# Trip To An Event Horizon

1. Mar 13, 2012

### Spin-Analyser

I'm not sure why the other thread was locked, unless you're banning questions? I'll ask it in as clear a way as possible.

When an observer approaches an event horizon to one plank length away one of two things must happen:

1). Any observers previously falling towards the black hole would be right along side you.

2). There is distance between you and the light from previous in-falling observers when you're right on the horizon. That would mean you would have to be seeing light from inside the horizon, although it's light that hasn't reached the horizon yet, which is paradoxical because you can't tell if they crossed. How can you be hovering the same distance away from the horizon as a previous observer and have space between you?

Please define the parameters of the universe in which that's supposed to make sense.

I was told that if you decide at the last minute to brake and hover at a small fixed distance outside the event horizon, you see your partner's image slow down, red-shift and darken and never actually cross the horizon. But if you're right next to the horizon and you never see them cross the horizon then how can there be space between you?

Please don't lock or delete this thread. These are legit questions. I honestly don't think anyone's explained how an image can be coming from inside the horizon of an object that might not have crossed the horizon yet? This isn't "word soup", it's a fair question.

2. Mar 14, 2012

### Staff: Mentor

The other thread (https://www.physicsforums.com/showthread.php?t=583903) was locked, not because you were asking questions, but because you were simply ignoring the answers and repeating the same thing. Don't let that happen here. Threads that simply go round in circles will be locked.

3. Mar 14, 2012

### Sam Gralla

Point 1) is wrong, and I don't understand point 2). It all sounds very suspicious because you're talking about being "1 Planck length away". Classical physics (such as general relativity) is based on a continuous description of spacetime and you won't find any qualitatively different phenomena at any two different finite distances from the horizon. Of course, if you're saying GR isn't going to correctly describe physics within a Planck length, few will disagree with you.

4. Mar 14, 2012

### Naty1

I don't believe this perspective was discussed in the prior [now locked] discussion:

http://en.wikipedia.org/wiki/Black_hole_complementarity

note to PeterDonis: This short article addresses your prior question we discussed in yet another thread about the 'no cloning theorem'. The explanation in Wiki confirms my recollection of what Susskind says in his BLACK HOLE WAR.

Spin:

"..I accept the concept that, as almost infinite energy is required to hover near the event horizon, gravity can be inferred to become infinite at the EH....."

from #67, PeterDonis:
""....Gravity" in the sense of "the proper acceleration required to hover at a constant radius". But there are other senses of "gravity" that are *not* infinite at the horizon, as several posters have pointed out. Curvature in the sense of the Riemann curvature tensor, for example, or various scalars derived from it, is perfectly finite at the horizon...

Conditions at or near a horizon are indeed 'weird'.....contrary to our normal 'intuition'....Gravity IS the curvature of space and time....yet acceleration layers further curvature in addition to this 'spacetime curvature' of gravity.....Peter's explanation is CRAZY...but I believe accurate!!!!!!

you are assuming light approaches you 'normally' when it is in fact ..."takes a while (up to forever) to reach you.....[PAllen]

I can see two issues perhaps 'fogging' your interpretation[one's we have all probbaly faced] : one is that you are perhaps not thinking very carefully about what has been posted, perhaps applying traditional intuition, [and this stuff is not easy to 'visualize] and secondly, it may not be clear which frame of reference is under discussion, sentence by sentence.

For example when you wrote:
I thought "oh boy" nobody said THAT!!!! A hovering observer is causally disconnected from all events INSIDE the horizon....while a free falling observer, in sharp contrast, can 'see inside'....if those two observers pass each other they observe very different things and, for example, time passes very differently for them. It's analogous, if not identical, to an Unruh [accelerated] observer making different observations than an inertial observer.

As an impartial observer, I don't think you'll get any clearer explanations than were provided in the prior discussion.....but don't expect they will all make immediate 'intuitive sense'....you have to think differently as a result of what the mathematics reveals. Have you considered rereading that last thread??? You should not expect immediate clarity the first time you read some of these perspectives. I have sometimes have to read four or five times before the 'light goes on'....

