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Is the "event horizon" of a black hole potentially violable? Black holes are commonly described as being unidirectional (matter / energy goes in, but doesn't come out), but is the event horizon of a black hole potentially deformable? That is, in the case of two black holes colliding (let's say), is there not a point when the effective gravitational field between the two black holes "cancel out", thereby causing a potential "escape path" for light traveling perpendicular to the axis the two black holes are on?

In a more complicated case, consider two black holes of similar mass orbiting each other in a stable configuration; since no matter or energy actually pass the event horizon of a black hole, would the area between the black holes be a potential zone for escape of energy? (I'm assuming minimal frame dragging effects here.) And since the black holes would need to rotate at a very high rate of speed to avoid actually colliding, would that not restrict the emission of energy to be along the axis of rotation?

A bit of clarification here; the primary question here is whether or not the gravitational force vectors of the black holes can properly cancel each other out at an appropriate point between them. If that's the case, then (in a static system) it can be easily determined that there is a plane through that point perpendicular to the axis along which the two black holes exist where the gravitational force vectors are deeply reduced (not zero, of course, except at the precise point where the masses perfectly balance; however, reduced significantly based upon the inverse distance from said axis).

In an (again static) two black hole system, orbiting material could enter a "figure eight" orbit; much of this material would pass through this "nodal point" (or very nearby it). The high speeds and volumes of the material would indicate a large number of collisions; some resultant (high-energy) ejecta from these collisions would result in some (small) quantity of ejecta traveling along the plane of reduced gravitational force. Now consider the (more dynamically stable) case of stably orbiting black holes; the plane of reduced gravitational force is itself rotating (about the axis of rotation of the black holes); this turns the "escape plane" (not entirely just a plane, it having some (small) quantity of thickness) into a line (again, not entirely just a line, but a very thin cone).

The primary point is that that the cancellation of force vectors effectively makes for a "dimple" in the event horizon of each of the black holes; that is, the existence of another strong gravitational attractor in close proximity deforms the event horizon of a black hole.

The question is, is there anything which invalidates this analysis? As mentioned before, frame dragging is a concern; is there anything else which would be a concern? (I suspect the frame dragging concern would actually be resolved by the mutual forced frame dragging, i.e. the holes' spins would "synchronize" because of the mutual frame drag effect.)

If this is a possibility, then does this not imply another means of black hole mass "evaporation" other than Hawking radiation? (And, in fact, wouldn't it imply the type of "polar jet" that has been observed?)

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Well, there is (proposed to be) the Hawking mechanism...but that is not what you are asking about. –  dmckee Nov 20 '10 at 1:13
I thought an event horizon of a black whole, wasn't a physical surface. Just the point/radius at which the escape velocity is greater than light, and therefore no light that can get out if it is beyond that point. –  Jonathan. Nov 20 '10 at 21:41

5 Answers 5

Well, if I'm being pedantic, your question is self-contradicting. The mathematical definition of a black hole is determined by tracing back the points that escape to infinity at late times, and then taking their boundary--thus, the event horizon is PRECISELY those points that cannot escape to infinity. But that's a boring answer that doesn't tell you anything about physics.

A less boring answer states that there is a very strong theorem that says that the area of a black hole is necessarily nondecreasing (so long as you have ordinary matter and no quantum mechanics), and as a corollary, you can prove that the null generators* of the hole either are destined to sit on the black hole's horizon forever, or to fall into the hole. They cannot escape. And since these light rays can't escape, then neither can particulate matter nor rays farther into the hole.

*The null generators are the light beams that are the boundary between the outgoing light rays that can freely leave the hole outside the horizon, and the ones forced to fall in that like inside the horizon. They simply float in space, neither able to go out or go in.

