# I Coordinate singularity at Schwarzschild radius

1. Dec 14, 2017

### Staff: Mentor

No, the reason why we should seriously consider the interior of the Schwarzschild geometry is that all of the curvature invariants are finite at the event horizon, so it's obvious that geodesics can be extended beyond that point. The O-S solution is just one particular illustration of that general fact.

No, we don't. A manifold is an open set. No open set extended into the Schwarzschild interior contains a singularity; the geometry is smooth everywhere.

What indicates the presence of a "singularity" in the limit $r \rightarrow 0$ is that geodesics only reach a finite value of their affine parameters as that limit is approached. There is no discontinuity in the actual geometry; the limit point $r = 0$ is not part of the manifold. This is discussed in any textbook on global methods in GR, starting with the classic Hawking & Ellis and the papers that led up to it.

2. Dec 14, 2017

### Staff: Mentor

I am not talking about the Schwarzschild spacetime, I am talking strictly about the Rindler chart on Minkowski spacetime. What geometric clues might hint to the idea that the manifold doesn’t end at X=0?

Think about the lengths of geodesics. If they draw a geodesic that crosses X=2 the length is finite and the geodesic can continue on. If they draw a geodesic that crosses X=1 the length is finite and the geodesic continues on. If they draw a geodesic that crosses X=0 the length is finite, so wouldn’t they suspect that the geodesic continues on? What would you suspect given similar information?

3. Dec 15, 2017

### zonde

It's not obvious unless you emphasize that geodesics are incomplete there.
In other words if geodesics are complete finite curvature invariants gives no reason to extend them further.
Do you say that any set that includes singularity can't be considered open? How do you motivate this statement?

Can't argue with textbook.
Nonetheless it reminds me of an old anecdote.
Man brings pendulum to the clocksmith. Clocksmith asks where is the clock. Man says: "No, no, the clock itself is ok. It's the pendulum that does not swing."

Last edited: Dec 15, 2017
4. Dec 15, 2017

### Staff: Mentor

Why would they end there?

5. Dec 15, 2017

### Staff: Mentor

I did. That's what "geodesics can be extended" means.

Geodesics can't be complete if curvature invariants are finite.

No, I'm saying that there is no such thing as a "set that includes a singularity". The term "singularity" does not refer to a point in the manifold.

6. Dec 15, 2017

### Staff: Mentor

I may misunderstand the terminology. Isn’t a straight line in flat spacetime considered a complete geodesic. The affine parameter has an infinite range, I thought that was also complete.

7. Dec 15, 2017

### Staff: Mentor

I was a bit sloppy; I was referring specifically to the case at the horizon of the Schwarzschild geometry, where we have finite curvature invariants and finite values of affine parameters for geodesics; in that case the geodesics obviously aren't complete in the usual sense within the exterior region. You are right that if we already know we can extend geodesics indefinitely, i.e., we know they are complete, that doesn't mean curvature invariants must diverge; flat spacetime, and also extending geodesics out to timelike or null or spacelike infinity in any asymptotically flat spacetime, are all cases of complete geodesics with finite curvature invariants everywhere.

8. Dec 15, 2017

### Staff: Mentor

Thanks, that is helpful! I hope that my question doesn't cause a distraction for the thread

9. Dec 15, 2017

### Staff: Mentor

@zonde perhaps a good way to think of this is to think of a piece of paper with grid lines on one part of the paper but not the other. The paper doesn’t end just because the grid lines end.

Infinite curvature invariants are taken as an indication that the spacetime ends (or more likely that our model breaks down). Without that we have no reason to believe that the spacetime ends.

10. Dec 16, 2017

### zonde

The analogy with Rindler chart is clear to me. There is no problem. Conclusions are harder to accept because the correct answer is already known. I requires sort of backward reasoning - by what reasoning we could arrive at correct answer and which version of reasoning seems better. So if the reasoning can't be restricted to math only it's sort of philosophy.
Nonetheless the reasoning based on geodesic incompleteness seems to be convincing.
I certainly do not understand that "spacetime ends" option. As I see this it's geodesics that end there not spacetime. Maximum we could say that spacetime has a hole there.
And I don't understand why option "particular solution has somewhere sneaked in unphysical assumption" is not considered. Well I sort of understand, but it has nothing to do with science.

11. Dec 16, 2017

### Staff: Mentor

I agree. We also have the additional experimental fact that we have never observed anything that indicates an edge of spacetime with finite curvature.

If the spacetime didn’t end then the geodesic could be further extended. Anyway, consider the standard KS coordinates covering the maximally extended Schwarzschild vacuum. The singularity in region II is an edge of the manifold.

https://en.m.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

That option definitely is considered. That is one of the drivers of work on quantum gravity.

12. Dec 19, 2017

### zonde

I don't really follow. You said that we shouldn't consider Rindler chart as saying something about manifold, because geodesics can extend beyond Rindler horizon (when curvature invarinats are finite). And yes, this makes sense to me.
But now you sort of say that we should consider KS chart to get the idea that manifold ends at (near) singularity. These two explanation seem contradictory. If coordinate charts mean next to nothing in GR then it should apply equally to both scenarios considered.
PeterDonis point about infinite curvature invarinats seems at least self-consistent.