What I also found useful was writing down some 'principles' which I would accept and begin my reasoning from there....
like maybe "everything is relative...everything is frame dependent" and refining it near balck hole horizons to "A free falling observer sees no horizon, a hovering observer is fried [cooked] by it..."
and refining my understanding from there....

Last edited: Mar 14, 2012
5. Mar 14, 2012

### Naty1

Sorry, the above post got too long...

Spin:
Maybe it would be clearer if you thought of spaceTIME instead of space????.........

Peter #33:
PAllen #82

6. Mar 14, 2012

### PAllen

Not only accurate, but it need not be mysterious at all. There is an equivalent Newtonian analog to this:

Imagine you are on the surface of a planet enormous radius with enormous mass. The acceleration you will need to hover just above the surface will be enormous. Tidal forces will be small (because radius is enormous). The acceleration is due to depth of gravitational well - a global feature. The tidal forces due to gradient - a local feature.

The GR analog is conceptually similar. Curvature per se is local and corresponds roughly to tidal gravity. Very roughly, you can imagine the proper acceleration aspect of gravity as reflecting accumulated difference in geometry between near horizon and flat spacetime at infinity. More specifically, you can say that an (r,theta,phi)=constant world line is geodesic at infinity, but diverges ever more from the local geodesic as you approach the horizon.

7. Mar 14, 2012

### yuiop

If previous falling observers are "right alongside you" then they are exactly one Planck length away from the EH and therefore did not fall earlier but simultaneously. The sentence is a self contradiction. (Perhaps the letters got mixed up when you stirred the soup? ...hehe :P ) Previously falling observers will be less than a plank distance away from the EH. Of course you might be assuming that there cannot be a distance of less than one Plank length but that is unproven and unlikely and you seem to assuming that we are living in a spatially digital universe. Also bear in mind that due to gravitational length contraction (for want of a better term) what we call one Planck length from a coordinate point of view might well be measured as a distance of several Kilometers from the point of view of local observers. Where do we define the Planck length? Locally or in coordinate terms?

Also bear in mind you cannot see light emitted from inside the EH or even exactly at the EH. even if you one Planck length outside the EH.

This is looking soupy too. You said you are "right on the horizon" which to most people would mean you are exactly zero Plank lengths from the horizon but then you say you are hovering the same distance away from the horizon. If you are hovering you cannot be exactly zero Planck lengths away from the horizon so again you are contradicting yourself. Are you on the horizon on just outside it? You seem to be using one Planck length to mean a finite non zero quantity sometimes and at other times pretending that one Plank length is as good as zero, when in reality they are worlds apart. It is similar to saying that travelling at 99.9999999999% of the speed of light is as good as travelling at the the speed of light when they are worlds apart.

More soup. By "right next to the horizon" do you mean right on the horizon or do you mean some distance away from the horizon? If you are some small but non zero distance away from the horizon then by definition there is space between you and the horizon and also space between you and the previously infalling observer. If you mean you are zero Planck lengths away from the horizon (i.e. right on the horizon) then you can see light from other objects that fell through the horizon.

The only time you can see light from objects that see inside the horizon or exactly on the horizon is when you are also exactly on the horizon or below it. Even hovering one Planck length outside the horizon, you never see anything arrive exactly at the horizon. All you see are gradually red shifting, dimming but persistent ghostly images that become undetectable before they cross.

8. Mar 14, 2012

### yuiop

This is potentially confusing to the the OP and perhaps fuelling the fire.

You should perhaps stress that a free falling observer can "see inside" if and only if the free falling observer is at the event horizon or has passed through the event horizon.

They might see different things such as the degree of red shift but what they have in common is that neither of them see inside the EH while outside the EH.

9. Mar 14, 2012

### Staff: Mentor

Yes, what's on that wiki page seems consistent with what I remember of Susskind's The Black Hole War as well. His way of getting around the no-cloning theorem is basically to say that no single observer will ever see both "copies" of the information, so it's OK.

