The Arrow of Time: The Laws of Physics and the Concept of Time Reversal

[PAllen]

Right. So you’re saying when B crosses the horizon B sees the light from A (and so presumably all the other objects that went in before them) cross at the same time that they do, and then they all jump to whatever distance resembles the last frame that made sense, when they were all in a line?

If I'm on a train with a rows of seats ahead of me, at any moment I see light from all rows ahead of me. I image it so it looks likes the rows are spaced out in front of me (which is the correct reality). The moment of B crossing the horizon is no different. They see light from A (and prior infallers) and it images just like rows of seats on a train.

Then you agree that A cannot cross the horizon until B does?

No, I completely disagree. Light and the emitter of light (A) are not the same thing. The event of B crossing the horizon and receiving light from when A crossed, is not the same event as the event of A crossing the horizon. This is no different from any normal situation - my receiving your signal is different from you sending it. From A's point of view, A has moved on before B crosses the horizon. A can see when B crosses the horizon after (behind) him (if they are close enough). B also sees (visually interprets, correctly) that A crossed before.

If you’d replace the word singularity with the word horizon in those sentences then I’d agree. The singularity is the horizon at zero distance.

I was speaking of the actual singularity, not the horizon. See above for the horizon behavior. The actual singularity is really a time, not a place. B is seeing A some finite distance away the moment B ceases to exist 'when the singularity occurs' for B.

 

[A-wal]

If I'm on a train with a rows of seats ahead of me, at any moment I see light from all rows ahead of me. I image it so it looks likes the rows are spaced out in front of me (which is the correct reality). The moment of B crossing the horizon is no different. They see light from A (and prior infallers) and it images just like rows of seats on a train.

That doesn’t work! They can’t see the light from any prior in-fallers crossing the horizon before they themselves reach it because if they were then to pull away, they would observer any objects that they saw cross coming back out from inside the horizon.

No, I completely disagree. Light and the emitter of light (A) are not the same thing. The event of B crossing the horizon and receiving light from when A crossed, is not the same event as the event of A crossing the horizon. This is no different from any normal situation - my receiving your signal is different from you sending it. From A's point of view, A has moved on before B crosses the horizon. A can see when B crosses the horizon after (behind) him (if they are close enough). B also sees (visually interprets, correctly) that A crossed before.

How? A would have to jump forward after B reaches the horizon. ?

I was speaking of the actual singularity, not the horizon. See above for the horizon behavior. The actual singularity is really a time, not a place. B is seeing A some finite distance away the moment B ceases to exist 'when the singularity occurs' for B.

I know what you meant. It’s an infinitesimally small portion of space-time. The closest anything can ever get to it is arbitrarily close to the event horizon.

 

[Passionflower]

A-Wal you are completely mistaken.
Perhaps it will help if you study Gullstrand–Painlevé coordinates.

It seems you completely mix up that some observers cannot see an event with an event actually happening, which is the whole point of an event horizon.

Suit yourself, but you will not learn if you do not accept what people are trying to tell you here.

 

[PAllen]

That doesn’t work! They can’t see the light from any prior in-fallers crossing the horizon before they themselves reach it because if they were then to pull away, they would observer any objects that they saw cross coming back out from inside the horizon.

How? A would have to jump forward after B reaches the horizon. ?

I know what you meant. It’s an infinitesimally small portion of space-time. The closest anything can ever get to it is arbitrarily close to the event horizon.

Let me try one last(?) to explain. Let's introduce C as well as A and B. A, B and C are free falling directly towards a supermassive black hole, A starts out closer than B, who is closer than C. Let's also be clear they are quite close together (it makes things simpler). From A,B,C perspective, the black hole horizon is rushing ever faster toward them. When it reaches A, it is moving at the same speed as the light A emits towards B. When it (the horizon) reaches B, it is moving at the same speed as the light from B towards C. To clarify further what C sees, let's introduce a fanciful element - that anything crossing the event horizon turns pink. So what C sees is A and B some distance away, pink horizon rushing towards them. At some moment, though A and B continue to appear some distance away, perfectly normal, they both turn pink at the same time (and so does C), from C's point of view. The visual 'turning pink all at once' occurs simply because the horizon is moving at the speed of light past A, then B, then C.

[EDIT: Note that if C interprets what he sees in a normal way, then even though he 'sees' A and B turn pink simultaneously, he reasons that since A is further away than B, A must have turned pink slightly before B. If it were really simultaneous from C's point of view, C would expect to see B turn pink first.]

[EDIT2: And to show the similarity to a Rindler horizon, let's say, instead of a horizon, we simply have A, B, and C sitting in a row. A light source beyond A (in the same line) suddenly starts emitting pink light. C visually sees A, B and himself turn pink at the same time, but interprets that A turned pink, then B, then C. Up until the moment the pink light reaches C, C has the option of starting frantic acceleration away from B. Given sufficient acceleration, the pink light never reaches C, and C never sees A and B turn pink (though they do). This is a Rindler horizon and it is quite analogous to the black hole horizon.]

 

[PeterDonis]

It’s 3.

I’m saying that they have energy applied to them in such a way that any object or individual part of any of the objects has a velocity equal to c / the distance between that part of the object and the object they’re all heading towards, squared.

