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In differential geometry and general relativity space is said to be flat if the Riemann tensor $R=0$. If the Ricci tensor on manifold $M$ is zero, it doesn't mean that the manifold itself is flat. So what's the geometrical meaning of Ricci tensor since it's been defined with the Riemann tensor as

$$\mathrm{Ric}_{ij}=\sum_a R^a_{iaj}?$$

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    $\begingroup$ Are you looking for a geometrical meaning within a physical context? Otherwise, this might be more appropriate for math.SE? Also, you'll probably find the statements in this terse Wiki useful en.wikipedia.org/wiki/Ricci-flat_manifold in case you haven't looked at it yet. $\endgroup$ – joshphysics Jan 7 '14 at 18:40
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    $\begingroup$ The Ricci tensor tells you how the volume of some test element changes as you move it around. If the Ricci tensor is zero then as you move your test element around, the shape may change (due to tidal forces) but the volume will remain constant. $\endgroup$ – John Rennie Jan 7 '14 at 18:47
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    $\begingroup$ You have written down the Ricci scalar, $R = {g^{\mu \nu }}{R_{\mu \nu }=0}$ (${g^{\mu \nu }}$ is the metric) not the Ricci tensor, where ${R_{\mu \nu }} = 0$, is a system of equations. This is quite different. $\endgroup$ – Bruce Dean Jan 7 '14 at 18:47
  • $\begingroup$ In my textbook Ricci tensor is written as I have. R is the Riemann tensor. $\endgroup$ – Apogee Jan 7 '14 at 18:53
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    $\begingroup$ You might also like: math.stackexchange.com/q/468651 . And $R$ is in fact, the Riemman tensor, the Ricci tensor and the Ricci scalar. But if you express their components then you get $R_{\mu \nu \lambda \sigma}$,$R_{\mu \nu}$ and $R$, respectively. $\endgroup$ – jinawee Jan 7 '14 at 19:56
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The local geometric structure of a pseudo-Riemannian manifiold $M$ is completely described by the Riemann tensor $R_{abcd}$. The local structure of a manifold is affected by two possible sources

  1. Matter sources in $M$: The matter distribution on a manifold is described by the stress tensor $T_{ab}$. By Einstein's equations, this can be related to the Ricci tensor (which is the trace of the Riemann tensor = $R_{ab} = R^c{}_{acb}$. $$ R_{ab} = 8 \pi G \left( T_{ab} + \frac{g_{ab} T}{2-d} \right) $$

  2. Gravitational waves on $M$. This is described by the Weyl tensor $C_{abcd}$ which is the trace-free part of the Riemann tensor.

Thus, the local structure of $M$ is completely described by two tensors

  1. $R_{ab}$: This is related to the matter distribution. If one includes a cosmological constant, this tensor comprises the information of both matter and curvature due to the cosmological constant.

  2. $C_{abcd}$: This describes gravitational waves in $M$. A study of Weyl tensor is required when describing quantum gravity theories.

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    $\begingroup$ I'd like to propose that the first time you write the word "manifold," you include a qualifier such as "Riemannian" or "pseudo-Riemannian" since a manifold need not possess a metric. Further, do your statements implicitly assume that the Levi-Civita connection has been chosen? $\endgroup$ – joshphysics Jan 7 '14 at 21:38
  • $\begingroup$ OK. I am talking about Riemannian Manifolds. I am not sure about the Levi-Civita connection. Can you elaborate how that affects the discussion? $\endgroup$ – Prahar Jan 7 '14 at 21:40
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    $\begingroup$ This is incorrect. The Schwarzschild spacetime is ricci flat and without gravitational waves, but has a nonzero Weyl tensor. The same is true of the Taub spacetime, and other generalized solutions. $\endgroup$ – Jerry Schirmer Jan 7 '14 at 21:48
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    $\begingroup$ @Trimok, not familiar with Lanczos tensor, but Petrov Classification en.wikipedia.org/wiki/Petrov_classification does decribe gravitational waves through algebraic special types II, III and N. Algebraic special type D should describe "coulombian"-like gravitational field. O'Neill's The Geometry of Kerr Black Holes describe petrov classification in detail. Thus Weyl tensor should be analogous to a gravitational field, the Ricci tensor describing separately the action of matter $\endgroup$ – cesaruliana Feb 5 '14 at 11:36
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    $\begingroup$ @Trimok: O'Neill is a great place to learn Petrov Class., but for a precise statement or its relation to grav. waves I should suggest Penrose and Rindler's "Spinors and Spacet-ime". Chp. 8 in Vol. 2 discuss Petrov Class. for both Weyl tensor and the eletromagnetic Faraday tensor, showing that in the later it decomposes in eletromag. radiation and coulomb. Chp. 9 discuss the decay of algebraic types that confirm what is waves and waht is coulomb-like $\endgroup$ – cesaruliana Feb 5 '14 at 16:18
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I've always liked the interpretation you get from the Raychaudhuri equation. It shows you that the Ricci tensor tends to cause geodesics to focus together. If you begin with a family of geodesics with tangent vector $u^a$, you can define the expansion $\theta\equiv \nabla_a u^a$ which measures the rate at which geodesics are spreading out or converging together. As you move along a an integral curve of $u^a$, the Raychaudhuri equation tells you how the expansion changes as a function of curve's parameter, $\lambda$: $$ \frac{d}{d\lambda}\theta = -\frac13\theta^2-\sigma_{ab}\sigma^{ab}+\omega_{ab}\omega^{ab}-R_{ab}u^au^b.$$ $\sigma_{ab}$ is called the shear and is related to the tendency of a cross section of the curves to distort toward and ellipsoid, and $\omega_{ab}$ is the vorticity and describes how the curves twist around each other. The Ricci tensor appears in this equation with a minus sign, so that when $R_{ab}u^au^b$ is positive, it tends to decrease the expansion, which describes focusing of the geodesics.

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