Schwartzschild exterior and interior solutions

[PAllen]

As I understood the references you gave before about a type of conformal transform that changed geometry, the meaning was completely differfent: these were not just coordinate transforms, they introduced 'points at infinity' which certainly does change the geometry.

 

[TrickyDicky]

Ok, fine, so they define a concept of a 2 space that has curvature but also has a conformally flat metric by their definition. So what? There is not the slightest hint of a claim or argument that any of the transforms they have done change geometry. All they are saying is the metric takes this form in these coordinates. And...

Note, especially, they say things like:

"We can find lots of different coordinate transformations that preserve the conformal nature
of the 2-space defined by equation (5)."

This is just a nice metric form that can be achieved by coordinate transforms. Reading any of this to say these coordinate transforms change the geometry is a false reading of this material.

Let's not jump into conclusions yet (me included), do you agree that the line element in (9), when considering only null geodesics (light), in other words considering the angular components of the metric constant, is that of a conformally flat spacetime?

Conformally flat spacetimes have a vanishing Weyl curvature, any vacuum solution is incompatible with a vanishing Weyl curvature.

If so (and I insist considering a radiation only situation), how could this line element be a vacuum solution for light?

 

[PAllen]

Let's not jump into conclusions yet (me included), do you agree that the line element in (9), when considering only null geodesics (light), in other words considering the angular components of the metric constant, is that of a conformally flat spacetime?

Conformally flat spacetimes have a vanishing Weyl curvature, any vacuum solution is incompatible with a vanishing Weyl curvature.

If so (and I insist considering a radiation only situation), how could this line element be a vacuum solution for light?

I am not familiar with the formal definition and properties of conformal flatness. Your reference defines that this metric is conformally flat, noting carefully, that it still has curvature using the precisely the same argument about derivatives that I gave in my post #94.

I am not very familiar with the Weyl curvature tensor. A quick look up in Wikipedia suggests that in dimension >=4, vanishing Weyl tensor is necessary and sufficient for conformal flatness. However, we are talking about conformal flatness of a 2-d slice of a manifold. The same Wiki article (on Weyl tensor) observes that *any* 2-d manifold is conformally flat, so all your other conclusions are unfounded. That is, I conclude from all this that you can take any 2-d slice you want of any 4-d manifold, and you can find coordinates that express the metric in a conformally flat form on that slice.

Obviously, the conclusion that KS can't be a vacuum solution is wrong, as I gave you a reference to a complete derivation of it directly from the vacuum field equations.

 

[TrickyDicky]

I am not familiar with the formal definition and properties of conformal flatness.
I am not very familiar with the Weyl curvature tensor.

That says a lot.

Obviously, the conclusion that KS can't be a vacuum solution is wrong, as I gave you a reference to a complete derivation of it directly from the vacuum field equations.

Your reference doesn't even derive the conformal nature of the line element, is put by hand in the very first mathematical expression by saying the metric must have a diagonal form.

 

[George Jones]

Let's not jump into conclusions yet (me included), do you agree that the line element in (9), when considering only null geodesics (light), in other words considering the angular components of the metric constant, is that of a conformally flat spacetime?

No, (9) does not give a conformally flat spacetime. Why should the angular components be constant for null geodesics?

 

[PAllen]

That says a lot.

Yes, but so what? None of this is really relevant to questions of whether a coordinate transform can change geometry (in which case almost all books on GR are wrong), or what it means to apply the coordinae definition of AF.

Be that as it may, I looked information up and found, I think, a key point.

Your reference doesn't even derive the conformal nature of the line element, is put by hand in the very first mathematical expression by saying the metric must have a diagonal form.

So what? The point is that it verifies that the KS metric form is vacuum solution of the field equations, and can be directly derived from them. Putting the metric in a desired general form and then seeing if you can find a solution of that form is a standard GR technique. Further, given my finding on conformal flatness of any 2-d manifold, I think it follows that putting the metric in this form is not restrictive of possible geometries of the solutions.

