0
$\begingroup$

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?
$\endgroup$
2
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/87332/174766 $\endgroup$ Commented May 24 at 22:46
  • 5
    $\begingroup$ It is bad enough to think that an artificial idiot could possibly know anything about GR, it is quite another to refer its answer to this site. $\endgroup$
    – m4r35n357
    Commented May 25 at 8:52

5 Answers 5

5
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ To further the anology, consider Gauss Law: no flux at a point does not mean there are no charges nearby. It just means there are no sources in the region considered, but there can well be non-zero $E$ and $B$ fields etc. $\endgroup$
    – Craig
    Commented May 29 at 12:20
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Isn't it possible that some mass is on the black hole "surface", that is, event horizon? $\endgroup$ Commented May 27 at 12:56
  • $\begingroup$ @JánLalinský Most black holes are actively absorbing material and there is continuous flow of matter through the event horizon, so at any given time there matter at the event horizon. However, if you are suggesting there is stationary matter at the 'surface' that is a very controversial position to take. However, there was an interpretation called the gravastar which suggests the matter is frozen in time at the event horizon and it would look very much like a traditional black hole externally. $\endgroup$
    – KDP
    Commented May 27 at 17:39
  • $\begingroup$ Re continuous flow through, that may happen in the frame of a freely falling observer' who crosses the event horizon, but is that really what happens in the frame of the outside observer? I thought from their perspective, all processes in any body that approaches the event horizon slow down, and at the event horizon they stop, and even the fall itself stops. So anything that fell there after the black hole horizon is known to exist, must be still approaching the horizon and will never get through. So all such mass should be just above the horizon. $\endgroup$ Commented May 27 at 22:11
  • $\begingroup$ @JánLalinský That is the coordinate point of view. The universe will die a heat death and all other black holes will have evaporated by the time the falling observer arrives at the event horizon. The proper time of the falling observer is less than a few minutes. The proper time is considered more valid. Suggesting the coordinate time in more valid gets the same reception as Galileo got for suggesting the planets orbits the Sun and not the Earth. Its heresy. $\endgroup$
    – KDP
    Commented May 28 at 2:35
  • $\begingroup$ > "The universe will die a heat death and all other black holes will have evaporated by the time the falling observer arrives at the event horizon. " I wouldn't be so sure about universe death and black hole evaporation (they seem to be based on poorly supported assumptions/nonexistent theories), but the time of fall of anything falling head-on towards the horizon is infinite in frames outside the black hole and at rest with respect to its center. $\endgroup$ Commented May 28 at 11:11
0
$\begingroup$

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 $\square h_{\mu\nu} =0$.

$\endgroup$
3
  • $\begingroup$ Would Minkowski space be physically possible? Or is it just a mathematical abstraction? $\endgroup$
    – Elvis
    Commented May 25 at 13:46
  • $\begingroup$ The fundamental assumption of General Relativity of considering spacetime as a smooth manifold implies that at sufficiently small scales one can always find a reference frame in which the components of the generally curved metric reduces to Minkowski e.g., local flatness theorem. For this reason, the Standard Model of particle physics -through the framework of quantum field theory- works very well only supposing Minkowski as the background spacetime. That is enough to make physical predictions. At large scales the situation is quite different, and to describe our Universe curvature is needed. $\endgroup$
    – ouroboros
    Commented May 25 at 22:26
  • $\begingroup$ This is incorrect. The Schwarzschild spacetime on its own describes the entire black hole. $T$ is zero everywhere except the singularity. The event horizon is only a coordinate singularity where Schwarzschild coordinates fail. But other coordinate systems work there. On the other hand, the interior solution is different. It refers to the solution within an actual physical spherical object itself, where there is mass, which is much larger than the event horizon. The "interior" here does NOT refer to the region inside the black hole horizon. $\endgroup$ Commented May 26 at 11:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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