# I Coordinate singularity at Schwarzschild radius

1. Dec 12, 2017

### zonde

I would like to ask how rigorous is the statement that Schwarzschild metric has coordinate singularity at Schwarzschild radius.
The argument is that singularity at Schwarzschild radius appears because of bad choice of coordinates and can be removed by different choice of coordinates.
However removable singularity has closed neighborhood around it. But singularity surface at Schwarzschild radius is open toward infinite future and infinite past.
Then if we consider gravitating body that has formed at finite past it would be open toward infinite future only. And considering that Schwarzschild metric is vacuum solution in such a case we have to glue it to some other metric that has non-zero stress-energy tensor. So this coordinate singularity has to first appear in some non-vacuum solution and only then it can be glued to Schwarzschild metric's illusory singularity. So in case of gravitating body with finite past the argument seems to depend on existence of such valid non-vacuum solution.

Another argument is that coordinate independent quantities are finite at Schwarzschild radius so it should not be real. But that argument can be turned upside down and stated that coordinate independent quantities should not approach finite values as space-time is extended toward timelike infinity. And taken way the argument is less obvious.

2. Dec 12, 2017

### Staff: Mentor

Such solutions do exist. https://journals.aps.org/pr/abstract/10.1103/PhysRev.56.455

Why not? Your reasoning here does not seem valid at all, and I have never seen a scientific reference which uses it. Have you?

Last edited: Dec 12, 2017
3. Dec 12, 2017

### zonde

Well, how this is relevant to the point I was trying to make? That surface of singularity in Schwarzschild metric does not have closed neighborhood.
Hmm, not sure what do you mean. Do you mean that the argument can't be generalized as I was trying to do it?
I am fairly sure that I have seen the argument about curvature being finite at Schwarzschild radius as an argument about singularity not being "real", but I can check if you are questioning that part.

4. Dec 12, 2017

### Staff: Mentor

I don't know. You said "the argument seems to depend on existence of such valid non-vacuum solution", and I was just pointing out that such solutions do exist so any dependence on their existence can be resolved.

Yes, I have seen that. What I have not seen and what I do not think is correct is your suggestion that that argument can "be turned upside down". I have never seen anyone who turned it upside down and came out with the resulting restatement you mentioned nor anything equivalent.

5. Dec 13, 2017

### Staff: Mentor

By "Schwarzschild metric" I assume you mean "Schwarzschild coordinate chart on the Schwarzschild spacetime geometry". The statement that the singularity in this coordinate chart at the Schwarzschild radius for this geometry is a coordinate singularity is certainly "rigorous". There are a number of ways to see this.

That's one possible argument, but not the only one. A better one is to compute curvature invariants as functions of the Schwarzschild coordinates, and then take their limits as $r \rightarrow 2M$. You will find that all the limits exist and are finite. That proves that the singularity is just a coordinate singularity without having to find an explicit transformation to a non-singular coordinate chart.

Why do you think this?

Only in the maximally extended geometry, where the stress-energy tensor is zero everywhere. But any real spacetime won't be described entirely by this geometry.

Then you are not considering the maximally extended Schwarzschild geometry, so your claim about the singularity surface being open to the infinite past no longer holds.

Yes. The simplest such solution that contains the Schwarzschild coordinate singularity in the vacuum region is the Oppenheimer-Snyder solution describing a spherically symmetric collapse of dust (zero pressure matter).

If you describe the non-vacuum region of the Oppenheimer-Snyder solution in Schwarzschild coordinates, there will indeed be a coordinate singularity there.

I don't see how this has anything to do with the coordinate singularity in Schwarzschild coordinates, since that singularity does not occur at timelike infinity.

Are you under the misapprehension that the event horizon in Schwarzschild spacetime ends at future timelike infinity? It doesn't. Don't be misled by Penrose diagrams; on the boundaries of the diagram adjacent points are not always connected.

6. Dec 13, 2017

### zonde

Well yes, existence of Oppenheimer-Snyder solution is sufficient condition to conclude that this coordinate singularity is removable. But is it necessary condition?
Say can another person which does not know about existence of Oppenheimer-Snyder solution conclude that this coordinate singularity is removable? And indirectly conclude that such Oppenheimer-Snyder solution should exist. Or he can't conclude any such thing and necessarily has to go look for Oppenheimer-Snyder type solution?

7. Dec 13, 2017

### zonde

Yes, thanks for correcting me.

I do not understand this "that proves" statement.
Say we have solution describing spacetime with gravitating body that is small enough so that according to our knowledge it can't turn into black hole. Now we look what happens to curvature as we reach infinity in some physically motivated coordinate chart. Well, I'm not sure but I guess that curvature invariants will have finite limits. Can you confirm that before I go any further?

