Suppose two SMBHs (supermassive black holes) are brought close (e.g. during a galactic merger). Suppose that a "normal" black hole drifts into the space between the SMBHs. Could the gravitational pull of the SMBHs be sufficient to tear the normal black hole to pieces?

If so, will we be able to examine the fragments, free from being obscured by the black hole's event horizon?

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    $\begingroup$ The idea of fragments of a black hole is fundamentally wrong. A black hole is a region of space defined by an event horizon. The idea of an internal structure is meaningless from the outside - see the No Hair Theorem $\endgroup$ – StephenG Jun 16 '18 at 22:27

@peterh's answer contains the essential point, but I think lacks enough background. Let me try to go a bit deeper.

First of all, let's be clear that we're talking about the classical Black Holes (BHs) of General Relativity (GR), and not the true BHs of the as-yet unknown theory of quantum gravity (TQG). To answer your question, TQG does not have a role, because your question is about a regime which is far from where TQG effects would be relevant.

It's contrary to intuition, but the vicinity of a SMBH is a less exotic and less stressed environment than the vicinity of an ordinary BH. If you fell toward a solar-mass BH, the tidal forces would have torn you apart well before you got to the event horizon, while in a SMBH, it happens well after you pass the event horizon. Likewise the Hawking radiation (which can be taken as a measure of the stress in spacetime) is greater outside a solar mass BH than it is outside a SMBH. Wikipedia has a decent article on this which, while it contains at least one howler, on the whole seems to be reliable.

Bottom line: the tidal forces which tear things apart are weaker outside a SMBH than outside a smaller BH. But we've already observed several smaller BHs collide using LIGO, and they don't shred each other, but merge. So we can confidently say that a SMBH will not shred a smaller BH.

Another way to understand why a BH does not shred is to think about what's actually there. The answer is nothing. If you could survive the tidal forces (perhaps by falling into a SMBH) and could avoid the matter which generally forms an accretion disk around a BH (maybe by finding one in intergalactic space where there's no matter to pick up) and fall through the event horizon (EH), you'd notice nothing at all. It's just ordinary empty space outside the hole, at the EH, and inside the EH. Locally (i.e., it the cabin of your (now doomed) spaceship), there's nothing to see and nothing to detect, other than the ever-increasing tidal forces which will eventually tear you apart. The BH is nothing but curved spacetime and there's nothing else there. What's to tear apart?

I'm using words. The mathematics of GR, which is very difficult but is also well-understood, describes exactly what happens when to BHs approach one another and Hawking's Area Theorem does prove that under GR, merger is the inevitable result.

What about the Theory of Quantum Gravity? Obviously we don't know much about it, since it hasn't been discovered yet, but one of the few things we're sure of is that it will give results exactly* the same as GR at "large" distances. And "large" is this case is probably anything bigger than an atom. Even if the limit is "bigger than a bread box", the two BHs are far too large for their collision to have any TQG effects. The TQG effects all happen down at the singularity the GR predicts at the center of the BH -- but whatever happens there is lost to the outside world.

  • $\begingroup$ I think one of the sources of confusion is that the event horizon can be deformed. Spinning black holes have a different shape to their EH than non-spinning ones. And merging black holes, going by simulations thereof, have a protruding portion of their EHs that meet and becomes a somewhat cylindrical bridge. So while there's nothing to deform "inside", a deformation still happens. It visually looks rather like a piece of the EH is being pulled up and away from the rest of the EH, and one is inclined to wonder "can it be pulled hard enough that that piece breaks off"? $\endgroup$ – zibadawa timmy Jun 17 '18 at 17:01
  • $\begingroup$ It does not strike me as particularly obvious that it's impossible for the spacetime to warp so severely that there is a "pinch" that, at least temporarily, separates one connected EH into two (or more) components, though perhaps this is a theorem somewhere. $\endgroup$ – zibadawa timmy Jun 17 '18 at 17:05
  • $\begingroup$ @zibadawa timmy It's certainly not obvious, and many expert physicists felt as you do, but Hawking and Roger Penrose showed that General Relativity forces it. (This was the work which made his reputation among scientists, and Penrose's also.) $\endgroup$ – Mark Olson Jun 17 '18 at 19:45

No. The total mass of the Black Hole is in the point in its center, but there is no return from passing the event horizon.

The Area theorem (from Hawking) says, that the total area of the event horizons never decreases.

You can't break up the hole, there is nothing on the event horizon, only the geometry of the spacetime looks so, that there is no way out any more. The mass of the black hole is (probably) in a small point in the centre. What you call "black hole", is mainly vacuum. It doesn't matter, how do we play snooker with the black holes, the result will be at most some black hole merging, and not a broken up hole.

