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.)