In field theory, the energy momentum defined as the functional derivative wrt the metric

$$T_{\mu\nu}=\frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu\nu}}$$

(up to a sign depending on conventions) For a theory in flat space, this has the advantage to give you directly an improved energy momentum tensor as the metric is symmetric, but it hold also for dynamical metrics.

Now, my problem is the following: If you have in your Lagrangian a term like $A^{\mu\nu}\nabla_\mu V^\mu$, where $V^\mu$ is a vector field (not needed to be a gauge field, for instance say it represents the velocity in hydrodynamics) and $A^{\mu\nu}$ an arbitrary tensor a priori depending on anything (metric, $V$, or any other field), you will have a Christoffel symbol. If you take a functional derivative, the expression will contain terms like

$\frac{\delta \Gamma^\rho_{\mu\nu}(x)}{\delta g^{\alpha\beta}(y)}\supset g^{\rho\sigma} \partial_\mu(g_{\nu\alpha}g_{\sigma\beta}\delta(x-y))$+permutations

those terms will not combine again to form a christoffel symbol. How can one take the functional derivative of something covariant ($\nabla V$) with respect to something covariant (the metric) and end up with something not covariant? For flat space cases, everything is well since those terms will vanish, but for curved space, we loose general covariance/diff invariance. What am I missing here?

  • $\begingroup$ Well, $g^{\mu\nu}\nabla_{\mu}V_{\nu}$ looks like a boundary term, so it wouldn't contribute to a Lagrangian. This is a stupid minor point, but perhaps $\nabla_{\mu}V_{\nu}\nabla^{\mu}V^{\nu}$ would be better... $\endgroup$ Commented Jul 26, 2013 at 14:51
  • $\begingroup$ yes indeed, I was thinking of a derivative of a vector appearing in the action. I edited $\endgroup$
    – Bulkilol
    Commented Jul 26, 2013 at 15:11
  • 5
    $\begingroup$ Note that $\delta \Gamma^\lambda_{\mu\nu}$ is tensor. Precisely $\delta \Gamma^\lambda_{\mu\nu} = \frac{1}{2} g^{\lambda \rho} \left( \nabla_\mu \delta g_{\rho \nu} + \nabla_\nu \delta g_{\mu \rho} - \nabla_\rho\delta g_{\mu\nu } \right)$. $\endgroup$
    – Prahar
    Commented Jul 26, 2013 at 15:15

1 Answer 1


@Prahar is right, the variation of the Christoffel symbol is a tensor, even if the Christoffel itself is not. We have

$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}\delta\bigg(g^{\rho\alpha}(2\partial_{(\mu}g_{\nu)\alpha}-\partial_\alpha g_{\mu\nu})\bigg)=\frac{1}{2}\delta g^{\rho\alpha}(2\partial_{(\mu}g_{\nu)\alpha}-\partial_\alpha g_{\mu\nu})+ \frac{1}{2}g^{\rho\alpha}(2\partial_{(\mu}\delta g_{\nu)\alpha}-\partial_\alpha \delta g_{\mu\nu})$

where $A_{(\mu\nu)}=\frac{1}{2}(A_{\mu\nu}+A_{\nu\mu})$. Using $\delta g^{\rho\alpha}=-g^{\rho\gamma}g^{\alpha\delta}\delta g_{\gamma\delta}$ we have:

$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}g^{\rho\alpha}(2\partial_{(\mu}\delta g_{\nu)\alpha}-\partial_\alpha \delta g_{\mu\nu}-2\Gamma_{\mu\nu}^\beta\delta g_{\alpha\beta})$

The Christoffel then combines nicely with the standard derivative to give a covariant tensor (the other Christoffel symbols cancel each other)

$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}g^{\rho\alpha}(2\nabla_{(\mu}\delta g_{\nu)\alpha}-\nabla_\alpha \delta g_{\mu\nu})$.

So to answer the original question, we finally have:

$\nabla_\mu V_\nu=\nabla_\mu \delta V_\nu-\frac{1}{2}g^{\rho\alpha}(2\nabla_{(\mu}\delta g_{\nu)\alpha}-\nabla_\alpha \delta g_{\mu\nu})A_\rho$

Remember that we did not assume anything on $V_\mu$. Depending on the problem, it is then possible to integrate by parts to isolate $\delta g_{\mu\nu}$ and obtain the energy momentum tensor.


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