action with respect to metric conserved? $\endgroup$
    – G. Smith
    Feb 24, 2020 at 5:59
  • $\begingroup$ yeah it's not clear as to why this would help motivate a solution to my problem sadly. Can you give a little more detail as to why this could help? $\endgroup$ Feb 26, 2020 at 23:17
  • $\begingroup$ Did you see that part that says “we obtain in a single stroke ... the covariant conservation law” that you are trying to show? $\endgroup$
    – G. Smith
    Feb 26, 2020 at 23:35
  • $\begingroup$ And did you read the links in Qmechanics’ answer? $\endgroup$
    – G. Smith
    Feb 26, 2020 at 23:38
  • $\begingroup$ yes, and after reading it several times I think I know what I need to do. Thanks :) $\endgroup$ Feb 27, 2020 at 2:57