# Black-hole's foreknowlege

1. Dec 10, 2008

### lzkelley

I'm reading Hartle's "Gravity," for the general relativity class i'm taking; chapter 12-Gravitational Collapse and Black Holes says, "A horizon's locaton at any one moment depends on the geometry of spacetime to the future of that moment." (pg. 168) That statement is vague and prozaic -- but a figure on the next page elaborates: it shows a star which has collapsed to a singularity; followed by a spherical shell of mass that some time later falls into the black hole. The figure shows the event horizon beginning to expand BEFORE the shell enters it; i.e. the shell expands as the shell compacts towards it - with them eventually meeting at what becomes the final event horizon radius.

What the hell?! And it doesn't explain why this is; nor alleviate the inevitable terrors that arise after reading it.

Can anyone explain or elaborate at all? The radius of the event horizon is directly proportional to the mass INSIDE --> so how can it expand before the mass is accreted? If we say that at infinity, there is an infinitely large (and infinitely massive) spherical shell--> would the black hole "know" of it ahead of time, and be of infinite size already?

Im really confused...
Thanks!

2. Dec 10, 2008

### JesseM

There's a good discussion of the issue of the horizon of an evolving black hole in chapter 12 of Kip Thorne's Black Holes and Time Warps. He mentions that there are actually two ways of defining the event horizon, known as the "apparent horizon" and the "absolute horizon"--the "anticipatory" version you're talking about is the absolute version. The absolute horizon is defined in terms of the set of events from which signals moving at the speed of light will never be able to make it to the "distant universe" beyond the horizon; the reason it seems anticipatory is because whether a light signal transmitted from an event will ever reach the outside world or not depends on the future behavior of the matter forming the black hole as the signal moves outward. Here's Thorne on p. 417:

3. Dec 11, 2008

### yuiop

The radius of the event horizon is NOT proportional to just the mass INSIDE. The normal Schwarzschild solution is a vacuum solution. If you have a shell of material outside the black hole then the normal Schwarzschild solution does not apply, because the material in the shell disqualifies it from being a vacuum. The normal exterior Schwarzschild solution describes the spactime above the suface of a massive body located in an 'otherwise empty universe'. You have to use the interior Schwarzschild solution to describe anything that is is not empty space. Basically the spacetime and location of the event horizon is determined by ALL the mass everywhere. Every massive object such as the Earth has a Schwarschild radius that is determined by all the mass involved and should not be confused with the event horizon even though that is the normal location of the event horizon of a stable fully formed black hole. The Schwarzschild radius is fixed for a given mass and does not alter even if the object collapses. There is no event horizon at the Schwarzschild radius of the Earth, but if the Earth were to suddenly gravitationally collapse an event horizon would appear at the centre of the Earth when it has collapsed to 9/8 of its Schwarzschild radius and then move outwards to meet the infalling mass at the Schwarzschild radius and form a black hole.

As to the question of how the black hole knows of the outside the shell off mass ahead of time you first have to ask yourself how you place the outside mass in that location 'instantaneously'. Mass can not be created instaneously. Even if the mass was in the form of energy it would have a mass equivalent and act on the spacetime gravitationally. Since the mass has always existed in one form or another since the beginning of the universe, the black hole has had a few biilion years advance notice of the shell's arrival.

If you imagine the spacetime above and below a black hole as being made of imaginary concentric shells, then a clock in any given shell is continuously gravitationally 'aware' of all the shells above and below it. All an event horizon is, is the location where coordinate time stops. Coordinate time is determined by gravitational potential and this in turn is determined by the distribution of mass 'everywhere'.

Last edited: Dec 11, 2008
4. Dec 11, 2008

### JesseM

Kip Thorne's comments make it sound like the behavior of the absolute horizon is genuinely anticipatory though. Suppose you have two scenarios where some mass is being dropped towards a black hole, and both scenarios are identical up to some time t, but in scenario #1 the mass fires powerful rockets at time t to escape before crossing the horizon, while in scenario #2 the mass keeps falling after time t and enters the horizon. In this situation, wouldn't the behavior of the absolute horizon be different in the two scenarios before time t, because light emitted from an event closer to the horizon than the mass before time t will later be able to escape in one scenario but not in the other? If the absolute horizon's behavior doesn't depend on future events in this way, why would Thorne talk about how it "violates our cherished notion that an effect should not precede its cause" and is defined in a "teleological" way where "the very definition of the absolute horizon depends on what will happen in the future"?

