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decafdave
2012-Jan-08, 08:59 PM
Hello everyone,

I found out about this forum from the astronomy cast program. I've heard speculation about what occurs when 2 singularities merge-gravity waves and 2 black holes becoming one. But what happens to the space between the 2 singularities when the 2 event horizons intersect?

If our spacecraft passes the point of no return in supermassive black hole A, it might take hours to actually reach the singularity. However, if during this time a large supermassive black hole B starts to collide with A, and the spacecraft is caught in 2 event horizons, will it leave the weaker one? Has anyone done the math on this? I thought that it would violate the laws of physics to leave an event horizon, regardless if it's a stronger black hole pulling it out.

Shaula
2012-Jan-08, 10:36 PM
There isn't really such a thing as a weaker event horizon. The event horizon is just a line in space at which things become causally disconnected from the rest of the universe. So as two black holes get near each other a composite event horizon forms. It is not correct to say that an object 'left' one and entered another as the composite event horizon is the only event horizon of the system.

antoniseb
2012-Jan-08, 10:48 PM
... what happens to the space between the 2 singularities when the 2 event horizons intersect? ...

This question comes up now and again, and the answer appears in several other threads, but making you do research as the first thing you do on arriving seems cruel. Here's something that will help make sense of it almost right away for you:

Imagine BH1 with a mass of 106 solar masses, and another one with a mass of 4x106 solar masses. They have event horizons with radii of 3 and 12 million kilometers. They will form a BH with a radius of 15 million kilometers... so really, by the time the old view of the horizons touch, the smaller black hole center is already inside the new combined event horizon, so there is no place with an opportunity for escape.

WayneFrancis
2012-Jan-10, 03:08 PM
Hello everyone,

I found out about this forum from the astronomy cast program. I've heard speculation about what occurs when 2 singularities merge-gravity waves and 2 black holes becoming one. But what happens to the space between the 2 singularities when the 2 event horizons intersect?

If our spacecraft passes the point of no return in supermassive black hole A, it might take hours to actually reach the singularity. However, if during this time a large supermassive black hole B starts to collide with A, and the spacecraft is caught in 2 event horizons, will it leave the weaker one? Has anyone done the math on this? I thought that it would violate the laws of physics to leave an event horizon, regardless if it's a stronger black hole pulling it out.

The maths will have you hit the singularity of the black hole who's event horizon you passed through first. From your proper time you wouldn't even know about the 2nd black hole irrespective of the 2nd black holes size. Once you pass a black hole's event horizon the only thing in your future is that black hole's singularity.

Actually "passed through first" is a bad phrase since it leaves open the possibility of passing through some other black hole's event horizon "second". A better way of saying what will happen is that you'll hit the singularity of the black hole whose event horizon you passed through.

hypergreatthing
2012-Jan-10, 05:44 PM
This question comes up now and again, and the answer appears in several other threads, but making you do research as the first thing you do on arriving seems cruel. Here's something that will help make sense of it almost right away for you:

Imagine BH1 with a mass of 106 solar masses, and another one with a mass of 4x106 solar masses. They have event horizons with radii of 3 and 12 million kilometers. They will form a BH with a radius of 15 million kilometers... so really, by the time the old view of the horizons touch, the smaller black hole center is already inside the new combined event horizon, so there is no place with an opportunity for escape.
I'm not an astrophysicist, but I've always thought that physics cannot describe the space inside a black hole because they operate outside of our understanding of physics. So in your example, is that what is main stream theorized or has that been actually observed?

Jeff Root
2012-Jan-10, 08:53 PM
The space inside the event horizon of a black hole is described
by general relativity just fine. What cannot be described by any
known physics is the nature of the matter at the singularity, at
the very center of the black hole.

If two black holes near each other, their event horizons reach
out toward each other and merge, becoming a single horizon
surrounding both singularities. The singularities continue to
fall together until they become singular!

The horizons do not "intersect" so much as merge.

Remember that an event horizon is only a location in space:
It is the location where the gravitational field has a particular
strength, or the location where spacetime has a particular
amount of curvature. Inside the horizon, the gravitational
field strength / spacetime curvature increases as you get
closer to the singularity. When two black holes near each
other, the region within which the gravitational field has a
particular strength increases in size. So the event horizons
move farther out, and merge into one.

If an observer falls past the event horizon of one black hole
while another black hole is nearing, the observer will first fall
to the singularity of the first black hole, then the singularities
fall together to become a single singularity.

If the observer falls past the event horizon right after the
horizons of the two black holes have merged, then the two
singularities will fall together and merge before the observer
falls into the resulting singularity.

