Can another blackhole pull you out of the event horizon of a blackhole

[maa105]

say you fell beyond the event horizon of a black hole can another black hole passing close enough pull you out?

 

[mathman]

No. Most likely the two black holes would merge to form a bigger one.

 

[maa105]

I understand that fact, but I am interested if say it wasnt a head on collision where the other black hole passes by close enough where its gravitational pull pulls you out of the other black hole's grip and not pull you into it's own. is that scenario possible?

 

[PeterDonis]

I understand that fact, but I am interested if say it wasnt a head on collision where the other black hole passes by close enough where its gravitational pull pulls you out of the other black hole's grip and not pull you into it's own. is that scenario possible?

No. Once you are inside the event horizon of a black hole, nothing can get you out.

 

[Chronos]

Black hole event horizons can merge, but, confinement inside an event horizon is the gift that keeps on giving.

 

[Algr]

Does this mean that if two event horizons intersected, it would be impossible to prevent the black holes from merging?

It seems to me that two event horizons would repel each other. You can't escape the horizon, but could the horizon be pushed past you, thus releasing you?

 

[fobos]

In my understanding, If black holes are objects with a huge gravitational pull, then both will try to pull each other; in the end the larger one will actually succeed; in the same way that would actually happen with any other large objects in space with different masses.

That means that looking with a slow motion camera I will actually see what is after the event horizon coming out from the first black hole and entering into the second one ?? I have difficulty understanding merging when it comes to black holes.

 

[phinds]

Does this mean that if two event horizons intersected, it would be impossible to prevent the black holes from merging?

yes

It seems to me that two event horizons would repel each other. You can't escape the horizon, but could the horizon be pushed past you, thus releasing you?

No. Why would they repel?

First off, the EH of a black hole is not a physical thing, it's just a DISTANCE. The distance from the singularity below which light cannot escape the gravitational pull of the singularity. EH's don't attract or repel. What is involved is gravitational attraction. There is no repulsion to be found anywhere in the scenario of two BH's coming close and then merging.

 

[PeterDonis]

Does this mean that if two event horizons intersected, it would be impossible to prevent the black holes from merging?

Two event horizons intersecting *is* the black holes merging. (Actually, even that's not quite right: if you have a spacetime with two black holes merging, then there is only one event horizon, but it's shaped like a pair of trousers rather than a cylinder, so to speak--if you imagine time being vertical and two of the three space dimensions being horizontal. The single EH has two "branches" when you follow it into the past.)

could the horizon be pushed past you, thus releasing you?

An event horizon isn't a thing that can get "pushed". It's the boundary of a region of spacetime that can't send light signals out to infinity. In other words, it's part of the spacetime geometry, not a "thing" that gets added to the spacetime geometry. "Pushing" the EH would be like "pushing" the Earth's equator to occupy some other place on the Earth's surface--the idea makes no sense.

 

[PeterDonis]

the EH of a black hole is not a physical thing, it's just a DISTANCE.

Actually, the best way to think of the geometric relationship between the EH and the singularity is not as a distance, but as a time--the singularity is a certain amount of time to the future of the EH.

 

[Algr]

Event horizons would repel each other because they would tip an object's light cone in opposite directions, thus canceling out the effect of their gravity. It's rather like how you would be weightless at the center of the Earth.

 

[phinds]

Actually, the best way to think of the geometric relationship between the EH and the singularity is not as a distance, but as a time--the singularity is a certain amount of time to the future of the EH.

Fair enough. I DO tend to not understand that time-like vs space-like business inside an EH.

 

[PeterDonis]

Event horizons would repel each other because they would tip an object's light cone in opposite directions, thus canceling out the effect of their gravity. It's rather like how you would be weightless at the center of the Earth.

The claim here is not correct, but it does raise a good question: given that the light cones *are* tipped in opposite directions by two black holes that are falling towards each other, what do the light cones look like when the horizons merge?

To answer this, it's important to bear in mind that there are actually two types of horizons: apparent horizons and absolute horizons. The apparent horizon is the point at which outgoing light no longer moves outward--it just stays at the same radius. The absolute horizon is the boundary of the black hole, i.e., the boundary of the region that can't send light signals to future null infinity.