 

[TrickyDicky]

No, (9) does not give a conformally flat spacetime. Why should the angular components be constant for null geodesics?

I'm considering a specific case, might look unnatural but is the case used by GR textbooks to introduce the concept of conformal flatness and it is the line element structure that appears when the EF coordinates and the tortoise r coordinate are introduced in the Schwartzschild metric:
From Hobson GRtext: "The form of the line element (9)has an important consequence for studying the paths of radially moving photons (for which dphi=dtheta=0). Since the conformal factor that multiplies (dv^2−du^2) is just a scaling, it does not change the lightcone structure and so the latter should just look like that in Minkowski space."
Further every diagram to understand the Kruskal space obviates the angular components so that every point in the diagram represents a 2-sphere.
To derive the EF coordinates p and q (which are just the integration constants of the null geodesics in Schwartzschild metric)from which later the tortoise coordinate make use to derive the Kruskal line element, only null geodesics are used, the advanced or ingoing photon and the retarded or outgoing photon.

So in this context, considering only radially moving light, I guess it is fine to make this simplification in which the (9) is conformally flat for radiation.

 

[TrickyDicky]

Not sure if absence of rebuttal to my last post means compliance or lack of interest in answering, which would be kind of odd in a forum that is all about debating and learning.
To be clear, I'm not saying the Kruskal line element is conformally flat per se, only that when the simplification made by textbooks (see also Ryder General relativity chapter7 on Kruskal coordinates) to introduce the change of coordinates of the Scwartzschild line element, namely that for radially ingoing and outgoing null geodesics we can neglect the angular components.
If that simplification is valid to introduce the new coordinates, it must be valid to say that using that same working assumption the Kruskal line element is conformally flat with respect to radial radiation, and since the solution should be spherically symmetric wrt any radiation.
But then how can we deal with light bending effects,etc?

 

[TrickyDicky]

An alternative way to view this conformal flatness is to remember that in the Kruskal spacetime describing the eternal symmeric black hole, hypersurfaces of constant time t are straight lines through the origin.
To better understand this t=constant hypersurfaces we make use of the isotropic transformation of coordinates from the Schwartzschild metric where it is easy to see that the metric in isotropic coordinates is conformally flat for the t= constant hypersurfaces.
And being the Kruskal an static spacetime and therefore invarian for t, wouldn't be valid to consider the manifold conformally flat?
Can someone help with this?

 

[PAllen]

Not sure if absence of rebuttal to my last post means compliance or lack of interest in answering, which would be kind of odd in a forum that is all about debating and learning.
To be clear, I'm not saying the Kruskal line element is conformally flat per se, only that when the simplification made by textbooks (see also Ryder General relativity chapter7 on Kruskal coordinates) to introduce the change of coordinates of the Scwartzschild line element, namely that for radially ingoing and outgoing null geodesics we can neglect the angular components.
If that simplification is valid to introduce the new coordinates, it must be valid to say that using that same working assumption the Kruskal line element is conformally flat with respect to radial radiation, and since the solution should be spherically symmetric wrt any radiation.
But then how can we deal with light bending effects,etc?

No, it means I think I have already answered all your objections and there was nothing more to say since you haven't provided any new substantive responses. The observations about the ability to make any chosen 2-d slice have conformally flat metric form with no loss of generality is fundamental, and all other points have been answered previously.

As to your new point above, your difficulty seems hard to understand. Radial light rays don't bend. Off radial light rays bend. The former can be treated ignoring theta and phi, the latter cannot. What's the issue here?

Maybe you are thinking that conformal flatness of metric form in two coordinates of a spherically symmetric solution means the whole metric is conformally flat? This is simply wrong from what I read about conformal flatness. Even intuitively, all you can conclude is that radial rays from any direction go straight through to the singularity; this says nothing about off radial rays, that start out with a direction involving changing theta/phi.