-- Jeff, in Minneapolis

antoniseb
2012-Jan-10, 10:06 PM
I'm not an astrophysicist, but I've always thought that physics cannot describe the space inside a black hole because they operate outside of our understanding of physics. So in your example, is that what is main stream theorized or has that been actually observed?

No work so far has directly observed and measured the size of an event horizon, though some secondary features of black hole physics have been measured indirectly, such as the innermost stable circular orbit (ISCO). This is the case for black holes just sitting there waiting to be observed. The extremely rare events of two black holes merging has never been seen with anything that can measure the size and shapes of the event horizons.

That being said, in the next several years we should be able to observe the Giant black holes in the center of the Milky Way, and in M87 with enough detail to measure the size of various features, and confirm (or not) General Relativity's description of these environments. There is also some possibility that in the next twenty to fifty years, we will be able to make observations of Gravitational Waves associated with short duration GRBs that might tell us some details about the changing shapes of the event horizons.

So, to your point, yes. The statements made in this thread are all based on the extrapolation of model that works very well elsewhere, and not on direct measurement.

caveman1917
2012-Jan-11, 12:46 AM
There isn't really such a thing as a weaker event horizon. The event horizon is just a line in space at which things become causally disconnected from the rest of the universe. So as two black holes get near each other a composite event horizon forms. It is not correct to say that an object 'left' one and entered another as the composite event horizon is the only event horizon of the system.

Going a bit deeper into this, there is actually only one event horizon all the time. It's just because we see consecutive spatial-slices that we consider them to be two "different" event horizons at the start and a single "combined" event horizon at the end. But remember that we are working in a spacetime background here, and event horizons are surfaces in spacetime (not just space). If you look at it that way, the entire thing will resemble something like a pair of trousers. With time going upwards, if you take horizontal slices, first you will have two seperate pipes (circles) and more upwards the pipes will join in one big pipe/circle. Yet you still say it's just one single pair of trousers all along.

Jeff Root
2012-Jan-11, 02:53 AM
I understand what you just said, caveman, but it does not
explain anything relevant to the original question. It is just
an interesting way of describing change. It doesn't tell us
anything about black holes or their event horizons. Its only
feature is that it sounds astonishing.

The two seemingly separate glass bottles I have here, which
I acquired in different grocery stores at different times, are
actually one piece of glass. Someday I'm going to put them
in the recycling, and they will go into a furnace and be melted
to become part of a single glass object.

Logical, but not sensible, and not useful. It doesn't apply to
black holes any more than it applies to glass bottles.

-- Jeff, in Minneapolis

m74z00219
2012-Jan-11, 04:12 AM
Hello everyone,

I found out about this forum from the astronomy cast program. I've heard speculation about what occurs when 2 singularities merge-gravity waves and 2 black holes becoming one. But what happens to the space between the 2 singularities when the 2 event horizons intersect?

If our spacecraft passes the point of no return in supermassive black hole A, it might take hours to actually reach the singularity. However, if during this time a large supermassive black hole B starts to collide with A, and the spacecraft is caught in 2 event horizons, will it leave the weaker one? Has anyone done the math on this? I thought that it would violate the laws of physics to leave an event horizon, regardless if it's a stronger black hole pulling it out.

Hi decafdave,

Recently, Physics Today had a great article on the solutions of the equations of General Relativity that focused on black hole mergers. If you're a student, I highly recommend you join APS: for one, they give you a year long subscription to PT.

M74

Tensor
2012-Jan-11, 04:30 AM
I understand what you just said, caveman, but it does not
explain anything relevant to the original question. It is just
an interesting way of describing change. It doesn't tell us
anything about black holes or their event horizons. Its only
feature is that it sounds astonishing.

It sure does explain it Jeff, As soon as the individual event horizons touch, there is then one event horizon. Both of the previous horizons are now within the new one. In caveman's analogy, the two round legs of the trousers end up as the round waist of the trousers. The smaller legs become the larger waist.

Selfsim
2012-Jan-11, 05:06 AM
Remember that an event horizon is only a location in space:
It is the location where the gravitational field has a particular
strength, or the location where spacetime has a particular
amount of curvature. Inside the horizon, the gravitational
field strength / spacetime curvature increases as you get
closer to the singularity. When two black holes near each
other, the region within which the gravitational field has a
particular strength increases in size. So the event horizons
move farther out, and merge into one.

If an observer falls past the event horizon of one black hole
while another black hole is nearing, the observer will first fall
to the singularity of the first black hole, then the singularities
fall together to become a single singularity.

If the observer falls past the event horizon right after the
horizons of the two black holes have merged, then the two
singularities will fall together and merge before the observer
falls into the resulting singularity.
Hi Jeff;
I was under the impression that the Event Horizon, (EH), is independent of space-time distortion in the merger process ? (Ie: one can create an EH by accelerating a clock in flat (Minkowski) spacetime. In math terms, it is the time component of the Schwarzchild metric ?)

