# Relation between curvature and metric determinant

This is my first question so please let me know if I'm asking my question appropriately!

I've been trying to build up an intuition for what curvature of spacetime represents physically. Mostly I have been trying to use Regge Calculus to inform my intuition because it is easier to visually see the difference between a curved and not curved spacetime (at least in two dimensions).

What I have gathered from Regge Calculus is that when your space is positively curved at some point, in some sense there is "less" space there, while when it is negatively curved there is more. You can demonstrate this by taking a flat piece of paper, cutting a wedge out and taping the remaining paper back together. You will end up with a cone, with a positive curvature concentrated at the vertex. Similarly if you add paper to your flat piece (cut the paper partially and tape a wedge onto it), and tape the paper back together, you will end up with a saddle point, indicative of negative curvature.

From my understanding, the square root of the metric determinant $\sqrt{-g}$ can unequivocally be interpreted as the density of spacetime, because $\sqrt{-g}d^4x$ is the invariant volume of spacetime, where $d^4x$ is the volume if the spacetime were flat.

My question is, is $\sqrt{-g}$ somehow related to the curvature of spacetime? If it is, my guess would be that the second partial derivative of some function of $\sqrt{-g}$ is equal to the Ricci tensor:$$\frac{\partial^2f(\sqrt{-g})}{\partial x_\mu\partial x_\nu}= R_{\mu\nu}$$ If only to satisfy dimensions. $f$ would need to have the appropriate transformation properties to make the lhs a tensor. Again, if there is a relationship between the two, it should depend on the variation of the density, because if $\sqrt{-g}>1$ but constant, the spacetime is still flat.

I have attempted to write $R_{\mu\nu}$ solely in terms of $\sqrt{-g}$ but I got bogged down with the large number of terms.

If there is a reference that would be helpful, that would also be greatly appreciated!

Edit: I found an old post Does curved spacetime change the volume of the space?

Where one of the answers states that the volume element is equal to

$$dV=(1-\frac{1}{6}R_{ij}x^ix^j+\mathcal{O}(x^3))dV_f$$

I'm not really sure how they arrived at this relationship, but it has some promise for a relation similar to what I mentioned above. Does anyone know how one can derive this relationship?

Unfortunately, there is no simple relation as,

$$\partial_a \partial_b f(\sqrt{g}) \sim R_{ab}.$$

If there were, it would make many computations simpler, and the curvature much faster to calculate by hand. As far as I am aware the closest link between the determinant and the Ricci curvature tensor is given by,

$$R_{ab} = \partial_c \Gamma^c_{ab} - \Gamma^d_{ac}\Gamma^c_{db} - \nabla_b(\partial_a \log \sqrt{g})$$

which can be obtained from usual formulas by noting $\Gamma^a_{ba} = \partial_a \log \sqrt{g}$. This necessitates calculating the Christoffel symbols as well as a covariant derivative, so it is not really a shortcut.

One thing that can be said in $d= 2$ is that there is only one independent component of the Riemann curvature tensor, say, $R_0$ and $R_0 \propto \det g$ with scalar curvature as the proportionality constant.

As for the relation mentioned in the question, it is a consequence of the fact that, in normal coordinates, in a neighbourhood about the point $p$,

$$g_{ab} = \delta_{ab} - \frac13 R_{acdb}x^cx^d -\frac16 (\nabla_eR_{acdb})x^cx^dx^e + \mathcal O(x^4).$$

See here and here. For $\mathbb R^{1,d-1}$ as opposed to $\mathbb R^d$, $\delta_{ab}$ is replaced with $\eta_{ab}$.

• Thanks for your answer. I edited my question with a result that I found after posting, are you familiar with it? Jun 22, 2018 at 18:03
• @LucashWindowWasher Dear questioner, maybe you should look at Kahler manifolds where the ricci tensor is proportional to the double derivative of log the metric determinant. Consult nakahara chapter 8. May 26, 2020 at 7:23