Schwartzschild exterior and interior solutions

[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 definition 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 definition 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.