Isn't the distortion of the EHs, (to become one), a property of angular momentum .. not gravity (spacetime distortion) ?

Also, is the 'observer' you mention from the spacecraft's perspective, or an observer of the infalling spacecraft ? :confused:

Regards

Jeff Root
2012-Jan-11, 05:07 AM
It doesn't explain anything.

As soon as the glass from the two bottles fuse, there is then
one piece of glass. Both of the previous bottles are now within
the new one. The analogy of the trousers applies to the bottles
no less than to black holes or their event horizons, but it doesn't
do anything useful in either case. It doesn't tell us anything new.
Two separate bottles join into one piece of glass. Two separate
event horizons join into one. It was one single event horizon
all along. It was one single piece of glass all along. Bleah.

-- Jeff, in Minneapolis

Jeff Root
2012-Jan-11, 05:32 AM
Remember that an event horizon is only a location in space:
It is the location where the gravitational field has a particular
strength, or the location where spacetime has a particular
amount of curvature. Inside the horizon, the gravitational
field strength / spacetime curvature increases as you get
closer to the singularity. When two black holes near each
other, the region within which the gravitational field has a
particular strength increases in size. So the event horizons
move farther out, and merge into one.

If an observer falls past the event horizon of one black hole
while another black hole is nearing, the observer will first fall
to the singularity of the first black hole, then the singularities
fall together to become a single singularity.

If the observer falls past the event horizon right after the
horizons of the two black holes have merged, then the two
singularities will fall together and merge before the observer
falls into the resulting singularity.
I was under the impression that the Event Horizon, (EH), is
independent of space-time distortion in the merger process ?
(Ie: one can create an EH by accelerating a clock in flat
(Minkowski) spacetime. In math terms, it is the time
component of the Schwarzchild metric ?)
I would say that the event horizon is a product of the
space-time distortion. So very much not independent of it.

Off the top of my head, it seems to me that in order to
create something resembling an event horizon in flat
spacetime by acceleration, one would have to accelerate
forever. Otherwise the "event horizon" so created would
only be temporary. It would also only exist for the observer
who was accelerating. The event horizon of a black hole
exists for *all* observers.

Isn't the distortion of the EHs, (to become one), a property
of angular momentum .. not gravity (spacetime distortion) ?
I've never heard anything like that. There is a nonspherical
"ergosphere" surrounding a rotating black hole, but as I
understand it (and somewhat surprisingly) the event horizon
of even a rotating black hole is spherical.

The reason the event horizons of two approaching black
holes "reach out to each other" is that the distortion is a
result of the amount of matter nearby, and just how near
that matter is. Between two black holes that are not far
apart, there is a lot of matter on both sides of you, close
by, so you may already be inside the event horizon of the
nearer one without realizing it, if you didn't know a second
black hole was right behind you. You're doomed!

Also, is the 'observer' you mention from the spacecraft's
perspective, or an observer of the infalling spacecraft ? :confused:
My observer is the same as the spacecraft of the original post.
It really shouldn't make any difference, though, since I made
clear that the events were happening to the observer.

-- Jeff, in Minneapolis

WayneFrancis
2012-Jan-11, 05:34 AM
I'm not an astrophysicist, but I've always thought that physics cannot describe the space inside a black hole because they operate outside of our understanding of physics. So in your example, is that what is main stream theorized or has that been actually observed?

Physics can not describe what happens at the singularity. Inside the event horizon is not a problem.

WayneFrancis
2012-Jan-11, 05:38 AM
Going a bit deeper into this, there is actually only one event horizon all the time. It's just because we see consecutive spatial-slices that we consider them to be two "different" event horizons at the start and a single "combined" event horizon at the end. But remember that we are working in a spacetime background here, and event horizons are surfaces in spacetime (not just space). If you look at it that way, the entire thing will resemble something like a pair of trousers. With time going upwards, if you take horizontal slices, first you will have two seperate pipes (circles) and more upwards the pipes will join in one big pipe/circle. Yet you still say it's just one single pair of trousers all along.

Great analogy.

WayneFrancis
2012-Jan-11, 05:43 AM
I understand what you just said, caveman, but it does not
explain anything relevant to the original question. It is just
an interesting way of describing change. It doesn't tell us
anything about black holes or their event horizons. Its only
feature is that it sounds astonishing.

The two seemingly separate glass bottles I have here, which
I acquired in different grocery stores at different times, are
actually one piece of glass. Someday I'm going to put them
in the recycling, and they will go into a furnace and be melted
to become part of a single glass object.

Logical, but not sensible, and not useful. It doesn't apply to
black holes any more than it applies to glass bottles.