For a single, stationary black hole, the two horizons coincide; but in a spacetime in which two black holes merge, they don't. In the case of a black hole merger (or more generally in any case where a black hole gains mass by something falling into it), the absolute horizon will be *outside* the apparent horizon. This is important because the degree of "tipping" of the light cones corresponds to the *apparent* horizon, not the absolute horizon, which means that a point where the light cones are tipped enough so that the outgoing side is vertical (i.e., outgoing light stays at the same radius--the apparent horizon) is already *inside* the absolute horizon, i.e., already inside the black hole.

So, for example, suppose you are exactly halfway between two black holes of equal mass that are falling towards each other. Initially, the two holes are far apart, and you are outside both of them: you are able to send light signals to infinity. In your vicinity, the light cones are not tipped at all; you can indeed view this as the gravity of the two holes cancelling. As the two holes fall towards each other, the light cones in your vicinity still are not tipped; but the light signals you send out towards infinity will become more and more redshifted, as seen by an observer very, very far away from you and from both holes. At some point, you will become unable to send light signals to infinity at all; in other words, you will be inside the event horizon (which you would view as the horizon of the combined black hole formed by the two original holes joining). But at this point, there will be *no* apparent horizon anywhere in your vicinity; you will still have seen no tipping of the light cones near you.

Once you are inside the event horizon, you will eventually reach an apparent horizon; you will eventually see outgoing light in your vicinity no longer move outward. But at the point, you are already inside the event horizon of the combined hole, and there will be no simple way to relate the location of the apparent horizon you reach to any apparent horizons that someone falling into one of the original holes would have encountered.

 

[Algr]

Edit:
I think I get it now. A straight light cone does not necessarily mean that you are outside the (absolute) event horizon. There would be a zone that is "outside" the absolute horizon - but surrounded by it in every direction, and thus inside the apparent horizon. Is that right?

Hmmm... If the two black holes were orbiting each other - this zone could last for millennia. You couldn't escape, but neither would you be destroyed. You'd need to keep exerting energy to stay in the middle, but not too much if the zone was big enough. Then you could escape once the black holes evaporate.

Of course this depends on the two black holes NOT being within each other's absolute horizons - if they were, they'd need to orbit faster than light.

 

[PeterDonis]

A straight light cone does not necessarily mean that you are outside the (apparent) event horizon.

[Edit: this portion of your post was edited, but I think these comments are still relevant and useful to clarify the relationship between the two types of horizons.]
The apparent horizon is not an event horizon. The apparent horizon is the point at which the light cones are tipped over enough so that the outgoing side is "vertical" (i.e., outgoing light stays at the same radius). So if the light cones are straight (i.e., not tipped), then you *are* outside the apparent horizon.

Also, if the light cones are not tipped at all (i.e., they are like they are in flat spacetime), then you are most likely outside the event horizon at well. If two black holes are merging, the event horizon will be outside any apparent horizons, but not so far outside that the light cones won't be tipped at all. They still will be, just not to the point of the outgoing side being vertical.

There would be a zone that is "outside" the absolute horizon - but surrounded by it in every direction.

No, that's not possible. If there is a region inside the absolute (i.e., event) horizon in every direction, then there is no way to send light signals to infinity, so you must be inside the event horizon.

Of course this depends on the two black holes NOT being within each other's absolute horizons

There's no such thing. The event horizons *define* the boundary of a black hole; you can't have one hole inside another. You just have two holes merging into one, and a single event horizon that has two "branches" in the past (like a pair of trousers, if we view time as vertical).

 

[Algr]

Oops, I edited my post just as you posted yours.

 

[PeterDonis]

Responding to the edited portion of your post:

There would be a zone that is "outside" the absolute horizon - but surrounded by it in every direction, and thus inside the apparent horizon. Is that right?

No. The absolute horizon is outside the apparent horizon in the case we're discussing. Also see the comments in my previous post.

 

[Algr]

The event horizons *define* the boundary of a black hole;

Okay. I've been using "Black Hole" as synonymous with "Singularity". I seem to recall that if you enter the event horizon of a large quiet black hole, you won't notice anything unusual in your vicinity. Distant stars would look strange, but your ship would seem normal until you (inevitably) got close enough to the singularity for spaghettification to occur.

 

[PeterDonis]

I've been using "Black Hole" as synonymous with "Singularity".

Some people do this, but it's not really correct. The correct definition of "black hole" is the region that can't send light signals to infinity, i.e., the region inside the absolute horizon.

I seem to recall that if you enter the event horizon of a large quiet black hole, you won't notice anything unusual in your vicinity. Distant stars would look strange, but your ship would seem normal until you (inevitably) got close enough to the singularity for spaghettification to occur.

