Can spacetime be curved even in absence of any source? [duplicate]

Einstein's equation in absense of any source (i.e., $$T_{ab}=0$$) $$R_{ab}-\frac{1}{2}g_{ab}R=0$$ has the solution $$R_{ab}=0.$$

But I think $$R_{ab}=0$$ does not imply that all components of the Riemann-Christoffel curvature tensor $$R^c_{dab}$$ be zero (or does it?). From this can I conclude that spacetime can be curved even in absence of any source?

• Nov 20 '20 at 13:50
• Thanks, I think I get it. But I wonder under what condition $R_{\mu\nu}=0$ will also imply a flat space. No source anywhere? See my comment at Charlie. Nov 20 '20 at 13:59
• @Thiago Thanks. These are helpful. Nov 20 '20 at 14:11
• Yes, that is the reason why gravitation propagates from sources.... Nov 20 '20 at 19:06

What you're asking about is referred to as a vacuum solution to the field equations. This does not mean that there is no mass anywhere, rather that we are considering a region of our curved spacetime in which there is no mass.

The Schwarzschild solution for instance is a "vacuum solution" because we are considering the region outside of the central mass in which there is no matter, but in which the curvature is non-zero.

You are correct that the vanishing of the components of the Ricci tensor does not imply the vanishing of the components of the full Riemann tensor. $$R_{\mu\nu}=0$$ is a vacuum solution, $${R^\alpha}_{\beta\mu\nu}=0$$ is flat spacetime.

• I see. So the source enters via boundary conditions? Can I, therefore, draw the conclusion that if there is no source anywhere $R_{ab}=0\Rightarrow R^c_{dab}=0$? Nov 20 '20 at 13:56
• @mithusengupta123: You'd also have to make some kind of assumption about the behavior of the metric at infinity (timelike, spacelike and null.) For example, a solution with a plane gravitational wave traveling through space is a valid vacuum solution (just like a plane EM wave is a solution to Maxwell's equations with $J_a = 0$.) Nov 20 '20 at 14:02
• I think I have to delve deeper into the details of it to understand things better. Thanks for the answer and comments though :-) Nov 20 '20 at 14:04

I would view this in the same light as the following question:

Does

$${\bf \nabla \cdot E} = \frac{\rho}{\epsilon_0}$$

imply zero electric field in region with no charge density?

To which the answer is clearly, "No".

And as an example: The astronauts on the moon. They were there in a pretty good vacuum dropping feathers and hammers, which then took off on like geodesics.

You are right. $$R_{ab}=0$$ does not imply $$R^{a}_{bcd}=0$$. For one thing, $$R_{ab}$$ has 10 components (in $$n=4$$ dimensions), whereas $$R^{a}_{bcd}$$ has $$20$$ components. The simplest example I can think of is Schwarzschild solution, which has $$R_{ab}=0$$ everywhere but $$R^{a}_{bcd}\neq0$$. If you allow the inclusion of a cosmological constant, then the de Sitter metric is an example of an empty solution with non-trivial spacetime curvature. As pointed out here

https://physics.stackexchange.com/a/105336/96768

A spacetime containing gravitational waves is empty but with non-trivial Riemann tensor.

That's right. But it doesn't mean that the curvature is from nowhere. The Field equation describes the curvature (locally) at a point only from $$T_{\mu \nu}$$ at the same point (Since it's all built in a differential manifold and tangent spaces at each points aren't related to each other). If $$T_{\mu \nu}$$ is zero at a point, then you end up deriving a vacuum solution.