-- Jeff, in Minneapolis

There you had to poke at the analogy to much. I do think it applys to the OP because the OP is talking about 2 merging black holes. And if you can warp your head around the 3+1 dimensions of space time EH's are like he describes but you are right that as far as someone falling in...well let me use the pants. If you put a needle through the fabric in a pair of pants in the leg then you are sewing up the leg. if you put it in the butt then well that is "after" the 2 separate legs and you are stitching up the butt. Ok that didn't work out as well as I thought :)

Jeff Root
2012-Jan-11, 05:56 AM
The analogy of the pants is fine. It just doesn't cover anything
that needs covering.

-- Jeff, in Minneapolis

Selfsim
2012-Jan-11, 06:24 AM
Thanks Jeff .. much appreciated.

The event horizon of a black hole
exists for *all* observers.

Does an Event Horizon 'exist' from the BHs frame of reference ?

Regards

WayneFrancis
2012-Jan-11, 06:38 AM
Thanks Jeff .. much appreciated.

Does an Event Horizon 'exist' from the BHs frame of reference ?

Regards

Do you mean the singularity's frame of reference?

All external observers agree to the location of the EH. Even something that has dropped through the event horizon can say when that happened for them. What they can not do is point you to where that EH actually is any more. Once past the EH all paths lead to the singularity. The EH is no longer in their future light cone. To be able to say where something is in relation to you I'd say requires you to have it within your future light cone. Since the EH is and never will be in your future light cone once you pass through it then it is "no where" to you.

Jeff Root
2012-Jan-11, 06:55 AM
Very good answer Wayne. Thank you!

-- Jeff, in Minneapolis

Selfsim
2012-Jan-11, 07:06 AM
Hi Wayne !

Do you mean the singularity's frame of reference?
Yes.

I'm not sure that I can see any physical significance, (ie: 'exists'), from the singularity's frame of reference (FOR) .. ie: I don't think we can say that it 'exists' from this FOR. (Sure .. it exists from an observer's FOR ..)

Also, if two BHs merge, would they rotate ?

If so, then how does their angular momentum effect the EHs ?
How can spacetime distortion increase the distortion of the EHs, (as the BHs merge), if the EH doesn't exist (physically) from the singularity's FOR ?

Regards

Jeff Root
2012-Jan-11, 07:48 AM
I'd say the singularity doesn't have a frame of reference.
It is too weird. Too extreme.

For any falling observer inside the event horizon who has not
yet reached the singularity, I agree with Wayne: the horizon is
in their past. Any change in the horizon will not affect them,
just as an explosion in the core of the Milky Way five years
ago could never have any effect on me. I'd be long gone by
the time the blast wave reached Earth.

But an infaller could track his fall and know when he had
crossed the event horizon, and know that no message from
him after that instant will ever reach his friends on the
outside. The event horizon doesn't have quite the same
significance for him as it does for outside observers, but
its significance is at least as great.

Two colliding black holes would undoubtedly result in a
rotating merged black hole. I don't see any reason why
they couldn't collide exactly head-on, but that just isn't
reasonably likely.

You might think the collision would have to be precise
to a ridiculous number of significant digits in order to be
exactly head-on enough not to result in a large rotation,
but I suspect that the kinetic energy is radiated away so
strongly that the angle of the spiral actually steepens
after the event horizons merge.

I'm not the person to propose such an advanced idea, but
I'll wonder anyway: Could the energy of rotation always be
radiated away completely, so that all black holes end up with
zero rotation? Doesn't seem likely....

As I said, the rotation of a BH does not appear to affect the
event horizon, to the best of my understanding. It does drag
the space around it, though, which is what the "ergosphere"

A dense concentration of matter is the cause of a black hole.
It distorts the space it is in. The amount of matter determines
the degree of distortion and thus the size of the event horizon.
The fact that the matter can never "know" that the horizon
exists isn't relevant.

-- Jeff, in Minneapolis

Selfsim
2012-Jan-11, 08:31 AM
If gravity distorts the event horizon, then how could one define the Schwarzschild radius, (ie: the radius from the center where the escape speed from the surface is equal to the speed of light), for a low mass black hole ? .. (as the event horizon would be well within the tidal forces of the BH, and is therefore distorted ?)

Cheers

Jeff Root
2012-Jan-11, 09:04 AM
The Schwarzschild radius is the location of the event horizon
of an isolated black hole. Bringing in a second large mass
nearby, such as a second black hole, means that the location
of the event horizon no longer has a simple description, and
must be modelled by quite complex computer programs.

When you bring up "tidal forces", are you talking about the
force from a second mass, perhaps a second black hole?
If you are talking about a solitary black hole, then your
question seems very odd. The event horizon is a result of
the distortion in spacetime caused by the presence of the
large mass. Naturally there is a tidal force at the horizon,
and it is, as you appear to imply, stronger for low-mass
black holes. The strength of the tidal force has no effect
on the horizon.

