While there are many explanations about how to derive the Ricci curvature from a given metric, I can't recall having seen the reverse approach, ie. given that the Ricci curvature numbers already supplied to us, show then how these numbers can be useful in concrete calculations.
There have been many processes in mathematics and physics which are difficult to derive, but easy to use (eg. a Fourier transform will take forever to compute by hand, but the results of the frequency bin values computed laboriously are immediately useful, for example to decode a radio message.) Since it is also quite laborious to derive the Ricci 4×4 curvature numbers for a spacetime, one might expect that the use of these numbers, once obtained, be easy, after all our hard work calculating them?
Some Ricci curvature results I have come across are:
- Ricci tensor values in flat space are all zeroes.
$$
\begin{matrix}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{matrix}
$$
- Ricci values for a 3D spherical surface are:
$$
\begin{matrix}
1 & 0\\
0 & (sinx)^2\\
\end{matrix}
$$
- Ricci values for the Schwartzchild metric are zeroes, ref. this
$$
\begin{matrix}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{matrix}
$$
- Ricci values for the Kerr metric are also zeroes.
$$
\begin{matrix}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{matrix}
$$
- Ricci values for the Robertson-Walker (FRW) metric are:
$$
\left(\begin{matrix}
-\frac{3}{c^2}\frac{\ddot{a}}{a} & 0 & 0 & 0\\
0 & \frac{a\ddot{a} + 2\dot{a}^2 + kc^2}{1-kr^2} & 0 & 0\\
0 & 0 & (a\ddot{a} + 2\dot{a}^2 + kc^2)r^2 & 0\\
0 & 0 & 0 & (a\ddot{a} + 2\dot{a}^2 + kc^2)r^2\sin^2\theta\\
\end{matrix}\right)
$$
- Ricci values for the Reissner-Nordstrom metric are:
$$
\left(\begin{matrix}
\frac{r_Q^2}{r^4} – \frac{r_Q^2 r_s}{r^5} +\frac{r_Q^4}{r^6} & 0 & 0 & 0\\
0 & \frac{r_Q^2}{ r^3 r_s- r^4 – r^2r_Q^2} & 0 & 0\\
0 & 0 & \frac{r_Q^2}{r^2} & 0\\
0 & 0 & 0 & \frac{r_Q^2}{r^2}\sin^2\theta\\
\end{matrix}\right)
$$
Looking at the common analytically known metrics above, one is possibly impressed by the number of zero entries in these matrices. If there are more zeroes than useful values in a matrices representation of a theory, does it not imply that at some level we are perhaps "over-complicating" the concept? For example a 4×4 with only diagonal non-zero entries can be more simply represented as a [4×1] vector with modified operational rules for manipulation.
My question is, given this ingenious Ricci values that I have divined in my head:
$$
\begin{matrix}
0 & 1 & 2 & 3\\
1 & 5 & 6 & 7\\
2 & 6 & 8 & 7\\
3 & 7 & 7 & 3\\
\end{matrix}
$$
What useful, conrete calculations can I do now with these values?
Thank you.
Best Answer
It seems you are after some sort of geometric understanding of what the components of the Ricci tensor "tell us", so I'll try to give some indications in this regard. Intrinsic curvature is described by the Riemann tensor, and its effects can be qualitatively decomposed into
Let us focus on the first. By "volumetric" I mean that it distorts spacetime in a way which "expands" or "contracts" volumes compared to the corresponding Euclidean manifold. In this wikipedia page you can find an expression for the volume element expanded around a point that uses the Ricci tensor. This can be used to understand what the $ij$ components of the Ricci tensor "do".
If all of them are zero, the space may not be Euclidean, but its volume element is (try multiplying all the diagonal entries in the Schwarzschild metric, for example). There still can be curvature, though, and it will be described by the Weyl tensor.
In the sphere case, the fact that $R_{\phi \phi} = \sin^2 \theta$ means that volume (or area, in this case) along the $\phi$ direction is "squished" as $\theta$ decreases. If you are near a Pole, for example, a small coordinate rectangle with sides $\Delta \theta$ and $\Delta \phi$ will have a smaller area than a rectangle with the same parameters drawn at the equator, and it will be compressed along the $\phi$ direction, while the $\theta$ direction will remain untouched.