# How do we know we cannot escape the event horizon?

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?

Sources:

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

• But we did see the horizon changing with the merging of the two black holes in LIGO observation, no? – anna v Jul 28 at 6:24
• 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.. – anna v Jul 28 at 7:39
• 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? – PM 2Ring Jul 28 at 8:19
• 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. – Alessandro Rovetta Jul 28 at 8:38
• 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. – Alessandro Rovetta Jul 28 at 8:44

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.

• 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. – Alessandro Rovetta Jul 28 at 12:01
• @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 :-) – John Rennie Jul 28 at 14:28
• 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! – John Duffield Jul 29 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.

• Wow that's a very interesting point. I will deepen it, thank you! – Alessandro Rovetta Jul 29 at 15:48

How do we know we cannot escape the event horizon?

Because a black hole is black. It's black because light can't get out. Light can't get out because at the event horizon the "coordinate" speed of light is zero. And we can't move faster than light. So we can't get out either.

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.

It might be beautiful, but it's wrong. It says this:

"The metric, in these coordinates, is singular at r = 0 and at r = 2M. However, the singularity r = 2M is only an artifact of the coordinate choice, and can be removed by changing coordinates (and then it is called “coordinate singularity”); the singularity r = 0, instead, is a true singularity of the metric (and is called “curvature singularity”)".

Why is it wrong? See Einstein's 1939 paper on a stationary system with spherical symmetry consisting of many gravitating masses. He said “g44 = (1 – μ/2r / 1 + μ/2r)² vanishes for r = μ/2. This means that a clock kept at this place would go at the rate zero". He also said “In this sense the sphere r = μ/2 constitutes a place where the field is singular”. Do you know how the singularity at the event horizon is removed? By using seconds of infinite length! I'm afraid it's "unphysical". See the Wikipedia article on Eddington-Finkelstein coordinates. They were invented by Roger Penrose. Then take a look at the section on tortoise coordinates.

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.

This is unphysical too. See what Einstein said in 1920: “the curvature of light rays occurs only in spaces where the speed of light is spatially variable”. At the event horizon, the speed of light is zero, and it can't go lower than that. Hence it's clear that Oppenheimer and Snyder’s 1939 “frozen star” paper on continued gravitational contraction describes the black hole better than your article. It's like Ruffini and Wheeler said in their 1971 article introducing the black hole, “in this sense the system is a frozen star”. That means the r=0 curvature singularity doesn't exist.

As for the strong equivalence principle, see the mathspages article on the many principles of equivalence where you can read this: “the modern statement of the strong equivalence principle, of the assertion that the laws of physics are the same for all frames of reference (i.e. independent of velocity) is also conceptually quite distinct from the original meaning of Einstein’s equivalence principle”. The strong equivalence principle is nothing to do with Einstein, and gamma ray bursters make it clear that it doesn't apply absolutely. So you shouldn't use it to speculate about the interior of a black hole.

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?

Because gravity isn't a geometric property of space-time. You won't find Einstein saying that. Instead you'll find him saying a gravitational field is a place where space is neither homogeneous nor isotropic. That's caused by a concentration of matter in the guise of a massive star "conditioning" the surrounding space. Because of this the speed of light varies. Look again at this: “the curvature of light rays occurs only in spaces where the speed of light is spatially variable”. Light doesn't curve because it follows a geodesic. It doesn't follow the curvature of spacetime. It curves because the speed of light varies. The important thing to note about all this, is this: at the event horizon the speed of light is zero, and it can't go lower than that. So there's no gradient in the speed of light, and no gravitational field.

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?

Because when you add a concentration of energy to a region of space, you alter the surrounding space such that you set up a gradient in the speed of light. That's what a gravitational field is - a region of space where the speed of light is spatially variable. Light goes slower at a lower elevation. That's why optical clocks run slower when they're lower. However clocks can't go slower than stopped. The event horizon is a place where the speed of light is zero, and you can't make it go slower than that.

On page 50 of your second reference the author says "this situation is generally considered unphysical". When you read the original general relativity in the Einstein digital papers, you come to realise that some other situations invented in the 1960s are unphysical too.

• Alessandro is asking about the back reaction of the test mass, i.e. the perturbation of the geometry by the test mass, and how this affects the calculation of the null geodesics. You have missed the point. – John Rennie Jul 28 at 10:37
• @John Rennie: I haven't missed the point. I answered his question step by step. My answer gives a careful explanation that ought to clear up his confusion. The doubt that assails him stems from reading a "beautiful" article which flatly contradicts Einstein's general relativity, and is wrong. Again, see my answer here where I say some more about Schwarzschild coordinates and Kruskal-Szekeres coordinates. – John Duffield Jul 29 at 9:35
• @safesphere that's true, but the contention here is that the coordinate transformation is unphysical. It's in this respect that the OP differs from every relativity book ever written. – John Rennie Jul 29 at 11:23
• Comments are not for extended discussion; this conversation has been moved to chat. – tpg2114 Jul 30 at 22:55
• Sic transit gloria mundi – John Rennie Aug 1 at 13:55