The claim that no observer can see both copies, in and of itself, seems correct to me because once you're inside the horizon, not only can you never get back out, but you can't even "hover" at a constant distance from the singularity--you will inevitably reach the singularity and any information you carry will be destroyed there along with you; even light rays can't "hover" at a constant distance from the singularity any more, so they will also be sucked into the singularity and destroyed.

This does seem like a lot of quirky claims in QM--yes, this can in principle happen, but nobody will ever see it, so it's doesn't violate the rules. Kind of like EPR experiments show that nonlocal correlations can be mediated faster than light, but no actual information can ever be exchanged that way, so it's OK. Einstein would probably be tearing his hair if he were still around to read Susskind's book.

10. Mar 15, 2012

### Staff: Mentor

First, please note that Sam Gralla's comments are correct: if we are talking about classical GR, there is no "minimum distance" that something can be above the horizon, so the Planck length is irrelevant; above the horizon is above the horizon. Everything I say in what follows will assume that the questions above have been modified appropriately.

There's a model of Schwarzschild black holes that might be helpful in visualizing how all this works: the "river model" of black holes. See this paper:

http://arxiv.org/abs/gr-qc/0411060

Basically, the idea is that space is flowing inward into the black hole's singularity; at any given radius r, the rate at which space flows inward is the "escape velocity" at that radius. So space is like a "river" that continually flows inward, carrying things along with it.

If you are free-falling into the hole, you are basically at rest in the river, so you are moving inward at the same rate the river is. So suppose you are free-falling inward, and you have a blinker below you that sends repeated flashes of light back up to you. The blinker is attached to you so it stays at the same physical distance from you; we assume that the hole is large enough that tidal gravity is negligible until you are almost at the singularity, so the blinker can remain attached to you safely. The light from the blinker is moving "against the stream", so to speak, at the speed of light--that is, it moves outward at the speed of light *relative to the river*.

But relative to the "river bed", that is, relative to observers "hovering" at rest at a constant radius r, the light moves outward *more slowly* as you get closer to the hole (because the river flows inward faster). At the horizon, the river is flowing inward at the speed of light, so outgoing light is just able to cancel out the flow and remain at the same radius, the horizon radius, forever. Inside the horizon, the river is flowing inward faster than light (this is OK because the "river bed" doesn't actually exist physically, it's just an aid to visualization), so even outgoing light rays are dragged inward (just more slowly than anything else in the river).

Now consider flashes of light emitted by the blinker at various points as you fall. To you, they all look the same: they rise toward you at the speed of light, and take the same time (relative to your clock) to reach you. However:

(1) A flash emitted far away from the hole will be moving outward, relative to the river bed, at almost the speed of light (since the river is flowing inward very slowly). So an observer "hovering" at rest relative to the river bed will see things pretty much the same way you do.

(2) A flash emitted when you are getting fairly close to the hole will be moving outward, relative to the river bed, somewhat slower than the speed of light (since the river is now flowing inward at an appreciable fraction of the speed of light). So the observer will see the light rise more slowly than you do; to him, it will look as if you are falling to meet the light, to some extent, rather than the light rising to you.

(3) A flash emitted when the blinker is just above the horizon will be moving outward *very* slowly; and you yourself will be falling, relative to the river bed, at almost the speed of light. So to an observer "hovering" at rest there, the light hardly seems to move at all; instead, you fall to meet it at a point just slightly above the point where it was emitted, relative to the river bed. In fact, if the flash is emitted close enough to the horizon, the blinker itself might actually be *below* the horizon by the time you have fallen to meet the flash of light.

(4) A flash emitted when the blinker is exactly *at* the horizon stays at the horizon; as noted above, the outward speed of the light at the horizon just cancels the inward speed of the river, so the light stays at the same place relative to the river bed. So you will see the flash exactly when you, yourself, cross the horizon. By that time, though, the blinker will have fallen below the horizon; the blinker itself does not "wait" at the horizon for you, only the flash it emitted at the horizon does. So the blinker will not be "next to you" when you cross the horizon; it will be further inward, just as it is throughout your fall.

