Schwartzschild exterior and interior solutions

[TrickyDicky]

Are the interior and exterior solutions described on a common manifold?
I mention it because the exterior one is an asymptotically flat space while the interior solution is conformally flat. I'm not sure if a single physical scenario (the vacuum soulution) can be described by two different geometries (manifolds) when according to GR, the geometry describes the physics.

 

[Mike_Fontenot]

Are the interior and exterior solutions described on a common manifold?
I mention it because the exterior one is an asymptotically flat space while the interior solution is conformally flat. I'm not sure if a single physical scenario (the vacuum soulution) can be described by two different geometries (manifolds) when according to GR, the geometry describes the physics.

I spent some time trying to understand the "within the event horizon" solution, and eventually came to suspect that it has no physical significance ... i.e., that only the "outside the event horizon" solution exists in our universe.

That opinion is in the extreme minority, but I found out, AFTER I had arrived at my conclusion, that Dirac came to the same conclusion. So I'm definitely in the minority, but I like the company that I'm keeping!

Mike Fontenot

 

[Mentz114]

If I remember correctly the two solutions are pasted together and there is a coordinate singularity at the join. I don't know how to describe that technically, but the coordinates are different and together cover the whole space.

 

[JustinLevy]

I spent some time trying to understand the "within the event horizon" solution, and eventually came to suspect that it has no physical significance ... i.e., that only the "outside the event horizon" solution exists in our universe.

I'm truly curious about this position, as someone here states this every once in awhile. Is it okay for me to ask questions about your views here, or is that against the forum rules?

If someone decided to free fall into the black hole, do you think they will hit a physical wall at the event horizon or something?

Note that we can choose a coordinate system which smoothly covers all of the manifold. So there clearly isn't a real singularity at the event horizon. The event horizon is a global concept and can't be noticed / measured locally.

 

[atyy]

I think the OP is about the non-vacuum interior solution.

 

[TrickyDicky]

If I remember correctly the two solutions are pasted together and there is a coordinate singularity at the join. I don't know how to describe that technically, but the coordinates are different and together cover the whole space.

Yes, I know the two solutions are pasted together, what I'm saying is that we have an exterior exact vacuum solution that is used to test GR (Mercury precession, deflection of light and gravitational redshift) and that uses a metric (Schwartzschild's metric) that describes an static geometry, and then we have an inner vacuum solution (describing a different type of manifold,that is not asymptotically flat) that we paste to the outer solution, thru a convenient change of coordinates, and together cover the whole space.

My question comes here, can we really cover the whole space with two geometries that are so different, shouldn't we describe a unique physical situation with just one type of manifold?
According to Einstein, coordinate transformations are allowed in GR as long as they don't change the intrinsic geometry, or that is what general covariance seems to imply, otherwise we would be free to use whatever manifold we chose to describe the curvature in a vacuum outside a spherical object or in the universe, but I believe the physics can be described with many different coordinates systems (that's what general covariance is about) but all of them referring to a unique geometry (manifold). Or is this not correct?

 

[PeterDonis]

If I remember correctly the two solutions are pasted together and there is a coordinate singularity at the join. I don't know how to describe that technically, but the coordinates are different and together cover the whole space.

The full "extended Schwarzschild spacetime" is a single manifold (a single "geometry" in TrickyDicky's terminology), and Kruskal-Szeres coordinates cover all of it with a single coordinate patch. See this thread for a discussion of those coordinates:

https://www.physicsforums.com/showthread.php?t=239378

You can also Google to find plenty of other references; however, note that the Wikipedia page says that the article is in need of attention from an expert, so it may contain errors or misstatements.

I find Kruskal coordinates, and the diagram of the extended Schwarzschild spacetime in terms of them, very useful in keeping straight what's going on in that spacetime, and avoiding many common confusions that can arise from looking only at Schwarzschild coordinates.

 

[atyy]

Yes, I know the two solutions are pasted together, what I'm saying is that we have an exterior exact vacuum solution that is used to test GR (Mercury precession, deflection of light and gravitational redshift) and that uses a metric (Schwartzschild's metric) that describes an static geometry, and then we have an inner vacuum solution (describing a different type of manifold,that is not asymptotically flat) that we paste to the outer solution, thru a convenient change of coordinates, and together cover the whole space.

