Mathematical problem with Schwarzschild spacetime?

[Bosko]

{Mentor’s note: split from https://www.physicsforums.com/threa...ingularities-apply-to-all-black-holes.1061139 }

However, a simple mathematical-logical problem remains:
Einstein's field equations are a system of differential equations. Their solution should be differentiable (smooth).
Schwarzschild's solution is not differentiable for $r=0$ (central singularity) and $r=r_s$ (event horizon)
Doubts about the correctness of such a solution are justified.

 

[PeterDonis]

However, a simple mathematical-logical problem remains:

No, there is no problem, just a misunderstanding on your part.

Einstein's field equations are a system of differential equations. Their solution should be differentiable (smooth).

More precisely, they must be smooth on an open manifold.

Schwarzschild's solution is not differentiable for $r=0$ (central singularity)

$r = 0$ is not part of the open manifold. So this claim does not even make sense, since differentiability cannot even be evaluated for points that are not part of the manifold.

and $r = r_s$ (event horizon)

This is wrong. Schwarzschild coordinates are invalid on the horizon, but the spacetime geometry is still smooth. Multiple coordinate charts are known that remove the coordinate singularity on the horizon and explicitly show the smoothness of the solution there.

Doubts about the correctness of such a solution are justified.

No, they are not. See above.

What is an open question is whether the full Schwarzschild solution, all the way down to the limit $r \to 0$, is actually realized in our universe. But that question has nothing to do with its mathematical consistency.

 

[Dale]

Schwarzschild's solution is not differentiable for … r=rs (event horizon)

As @PeterDonis said, this is not correct. The Schwarzschild spacetime metric is differentiable at the horizon. The Schwarzschild coordinate chart is a different thing from the manifold with its metric. The non-differentiability of that chart is not an issue, you can make non differentiable coordinate charts at any event in any spacetime.

 

[Bosko]

$r = 0$ is not part of the open manifold.

Why is not ?

This is wrong. Schwarzschild coordinates are invalid on the horizon,

Ok

but the spacetime geometry is still smooth.

It should be smooth everywhere

Multiple coordinate charts are known that remove the coordinate singularity on the horizon and explicitly show the smoothness of the solution there.

You are using a circular argument.

 

[Dale]

You are using a circular argument.

The existence of any smooth chart at the horizon shows that the manifold is smooth at the horizon. I am not sure what you think is circular.

 

[Bosko]

As @PeterDonis said, this is not correct. The Schwarzschild spacetime metric is differentiable at the horizon.

Is not with the Schwarzschild coordinates.

The Schwarzschild coordinate chart is a different thing from the manifold with its metric. The non-differentiability of that chart is not an issue, you can make non differentiable coordinate charts at any event in any spacetime.

If I understand you correctly, this would violate the very definition of the coordinate chart because of the diffeomorphism property .

 

[Dale]

this would violate the very definition of the coordinate chart because of diffeomorphism

How so? Please justify explicitly this violation you are claiming

 

[Bosko]

How so? Please justify explicitly this violation you are claiming

Look definition of the coordinate chart .

 

[Dale]

Look definition of the coordinate chart .

This is completely insufficient. No amount of my looking at a definition can possibly tell me what is in your head. Specifically what you incorrectly think is violated here.

Let’s try again: Please justify explicitly this violation you are claiming

 

[PeterDonis]

Why is not ?

One way of seeing why $r = 0$ is not part of the manifold for Schwarzschild spacetime is to look at a Kruskal diagram, like the one at the top right of this page:

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

The locus $r = 0$ is the two hyperbolas in bold (blue at the top, red at the bottom). It should be obvious from the diagram that the region bounded by those hyperbolas, not including the hyperbolas themselves, is an open region--and that if we do include the hyperbolas, the region is no longer open.

 

[PeterDonis]

It should be smooth everywhere

That's exactly what I said: the spacetime geometry of Schwarzschild spacetime is smooth everywhere.

You are using a circular argument.

Tell that to the authors of all the GR textbooks that say precisely the same thing I am saying.

Is not with the Schwarzschild coordinates.

This is wrong. Differentiability, or lack thereof, is an invariant property of the manifold. You cannot disprove it by showing one coordinate chart which is singular at the horizon. You have to prove that all possible coordinate charts are singular at the horizon. Which is obviously impossible for Schwarzschild spacetime since there are multiple coordinate charts known which are not singular at the horizon. One of them is the Kruskal chart, which is used to draw the diagram I referred you to in a previous post.