(5) A flash emitted when the blinker is below the horizon moves *inward*, relative to the river bed, even though it is emitted outward; as noted above, even outgoing light is dragged inward in the region below the horizon. But the outgoing light is dragged inward more slowly than you are, since you are moving with the river; so you will still catch up to each light flash emitted by the blinker. When you do catch up to each flash, the blinker will have fallen further than it was when it emitted the flash, since it is also moving inward faster than the outgoing light is.

11. Mar 15, 2012

### yuiop

The trouble with the river analogy is that it suggests that to a stationary observer hovering just outside the EH, up going light will be almost stationary and down going light will be travelling at almost twice the speed of light. In reality such a local observer see both up going light and down going light travelling at exactly the normal speed of light. The river model does not seem very good to me or I am missing the point somewhere.

12. Mar 15, 2012

### PAllen

Actually, a near horizon hovering observer is extremely non-inertial, and there is no expectation of light speed isotropy for such an observer (as there is for a free falling observer). In a thread here, I think about a year ago, Passionflower presented rigorous calculation that such an observer measuring light speed over a finite distance (rather than infinitesimal) will, indeed, measure a difference in upward versus downward speed. There was even anisotropy in two way light speed for sucn an observer over finite distances.

Last edited: Mar 15, 2012
13. Mar 15, 2012

### yuiop

I was initially using the infinitesimal assumption. However I disagree with the assertion that the upward speed of light differs from the downward speed of light over the same distance at the same location. Let me make this clearer. Consider the case of an observer with a mirror one metre below him and another ruler one metre above him. He finds that the time for a light signal to make the two way trip (down then up) to the mirror below him takes longer than the two trip way trip (up then down) to the mirror above him. Note that these are two way trips and so say nothing about the difference between upgoing light and down going light. Another difficulty is in how we define a metre. A metre is officially defined in terms of light speed or wavelength. If there is a difference in the time time to the lower mirror and back and to the upper mirror and back then we have to conclude that the mirrors are a metre above and below and have to adjust them until their times are equal. If we are more fussy then we can measure circumferences and divide by 2 Pi to get the radius and this will result in a different definition of the metre from the international CGPM definition and an anisotropic result will be obtained but all this tells us is that the speed of light lower down is slower than the speed of light higher up.

Now we can alter the experiment so that the we have a single ruler. Initially we have a source and clock at the bottom and a mirror at the top and record the two way time (up then down). Next we place the clock and source at the top and the mirror at the bottom and record this two way time (down then up). We note that we obtain different times. We also note that the light paths are exactly the same (only differing in order) and by symmetry arguments we conclude that the anisotropy in times is purely due to differences in time dilation on the clock depending upon whether they are placed at the top or bottom. None of these experiments say anything about differences in the one way speed of light going up or down, only that light speed id slower lower down (in either direction) than it is higher up.

OK, now we attempt to measure the one way speed (up or down) over the same vertical distance. First difficulty is synchronising the clocks which is a problem because we know clocks higher up run faster than clocks lower down. We can speed up the lower clock so that it runs at the same rate as the upper clock. Now we still have the standard difficulty that synchronising clocks requires we make certain assumptions about the one way speed of light in the first place so any conclusions are subject to circular reasoning. We cannot measure the one way speed of light in GR any more than we can measure the one way speed of light in SR for very similar reasons.

If we look at the Schwarszchild metric we can easily calculate that the vertical speed of light is the exactly the same going up or going down at a given radial coordinate. I can only assume that what Passionflower actually calculated was an anisotropy between the speed of light lower down compared to the speed of light higher up which can indeed be measure over finite local distances. I am pretty sure the calculation said nothing about an anisotropy between the up going one way speed of light and the down going one way speed of light over the same finite vertical distance.