This is true, but it doesn't change anything I said. The "tipping" of the light cones is not really a local phenomenon, nor is the absolute horizon. For example, if your ship is falling into a black hole, there is no way for you to tell just by local measurements that you have crossed the absolute horizon--to know for sure, you would have to know the entire future of the spacetime, so you could know exactly which light signals will eventually reach infinity. You don't need to know quite that much to know whether you've crossed the apparent horizon; but you would need to measure the trajectories of light signals that you emitted in different directions, for long enough to see whether they were converging or diverging, and that's not really a local measurement either.

 

[Algr]

So in those terms, I am describing a stable area within the black hole where regular mater and information could exist.

In this one dimensional diagram, the red time cones mark the event horizon as I understand it. The green time cone is not tipped, but any light from it must enter one of the two event horizons. In one dimension I can see that the two singularities must merge, but in two dimensions they could instead orbit each other. Then the area near the green time cone would be stable, yet unable to reach the outside world.

 

[PeterDonis]

So in those terms, I am describing a stable area within the black hole where regular mater and information could exist.

"Regular matter and information" can exist anywhere inside a black hole, within the limitations of tidal gravity--i.e., whatever substance the matter and information is made of must be able to withstand the tidal gravity in its vicinity. But for large enough black holes there is plenty of room in the interior for that.

In this one dimensional diagram

There is a fundamental problem with this diagram: it attempts to depict "space" at an instant of "time". There is no way to depict the interior of a black hole (or multiple black holes) like that. You have to include time, because, for example, the two singularities are not at some "spatial location"; they are in the future. (More precisely, they are in the future for anyone inside the event horizon.) The same goes for the light cones: the "tipping" of the light cones is not purely spatial, because the light cones themselves are not purely spatial. They don't describe spatial paths of light: they describe *spacetime* paths of light.

You may be thinking of diagrams that show the exterior of a black hole as a sort of funnel shape, so that "gravity" causes things to fall towards the interior of the funnel. This is a permissible way of looking at the *exterior* of a black hole (the region outside the horizon) at an instant of time--although you can't draw light cones on this type of diagram even for the exterior. But you can't "extend the funnel" through the horizon to the interior; there is no way to represent the interior by that kind of diagram.

 

[Algr]

Have I explained why I think that event horizons would repel each other?

 

[Nugatory]

Have I explained why I think that event horizons would repel each other?

You have. However, it's not correct - event horizons do not repel each other.

(and please do remember that this site is not for arguing with established science or developing alternate theories)

 

[Algr]

Dude! I was just asking a question.

 

[ChrisVer]

Another way to think of it maybe is this:
just have the black holes 1 and 2. If some part of the event horizon of 1 enters 2, then it can't leave 1, and it can't also leave 2's (by definition). Then what's the result? the result will be the two horizons to merge into a single larger one (because the part of 1 will start falling into 2 and vice versa). However I am not sure if that's a good explanation because then the resulting thing will have to be a single black hole with two singularities that will forever fall in each other.

 

[PeterDonis]

If some part of the event horizon of 1 enters 2

This is already wrong. There is only one event horizon, not two. I've described this before, but I'll try again:

First, consider the case of a single black hole. If we think of a spacetime diagram, with time as vertical and "space" at a given instant of time as a horizontal plane (we are suppressing one space dimension), the event horizon is a cylinder; its intersection with each plane of "space at an instant of time" is a circle (a sphere once we add back the suppressed space dimension), and the horizon itself, as a surface in spacetime, is all of those circles stacked one on top of the other, i.e., a cylinder.

Now consider the case of two black holes that merge. Then the event horizon looks like a pair of trousers, so to speak: in the far past, it's two cylinders, but the two cylinders join into one, and in the far future, there's just one cylinder. There is still only one event horizon--one unbroken surface in spacetime. The "merging" of the black holes is just the joining of the two cylinders; one doesn't "enter" the other, they just join like the legs of a pair of trousers.

the resulting thing will have to be a single black hole with two singularities

No, there is only one singularity. Remember that the singularity is not a "place in space"; it's not located at some spatial point. It's located in the future--more specifically, it's in the future of every event inside the event horizon. The fact that the horizon happens to be shaped differently doesn't change that.

