# Variation of the metric with respect to the metric

For a variation of the metric $g^{\mu\nu}$ with respect to $g^{\alpha\beta}$ you might expect the result (at least I did):

$$\frac{\delta g^{\mu\nu}}{\delta g^{\alpha\beta}}= \delta^\mu_\alpha\delta^\nu_\beta.$$

but then to preserve the fact that $g^{\mu\nu}$ is symmetric under interchange of $\mu$ and $\nu$ we should probably symmetrise the right hand side like this:

$$\frac{\delta g^{\mu\nu}}{\delta g^{\alpha\beta}}= \delta^\mu_\alpha\delta^\nu_\beta + \delta^\mu_\beta\delta^\nu_\alpha.$$

Is this reasonable/correct? If not, why not?

It seems that I can derive some weird results if this is right (or maybe I'm just making other mistakes).

Since the metric $g_{\mu\nu}=g_{\nu\mu}$ is symmetric, we must demand that

$$\tag{1} \delta g_{\mu\nu}~=~\delta g_{\nu\mu}~=~\frac{1}{2}\left(\delta g_{\mu\nu}+\delta g_{\nu\mu}\right)~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\delta g_{\alpha\beta},$$

and therefore

$$\tag{2} \frac{\delta g_{\mu\nu}}{\delta g_{\alpha\beta}} ~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right).$$

The price we pay to treat the matrix entries $g_{\alpha\beta}$ as $n^2$ independent variables (as opposed to $\frac{n(n+1)}{2}$ symmetric elements) is that there appears a half in the off-diagonal variations.

Another check of the formalism is that the RHS and LHS of eq. (2) should be idempotents because of the chain rule. For further motivation, see e.g. this Phys.SE post.

• I also considered this, but consider $\frac{\delta g^{12}}{\delta g^{12}} = \frac{1}{2}(\delta^1_1\delta^2_2 + \delta^1_2\delta^2_1) = \frac{1}{2}$ Would I not expect to get 1? – Sam Nov 26 '14 at 23:57
• Thank you, I had to think about it for a while but it makes sense now :) – Sam Nov 27 '14 at 21:48
• Hey, Sam. I got linked to this question from a similar one I had: physics.stackexchange.com/q/335173. I still don't quite get the issue with the factors of $\frac{1}{2}$. Why did it eventually make sense to you that $\frac{\partial g^{12}}{\partial g^{12}} = \frac{1}{2}$? That just looks so wrong to me. – Klein Four May 26 '17 at 20:34
• The point is that the two elements $g_{12}=g_{21}$ are equal, so they cannot be varied independently. If you (wrongly) think that $\frac{\partial g_{12}}{\partial g_{12}}$ should be $1$, while $\frac{\partial g_{21}}{\partial g_{12}}$ should be $0$, then recall that they are actually equal $\frac{\partial g_{12}}{\partial g_{12}}=\frac{\partial g_{21}}{\partial g_{12}}$! – Qmechanic May 26 '17 at 20:50