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?)


1 Answer 1


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).

  • 1
    $\begingroup$ what are " important geomeric objects which are Riemann flat"? I only see one, minkowski space time $\endgroup$
    – magma
    Commented Dec 8, 2017 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
    Commented Dec 8, 2017 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.