Vacuum solutions in presence of mass?

Question

Here is the page I will be referencing: Vacuum solution (general relativity) - Wikipedia

My point is: if $T_{\mu\nu}=0$ implies that there is no mass, how can Schwarzschild vacuum be a solution, if its whole point is the description of the geometry of the spacetime around a spherical mass?

I'm quite ashamed to say I asked ChatGPT about it, and it told me that vacuum is a local thing, it's about a specific region of spacetime which is considered, and that the whole universe can have nonzero mass or energy. My questions are:

1. Is this right?
2. If it is, is it mathematically possible to have a solution where no mass nor energy is present in all of the universe? How?

Answer 1

The Einstein field equations hold pointwise, i.e. the value of $G_{\mu\nu}$ at one point is $8\pi G$ times the value of $T_{\mu\nu}$ at that point. Having $G_{\mu\nu}=0$ or $T_{\mu\nu}=0$ at one point does not mean that there cannot be any mass elsewhere. The Schwarzschild solution is a vacuum solution by definition because it describes the vacuum region surrounding a single spherically-symmetric object. $T_{\mu\nu}=0$ there means that $G_{\mu\nu}=0$ there too. But again, it does not necessarily extend to all regions. If we move into a region with mass, such as within the object itself, the equations and solutions for that region indeed become different. It's like solving a differential equation in two adjacent regions with different sources and matching the boundary conditions at their common boundary to obtain the final solution. In the case where $T_{\mu\nu}=0$ everywhere except a single point, the solution extends everywhere else and describes a Schwarzschild black hole.

In addition, $T_{\mu\nu}=0$ only means that the Ricci tensor is zero. It does not imply that the Riemann tensor is zero. So spacetime curvature is very much possible in vacuum.

It is certainly mathematically possible to have a solution where no mass nor energy is present in all of the universe. Minkowski spacetime is such a solution.

Answer 2

In general relativity, mass doesn't always mean matter. In the maximally extended Schwarzschild solution, which represents a black hole, you don't have matter anywhere: $T_{\mu\nu} = 0$ at all points in spacetime. Yet, you have mass. How can this be?

The answer is that in general relativity gravity also gravitates. The very curvature of spacetime gravitates due to the nonlinearities in Einstein's equation, and leads to more gravity. At infinity, we can measure the gravitational field and attribute it to a certain amount of mass. This is essentially how you define the mass of a black hole: by its gravitational field. Yet, there is no matter to be seen inside the black hole. All there is is gravity.

In other cases, in which there is some matter somewhere in the spacetime, you get the aspects of gravitation that were mentioned in other answers. Namely, while the Einstein equations enforce that the Ricci tensor vanish on vacuum, they still leave room for other components of the Riemann tensor to be finite. These components are determined based on the presence of matter on all of spacetime, not only at the point in which there is vacuum. This is similar to how the electromagnetic field can be nonvanishing at points in which there are no charges or currents.

Answer 3

Energy distorts spacetime is a similar way to mass, so mass is not strictly required to have the curved spacetime of the Schwarzschild metric. If there is neither mass nor energy, then the space would be flat and described by the Minkowski metric.

I suspect you are possibly confused by the the Schwarzschild metric being called a "vacuum" solution, implying no mass. In the Schwarzschild metric, the region covered extends from zero to infinity. The mass is all located at the central singularity and occupies no space at all, so the vacuum extends from zero to infinity, so calling it a vacuum solution is reasonable, but there is mass present.

Answer 4

ChatGPT's answer is not wrong, but it does not answer your question. The problem lies in a misunderstanding of what a solution to Einstein's field equations is. In the case of static spherically symmetric spacetime, they are simply linear ordinary differential equations of first and second order, see reference here. Mathematically, a differential equation has an infinite number of solutions. Physically, however, there is only one solution, a function satisfying the equation that also satisfies the corresponding physical boundary conditions, for example, the values of pressure and energy density at a given boundary. Now, if you start from a "universe" consisting only of spacetime, you cannot set any physical boundary conditions, because spacetime is immaterial. When you solve the corresponding differential equations for Schwarzschild vacuum spacetime, you will get the result:

$$g_{tt}(r)~\cdot g_{rr}(r)=1,~~~g_{tt}=1+\frac{C}{r}~.$$

The obove solution apparently depends on an arbitrary constant. How to determine it without the presence of matter is not obvious and has been the subject of some conceptual contortions. Usually the requirement of consistency with Newton's theory of gravity is used. It is $$g_{tt}(r)=1+2 \frac{\Phi(r)}{c^2}+...=1-\frac{2 G M}{ c^2}\frac{1}{r}+...=1+\frac{C}{r}~.$$ In this way the constant $C$ can be determined as $$-C=\frac{2GM}{c^2}\equiv r_{S}~,$$ the famous Schwarzschild radius. However, Newton's theory implies the presence of mass, which would contradict the assumption of an empty universe.

Answer 5

Notice that the Schwarzscshild metric is valid only for the region $r\ge 2GM$, as in its derivation it's assumed that outside of the spherically symmetric distribution of mass there is only vacuum and no other matter sources. If you want to extend the description of the metric to $r<2GM$, then you need to consider a particular form for $T_{\mu\nu}$. See for example the Schwarzscshild interior solution.

For your second question, General Relativity allows a vacuum solution of the Einstein Field Equations (EFEs) where $T_{\mu\nu}=0$ throughout the whole manifold, e.g., Minkowski (so no matter/energy is present in "all of the Universe), and it's gravitational waves, as they are solutions of the linearized EFEs $\square h_{\mu\nu} =0$.