homework and exercises - Variation of square root of determinant of metric, $delta g$

I am trying to calculate

$$ \frac{\partial \sqrt{- g}}{\partial g^{\mu \nu}},$$

where $g = \text{det} g_{\mu \nu}$. We have

$$ \frac{\partial \sqrt{- g}}{\partial g^{\mu \nu}} = - \frac{1}{2 \sqrt{-g}}\frac{\partial g}{\partial g^{\mu \nu}},$$ so the problem becomes how to calculate $\frac{\partial g}{\partial g^{\mu \nu}}.$

I have used the identity $\text{Tr}( \text{ln} M) = \text{ln} (\text{det} M)$ to obtain, applying it with $M = g^{\mu \nu}$ and varying it:

$$\delta (\text{Tr} (\text{ln} (g^{\mu \nu}))) = \frac{\delta g}{g} $$

but then I am stuck. How can I go on? I know the result should be $ -\frac{1}{2} g_{\mu \nu} \sqrt{-g}$

Answer

Use the identity that if $M$ is invertible and $\delta M$ is "small" compared to $M$, then we have $$ \det (M + \delta M) = \det(M) \det( 1 + M^{-1} \delta M) \approx \det(M) \left[ 1 + \text{tr} (M^{-1} \delta M) \right]. $$ In the case of the metric, this implies that $$ -\det(g + \delta g) \approx -\det(g) \left[ 1 + g^{ab} \delta g_{ab} \right] $$ and so $\delta (-g) = (-g) g^{ab} \delta g_{ab}$.

To complete the calculation you'll then have to relate $\delta g^{ab}$ to $\delta g_{ab}$, but this should get you on your way. If this isn't a homework problem or the like, let me know and I can expand on this latter part.