My question comes here, can we really cover the whole space with two geometries that are so different, shouldn't we describe a unique physical situation with just one type of manifold?
According to Einstein, coordinate transformations are allowed in GR as long as they don't change the intrinsic geometry, or that is what general covariance seems to imply, otherwise we would be free to use whatever manifold we chose to describe the curvature in a vacuum outside a spherical object or in the universe, but I believe the physics can be described with many different coordinates systems (that's what general covariance is about) but all of them referring to a unique geometry (manifold). Or is this not correct?

The interior solution is not vacuum.

We don't use the full manifold of either exterior (maximally extended Schwarzschild solution) nor the full interior solution. We cut the parts we want and join them. The joining is determined by boundary conditions, which are meant to ensure that things look ok as we pass from one region to another.

 

[TrickyDicky]

The interior solution is not vacuum.

The BH interior solution is not vacuum?

We don't use the full manifold of either exterior (maximally extended Schwarzschild solution) nor the full interior solution. We cut the parts we want and join them. The joining is determined by boundary conditions, which are meant to ensure that things look ok as we pass from one region to another.

This is what I believe I asserted.
I'll try to be more specific, I'm referring to the Kruskal coordinates system ad the line element we obtain from it. It is my understanding (please correct me if I'm wrong here)
that it determines a manifold that becomes flat thru a conformal transformation (conformally flat).
So it would look as if the boundary conditions are different from those who demand an asymptotically flat manifold as was the case with the initial Schwartzschild metric.
So it seems natural to ask, on what grounds are the boundary conditions for the vacuum solution of the Einstein fied equations changed?
Doesn't a change of the boundary conditions produce a change of the geometry and thus different physical consequences?

 

[atyy]

Conformal flatness is not flatness.

Asymptotic flatness is how the metric behaves at infinity, but the part that we cut and paste for joining the interior and exterior solutions is not at infinity.

KS coordinates usually refer to the maximally extended Schwarzschild solution, which includes the black hole. OTOH, the interior Schwarzschild solution has no black hole. Which are talking about?

 

[PeterDonis]

KS coordinates usually refer to the maximally extended Schwarzschild solution, which includes the black hole. OTOH, the interior Schwarzschild solution has no black hole.

It appears that by "interior Schwarzschild solution" you mean "an FRW solution modeling the interior of a collapsing star, which is patched onto an exterior vacuum solution that is a portion of the extended Schwarzschild spacetime". The term "interior Schwarzschild solution" is also used to refer to the portion of extended Schwarzschild spacetime which is inside the horizon, in which case the "interior solution" *is* a vacuum solution (but in Schwarzschild coordinates, it's a separate, disconnected patch from the exterior solution, because of the coordinate singularity at the horizon).

If you are using the term as I just noted, then it's true that the interior solution has no "black hole", since it is modeling the non-vacuum collapsing star. But it's also true that that "interior solution" does not last for all time, because the star is collapsing. When the collapse reaches the center, radius r = 0, the "interior solution" disappears and a singularity is formed (at least according to the model of straightforward classical GR, without including quantum gravity effects, which we don't have an accepted theory for). At any time slice after the surface of the collapsing star passes through radius r = 2M, where M is the total mass of the star as measured at infinity (for example, by looking at Keplerian orbits very far away), a horizon is present at r = 2M, so it includes a "black hole" by the usual definition. And at any time slice after the singularity is formed at r = 0, the entire slice is isomorphic to a slice of the extended Schwarzschild spacetime that passes through the future horizon and hits the future singularity.

 

[TrickyDicky]

Conformal flatness is not flatness.

Of course, and? Don't know what this assertion has to do with anything. Conformally flat here refers to a manifold that is related to minkowki manifold by a conformal factor.

flatness is how the metric behaves at infinity, but the part that we cut and paste for joining the interior and exterior solutions is not at infinity.

This is the point of my question, can we cut and paste different parts so blithely?
disregarding the fact that we are changing the boundary conditions?, It seems clear that if we set as a boundary condition that the manifold must be minkowskian at infinity (g=1), we can't use the KS line element, and viceversa if we choose as a boundary condition that the manifold is minkowskian thru a conformal transformation we are rejecting the Schartzschild static manifold. I guess we must choose one of the two but not both a the same time if we are describing a single physical phenomenon (gravitation in empty space outside a spherically symmetric object).

KS coordinates usually refer to the maximally extended Schwarzschild solution, which includes the black hole. Which are talking about?

