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I'm just following my curiosity here. Feel free to correct me and point me in a better direction. (Good references and explaining what is and what is not known greatly increases the odds that an answer is accepted.) Below, when I say "singularity" I mean a gravitational singularity, not an event horizon, which is roughly characterized by:

The Penrose–Hawking singularity theorems [which] define a singularity to have geodesics that cannot be extended in a smooth manner. The termination of such a geodesic is considered to be the singularity.

Furthermore, regarding curvature singularities:

More generally, a spacetime is considered singular if it is geodesically incomplete, meaning that there are freely-falling particles whose motion cannot be determined beyond a finite time, being after the point of reaching the singularity.

Feel free to specify answers to curvature singularities, rather than "gravitational singularities" as a whole.

My questions:

  1. Is there a formal distinction between intrinsic and extrinsic geometry for a gravitational singularity? Does this require topology, rather than differential geometry? Is it by definition that it is impossible to obtain the intrinsic geometry of a singularity?

  2. Why can the metric of a spacetime inform us about the extrinsic geometry of a gravitational singularity, but not the intrinsic geometry? Is this related to the difference between properties that depend only on the "abstract manifold" vs. properties that depend on an imbedding of the manifold (see last para. of Michael Weiss' answer)?

The example that got me asking is the Kerr metric, the solution to the Einstein equations with axisymmetry, which possesses mass and spin angular momentum. It is a common result in GR textbooks that the gravitational singularity has a ring structure.

At the bottom of page 8 of this paper, they state:

Of course in terms of the “full” geometry, specified by the physical metric $g_{ab}$, the intrinsic geometry of the curvature singularity is, unavoidably and by definition, singular.

and on page 28:

But like the statement that the singularity is a “ring”, this is not a statement about the intrinsic geometry — it is instead a statement about the mathematically convenient but fictitious flat Minkowski space that is so useful in analyzing the Kerr–Schild form of the Kerr spacetime.

So, is the reason why the authors, quoted above, state that the ring shows the extrinsic geometry of the singularity because the ring is a property of the manifold that depends on the embedding utilized? If yes, then does any embedding show the ring, or do only some? What is the embedding in this case that shows the ring?

  1. Lastly, what is the role of a maximal extension of an analytic continuation of a spacetime here?

To take Kerr as an example again, the singular disk region of the ring can be made regular, reducing the gravitational singularity to a smaller region (not necessarily a point, it's complicated). Is this analytic continuation an example of choosing a different embedding where the disk region is no longer regular? OR am I totally butchering this? I apologize if this question is crude, but I'm reaching the limits of my mathematical knowledge.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro Tamaroff
    Jul 20 at 15:35
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I am not sure if there is a universally accepted definition of singularity and when it matters it is probably being specified carefully.

However generally a gravitational singularity is a point of the spacetime where some scalar quantity constructed out of the curvature tensor diverges (you want a scalar quantity so that you don't get singularities due to a bad choice of coordinates).

From a mathematical point of view instead of working with diverging quantities it is often preferred to remove from the spacetime $M$ the point at which the divergence occurs. This generally leads to a (geodesically) incomplete manifold: there are geodesics which terminate in finite proper time.

Incompleteness seems a good way to encode the notion of singularity, however you don't want to include those cases in which the manifold is incomplete just because you're omitting a part of it. Inextendible but incomplete may capture at least a good part of those spacetimes you want to consider singular. I am not sure to what extent it succeeds but it seems a good starting point.

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  • $\begingroup$ cases in which the manifold is incomplete just because you're omitting a part of it” - When spacetime can be mathematically extended through a coordinate singularity, it is not given that the extension exists in reality. There is little doubt that the maximal Kruskal extension of the Schwarzschild spacetime or alternate universes inside a Kerr black hole are not real. $\endgroup$
    – safesphere
    Jul 17 at 8:28
  • $\begingroup$ @safesphere Is your "There is little doubt..." sentence perhaps rather strong? I might have expected "There is no current physical interpretation of..." or something like that... $\endgroup$
    – Lee Mosher
    Jul 18 at 14:06
  • $\begingroup$ I think the maximal extension of Schwarzschild and Kerr are considered unphysical since black holes are formed in star collapse, while these solutions would correspond to black holes which have "always been there". Also they are a bit weird (e.g. for Schwarzschild case you get a corresponding white hole) so it "feels good" to exclude the weird parts. But I agree that "there is little doubt" feels strong, I know of no result excluding them. $\endgroup$
    – GFR
    Jul 18 at 15:30
  • $\begingroup$ (continued) also I imagine that any physically relevant solutions of Einstein equations would be expected to be inextendible as abruptly terminating something which can be extended feels unnatural both on mathematical and physical grounds - a principle often invoked in physics is that everything which is not explicitly forbidden by some symmetry or conservation law will occur". Anyway, I am not an expert on this topic. $\endgroup$
    – GFR
    Jul 18 at 15:34
  • $\begingroup$ I don't see how this answers any of my questions, with all due respect. This really seems like a comment rather than an answer $\endgroup$ Jul 19 at 13:11

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