As you've stated it, this makes no sense. The units don't even work; c is a velocity, so you can't divide it by a distance to get another velocity.

Anyway, in flat spacetime it's impossible for an object to be accelerating towards another object which is not accelerating itself, and not reach the other object eventually. So whatever scenario you think you're constructing as an analogy, it won't work.

 

[Dale]

I’m assuming that if no object in front of you can reach the horizon before you do from the perspective of an in-falling observer, then obviously if you can’t reach the horizon before they do neither then they all reach at the same time. To say that they reach the horizon separately in a finite time own their own clocks doesn’t make sense. Either make it make sense or admit that you’ve all made a terrible mistake.

I thought you were basing your conclusions on Schwarzschild coordinates, which definitely do not represent the perspective of an in-falling observer. If you want to do that then you would be much better off using GP coordinates as Passionflower suggested. However, those results would still be coordinate-dependent, but at least they would be coordinates that reflect the persepective of an in-falling observer.

 

[Passionflower]

However, those results would still be coordinate-dependent, but at least they would be coordinates that reflect the perspective of an in-falling observer.

Indeed, radially in-falling at escape velocity to be exact.

 

[Dale]

Good point, you could certainly consider radial free-falling observers that were falling at velocities other than escape velocity. The GP coordinates would not reflect their perspective.

 

[A-wal]

A-Wal you are completely mistaken.
Perhaps it will help if you study Gullstrand–Painlevé coordinates.

It seems you completely mix up that some observers cannot see an event with an event actually happening, which is the whole point of an event horizon.

Suit yourself, but you will not learn if you do not accept what people are trying to tell you here.

What people are trying to tell me here is that I need to use coordinate systems to put it into terms they understand. I’m not talking about what they see. I’m talking about crossing an event horizon. You can account for the delay in what they see.

Let me try one last(?) to explain. Let's introduce C as well as A and B. A, B and C are free falling directly towards a supermassive black hole, A starts out closer than B, who is closer than C. Let's also be clear they are quite close together (it makes things simpler). From A,B,C perspective, the black hole horizon is rushing ever faster toward them. When it reaches A, it is moving at the same speed as the light A emits towards B. When it (the horizon) reaches B, it is moving at the same speed as the light from B towards C. To clarify further what C sees, let's introduce a fanciful element - that anything crossing the event horizon turns pink. So what C sees is A and B some distance away, pink horizon rushing towards them. At some moment, though A and B continue to appear some distance away, perfectly normal, they both turn pink at the same time (and so does C), from C's point of view. The visual 'turning pink all at once' occurs simply because the horizon is moving at the speed of light past A, then B, then C.

[EDIT: Note that if C interprets what he sees in a normal way, then even though he 'sees' A and B turn pink simultaneously, he reasons that since A is further away than B, A must have turned pink slightly before B. If it were really simultaneous from C's point of view, C would expect to see B turn pink first.]

C sees that A is further away than B? For what you’re saying to make any sense you must mean that C is able to see A and B cross the event horizon? If the answer is yes then that’s not what I’ve been told before, and if it’s no then the horizon can’t be reached by A or B before it’s reached by C, so everything reaches it at the same time, never.

[EDIT2: And to show the similarity to a Rindler horizon, let's say, instead of a horizon, we simply have A, B, and C sitting in a row. A light source beyond A (in the same line) suddenly starts emitting pink light. C visually sees A, B and himself turn pink at the same time, but interprets that A turned pink, then B, then C. Up until the moment the pink light reaches C, C has the option of starting frantic acceleration away from B. Given sufficient acceleration, the pink light never reaches C, and C never sees A and B turn pink (though they do). This is a Rindler horizon and it is quite analogous to the black hole horizon.]

That’s the equivalent to the point when no signal sent will be able to catch a free-faller.

As you've stated it, this makes no sense. The units don't even work; c is a velocity, so you can't divide it by a distance to get another velocity.

Of course you can.

Anyway, in flat spacetime it's impossible for an object to be accelerating towards another object which is not accelerating itself, and not reach the other object eventually. So whatever scenario you think you're constructing as an analogy, it won't work.

Okay, so you tell me how much proper time it would take me to reach an object 100 light years away if my velocity were equal to c / the distance between me and it, squared?

I thought you were basing your conclusions on Schwarzschild coordinates, which definitely do not represent the perspective of an in-falling observer. If you want to do that then you would be much better off using GP coordinates as Passionflower suggested. However, those results would still be coordinate-dependent, but at least they would be coordinates that reflect the persepective of an in-falling observer.

I'm simply talking from the perspective of the free-faller. I don't know enough about coordinate systems to tell you which ones you should be using.

 

[PAllen]

C sees that A is further away than B? For what you’re saying to make any sense you must mean that C is able to see A and B cross the event horizon? If the answer is yes then that’s not what I’ve been told before, and if it’s no then the horizon can’t be reached by A or B before it’s reached by C, so everything reaches it at the same time, never.

You have a chain of infallers. At all times C sees A further away than B. What I described is correct. If you want to simply deny the factual predictions of GR, there is obviously nothing to discuss. Please read and think about what I wrote. The moment C crosses, they see that A and B crossed before. They can't see that A and B crossed until they crossed for the simple reason that the horizon is keeping pace with the light from A at the point where A crossed, and also in step with the light from B when B crossed.