14. Mar 15, 2012

### PAllen

You have to adopt some conventions. Fermi-Normal coordinates say you use spacelike hypersurfaces 4-orthogonal to an observer's world line. Time, anywhere on each such surface is taken to be proper time along the time axis world line. Distances are proper lengths in the surfaces. Then, using this synch convention, you can define a one way light speed for such a non-inertial observer. There is no general expectation it is c, nor that it has any particular isotropies. In SC, it is trivial to compute that for a hovering observer, such a surface restricted to (r,t) is simply constant r. That is, unit vector in r direction is always 4-orthogonal to unit vector in t direction in SC geometry.

However, reviewing the calculation I see that I mis-remembered, and it is clear that up an down speed between r and r+h (for r taken as observer coordinates) will be the same. This is, in fact, obvious due to SC geometry being static (therefore time symmetric). What is true is that two way speed away from the horizon is different from two way speed toward the horizon, measured over any finite distance, for a hovering observer.

Last edited: Mar 15, 2012
15. Mar 15, 2012

### Staff: Mentor

The river model is derived from Painleve coordinates, so it does look a little strange if you're used to thinking of everything in Schwarzschild coordinates. Consider that the Painleve observer, who is free-falling into the hole "from infinity" (so his inward velocity is always the "escape velocity") also sees ingoing and outgoing light traveling, relative to him, at the speed of light. If he is just a little bit above the horizon, and then adjusts for the speed of the river, then he will say that relative to the "river bed", the ingoing light *is* moving inward at almost twice the speed of light, and outgoing light *is* almost stationary, just barely crawling outward.

To see what this situation looks like from the viewpoint of the "hovering" observer, we take the free-falling observer's picture and boost it, outward, by the "escape velocity", which in this case (just outside the EH) is almost the speed of light. (Strictly speaking, this tells us what things look like in the instantaneous local inertial frame of the hovering observer, at the instant the free-falling observer passes him. But since SR is locally valid, we can deduce all the local physics as seen by the "hovering" observer from this.) What do we find? The ingoing and outgoing light are still moving at the speed of light, relative to the "hovering" observer; but now the ingoing light is strongly blueshifted and the outgoing light is strongly redshifted.

So it is true that, to the "hovering" observer, the light is *not* moving relative to the "river bed" the way the free-falling observer says it is. That's OK because the "river bed" is not physically real anyway; it's just a visualization tool. I suggested the river model because I thought it might help the OP to see how things could look perfectly "normal" to the free-falling observer even as he and the blinker cross the horizon, while still obeying the global requirement that light cannot escape from inside the horizon to outside. I did not mean to claim that the river model is helpful for *all* aspects of the physics of black holes, or that it captures all aspects of that physics in a "natural" way.

16. Apr 21, 2012

### Rock2shark

In my mind any discussion about the nature of and physical properties present at a distant relativistic type of condition requires a specification of the reference point from which the observation is made?

If we are to analyse the physical circumstances and observations of the observer hovering above the event horizon of a black hole, is it not helpful to proceed in the following way:

Using identical rigid rods of equal length and identical accurate clocks we construct a 3 dimensional rectangular frame, starting from our own location in space and extending sufficiently to contain the whole of our galaxy. We maintain synchronisation of the clocks so that the frame serves as a physical realisation of the reference frame in which we are stationary. The reference frame is an extension of the SR of our own location in space and serves to provide a basis for measuring the motion of distant bodies relative to ourselves as primary observers.

When we place an observer just above the event horizon, the observer must be stationary with respect to the reference frame, otherwise the statement ‘hovering above the event horizon’ is ambiguous in its meaning?

17. Apr 21, 2012

### Staff: Mentor

This construction, which is familiar from introductory texts on SR, works globally for flat spacetime, where there is no gravity; it does *not* work globally in the presence of gravity. In the presence of gravity, you can only do this construction *locally*--i.e., for a small patch of spacetime centered on some particular event.

