I'm trying to find the field equations for some particular Lagrangian. In the middle I faced the term
$$\frac{\delta \Gamma_{\beta\gamma}^{\alpha}}{\delta g^{\mu\nu}} \, .$$
I know that
$$\delta \Gamma_{\beta\gamma}^{\alpha} = \frac{1}{2}g^{\sigma\alpha}(\nabla_{\beta}(\delta g_{\sigma\gamma}) + \nabla_{\gamma}(\delta g_{\sigma\beta}) – \nabla_{\sigma}(\delta g_{\beta\gamma})) \, .$$
I have two questions:
-
Is the expression for $\delta \Gamma_{\beta\gamma}^{\alpha}$ somehow related to $\frac{\delta \Gamma_{\beta\gamma}^{\alpha}}{\delta g^{\mu\nu}}$?
-
The idea at the end is to have terms like
$$\frac{\delta \mathcal{L}}{\delta g^{\alpha\beta}} = 0$$ and thus make the variation of the action invariant under $\delta g^{\alpha\beta}$. So in simple words, is there is any way to have the term $\delta g^{\alpha\beta}$ put of the variation of the Christoffel symbol?
Best Answer
Possible hint with regards to your first question:
Notice that $$\frac{\delta \Gamma^a_{bc}}{\delta g^{\mu \nu}} \equiv \frac{\delta \Gamma^a_{bc}}{\delta\sqrt{-g}}\frac{\delta\sqrt{-g}}{\delta g^{\mu\nu}}.$$
It then follows that (see this answer and this wiki)
$$ \frac{\delta\sqrt{-g}}{\delta g^{\mu\nu}} = -\frac{1}{2}g_{\mu\nu}\sqrt{-g}.$$
Hopefully this can project the answer in the right direction. Or at least bring the question up on the feed again.