Relativity and Black Hole Question

[Salamon]

Assume that there exist two people, person A and person B. Person B is falling into a black hole.

I understand that due to relativity and time dilation, person A will have to spend an infinite amount of time to watch person B cross the event horizon. Person B will appear to move at a slower and slower rate towards the event horizon. (I know that person B will be gravitationally red shifted).

So is it true, that relative to us in our universe, nothing ever falls into a black hole?

If this is true, how is it possible that black holes ever form in the first place?

 

[Dale]

There is no requirement in physics that says that if something is not visually seen to happen by a particular observer then it must not have happened. Person A's visual experience has nothing to do with what happens or doesn't happen to B.

 

[phinds]

If this is true, how is it possible that black holes ever form in the first place?

Because relative to the black hole, stuff falls in. It doesn't care what things look like from our reference frame.

 

[BruceW]

well, each coordinate system should give a consistent picture, right? So with the Schwarzschild metric, any test particle which starts outside the event horizon will end up at the event horizon at $t \rightarrow \infty$. So I guess the OP's concern is how matter would end up inside the event horizon to form the black hole.

I don't know much about GR, so I'm not really sure about this kind of stuff. But I would guess that the reason is related to the fact that it is a test particle. And if we consider the 'true' metric due to some spherical mass distribution, then it is possible for that matter to fall inwards to create the black hole. (i.e. to create the situation which is given by the Schwarzschild metric, where all the significant matter ends up inside the event horizon). I mean, when there is significant matter outside the event horizon, then that situation is not really modeled by a Schwarzschild solution, if I understand it right.

 

[Nugatory]

well, each coordinate system should give a consistent picture, right?

For the portions of spacetime that the coordinate systems covers, yes. But there's an entire region of spacetime that isn't covered by the external Schwarzschild solution at all. I can't assign a Schwarzschild $t$ coordinate to a point on the event horizon, but that doesn't mean that a particle cannot follow a worldline to and through the event horizon, it just means that I can't use Schwarzschild coordinates to describe that worldline.

 

[BruceW]

hmm. OK. But in the Schwarzschild coordinate system, it is true that the test particle approaches the event horizon, and gets 'stuck' just outside it, right? You can say this is not the best coordinate system to choose, but that's what happens according to this coordinate system. (unless I've not understood correctly).

 

[PeterDonis]

in the Schwarzschild coordinate system, it is true that the test particle approaches the event horizon, and gets 'stuck' just outside it, right?

No, in the Schwarzschild coordinate system, the test particle asymptotically approaches the horizon as $t$ goes to infinity; it never gets "stuck". But even the asymptotic approach is because the coordinates are singular at the horizon, not because the particle doesn't reach the horizon. The closer you get to the horizon, the more distorted Schwarzschild coordinates are, to the point of infinite distortion at the horizon.

that's what happens according to this coordinate system

No, this isn't right, because you can't use coordinate-dependent quantities to tell "what happens". You have to look at invariants--things that don't depend on what coordinates you choose. The appropriate invariant here is the proper time along the infalling object's worldline from some large radius to the horizon, and that is finite.

 

[PeterDonis]

when there is significant matter outside the event horizon, then that situation is not really modeled by a Schwarzschild solution, if I understand it right.

In the idealized case of a perfectly spherically symmetric collapse, the vacuum portion of the spacetime (i.e., outside the collapsing matter) is modeled by a Schwarzschild solution (which contains both a region outside the horizon, and a region inside the horizon, once the collapsing matter falls inside the horizon radius for its mass). The region occupied by the collapsing matter is modeled by a collapsing FRW solution (the time reverse of the sort of solution that's used to model the expanding universe), which is matched to the Schwarzschild solution along the boundary at the surface of the collapsing matter. Oppenheimer and Snyder first developed this idealized solution in a 1939 paper; MTW give a good discussion of it.

 

[Dale]

well, each coordinate system should give a consistent picture, right?