5. Dec 11, 2008

### yuiop

If what you are describing really happened then it would be a genuine miracle. What Kip is describing is an illusion or artifact of the seemingly instaneous behavior of gravity. The same phenomenum is observed in normal orbiting bodies. The Earth is about 8 light seconds away from the Sun. If the Earth fell towards where the Sun was 8 seconds ago and if the Sun fell towards where the Earth was 8 seconds ago then the orbit would not be what we actually observe. The Earth and Sun appear to anticipate where each other will be in 8 seconds in the future or gravity appears to act instantaneously. General Relativity explains this apparently predictive behavior by gravity by not requiring gravity to act by interchange of particles but by bodies following geodesics in spacetime. No actual predictive behavior actually occurs and effect never precedes cause. That is purely an artifact of creative interpretation.

6. Dec 11, 2008

### xantox

The definition of an (absolute) event horizon is dependent over the entire spacetime and is thus nonlocal. As a consequence, and like Kip Thorne indicates, it is a teleological entity, meaning that an event horizon could form now around the earth because of some collapse which will happen somewhere some billion years later.

Last edited: Dec 11, 2008
7. Dec 11, 2008

### George Jones

Staff Emeritus
8. Dec 11, 2008

### JesseM

Why would it be a genuine miracle? The absolute horizon is defined in terms of whether light from an event emitted at one point will in the future escape to infinity or hit the singularity--why should it be surprising that this depends on what happens to the matter falling into the black hole in the future? Again, I think this is Thorne's point when he writes "The very definition of the absolute horizon depends on what will happen in the future: on whether or not signals will ultimately escape to the distant Universe." And keep in mind it's not like there's any physical way to measure the position of the absolute horizon at a given moment and thus gain information about the future, you can only reconstruct the position of this theoretical entity in retrospect.

Thorne provides a diagram on p. 415 of both the apparent horizon and the absolute horizon of a star collapsing into a black hole. The apparent horizon does not "switch on" until the entire star has shrunk to the Swarzschild radius for its mass, like the image here:

On the other hand, the absolute horizon actually appears in the center of the star before the star has shrunk to the Schwarzschild radius, and grows outward until it meets the surface of the star at the star's Schwarzschild radius, like the pink line in the diagram here (the outer blue lines represent the outer edge of the collapsing star):

Assuming they are modeling a star of uniform density collapsing--and I think that's normally the simple model that would be used--then the absolute horizon inside the star at a time before the moment the apparent horizon forms should enclose a collection of mass with a lower density than the density of the star at the moment the apparent horizon forms. But smaller black holes are supposed to have a higher average density inside (mass/volume) than larger black holes. So this also suggests to me that the absolute horizon is only appearing inside the star because of what will happen in the future--if the star were actually made of a huge collection of tiny rockets, then at any time before the apparent horizon forms I think all the rockets could fire and the whole collection could avoid forming a black hole, in which case no absolute horizon would ever form.

Last edited: Dec 11, 2008
9. Dec 12, 2008

### yuiop

Your reasoning about the density is good but you appear to be locked into thinking in terms of the well known exterior Schwarzschild solution rather than the interior Schwarzschild solution that is applicable in this case. The average density of the expanding 'inner black hole' is indeed less than an isolated black hole of the same radius without the outer unenclosed mass, but the absolute horizon is not determined purely by the mass within its enclosing radius but by all the mass involved.

If you draw a horizontal line just after the absolute horizon forms you will find that its radius is completely determined by all the mass and its distrubution at that instant (defined by the horizontal line.