-- Jeff, in Minneapolis

Selfsim
2012-Jan-11, 10:26 AM
The strength of the tidal force has no effect
on the horizon.

Ok .. (agreed). So, by the same reasoning, 'tidal forces', (presumably we're talking about gravitationally induced, spacetime distortionally causing tidal forces), should also not have any effect on the topology of the combined EHs, as the BHs merge ?

Whereas, I think originally you mentioned:

I would say that the event horizon is a product of the
space-time distortion. So very much not independent of it.

So, it would seem that the EHs are caused by something other than spacetime distortion ?

If you are talking about a solitary black hole, then your
question seems very odd. The event horizon is a result of
the distortion in spacetime caused by the presence of the
large mass.
So this might imply that our Sun would behave in the same way as a one solar mass black hole ? What happens for BHs smaller than one solar mass ?
:confused:
Regards

WayneFrancis
2012-Jan-11, 04:41 PM
If gravity distorts the event horizon, then how could one define the Schwarzschild radius, (ie: the radius from the center where the escape speed from the surface is equal to the speed of light), for a low mass black hole ? .. (as the event horizon would be well within the tidal forces of the BH, and is therefore distorted ?)

Cheers

The EH is always given in terms of Schwarzschild coordinates. It takes into account to the curvature of space. It does not work inside the EH because at that point the polar coordinate system breaks down do to the extreme warping of space. I'm not wording this well but its not an easy concept.

Essentially if you take a flat Minkowski space time you can use normal Euclidean geometry to accurately calculate the volume of a sphere. If you introduce mass/energy into that space time then the calculation becomes non Euclidean because of the amount of warping of space time as you get closer to the mass/energy. If a black hole's singularity is a true singularity then any black hole of any size actually has an infinite volume within its event horizon. The issue of the EH growing proportionally to the mass/energy inside of it is interesting but doesn't change the fact that you could dump any amount of energy into that original volume in some respects. In others you can say "no, you can't" because of the way the EH works. Anyway I'm getting off track here. The metric used is unsurprisingly in the name. Its a metric that doesn't have the issue you discuss. This is why we use the correct metric for the question being asked. So we don't run into problems that a different metric may have with a given situation.

WayneFrancis
2012-Jan-11, 04:54 PM
Ok .. (agreed). So, by the same reasoning, 'tidal forces', (presumably we're talking about gravitationally induced, spacetime distortionally causing tidal forces), should also not have any effect on the topology of the combined EHs, as the BHs merge ?

Whereas, I think originally you mentioned:

So, it would seem that the EHs are caused by something other than spacetime distortion ?

So this might imply that our Sun would behave in the same way as a one solar mass black hole ? What happens for BHs smaller than one solar mass ?
:confused:
Regards

behave in what way? To the Earth...our sun behaves in the same way as a 1 solar mass black hole with regard to gravity at our distance. This is roughly true in the same way as someone inside an elevator can't tell if they are accelerating in deep space or standing still in the gravitational field of the Earth. Given good enough equipment you can tell which is which just by the tidal forces. This is because there is no true uniform gravitational field. It is just the shorter the distance and the shorter the time period involved the harder it is to measure the different scenarios.

But as Jeff points out the Schwarzschild metric is an idealised solution. Start introducing other bodies into the metric and you'll start getting weird results because that metric wasn't meant for that. You can still get some approximations but the more extreme the situation the less accurate the results will be. This is why we turn to modelling and use computers. Because no one has invented/discovered a metric that = reality in all cases. What we have are different metrics that are more applicable in some situations then in others. If you come up with a model that is exactly like reality then...well is it a model or is it just reality :)

caveman1917
2012-Jan-12, 12:14 AM
I understand what you just said, caveman, but it does not
explain anything relevant to the original question.
{...}
It doesn't apply to
black holes any more than it applies to glass bottles.

It certainly does, though perhaps i didn't explain fully.

Remember the definition of the event horizon, it is the boundary of the set of points in the causal past of future null infinity. Since we usually consider a 4d spacetime, the boundary of a subset of that spacetime will be 3d. In other words, the event horizon is a 3d surface, not a 2d surface as you (and the OP) seem to be considering it. There is only one black hole region (\mathcal{B} = \mathbb{M} \backslash I^-(\mathcal{J^+}) ) and only one event horizon (\mathcal{\epsilon} = \partial \mathcal{B}) in the universe. When we take constant-time slices this black hole region (and event horizon) may be a disjoint union at some slices, each of which we would call a "black hole" (but note that we are somewhat free in how exactly we slice a spacetime, the proper view is to consider the geometrical properties of the entire spacetime).