I would take the complete opposite approach. I would say that no coordinate system gives the physical "picture". The physics is in the invariants, not the coordinates.

Each coordinate system will calculate the same invariants, but it is those invariants and not the coordinates which are important.

 

[DrGreg]

hmm. OK. But in the Schwarzschild coordinate system, it is true that the test particle approaches the event horizon, and gets 'stuck' just outside it, right? You can say this is not the best coordinate system to choose, but that's what happens according to this coordinate system. (unless I've not understood correctly).

Consider as an analogy a map of the world drawn with the Mercator projection. On such a map you have to travel an infinite distance (on the map) upwards to reach the North Pole. Yet people have traveled to the North Pole, and beyond, covering a finite distance (on the Earth's surface).

 

[Khashishi]

I think it is necessary to include the mass of the infalling person to the mass of the black hole. If you treat the black hole as static, and the person's mass as negligible, then the person never crosses the event horizon in the Schwarzschild metric. But a real person has mass, so the event horizon increases in size when a person falls in, so the static Schwarzschild metric isn't valid.

 

[PeterDonis]

If you treat the black hole as static, and the person's mass as negligible, then the person never crosses the event horizon in the Schwarzschild metric.

This is incorrect. Why do you think this?

But a real person has mass, so the event horizon increases in size when a person falls in, so the static Schwarzschild metric isn't valid.

Technically this is true, but for any black hole of practical interest, i.e., of stellar mass or larger, its mass will be so much larger than the mass of a typical infalling object that the static Schwarzschild metric works fine for analyzing the infall process.

 

[WannabeNewton]

I think it is necessary to include the mass of the infalling person to the mass of the black hole. If you treat the black hole as static, and the person's mass as negligible, then the person never crosses the event horizon in the Schwarzschild metric. But a real person has mass, so the event horizon increases in size when a person falls in, so the static Schwarzschild metric isn't valid.

No that's not really what's going on. In general we go from stationary state to stationary state of the black hole i.e. a certain multi-parameter family of stationary black hole solutions and infalling mass perturbations can be taken into account using the equations of black hole dynamics but if the infalling mass is negligible in comparison to the black hole mass then the stationary state solutions are essentially the same.

In the freely falling frame of the radially infalling test particle, there are no "issues": the Schwarzschild metric can perfectly well describe the physics as observed in this freely falling frame and we simply find that the freely falling particle falls past the event horizon in finite time as read by a comoving clock. For static observers there is an infinite redshift of light signals being sent from the infalling particle, simple as that-it's only ostensibly an "issue".

 

[BruceW]

In the idealized case of a perfectly spherically symmetric collapse, the vacuum portion of the spacetime (i.e., outside the collapsing matter) is modeled by a Schwarzschild solution (which contains both a region outside the horizon, and a region inside the horizon, once the collapsing matter falls inside the horizon radius for its mass). The region occupied by the collapsing matter is modeled by a collapsing FRW solution (the time reverse of the sort of solution that's used to model the expanding universe), which is matched to the Schwarzschild solution along the boundary at the surface of the collapsing matter. Oppenheimer and Snyder first developed this idealized solution in a 1939 paper; MTW give a good discussion of it.

Awesome, thanks. I believe this is the kind of answer the OP (and I) was looking for. I don't have that book, but I found this webpage http://www.aei.mpg.de/~rezzolla/lnotes/mondragone/collapse.pdf Which gives more detail on this kind of stuff (although probably not as much detail as that book gives, but oh well).

So anyway, the Oppenheimer and Snyder solution seems to be a specific case for when the matter distribution is uniform within some region, and zero outside. But that's pretty good. I think that gives a good idea of the sort of behaviour to expect. And as you say, in this case, the region occupied by the collapsing matter is modeled by an FRW metric, not a Schwarzschild metric.

