# Does a vacuum solution to the Einstein equation imply flat spacetime?

I have read that a solution to the vacuum Einstein equation has a vanishing Einstein tensor, and therefore a vanishing stress-energy tensor. This means that there is no matter to generate spacetime curvature. If there is no matter to generate spacetime curvature, I assumed (apparently naively) that spacetime is flat.

But the Schwarzchild metric is a solution to the vacuum Einstein equation and is obviously not flat. For instance, one can calculate the deflection of light and Shapiro time delay which are obviously curvature effects. Moreover, the Schwarzchild metric describes spacetime outside a spherically symmetric mass $$M$$ and so there is necessarily some mass present.

How does one reconcile the fact that the Schwarzchild metric depends explicitly on a mass $$M$$ yet is also a solution to the vacuum equations? More generally, how can solutions to the vacuum Einstein equations have curvature if "mass tells spacetime how to curve"?

• Read the answer here: physics.stackexchange.com/a/439754/157704 Mar 28, 2019 at 3:48
• For a concrete example, think of gravitational waves in vacuum. Mar 28, 2019 at 5:11

## 2 Answers

You must be careful with the definitions. A vacuum solution to Einstein's equations might mean vacuum on the entire manifold or a subset of it. In the case of Schwarzschild solution, vacuum solution means that $$G_{\mu \nu}$$ vanishes everywhere except at the origin $$r=0$$. The origin is excluded from the definition of Schwarzschild solution. So when it is said that Schwarzschild is a Ricci flat solution, it really means that $$R_{\mu \nu} = 0$$ in whole spacetime, excluding the origin. This is very similar to the case of electrostatics where the a point charge (delta function) at $$r=0$$ produces an electric field everywhere outside, while blowing up at the origin. Similarly, for GR, there is a source that produces the Schwarzschild metric: it's a delta function source at $$r=0$$. But this point is excluded in the domain of validity of Schwarzschild metric because we are often concerned with the curvature outside of a spherically symmetric object (as you correctly stated in your question).

Also, the correct measure of curvature is not Ricci tensor but the Riemann tensor $$R_{\mu \nu \rho \sigma}$$. This is non-zero for Schwarzschild everywhere outside the point source/spherical object.

And lastly, it is entirely possible for spacetime to have curvature without any matter: freely propagating gravitational waves carry energy and momentum themselves, producing non-zero curvature.

Reference: pages $$43,78$$: General Relativity and the Einstein Equations, Yvonne Choquet-Bruhat

No, a vacuum solution does not imply flat spacetime.

It is possible, as in a Schwarzschild metric, to have a zero Einstein tensor but a nonzero Riemann tensor. The Riemann tensor is the most detailed indicator of curvature, with 20 independent components (out of 256 nominal components) at each point in spacetime. The Einstein tensor is more like a curvature average, because each of its components is a sum over multiple components of the Riemann tensor. It has only 10 independent components (out of 16 nominal components) at each point.

“Mass tells spacetime how to curve” is oversimplified. “The density and flow of energy and momentum tells spacetime how to curve on average” is more accurate.