My point is that it makes no sense to ask "which" event horizon something went through, as there is only one. The entire trousers is the (only) event horizon. When you stick a needle through you wouldn't ask which trousers you stuck the needle through, there is only one.

It certainly applies more to event horizons than glass bottles since this is how the event horizon is defined.

* ETA: \mathbb{M} is the spacetime, I^- denotes "the past of" and \mathcal{J^+} is future null infinity.

caveman1917
2012-Jan-12, 12:38 AM
I was under the impression that the Event Horizon, (EH), is independent of space-time distortion in the merger process ? (Ie: one can create an EH by accelerating a clock in flat (Minkowski) spacetime. In math terms, it is the time component of the Schwarzchild metric ?)

Confusingly, there are several types of event horizons. The one you can create by accelerating in minkowski space is called an apparent horizon. Suppose you have some constant acceleration a towards the right on a line. Let's use coordinates where you are stationary (you carry the origin of the coordinates "with you"), so staying at a stationary place means you're accelerating at a. In relativity something funny comes up that doesn't happen in newtonian physics, someone that is some distance behind you must accelerate harder in order to keep that same distance from you (ie remain stationary in those accelerating coordinates). This is because of length contraction/time dilation, see for example bell's spaceship paradox for more information on this.

Now it turns out that if you look further and further to your back, other people must accelerate harder and harder to keep up, and this acceleration required becomes infinite at some distance behind you. That is where an apparent horizon forms. This situation is not all that different from a black hole, where the same thing goes on. If you hover above a black hole you must constantly accelerate (keep your rocket engines firing) in order to remain stationary and not fall in. If you try to do this closer and closer to the black hole, you must accelerate harder and harder for this, and at the apparent horizon of the black hole this acceleration required becomes infinite.

This is however not the type of event horizon that defines the black hole, that would be the absolute horizon. This horizon is defined as that boundary beyond which you cannot send signals to the outside world anymore. For a quiet stationary black hole they are at the same position, so we can use either as a subsitute for the other, but in dynamic cases (such as mergers) these will not coincide. An absolute horizon is only there with a black hole, accelerating in minkowski only gets you an apparent horizon, not an absolute one. There is however no way to locate an absolute horizon, only apparent horizons can be located (by checking the acceleration required to stay stationary at different places), so we could even be inside one without knowing (and without ever knowing).

The absolute horizon is a function of the spacetime distortion in the merger process, it is that horizon that "stretches out" to meet its counterpart. The apparent horizon is not.

caveman1917
2012-Jan-12, 12:48 AM
If gravity distorts the event horizon, then how could one define the Schwarzschild radius, (ie: the radius from the center where the escape speed from the surface is equal to the speed of light), for a low mass black hole ? .. (as the event horizon would be well within the tidal forces of the BH, and is therefore distorted ?)

The event horizon (well, going by my previous argument i should be strict and say "a spatial slice of the event horizon" :)) is spherical, and thus the tidal force is equal on all points of it and it won't deform.

But to answer the first question, the schwarzschild radius is defined as the circumference of the event horizon divided by 2\pi. This may seem a bit odd since usually we would define the circumference of a circle by the radius, not the other way around. But consider the following, in a very real sense you could consider the event horizon as the "boundary" of "our" universe. So how would you extend your notions of length outside the universe? The answer is that you can't do that unambiguously, so we must define the schwarzschild radius by the circumference of the horizon (which is the last bit that's "accessible" to us).

caveman1917
2012-Jan-12, 12:59 AM
I'm not the person to propose such an advanced idea, but
I'll wonder anyway: Could the energy of rotation always be
radiated away completely, so that all black holes end up with
zero rotation? Doesn't seem likely....

Angular momentum cannot be "radiated away" from a black hole, although the penrose process allows a way to extract angular momentum from a black hole. But this is a very specific process, and can work just as well the opposite way, so it doesn't seem likely indeed.

As I said, the rotation of a BH does not appear to affect the
event horizon, to the best of my understanding.

It does, it makes it smaller (still spherical though) the faster the black hole rotates. In the limiting case where you let the rotation approach the "critical" value, the size of the event horizon goes to zero [correction: the size goes to r_s/2 at which point it disappears] and you get a naked singularity (not that it is physically possible to spin up a black hole to the critical value).

Selfsim
2012-Jan-12, 01:34 AM
Hi caveman1917 … pleased to meet you ! :)

Thank you for the concepts in your previous posts .. I'll have a ponder for a while, to absorb them in the context of the OP topic.

Having thought about it, and had a few conversations with others, I think the phenomenological observation which uniquely characterises the EH we were talking about originally, (ie: defined by the Schwarzschild radius), is gravitational redshift.

