1
$\begingroup$

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?

$\endgroup$
5
6
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I see. So the source enters via boundary conditions? Can I, therefore, draw the conclusion that if there is no source anywhere $R_{ab}=0\Rightarrow R^c_{dab}=0$? $\endgroup$ Nov 20 '20 at 13:56
  • 2
    $\begingroup$ @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 $J_a = 0$.) $\endgroup$ Nov 20 '20 at 14:02
  • $\begingroup$ I think I have to delve deeper into the details of it to understand things better. Thanks for the answer and comments though :-) $\endgroup$ Nov 20 '20 at 14:04
4
$\begingroup$

This is a simple answer:

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

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