That’s the equivalent to the point when no signal sent will be able to catch a free-faller.

No, it's equivalent to the fact that until the infaller crosses the horizon (from the free faller's point of view: until the horizon passes them at the speed of light), they have the choice to start accelerating frantically to stay 'ahead' of the event horizon that would pass them at the speed of light if they did not accelerate away from it. If they do so, they will never see light from beyond the horizon. The every day scenario I described of an expanding light front captures all essential features of a chain of free fallers approaching (being approached by) the event horizon.

 

[PeterDonis]

Of course you can.

You have got to be either kidding or extremely confused. See next comment.

Okay, so you tell me how much proper time it would take me to reach an object 100 light years away if my velocity were equal to c / the distance between me and it, squared?

I can't because, as I said, if I divide c by a distance I don't get a velocity. c is in meters per second, or m s^-1; distance is in meters. If I divide c by a distance I get s^-1, which is a frequency, not a velocity. If I square it, I get a frequency squared, not a velocity. Or if I interpret the "squared" as applying to the "distance" part only, I divide m s^-1 by m^2, which gives me m^-1 s^-1, which is something I don't even think there's a common word for. It's certainly not a velocity. So what you're asking makes no sense.

 

[A-wal]

You have a chain of infallers. At all times C sees A further away than B. What I described is correct. If you want to simply deny the factual predictions of GR, there is obviously nothing to discuss. Please read and think about what I wrote. The moment C crosses, they see that A and B crossed before. They can't see that A and B crossed until they crossed for the simple reason that the horizon is keeping pace with the light from A at the point where A crossed, and also in step with the light from B when B crossed.

Your explanation doesn’t work. You say that at the moment an in-falling observer reaches the horizon they see all the objects in front of them at the exact instant that they reached the horizon? The horizon’s moving outwards, so you’re seeing them as they were crossing the horizon when the horizon was where they are now? So the horizon is everywhere inside the black hole all at once? Right, so the horizon moves outwards (wrong btw, it’s moving inwards by the time you ‘see’ it) at c locally, and obviously slower the further you are away from it? Two problems.

First the event horizon can only move slower than the speed of the light from the objects in front of you until it reaches you, so you would still only be able to see them cross the horizon before the horizon reaches you, which is wrong.

And also if as you cross you’re seeing everything as it crosses the horizon then the horizon moves depending on what you’re looking at. It can only be in one place at a time. If it’s right in front of you for example and the in-falling objects are all still neatly lined up at varying distances then you’ve seen the ones in front of you cross the horizon! Stop trying to have it both ways. You’re contradicting yourself.

No, it's equivalent to the fact that until the infaller crosses the horizon (from the free faller's point of view: until the horizon passes them at the speed of light), they have the choice to start accelerating frantically to stay 'ahead' of the event horizon that would pass them at the speed of light if they did not accelerate away from it. If they do so, they will never see light from beyond the horizon. The every day scenario I described of an expanding light front captures all essential features of a chain of free fallers approaching (being approached by) the event horizon.

No it isn’t. That happens with free-fallers too. It’s exactly the same. There’s a point when no signal sent will reach a free-faller. That horizon is equivalent the Rindler horizon and it’s further away the further away the free-faller is. It gets closer to them as they fall in, but it never overtakes or reaches them.

You have got to be either kidding or extremely confused. See next comment.

I don't think I'm the one getting confused.

I can't because, as I said, if I divide c by a distance I don't get a velocity. c is in meters per second, or m s^-1; distance is in meters. If I divide c by a distance I get s^-1, which is a frequency, not a velocity. If I square it, I get a frequency squared, not a velocity. Or if I interpret the "squared" as applying to the "distance" part only, I divide m s^-1 by m^2, which gives me m^-1 s^-1, which is something I don't even think there's a common word for. It's certainly not a velocity. So what you're asking makes no sense.

Just give them a starting velocity of zero and make them accelerate smoothly so that their velocity stays equal to c / the distance between them and the object they’re heading towards, squared. If you halve the distance you multiply it by four, because it’s squared. WTF wouldn’t you be able to do that? Start them at c and work backwards to zero where they start. Now work out the proper time it would take to reach the object. It's infinite! I thought I was the one who was crap at maths.

 

[PAllen]

Your explanation doesn’t work. You say that at the moment an in-falling observer reaches the horizon they see all the objects in front of them at the exact instant that they reached the horizon? The horizon’s moving outwards, so you’re seeing them as they were crossing the horizon when the horizon was where they are now? So the horizon is everywhere inside the black hole all at once? Right, so the horizon moves outwards (wrong btw, it’s moving inwards by the time you ‘see’ it) at c locally, and obviously slower the further you are away from it? Two problems.

First the event horizon can only move slower than the speed of the light from the objects in front of you until it reaches you, so you would still only be able to see them cross the horizon before the horizon reaches you, which is wrong.

And also if as you cross you’re seeing everything as it crosses the horizon then the horizon moves depending on what you’re looking at. It can only be in one place at a time. If it’s right in front of you for example and the in-falling objects are all still neatly lined up at varying distances then you’ve seen the ones in front of you cross the horizon! Stop trying to have it both ways. You’re contradicting yourself.