 

[PAllen]

An alternative way to view this conformal flatness is to remember that in the Kruskal spacetime describing the eternal symmeric black hole, hypersurfaces of constant time t are straight lines through the origin.
To better understand this t=constant hypersurfaces we make use of the isotropic transformation of coordinates from the Schwartzschild metric where it is easy to see that the metric in isotropic coordinates is conformally flat for the t= constant hypersurfaces.
And being the Kruskal an static spacetime and therefore invarian for t, wouldn't be valid to consider the manifold conformally flat?
Can someone help with this?

I'll try one more time. Since any 2-d manifold is conformally flat (see the Wikipedia article on the Weyl tensor), and this can be made manifest by change of coordinates, the it follows that a coordinate change (with no loss of generality) can give a conformally flat metric form to the family of 2-d slices obtained by holding two coordinates of a 4-d metric constant. The ability to do this says nothing about the conformal flatness of the 4-d manifold.

This is similar to the fact that lines have no intrinsic curvature, so all lines embedded in a 2-manifold have no intrinsic curvature. Yet that says nothing about the curvature of the 2-manifold.

Note also that the reference you gave for conformal transforms that changed geometry were talking about something completely different: a mapping *plus* adding points at infinity. It is specifically adding points at inifinity that change geometry. Schwarzschild to Kruskal is just a coordinte change; further, you can arrive at Kruskal directly, as in the reference I gave. As to your objections to that derivation, I already answered them.

 

[TrickyDicky]

I'll try one more time. Since any 2-d manifold is conformally flat

My last example is of 3-d hypersurface.

 

[TrickyDicky]

Radial light rays don't bend. Off radial light rays bend. The former can be treated ignoring theta and phi, the latter cannot.

Correct, all I'm saying is the change of cordinates is done under the assumption that we can ignore theta and phi, so it only refers to radial light rays, it can't be applied to off radial light rays.

 

[TrickyDicky]

further, you can arrive at Kruskal directly, as in the reference I gave.

That derivation is anything but a rigorous mathematical proof, the quantity of unwarranted assumptions thru the long and winding path it takes to the KS line element can be used to demonstrate just about anything.

 

[PAllen]

My last example is of 3-d hypersurface.

Then I am confused. The V=constant hypersurfaces of KS coords are certainly not conformally flat in metric form (and obviously in geometry). If you are claiming that a surface of constant Schwarzschild t is conformally flat in KS cordinates, I don't see this either. Constant t is U/V constant, so expressing one in terms of the other to get a 3-metric form, the result is not at all conformally flat.

Ok, I think you are referring to istotropic Schwarzschild coordinates, where t=constant hypersurface is conformally flat. That is interesting. However, I don't think it is true that the existence of conformally flat embedded 3-surfaces implies anything at all about the geometry of the 4-manifold. I can fill Euclidean 3-space with conscentric embedded 2-spheres. Does that make the 3-space spherical in geometry? Obviously not. I think what is going on here is equivalent. In general, you can't conclude very much about the geometry of a manifold from the geometry of embedded lower dimensional manifolds.

 

[TrickyDicky]

Ok, I think you are referring to istotropic Schwarzschild coordinates, where t=constant hypersurface is conformally flat. That is interesting. However, I don't think it is true that the existence of conformally flat embedded 3-surfaces implies anything at all about the geometry of the 4-manifold. I can fill Euclidean 3-space with conscentric embedded 2-spheres. Does that make the 3-space spherical in geometry? Obviously not. I think what is going on here is equivalent. In general, you can't conclude very much about the geometry of a manifold from the geometry of embedded lower dimensional manifolds.

Only that is not what I'm saying as I specifically stressed in my last posts, I'm not claiming the KS is conformally flat, only that if we maintain the assumptions used to introduce the new coordinates, that is that we can neglect angular components, but then if you consider spherical symmetry, all radial UV planes are included and the assumption extends to all radial light.