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See my clarification; in particular, the bit about the "dimpling" of the event horizon due to the effect of a nearby extremely strong gravitational force. In particular, it would appear that for the "null generators" you describe, an applied (transitory) force could allow them to escape. Consider the orbital mechanics counterpart (not the same, I know); an asteroid in stable orbit can be perturbed very slightly by a passing gravitational attractor, and be perturbed into an unstable orbit, even achieving escape velocity. –  Paul Sonier Nov 20 '10 at 5:18
You have to be very, very careful when applying Newtonian intuition to a problem in strong-field General Relativity. Thinking about a problem like this in terms of Newtonian force is very very dangerous, and likely to be wrong. The area increase theorem is very general. My intuition leads me to think that (but I can't necessarily prove), in this case, if the holes are sufficiently distorted to allow escape from a local apparent horizon, you will already have formed a common apparent horizon around the two holes, and "escape" will still mean you're trapped inside an apparent horizon. –  Jerry Schirmer Nov 21 '10 at 5:03

The form of a black hole horizon behaves as a viscous liquid. It flattens at poles when BH is rotating, there can be waves on the surface of a black hole when it eats something and the energy of waves then radiated as gravitational waves so the surface becomes smooth again.

As something falls on a BH, its diameter increases, but not simultaneously everywhere (the BH can be so large that even light takes some time to reach the other side, so are the gravitational waves). When two BHs are approaching each other they also distend towards each other as two drops of attracting liquids, and then they fuse. The energy of the surface waves is again radiated as gravitational rays.

Closer to your question, being between two black holes would not help to escape, just the opposite, the surfaces of the both BHs expand in the direction of each other so you would be in even greater trouble. To escape from between two BHs to the infinity you'd have to fight the gravitation of both.

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Well, by definition, an event horizon is a surface such that any event inside cannot affect an outside observer. (I think I've heard it defined as the boundary of a region in spacetime excluded from the causal past of future null infinity.) So the simple answer to your question has to be no, the event horizon cannot be "violated," otherwise it would not actually be an event horizon.

But I think you're getting at something interesting by considering the "cancellation" of gravitational fields between two black holes. I guess the question is, if you take an isolated black hole and combine it with another one to form a binary BH, are there any points that would be inside the event horizon of the single BH that are outside the horizon of the binary? That I'm not sure about, at least not without doing some analysis of spacetime around a black hole collision (and I don't think that's an analysis I would know how to do in detail). Besides, you'd have to define some physically meaningful procedure for identifying spacetime points around the original black hole with spacetime points around the binary.

I suspect that the answer to this is also "no," though; that is, every point that is inside the event horizon of a black hole cannot be placed outside an event horizon by the addition of additional matter.

Of course, if you had two orbiting black holes with a region between them that was not inside the event horizon, energy (in the form of light, for example) should certainly be able to escape from that region. Again, this stems from the definition of the event horizon: if light had no way to escape, then the region would be inside the horizon.

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Yes, I understand the event horizon cannot be violated, by definition; but the event horizon is just a convenient term. Let's remember that nothing can actually pass the event horizon; everything just gets closer and closer, but never passes (within non-infinite time). And that's the crux of the point; since nothing crosses the event horizon, it's perpetually orbiting (even in a deeply degenerate orbit), and therefore can be "liberated" by traveling along a (admittedly small) exit vector. –  Paul Sonier Nov 20 '10 at 4:32
Event horizon is not just a convenient term. It is a precisely defined mathematical object. It's fine to just say that you misspoke. You mean apparent horizon, not event horizon. –  Jerry Schirmer Nov 21 '10 at 5:05
@JerrySchirmer: ah, thank you for the clarification. I think the point I was trying to make is that as a location (with a precise definition) it is fundamentally unreachable and unobservable. –  Paul Sonier Nov 29 '10 at 17:05

In the case of colliding black holes there is no line where the gravitational force acts the opposite direction and cancels out: only one point. There is a line where the gravitational force is equal magnitude, but the force vectors add up to a non-zero value. Thus there is no escape route. Also, when the two colliding black holes are so close that the event horizon overlaps, they are going to merge in a very short time.