In fact, that diagram can also be used to see why Schwarzschild coordinates are singular on the horizon while the manifold itself is not. The diagram shows "grid lines" of Schwarzschild coordinates: they are the straight lines radiating in all directions from the origin (curves of constant $t$), and the hyperbolas orthogonal to those lines (curves of constant $r$)--except for the horizon itself, which is the crossed 45 degree lines that intersect at the center of the diagram. Those lines are lines of constant $r$, but they are also lines of constant $t$ ("constant" in the sense of $t$ being infinite/undefined)--in other words, on the horizon the Schwarzschild coordinate grid "collapses" and is no longer valid. But that is a problem with the coordinate grid, not the manifold.

 

[Dale]

an open region

I think that this may be the thing that @Bosko is missing. The manifold and charts on the manifold are all based on open sets and unions of open sets. So the boundaries are not part of the chart or the manifold. Meaning that non-differentiability on the boundary is simply not relevant because the boundary is not part of the chart or manifold

 

[PeterDonis]

I think that this may be the thing that @Bosko is missing. The manifold and charts on the manifold are all based on open sets and unions of open sets. So the boundaries are not part of the chart or the manifold.

Yes, that's true, but it's also true that in other cases where a radial coordinate $r$ is defined (such as a spherically symmetric spacetime describing a non-rotating planet or star surrounded by vacuum), the locus $r = 0$ is part of the manifold, even though the coordinate interval $0 \le r < \infty$ is only half-open. So it might seem confusing that in the case of maximally extended Schwarzschild spacetime, the locus $r = 0$ is not part of the manifold; only the fully open coordinate interval $0 < r < \infty$ is.

Of course none of this applies at the horizon $r = 2M$; there the issue @Bosko seems to be missing is different.

 

[Dale]

the locus $r = 0$ is part of the manifold, even though the coordinate interval $0 \le r < \infty$ is only half-open.

Well, that is actually a bit of very common and forgivable laziness. A coordinate chart is a smooth and invertible mapping between an open subset of the manifold and an open subset of $\mathbb{R}^N$. So you cannot have $0 \le r < \infty$, or it would not be open in $\mathbb{R}^N$. Instead the chart must exclude $r=0$ even though it is part of the manifold. Keeping $r=0$ would also make the map non-invertible.

 

[PeterDonis]

Well, that is actually a bit of very common and forgivable laziness.

Yes, agreed, $r = 0$ is a coordinate singularity, strictly speaking, even in ordinary spherical polar coordinates on a spacetime describing a non-rotating planet or star. But as you say, glossing over this is common and forgivable in that case, because we know there are other charts, e.g., a Cartesian chart, that show that locus to be a nonsingular part of the manifold in such a case. (We commonly gloss over the coordinate singularities at the poles in ordinary latitude/longitude coordinates on a 2-sphere for similar reasons.)

In Schwarzschild spacetime, however, the locus $r = 0$ is not a nonsingular part of the manifold, and so cannot be covered in any valid chart.

 

[Dale]

Agreed. Now hopefully @Bosko can clarify their concerns with some detail as neither topic is mathematically problematic, but from the little they have said I cannot tell for sure why they think those are issues.

 

[Bosko]

@PeterDonis, @Dale Thank you.

 

[cianfa72]

We commonly gloss over the coordinate singularities at the poles in ordinary latitude/longitude coordinates on a 2-sphere for similar reasons.

Yes, in a sense it is just an "extension" of coordinate chart definition to an non open region of latitude/longitude "plane" in which the mapping with the globe is not one-to-one.

 

[Bosko]

Agreed. Now hopefully @Bosko can clarify their concerns with some detail as neither topic is mathematically problematic, but from the little they have said I cannot tell for sure why they think those are issues.

In general, if you have a system of differential equations with the property that space-time must be smooth everywhere, then a solution that is not smooth everywhere is suspicious.

The only thing I found that has no proof in Schwarzschild's solution is:

That a massive spherical object can be replaced by a material point.
There is a clear mathematical proof for flat but not for curved space-time.

 

[Dale]

a system of differential equations with the property that space-time must be smooth everywhere

What does everywhere mean? Not verbally, but in precise mathematical terms?