  • $\begingroup$ The second law of black hole thermodynamics doesn't apply if Hawking radiation exists. $\endgroup$ – Chappo Jun 17 '18 at 3:53
  • $\begingroup$ But wouldn't the combined gravitational pull of 2 SMBHs be enough to overcome the black hole's own gravity and break it apart. This is an external influence that the laws do not cover $\endgroup$ – RonS Jun 17 '18 at 6:41
  • $\begingroup$ @RonS Hawking's Area Theorem covers it. You can't break up the hole, there is nothing on the event horizon, only the geometry of the spacetime looks so, that there is no way out any more. The mass of the black hole is (probably) in a small point in the centre. What you call "black hole", is mainly vacuum. It doesn't matter, how do we play snooker with the black holes, the result will be at most some black hole merging, and not a broken up hole. $\endgroup$ – user259412 Jun 17 '18 at 12:18
  • $\begingroup$ @RonS Gravity in the Black Holes doesn't work as we imagine in smaller scales. It is not a force, it is a curvature on the spacetime. Not that happens that the black hole pulls very strongly, what you could compensate with a much more stronger force. It works so, that it curves the spacetime on a way, that there won't be way out any more. What exactly happens if multiple black hole meet, it can be calculated by math. Hawking's result is that the surface area of the event horizons never decreases (i.e. you can't break it apart). $\endgroup$ – user259412 Jun 18 '18 at 18:15
  • $\begingroup$ @RonS Btw, probably there is escape from the Black Hole, at least a single effect is known which makes it possible (Hawking radiation), it is only a theoretical construction and it is very small, but it exists. From the other side, escaping/breaking up a black hole seems strongly correlating to faster than light travel. FTL leads to logical contradictions (google for: "tachyonic anti-telephone"). No one knows, what is the solution for this problem. $\endgroup$ – user259412 Jun 18 '18 at 18:18

Taking this question from a different perspective. Lets forget the black hole. Black holes run into the problem with having unknown interiors inside their inescapable event horizons with the possibility of a near singularity in size at their center due to no known force being able to prevent their continued collapse, so forget black holes. A neutron star is easier to work with as a theoretical scenario.

Imagine, instead, that we have a very high density Neutron star on the verge of sufficient mass and density to collapse. The Neutron star equivalent of the Chandrasekhar limit, aka the Tolman–Oppenheimer–Volkoff limit. This isn't a black hole, but it's as close as you can get to a black hole without being a black hole, with an escape velocity approaching to the speed of light.

Now lets take your scenario:

Suppose two SMBHs (supermassive black holes) are brought close (e.g. during a galactic merger). Suppose that a "normal" black hole drifts into the space between the SMBH

I like your scenario because galaxies and their supermassive central black holes do merge and as a result it's theoretically possible that this could happen. Keep in mind, everything in space is in free-fall and objects in free fall don't experience the gravity of the object they orbit. What they experience are tidal forces and the tidal forces of super-massive black holes is comparatively pretty weak. You get the highest tidal forces (just outside the event horizon) from the smallest stellar black holes not supermassive ones. See here or here or here. A Neutron star would withstand any tidal forces from two colliding supermassive black holes with ease.

What would happen is some tidal stretching like Europa experiences from Io and Ganymede as it orbits Jupiter. In the case of Io, this is often called tidal heating, but the heat comes from stretching and squashing and it's continuously generated every orbit due to the 3 body dynamics. Super-dense objects like Neutron stars are much more resistant to stretching and squashing, so the gravitational tidal forces would need to be enormous to have any effect.

3 body tidal stretching outside the event horizon of super-massive black holes is comparatively weak. Much too weak to pull apart a neutron star. You'd get the strongest theoretical tidal stretching from stellar mass black holes and the strongest tidal forces possible from the smallest stellar mass black holes.

For this theoretical scenario to work, you'd need two stellar mass black holes about as small as black holes can form, which would have maximum density in relation to the volume of their event horizon and a Neutron star to fall perfectly in between them, nearly touching them as it falls between them and in such a scenario, the tidal forces might pull apart the high density Neutron star . . . if it was perfectly timed. The improbability of a 3 body scenario like this actually happening is essentially zero.

I should add, I'm not sure how the relativistic factors would affect the tidal forces. Weird things happen with relativity, for example, the escape velocity becomes lower than the orbital velocity and precisely how the relativistic effects would affect orbital velocity and tidal stretching, if they would affect it at all, I'm not sure.

Pulling apart a black hole raises a whole set of impossibilities and to my understanding, covered by @Peterh answer. Pulling apart a dense Neutron star between two small, close to maximum density stellar mass black holes if the Neutron star threads the needle between them as they tightly orbit each other, just barely, maybe, but as noted, it's a prohibitively improbable scenario.

That's not the question you asked, but I think it touches on the idea behind your question.


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