THis has bearing on your second question too. In the case of orbitin black holes there are the usual Lagrangian points of the orbit, In a nutshell, if two very massive bodies are on a stable orbit around each other, there are 5 points where you could place objects with a very small mass (compared to the orbiting bodies) so that they stay there relatively to the other two because the gravitational pull of the two "big ones" kinda cancel each other out, together with the orbiting. But this also means, that there is no escape route: if the object leaves the Lagrangian point, it will start to orbit the two bodies or likely fall into one of them in the case of black holes. These orbits are also not possible within the event horizon, then the black holes will just merge.

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It stands to reason that along a plane through the point where the gravitational forces cancel out (the plane being perpendicular to the axis defined by the centers of mass of the two black holes), that there is a path of escape. In a static system, this plane is the exit path; in the case of rotating black holes, this would become restricted to a line (the axis of rotation). The question is, it appears that this exit path makes the event horizon of either of the black holes asymmetrical; is that a valid analysis? –  Paul Sonier Nov 20 '10 at 4:42
@McWafflestix: the point Greg was making was that the gravitational forces do not cancel out, except at a single point. Not a whole plane. –  David Z Nov 21 '10 at 5:35
That's exactly what I meant, thanks. –  Greg Nov 21 '10 at 12:50
@DavidZaslavsky: yes, I understand that; what I'm saying is that the gravitational force vectors from each of the black holes effectively cancel MOST of each other out; not entirely, but mostly. Through the point where the forces DO fully cancel out, there is a plane along which the forces cancel each other out so as to provide a combined force vector which remains within the plane; energy traveling within that plane will NOT begin to orbit either of the two bodies (as Greg said). So long as the force vector remains less than the escape force, energy can escape along that plane. –  Paul Sonier Nov 25 '10 at 0:06

I'm not sure I understood your questions properly, but the black hole region of a (strongly asymptotically predictable*) spacetime is a well-defined topological subspace (the complement of the causal past of future null infinity). You have a foliation of spacetime by Cauchy surfaces, each one corresponding to an "instant of time". Then a black hole in time T is just a connected component of the black hole region, restricted to the corresponding Cauchy surface, and its boundary is the event horizon. It is therefore clear what it means to be inside or outside the black hole, and those notions are mutually exclusive.

Now, classically (i.e., in the context of general relativity) it is impossible to escape the event horizon by its own definition; no causal curve starting inside of it can get to future null/timelike infinity.

You asked if the event horizon can be "deformed". I'm not sure what you mean, but there are very general theorems which restrict the way black holes and event horizons behave and evolve. For instance, the event horizon must be topologically a 2-sphere; A black hole cannot bifurcate into 2 differente black holes, nor "evaporate" (note that Hawking radiation is a quantum effect, so it does not belong in this context). Finally, the event horizon area of all black holes never decreases in time. These are precise theorems of topology/Lorentzian geometry which are true in any physically reasonable spacetime. Of course, I'm not stating then precisely here, there are hypothesis about "energy conditions" and about the asymptotic behaviour of spacetime.

You can see all these results stated and proved in Hawking, Ellis - The large scale structure of spacetime.

*this is a technical condition for the definition to make sense.

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Hmm. Again, consider the case of two separate black holes; clearly, they can collide and become one black hole. At some point before the collision, there are two event horizons; at some point after the collision, there is one event horizon. The question is, at some point between two and one event horizons, there is (by necessity) a situation where the topological surface which defines the point of "inescapability" (for each of the black holes) is not spherical. –  Paul Sonier Nov 20 '10 at 5:26
By topological sphere I mean a surface which is homeomorphic to a sphere. It does not need to be precisely spherical, all that is needed is that the Euler characteristic be 2. Of course, during the merging of the black holes the event horizon might deform, but at any time it is homeomorphic to a sphere, i.e., you can't create "holes" in it. –  Rodrigo Barbosa Nov 20 '10 at 14:13

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