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The following question is based upon elementary concepts of General Relativity, and Tensor Calculus

The condition for a space-time manifold to be flat is:

$$R^{a}_{bcd} \equiv 0 $$

I.e., the Riemann Tensor is must be identically null.

And, Einstein Field Equations are written in terms of Ricci tensor ($R_{bd} = R^{a}_{bad}$) as follows:

\begin{equation} \tag{1} R_{bd} - \frac{1}{2}Rg_{bd} = 8\pi T_{bd} \end{equation}

My question is: If a vaccum solutions of (1) is the one that: \begin{equation} \tag{2} R_{bd} = 0 \end{equation}

And, since Ricci tensor is just a contraction of Riemann tensor, vaccum solutions describe a flat or curved manifold? Ok, I know that (2) gathers both kinds of solutions (even still confused), but what is the whole of Energy-Momentum tensor and Vaccum Solutions then?(e.g. what happens if I put Dust Energy-Momentum tensor on Schwarzschild metric. I mean if Schwarzschild solution is derived from (2) what suppose to mean a solution with Energy-momentum tensor?)

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First, ${R^a}_{bcd} = 0$ does imply that $R_{bd} = 0$, but $R_{bd} = 0$ does not imply that ${R^a}_{bcd} = 0$.

Also, in General Relativity specifically, we're not usually too interested in metrics that satisfy ${R^a}_{bcd} = 0$. These are called Riemann-flat metrics. Some important geometric objects are Riemann-flat, but they don't generally make the most interesting model spacetime.

Now, if you know that your spacetime has stress-energy tensor $T_{bd}$ that is not equal to zero, but the metric of the spacetime is exactly Schwarzschild, then you don't have a solution of Einstein's equations. Maybe it's almost a solution, or maybe it's a solution to some modified form of gravity, but it just doesn't satisfy your equation (1). That's all that says. It's still a geometrically possible manifold, it just doesn't correspond to what we believe is physical reality.

On the other hand, if you know that your spacetime has stress-energy tensor $T_{bd}$ that is not equal to zero and you know that your spacetime satisfies Einstein's equations, then you know that your metric is not the same as Schwarzschild — you need to look for some other metric to satisfy equation (1).

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    $\begingroup$ what are " important geomeric objects which are Riemann flat"? I only see one, minkowski space time $\endgroup$ – magma Dec 8 '17 at 6:08
  • $\begingroup$ The Riemann tensor is not only used for spacetimes in GR; it can also be computed for other manifolds. The line, the plane, the circle, and many other geometrical objects can be endowed with a flat metric. But as I said, Riemann-flat manifolds do not make for interesting model spacetimes in GR. Of course, this is not to mention Minkowski with non-trivial topological identifications, which is a non-trivial object of study for some relativists. $\endgroup$ – Mike Dec 8 '17 at 13:46

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