Undergrad Schwarzschild Geometry: Einstein Tensor & Mass Density

[Joe Prendergast]

TL;DR

Schwarzschild Geometry/Einstein Tensor. Why equal to zero?

The Einstein tensors for the Schwarzschild Geometry equal zero. Why do they not equal something that has to do with the central mass, given that the Einstein equations are of the form: Curvature Measure = Measure of Energy/Matter Density?

 

[Ibix]

Your summarisation of the meaning of the Einstein field equations is over-simplified. A better statement would be that they relate the local Ricci curvature to the local stress-energy.

The Schwarzschild solution is a vacuum solution the stress-energy tensor is everywhere zero; hence so is the Einstein tensor. Note that the Einstein tensor depends on the Ricci tensor, which is a summarisation of the Riemann curvature tensor and does not include all parts of the Riemann. The remaining parts are the Weyl curvature, which is non-zero.

 

[Joe Prendergast]

Thank you, that helps. So even though there is a rest mass in the Schwarzschild solution, it is still considered a vacuum solution?

 

[Ibix]

There isn't any mass anywhere in Schwarzschild spacetime. The $M$ term is a mass, but it isn't the mass of anything. To the extent that there's a source of gravity in this solution it's the singularity, which isn't a part of spacetime. This kind of thing is almost certainly a result of pushing an idealised model beyond the range of validity of GR, but it's a useful model away from the singularity (you can prove that spacetime outside any spherically symmetric and non-time-varying mass distribution is Schwarzschild, with something else in the middle where the matter is) and we don't have a better theory of gravity.

There are vaguely realistic stellar collapse models (for example the Oppenheimer-Snyder solution) in which you can argue that the stellar matter is the source of gravity. Those spacetimes are not vacuum everywhere.

 

[vanhees71]

Would you say that in the Coulomb-field solution of classical electrodynamics there's no charge?

 

[Ibix]

Would you say that in the Coulomb-field solution of classical electrodynamics there's no charge?

Who, me? No. Nor would I say a pointlike Newtonian mass has no mass. And in the case of Schwarzschild spacetime I'm not saying there's no mass, just that there is no thing to which you can ascribe the mass in the spacetime. At least not within GR, which is probably because GR goes wrong somewhere between here and the core (for want of a better word) of a black hole.

 

[Joe Prendergast]

In Schwarzschild, I believe there is a coordinate singularity at r = 2M and a physical singularity at r=0

 

[Ibix]

In Schwarzschild, I believe there is a coordinate singularity at 2M and a physical singularity at M=0

There's only a coordinate singularity at $r=2M$ if you use coordinates that break there, as Schwarzschild's original ones do. You can pick better ones, and there are at least three options (Eddington-Finkelstein, Gullstrand-Painleve and Kruskal-Szekeres). The singularity at $r=0$ is a true curvature singularity, and if someone's talking about "the" singularity that's the one they mean.

 

[vanhees71]

Who, me? No. Nor would I say a pointlike Newtonian mass has no mass. And in the case of Schwarzschild spacetime I'm not saying there's no mass, just that there is no thing to which you can ascribe the mass in the spacetime. At least not within GR, which is probably because GR goes wrong somewhere between here and the core (for want of a better word) of a black hole.

I'd say, a Schwarzschild black hole indeed has a mass in GR, and from the point of view of an observer at $r \rightarrow \infty$ in standard Schwarzschild coordinates, it has the same meaning as in Newton's gravitational force law.

Also the vacuum Schwarzschild solution holds for any spherically symmetric distribution of matter, where there are no singularities.

 

[Dale]

I think it is more natural to describe the Schwarzschild spacetime in terms of a length scale (the Schwarzschild radius) rather than in terms of a mass scale. In geometrized units, of course, you can convert a mass into a length geometrically. So you can convert between the two easily. But regardless of whether or not there is any mass anywhere in the spacetime the length scale applies.

 

[haushofer]

TL;DR Summary: Schwarzschild Geometry/Einstein Tensor. Why equal to zero?

The Einstein tensors for the Schwarzschild Geometry equal zero. Why do they not equal something that has to do with the central mass, given that the Einstein equations are of the form: Curvature Measure = Measure of Energy/Matter Density?

The energy momentum tensor is, strictly speaking, not zero, but contains a delta function such that at r=0 the mass density is infinite, comparable to the charge source of an electric field. Outside r=0 you then have a vacuum solution (or outside r=R if you consider a spherical object with mass R). So the mass is there; just write down the energy momentum tensor describing a point mass with mass M at r=0.

