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Is there any necessary and sufficient mathematical condition(s) so that a (general) energy-momentum tensor can possess an assemblage of black holes? Or in other words, if I'm given a general energy momentum tensor ($T_{\mu\nu}$) and minimum amount of some other property of the spacetime (like topology), can I determine whether it can possess assemblage of black holes or not?

R. Wald says

A strongly asymptotically predictable spacetime is said to contain a black hole if $M$ is not contained in $J^-(\mathscr{I}^+)$. The black hole region, $B$, of such a spacetime the set $B = [M - J^-(\mathscr{I}^+)]$ and the boundary of $B$ in $M$, $H = M \cap J^-(\mathscr{I}^+)$, is called the event horizon.

(1) Now as the General Theory of Relativity predicts, local geometry, curvature etc. of the spacetime is determined by the content in that region. So how can we achieve the aforementioned definition in terms of the energy momentum tensor of the content?

(2) Also, the aforementioned definition seems to be corresponded the existence of a single black hole in the spacetime. How can this definition be modified for corresponding an assemblage of blackholes?

Qmechanic
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SCh
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    For Question 2, $B$ need not correspond to a single black hole. If it contained two black holes (that never merged) then you'd just have a situation where $B$ was not a connected region of spacetime. – Michael Seifert Nov 22 '22 at 01:40
  • For Question 1, you may want to look into the hoop conjecture. It is also worth noting that it is still a conjecture 50 years after Thorne first thought it up. – Michael Seifert Nov 22 '22 at 01:42
  • Actually I am looking for some modifications in Wald's definition to incorporate the notion of energy momentum tensor of the system (instead of the geometry, topology etc. of the spacetime). – SCh Nov 22 '22 at 01:45
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    “as the General Theory of Relativity predicts, local geometry, curvature etc. of the spacetime is determined by the content in that region” - This is incorrect. The equations are differential, so the curvature and geometry also are defined by the initial conditions and by the constants of integration. For example, the Schwarzschild black hole has no “content”, meaning the stress-energy tensor is zero everywhere. – safesphere Nov 22 '22 at 03:07
  • The left side of EFE ($G_{\mu \nu}$) describes gravity and the left side ($T_{\mu \nu}$) how it is affected by matter. You may like to read https://arxiv.org/abs/1803.09872v1, Dennis Lehmkuhl, "How Einstein saw the role of the energy-momentum tensor in GR", page 5. For gravity matter is merely an additional distortion of spacetime curvature. – JanG Nov 22 '22 at 08:22
  • @JanGogolin “The left side […] describes gravity and the [right] side […] [describes] how [gravity] is affected by matter.” - Indeed plus it may be worth clarifying that the left side still describes gravity, even if the right side is zero, such as in case of Schwarzschild or Kerr black holes. – safesphere Nov 23 '22 at 02:52
  • @safesphere thanks for correction. What about that: for the right side zero the gravity equation reduces to $R_{\mu \nu}=1/2 g_{\mu \nu} R$. The solution (metric) to this equation describes the pure spacetime without energy or stresses in it. – JanG Nov 23 '22 at 11:47

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Black holes are vacuum solutions, so the stress energy tensor is zero. It's possible to have a spacetime with a black hole and non-zero stress-energy, but then you have a black hole plus something else, not just a black hole.

Andrew
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