Assuming even density dstribution the radius of the absolute event horizon Ro is determined by this equation:

$$R_o = \sqrt{9R_m^2-8R_m^3/R_s}$$

where Rm is all the mass involved and Rs is the Schwarzschild radius.

Another useful equation is:

$$\frac{dtau}{dt} = \frac{3}{2}\left(1-\frac{2GM_{total}}{R_mc^2}\right)^{0.5}-\frac{1}{2}\left(1-\frac{2GM_{enclosed}}{Rc^2}\right)^{0.5}$$

where R is the location of the clock measuring the proper time.

The only apparently magical thing going on is that the event horizon is aware of the location of all the mass at any given instant. In other words, gravitational information appears to be communicated instantaneously. If as you correctly pointed out the small rockets turned around the absolute event horizon responds instantaneously and does not wait for the mass distribution to be communicated at the speed of light. This is not too surprising when you consider that the mass inside a black hole acts on matter outside the black hole despite the fact it is known that nothing going at less than the speed of light can exit the black hole. What is very bothersome about all this is that it does appear to give a mechanism by which information can transmitted faster than light. If a clock is placed at the centre before any event horizon has formed, an observer just outside the centre could observe the clock stop or stop seeing the clock, because it is behind an event horizon and be aware of whether the rockets had turned around or not before he/she could possiby be aware of that information by means of light speed signals. Hmmm.. I'll have to think about this some more. Einstein wouldn't like this....

[EDIT] I am guessing the solution to all this is that the Schwarzschild solutions are static solutions and when the all the mass is moving simultaneously the equations might be different and be based on the distribution of mass as it was at an earlier time according to the know positions of the mass communicated at the speed of light. A FRW type metric might be more appropriate. On the other hand Birkhoff's theorem states that the spacetime outside a massive body is unaffected by the motion of the particles that make up the body. For example a neutron star that this rapidly expanding and contracting would not send out gravitational waves. However Birkhoff's theorem does not seem to say anything about whether the spacetime inside a cloud of collapsing particles is effected by the motion of the infalling particles.

Last edited: Dec 12, 2008
10. Dec 12, 2008

### xantox

There is no way to locate an (absolute) event horizon experimentally. Note that an event horizon may form in a completely flat region of spacetime, without any deep gravitational wells there and time dilation effects.

11. Dec 13, 2008

### yuiop

Are you saying that light can escape from an absolute event horizon? I beg to differ.

I assume here you are talking about the apparent event horizon that is seen by accelerating observers in Minkowski spacetime. It is true that free falling observers in that spacetime would not see anything unusual about the event horizon but it is also true that the free falling observers would not be shredded by tidal forces and nor do they meet a central singularity, so the apparent black hole of Minkowski spacetime is not exactly eqivalent to a real black hole from the point of view of free falling observers. On the other hand, the accelerating observers are not able to see any light coming from behind the the apparent event horizon and they even see Unrah radiation which is equivalent to Hawking radiation from a real black hole from the point of view of free falling observers. To the accelerating observers, the apparent event horizon looks very much like a real event horizon.

In the example I gave, I said "an observer just outside the centre could observe the clock stop or stop seeing the clock". Perhaps I sould have been more explicit and said {an observer hovering just outside the centre could observe the clock stop or stop seeing the clock.} When everything else around him is gravitationally collapsing, the observer would have to be accelerating outwards using some sort of propulsion system in order to remain stationary. Because said observer is accelerating, the absolute event horizon will appear real to him. Clocks near the absolute event horizon will appear to slow down, clocks exactly at the absolute event horizon will appear to stop and clocks behind the absolute event horizon will not be visible to the hovering observer.

12. Dec 13, 2008

### yuiop

OK, I given this some more thought.

The diagram posted by JesseM would seem not to be to scale. To a first order aproximation using a static solution the absolute event horizon does not appear until the the mass is enclosed with a radius of 9/8 Rs where as the diagram shows the absolute event horizon appearing much earlier when the total mass is enclosed with a radius of about 2 Rs. The situation shown is dynamic and presumably (if we assume gravity is communicated at the speed of light) the absolute horizon would not appear at the centre until the mass is enclosed well within 9/8 Rs.