13. Dec 19, 2017

### zonde

How do I know there is vacuum up to the horizon? Say I look at spherically symmetric non rotating gravitating body. SC solution describes vacuum portion outside gravitating body. So the SC solution is not valid at radius less than radius of the body.
Let's imagine we have three bodies with the same SC solution's parameter M. They have radii r1<r2<r3. At r>r3 SC solution describes all three bodies. As we reduce the radius below r3 this solution is no longer valid for third body. As we reduce radius below r2 it is no longer valid for second body as well. And as we reduce radius below r1 it is no longer valid for any of the three bodies. So as we reduce the radius in SC solution, number of possible physical configurations for which SC solution is still valid is reduced. So we can ask if there is radius at which there is no corresponding physical configurations for which SC solution is still valid. And answer to this question can't be found out by looking at SC solution because it is not constructed for description of regions with non vacuum SET.

14. Dec 19, 2017

### deRoy

Dear @zonde I am following this thread for a while and you seem to have problem understanding the nature of infinite curvature.

Here is an example:

Take line $t = \sqrt {x^2 - \frac {1} {x^2}}$ It looks like $t = \sqrt { x^2 - 1}$ but is trickier like an helix is to a circle.

Now embed this line in usual Minkowski spacetime and rotate along t-axis. You are getting a surface resembling an hyperboloid. Clearly line $t = \sqrt {x^2 - \frac {1} {x^2}}$ is a geodesic. The tricky part about it however is although it extends to infinity, an observer tracing this geodesic reaches infinity at finite proper time. Because the usual Minkowskian metric indicates so. And what's more there is a singularity at infinity!

Now you free-falling observer have no idea as to what space this surface is embedded into. (You don't have to and it's not necessary.) You are concluding since you are reaching a singularity at finite proper time that there is an edge or vertex to your spacetime there. But as this example is showing you spacetime doesn't have to end somewhere and there are no edges!

I hope this clarifies matters a bit.

15. Dec 19, 2017

### Staff: Mentor

The difference is that the KS chart covers the full maximally extended manifold where the Schwarzschild chart does not. If you want to talk about a whole manifold then it is convenient to use a chart which covers the whole manifold.

I agree, but you seemed to be rejecting that point.

Last edited: Dec 19, 2017
16. Dec 19, 2017

### deRoy

Take this with a grain of salt. I have to do the maths for the particular surface as there is a possibility for the curvature to be finite at infinity but the main chain of reasoning is there.

17. Dec 19, 2017

### PAllen

It seems two separate things are mixed up in this discussion. One theme, as per the title of this thread, is about what follows from assuming the SC solution up to its horizon coordinate singularity. The post you respond to was responding to this thread titular context. It discussed that it is worth asking about both matter and vacuum extension, motivating that some extension is mandated by physical considerations, and that it couldn’t be stable matter, and thus that more vacuum was the only possibility per GR, after at most a a short collapse time. The specific statement you quote is indisputable - the SC solution up to the coordinate singularity is everywhere vacuum. Given the hypothesis of the SC solution to horizon, the post referenced also established that answering regional questions about extension need no information about the distant past or global issues.

You now shift to the completely different question of whether there is a process in our universe that can produce the SC solution up to the horizon. That, indeed, is a question about theories of matter and cosmology along with GR. A trivial observation would be that if our universe happened to have no stars much bigger than a solar mass, and there were no apparent supermassive BH in galactic centers, then current matter theories along with GR would suggest there were no BH and never would be any until the incomprehensible future when galaxies radiated their angular momentum away via GW, and galaxy’s effectively collapsed as per the Oppenheimer Snyder collapse.

Of course, that doesn’t describe our universe at all. And LIGO detects GW waveform that, per GR, are only explicable as merging BH. This makes my post you quote absolutely relevant to our universe, and your shift of discussion irrelevant because it is empirically refuted.

Last edited: Dec 19, 2017
18. Dec 19, 2017

### Staff: Mentor

Because you started this thread by talking about the Schwarzschild geometry and the coordinate singularity at the Schwarzschild radius. That coordinate singularity is only present if there is vacuum up to the horizon (and if there is a horizon in the first place). If you're talking about a case like a planet or star, where the Schwarzschild geometry only applies in the vacuum region outside the body, there is no coordinate singularity anywhere; even if you extend the Schwarzschild coordinates into the interior of the body, you will never reach a point where $g_{tt}$ vanishes or $g_{rr}$ diverges.

Wrong. It can, by analytically extending the solution and showing that all invariants are finite at and below the horizon, and combining that with Birkhoff's theorem, which says that the SC vacuum geometry is the geometry of any spherically symmetric vacuum region, even one outside a non-static object (such as a region of matter that is collapsing). The Oppenheimer-Snyder solution adds to this by telling you the spacetime geometry inside the spherically symmetric collapsing region of matter, but you don't need the O-S solution to tell you that the SC geometry of the vacuum region is viable at and below the horizon.