I'm confronting KS and Schwartzschild line elements.

 

[atyy]

@PeterDonis: Yes, I understood you perfectly. My question was meant for the OP. I think it is the non-vacuum interior solution which is conformally flat. I don't know about the interior of the vacuum solution - is it?

 

[atyy]

p413 of http://books.google.com/books?id=Si...y+flat+vacuum+solutions&source=gbs_navlinks_s states that the only conformally flat vacuum solution is flat. So the interior of the Schwarzschild vacuum solution is not conformally flat. It is the interior Schwarzschild solution containing matter which is conformally flat.

 

[Mike_Fontenot]

I'm truly curious about this position, as someone here states this every once in awhile. Is it okay for me to ask questions about your views here, or is that against the forum rules?

The details are in a thread on the "sci.physics.foundations" newsgroup. The title of the thread is "Schwarzschild Inside the Event Horizon". (Note that "Schwarzschild" is misspelled in the title of that thread). You should be able to find that thread via Google Groups at

http://groups.google.com/advanced_search?q=& .

Mike Fontenot

 

[TrickyDicky]

p413 of http://books.google.com/books?id=Si...y+flat+vacuum+solutions&source=gbs_navlinks_s states that the only conformally flat vacuum solution is flat. So the interior of the Schwarzschild vacuum solution is not conformally flat. It is the interior Schwarzschild solution containing matter which is conformally flat.

Please define "flat" in this context.

Is the KS line element conformally flat or not?, perhaps someone can clarify this

 

[atyy]

Kruskal-Szekeres coordinates cover the maximally extended Schwarzscild vacuum solution and are not conformally flat.

 

[TrickyDicky]

Kruskal-Szekeres coordinates cover the maximally extended Schwarzscild vacuum solution and are not conformally flat.

And they are not asymptotically flat either, right?

 

[atyy]

K-S coordinates are asymptotically flat.

 

[atyy]

Also useful for pasting solutions is section 3.7 of Eric Poisson's http://www.physics.uoguelph.ca/poisson/research/agr.pdf

 

[TrickyDicky]

The bottom line is still that the geometry described by the K-S coordinates and the geometry described by the schwartzschild line element seem really different. They certainly look like different manifolds, are they really the same?

 

[PeterDonis]

The bottom line is still that the geometry described by the K-S coordinates and the geometry described by the schwartzschild line element seem really different. They certainly look like different manifolds, are they really the same?

The exterior Schwarzschild line element covers "Region I" of the K-S coordinates (the "right-hand wedge", U > 0, |V| < U using the coordinate definitions on the Wikipedia page http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates for r > 2M). The interior Schwarzschild line element (using the alternate definition I gave in an earlier post--the same line element as the exterior, but with r < 2M) covers "Region II", the "upper wedge" bounded by the future singularity (V > 0, |U| < V). Neither patch covers the future horizon (U = V >= 0), because of the coordinate singularity in the Schwarzschild line element at r = 2M; this is why the interior and exterior Schwarzschild coordinate patches are disconnected. So the underlying geometry is the same, but the K-S coordinates cover all of it, while the Schwarzschild coordinates only cover two disconnected portions of it.

 

[TrickyDicky]

K-S coordinates are asymptotically flat.

Well, yes, according to the modern coordinate-free definition of AF spacetime due to Penrose, Hawking and Ellis, and others that was changed to allow black holes in the 70's and instead of requiring an asymptotically simple and empty manifold, requires jus a "weakly asymptotically simple and empty(WASE) spacetime". see wikipedia page under formal definitions: http://en.wikipedia.org/wiki/Asymptotically_flat_spacetime

But K-S metric is not asymptotically flat according to the historically first coordinate-dependent definition of AF or one which requires an asymptotically simple and empty manifold.

 

[PAllen]

Well, yes, according to the modern coordinate-free definition of AF spacetime due to Penrose, Hawking and Ellis, and others that was changed to allow black holes in the 70's and instead of requiring an asymptotically simple and empty manifold, requires jus a "weakly asymptotically simple and empty(WASE) spacetime". see wikipedia page under formal definitions: http://en.wikipedia.org/wiki/Asymptotically_flat_spacetime

But K-S metric is not asymptotically flat according to the historically first coordinate-dependent definition of AF or one which requires an asymptotically simple and empty manifold.