The diagram comes from Ned Wright's Cosmology website and the accompanying text states "The coordinates used in this diagram are the Kerr metric coordinates for times later than the creation of the singularity. The Kerr metric describes a rotating black hole, but the rotation is set to zero here. Before the creation of the singularity a simple analytic form is used instead of the exact solution for a collapsing sphere of pressureless matter."

One odd thing about that statement is that he says he uses the Kerr metric with the rotation set to zero. The Kerr metric with rotation set to zero is the Schwarzchild metric so why not use that in the first place? There are variables for motion of the test particle and the angular momentum of the gravitational body in the Kerr metric, but there are no variables for the change in mass or change in density of the gravitational body in the Kerr metric so the Kerr metric is radially static just like the Schwarzschild metric.

He then adds that for the creation of the singularity he uses a method that is not an exact solution.. Is it safe to conclude that super luminal effects and that a black hole can effectively read minds on the basis of a non-exact solution?

The correct approach is to use a Kerr-Vaidya type metric that is dynamic rather than static with the angular momentum set to zero to simplify things.

13. Dec 13, 2008

### xantox

Light cannot escape from an event horizon - still, there is no way to determine its location experimentally in the dynamical regime - only the total knowledge of the entire spacetime may allow to determine whether and where it does exist. This is clear from the mathematical definition of the event horizon (Hawking and Ellis, 1973).

I was referring to the absolute event horizon (the apparent horizon is not usually considered an event horizon in the literature). The event horizon is the absolute one, and it is observer-independent. It is possible to construct event horizons forming in empty flat spacetime long before any gravitational collapse exists at all.

14. Dec 13, 2008

### JesseM

Doesn't the Schwarzschild solution only describe the case of an eternal nonrotating black hole, rather than one that forms from a collapsing star (idealized as a uniform dust sphere)? Of course this also somewhat undermines my argument about the radius of the absolute horizon at an early moment being larger than the Schwarzschild radius for the amount of mass contained within the absolute horizon at that moment, since in this dynamic situation the Schwarzschild radius may no longer describe the point at which a collection of mass can no longer avoid collapsing to a singularity. Still, from Kip Thorne's straightforward description, and from xantox's comments, I'm still fairly certain that whether or not there's an absolute horizon at a particular point in spacetime cannot be determined solely by events in the past light cone of that point. And I would still guess that, prior to the point where the entire collapsing dust sphere has passed the point where the "apparent horizon" would form (which I think is still the Schwarzschild radius), it would in principle be possible to blow it apart completely using non-gravitational forces and prevent any horizons from forming at all (by setting off a powerful bomb at the center, say, or by imaging the sphere to be made of a giant collection of tiny rockets, which all begin to fire away from the center at some time prior to the whole sphere shrinking to a radius smaller than the Schwarzschild radius).
It's probably true that if you assume no non-gravitational forces are applied and all the dust particles just follow geodesics, then with that assumption you can determine the absolute horizon's radius at any instant just based on the coordinate position of all the dust particles at that instant. But like I said, I would guess that before the whole sphere crosses the point where the apparent horizon forms, then the entire sphere can be disrupted by non-gravitational forces and this retroactively means that no absolute horizon formed at all. In contrast, once the sphere has passed the point where the apparent horizon forms (again, I think that's just the Schwarzschild radius) then no non-gravitational forces can prevent it from collapsing into a singularity.
But the absolute horizon is defined in terms of whether light from an event will ever reach the outside. The mere fact that light from a certain tick of a clock hasn't reached you yet doesn't prove that this event happened behind an absolute horizon, because it still might reach you at a later time. Imagine two collapsing dust spheres made out of tiny rockets, and in sphere A the rockets never fire and the entire sphere collapses to a black hole as in the diagram, while in sphere B all the rockets fire at some time t1 prior to the point when the whole sphere has crossed the radius of the apparent horizon. Suppose in sphere A, the absolute horizon begins to form at time t0. If a tick of a clock is behind this growing absolute horizon in sphere A, it will never reach the eyes of an observer outside the absolute horizon. Now, my guess is that in sphere B, in spite of the fact that the behavior of all the rockets is identical to those of sphere A prior to time t1, no absolute horizon will ever form. Nevertheless, if a certain clock tick happens in sphere B prior to t1 at a point where it would be inside the absolute horizon of the identical sphere A, I would assume that the light from this event cannot reach any observer at a position that would be outside where the absolute horizon would be in sphere A until after time t1 when the rockets have fired and the behavior of the two spheres has begun to diverge. So up until t1, this external observer won't see the clock tick and might be tempted to think it's behind a horizon, but then sometime after t1 he will see it and he'll realize that the light was just delayed but not permanently trapped. This is of course just my guess, I don't know enough to actually do the math to prove or disprove it, but the idea seems consistent enough and doesn't lead to any physical paradoxes.