An observer sees an infalling spacecraft 'frozen' at the EH. The measurement is extreme redshift. Redshift is derived from the Schwarzschild metric specifically excluding the spatial components of spacetime. So, when the observer measures this redshift, it is completely explained by the derivation which excludes the spatial components. This therefore demonstrates that the EH is independent of the spatial geometry of spacetime (???).

(The above is a question, really … feedback/comments welcomed).
:)
Best Regards .. and thanks again for the great explanations ! :)

caveman1917
2012-Jan-12, 02:04 AM
Hi caveman1917 … pleased to meet you ! :)

Nice to meet you too.

Having thought about it, and had a few conversations with others, I think the phenomenological observation which uniquely characterises the EH we were talking about originally, (ie: defined by the Schwarzschild radius), is gravitational redshift.

That's the apparent horizon. You can of course consider the apparent horizon if you want, but consider a few things. This horizon doesn't "stretch out" in a merger, nor do many other things you usually consider to be a "black hole's behaviour". In short, this horizon is not what is meant by the event horizon of a black hole, so if you don't specify you'd talk about the apparent horizon there can be misunderstandings. It has everything to do with coordinates and nothing with the actual properties of spacetime, you can make it go away with a simple coordinate change.

An observer sees an infalling spacecraft 'frozen' at the EH. The measurement is extreme redshift. Redshift is derived from the Schwarzschild metric specifically excluding the spatial components of spacetime. So, when the observer measures this redshift, it is completely explained by the derivation which excludes the spatial components. This therefore demonstrates that the EH is independent of the spatial geometry of spacetime (???).

Yes (well, other than its location of course :)). In fact very near to the horizon you can approximate the schwarzschild metric with rindler coordinates (those of that accelerating clock in minkowski space) which have simple euclidean spatial components.

But remember this doesn't say anything about black holes per se, only about schwarzschild coordinates.

Tensor
2012-Jan-12, 06:35 AM
This horizon doesn't "stretch out" in a merger, nor do many other things you usually consider to be a "black hole's behaviour". In short, this horizon is not what is meant by the event horizon of a black hole, so if you don't specify you'd talk about the apparent horizon there can be misunderstandings.

Hey, just to bounce something off of you, would Thorne's hoop conjecture (C ≤ 4πM) be something to consider when talking about merging black holes?

whimsyfree
2012-Jan-12, 11:13 AM
Physics can not describe what happens at the singularity. Inside the event horizon is not a problem.

Have these predictions been tested by observation?

WayneFrancis
2012-Jan-12, 02:42 PM
Have these predictions been tested by observation?

Are you asking if we've observed in side of an event horizon? If you are the answer is "no" and if you ever do...you won't be able to tell anyone about it anyway.

This issue is that we have no reason to believe that anything drastic happens in the the reference frame of a free falling observer as they pass through an event horizon. We've never "observed" the centre of a star but we can work out what is going on pretty well.

You've probably heard that you'll get spaghettified when you get close to a black hole but in truth it depends on the size of the black hole. The bigger the black hole the closer you can get. In fact with really big black holes you can get well inside the event horizon before feeling those tidal forces. Modern physics doesn't have a problem describing the things until you get very close to the singularity....even if you'd be ripped apart before then.

Grey
2012-Jan-12, 02:59 PM
Physics can not describe what happens at the singularity. Inside the event horizon is not a problem.

Have these predictions been tested by observation?No, and if our predictions about black holes are correct, they never can be. That is, anything that crosses the event horizon can never have an influence on anything outside, so even if we had a convenient nearby black hole to do any experiment we wanted on, we'd never be able to test what happens inside. Well, at least we'd never be able to test that and then report back to anyone who was still outside the black hole.

That said, observations of things happening outside of a black hole seem to match very well with our predictions based on those same theoretical models. So that gives them some credibility, at least.

Grey
2012-Jan-12, 03:09 PM
But to answer the first question, the schwarzschild radius is defined as the circumference of the event horizon divided by 2\pi. This may seem a bit odd since usually we would define the circumference of a circle by the radius, not the other way around.Heck, even if you don't cross the event horizon, measurements near a black hole won't necessarily give you the results you expect. For example, if you were to measure* the circumferences of two concentric circles centered on a black hole, one of which is 1 meter further out, you'd expect that the circumference of the larger one should be 2\pi meters larger, but that is not actually the case.

* Actual measurement procedure left as an exercise for the reader. ;)

Tensor
2012-Jan-12, 03:13 PM
* Actual measurement procedure left as an exercise for the reader. ;)

I had it done, but the sxihuud ate my homework.

caveman1917
2012-Jan-12, 04:14 PM
Hey, just to bounce something off of you, would Thorne's hoop conjecture (C ≤ 4πM) be something to consider when talking about merging black holes?