If $(M,g)$ is a pseudo-Riemannian manifold then “everywhere” means for all $X \in M$. A differentiable manifold is therefore smooth for all such $X$. And an assertion that $M$ is not smooth for some $Y \notin M$ does not contradict the fact that $M$ is smooth everywhere.

You are “suspicious” because you claim that $M$ is not smooth at $Y$. And we are pointing out the fact that $Y \notin M$. So $M$ is still smooth for all $X \in M$. In other words, $M$ is smooth everywhere. The fact that $Y\notin M$ makes your claim that $M$ is not smooth at $Y$ irrelevant and probably meaningless.

Similarly for charts.

The only thing I found that has no proof in Schwarzschild's solution is:

That a massive spherical object can be replaced by a material point.

Indeed, that is neither proved nor claimed.

The related concept that has been proved and is claimed is Birkhoff’s theorem. https://en.m.wikipedia.org/wiki/Birkhoff's_theorem_(relativity)

It basically says that the Schwarzschild spacetime is the unique spherically symmetric vacuum spacetime. So this means that in the vacuum outside of a spherical gravitating object you can use the Schwarzschild metric. It does not make any claim whatsoever about the non-vacuum region. In particular it makes no claims about replacing a spherical object by a material point.

 

[PeterDonis]

Yes, in a sense it is just an "extension" of coordinate chart definition to an non open region of latitude/longitude "plane" in which the mapping with the globe is not one-to-one.

I'm not sure what you mean by this.

 

[PeterDonis]

The only thing I found that has no proof in Schwarzschild's solution is:

That a massive spherical object can be replaced by a material point.

There is no proof of this because it is false. There is no "material point" in the maximally extended Schwarzschild spacetime; the spacetime is entirely vacuum. For example, if you look at the Kruskal diagram I referred to earlier, there is no "material point" anywhere.

There is a clear mathematical proof for flat but not for curved space-time.

Flat spacetime would mean there is no gravitating object present at all, so there would be no "massive spherical object" to be replaced.

 

[cianfa72]

I'm not sure what you mean by this.

I meant that a chart of the Earth based on latitude/longitude is not one-to-one (consider for instance the longitude of Poles) and maps it to a non-open set of latitude/longitude coordinates.

 

[PeterDonis]

I meant that a chart of the Earth based on latitude/longitude is not one-to-one (consider for instance the longitude of Poles)

This is wrong. The chart you refer to does not cover the poles. This is the fact that is commonly glossed over when discussing such charts; we treat the poles as though they were covered by the chart, but they actually are not. The actual latitude/longitude chart only covers the open region that consists of the rest of the 2-sphere other than the poles, and is one-to-one. Being one-to-one and covering an open region are part of the definition of a coordinate chart; it is impossible to have a valid chart that does not have those properties.

and maps it to a non-open set of latitude/longitude coordinates.

This is wrong as well. See above.

Please do not clutter someone else's thread with mistaken comments.

 

[Dale]

I meant that a chart of the Earth based on latitude/longitude is not one-to-one (consider for instance the longitude of Poles) and maps it to a non-open set of latitude/longitude coordinates.

This is wrong. The chart you refer to does not cover the poles.

I think that we need to distinguish the mathematical meaning of "chart" from the common meaning of "chart". The mathematical meaning of "chart" is a smooth and invertible mapping between an open subset of $\mathbb{R}^N$ and an open subset of a manifold. The common meaning of "chart" is a drawing on a piece of paper that represents the world.

Indeed a common "chart" of the entire earth is not isomorphic to a mathematical "chart" for exactly the reason that @cianfa72 identified: it is not one-to-one.

 

[PeterDonis]

I think that we need to distinguish the mathematical meaning of "chart" from the common meaning of "chart".

In this thread (and indeed this forum) I had assumed we would be using the physicist's meaning of "chart", which is the same as the mathematical meaning you describe. Indeed, that meaning, and its implications, are part of what we are trying to get the OP to understand.

 

[Dale]

In this thread (and indeed this forum) I had assumed we would be using the physicist's meaning of "chart", which is the same as the mathematical meaning you describe. Indeed, that meaning, and its implications, are part of what we are trying to get the OP to understand.

Yes, good point.

@cianfa72 it would be better to stick with the precise technical meaning of the specific technical terms being discussed in any given thread. It just invites confusion otherwise, on topics that are already inherently confusing on their own.