Last edited: Dec 14, 2008
15. Dec 14, 2008

### MeJennifer

Yes.

It is the solution of a static, and thus implicitly stationary, spacetime.

16. Dec 14, 2008

### yuiop

That is why I mentioned in post 9 that "[EDIT] I am guessing the solution to all this is that the Schwarzschild solutions are static solutions and when the all the mass is moving simultaneously the equations might be different " and later said in post 12 "The correct approach is to use a Kerr-Vaidya type metric that is dynamic rather than static".

Part of the diagram produced by Ned Wright has been produced using a "non-rotating Kerr metric" which is effectively a (static) Shwarzchild metric. Ned clearly shows light paths entering the event horizon as straight diagonal lines and that is not true for a Schwarzschild metric and he is clearly using some form of Eddington-Finkelstein coordinate system, so that is an inconsistency in his diagram.

Any delay in seeing the clock (or a regularly flashing light beacon) would be a detectable physical effect that the observer could use that observation to infer information about the motion of the rockets on the outside (whether they have changed direction or not) of the collapsing sphere and therefore he would have obtained information about an event, before he could have obtained that information by signal limited to the speed of light.

A lot depends on whether you think being able obtain information about a distant event faster than speed of light, is a physical paradox or not.

I would suggest that is impossible to have an event horizon with flat space on either side of it and by definition flat space does not have event horizons detectable by inertial observers.

I can imagine a situation in static spacetime that contains a sphere of flat spacetime containing no mass at all, bounded by an event horizon that has mass and curved space outside of it. Is that what you are hinting at? That is an interesting example of a black hole that contains no mass. It would be created by a non-rotating static hollow shell of matter with an internal radius of Rs and and external radius of 1.125 Rs. Everywhere inside such a shell would be flat space with a gravitational time dilation factor of zero.

17. Dec 14, 2008

### JesseM

How can he use the delay to obtain information faster than light? My scenarios A and B are supposed to be physically identical up until time t1 when all the rockets fire in scenario B (while they don't fire in A), and if there's a pair of identical events E in each scenario at the same time (prior to t1) and position in the sphere, a time and position which in scenario A is inside the absolute horizon, I'm suggesting the outside observer in scenario B won't get any light from E until sometime after the time t1 when all the rockets begin to fire (after the light from the event of the outermost rockets beginning to fire has already reached him, so he knows the rockets aren't just falling along geodesics like in A). I don't see how this would violate the speed-of-light limit on information transmission, can you elaborate?

18. Dec 14, 2008

### xantox

Since you cited the Vaidya metric, check the classic example given in A. Ashtekar, B. Krishnan, "http://relativity.livingreviews.org/Articles/lrr-2004-10/ [Broken]", LRR 7, 10 (2004), section 2.2.2 fig. 4. The Penrose diagram shows that the event horizon originates in the green flat space-time region in anticipation of the null fluid that is going to fall in.

Last edited by a moderator: May 3, 2017
19. Dec 15, 2008

### lzkelley

Thanks everybody for your posts; all quite insightful.
I think xantox's above post is getting there--but what metric can be used to show the anticipatory, absolute horizon growth?