What is the intuitive meaning of the scalar curvature R?

Question

Background:

Let $M$ be a smooth, Riemannian manifold with metric $g$ and dimension $n$. Let $R^a_{bcd}$ be the Riemann tensor with respect to the Levi-Civita connection for $g$.

Question:

Is there any rigorous result that gives a good intuitive sense of the meaning of the scalar curvature $R = R^{ab} R_{ab}$?

Discussion:

What I have in mind is something like the following:

For $n =2$, the volume of a geodesic ball of radius $\epsilon$ in $M$ is $\pi \epsilon^2 [1 - (R / 48) \epsilon ^2 + O(\epsilon^4) ]$. I may have the numerical factors wrong, but the point is this: $R$ tells you the difference between the volume of a geodesic ball and an ordinary Euclidean ball (for small radius). That's the kind of result I'm looking for.

My problem with this result is that it holds in normal geodesic coordinates but not in general coordinates. (Note that a choice of coordinates is necessary to define 'a geodesic ball of radius $\epsilon$'). If you know how to generalize this result to arbitrary coordinates, or you know another result that gives some intuition for $R$, please let me know.

Thanks for any help!

Answer 1

What you have mentioned for $n=2$ is in fact true for higher dimension too: Quoted from here, we know that the ratio of the $n$-dimensional volume of a ball of radius $\epsilon$ in the manifold $M$ to that of a corresponding ball in Euclidean space $\mathbb{R}^n$ is given by, for small $\epsilon$, $$\frac{\mbox{Vol}(B_\epsilon(p)\subset M)}{\mbox{Vol}(B_\epsilon(p)\subset\mathbb{R}^n)}=1-\frac{R}{6(n+2)}\epsilon^2+O(\epsilon^4).$$ (Therefore, you are right with the constant for the case when $n=2$).

Answer 2

Let us start with two familiar settings. On the sphere of radius 1, a spherical cap of geodesic radius $\rho$ has area $2 \pi (1 - \cos \rho).$ In the hyperbolic plane of curvature $-1,$ the area of a disk of radius $\rho$ has area $2 \pi ( \cosh \rho - 1).$ Writing out just a few terms of the power series for $\cos$ and $\cosh$ tells you the comparison to $\pi \rho^2.$

Actually, it is not necessary to specify a coordinate system of any sort to define distance on a Riemannian manifold, that is what geodesics are for. As a result, for small radii, the geodesic ball is also well-defined. Furthermore, in a pleasant feature, for small enough radii the geodesics do meet the sphere bounding the ball orthogonally, this is usually called the Gauss Lemma

Other than that, see Paul's answer...