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Given a metric $g_{\mu \nu}(x)$, can we identify whether it corresponds to a black hole? To be more precise, can we perform some calculations or define certain parameters of the metric which can help us identify it as a black hole solution?

I understand that there are two essential ingredients to identify if a solution to the Einstein field equations is a black hole solution -- (1) presence of a singularity, and (2) presence of a global event horizon. Please correct me if I am wrong -- Penrose-Hawking singularity theorem does not help us differentiate one type of singularity from another, what I mean by this is that from geodesic incompleteness argument I cannot differentiate a naked singularity from a black hole singularity. So to correctly identify whether a given solution corresponds to a black hole I need to be able to describe the presence of a global event horizon.

There might be some argument via gravitational path integrals made by Hawking and Gibbons in their 1977 PRD paper titled "Action integrals and partition functions in quantum gravity" (https://journals.aps.org/prd/abstract/10.1103/PhysRevD.15.2752), but I am not yet very comfortable with gravitational path integrals. I would like to know if there is any result which says that if the metric has "certain properties" we can identify it as a black hole solution?

The discussion in Identifying Black Hole Horizon from metric tensor makes some remarks relevant to my question, but I do not restrict myself to a spherically symmetric geometry.

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A singularity is not required for something to be a black hole (in the literature, people sometimes talk about regular black holes). As you can check on more mathematical books on GR (such as Hawking & Ellis or Wald), the definition of a black hole is

Let $M$ be a strongly asymptotically predictable spacetime. If the set $B = M \setminus J^-(\mathscr{I}^+)$ is non-empty, it is said to be a black hole.

"Strongly asymptotically predictable" means the spacetime is sufficiently well-behaved at infinity so that $\mathscr{I}^+$ (the future null infinity, which you can roughly think of as all of the observers that go infinitely far away in infinite time, or where the light rays go to in infinite time) has some nice properties. $J^-(S)$ is the causal past of the set $S$, i.e., it is a generalization of the past light cone (including the cone's surface) for a set in a general curved spacetime.

A less mathematical way of stating the previous definition is

Let $M$ be a "well-behaved" spacetime. If there is a region $B$ in the spacetime which does not lie in the causal past of any observers that go infinitely far in infinite time, then $B$ is said to be a black hole.

The condition of the observers going to infinity in infinite time is to ensure they don't end up trapped in a black hole, for example. Notice that you have to ask all observers if each event in the spacetime is in their causal past. If some point is in the causal past of a single observer, then it is not in the black hole.

Notice then that this means that the definition of a black hole is global. You can't use only local properties to characterize whether black hole. In particular, you can't pinpoint locally where is the event horizon of a black hole.

With all this in mind, notice that the answer to your question is no. The metric alone is not enough to characterize the presence of a black hole, since it is a local construction. You need at least some other source of information. For example, you might need to clarify what is the topology of the spacetime you are considering. Notice that the Schwarzschild metric, for example, might or not lead to a black hole: depending on how you set up the spacetime, the metric might correspond to a black hole or to the exterior region of a star. It ends up depending on the ranges of the parameters that go in the metric (i.e., in the topology, a global property).

However, there is an interesting addition. If you happen to know you are dealing with a solution to the Einstein equations that has some nice initial properties (for example, it starts as a star and collapses to a singularity), then it is conjectured that no singularity can form without the formation of an event horizon. I.e., there is some belief that many problems of physical interest actually can't present a naked singularity. These are known as the Cosmic Censorship Conjectures. If you assume a conjecture of this form (and its hypothesis), then Penrose's theorem implies the presence of a black hole. Notice the Nobel committee did assume this when saying that "black hole formation is a robust prediction of the general theory of relativity". (Notice global information is already being assumed in the hypothesis to the conjectures and in whatever singularity theorem you use, and therefore this comment doesn't disagree with what I stated previously.)


Comments

Q: Does thermodynamic properties imply that a spacetime has a black hole?

A: No, thermodynamics does not imply a black hole: de Sitter spacetime also has thermodynamic properties, as I discussed in this answer, with some further comments in this other answer. I can't think of any thermodynamic properties that are specific to black holes, specially since their behavior is pretty much dictated by the usual laws of thermodynamics.

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  • $\begingroup$ @Nikolas thanks for the detailed answer. I have not read on BTZ solutions in much detail, but it is a non-trivial black hole (BH) solution in 3 spacetime dimensions. Now a naive question in this case would be the following : how do I conclude that the BTZ solution corresponds to a BH? They claim that the solution admits a no hair theorem and has non-trivial thermodynamic properties. So by looking at thermodynamic properties can we conclude whether a solution is a BH solution? Is there some connection between thermodynamics and global properties of solutions you mention? $\endgroup$
    – physmath17
    Nov 19, 2022 at 6:14
  • $\begingroup$ @physmath17 I also don't know much about BTZ black holes, but I'd say you have to show there is a region from which nothing escapes to infinity, and hence it it that set $B$ I defined. The calculation would likely be similar to how one shows that once you go past $r = 2M$ in Schwarzschild, you can't go back (plus some details on the structure at infinity, which can sometimes be overlooked because it is clearly flat in the spherical coordinate system). As for thermodynamics, I added a new remark to my answer. $\endgroup$ Nov 19, 2022 at 6:31
  • $\begingroup$ @Nikolas thanks for your comments. It helped me clarify some things. $\endgroup$
    – physmath17
    Nov 19, 2022 at 11:02
  • $\begingroup$ @safesphere could you refer me a book or article where I can read more about the comments you made? $\endgroup$
    – physmath17
    Nov 19, 2022 at 11:04

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