A doubt assails me: deepening the Schwarzschild and Kerr space-times I found a beautiful article of the University of Rome, which provides a method to identify an event horizon based on the analysis of the square norm of constant "radial" coordinate hypersurfaces normal vector ¹ ².

In particular, taking advantage of the Strong Equivalence Principle, it is easily demonstrated that if the normal vectors are timelike the tangent vectors are spacelike i.e. once we enter the horizon we can no longer go out.

The question is: since gravity is a geometric property of space-time, how can we be sure that by heavily disturbing the region of the event horizon this effect is not destroyed i.e. it is possible to escape from the horizon itself?

Let me explain better: by introducing any energy source, gravity changes i.e. space-time changes. Considering that the horizon manifests itself through a geometric property (gravity), how can we be sure that the prediction "is impossible to get out of an event horizon" is valid even if on the horizon there is a star, an asteroid or, in general, an object with non-negligible energy effects capable of significantly altering the geometry of space-time?


¹http://www.roma1.infn.it/teongrav/leonardo/bh/bhcap12.pdf Page 4 chapter 1.2
²http://www.roma1.infn.it/teongrav/leonardo/bh/bhcap3.pdf Page 50 chapter 3.4.2

  • $\begingroup$ But we did see the horizon changing with the merging of the two black holes in LIGO observation, no? $\endgroup$
    – anna v
    Jul 28, 2019 at 6:24
  • $\begingroup$ the title and the content do not say the same thing. The mathematics of the even horizon is such that once a particle falls through, thats it for those boundary conditions of the solution of the problem Bringing in stars etc changes the boundary conditions of the problem andthus the solutions, one of which is the horizon.. $\endgroup$
    – anna v
    Jul 28, 2019 at 7:39
  • $\begingroup$ Any energy source introduced into the vicinity of a BH is going to add to the existing spacetime curvature, so why do you think that could make it easier to escape? $\endgroup$
    – PM 2Ring
    Jul 28, 2019 at 8:19
  • $\begingroup$ Thank you StephenG, i did it. Dear Anna, that's exactly the point: if we have an initial situation with a (real) black hole such that in a neighborhood of the event horizon we have $g_{\mu \nu}$ ~ Kerr metric, and a final situation where a star is on the event horizon such that $g_{\mu \nu} \neq$ Kerr metric, how can we mathematically know the event horizon is conserved near the star? PM 2Ring, Ok, theoretically the curvature increases, but if we introduce an important energy source the metric $g_{\mu \nu}$ should be recalculated and maybe the situation would be completely different. $\endgroup$ Jul 28, 2019 at 8:38
  • $\begingroup$ Thanks Qmechanic, i'll do it. Just to be clear, for "SEP" I mean the Strong Equivalence Principle. Trying to summarize my question is: is there a way to analytically study very heavy perturbations and observe that the event horizon is preserved? Perhaps it is sufficient to show what PM 2Ring states and in that case I would be very grateful if someone would give me some indications on how to do it analytically. Thank you. $\endgroup$ Jul 28, 2019 at 8:44

2 Answers 2


I can't give you an authoritative answer, but in the absence of one I'll attempt to at least clarify the problem.

The situation you describe is complex because there are no known analytic solutions for the two body system in GR. So there is no simple way of describing the modification of the Schwarzschild or Kerr geometry caused by your test mass. The test mass will certainly perturb the horizon but not in a way that is easy to calculate.

For large masses this is the merging black hole problem and numerical techniques for calculating the geometry are now routine courtesy of all the work done in support of the LIGO experiment. The horizon evolves in the way you'd intuitively expect i.e. it evolves from two separate Kerr geometries through to a final Kerr geometry passing through various shapes on the way.

What would be more interesting is if there was a way of analytically approximating the perturbation in the limit of a small but not negligible test mass. However I have never seen a calculation of this type.

  • $\begingroup$ Thank you very much John, you got a point! I will try to answer the question you raised by contacting other professors and/or researchers. If I find something interesting I'll post it below. $\endgroup$ Jul 28, 2019 at 12:01
  • 2
    $\begingroup$ @AlessandroRovetta yes, if you find an answer, or even a partial answer then please do post it as an answer to your own question. The site encourages this as it will be helpful to future visitors, and indeed I'd like to see it :-) $\endgroup$ Jul 28, 2019 at 14:28
  • $\begingroup$ See what Einstein said here: “As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable”. A gravitational field is a place where there's a gradient in the speed of light. A black hole however is a place where the speed of light is zero. So there's no gradient in the speed of light, and no dynamical mechanism by which a black hole can fall down. This creates something of an issue for LIGO. Interesting stuff! $\endgroup$ Jul 29, 2019 at 9:49

The second law of black hole thermodynamics can be proved in GR, and it says that the area of the event horizon always increases. So you cannot eliminate the event horizon by introducing the gravitational field of some second object. Furthermore, area can't be transferred from one black hole to another, and black holes can't bifurcate (Hawking and Ellis, prop. 9.2.5, pp. 315-316). Intuitively, what this is telling us is sort of that once a black hole grabs a particular region of space, it will never relinquish any of it.

  • $\begingroup$ Wow that's a very interesting point. I will deepen it, thank you! $\endgroup$ Jul 29, 2019 at 15:48

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