In the case of the Schwarzschild metric in isotropic coordinates, the fact that the solution must be static implies that all 3d hypersurfaces are t=constant and thus the whole manifold must be conformally flat.

 

[PAllen]

Only that is not what I'm saying as I specifically stressed in my last posts, I'm not claiming the KS is conformally flat, only that if we maintain the assumptions used to introduce the new coordinates, that is that we can neglect angular components, but then if you consider spherical symmetry, all radial UV planes are included and the assumption extends to all radial light.

I don't understand what you are getting at. What is it, exactly, you are saying about radial light? And what are you claiming that might imply about a claim that KS coordinates are geometrically different fomr Schwarzschild when restricted to the regions covered by the latter (say, regions I and II)?

In the case of the Schwarzschild metric in isotropic coordinates, the fact that the solution must be static implies that all 3d hypersurfaces are t=constant and thus the whole manifold must be conformally flat.

I don't see this at all. I see that all t=constant 3-surfaces in these coordinates are conformally flat. That clearly does not say the 4-manifold is conformally flat, because if you compute the Weyl tensor in these coordinates it does not vanish (obviously, it is actually sufficient to compute the Weyl tensor in any coordinates, because its vanishing / non-vanishing character is invariant).

 

[TrickyDicky]

I don't see this at all. I see that all t=constant 3-surfaces in these coordinates are conformally flat. That clearly does not say the 4-manifold is conformally flat

You seem a bright chap so I'm sure you'll understand this last reflection:
In isotropic coordinates, the Schwarzschild metric time 3d-slices (t constant) are conformal to 3d-Euclidean space, right so far?
The Schwarzschild 4d-manifold in isotropic coordinates must also be a static spacetime to be a vacuum solution so it must be time-invariant, right so far?
If a 4d manifold that is time invariant has time slices that are conformally flat, does that tell us anything about the conformally flat nature of the 4d-manifold?
That is my question.

 

[PAllen]

You seem a bright chap so I'm sure you'll understand this last reflection:
In isotropic coordinates, the Schwarzschild metric time 3d-slices (t constant) are conformal to 3d-Euclidean space, right so far?
The Schwarzschild 4d-manifold in isotropic coordinates must also be a static spacetime to be a vacuum solution so it must be time-invariant, right so far?
If a 4d manifold that is time invariant has time slices that are conformally flat, does that tell us anything about the conformally flat nature of the 4d-manifold?
That is my question.

I I've answered it several times, to my satisfaction but apperently not yours.

1) compute Weyl tensor in these coordinates; it does not vanish; therefore, by established theorems, the conclusion does not follow. That is really all that needs to be said. You have an intuition about what 'ought to be true'; it is wrong. I've been there countless times in my life. I respond by educating my intuition when it conflicts with proofs, or measurements. (In fact, as I freely admitted, I was not familiar with conformal flatness theory and the properties of the Weyl tensor before you raised them; so I educated myself and read about them).

2) Intuitive explanation: filling a manifold with embedded lower dimensional manifolds does not allow you to claim the geometry of lower dimensional manifolds carries over the the higher dimensional one. My example with concentric 2-spheres in flat euclidean 3-space makes the conclusion in (1) seem *not* counter-intuitive.

 

[TrickyDicky]

I I've answered it several times, to my satisfaction but apperently not yours.

1) compute Weyl tensor in these coordinates; it does not vanish; therefore, by established theorems, the conclusion does not follow. That is really all that needs to be said. You have an intuition about what 'ought to be true'; it is wrong. I've been there countless times in my life. I respond by educating my intuition when it conflicts with proofs, or measurements. (In fact, as I freely admitted, I was not familiar with conformal flatness theory and the properties of the Weyl tensor before you raised them; so I educated myself and read about them).

Fine, would you be so kind to show me the non-zero components of the weyl tensor in terms of the isotropic coordinates? I assume you've done or have access to that computation. It'd help me a lot.