I guess my terminology is off. Sorry.
Let me explain it this way. If I look at the coordinate singularity at the pole in polar coordinates I can reach any point trough continuous change of coordinates. The only thing I can't do is cross the pole. On the other hand in Schwarzschild coordinate chart this does not work. To reach some point on the inside of event horizon I have to cross the horizon. I can't go around the horizon to reach that point. So there are two disconnected regions on each side of singularity surface in this coordinate chart.

Let me illustrate my concerns with polar coordinates. Lets say that in polar coordinates latitude has coordinate 90 at the pole. Now I claim that pole is not actually a point but rather a circle and I can go inside that circle by extending maximum coordinate value of latitude from 90 to 100. So I get extra region at pole. It does not exist but mathematically I can get extra points with different coordinates. So where I went wrong with my extra region? Well it seems I ignored that distance in direction of longitude goes to zero as I approach pole. So I need physically motivated idea of distance to avoid unphysical conclusions.

8. Dec 13, 2017

### A.T.

Imagine you express the position on the upper half of a circle by the x-coordinate of the intersection between the point's tangent and the x-axis:

When you let xtangent -> infininty, you approach the point (0,1), but you never cross it, despite the fact that there is an entire region behind it, and nothing special happens with the curvature at this point.

9. Dec 13, 2017

### zonde

Imagine line x=y, now let x -> infinity. Noting special happens with the curvature. Should I believe that therefore the line should be extendable beyond x-infinity?

10. Dec 13, 2017

### A.T.

That coordinate doesn't have a singularity. I gave you an example that has.

11. Dec 13, 2017

### martinbn

I don't understand your concern here. You can say the exact same thing about the light cone of an event in Minkowski space-time. To reach a point on the inside from the outside you need to cross the cone. It divides the space-time into disconnected components as well. And you can choose coordinates so that the cone is singular surface for those coordinates. But I am sure you can easily accept that it is just a coordinate singularity in this case.

12. Dec 13, 2017

### zonde

Well, my point is that like you can't say that region does not exist when you can't go continuously in coordinate chart beyond some point you can't claim the opposite as well - you can't say that region exists just because you can came up with some coordinate chart where there are coordinates for that region. You have to invoke physical reasoning to say that something exists or something does not exist.

13. Dec 13, 2017

### zonde

So you gave example where certain point in one coordinate chart can't be reached in other coordinate chart.
Then, let me introduce additional coordinate chart to the one I gave. Define x'=1-1/x^2 and y'=1-1/y^2 for x,y>0. In coordinate chart x',y' point x'=1 (and any point beyond) can't be reached with coordinates x,y.

14. Dec 13, 2017

### pervect

Staff Emeritus
https://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates gives a transformation which explicitly removes the singularity at the event horizon. And more detail can be found in many GR textbooks. So I'm not sure why we're arguing. (I do have the feeling that this is an argument, not an attempt by the OP to figure out where he went wrong. But I suppose I could be wrong - it just doesn't seem that way to me.)

15. Dec 13, 2017

### martinbn

What does this have to do with the singularity being a coordinate singularity? Whether something exists or not in the real world doesn't change anything about the mathematical properties of the model.

16. Dec 13, 2017

### zonde

The question I'm trying to resolve is why I can arrive at stupid results when I try to use the type of reasoning used in GR context.
Say what is wrong with my attempt at removing coordinate singularity at latitude 90 of polar coordinates by allowing latitude to go beyond value 90:

17. Dec 13, 2017

### Staff: Mentor

A coordinate chart is a differentiable one to one mapping between events in a open subset of the manifold and points in a open subset of R4. What you described is not a coordinate chart beyond 90 deg.

18. Dec 13, 2017

### PAllen

The physical reasoning is that if a time-like geodesic ends wth finite proper time from some starting point, there better be some some physical reason for it, or it makes no sense. My coordinates become singular is no reason to say that a free falling clock ceases to exist when it reads 3:00. Generalization of this argument leads to the modern, physically motivated definition of a true singularity - that there are incomplete geodesics, and there is no smooth way to extend the manifold to complete them.

19. Dec 13, 2017

### zonde

Well, the way I understand the term "removable singularity" is when we can approach some discontinuity from opposite sides and show that in neighborhood of that discontinuity everything seems fine except at that particular point. But if we can approach the discontinuity only from one side it can be the type of "discontinuity" we have at infinity. So in order to say that there is removable singularity at Schwarzschild radius we have to conclude that there is "normal" spacetime at the other side of singularity. And we have to reach that conclusion by some physical reasoning, say by considering Oppenheimer-Snyder solution.

20. Dec 13, 2017

### PAllen

Geometry prevents this. The circumference of circumpolar circles goes to zero as latitude goes to 90. Without changing the geometry there is no place to add anything.