No it isn’t. That happens with free-fallers too. It’s exactly the same. There’s a point when no signal sent will reach a free-faller. That horizon is equivalent the Rindler horizon and it’s further away the further away the free-faller is. It gets closer to them as they fall in, but it never overtakes or reaches them.

We have reached the point where you are simply making false statements with no basis. You are doing the equivalent of saying arithmetic does not state that 1+1=2 because you disagree. I don't know how to proceed from here because lack of knowledge or understanding is no longer the issue - denial of facts is impossible to discuss.

 

[PeterDonis]

Just give them a starting velocity of zero and make them accelerate smoothly so that their velocity stays equal to c / the distance between them and the object they’re heading towards, squared.

Now you're saying something *different*, which still doesn't make sense. Call the moving object O and the target object T. Suppose O starts out 1000 meters from T. You say the starting velocity is zero; but c divided by 1000 meters, or c divided by 1000 meters squared, is not zero (even if I ignore the fact that the units don't work). So I can't start O out consistently with what you're saying, even if I ignore the units (which I can't because the units are part of the physics, and if they aren't balancing, there is something wrong).

If you halve the distance you multiply it by four, because it’s squared.

In which case the velocity would go to infinity as O approached T (see below). This is inconsistent with SR, which you have said you accept.

WTF wouldn’t you be able to do that? Start them at c and work backwards to zero where they start.

*Now* you're saying something different from the above *again*. You're saying that the velocity of O should be c when it is co-located with T. But when it is co-located with T, its distance from T is zero, so going by your previous prescription, its velocity would be c / 0 = infinite.

It's infinite! I thought I was the one who was crap at maths.

You are certainly not winning any prizes for logic or clarity.

 

[PeterDonis]

I don't know how to proceed from here because lack of knowledge or understanding is no longer the issue - denial of facts is impossible to discuss.

Welcome to the reason why this thread is 540 posts long. :sigh:

 

[PeterDonis]

Call the moving object O and the target object T. Suppose O starts out 1000 meters from T.

I suppose I ought to try and expedite this discussion by laying out what SR actually says about this type of scenario, for what it's worth. A-wal appears to be saying that O should start out with velocity zero, and should accelerate towards T. That, at least, is a scenario capable of consistent analysis. However, he also appears to want to stipulate that O should accelerate in such a way that its velocity is c when its distance from T is zero, which of course is impossible. I'm pretty sure A-wal knows *that*, anyway.

A-wal appears to want to draw the conclusion, however, that all this proves somehow that O can't reach T. In SR, of course, it proves no such thing. O can accelerate towards T with any acceleration profile he likes, and two things still will be true: (1) O will reach T in some finite proper time T by O's clock, and also in some finite time t by T's clock (which is the time in the inertial frame in which T is at rest, and in which O is initially at rest--call this frame T's frame); (2) O's velocity, as seen in T's frame, will always be *less* than c (if he maintains a non-zero acceleration, regardless of the specific acceleration profile, his velocity in T's frame will continually increase and approach c more and more closely, but will never reach it).

 

[Passionflower]

Notice that when O is accelerating away from T with an increasing proper acceleration the distance between O and T only approach zero in the limit but never become zero.

 

[PeterDonis]

Notice that when O is accelerating away from T with an increasing proper acceleration the distance between O and T only approach zero in the limit but never become zero.

I assume you mean "distance between O and T, as seen in O's comoving frame"? If O's proper acceleration increases without bound, yes, this distance will approach zero asymptotically if O is accelerating away from T.

However, A-wal has explicitly used the word "towards" several times (see, for example, post #525). So I really think he has in mind a scenario where O's proper acceleration is *towards* T.

 

[Passionflower]

However, A-wal has explicitly used the word "towards" several times (see, for example, post #525). So I really think he has in mind a scenario where O's proper acceleration is *towards* T.

Of course, I just thought it'd be interesting to highlight.

 

[Dale]

I'm simply talking from the perspective of the free-faller. I don't know enough about coordinate systems to tell you which ones you should be using.

GP would definitely be better than Schwarzschild. In GP coordinates the observers fall through the horizon in a finite coordinate time as well as a finite proper time.

 

[Dale]

What people are trying to tell me here is that I need to use coordinate systems to put it into terms they understand. I’m not talking about what they see. I’m talking about crossing an event horizon.

In relativity, the words "X's perspective" is shorthand for "the coordinate system where X is at rest", usually X is an inertial observer and it is implied that they are using an inertial coordinate system. It isn't what they visually see, but what they determine happens after accounting for visual effects like the finite speed of light.

So what I am trying to tell you is that since you are interested in a free fallers perspective you are interested in a coordinate system like GP where free fallers are at rest. They mean the same thing.

 

[Lost in Space]

What puzzles me is the time dilation thing. I mean, from an outsider's perspective (us in other words) no object can actually pass over the event horizon but just appears to hover on the edge forever. But black holes feed, and their event horizons grow as they swallow more mass, don't they? We can observe and measure their gravitational influence caused by their mass even if we can't see them directly so does this mean that effectively all of the mass they have previously absorbed is (as far as we're concerned) piled up on the event horizon, when in reality it has already crossed over? And also, if this is correct, shouldn't black holes then be extremely bright objects?