Again, you can only do this locally in the presence of gravity; clocks which start out synchronized at different spatial locations in the presence of gravity will not stay synchronized.

The statement "hovering above the event horizon" does not require the construction of a local inertial frame to be given an unambiguous meaning. It simply means staying at the same radial coordinate r; the radial coordinate is defined as the "reduced circumference" of a 2-sphere at that radius, i.e., the circumference of the 2-sphere at r is 2 pi r and its area is 4 pi r^2. In principle r could therefore be measured without ever having to use the construction of a local inertial frame.

Also note that a local inertial frame, constructed as above, will only stay "hovering" at a given radius above the hole for an instant. For that instant, the "hovering" observer is at rest relative to this frame, yes. But a bit later, that particular inertial frame will have fallen towards the hole; the "hovering" observer, who is firing rockets to stay at a constant radius for all time, will now be at rest relative to a *different* local inertial frame, which is momentarily at rest at the appropriate radius at that later instant.

18. Apr 22, 2012

### Rock2shark

I understand in principle that the physical reference frame that I described is instantly distorted and wrecked by the bending of spacetime caused by the presence of gravitatoinal masses in the region of the framework. What I wished to describe was a frame that is not real, only imaginary, one that, because it is imaginary, is also immune to the effects of the massive bodies and the resulting accelerations of actual bodies and spacetime within the region covered by this imaginary reference frame.

In my mind I can picture this reference grid and the frame it provides within which an observer ‘O’ at a distance from the black hole is stationary. This is an imaginary grid that it would seem would provide a reference tool by which it seems it should be possible to measure the net effect relative to the observer at ‘O’ of gravitating bodies relative to the flat spacetime that would exist if those bodies and all gravitation were absent from the region. This would be done througth the eyes of a second observer placed at a point ‘A’ which is stationary relative to the frame of 'O' and which is also located corresponding close to or actually on the event horizon whilst fixed in relation to the imaginary 'special' frame may therefore (theoretically) at leisure take measurements of, say, the speed and other properties of bodies that are in real spacetime and which are therefore free falling towards the black hole and approaching and crossing the event horizon.

Does a flat spacetime of the kind described by the imaginary gridded reference frame in which 'O' and 'A' are stationary not provide a suitable kind of absolute reference?

Would the measurements taken by the observer at ‘A’ of the properties of bodies in real ‘actual’ spacetime not reflect the predictions of GR and the field equations?

I am intrigued by this question, and am endeavouring to gain as much understanding of GR and the tensor mathematics of GR and the Schwarzschild phenomena as I possibly can.

Last edited: Apr 22, 2012
19. Apr 22, 2012

### Staff: Mentor

There is no such thing. The reason such an "imaginary" frame works in SR, in flat spacetime, is that we can pick out geometric objects in the flat spacetime that correspond to the imaginary reference frame. In curved spacetime, we can't; there are no such geometric objects, not even imaginary ones. In other words, in curved spacetime, even imaginary rods are bent and distorted by gravity, and even imaginary clocks run at different rates at different points.

Another way to put this is, in SR, in flat spacetime, we can use the math of the theory to tell us how imaginary rods and clocks behave. In curved spacetime, we can't; the math of the theory does not describe *any* objects that are not affected by gravity. So even if we were to postulate imaginary rods and clocks that were somehow unaffected by gravity, we have no way of making any statements at all about their behavior, so it's the same as if such objects did not exist.

You may think you can imagine this, but if you try to work out a consistent set of predictions from it, you will find that that requires extra assumptions that are not always valid. For example:

It's not just that such an observer is imaginary; an observer who "hovers" at or inside the event horizon would have to travel at or faster than the speed of light. But that is inconsistent with the way you wanted to set up the imaginary reference frame; setting up that frame requires that observers at rest in the frame travel slower than light.

No. See above.

There can't be any such observer at or inside the horizon. Outside the horizon, yes, you can have an observer that "hovers" at a constant radius, and that observer's measurements will match the predictions of GR. But of course we do not need to set up any imaginary frame to derive those predictions. We can derive them, if we really want to, without using any frame of reference, any coordinates, at all; we can express everything entirely in terms of invariants, things that are independent of all coordinates.

