Coordinate singularity at Schwarzschild radius

[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.

 

[Dale]

So in case of gravitating body with finite past the argument seems to depend on existence of such valid non-vacuum solution.

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

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

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

 

[zonde]

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

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.

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

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.

 

[Dale]

Well, how this is relevant to the point I was trying to make?

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.

That said, I don't understand your concern about neighborhoods.

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",

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.

 

[PeterDonis]

I would like to ask how rigorous is the statement that Schwarzschild metric has coordinate singularity at Schwarzschild radius.

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.

The argument is that singularity at Schwarzschild radius appears because of bad choice of coordinates and can be removed by different choice of coordinates.

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.

removable singularity has closed neighborhood around it.

Why do you think this?

singularity surface at Schwarzschild radius is open toward infinite future and infinite past.

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.

if we consider gravitating body that has formed at finite past

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.

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.

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).

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.

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

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.

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.

 

[zonde]

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.

That said, I don't understand your concern about neighborhoods.

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?

 

[zonde]

By "Schwarzschild metric" I assume you mean "Schwarzschild coordinate chart on the Schwarzschild spacetime geometry".

Yes, thanks for correcting me.

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.

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?

Why do you think this?

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. Let's 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.

 

[A.T.]

Let me illustrate my concerns with polar coordinates.

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 x_tangent -> 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.

 

[zonde]

and nothing special happens with the curvature at this point.

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?

 

[A.T.]

Imagine line x=y, now let x -> infinity.

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

 

[martinbn]

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.

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.

 

[zonde]

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.

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.

 

[zonde]

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

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.

 

[pervect]

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.)

 

[martinbn]

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.

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.

 

[zonde]

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.)

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:

Let me illustrate my concerns with polar coordinates. Let's 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.

 

[Dale]

So where I went wrong with my extra region?

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.

 

[PAllen]

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.

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.

 

[zonde]

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.

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.

 

[PAllen]

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:

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.

 

[zonde]

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.

Yes, in my example I used two dimensional closed manifold. But it seems confusing to me when we consider open manifolds. Does an open manifold has some limit? Say can I say that some coordinate chart covers whole manifold i.e. manifold can't be extended beyond infinite coordinate of some "good" coordinate chart?

 

[Dale]

Yes, in my example I used two dimensional closed manifold. But it seems confusing to me when we consider open manifolds. Does an open manifold has some limit? Say can I say that some coordinate chart covers whole manifold i.e. manifold can't be extended beyond infinite coordinate of some "good" coordinate chart?

E.g. a standard flat Lorentzian spacetime is open and is completely covered by the usual Minkowski coordinate chart but it is not completely covered by the Rindler chart.

 

[zonde]

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.

Yes, this is physical argument. But it is a bit circular. Schwarzschild metric is derived assuming unphysical stress-energy tensor. So we necessarily have to glue it to some other metric with physically reasonable stress-energy tensor. So in order to say that clock in free fall can fall beyond singular coordinate we have to now that we can glue metric with physically reasonable stress-energy tensor on the inside of that singularity.

 

[PAllen]

Yes, this is physical argument. But it is a bit circular. Schwarzschild metric is derived assuming unphysical stress-energy tensor. So we necessarily have to glue it to some other metric with physically reasonable stress-energy tensor. So in order to say that clock in free fall can fall beyond singular coordinate we have to now that we can glue metric with physically reasonable stress-energy tensor on the inside of that singularity.

It’s not circular at all. You don’t need to even address the plausibility of eternal vacuum everywhere. You find that within your coordinates clock ends at 3:00. You conclude that is not reasonable, so you see if smooth geometric extension is possible. You find it is possible, so it is not a true singularity. Then you follow clock further, and it ends at 4:00. This time, precisely because curvature invariants become infinite, you find that there is no extension possible. This leads you to suspect that this true singularity is a breakdown of GR, because it still makes no physical sense to have incomplete geodesics, but there is nothing you can do about it within GR

 

[PeterDonis]

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.

In that case there is no event horizon, so there is not a coordinate singularity. So the issue you raise doesn't even arise.

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?

Which "infinity" are you talking about? Spatial infinity? Future timelike infinity? Future null infinity?

For your particular question it so happens that the answer to your question is "yes" for all of the above. But it's important to realize that "infinity" is not enough to pin down what you are referring to when we are talking about asymptotically flat spacetimes.

 

[PeterDonis]

Schwarzschild metric is derived assuming unphysical stress-energy tensor.

No, it isn't. A vacuum SET is not "unphysical".

What you should be saying here is that a physically reasonable spacetime with a black hole in it cannot be vacuum everywhere. That's true. But it doesn't justify the deductions you are drawing from it. See below.

in order to say that clock in free fall can fall beyond singular coordinate we have to now that we can glue metric with physically reasonable stress-energy tensor on the inside of that singularity.

No, we don't. A physically reasonable black hole interior is vacuum all the way to the singularity. The non-vacuum region in the spacetime is in the past, when some object originally collapsed to form the hole. So the only "gluing" that needs to be done is in the past; an object free-falling into the hole will not encounter any portion of the non-vacuum region that is "glued" on in the past.