2) Intuitive explanation: filling a manifold with embedded lower dimensional manifolds does not allow you to claim the geometry of lower dimensional manifolds carries over the the higher dimensional one. My example with concentric 2-spheres in flat euclidean 3-space makes the conclusion in (1) seem *not* counter-intuitive.

Your example of the concentric spheres in a 3-space is silly in this context, and has nothing to do with what I'm saying about the Schwarzschild manifold in isotropic coordinates.
First of all Euclidean 3-space is not invariant for any of their 3 components. While Schwarzschild spacetime is static and therefore invarian with respect to one of its 4 components.
The fact that a flat 3-space allows curved surfaces to exist,and is not itself curved is trivial but in fact you cannot have an x constant slice of 3-d euclidean that defines a curved surface because an x constant slice is a 2d-flat plane. But in my example you can have a time constant 3d-slice of the 4d-manifold that is conformally flat.

 

[PAllen]

Fine, would you be so kind to show me the non-zero components of the weyl tensor in terms of the isotropic coordinates? I assume you've done or have access to that computation. It'd help me a lot.

Ok, I will provide.

Your example of the concentric spheres in a 3-space is silly in this context, and has nothing to do with what I'm saying about the Schwarzschild manifold in isotropic coordinates.
First of all Euclidean 3-space is not invariant for any of their 3 components. While Schwarzschild spacetime is static and therefore invarian with respect to one of its 4 components.
The fact that a flat 3-space allows curved surfaces to exist,and is not itself curved is trivial but in fact you cannot have an x constant slice of 3-d euclidean that defines a curved surface because an x constant slice is a 2d-flat plane. But in my example you can have a time constant 3d-slice of the 4d-manifold that is conformally flat.

It's an analogy, not an exact correspondence. However also note the following:

1) You can change coordinates in Euclidean flat space such that r=constant slices are 2-spheres. This is in fact similar to the fact that non-isotropic coordinates don't have conformally flat t=constant slices, while the isotropic coordinates do.

2) The Euclidean metric is completely static in its ordinary form: it is the identity matrix. No component depends on any coordinate values.

 

[TrickyDicky]

However also note the following:

1) You can change coordinates in Euclidean flat space such that r=constant slices are 2-spheres. This is in fact similar to the fact that non-isotropic coordinates don't have conformally flat t=constant slices, while the isotropic coordinates do.

Nope, r= constant slices are not 2-spheres,think of a cylinder, it has constant radius but no intrinsic curavature, whereas a 2-sphere has intrinsic gaussian curvature.

2) The Euclidean metric is completely static in its ordinary form: it is the identity matrix. No component depends on any coordinate values.

I can't make sense out of this in the discussed context. The 3d-Euclidean geometry is not a spacetime, it can't be defined in terms of staticity that refer to pseudo-riemannian manifolds.

 

[PAllen]

Nope, r= constant slices are not 2-spheres,think of a cylinder, it has constant radius but no intrinsic curavature, whereas a 2-sphere has intrinsic gaussian curvature.

I thought my suggestion would be obvious. I think assumptions like that have interfered with our communication. Anyway, I simply define a new coordinate r=x^2+y^2+z^2, and theta and phi as traditionally defined for spherical coordinates. This coordinate patch has missing points, but that is fine. Anyway, r=constant slices of this coordinate system define 2-shperes. As long as transform the metric properly, the Euclidean geometry of the 3-space has not been changed.

I can't make sense out of this in the discussed context. The 3d-Euclidean geometry is not a spacetime, it can't be defined in terms of staticity that refer to pseudo-riemannian manifolds.

We really have trouble communicating. Things I think should be obvious, are not at all to you, and seemingly vice versa. In spacetime, time is just a coordinate. A static metric form has the feature that metric does not depend on the t coordinate. The Euclidean metric is constant for all of its coordinates (in standard coordinates). This is a perfect analogy in my mind.