 

[A-wal]

We have reached the point where you are simply making false statements with no basis. You are doing the equivalent of saying arithmetic does not state that 1+1=2 because you disagree. I don't know how to proceed from here because lack of knowledge or understanding is no longer the issue - denial of facts is impossible to discuss.

No we’ve reached the point were you don’t have an answer so you’re just claiming I don’t understand. Either answer the question or admit that you can’t. A lack of understanding on my part has never been the issue, it’s why I first came here. I think you should try to answer my question. Are you saying that before an object crosses an event horizon that object observes any of the objects in front of them that they can see as being on the other side of the event horizon but they’re seeing them as they were before they reached the horizon, and as the object crosses the horizon it sees them as they were the moment they crossed the horizon? Just trying to be clear because I don’t think any of you even know what you think would happen. So how far in can they observe the objects in front of them? All the way to the singularity? How close do they have to be to observe objects on the other side of the horizon?

Now you're saying something *different*, which still doesn't make sense. Call the moving object O and the target object T. Suppose O starts out 1000 meters from T. You say the starting velocity is zero; but c divided by 1000 meters, or c divided by 1000 meters squared, is not zero (even if I ignore the fact that the units don't work). So I can't start O out consistently with what you're saying, even if I ignore the units (which I can't because the units are part of the physics, and if they aren't balancing, there is something wrong).
...
In which case the velocity would go to infinity as O approached T (see below). This is inconsistent with SR, which you have said you accept.
...
*Now* you're saying something different from the above *again*. You're saying that the velocity of O should be c when it is co-located with T. But when it is co-located with T, its distance from T is zero, so going by your previous prescription, its velocity would be c / 0 = infinite.
...
You are certainly not winning any prizes for logic or clarity.

I’ve been crystal clear about the scenario. Your lack of understanding at this point is a reflection on you, not me. It’s really not that complicated, you’re just making it seem that way. To work out the velocity of any of the objects in the second scenario simply divide c by the distance squared, so that for example if they’re half way in from their starting point then they would be moving at .25c relative to the object that they’re heading towards. Clear? Nothing ever reaches c and I never suggested it did. That’s the whole point.

Welcome to the reason why this thread is 540 posts long. :sigh:

That’s not the reason. It’s taken me this long to get to the point where I can show any who’s actually willing to listen exactly why it works the way it does and confidently challenge you, Dalspam and anyone else who thinks they know better. I’m a slow learner.

I suppose I ought to try and expedite this discussion by laying out what SR actually says about this type of scenario, for what it's worth. A-wal appears to be saying that O should start out with velocity zero, and should accelerate towards T. That, at least, is a scenario capable of consistent analysis. However, he also appears to want to stipulate that O should accelerate in such a way that its velocity is c when its distance from T is zero, which of course is impossible. I'm pretty sure A-wal knows *that*, anyway.

A-wal appears to want to draw the conclusion, however, that all this proves somehow that O can't reach T. In SR, of course, it proves no such thing. O can accelerate towards T with any acceleration profile he likes, and two things still will be true: (1) O will reach T in some finite proper time T by O's clock, and also in some finite time t by T's clock (which is the time in the inertial frame in which T is at rest, and in which O is initially at rest--call this frame T's frame); (2) O's velocity, as seen in T's frame, will always be *less* than c (if he maintains a non-zero acceleration, regardless of the specific acceleration profile, his velocity in T's frame will continually increase and approach c more and more closely, but will never reach it).

Start with a THEORETICAL velocity of c at zero distance then use the distance squared to work out the relative velocity at that time. This isn’t difficult or complicated.

GP would definitely be better than Schwarzschild. In GP coordinates the observers fall through the horizon in a finite coordinate time as well as a finite proper time.

In relativity, the words "X's perspective" is shorthand for "the coordinate system where X is at rest", usually X is an inertial observer and it is implied that they are using an inertial coordinate system. It isn't what they visually see, but what they determine happens after accounting for visual effects like the finite speed of light.

So what I am trying to tell you is that since you are interested in a free fallers perspective you are interested in a coordinate system like GP where free fallers are at rest. They mean the same thing.

Okay and what does GP tell you about seeing objects in front of you crossing the event horizon?

What puzzles me is the time dilation thing. I mean, from an outsider's perspective (us in other words) no object can actually pass over the event horizon but just appears to hover on the edge forever. But black holes feed, and their event horizons grow as they swallow more mass, don't they? We can observe and measure their gravitational influence caused by their mass even if we can't see them directly so does this mean that effectively all of the mass they have previously absorbed is (as far as we're concerned) piled up on the event horizon, when in reality it has already crossed over? And also, if this is correct, shouldn't black holes then be extremely bright objects?

They can't answer you.

 

[Dale]

Okay and what does GP tell you about seeing objects in front of you crossing the event horizon?

That they each cross in a finite coordinate time as well as a finite proper time.