I think this is the key point, because in the FRW metric, the geodesics will fall through the event horizon, to the centre of the mass distribution. So this gives an answer as to how matter can form the black hole in the first place. (Because the metric is not a Schwarzschild metric in the first place). When the matter distribution is spread out, the event horizon is well within the region of matter. Therefore, this matter can pass inwards through the event horizon (which is contained in the region described by FRW metric). And as matter passes inside the event horizon, the event horizon grows, until eventually, the event horizon is greater than the matter region. And it is only now, that the event horizon is in the region described by the Schwarzschild metric.

Also, the region which is described by the Schwarzschild metric is only outside of any non-negligible amount of matter. So, if we have formed a black hole, yet there is more material which is outside the event horizon, then the metric describing the larger region of matter is not a Schwarzschild metric. And this is the reason that the black hole is able to pull the extra material inside the event horizon (even according to some far-away stationary observer).

 

[BruceW]

Just to reiterate the point I'm trying to say here: I agree that we can just use Invariants, or some other coordinate system, and everything will work out fine. What I was trying to figure out is how the Schwarzschild coordinates are reconcilable to what is actually going on. In other words, how it can be reconciled with the fact that matter must be inside the black hole, yet in the Schwarzschild coordinates, geodesics do not pass through the event horizon. And what I think is the reconciling answer, is that the FRW metric is the correct metric, when there is still infalling, non-negligible matter. (And I think this is the kind of answer the OP was looking for). And thanks to PeterDonis, who actually explained this first.

edit: well, the FRW metric is only for a specific case when the infalling matter is homogeneous within some region. For a more general matter distribution, there would be a more general metric. But anyway, the FRW metric is a good example of the kind of metric tensor which would apply to the infalling matter.

 

[dauto]

I think the best way to try to understand that is to forget about the black hole and study Rindler coordinates instead. Rindler coordinates is a set of coordinates for flat Minkosky space (special relativity) which exhibits a horizon (with hawking radiation and all). It comes to show how nothing special really happens at a horizon. Rindler coordinates are useful for uniformly accelerated motion (constant proper acceleration). It turns out that if you accelerate, part of minkowsky space becomes invisible to you (because it is behind the horizon).

 

[PeterDonis]

in the Schwarzschild coordinates, geodesics do not pass through the event horizon

This is not correct. The correct statement is that geodesics reach the "edge" of Schwarzschild coordinates (i.e, the limit as $t \rightarrow \infty$) in finite proper time, and at that point, Schwarzschild coordinates can't describe what happens. That's not the same as what you said in the quote above; what you said in the quote above would only be valid if Schwarzschild coordinates could actually cover the horizon, and they can't.

And what I think is the reconciling answer, is that the FRW metric is the correct metric, when there is still infalling, non-negligible matter.

Obviously the Schwarzschild metric is not valid when there is non-negligible matter present, since the Schwarzschild solution is a vacuum solution. However, you are wrong in thinking that this means there is anything to "reconcile". See above.

 

[BruceW]

according to a far-away observer, if we used Schwarzschild solution, infalling matter would stop at the event horizon. This is a problem. But Schwarzschild solution is not the correct solution when there is still infalling matter. Therefore, no problem. This is what I would say to the OP. Am I right in saying this? Or do I still have it wrong?

 

[WannabeNewton]

according to a far-away observer, if we used Schwarzschild solution, infalling matter would stop at the event horizon.

This is wrong. It doesn't stop at the event horizon. The distant static observer simply asymptotically loses communication with it because incoming light signals from it to the distant static observer get asymptotically infinitely redshifted.

 

[PeterDonis]

according to a far-away observer, if we used Schwarzschild solution, infalling matter would stop at the event horizon.

No, this is *not* correct. You are mistaking an artifact of coordinates for an actual physical phenomenon.

Schwarzschild solution is not the correct solution when there is still infalling matter

If the infalling matter is negligible in mass compared to the black hole, the Schwarzschild solution is not *exactly* correct, but it's a very, very good approximation, certainly good enough that if it did in fact predict that matter could not fall through the horizon, that would be a problem. There is no problem because it does *not* predict that.

