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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 4x4 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:

  1. Ricci tensor values in flat space are all zeroes.

$$ \begin{matrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{matrix} $$

  1. Ricci values for a 3D spherical surface are:

$$ \begin{matrix} 1 & 0\\ 0 & (sinx)^2\\ \end{matrix} $$

  1. Ricci values for the Schwartzchild metric are zeroes, ref. this

$$ \begin{matrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{matrix} $$

  1. Ricci values for the Kerr metric are also zeroes.

$$ \begin{matrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{matrix} $$

  1. Ricci values for the Robertson-Walker (FRW) metric are:

$$ \left(\begin{matrix} -\frac{3}{c^2}\frac{\ddot{a}}{a} & 0 & 0 & 0\\ 0 & \frac{a\ddot{a} + 2\dot{a}^2 + kc^2}{1-kr^2} & 0 & 0\\ 0 & 0 & (a\ddot{a} + 2\dot{a}^2 + kc^2)r^2 & 0\\ 0 & 0 & 0 & (a\ddot{a} + 2\dot{a}^2 + kc^2)r^2\sin^2\theta\\ \end{matrix}\right) $$

  1. 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 4x4 with only diagonal non-zero entries can be more simply represented as a [4x1] 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.

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    $\begingroup$ The Ricci tensor is symmetric, so the components you propose are not a valid set of components for the Ricci tensor. $\endgroup$ Oct 19, 2021 at 4:44
  • $\begingroup$ @VincentThacker thanks for pointing this out, I have edited the matrice to be symmetric around the diagonal axis :) $\endgroup$
    – James
    Oct 19, 2021 at 4:48
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    $\begingroup$ The Ricci tensor contains the exact same information as the stress-energy tensor. $\endgroup$ Oct 22, 2021 at 1:30
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    $\begingroup$ @James: you don't need all of the curvature of space for the geodesic path. The metric and the christoffel symbols are enough. In fact, knowing only the curvature means that you have to do some sort of integration to find everything else before you can compute the geodesic path. $\endgroup$ Oct 22, 2021 at 3:24
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    $\begingroup$ But all I'm saying is, if the einstein equaiton is true, then so is, $-R = 8\pi g^{ab}T_{ab}$, which then means $R_{ab} = 8\pi\left(T_{ab} - \frac{1}{2} g^{ab}T_{ab}\right)$. If you know the ricci tensor, you know the stress- energy tensor and vice-versa. They are direct functions of each other. $\endgroup$ Oct 22, 2021 at 3:27

3 Answers 3

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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

  • "volumetric" curvature, described by the Ricci tensor, the trace of the Riemann one,
  • "deviatoric" curvature, described by the Weyl tensor, the traceless part of the Riemann one.

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.

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  • $\begingroup$ @ Jacopo thanks for answering! I had not known about Ricci and Weyl tensor being subcomponents of the Riemann tensor. I will first read up on the Riemann tensor next. $\endgroup$
    – James
    Oct 19, 2021 at 8:29
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Since computers are very good at number crunching, I would leave that up to them. One of the major motivations of physics is to understand the physics and not merely calculate. Thus one should not 'shut up and calculate' but stop calculating and think, contemplate, understand etc etc.

For example, Ricci curvature, though termed a curvature, actually measures the deviation of a small geodesic ball from the standard ball. Thus in a Ricci-flat manifold where the Ricci curvature vanishes, small geodesic balls display no volume deviation but do have shape deviation. They are the special case of Einstein manifolds where the cosmological force vanishes, hence they are vacuum solutions of GR where the cosmological force vanishes. Arbitrary Einstein manifolds are vacuum solutions of GR with cosmological force, whether that be positive, vanishing, or negative.

Furthermore, Calabi-Yau manifolds which are used in the fibring of higher dimensional spacetime over our own 4d spacetime in string theory are Ricci-flat. We can alternatively see them as fattened spacetime points, or 'spacetime atoms'. Thus we see the 'atoms' of spacetime are vacuum solutions of GR and whose volume does not change but whose shape may do so.

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  • $\begingroup$ @Mozibur_Ullah thank you for answering! yes, but I have never seen how exactly to use computers to "calculate" usable results from the Ricci tensor. Maybe it's how the Ricci tensor is traditionally presented in textbooks. But instictively, since Ricci numbers describe curvature, I should be able to generate a computer graphics simulation of the manifold just be looking/processing the Ricci values at every single point, right? What is the concrete algorithm for doing this? $\endgroup$
    – James
    Oct 19, 2021 at 4:45
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    $\begingroup$ @James: No, I don't think you have 'understood' what I'm saying. Computers aren't the way to understand curvature tensors. It's because people have understood such tensors that they have been able to design computer packages of tensors. There are plenty of them. I don't know of any specific one - or rather, I did, but I've forgotten - a seach engine query will find you plenty $\endgroup$ Oct 19, 2021 at 4:59
  • $\begingroup$ @Mozibur_Ullah thanks. I have done another round of search using the term "algorithm to render manifold from ricci tensor" but there seems no result that presents the algorithm. Maybe it's the wrong keyword i used, if anyone more experienced in tensors can point out the right links please. $\endgroup$
    – James
    Oct 19, 2021 at 5:03
  • $\begingroup$ A search I did showed that mathematica has such a toolkit. Perhaps your seach request wasn't the most perspicacious? As old timer programmers used to say: garbage in, garbage out ... $\endgroup$ Oct 19, 2021 at 5:14
  • $\begingroup$ @Mozibur_Ullah thanks. I am using google search. Can you link to the matematica toolkit page you found, please? $\endgroup$
    – James
    Oct 19, 2021 at 5:17
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If you would like to do real practical calculations and know Python try Einstein.py or GraviPy.py. They are very useful.

References: Einstein.py GraviPy.py

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  • $\begingroup$ @ Samson thank you for the links! $\endgroup$
    – James
    Oct 19, 2021 at 7:23

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