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Minkowski space describes a flat spacetime and a Ricci-flat manifold is a solution of GRT without curvature. Shouldn't they be the same? So, what's the difference?

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    $\begingroup$ A Ricci-flat manifold is a solution of GRT without curvature. No, it isn’t. $\endgroup$ – G. Smith Jul 11 at 17:17
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Jul 11 at 17:26
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A flat spacetime has a zero Riemann tensor, while a Ricci flat spacetime has a zero Ricci tensor.

The Ricci tensor is a contraction of the Riemann tensor, and it is possible for the Ricci tensor to be zero when the Riemann tensor is not. An obvious example of this is the Schwarzschild geometry that describes a static black hole. In this geometry the Ricci tensor is zero everywhere but the spacetime is most certainly not flat.

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  • $\begingroup$ Is this somewhat similiar to a tube that has no inner curvature but an outer curved topology? $\endgroup$ – gabr1eL44 Jul 11 at 16:05
  • $\begingroup$ No. The Ricci tensor on its own does not fully describe a geometry. It describes only one aspect of the geometry - roughly it tells you what happens to the volume of an element transported along a geodesic. A zero Ricci tensor simply does not mean the spacetime is flat. $\endgroup$ – John Rennie Jul 11 at 16:13
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Ricci flat is not necessarily "truly" flat. Minkowski space has the entire Riemann curvature tensor vanishing, ie, $R_{\mu\nu\rho\sigma}=0$. However, being just Ricci flat is a weaker condition that only requires that the Ricci tensor $R_{\mu\nu}$ vanish.

For example, Calabi-Yau manifolds are Ricci flat, but they are certainly not the same as Minkowski space.

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  • $\begingroup$ Interestingly, for 2-dimensional manifolds, Ricci flatness and flatness are equivalent concepts. $\endgroup$ – Prof. Legolasov Jul 12 at 6:28

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