If the infalling matter is *not* negligible in mass compared to the black hole, then the solution is certainly more complicated, yes. But if such a solution were the only way to describe matter falling into a black hole's horizon, that would be a problem, since there are many cases of interest in which the mass of the infalling matter *is* negligible. There is no problem because, as above, the Schwarzschild solution works fine for such cases.

 

[PAllen]

according to a far-away observer, if we used Schwarzschild solution, infalling matter would stop at the event horizon. This is a problem. But Schwarzschild solution is not the correct solution when there is still infalling matter. Therefore, no problem. This is what I would say to the OP. Am I right in saying this? Or do I still have it wrong?

No, using SC coordinates, you readily calculate that a falling body reaches the end of the coordinates in finite proper time, with all local processes proceeding normally. The coordinates then cannot say anything more. You want to say more, you must use coordinates that cover more. As for 'stopping', using SC coordinates you can compute the relative velocity of an infaller with respect to a near horizon static observer. This speed is not asymptotically zero, it is near light speed instead.

The only 'takes forever' aspect to SC coordinates is that the events described in the paragraph above are given t coordinate values approaching infinity. This infinite value has some physical basis because it corresponds to natural simultaneity surfaces for an observer at infinity. Natural simultaneity reflects two way causal connection - ability to exchange signals. Because signals have increasing difficulty escaping the near horizon, the closer you get to it, normal simultaneity makes these correspond to events approaching infinite time for the distant observer. But even in SC coordinates, this says nothing different about the near horizon physics than any other coordinates.

 

[BruceW]

hmm. but according to a far-away observer using SC coordinates, any matter that falls towards the event horizon would only approach the event horizon, as $t \rightarrow \infty$ So any infalling matter would never reach the event horizon, even if $t$ was really large. yes, the infalling matter would pass through the event horizon within finite proper time. But this isn't important to me (the far-away observer), because in reality, according to me, the matter will never reach the event horizon, even if I wait an arbitrarily long amount of time.

 

[WannabeNewton]

But this isn't important to me (the far-away observer), because in reality, according to me, the matter will never reach the event horizon, even if I wait an arbitrarily long amount of time.

No, as has already been stated you simply lose communication with the particle. Two-way light signaling asymptotically approaches infinite round-trip time. The same exact thing happens in Rindler space-time.

This is not the same thing as the particle "stopping" at the event horizon. In order for the particle to "stop" it would have to have zero relative velocity. Relative velocity is local. You cannot have local static observers at the event horizon.

 

[PAllen]

hmm. but according to a far-away observer using SC coordinates, any matter that falls towards the event horizon would only approach the event horizon, as $t \rightarrow \infty$ So any infalling matter would never reach the event horizon, even if $t$ was really large. yes, the infalling matter would pass through the event horizon within finite proper time. But this isn't important to me (the far-away observer), because in reality, according to me, the matter will never reach the event horizon, even if I wait an arbitrarily long amount of time.

So you equate reality with ability to communicate? If you uniformly accelerate away from earth, you will find it approaching a Rindler horizon. In exactly the sense in which SC coordinates give t=∞ to an infaller reaching the horizon (at 2pm, on their watch, ordinary world around them), Rindler coordinates would give t=∞ to Earth's approach to 2pm. Would really believe, if you were on this rocket, the Earth froze at 2pm?

 

[PeterDonis]

according to a far-away observer using SC coordinates, any matter that falls towards the event horizon would only approach the event horizon, as $t \rightarrow \infty$ So any infalling matter would never reach the event horizon, even if $t$ was really large.

This doesn't mean what you are claiming it means. See below.

in reality, according to me, the matter will never reach the event horizon, even if I wait an arbitrarily long amount of time.

No, the words "in reality" are not correct, because, once more, coordinates do not tell you what happens "in reality". Only invariants can do that. And there is no invariant that says it takes an infinite amount of time for an object to fall to the horizon.