I'm not sure. The problem with the hoop conjecture is that it is quite ambiguous and, under the usual extra assumptions (ie M is ADM mass and the black hole that forms is to be taken as a region inside a marginally trapped null surface), has only been proven (as far as i know) in the spherically symmetric case (though there may be results in the Kerr-Newman case). There exist extra problems when trying to prove it in more general mass distributions.

caveman1917
2012-Jan-12, 05:26 PM
Heck, even if you don't cross the event horizon, measurements near a black hole won't necessarily give you the results you expect. For example, if you were to measure* the circumferences of two concentric circles centered on a black hole, one of which is 1 meter further out, you'd expect that the circumference of the larger one should be 2\pi meters larger, but that is not actually the case.

* Actual measurement procedure left as an exercise for the reader. ;)

True, but this is where things get rather subtle. It depends on what exactly you mean by "1 meter further out".

If you mean that you take two circles, one at r_0 and one at r_0 + 1, then they do have the expected circumferences. This is how the radial coordinate is defined. Well, strictly speaking it is defined as the radius of curvature of the sphere at that radial coordinate (it can be proven that the surfaces of constant t and r are in fact geometric spheres), but from that you can derive the area of the sphere and from that the circumference of the circle.

However by "1 meter further out" you probably mean "the length of the spacelike geodesic between the two radial coordinates of the concentric circles", ie proper spatial distance. In that case you need to integrate over the geodesic
\Delta s = 1 = \int_{r_0}^{r_0+x}\frac{dr}{\sqrt{1-r_s/r}}

and solve for x.

We can easily see that the proper distance is going to be larger than the coordinate distance (this is how you know you have spatial curvature btw) by taking the reduced spatial schwarzschild metric
ds^2 = \frac{1}{1 - r_s/r} dr^2
and noting that the factor will be larger than 1 whenever r > r_s (ie outside the black hole).

So while you are correct that the existence of spatial curvature means you have to be careful about notions of "1 meter further out", the schwarzschild radial coordinate (and thus the schwarzschild radius) is still unambiguously defined by the circumference of concentric circles (or curvature of concentric spheres to be exact).

extremedave
2014-May-02, 12:41 AM
Hi All - I just joined because this idea has been plaguing me but with an additional nuance. Namely, suppose two black holes are flying by one another on parallel courses at relativistic speeds. Black hole A and B both have similar mass/radii. They pass by at 1.9x the Schwarzschild radius away from one another, such that neither singularity is caught inside the event horizon of the opposite black hole, but the two event horizons do "merge" as described above for some 10% of the radius. Can the black holes then escape one another since the masses never actually crossed the event horizon, if they have sufficient speeds?
If so, then question 2: what happens to an object that just fell into black hole A and is still within the region of space that will be overlapped by the two radii. When/if the holes move apart, can this object have any of the following fates: a) - still locked in BH A, b) actually transfer to BH B, or c), somehow manage to move to the perfect location between BHs when they let each other go, thereby escaping back out into space?

Jeff Root
2014-May-02, 12:05 PM
Hello, extremedave!

I will make a wild guess that if the two black holes get close
enough to each other for their event horizons to meet, they
will lose angular momentum via gravitational radiation at such
a high rate that they will immediately be captured and spiral
together.

I expect that *any* and *every* encounter between two black
holes will end up with them having relativistic relative speeds
as their event horizons approach each other, so the situation
you describe would *always* be what happens!

-- Jeff, in Minneapolis

.

Hornblower
2014-May-02, 12:20 PM
Hello, extremedave!

I will make a wild guess that if the two black holes get close
enough to each other for their event horizons to meet, that
they will lose angular momentum via gravitational radiation
at such a high rate that they will immediately be captured
and spiral together.

I expect that *any* and *every* encounter between two black
holes will end up with them having relativistic relative speeds
as their event horizons approach each other, so the situation
you describe would *always* be what happens!

-- Jeff, in Minneapolis

That is consistent with a article I saw in Sky and Telescope a few years ago. What was interesting there was that they concluded that the energy carried away in the gravitational wave radiation is reflected in a reduction of the mass of the resultant black hole from the sum of the original masses. In a sense something escapes, but not in the form of any discrete body that fell in during or just before the merger.

Strange
2014-May-02, 12:48 PM
Note that the Schwarzschild radius is proportional to mass. So, for two equal mass black holes, the Schwarzschild radius will be twice as great. This means that as soon as the event horizons touch, the combined mass will be within the new Schwarzschild radius; therefore: black hole, no escape.

Before this though, the event horizons will have distorted as they start to "reach out" to one another. I believe the merger is inevitable from that point.