Ricci tensor

In differential geometry, the Ricci curvature tensor, named after Gregorio Ricci-Curbastro, is a geometric object which is determined by a choice of Riemannian or pseudo-Riemannian metric on a manifold. It can be considered, broadly, as a measure of the degree to which the geometry of a given metric tensor differs locally from that of ordinary Euclidean space or pseudo-Euclidean space.

In a local chart

R i c (X, Y) := R_{a m b}^{m} X^{a} Y^{b}

where $R_{j a b}^{i}$ are the components of the Riemann curvature tensor.

It has components $R_{a b} = R i c (e_{a}, e_{b}) = R_{a m b}^{m}$ .

Related: Ricci scalar curvature.

The expression for the Ricci tensor component $R i c_{i j}$ in terms of the Christoffel symbols is:

R i c_{i j} = \partial_{k} Γ_{i j}^{k} - \partial_{j} Γ_{i k}^{k} + Γ_{i j}^{k} Γ_{k l}^{l} - Γ_{i l}^{k} Γ_{j k}^{l}

Interpretation

Idea: see the introduction of this article.

From this paper.

Sum of Gaussian Curvatures

Whenever I've asked a mathematician what the Ricci tensor means, they've explained the meaning of the Riemann tensor as a collection of Gaussian curvatures and simply stated that the Ricci tensor was an average. While I
do not find this explanation very satisfying, it bears further investigation. Suppose you wanted to find the average curvature of all planes involving the vector $S$ . You could start by taking a collection of orthonormal vectors $t_{i}$ and saying

{\bar{κ}}_{S} = \frac{1}{D} \sum_{i = 1}^{D} R (S, t_{i}, S, t_{i}) = \frac{1}{D} R (S, S) .

We should note that these are actually area weighted curvatures. That is they contain a factor of the area squared of the parallelogram formed by $S$ and $t_{i}$ .

One may ask whether summing over one orthonormal set is sufficient. Perhaps we should average over all such orthonormal sets. The Ricci tensor does not depend on your basis, so no further averaging is required. So while the Riemann tensor told us the Gaussian curvature of any given sub plane, the Ricci tensor gives us the average of all
sub planes involving a given vector.

Volume Deviation

I find the previous description lacking because instead of describing the behavior of a single physical object, it describes an average of the behavior of several objects. There is a way to describe the physical meaning of the Ricci tensor without invoking the notion of an average.
Suppose instead of looking at two small objects in space, we considered a volume filling collection of small objects in space. Describing the relative acceleration of any two of them would require the geodesic deviation equation, but to describe the evolution of their volume, we would have to average over several different versions of the equation. These have roughly the result of averaging the Riemann tensor into a Ricci tensor.
So in roughly the same sense that the Riemann tensor governs the evolution of a vector or a displacement parallel propagated along a geodesic, the Ricci tensor governs the evolution of a small volume parallel propagated along a geodesic. We must be careful though. Unlike vectors, volumes may change along geodesics even in a flat space. We must therefore subtract off any change that would occur in a flat space. Suppose then that we have a small volume
$δ V$ of dust near a point $x_{0}^{μ}$ . If we allow that volume to move along a direction $T^{μ} = \frac{d x^{μ}}{d τ}$ we find the following
equation:

\frac{D^{2}}{{d τ}^{2}} δ V - \frac{D_{f l a t}^{2}}{{d τ}^{2}} δ V = - δ V R_{μ ν} T^{μ} T^{ν} .

One may ask whether this volume should be a $D$ -dimensional volume (in relativity a space time volume) or a $D - 1$ -dimensional volume (a space only volume). It turns out that it can be either, so long as the $D - 1$ -dimensional
volume is transverse to the vector $T$ . Thus, we may apply the equation to the deviation of a spacelike volume as it propagates through time.