Is GR Considered a Gauge Theory?

In summary: If you don't specify it, there are an infinite number of solutions. The same is true for general relativity.In summary, the gauge invariance in GR is a form of invariance, but it is accompanied by background independence, which makes GR's manifold abstractly as a simple differentiable manifold. However, this gauge invariance is not physical. There is no unique global geometry/topology that can be inferred from the solutions to the Einstein-Hilbert equations.
  • #36
Normally, nonanalytic and even discontinuous extensions would be possible, but with the assumptions of vacuum and spherical symmetry I guess Birkhoff's Theorem is what prevents them here.
 
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  • #37
Bill_K said:
I'm just saying that these assumptions are fine mathematically, but there is no physical justification for making them. Regions III and IV are terra incognita, and could plausibly be the source of matter and/or gravitational waves. Especially IV, where we're assuming that an exploding singularity produces nothing but vacuum.

Agreed. I realize the physical argument I made only motivates that some extension should be made if you have geodesics beginning or ending arbitrarily. It does not help decide what constraints should be applied.

My point is purely mathemtical: in the same sense that one says a 2-sphere is 'the' surface constant curvature (meaning that it contains any other such surface as a subset), KS geometry is 'the' manifold that is spherically symmetric and vaccuum.
 
  • #38
Bill_K said:
Normally, nonanalytic and even discontinuous extensions would be possible, but with the assumptions of vacuum and spherical symmetry I guess Birkhoff's Theorem is what prevents them here.

Right.
 
  • #39
PAllen said:
No additional assumption is needed, for uniqueness in the sense I gave to follow.

Bill_K said:
Normally, nonanalytic and even discontinuous extensions would be possible, but with the assumptions of vacuum and spherical symmetry I guess Birkhoff's Theorem is what prevents them here.

PAllen said:
Right.

There are many ways to state Birkhoff's theorem, this is the one used in WP:"Birkhoff's theorem states that any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat. This means that the exterior solution must be given by the Schwarzschild metric." Is this one ok to you? if not please choose another one you like better.
I would say that this one doesn't imply no additional assumption is needed for uniqueness. But since I' not sure you accept this version of the theorem I await yor answer for further discussion.
 
  • #40
TrickyDicky said:
There are many ways to state Birkhoff's theorem, this is the one used in WP:"Birkhoff's theorem states that any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat. This means that the exterior solution must be given by the Schwarzschild metric." Is this one ok to you? if not please choose another one you like better.
I would say that this one doesn't imply no additional assumption is needed for uniqueness. But since I' not sure you accept this version of the theorem I await yor answer for further discussion.

MTW demonstrates a more general version, that any manifold or section thereof that satisfies Einstein tensor = 0 and spherical symmetry, must be part of the KS geometry. This establishes there is no way to extend other than to add more of the KS geometry; and the KS geometry is the unique maximal extension.
 
  • #41
PAllen said:
MTW demonstrates a more general version, that any manifold or section thereof that satisfies Einstein tensor = 0 and spherical symmetry, must be part of the KS geometry. This establishes there is no way to extend other than to add more of the KS geometry; and the KS geometry is the unique maximal extension.

I suspected you were using a variant that named the KS geometry. The original statement said Schwarzschild metric, long before the KS geometry was found. Only if one previously assumes that the KS extension is unique, ignoring the mathematical requirement that it be analytic, can state the theorem in that way. But this thread is precisely about not taking for granted that assumption unless there is a clear physical justification for analyticity tht no one has so far provided.
 
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  • #42
TrickyDicky said:
I suspected you were using a variant that named the KS geometry. The original statement said Schwarzschild metric, long before the KS geometry was found. Only if one previously assumes that the KS extension is unique, ignoring the mathematical requirement that it be analytic, can state the theorem in that way. But this thread is precisely about not taking for granted that assumption unless there is a clear physical justification for analyticity tht no one has so far provided.

There is no requirement of analyticity. KS geometry is not assumed, it is discovered. You need only the following to arrive at full KS geometry:

(1) Einstein tensor is defined and zero everywhere on the manifold.
(2) spherical symmetry
(3) there is no other manifold meeting (1) and (2) of which this is a subset (i.e. this can't be extended further in any way meeting (1) and (2).

There is only one manifold meeting these conditions, and it is KS. Further, every manifold meeting these conditions with (3) left off, is an open subset of KS.

I am not going to dispute facts any more on this. If you have something different to say, then I might have further comment. If you re-dispute facts, this has become pointless.
 
  • #43
PAllen said:
There is no requirement of analyticity. KS geometry is not assumed, it is discovered. You need only the following to arrive at full KS geometry:

(1) Einstein tensor is defined and zero everywhere on the manifold.
(2) spherical symmetry
(3) there is no other manifold meeting (1) and (2) of which this is a subset (i.e. this can't be extended further in any way meeting (1) and (2).

There is only one manifold meeting these conditions, and it is KS. Further, every manifold meeting these conditions with (3) left off, is an open subset of KS.

I am not going to dispute facts any more on this. If you have something different to say, then I might have further comment. If you re-dispute facts, this has become pointless.
No need to dispute anything. If you want to claim that what is known as the "maximal analytic extensión" of the schwarzschild solution is not analytic...
 