Can you explain this? I read the wikipedia reference and I can't find any support for the idea that Kruskal is not coordinate assymptotically flat. Kruskal coordinates are not even mentioned. It would be strange that the same geometry is AF in one coordinate system and not in another. All books I have claim kruskal is just another coordinate map for the same geometry Schwarzschild (well, really, that Schwarzschild exterior / interior are two coordinate maps on regions of the geometry represented Kruskal in one coordinate patch).

 

[JesseM]

The bottom line is still that the geometry described by the K-S coordinates and the geometry described by the schwartzschild line element seem really different. They certainly look like different manifolds, are they really the same?

Presumably if you have any worldline (timelike, spacelike or lightlike) defined in terms of Schwarzschild coordinates, you can then use the coordinate transformation between Schwarzschild and Kruskal-Szekeres coordinates to find the description of the same worldline in KS coordinates. Then if you use the Schwarzschild line element to integrate ds along the path in Schwarzschild coordinates, and use the KS line element to integrate ds along the same path in KS coordinates (between a pair of points which also map to one another by the coordinate transformation), you should get the same answer. (isn't the KS line element derived by doing a coordinate transformation on the Schwarzschild line element, ensuring that this will be the case?) As I understand it, "the geometry" is defined entirely in terms of path lengths along arbitrary paths, so this is all that is required for them to both be describing the same geometry.

 

[TrickyDicky]

Can you explain this? I read the wikipedia reference and I can't find any support for the idea that Kruskal is not coordinate assymptotically flat. Kruskal coordinates are not even mentioned.

Read again my post, I say there that Kruskal metric is asymptotically flat according to the standard definition of AF that is currently used, but that according to the mentioned wikipedia reference (in the subsections "Formal definitions" and "A coordinate-free definition"), the definition was changed in the 60's to accommodate black holes by people like Penrose, by introducing the concept of "conformal compactification" , as I undertand from the same wikipedia page:subsection "A coordinate-dependent definition" historically asymptotic flatness was coordinate-dependent and therefore it only allowed unimodular transformations of the coordinates (equations that only holds if g=1). This was the case at the time Scwartzschild derived his solution, see 't Hooft comment at the bottom of page 49 in http://www.phys.uu.nl/~thooft/lectures/genrel_2010.pdf
Specifically is asserted in the wikipedia quote "A manifold M is asymptotically simple if it admits a conformal compactification {M} such that every null geodesic in M has a future and past endpoints on the boundary of{M}. Since the latter excludes black holes, one defines a weakly asymptotically simple manifold as a manifold M with an open set U⊂M isometric to a neighbourhood of the boundary of {M}, where {M} is the conformal compactification of some asymptotically simple manifold. A manifold is asymptotically flat if it is weakly asymptotically simple and asymptotically empty in the sense that its Ricci tensor vanishes in a neighbourhood of the boundary of {M}." End wikipedia quote.
This quote is actually taken from the reference 2 cited by the wikipedia page: http://arxiv.org/abs/gr-qc/9707012 I think the relevant pages are from pages 45-50.
According to the so modified definition of asymptotic flatness: Kruskal is an example of an asymptotically flat spacetime since it approaches the metric of compactified Minkowski spacetime as r → ∞.
Whereas before the introduction of the "conformal compactification" concept asymptotic flatness required to approach the metric of Minkowski spacetime (not just its conformal compactification) as r → ∞.

It would be strange that the same geometry is AF in one coordinate system and not in another.

It's not only strange, it's impossible, in fact both the Schwartzschild metric and Kruskal are AF according to the modern definition. But only the Schwartzschild metric is AF according to the original one. Whether this means they are actually the same geometry or different manifolds in fact is what I'm trying to ascertain here.

All books I have claim kruskal is just another coordinate map for the same geometry Schwarzschild

Not necessarily this case, but I've heard that textbooks have been wrong in the past about certain points.

 

[TrickyDicky]

Presumably if you have any worldline (timelike, spacelike or lightlike) defined in terms of Schwarzschild coordinates, you can then use the coordinate transformation between Schwarzschild and Kruskal-Szekeres coordinates to find the description of the same worldline in KS coordinates. Then if you use the Schwarzschild line element to integrate ds along the path in Schwarzschild coordinates, and use the KS line element to integrate ds along the same path in KS coordinates (between a pair of points which also map to one another by the coordinate transformation), you should get the same answer. (isn't the KS line element derived by doing a coordinate transformation on the Schwarzschild line element, ensuring that this will be the case?) As I understand it, "the geometry" is defined entirely in terms of path lengths along arbitrary paths, so this is all that is required for them to both be describing the same geometry.