 

[PAllen]

Here are components of the Weyl tensor in standard Schwarzschild coordinates (with two indexes raised). If you think a transform to isotropic coordinates can make a tensor vanish, you are questioning the whole theory of differential geometry, and there is little to discuss. If you have maple, you can add the grtensor package from here http://grtensor.phy.queensu.ca/ and readily compute in isometric coordinates. I thought I would be able to find something standalone, but this requires maple, which I don't have. I, however, have 100% confidence that if the Weyl tensor doesn't vanish in one coordinate system, it doesn't vanish in any other.


C^`(1) (2)`*``[`(1) (2)`] = m/(r^3)

C^`(1) (3)`*``[`(1) (3)`] = m/(r^3)

C^`(1) (4)`*``[`(1) (4)`] = -2*m/(r^3)

C^`(2) (3)`*``[`(2) (3)`] = -2*m/(r^3)

C^`(2) (4)`*``[`(2) (4)`] = m/(r^3)

C^`(3) (4)`*``[`(3) (4)`] = m/(r^3)

 

[TrickyDicky]

I thought my suggestion would be obvious. I think assumptions like that have interfered with our communication. Anyway, I simply define a new coordinate r=x^2+y^2+z^2, and theta and phi as traditionally defined for spherical coordinates. This coordinate patch has missing points, but that is fine. Anyway, r=constant slices of this coordinate system define 2-shperes. As long as transform the metric properly, the Euclidean geometry of the 3-space has not been changed.

Once again you can't define 2-spheres which are the tridimensional balls we all are familiar with in a 2 dimensional slice plane, they need 3-domensional space. This is a simple fact so your not getting it has little to do with trouble communicating and a lot to do with obfuscation.

We really have trouble communicating. Things I think should be obvious, are not at all to you, and seemingly vice versa. In spacetime, time is just a coordinate. A static metric form has the feature that metric does not depend on the t coordinate. The Euclidean metric is constant for all of its coordinates (in standard coordinates). This is a perfect analogy in my mind.

But, it's not. Euclidean metric has definite positive signature, the static definiton is referred to spacetimes, therefore to pseudo-riemannian signature (1,3 or 3,1). Check it before answering.

 

[TrickyDicky]

Here are components of the Weyl tensor in standard Schwarzschild coordinates (with two indexes raised). If you think a transform to isotropic coordinates can make a tensor vanish, you are questioning the whole theory of differential geometry, and there is little to discuss. I, however, have 100% confidence that if the Weyl tensor doesn't vanish in one coordinate system, it doesn't vanish in any other.

I already had those, I was hoping you could give me the components in isotropic coordinates, perhaps someone else can, maybe no one has computed them yet.

Differential geometry is alright, tensors are invariant for coordinate transforms.
But in the case we were dealing with different line elements and thus different geometries, we wouldn't be talking about the same tensor.

 

[PAllen]

Once again you can't define 2-spheres which are the tridimensional balls we all are familiar with in a 2 dimensional slice plane, they need 3-domensional space. This is a simple fact so your not getting it has little to do with trouble communicating and a lot to do with obfuscation.

What do you mean? I defined a coordinate transform from (x,y,z)->(r,theta,phi). In these coordinates, r=constant surfaces are 2-spheres. This is a perfectly good coordinate system for flat Euclidean 3-space If the metric is properly transformed. If you don't accept this, then I suspect there is very little we can effectively discuss.

But, it's not. Euclidean metric has definite positive signature, the static definiton is referred to spacetimes, therefore to pseudo-riemannian signature (1,3 or 3,1). Check it before answering.

Analogy, analogy. I know the signature is different. Analogies aren't exact. This is all part of an intuitive justification. As such, it may work for me and not for you. For me it motivates that I don't expect the existence of coordinates whose constant surfaces have some geometry to imply very much about the geometry of the overall space or spacetime.