 

[Passionflower]

Are you saying that before an object crosses an event horizon that object observes any of the objects in front of them that they can see as being on the other side of the event horizon but they’re seeing them as they were before they reached the horizon, and as the object crosses the horizon it sees them as they were the moment they crossed the horizon? Just trying to be clear because I don’t think any of you even know what you think would happen.

When two objects, close enough to each other, approach the EH then they will not lose contact because when, after the leading objects passes the event horizon, it sends a light signal back to the trailing object it will actually reach the trailing object only after the trailing object passes the EH as well due to the finite speed of light. However if the trailing object is too far removed from the leading object then because the trailing object does not reach the EH in time to catch the signal from the leading object it will lose contact with the leading object. We can calculate how far the objects can be separated for this to happen or not.

 

[A-wal]

That they each cross in a finite coordinate time as well as a finite proper time.

And do they observe the objects in front of them crossing the horizon before they themselves reach it?

When two objects, close enough to each other, approach the EH then they will not lose contact because when, after the leading objects passes the event horizon, it sends a light signal back to the trailing object it will actually reach the trailing object only after the trailing object passes the EH as well due to the finite speed of light.

So you’re saying that if the object behind is close enough to the one in front then the one behind can observe the one in front crossing the event horizon? And if the trailing object pulls away?

However if the trailing object is too far removed from the leading object then because the trailing object does not reach the EH in time to catch the signal from the leading object it will lose contact with the leading object. We can calculate how far the objects can be separated for this to happen or not.

So the trailing object sees the one in front vanish as it passes the horizon?

 

[PAllen]

No we’ve reached the point were you don’t have an answer so you’re just claiming I don’t understand. Either answer the question or admit that you can’t. A lack of understanding on my part has never been the issue, it’s why I first came here. I think you should try to answer my question. Are you saying that before an object crosses an event horizon that object observes any of the objects in front of them that they can see as being on the other side of the event horizon but they’re seeing them as they were before they reached the horizon, and as the object crosses the horizon it sees them as they were the moment they crossed the horizon? Just trying to be clear because I don’t think any of you even know what you think would happen. So how far in can they observe the objects in front of them? All the way to the singularity? How close do they have to be to observe objects on the other side of the horizon?

It remains fruitless to discuss when factually false statements are claimed to be true. I shall simply list below the major false statements in your response that triggered this. These are not matters of opinion; each of these statements of yours is mathematically false:



1) "First the event horizon can only move slower than the speed of the light from the objects in front of you until it reaches you, so you would still only be able to see them cross the horizon before the horizon reaches you, which is wrong." Answer: factually wrong. The horizon keeps up with the outwardly emitted light of each object it passes.

2) "And also if as you cross you’re seeing everything as it crosses the horizon then the horizon moves depending on what you’re looking at. It can only be in one place at a time. If it’s right in front of you for example and the in-falling objects are all still neatly lined up at varying distances then you’ve seen the ones in front of you cross the horizon! Stop trying to have it both ways. You’re contradicting yourself." I'm not contradicting myself. Here, you may have a simple misunderstanding of light. A single wave front (remember, the horizon is lightlike surface moving past the infaller) has lots of information. If you imagine the infaller as having a number of cameras, each focused on a different distance, then at the moment the horizon reaches you, each camera will get a picture of the infaller at its focal distance ahead, as of when that object passed the horizon.

4) "No it isn’t. That happens with free-fallers too. It’s exactly the same. There’s a point when no signal sent will reach a free-faller. That horizon is equivalent the Rindler horizon and it’s further away the further away the free-faller is. It gets closer to them as they fall in, but it never overtakes or reaches them." Answer: here I admit this response of yours makes no sense to me. My example of the simple light front is precisely analogous the passing event horizon, and should help you see the issues how your have a choice to not go through the horizon up until it passes you by accelerating away. The local equivalence of these examples (passing horizon; passing light front); event horizon, Rindler horizon; are mathematical facts. Don't know what you are going on about signals. It is a fact that an infaller crossing the horizon can continue to get signals from outside (despite the fact that his signals will never get out - they can't 'catch' the horizon). Less well known, perhaps, is that two infallers at close radial separation can continue two way communication inside the horizon for a little while (until the true singularity interferes; all assuming the ideal Schwarzschild geometry, which presumably doesn't actually exist in nature).

 

[Passionflower]

So you’re saying that if the object behind is close enough to the one in front then the one behind can observe the one in front crossing the event horizon? And if the trailing object pulls away?

So the trailing object sees the one in front vanish as it passes the horizon?

I strongly suggest you first try to understand what I am saying before you keep asking more questions.

Sorry A-wal I do not get the impression you are trying to learn anything here, my impression is that you are trying to argue your incorrect view.

I am wasting my time with you.

 

[A-wal]

It remains fruitless to discuss when factually false statements are claimed to be true. I shall simply list below the major false statements in your response that triggered this. These are not matters of opinion; each of these statements of yours is mathematically false:



1) "First the event horizon can only move slower than the speed of the light from the objects in front of you until it reaches you, so you would still only be able to see them cross the horizon before the horizon reaches you, which is wrong." Answer: factually wrong. The horizon keeps up with the outwardly emitted light of each object it passes.