It all seems to depend on whether this particular coordinate transformation between the Schwarzschild line element and the KS line element is valid in the context of the boundary conditions of the vacuum solution of the Einstein field equations, I know that according to standard textbooks it is.
But as I explained in my previous post, there might be reasons that lead us to think that it is not such an assured fact: an ad hoc change of the definition of asymptotic flatness to allow black holes seems to have been made thru the introduction of "conformal compactification", it is not clear to me that the original Schwartzschild manifold admits such conformal compactification since it would mean the central mass of the vacuum solution acts as a test particle (it doesn't curve the manifold) and can be then considered a minkowskian point. It makes one wonder: how can it be a gravitational source in empty space then? and originate planet precession, or bending of light.

 

[PAllen]

Read again my post, I say there that Kruskal metric is asymptotically flat according to the standard definition of AF that is currently used, but that according to the mentioned wikipedia reference (in the subsections "Formal definitions" and "A coordinate-free definition"), the definition was changed in the 60's to accommodate black holes by people like Penrose, by introducing the concept of "conformal compactification" , as I undertand from the same wikipedia page:subsection "A coordinate-dependent definition" historically asymptotic flatness was coordinate-dependent and therefore it only allowed unimodular transformations of the coordinates (equations that only holds if g=1). This was the case at the time Scwartzschild derived his solution, see 't Hooft comment at the bottom of page 49 in http://www.phys.uu.nl/~thooft/lectures/genrel_2010.pdf
Specifically is asserted in the wikipedia quote "A manifold M is asymptotically simple if it admits a conformal compactification {M} such that every null geodesic in M has a future and past endpoints on the boundary of{M}. Since the latter excludes black holes, one defines a weakly asymptotically simple manifold as a manifold M with an open set U⊂M isometric to a neighbourhood of the boundary of {M}, where {M} is the conformal compactification of some asymptotically simple manifold. A manifold is asymptotically flat if it is weakly asymptotically simple and asymptotically empty in the sense that its Ricci tensor vanishes in a neighbourhood of the boundary of {M}." End wikipedia quote.
This quote is actually taken from the reference 2 cited by the wikipedia page: http://arxiv.org/abs/gr-qc/9707012 I think the relevant pages are from pages 45-50.
According to the so modified definition of asymptotic flatness: Kruskal is an example of an asymptotically flat spacetime since it approaches the metric of compactified Minkowski spacetime as r → ∞.
Whereas before the introduction of the "conformal compactification" concept asymptotic flatness required to approach the metric of Minkowski spacetime (not just its conformal compactification) as r → ∞.

I still don't follow you. To me, this [the wikipedia article] says, that in trying to arrive at a coordinate independent definition of AF, it was necessary to adopt a
derfinition of "weakly asymptotically simple manifold" to allow black holes. I do not see any statement or argument that implies that Kruskal fails the coordinate definition. I also looked at your other references and don't come up with any argument Kruskal fails the coordinate defintion. Can you provide a more direct explanation or reference that Kruskal fails the coordinate defintion of AF? (I am not trying to be difficult - I just don't understand how this conclusion follows from the citations so far).

Most satisfying would be a calculation (or reference theirto) showing how direct application of the coordinate AF definition fails for Kruskal.

[EDIT: In particular, the coordinate definition is all about coordinate behavior in the limit 'at infinity', and the ability to introduce coordinates with certian properties. If one takes one of the two 'outside' singularity regions of Kruskal, it seems trivial to introduce coordinates meeting the given conditions. ]

 

[TrickyDicky]

I still don't follow you. To me, this [the wikipedia article] says, that in trying to arrive at a coordinate independent definition of AF, it was necessary to adopt a
derfinition of "weakly asymptotically simple manifold" to allow black holes. I do not see any statement or argument that implies that Kruskal fails the coordinate definition. I also looked at your other references and don't come up with any argument Kruskal fails the coordinate defintion. Can you provide a more direct explanation or reference that Kruskal fails the coordinate defintion of AF? (I am not trying to be difficult - I just don't understand how this conclusion follows from the citations so far).

Most satisfying would be a calculation (or reference theirto) showing how direct application of the coordinate AF definition fails for Kruskal.

