# Doubt about the vacua equations of General Relativity

I'm facing a quite annoying conceptual problem concerning the Einstein Field Equations (EFE) in so called "vacuum". This problem is both physical and mathematical.

So, in a elementary point of view, when a manifold is flat we have:

$$R^{\delta}_{\mu \gamma \nu} = 0\tag{1}$$

i.e. the Riemannian curvature is zero.

Consider now the EFE in the form:

$$R_{\mu \nu} = 0.\tag{2}$$

This equation have both strong mathematical and physical meaning:

About the mathematical meaning this is a condition called Ricci flat and about the physical meaning this is the EFE in vaccum.

1. So, firstly, Ricci flat condition implies Riemann flat condition?

2. Secondly, I have a serious doubts about the concept of vacuum. What is vaccum in GR? I mean, Kerr metric is a vacuum solution but the Manifold is curved, so the absence of matter shouldn't implies a flat space?

• That's not Einsteins Field Equations. That's just an arbitrary condition you've just applied. – Mozibur Ullah Nov 8 '18 at 21:01
• @MoziburUllah In vacuum the Einstein field equations really do simplify to $R_{\mu\nu} = 0$. – Mike Nov 8 '18 at 21:07

The Kerr solution is a vacuum solution everywhere BUT the singularity! If you were to "smooth out" the singularity, rest assured there would be a very big $$R_{\mu \nu}$$ in that region.

The way you solve the Kerr solution is by saying "what if I have $$R_{\mu \nu} = 0$$ everwhere in a rotationally symmetric space time with the boundary conditions that my metric becomes Minkowski at infinity?"

In actuality, the only such solution is to have space be flat everywhere! However, when solving the equations people usually discount the origin, allowing for more interesting solutions.

Note that if $$R_{\mu \nu \rho \sigma} = 0$$ your space must be Minkowski everwhere. However, you can have $$R_{\mu \nu \rho \sigma} \neq 0$$ somewhere and still have $$R_{\mu \nu} = 0$$. This is the case in the Kerr solution. Kerr space time is not flat. It is curved (even outside the singularity) even though $$R_{\mu \nu} = 0$$ there. There is no contradiction. Ricci flat is much weaker than being completely flat.

• I like this answer, and I always appreciated the role of the Weyl Tensor: a nice way of viewing the difference between flatness of the Ricci tensor vs flatness of the Riemann tensor, since the Riemann tensor is zero precisely when the Weyl tensor is zero. en.wikipedia.org/wiki/Weyl_tensor – N. Steinle Nov 8 '18 at 20:26

The vanishing of the curvature tensor $$R_{abcd}=0$$ indeed implies that the Ricci tensor also vanishes, $$R_{ab}=0$$, but not the other way around.

It is possible to have curved spacetime in a region where the stress-energy tensor $$T_{\mu\nu}=0$$ (assuming it is non-zero elsewhere causing curvature in the first place).

• "It is possible to have curved spacetime in a region where the stress-energy tensor $T_{\mu\nu}=0$", ok, why? – M.N.Raia Nov 8 '18 at 20:12
• I've just edited my answer - as long as $T_{\mu\nu}$ is non-zero somewhere, this will give rise to curvature everywhere in spacetime. If however we're only interested in what's happening in the vacuum, we can just say we wish to solve $R_{\mu\nu}=0$. – Garf Nov 8 '18 at 20:13
• When the energy-momentum tensor is zero, this means that there's no matter or you are faraway from the souce of curvature? – M.N.Raia Nov 8 '18 at 20:13
• No I don't think this is correct - for example a very basic form of $T_{\mu\nu}$ is for a non-relativistic fluid $$T_{\mu\nu}=\rho u_\mu u_\nu$$ and if we say $\rho$ is non-zero in a finite region (say a sphere) and is $0$ elsewhere, then $T_{\mu\nu}$ behaves the same - however, spacetime is still curved in the region where $\rho=0$. – Garf Nov 8 '18 at 20:18

The answers already given are correct, but I will elaborate a little in hopes of giving some physical insight.

It is a common misconception to think that zero $$R_{ab}$$ implies no curvature, or that zero $$T_{ab}$$ implies no curvature and hence no tidal effects (geodesic deviation and the like). To avoid the misconception about $$R_{ab}$$, you can think of $$R_{ab}$$ as a statement about direction-averaged curvature, that is, curvature after averaging over direction. Then what we are saying for vacuum is not there is no curvature, but rather if there is some positive curvature in one direction, then there must be negative curvature in another direction.

Next, to get a feel for the role of $$T_{ab}$$, compare it to the simpler case of charge and current as a source of electromagnetic fields. If there is a region of space with no charge and current, it does not necessarily mean the electromagnetic field tensor has to be zero there. Rather, its divergence is zero. Similarly, if $$T_{ab}$$ is zero, that in itself does not say whether the Riemann curvature is zero, but it does restrict the way the Riemann tensor can vary from one place to another.