Coordinate singularity at Schwarzschild radius

[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.

 

[zonde]

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.

Well, it is a bit confusing. We have a tool to calculate manifold distances from coordinate chart (it is coordinate chart dependent of course). We call it metric as well, right? And what is then the "metric" that is not just a tool to calculate the distances? Well manifold has curvature as well. But do we describe it with metric or we rather use Riemann curvature tensor for it?

 

[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?

Speaking about Rindler chart and comparing it with Schwarzschild chart on Schwarzschild metric.
We can say that manifold is extendable beyond Rindler horizon. And this extension is still a proper manifold. But when we extend manifold beyond event horizon of Schwarzschild metric we get geometry with geometric singularity. We can't say this extension is proper manifold. Manifolds don't have geometric singularities.

 

[PAllen]

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.

There is no need to worry about the global situation, that has no bearing on resolution geodesic incompleteness regionally. However, you do raise a very good observation. You start, in the SC coordinates, knowing that there is vacuum up to the horizon simply because the Ricci curvature is zero everywhere. You can than ask what kinds of extensions there are to resolve geodesic incompleteness. If you back off from the horizon to greater than 9/8 the SC radius (Buchdahl radius), you can smoothly join with an idealized non vacuum solution. Anything less than this, and it has been shown mathematically that either collapse must occur or matter inside must be following spacelike trajectories, i.e. moving locally FTL (a particular severe violation of energy conditions). Meanwhile, it is really trivial to quasilocally extend vacuum. Just switch coordinates from some event near the horizon to Riemann normal coordinates. These effectively express the local metric as perturbations from Minkowski metric. You find, for a large radius SC solution that perturbations from Minkowski are small and you have what looks like ordinary Minkowski spacetime with the horizon being a light like surface. The transform from SC just covers one side of this light like surface, moving at c. To extend, just remove the limitation to one side of this surface.

The key point is you don’t need to worry about the distant past or far away regions. The resolution of possible extensions to resolve geodesic incompleteness in one region is a quasi local geometric issue, coupled to quasi local physical plausibility questions about the geometric possibilities. You might then have additional problems arise if you try to extend your regional resolution globally.

 

[PAllen]

A follow up to my prior post is that you can use the transform from SC to Riemann Normal coordinates around a near horizon event to address the problems with a stable matter extension. In any given choice of near Minkowski coordinates up to the horizon, the speed of constant r trajectory approaches light speed as r approaches the horizon. Once you extend the near Minkowski coordinates beyond the arbitrary horizon chop, a path that keeps up with horizon from the inside is spacelike, I.e. FTL. Note, of course, that the horizon coordinate singularity vanishes on transform to Riemann Normal coordinates.

On the other hand, near the true singularity, setting up Riemann Normal coordinates encounters two problems - they cannot be set up at all to reach the singularity, and the perturbations from Minkowski grow without bound on approach to it, no matter how tiny a region you attempt to use.

 

[PeterDonis]

How do I decide which is coordinate singularity and not infinity type discontinuity.

I already answered that way back in post #5: compute curvature invariants and their limits as the coordinate singularity is approached.

Say I can imagine that Schwarzschild metric describes vacuum part of frozen star.

What does this mean?

I was talking about future timelike infinity.

Ok.

the limit of approaching event horizon as approaching future infinity

There is no such limit. A curve can either go to future timelike infinity or it can go to the event horizon. But no single curve can do both. The event horizon does not "touch" future timelike infinity. I already addressed this, too, way back in post #5.

 

[PeterDonis]

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

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.

when we extend manifold beyond event horizon of Schwarzschild metric we get geometry with geometric singularity.

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.

 

[Dale]

Speaking about Rindler chart and comparing it with Schwarzschild chart on Schwarzschild metric.
We can say that manifold is extendable beyond Rindler horizon. And this extension is still a proper manifold. But when we extend manifold beyond event horizon of Schwarzschild metric we get geometry with geometric singularity. We can't say this extension is proper manifold. Manifolds don't have geometric singularities.

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?

 

[zonde]

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.

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.

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.

Do you say that any set that includes singularity can't be considered open? How do you motivate this statement?

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.

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."

 

[Dale]

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.

Why would they end there?

 

[PeterDonis]

It's not obvious unless you emphasize that geodesics are incomplete there.

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

if geodesics are complete finite curvature invariants gives no reason to extend them

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

Do you say that any set that includes singularity can't be considered open?

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.

 

[Dale]

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

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.

 

[PeterDonis]

I may misunderstand the terminology.

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.

 

[Dale]

I was referring specifically ...

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

 

[Dale]

@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.

 

[zonde]

@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.

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.

Infinite curvature invariants are taken as an indication that the spacetime ends (or more likely that our model breaks down).

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.

 

[Dale]

So if the reasoning can't be restricted to math only it's sort of philosophy

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

I certainly do not understand that "spacetime ends" option. As I see this it's geodesics that end there not spacetime.

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

I don't understand why option "particular solution has somewhere sneaked in unphysical assumption" is not considered

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

 

[zonde]

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

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.

 

[zonde]

You start, in the SC coordinates, knowing that there is vacuum up to the horizon simply because the Ricci curvature is zero everywhere.

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.

 

[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.

 

[Dale]

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.

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.

PeterDonis point about infinite curvature invarinats seems at least self-consistent.

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

 

[deRoy]

And what's more there is a singularity at infinity!

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.

 

[PAllen]

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.

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.

 

[PeterDonis]

How do I know there is vacuum up to the horizon?

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.

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

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.