I think I understand the issues perfectly well (going back to what the coordinate based AF criteria really means); and now I think I understand a reasonable amount about conformal flatness and the properties of the Weyl tensor. It does not seem I can effectively help you arrive at similar understandings. Maybe someone else can. At this point, if you get no responses, it may be that others have also concluded that everything has already been said.

 

[PAllen]

I already had those, I was hoping you could give me the components in isotropic coordinates, perhaps someone else can, maybe no one has computed them yet.

Differential geometry is alright, tensors are invariant for coordinate transforms.
But in the case we were dealing with different line elements and thus different geometries, we wouldn't be talking about the same tensor.

NO, if you change coordinates and the line element together, the geometry is the same. Isotropic Schwarzschild coordinates, with their associated line element, represent exactly the same geometry as the standard ones, with their line element. If you are disagreeing on this, then there is really nothing to discuss.

 

[TrickyDicky]

What do you mean? I defined a coordinate transform from (x,y,z)->(r,theta,phi). In these coordinates, r=constant surfaces are 2-spheres. This is a perfectly good coordinate system for flat Euclidean 3-space If the metric is properly transformed. If you don't accept this, then I suspect there is very little we can effectively discuss.

PAllen , I can see now that our disagreement about this specific point comes from the distinction I make between topological and geometrical dimensions, that confused me about what you meant, and in fact I agree with your statement about concentric 2-spheres being formed from constant radius slices. Sorry about the misunderstanding.

And I must admit that now I'm not sure at all about what I asked about the Schwarzschild metric in isotropic coordinates, but I still would like to clarify what I was trying to get across up to post #108.

 

[PAllen]

For the next two weeks I will have very little time here (prep for and follow up to business conference). However, I still have some interest in discussing this . If I weren't pressed for time, I would write a longer version of the following, with more detailed justification, but this will have to do for a while. This is based both on further thought and some research in my oldest relativity texts (1967, 1960, 1942, 1921).

Very briefly, what I would argue is that the question of whether singularities are allowed in asymptotically flat solutions is completely orthogonal to whether your definitions are coordinate based or modern conformal definitions. Instead, the admissibility of such solutions depends completely on topological restrictions you couple to the definition, whichever main flavor of definition you use. This is made very explicit in the conformal definitions. However, the restrictive analog of a coordinate definition is to require that you cover spacetime in one coordinate patch, with no holes or excluded regions, that is everywhere 'minkowski like'. Such a definition really just hides an a-priori restriction that the topology (but not geometry) must be exactly that of flat Minkowski space. Such a definition necessarily excludes even the complete exterior Schwarzschild solution (but would allow the a solution for a non-critical perfect fluid ball coupled to a part of the exterior Schwarzschild solution). Any reasonable generalization of the most restrictive coordinate definition to allow the complete exterior Schwarzschild, will also allow completions of it through the event horizon up to the true singularity. Such extensions simply take the form of allowing coordinate patches with proper gluing rules, and asserting the boundary condition on a particular coordinate patch which covers at least all of space time except for a 2 sphere extended in time, such that the area of the two sphere computed from outside is finite at all times.

A critical thing missing from the Wiki definition of coordinate based AF is the following from 1960s GR book that has no hint of the conformal defintions: "A manifold is AF if there exists any mapping such that ..." making very clear that AF is a property of the manifold if the condition holds for any single coordinate system on it.

Another interesting historic tidbit I found is that even in Bergmann's 1942 book meant as a first introduction to relativity for university physics students, the removable nature of the event horizon 'singularity' was already considered established enough to include (and this was 20 years before Kruskal coordinates were invented). This was done based on Robertson's ( of Robertson- Walker fame) demonstration that local coordinates of a free falling observer show no anomaly crossing the horizon.

That is really all I can contribute for a couple of weeks.

 

[TrickyDicky]

Thanks, I'll think about this a bit before replying.