 

[Anoop Koushik]

I think that there could be one scenario where the black hole B's gravitational pull will cancel black hole A's gravitation, thus giving you small open gateway to escape from black holes up to an extent ! (If there aren't any other forces acting. i.e net gravitational force other than black holes is zero)

 

[PeterDonis]

I think that there could be one scenario where the black hole B's gravitational pull will cancel black hole A's gravitation, thus giving you small open gateway to escape from black holes up to an extent !

No, this won't work. If you're inside a black hole's horizon, nothing can pull you out.

 

[Algr]

I was wondering where this thread went.

Okay. Imagine one planet alone in an endless void. No matter where you were in the void, you would feel a gravitational pull, however slight, toward the planet.

Now picture two identical planets orbiting each other. The range of one planet's gravitational pull is no longer infinite. If you get far enough from one planet, you will instead be pulled to the other. There is a point between the two planets where their gravitational pull is equal and opposite. (I can't find the name for this. Everything points to "barycentre", but I want the one that in the Earth-Moon system is closer to the moon. It was an important factor for Apollo 13.)

A probe in this location is not pulled towards either planet. It's time cone is straight, even though it would not be straight if only one of the planets existed.

Now imagine these planets gain mass and become heavy stars. (Still identical. And I'm adding energy to keep the size of the orbit the same.) The gravity from each star is now considerable, but nothing about the probe's area changes. Being in both gravitational fields is the same as being in neither of them.

The stars gain more mass and become black holes. Now they have (non intersecting) event horizons. But what is an event horizon except a description of the effect of gravity in an area? What is the effect of gravity on the probe? Still nothing!

No matter how much mass you add to the black holes, their event horizons can never reach the probe because each black hole is canceling out the effect of the other. Thus event horizons repel each other.

 

[PeterDonis]

No matter how much mass you add to the black holes, their event horizons can never reach the probe because each black hole is canceling out the effect of the other.

No, this is not correct. I explained why in previous posts. The only additional comment I have is that this...

Being in both gravitational fields is the same as being in neither of them.

...is also incorrect: if the observer exactly halfway between the two identical planets, or stars, or black holes, tried to send light signals out to infinity, an observer at infinity would see them getting more and more redshifted as the mass of the gravitating objects grew, until eventually (when the observer halfway between was just crossing the merged event horizon of the single black hole formed from the two original bodies), the observer at infinity would see the light signals disappear altogether. (I explained this also in previous posts, but I thought it was worth repeating since the incorrect statement I quoted above appears to be a crucial part of your argument.)

 

[Algr]

Your text with the "pair of trousers" describes the two event horizons merging, but does not explain why they would behave that way.

Suppose the black holes has a mass such that in the absence of other objects, their event horizons would be one million miles radius. The two radii are one foot apart at their closest points. Is there really a gravitational equation that would show gravity as inescapable at a million miles away, but zero six inches further? I'm sure gravity can't bend around corners like that when so far from any physical object.

=====================
Concerning two black holes merging.

Remember you said: "An event horizon isn't a thing that can get "pushed". It's the boundary of a region of spacetime.."

Even if two event horizons merge as you say, the horizons are still just empty space. No actual matter from one black hole has entered the area where the other event horizon would be. So there is no reason for the two imploding stars to merge - they can still orbit at sub-light speeds. (Yes the singularity is in the future, but the star was in the past, so there must be something in the present.) And if the speed of the orbit were increased, the two event horizons could even separate - since no actual matter is actually escaping any horizon.

=====================
Could you go over your "Apparent" and "Actual" horizons again? I think that one of these is the horizon that I am talking about, but I don't follow why there is a different one.

 

[Chronos]

The problem with your scenario, Algr, is the region of null gravity between two black holes always lies outside the event horizon of both black holes. Once the event horizons merge, the very concept of distance between the putative singularities loses any sensible meaning.

 

[Algr]

But that is my point. The horizons DON'T merge because the region of null gravity between two black holes always lies outside the event horizon.

 

[PeterDonis]

Your text with the "pair of trousers" describes the two event horizons merging, but does not explain why they would behave that way.

They behave that way because that's how the Einstein Field Equation says they behave. Strictly speaking, we would need to do a detailed numerical solution to get the exact shape of the "pair of trousers"; there is no closed-form analytical solution that describes this scenario the way the Schwarzschild metric describes a single black hole. What I'm describing is simply a simplified, heuristic version of what you would get if you did the detailed numerical solution.

If that doesn't count as "explaining why", then you haven't explained why the light cones tilt in the single black hole case either, because that tilting comes from solving the Einstein Field Equation too.

