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Does geodesic incompleteness in Penrose-Hawking theorems imply curvature blow up?

+ 6 like - 0 dislike
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The singularity theorems in General Realtivity roughly stated say that given:

  • A global causal condition
  • An energy condition
  • The existence of a closed trapped surface

then spacetime must be geodesically incomplete.

However, in all the physical relevant scenarios like black holes metrics and big bang models the relevant physical characteristic is the blow-up of some scalar curvature quantity.

Is there an example where the theorem is satisfied and yet there is no curvature blow-up? Maybe the formation of some mild regularity, like a quasi-regular singularity?

Is there any extension of the theorems where the conclusion of the theorem is curvature blow-up rather than geodesic incompleteness?

This post imported from StackExchange Physics at 2015-07-26 09:36 (UTC), posted by SE-user yess
asked Jul 6, 2015 in Theoretical Physics by yess (90 points) [ no revision ]
From Wald's General Relativity (§9.1): "Given the existence of an incomplete timelike or null geodesic, one would like to know more about the character of the singularity, e.g., whether it is a curvature singularity or a non-curvature singularity. Unfortunately, the singularity theorems give virtually no information about the nature of the singularities of which they prove existence." (That said, there may have been some progress in the subject since 1984.)

This post imported from StackExchange Physics at 2015-07-26 09:36 (UTC), posted by SE-user Michael Seifert

2 Answers

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The most recent textbook answer I could find would be from Choquet-Bruhat's 2009 book "General Relativity and the Einstein Equations". She writes on page 403:

Remark A curvature singularity does not imply geodesic incompleteness. The geodesic flow depends only on the $C^{1,1}$ structure of the metric. Conversely, does geodesic incompleteness imply a curvature singularity? This question is linked with the strong cosmic censorship conjecture defined in the previous chapter. ...

As the strong cosmic censorship conjecture is still - as far as I know - a conjecture, I'd say there is no such extension of these theorems, so far. However, I remember vaguely that there are a lot of different formulations for the cosmic censorship conjectures. Some of the stronger ones are, iirc, disproven. The weaker you get, the more open the question is.

Along similar lines was what I found in the slightly older book by Kriele, "Spacetime", roughly its chapter 9.

Regarding yess' comment below: I can only give more ressources, when it comes to pathological behaviour of the curvature at accessible events. In Hawking, Ellis on page 290ff there is an example involving Taub-NUT which I don't follow completely. Then, Curiel wrote both an article on plato and a paper along the same lines, which can be accessed on his website. The bibliography seems to be a good starting point for further literature search. For example, the "marketing excerpt" of Clarke's "The Analysis of Space-Time Singularities" sound already very promising - sadly I cannot get my hands on it. This paper by Ellis and Schmidt has some interesting examples in section 4 on non-scalar singularities. If I understand the Clarke excerpt, page 7, correctly, it can be traced back to matters of regularity - like the above remark of Choquet-Bruhat also alludes to.

This post imported from StackExchange Physics at 2015-07-26 09:36 (UTC), posted by SE-user Wraith of Seth
answered Jul 7, 2015 by Wraith of Seth (0 points) [ no revision ]
And, of course, geodesic incompleteness does not imply a curvature singularity; for example Minkowski spacetime with a "wedge" ($0 < \phi < \phi_0$) removed and the sides of wedge identified is geodesically incomplete at the resulting cone point, and is flat everywhere else.

This post imported from StackExchange Physics at 2015-07-26 09:36 (UTC), posted by SE-user Michael Seifert
Hi, would you mind expanding the comment about that curvature singularity doesnt imply geodesic incompleteness?

This post imported from StackExchange Physics at 2015-07-26 09:36 (UTC), posted by SE-user yess
MichaelSeifert: Thank you for the addition! @yess : I tried to answer your question with some additional links in the post. I hope that helps.

This post imported from StackExchange Physics at 2015-07-26 09:36 (UTC), posted by SE-user Wraith of Seth
+ 0 like - 3 dislike

Is there any extension of the theorems where the conclusion of the theorem is curvature blow-up rather than geodesic incompleteness?

No. Remember that spacetime is an abstract 3+1 dimensional mathematical space where we plot motion through space against time, which is in essence motion through space inside a clock. Also remember that a geodesic is in essence a light path, which is curved because the speed of light is spatially variable. The understand this: a curvature blow-up doesn't make sense because a gravitational field is a place where space is inhomogeneous, not curved. You can see Einstein talking about that here.

But since you can relate the geodesic to a light-path, you can backtrack to the Schwarzschild metric, where the event horizon is said to be "a mere coordinate artefact". Then take a look and Kevin Brown's Formation and Growth of Black Holes, and pay special attention to the frozen-star interpretation. At the event horizon the coordinate speed of light is zero, so you can justifiably say that from where we're standing, light stops and the geodesic grinds to a halt at the event horizon. That qualifies as geodesic incompleteness.

Yes, Kruskal-Szekeres coordinates attempt to get past this, but they result in an elephant that goes to the end of time and back, and is in two places at once. It's tosh I'm afraid, resulting from the schoolboy error wherein you sit a stopped observer in front of a stopped clock and claim that he sees it ticking normally "in his own frame". When you reject this and fall back to the Schawarzchild singularity at the event horizon, you end up saying there is geodesic incompleteness rather than curvature blow-up, but not because of anything Hawking or Penrose said.

This post imported from StackExchange Physics at 2015-07-26 09:36 (UTC), posted by SE-user John Duffield
answered Jul 6, 2015 by John Duffield (-55 points) [ no revision ]

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