What $t \rightarrow \infty$ means is that the distant observer has drawn a coordinate grid on spacetime, such that events closer and closer to the horizon on an infalling object's worldline have larger and larger $t$ values. But coordinate grids are abstractions. They don't tell you what's physically "real"; you have to compute invariants to see that.

Did you read DrGreg's earlier post drawing an analogy with Mercator coordinates on Earth? The Mercator "latitude" coordinate of the North Pole is infinite; that means that the Mercator projection draws a coordinate grid on the Earth's surface such that points closer and closer to the North Pole have larger and larger coordinate values, increasing without bound. Does that mean that "in reality" the distance to the North Pole is infinite? No. Does it mean that "in reality, according to an observer at the equator, the distance to the North Pole is infinite" (because Mercator coordinates match up with actual physical distances at the equator)? No. All it means is that you've drawn a coordinate grid that gets more and more distorted as you get closer and closer to the North Pole, so coordinate intervals diverge more and more from actual distances.

Similarly, the Schwarzschild coordinate grid on spacetime gets more and more distorted as you get closer and closer to the horizon, so coordinate intervals diverge more and more from actual physical distances and times. That's why you have to compute invariants to see what the actual physical distances and times are. The fact that, very far away from the horizon, coordinate intervals happen to match up with the distant observer's distances and times doesn't mean those intervals must have the same meaning close to the horizon, any more than the fact that Mercator coordinate intervals match up with actual distances at the equator means they must have the same meaning close to the North Pole.

If you still feel inclined to dispute the above, I strongly suggest stepping back and taking some time to review a good textbook on the subject. MTW, for example, has IIRC a good discussion of the limitations of coordinates. If all you can do is to keep on repeating what I quoted above (which you've repeated several times now in spite of being told repeatedly that it's wrong), then there's no point in continuing this discussion since the response isn't going to change.

 

[BruceW]

So you equate reality with ability to communicate? If you uniformly accelerate away from earth, you will find it approaching a Rindler horizon. In exactly the sense in which SC coordinates give t=∞ to an infaller reaching the horizon (at 2pm, on their watch, ordinary world around them), Rindler coordinates would give t=∞ to Earth's approach to 2pm. Would really believe, if you were on this rocket, the Earth froze at 2pm?

I don't really care about the ability to communicate. But I am expecting that each coordinate system is a consistent description of the physical situation. You've given a good example :) thanks. I would say yes, according to me on the rocket, if I keep accelerating with the same proper acceleration, then the Earth will never reach 2pm (according to me, in the rocket). What is wrong with that? I thought this was a standard part of relativity.

 

[pervect]

according to a far-away observer, if we used Schwarzschild solution, infalling matter would stop at the event horizon. This is a problem. But Schwarzschild solution is not the correct solution when there is still infalling matter. Therefore, no problem. This is what I would say to the OP. Am I right in saying this? Or do I still have it wrong?

You're still on the wrong track. IT's true that infalling matter slightly perturbs the black hole metric, but for a small mass the pertubation is not significant, it's not really the answer, it's just a distraction.

The situation is more like Achilles, the tortise, and Zeno, in Zeno's paradox.

The tortise is like the black hole event horizon, and the question is whether Achilles, the fast runner who starts out behind the tortoise, ever reaches the tortoise (the event horizon).

Every time Achilles halves the distance to his goal, Zeno increments his clock by 1 tick. Zeno, according to Zeno time, sees Achilles get closer and closer to the tortoise, but never sees him reach it.

Zeno concludes (incorrectly) that Achilles never reaches the tortoise, because he can't assign a time at which it happens. He also sees Achilles appear to slow down as he gets closer and closer to the tortoise, like the black hole case.

Meanwhile, Achilles, using his own watch, blows past the tortoise in a finite amount of time by Achilles watch.