Suppose the black holes has a mass such that in the absence of other objects, their event horizons would be one million miles radius. The two radii are one foot apart at their closest points.

You are continuing to make the mistake of thinking of the horizon as a "place". It isn't; it's an outgoing null surface, meaning it's a surface formed by radially outgoing light rays. Light rays can never "stay in the same place". Don't be misled by the fact that the horizon, at least for a single black hole, has a constant $r$ coordinate; the $r$ coordinate is not the same as "where" the horizon is--the latter concept has no meaning. (Outside the horizon, $r$ can be interpreted as "where" something is, but you still have to take into account that "space" in this chart is non-Euclidean.)

Also, the event horizon does not have a radius; it has a circumference and an area. The $r$ coordinate is sometimes referred to as the "horizon radius", but that's sloppy terminology; $r$ is really the circumference of the horizon divided by $2 \pi$, or the square root of the area divided by $4 \pi$. (Note that this is for a single black hole; the relationships are more complicated for the case of merging black holes that we're discussing, but the fundamental fact remains that the horizon does not have a radius.) There is no such thing as a "distance" from the horizon to the singularity; if you're at the horizon, the singularity is in the future, not at a different place in space.

Is there really a gravitational equation that would show gravity as inescapable at a million miles away, but zero six inches further?

No. To the extent that your picture of the situation makes sense (which is problematic to begin with given the caveats I stated above), the equation shows that gravity is inescapable at each horizon, and almost inescapable at the midpoint between. The fact that there is no "acceleration due to gravity" at that midpoint does not mean it is just like empty space at infinity; there is still a very large gravitational redshift for any light emitted outwards towards infinity at that point. You continue to make this mistake as well--thinking that "zero acceleration" means "zero redshift"--even though it's been pointed out to you several times.

Could you go over your "Apparent" and "Actual" horizons again? I think that one of these is the horizon that I am talking about, but I don't follow why there is a different one.

The apparent horizon is where radially outgoing light no longer moves outward; if a flash of light is emitted spherically in all directions radially outward from the apparent horizon, the area of the sphere does not increase with time--it stays the same.

The absolute horizon is the boundary of the region that cannot send light signals to infinity (this region is the black hole). For a single black hole whose mass stays the same forever (i.e., nothing ever falls in), the absolute horizon coincides with the apparent horizon. But for more complicated cases, where things fall into a single hole or two holes merge, that's not the case. For the case we're discussing, of a black hole merger, the absolute horizon will, in general, be outside the apparent horizon: that is, light emitted outward towards infinity from the absolute horizon might still move outward when it is emitted. But sooner or later that light, if it's emitted from the absolute horizon, will get caught and pulled back by the gravity of the merging (or merged) black holes--it will never reach infinity.

One other thing to keep in mind is that a spacetime with two black holes merging is not spherically symmetric, so defining the "outward" direction--the direction you need to emit light for it to have a chance of reaching infinity--requires some thought for the region in between the holes, which is the region you're concentrating on. If you think of the "pair of trousers", with time vertical, and concentrate on the region in between the legs, obviously light emitted to the left or right won't escape--it will fall into one of the holes. Light emitted "towards" you (meaning perpendicular to the line connecting the two merging holes) has a chance of escaping: but if the holes are close enough together, it still might not, because it can't get out of the region between the holes fast enough--their combined gravity pulls it back. This region is where the "legs" of the trousers merge. All of this can't be described by a simple diagram showing "space", as you tried to do before--you have to include time, and you have to include at least two "space" dimensions as well, because of the lack of spherical symmetry. In short, the intuitions you may have from considering the single black hole case, which is spherically symmetric, might lead you astray in this case.

The horizons DON'T merge because the region of null gravity between two black holes always lies outside the event horizon.

No, it doesn't. If you're going to just keep re-stating your incorrect statements without even addressing the responses that have been made to them, there's no point in continuing discussion. Do you have any response to the multiple posts I and others have made showing why this claim of yours is incorrect?

 

[PeterDonis]

the region of null gravity between two black holes always lies outside the event horizon of both black holes.

I don't think this is correct--at least, not with the meaning Algr is putting on "the region of null gravity", in terms of light cones tipping. The light cones along that particular worldline--the worldline of an observer who sits in free fall exactly between the two identical black holes as they merge--never "tip" at all; they eventually end up pointing inside the merged horizon even though their "direction" never changes.