At which point of the universe $R_{\mu \nu}=0$ if there is a source of gravitation (point mass)

Question

Schwarschild found his solution to Einstein's field equations for vacuum ($T_{\mu \nu}=0$) by placing a point-mass in the center of origin.

Since the Ricci tensor $R_{\mu \nu}$ and the Einstein tensor $G_{\mu \nu}$ are trace-reversed, meaning they vanish identically:

Which point of a universe with a mass in it would meet this equation: $R_{\mu \nu}=0$ (and therefore $G_{\mu \nu}=0$)?

(to make sure, infinity is not a point, i.e. if gravity is infinite than you can't find a particular point, where it vanishes down to zero due to distance)

Answer 1

Gravitational curvature is not fully described by the Einstein or stress-energy tensors. It is only fully described by the Riemann tensor.

Consider, by analogy, a simple example from electromagnetism: $\nabla \cdot E =\rho/\epsilon_0$, Gauss's law for the electric field. The function $\rho$ describes the distribution of charge density. The LHS is zero wherever there is no charge density, but that does not mean the electric field is zero.

Similarly, in regions of vacuum, the stress-energy tensor is zero, and thus so is the Einstein tensor, but that does not mean the Riemann tensor is zero!

Indeed, the Riemann tensor can be decomposed into two parts: a traceful part that corresponds to the Ricci or stress-energy tensors but also a traceless, self-dual part, the Weyl tensor, which can be nonzero even in empty space (space without energy-momentum).

So Einstein's equations in vacuum mean exactly that: that $G_{\mu \nu} = 8\pi T_{\mu \nu} = 0$ in a region without mass-energy. That is far from saying that there is no gravity, just as it would be silly to say there is no electric field in the exterior of a charged ball.

Answer 2

The vacuum equation $G_{\mu\nu} = 0$ holds wherever there is vacuum. This is tautological so I don't really see what the content of the question is.

If you are hesiant about point masses you can consider the Schwarschild interior solution, which describes a spherically symmetric non-rotating star with uniform density. The Schwarschild interior solution, matched with the exterior solution, has $R_{\mu\nu} = 0$ everywhere outside the star. I don't have nearby right now so I can't give a precise page number, but the standard reference here is Exact Solutions to Einstein's Field Equations by Stephani et al.

To deal with a point mass you need to weaken your demands on smoothness and such, but this is not unique to general relativity, it's the same in Newtonian gravitation and classical electrodynamics.

Answer 3

In 3+1 dimensions, the Ricci tensor vanishes when the stress-energy tensor vanishes indeed. Which means that, whenever there's a vacuum, the Ricci tensor vanishes as well.

But it also happens that, in that number of dimension, the metric is not entirely determined by the Ricci tensor. The full curvature tensor (the Riemann tensor) is a mixture of the Ricci tensor and the Weyl tensor. The Weyl tensor roughly describes how gravity propagates : if you have some energy, it will influence the metric of the vacuum regions around it.

I emphasize the 3+1 dimensions because that is not the case in 2+1 dimensions (gravity stops in a vacuum, because the Riemann tensor is $\propto$ the Ricci tensor) and even less so in 1+1 dimensions (the Einstein tensor is a topological invariant, and does not vary with the stress energy tensor).

In the case of Schwarzschild, what you have is a distribution of energy in the middle. There's several ways to describe it. You can use distributions, but Schwarze distributions handle non-linear equations poorly, and non-linear distributions are a rather thorny topic. But there's still some ressources talking about how to describe a singularity as a Dirac distribution. It has the benefit of being rather clear : The stress energy tensor is a Dirac, and so is the Ricci tensor.

You can show it indirectly, as well, by calculating the Komar mass. You find that, in an arbitrarily small region around the singularity, there's a mass M. The Ricci tensor is, as usual, 0 outside of the singularity and undefined at r = 0.

Of course you can also just have a plain old spherical body. In that case, the metric is just Schwarzschild outside (with energy density 0) and whatever may be inside, with, hopefully, a continuity in the metric between the inside and the outside.

A good way to illustrate how the curvature of space does not depend entirely on the Ricci tensor is to look at empty space. The simplest vacuum spacetime is Minkowski space, but it is not the only one. You can freely add sourceless gravitational radiations (using for instance pp-wave metrics), which will also be a vacuum solution.

In the case of the Schwarzschild spacetime, the stress energy tensor is not enough to get the full metric because you are not looking at all the spacetime. You are only looking at its vacuum regions without looking at its energy distributions, which can influence other regions. In the case of gravitational waves, it's a boundary value problem : solving the Einstein equation still requires, like all PDEs, some boundary value of the field.

Answer 4

There is no per se 'point'. The entire 'mass' of the universe defines itself, no matter how we might perceive it, as an infinite density with no locality. In other words, the entire universe is, always has been and always will be a pointless so-called 'singularity'. We can't perceive it or measure it as such, because we are part of it, so it appears to us to have dimensional extensiveness.