In abstract terms, how do we describe this? Achilles wordline intersects the tortoise's worldline. Achilles worldline does not stop at the point where it intercepts the tortoise worldline, it continues on beyond that point. However, Zeno cannot assign a time to this event, because of the timekeeping system he adopted - his coordinate system, as it were. If you regard a coordinate system as a map of space-time, Achilles goes off of Zeno's map. But this is nothing physical, its just that Zeno's map doesn't cover all of space-time.

 

[PAllen]

I don't really care about the ability to communicate. But I am expecting that each coordinate system is a consistent description of the physical situation. You've given a good example :) thanks. I would say yes, according to me on the rocket, if I keep accelerating with the same proper acceleration, then the Earth will never reach 2pm (according to me, in the rocket). What is wrong with that? I thought this was a standard part of relativity.

No it is not a standard part of relativity. It is a standard part of crank misinterpretation (I am in no way saying this applies to you).

The 'relative' part of relativity is all about the same observed facts having observer dependent explanations. It is never about the observer determining 'what is real' or even the 'scope of reality'. Those are crank notions. Instead, Einstein often regretted the term 'relativity'; he would have preferred to call it the theory of invariance - emphasizing that the universe doesn't care who observes from where, or what coordinates they use.

 

[BruceW]

I'm trying to think of an observed fact that shows that the time on Earth will really 'freeze' at 2pm (according to me), if I keep accelerating away at constant proper acceleration. It's difficult, but this is the best I've got: OK, suppose the shape of our universe is the 3-torus. The universe is still flat, right, since the 3-torus is just $R^3$ modulo the action of the integer lattice $Z^3$. Now, if I keep accelerating away from the Earth at constant proper acceleration, then according to me, the Earth will never get past 2pm. But also, since the shape of the universe is the 3-torus, I will eventually come back round, and go past the earth. So this is a definite, observable fact, right? And in principle, I could keep going around the universe, passing the Earth each time, and the people on Earth would never reach 2pm according to me. So, anyway, does this example work? Or did I miss something?

 

[PAllen]

I'm trying to think of an observed fact that shows that the time on Earth will really 'freeze' at 2pm (according to me), if I keep accelerating away at constant proper acceleration. It's difficult, but this is the best I've got: OK, suppose the shape of our universe is the 3-torus. The universe is still flat, right, since the 3-torus is just $R^3$ modulo the action of the integer lattice $Z^3$. Now, if I keep accelerating away from the Earth at constant proper acceleration, then according to me, the Earth will never get past 2pm. But also, since the shape of the universe is the 3-torus, I will eventually come back round, and go past the earth. So this is a definite, observable fact, right? And in principle, I could keep going around the universe, passing the Earth each time, and the people on Earth would never reach 2pm according to me. So, anyway, does this example work? Or did I miss something?

Rindler coordinates don't change topology. No coordinates change topology. The universe is still (for SR) a flat Minkowski 4-space. It is just that for a uniformly accelerating observer, setting up coordinates via any simultaneity convention involving sending and receiving signals, 'most' of the universe is inaccessible to two way signals just because the rocket outraces light from much of the universe (it never reaches c, but light from 'most' of the universe can't catch it in finite time because it is getting ever closer to c relative to the emitter). So, setting up coordinates as we normally do, the Rocket simply cannot cover most of the universe, and assigns infinite time coordinates to what inertial observers consider 'ordinary' events.

The SC coordinate situation is very similar. The origin of the infinite time coordinate is nothing other than the inability to exchange signals with the horizon in finite time. A free fall reference frame using two way signals readily covers the horizon and interior. It really is nothing more than the Mercator analogy Dr. Greg mentioned, or the Zeno coordinates Pervect mentioned. Using a Mercator projection has no bearing on what happens at the North pole. It does not mean, in any plausible sense, that the north pole doesn't exist for you. It just means if you choose to use the Mercator projection, you can't represent the North pole.

Now I have to echo Peter, that this seems to going in circles, with all points made by several people multiple times.