You are right that it isn't stated that way in the wikipedia page. That Kruskal fails the coordinate-dependent definition is my own inference from the fact that in order to get the KS line element from the Schwartzschild line element one has to introduce a coordinate transformation that implicitly assumes the conformal compactification of the space or in other words, the kruskal manifold is just "locally asymptotically flat", while coordinate-dependent asymptotic flatness is "global" so to speak, since coordinates are not just local by definition.
But I'm actually just trying to understand it too, so if I come up with a better reference or explanation I'll post it. Even better if someone around here with a better grasp of differential geometry can clarify this.

 

[TrickyDicky]

One hint pointing to the two line elements describing diffeent geometries is that the Schwartzschild line element defines a static spacetime(with all metric components independent of coordinate t and invariant for time reversal) while the KS line element is not that of a static spacetime. That's odd because if the two line elements are different coordinates descriptions of the same geometry, how can it be static and non-static at the same time?
Any help understanding this would be welcome.

 

[PAllen]

One hint pointing to the two line elements describing diffeent geometries is that the Schwartzschild line element defines a static spacetime(with all metric components independent of coordinate t and invariant for time reversal) while the KS line element is not that of a static spacetime. That's odd because if the two line elements are different coordinates descriptions of the same geometry, how can it be static and non-static at the same time?
Any help understanding this would be welcome.

This might mean nothing. You can call any coordinate 't'. In the interior Schwarzschild solution, coordinate r is the timelike coordinate. Dirac used SR coordinates u and v that were both mixtures of timelike and spacelike character. I don't know how to specify a test of static character for arbitrary coordinates, or a coordinate independent test for static character. If I can think of one, I'll post; hopefully someone more knowledgeble will post instead.

 

[TrickyDicky]

This might mean nothing. You can call any coordinate 't'. In the interior Schwarzschild solution, coordinate r is the timelike coordinate. Dirac used SR coordinates u and v that were both mixtures of timelike and spacelike character. I don't know how to specify a test of static character for arbitrary coordinates, or a coordinate independent test for static character. If I can think of one, I'll post; hopefully someone more knowledgeble will post instead.

Look at the second condition for staticity.The KS line element and actually any metric describing a BH is not time-symmetric, by definition.

 

[atyy]

One hint pointing to the two line elements describing diffeent geometries is that the Schwartzschild line element defines a static spacetime(with all metric components independent of coordinate t and invariant for time reversal) while the KS line element is not that of a static spacetime. That's odd because if the two line elements are different coordinates descriptions of the same geometry, how can it be static and non-static at the same time?
Any help understanding this would be welcome.

The Schwarzschld coordinates only describe a static spacetime if the coordinate radius is greater than the Schwarzschild radius. It has to be joined to another solution that describes static matter with coordinate radius greater than the Schwarzschild radius for the entire spacetime to be static.

 

[PAllen]

Look at the second condition for staticity.The KS line element and actually any metric describing a BH is not time-symmetric, by definition.

Why do you say an BH solution is non-time symmetric? I would think an eternal solution is time symmetric. Some of the references you gave earlier distinguished black hole solutions that could result from collapsing matter from eternal solutions with a wormhole. My guess would be the complete Kruskal geometry is eternal, time symmetric, and static.

 

[George Jones]

One hint pointing to the two line elements describing diffeent geometries is that the Schwartzschild line element defines a static spacetime(with all metric components independent of coordinate t and invariant for time reversal) while the KS line element is not that of a static spacetime. That's odd because if the two line elements are different coordinates descriptions of the same geometry, how can it be static and non-static at the same time?
Any help understanding this would be welcome.

Why do you say an BH solution is non-time symmetric? I would think an eternal solution is time symmetric. Some of the references you gave earlier distinguished black hole solutions that could result from collapsing matter from eternal solutions with a wormhole. My guess would be the complete Kruskal geometry is eternal, time symmetric, and static.

Kruskal-Szekeres spacetime is an extension of (external) Schwarzschild spacetime, and, as such, K-S spacetime is static everywhere Schwarzschild is, i.e., outside the event horizon, as atty noted. A region of spacetime is static if there is a hypersurface-orthogonal (gives non-rotating) timelike Killing vector field (gives stationary) in the region.

Below the event horizon, no timelike Killing field exists, so K-S is not even stationary there, let alone static.