  • #44
Ok, so for the rest of people that can read in any peer-reviewed paper on black holes or in any GR textbook that the KS metric is an analytic continuation of the Schwarzschild metric, we can go on with the argument in the OP concerning GR as a gauge theory: on the one hand diffeomorphism invariance with background independence of GR only assumes a differentiable manifold M that can be given a pseudoRiemannian metric g, mathematically it is true that all analytic real manifolds are infinitely differentiable, but the converse is not true, so any analytic extension needs the analyticity assumption.

Usually that assumption is made without further comment, but it would probably need some physical justification, in fact most relativists admit that the KS geometry with the eternal features and the white hole is probably not physical even if mathematically is clearly a solution, we know there are tons of unphysical solutions of the EFE anyway. If for instance GR's manifold was complex the justification would not be needed at all since holomorphic functions are analytic.But if for the sake of the argument one was to also consider for a moment that analyticity is not physically justified, one could not make the analytic continuation at the Schwarzschild radius because the Taylor expansion wouldn't necessarily converge at those points of the hypersurface making impossible gluing a different chart, then the Schwarzschild and the KS solutions would be two different solution of the EFE instead of one, the latter with static and non static regions, and the former only with a static region.

This is why I'm looking for compelling physical arguments to justify analitycity. It seems it would be important to find some given the above. The alternatives are either denying that mathematical requirement like a previous poster, or ignore it and consider it doesn't need any physical justification and just assume it. Neither seems very scientific.

See also the Cauchy problem and the Cauchy–Kowalevski theorem http://en.wikipedia.org/wiki/Cauchy_problem
 
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  • #45
TrickyDicky said:
No need to dispute anything. If you want to claim that what is known as the "maximal analytic extensión" of the schwarzschild solution is not analytic...

Oh come on! The issue isn't whether KS is analytic, the issue is whether you have to assume this a-priori. You do not, in this special case.
 
  • #46
TrickyDicky said:
Ok, so for the rest of people that can read in any peer-reviewed paper on black holes or in any GR textbook that the KS metric is an analytic continuation of the Schwarzschild metric, we can go on with the argument in the OP concerning GR as a gauge theory: on the one hand diffeomorphism invariance with background independence of GR only assumes a differentiable manifold M that can be given a pseudoRiemannian metric g, mathematically it is true that all analytic real manifolds are infinitely differentiable, but the converse is not true, so any analytic extension needs the analyticity assumption.

Ok, I see your confusion. The key is that the statent: "The Einstein tensor is defined and zero everywhere" is sufficient to enforce the required degree of differentiability to force uniqueness with the further assumption of spherical symmetry.

Of course it is true that quite generally physicists tend to assume, for all theories, that nature is sufficiently smooth that any degree of differentiability is ok to require for a given problem.
 
  • #47
PAllen said:
Oh come on! The issue isn't whether KS is analytic, the issue is whether you have to assume this a-priori. You do not, in this special case.

I don't mind if you assume it a-priori or a-posteriori, the issue I present is that you have to assume it.
 
  • #49
TrickyDicky said:
I don't mind if you assume it a-priori or a-posteriori, the issue I present is that you have to assume it.

There is no such thing as an a-posterior assumption. If something follows from you assumptions it is a consequence not an assumption.
 
  • #50
PAllen said:
Ok, I see your confusion. The key is that the statent: "The Einstein tensor is defined and zero everywhere" is sufficient to enforce the required degree of differentiability to force uniqueness with the further assumption of spherical symmetry.

Of course it is true that quite generally physicists tend to assume, for all theories, that nature is sufficiently smooth that any degree of differentiability is ok to require for a given problem.

Yo are confusing differentiability with analyticity. You should know that even smooth (infinitely differentiable) manifolds can fail to be analytic. Uniqueness demands analyticity, have you heard of the Cauchy problem in GR?
 
  • #51
PAllen said:
There is no such thing as an a-posterior assumption. If something follows from you assumptions it is a consequence not an assumption.

lol, can't you recognize a joke?
 
  • #52
TrickyDicky said:
Here is an interesting discussion about analyticity in physics, one of the answers, by unknown even refers to GR (in this case it makes reference to the no-hair theorem that is also valid only for real analytic manifolds).

http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions

General theorems of this type typically do required a number of technical assumptions to prove anything. Such theorems don't assume anything about the metric, thus they typically need smoothness assumptions to constrain the problem enough to accomplish the proof.

Again, in the case of uniquness of KS geometry, such additional assumption is not needed because the existence and vanishing of the Einstein tensor everywhere is already requiring a sufficient degree of smoothness.
 
  • #53
PAllen said:
General theorems of this type typically do required a number of technical assumptions to prove anything. Such theorems don't assume anything about the metric, thus they typically need smoothness assumptions to constrain the problem enough to accomplish the proof.

Again, in the case of uniquness of KS geometry, such additional assumption is not needed because the existence and vanishing of the Einstein tensor everywhere is already requiring a sufficient degree of smoothness.

You are again conflating smoothness and analyticity.
 
  • #54
I must issue a correction here. I've read too many mathematically sloppy treatments of Birkhoff. It turns out, that as strong as it is, you really do need additional assumptions to arrive uniquely at the KS geometry. Here is a reference discussing these issues:

http://arxiv.org/abs/0910.5194
 
  • #55
PAllen said:
I must issue a correction here. I've read too many mathematically sloppy treatments of Birkhoff. It turns out, that as strong as it is, you really do need additional assumptions to arrive uniquely at the KS geometry.

Finally, I was starting to suspect your account had been stolen by someone not very reasonable. ;-)
 

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