2) "And also if as you cross you’re seeing everything as it crosses the horizon then the horizon moves depending on what you’re looking at. It can only be in one place at a time. If it’s right in front of you for example and the in-falling objects are all still neatly lined up at varying distances then you’ve seen the ones in front of you cross the horizon! Stop trying to have it both ways. You’re contradicting yourself." I'm not contradicting myself. Here, you may have a simple misunderstanding of light. A single wave front (remember, the horizon is lightlike surface moving past the infaller) has lots of information. If you imagine the infaller as having a number of cameras, each focused on a different distance, then at the moment the horizon reaches you, each camera will get a picture of the infaller at its focal distance ahead, as of when that object passed the horizon.

4) "No it isn’t. That happens with free-fallers too. It’s exactly the same. There’s a point when no signal sent will reach a free-faller. That horizon is equivalent the Rindler horizon and it’s further away the further away the free-faller is. It gets closer to them as they fall in, but it never overtakes or reaches them." Answer: here I admit this response of yours makes no sense to me. My example of the simple light front is precisely analogous the passing event horizon, and should help you see the issues how your have a choice to not go through the horizon up until it passes you by accelerating away. The local equivalence of these examples (passing horizon; passing light front); event horizon, Rindler horizon; are mathematical facts. Don't know what you are going on about signals. It is a fact that an infaller crossing the horizon can continue to get signals from outside (despite the fact that his signals will never get out - they can't 'catch' the horizon). Less well known, perhaps, is that two infallers at close radial separation can continue two way communication inside the horizon for a little while (until the true singularity interferes; all assuming the ideal Schwarzschild geometry, which presumably doesn't actually exist in nature).

Just answer these questions then:

In scenario 1 will the in-falling objects observe the objects in front of them reaching the event horizon before they do?

In scenario 2 how much proper time would it take an object to reach their destination?

I strongly suggest you first try to understand what I am saying before you keep asking more questions.

Sorry A-wal I do not get the impression you are trying to learn anything here, my impression is that you are trying to argue your incorrect view.

I am wasting my time with you.

I'm not going to respond to condisending coments that that any more, I'm only going to make them. I'd like to know what each of you think. Maybe you should get your stories straight first. You could vote on what you're going with.

 

[PAllen]

Just answer these questions then:

In scenario 1 will the in-falling objects observe the objects in front of them reaching the event horizon before they do?

In scenario 2 how much proper time would it take an object to reach their destination?

No observer can ever see anything passing the event horizon before they cross it. I feel that I have explained exactly what would be seen and how it would naturally be interpreted over and over, and you either deny it or ask the same thing again.

As for specific scenarios, are you referring to yours or mine? Please specify what post gives the complete definition of what you are calling scenario 1 and scenario 2.

 

[Dale]

And do they observe the objects in front of them crossing the horizon before they themselves reach it?

They observe that the object crosses the horizon before they themselves do. They make the observation that leads to that conclusion when they themselves cross the horizon.

I.e. they are separated by some finite distance, so when the observer crosses the horizon they receive the light from the objects crossing, account for the finite speed of light, and conclude that the objects crossing happened earlier.

The same thing happens with observers falling across a Rindler horizon in flat spacetime.

 

[PAllen]

A-wal, please note there is no discrepancy at all between my #555 and Dalespam's #556. It would be very worthwhile for you to think about why this is so.

 

[PeterDonis]

I’ve been crystal clear about the scenario. Your lack of understanding at this point is a reflection on you, not me. It’s really not that complicated, you’re just making it seem that way.

LOL.

To work out the velocity of any of the objects in the second scenario simply divide c by the distance squared, so that for example if they’re half way in from their starting point then they would be moving at .25c relative to the object that they’re heading towards.

So basically I'm dividing c by the ratio of the current distance d to the starting distance D? That would mean if d = D/2 then v = c/4, as you say. But by that formula, if d = D, then v= c; in other words, v = c at the *starting* point. Which already violates SR. But then you say:

Start with a THEORETICAL velocity of c at zero distance then use the distance squared to work out the relative velocity at that time. This isn’t difficult or complicated.

So does "zero distance" mean "zero distance from the starting point" or "zero distance from the target object"? If it's distance from the starting point, then you're postulating an object that starts out moving at c, which violates SR, as I just said. (You're also requiring the object to accelerate *away* from the target, but you've repeatedly said that you are postulating an object accelerating *towards* the target.)

If it's distance from the target object, then what is the velocity at the starting point, a distance D from the target object? Is it zero? You can't get that out of your formula; to get a zero by dividing c by a distance squared, the distance would have to be infinite.

That’s not the reason. It’s taken me this long to get to the point where I can show any who’s actually willing to listen exactly why it works the way it does and confidently challenge you, Dalspam and anyone else who thinks they know better.

I think it was Bertrand Russell who said: "The whole problem with the world is that fools are so sure of themselves and wise men so full of doubts."

I’m a slow learner.

That's for sure.

Okay and what does GP tell you about seeing objects in front of you crossing the event horizon?

Object A and object B are falling into the black hole. A is a little bit ahead of B. A emits three light signals directly outward, towards B: #1 just before A crosses the horizon, when he is at a radial distance d above the horizon; #2 exactly when A crosses the horizon; and #3 just after A crosses the horizon, when he is at a radial distance d below the horizon.