Now, having said all that, I should note that, for the special case of a stationary black hole, there actually *is* a model that sort of has an "imaginary background reference frame that is static". It's called the "river model", and you can read about it here:

http://arxiv.org/abs/gr-qc/0411060

The key thing that allows the "background" (or "river bed") to work in this model is that it does *not* obey the laws of relativity--for example, things can move relative to the "river bed" faster than light. So it does *not* qualify as the sort of imaginary reference frame you were describing. But it does provide a way of visualizing what's going on in the spacetime around a black hole that may be easier to work with for that special case.

20. May 20, 2012

### Spin-Analyser

The last time I didn't understand something the thread got locked so I'm going to have to be careful. I can't understand it until I know what would happen if an object approaches the event horizon and an object that had previously headed towards it. Please just tell me what happens.

1). The first object appears to cross the event horizon from the perspective of the second object but doesn't actually cross it until the second object reaches the event horizon.

2.) The first object crosses the event horizon from the perspective of the second object before the second object reaches the event horizon.

3). The first object doesn't reach the event horizon until the second object also reaches the event horizon.

If it's 1 then there's two horizons and if the second object were to move away before reaching the event horizon it would see the first object move backwards so that it's outside the horizon from a distance.

2 seems to be the one that fits in with the river model and the view that nothing special would happen in you crossed the event horizon of a super massive black hole. Then what happens if they're attached to each other with an extremely strong rope and the second object moves away from the black hole and when the first object is no longer inside the event horizon from the second objects perspective the rope goes taut and the second object pulls the first object away?

If it's 3 then everything that ever happens happens outside of event horizons and there is no interior.

21. May 20, 2012

### Staff: Mentor

2 is the correct answer (with some clarifications--see below).

What you've just described, as it stands, is impossible. If the second object accelerates away from the hole hard enough to keep from falling in, before it reaches the horizon but after the first object has crossed the horizon, there is no way to pull the first object back; the rope will break. The reason the rope will break can be phrased in two ways, which are consistent but which focus on somewhat different aspects of the situation:

A) The rope breaks because the lower end of it, the one attached to the first object, would have to travel faster than the speed of light to keep up with the upper end. Since it can't, the two ends move apart and the rope breaks.

B) The rope breaks because relativity imposes a limit on the tensile strength of materials (basically, the limit is that the speed of sound in the material can't exceed the speed of light). In order to keep from crossing the horizon, the second object must accelerate hard enough that the force it exerts on the rope will exceed the limit on the rope's tensile strength. So the rope breaks.

22. May 20, 2012

### Spin-Analyser

And what's to stop information about the first object escaping from inside the event horizon?

23. May 20, 2012

### Staff: Mentor

Through what pathway? If you're thinking that a signal can be sent by tugging on the rope, or sending electrical impulses along the rope, or whatever.... No matter how soon after the object crosses the horizon, and no matter how fast (but below the speed of light, of course) the signal travels, it will be stopped by the broken end of the rope somewhere inside the event horizon.

24. May 20, 2012

### Staff: Mentor

The same thing that stops the first object itself from escaping from inside the horizon: information can't travel faster than light.

25. May 20, 2012

### Staff: Mentor

It's even stronger than that: no signal emitted at a given radius r < 2M (i.e., inside the horizon) can reach a "receiver" at any radius greater than, or even equal to, the radius at which it was emitted. For example, if the first observer tries to send a signal up the rope (by shaking it, for example), the rope will fall (decrease in radius) *faster* than the signal moves up the rope. So when the signal reaches the broken end of the rope, that end will be at a *smaller* radius than the first observer was when he emitted the signal.

In the "river model", this is easy to visualize: the "river" inside the horizon flows inward faster than light, so any signal, even light, is carried inward faster than it can move outward against the river--its net motion is inward.