 

[PeterDonis]

The energy momentum tensor is, strictly speaking, not zero, but contains a delta function such that at r=0 the mass density is infinite

This is wrong. The Schwarzschild solution is a vacuum solution; the SET is zero everywhere. The locus $r = 0$ is not part of the spacetime; it is a limit point that is not included in the manifold.

just write down the energy momentum tensor describing a point mass with mass M at r=0.

This is wrong as well; solving the EFE with this prescription will not give you Schwarzschild spacetime with mass parameter $M$.

 

[PAllen]

To add to what @PeterDonis said, in Schwarzschild geometry, the limiting geometric elements approaching "r=0" are hypercylinders RXS2, with the S2 radius approaching 0 (this is the r=0), and the axial R being infinite in proper distance for all the elements. Each of these hypercylinders is spacelike, with shrinkage of S2 radius being a timelike paramaterization.

This is obviously totally different from a point charge in classical EM.

 

[haushofer]

This is wrong. The Schwarzschild solution is a vacuum solution; the SET is zero everywhere. The locus $r = 0$ is not part of the spacetime; it is a limit point that is not included in the manifold.This is wrong as well; solving the EFE with this prescription will not give you Schwarzschild spacetime with mass parameter $M$.

Yes, I've been sloppy here. When I talk about delta functions, I implicitly assume some sort of smearing because of the singular nature of these delta functions. I understand that singularities are not part of spacetime.

 

[PeterDonis]

When I talk about delta functions, I implicitly assume some sort of smearing because of the singular nature of these delta functions.

While this may be true, it doesn't change the fact that there is no delta function, not even a smeared one, in the Schwarzschild solution; that solution is a pure vacuum solution. If you try to construct a solution using a delta function, even a smeared one, you will not get the Schwarzschild solution, not even if you "remove the singularity" by excising the part covered by the delta function from the manifold. This has been well known for decades.

 

[PeterDonis]

I understand that singularities are not part of spacetime.

It also is worth clarifying here that, in modern GR, the term "singularity" does not mean "there's a delta function of stress-energy or some smeared equivalent here". It means geodesic incompleteness: that there are geodesics that cannot be extended beyond a finite value of their affine parameter. It is not even the case that all such singularities correspond with curvature invariants increasing without bound as the finite limit of the affine parameter is approached, although that is the case for the singularity at $r = 0$ in all black hole solutions.

 

[haushofer]

While this may be true, it doesn't change the fact that there is no delta function, not even a smeared one, in the Schwarzschild solution; that solution is a pure vacuum solution. If you try to construct a solution using a delta function, even a smeared one, you will not get the Schwarzschild solution, not even if you "remove the singularity" by excising the part covered by the delta function from the manifold. This has been well known for decades.

How about taking a sphere with radius R, write down the energy momentum tensor for such an object, solving the Einstein equations and taking the limit of R going to zero? That should produce the Schwarzschild solution due to Birkhoff's theorem, right?

Anyway, you can tell I haven't done these kind of calculations for a while.

 

[haushofer]

It also is worth clarifying here that, in modern GR, the term "singularity" does not mean "there's a delta function of stress-energy or some smeared equivalent here". It means geodesic incompleteness: that there are geodesics that cannot be extended beyond a finite value of their affine parameter. It is not even the case that all such singularities correspond with curvature invariants increasing without bound as the finite limit of the affine parameter is approached, although that is the case for the singularity at $r = 0$ in all black hole solutions.

Yes, I totally agree, it's sloppiness.

 

[PeterDonis]

How about taking a sphere with radius R, write down the energy momentum tensor for such an object, solving the Einstein equations and taking the limit of R going to zero?

You can't, because once the radius of the sphere becomes less than 9/8 of the Schwarzschild radius for the sphere's mass, you no longer have a valid solution of the Einstein Field Equation if you require the sphere to be static. At any radius less than 9/8 of the Schwarzschild radius, the sphere must be collapsing. We already know what the collapsing solution looks like: it looks like the 1939 Oppenheimer-Snyder solution, which does indeed have a Schwarzschild exterior region--but that solution also has a region containing the collapsing matter, and there is no way to "take the limit as the collapsing matter goes to zero", because the matter is collapsing, not static.