 

[Dale]

Yes, this is physical argument. But it is a bit circular. Schwarzschild metric is derived assuming unphysical stress-energy tensor. So we necessarily have to glue it to some other metric with physically reasonable stress-energy tensor. So in order to say that clock in free fall can fall beyond singular coordinate we have to now that we can glue metric with physically reasonable stress-energy tensor on the inside of that singularity.

Personally, I think that you may find it beneficial to consider the Rindler chart on flat spacetime. Starting from the Rindler chart, what clues do you think someone might have that the Rindler chart does not cover all of spacetime?

 

[zonde]

It’s not circular at all. You don’t need to even address the plausibility of eternal vacuum everywhere. You find that within your coordinates clock ends at 3:00.

Of course I have to consider where there is vacuum. Free fall can happen only in vacuum. No vacuum no free fall. So for the free falling clock to reach event horizon at 3:00 there should be vacuum at the event horizon.
So no, you first have to give reason to believe that it is plausible for there to be vacuum at event horizon and only then you can proceed with your argument. And here we come back to Oppenheimer-Snyder solution.

 

[zonde]

E.g. a standard flat Lorentzian spacetime is open and is completely covered by the usual Minkowski coordinate chart but it is not completely covered by the Rindler chart.

Ok, but how do I know there is no coordinate chart in which Minkowski coordinate chart's infinite time coordinate is reached at finite time coordinate and which locally is approximately flat? Because if there is such a chart I might argue that manifold can be extended in that direction.

Personally, I think that you may find it beneficial to consider the Rindler chart on flat spacetime. Starting from the Rindler chart, what clues do you think someone might have that the Rindler chart does not cover all of spacetime?

Well, my reasoning would be that one can not sustain uniform acceleration for very long as it is related to significant changes in physical configuration (say rocket has to spend fuel to do that). So physical reality will catch up with me when I stop accelerating.
But I guess you might have something else on mind. Is your point that inertial coordinates are "better" than accelerated ones?

 

[zonde]

In that case there is no event horizon, so there is not a coordinate singularity. So the issue you raise doesn't even arise.

This is one of the points I am not certain. How do I decide which is coordinate singularity and not infinity type discontinuity. Say I can imagine that Schwarzschild metric describes vacuum part of frozen star. Then this coordinate singularity (at Schwarzschild radius) might be considered as "surface" of infinite future.

Which "infinity" are you talking about? Spatial infinity? Future timelike infinity? Future null infinity?

For your particular question it so happens that the answer to your question is "yes" for all of the above. But it's important to realize that "infinity" is not enough to pin down what you are referring to when we are talking about asymptotically flat spacetimes.

Sorry, I was talking about future timelike infinity. I assumed coordinate chart in which gravitating body is at rest and considered the limit when time coordinate goes to infinity while spatial coordinates remain constant.

So my point is that if we would look at the limit of approaching event horizon as approaching future infinity there would be no problem about curvature remaining finite.

 

[zonde]

No, we don't. A physically reasonable black hole interior is vacuum all the way to the singularity. The non-vacuum region in the spacetime is in the past, when some object originally collapsed to form the hole. So the only "gluing" that needs to be done is in the past; an object free-falling into the hole will not encounter any portion of the non-vacuum region that is "glued" on in the past.

So the reason why we should seriously consider interior of Schwarzschild metric is Oppenheimer-Snyder solution, right?

 

[Dale]

Ok, but how do I know there is no coordinate chart in which Minkowski coordinate chart's infinite time coordinate is reached at finite time coordinate and which locally is approximately flat?

Such as ##T=1/t##. It reaches infinite Minkowski ##t## at finite ##T##. Flatness doesn’t change regardless of the chart.

Well, my reasoning would be that one can not sustain uniform acceleration for very long as it is related to significant changes in physical configuration (say rocket has to spend fuel to do that).

That is just a practical/engineering concern. In principle there is no finite limit to how long we could accelerate.

Think just about the geometry. What geometric clues would he have that there is more spacetime than that which is covered by the Rindler chart?

 

[zonde]

Think just about the geometry. What geometric clues would he have that there is more spacetime than that which is covered by the Rindler chart?

You mean event horizon at finite spatial coordinate?

 

[zonde]

Flatness doesn’t change regardless of the chart.

That is if you equip the coordinate chart with metric and whenever you transform to other coordinate chart you transform the metric in such a way as to keep the distances invariant. But if you calculate the distances directly from coordinates certainly the flatness can change.

 

[Dale]

That is if you equip the coordinate chart with metric

Yes, of course, that is part of the definition of a Riemannian manifold (except that it is the manifold that is equipped with the metric, not a coordinate chart see below)

whenever you transform to other coordinate chart you transform the metric in such a way as to keep the distances invariant.

The are invariant. The metric is a geometric object attached to the manifold regardless of the definition or existence of some coordinate chart. It is not the coordinate chart that determines the metric or distances. Those are geometric features of the Riemannian manifold.