B will receive all three signals: he will receive #1 just before he himself crosses the horizon, at a radial distance above the horizon slightly *larger* than d (because the outgoing light beam is able to move outward slightly to meet up with B); he will receive #2 exactly as he crosses the horizon; and he will receive #3 after he crosses the horizon, at a radial distance *below* the horizon slightly larger than d (because even outgoing light rays inside the horizon fall inward, so B has to fall slightly farther to catch the light than A did when he emitted it).

So B will indeed see A cross the horizon; but the light signals that show him that will only reach him when he himself crosses the horizon.

Just for completeness, consider a third observer, C, who falls right along with B until B receives the first light signal from A (signal #1, emitted when A is a distance d above the horizon). At that point, C fires his rockets very, very hard, so that he "hovers" a distance d above the horizon. Then C, unlike B, will *not* see A cross the horizon, because the light signals #2 and #3 that would show him that never get outside the horizon, where he is.

 

[A-wal]

This is getting stupid now. It’s obvious purely through the simple fact that you have to use coordinate systems that contradict each other that something’s very wrong. A coordinate system is just look at it from a certain perspective. If you’ve got two that contradict each other then they can’t possibly both be right! Rindler and Schwarzschild coordinates show that the horizon can’t be reached. Any coordinate system that says it can is an alternative view, and a wrong one.

I strongly suggest you first try to understand what I am saying before you keep asking more questions.

I strongly suggest you first try to understand whet you’re saying before answering more questions. Maybe that’s why you stopped trying to?

Sorry A-wal I do not get the impression you are trying to learn anything here, my impression is that you are trying to argue your incorrect view.

I don’t get the impression you’re capable of teaching me anything worth knowing, and at least it’s my own view.

I am wasting my time with you.

Yes you’re wasting everyones time. I’m backing up what I’m saying and you’ve now stopped trying to back up what you’re saying because you’re confused and you don’t know what to do.

No observer can ever see anything passing the event horizon before they cross it. I feel that I have explained exactly what would be seen and how it would naturally be interpreted over and over, and you either deny it or ask the same thing again.

LOL No you haven’t. You’ve tried to explain it in a consistent way over and over, and failed every time. You say they would see in in-fallers in front of them as normal like a row of seats on a train but you also think that objects can’t be seen to cross the horizon from outside the horizon. I keep asking questions because I still haven’t been given a self-consistent answer.

As for specific scenarios, are you referring to yours or mine? Please specify what post gives the complete definition of what you are calling scenario 1 and scenario 2.

Mine. This one.

A-wal, please note there is no discrepancy at all between my #555 and Dalespam's #556. It would be very worthwhile for you to think about why this is so.

It’s because they’re both wrong.

It remains fruitless to discuss when factually false statements are claimed to be true. I shall simply list below the major false statements in your response that triggered this. These are not matters of opinion; each of these statements of yours is mathematically false:

I’m not using maths, so if anyones maths is dodgy it must be yours and you’re the one claiming that factually false statements are true, not me. I can back up what I’m saying with a sound argument and you can’t. The Rindler horizon is not equivalent to an event horizon. C is equivalent to an event horizon.

1) "First the event horizon can only move slower than the speed of the light from the objects in front of you until it reaches you, so you would still only be able to see them cross the horizon before the horizon reaches you, which is wrong." Answer: factually wrong. The horizon keeps up with the outwardly emitted light of each object it passes.

I’m talking about before the object reaches the horizon. You’re saying the light from a previous event (an earlier object crossing the horizon) can reaches an object at the same time as the event horizon itself does, but that would mean the event horizon is traveling at c relative to an in-faller before they reach it. You think the event horizon moves outwards at c at some distance away? How is that calculated?

2) "And also if as you cross you’re seeing everything as it crosses the horizon then the horizon moves depending on what you’re looking at. It can only be in one place at a time. If it’s right in front of you for example and the in-falling objects are all still neatly lined up at varying distances then you’ve seen the ones in front of you cross the horizon! Stop trying to have it both ways. You’re contradicting yourself." I'm not contradicting myself. Here, you may have a simple misunderstanding of light. A single wave front (remember, the horizon is lightlike surface moving past the infaller) has lots of information. If you imagine the infaller as having a number of cameras, each focused on a different distance, then at the moment the horizon reaches you, each camera will get a picture of the infaller at its focal distance ahead, as of when that object passed the horizon.

That really doesn’t make any sense. You see them at some distance away passing the horizon as the horizon’s passing you?

4) "No it isn’t. That happens with free-fallers too. It’s exactly the same. There’s a point when no signal sent will reach a free-faller. That horizon is equivalent the Rindler horizon and it’s further away the further away the free-faller is. It gets closer to them as they fall in, but it never overtakes or reaches them." Answer: here I admit this response of yours makes no sense to me. My example of the simple light front is precisely analogous the passing event horizon, and should help you see the issues how your have a choice to not go through the horizon up until it passes you by accelerating away. The local equivalence of these examples (passing horizon; passing light front); event horizon, Rindler horizon; are mathematical facts. Don't know what you are going on about signals. It is a fact that an infaller crossing the horizon can continue to get signals from outside (despite the fact that his signals will never get out - they can't 'catch' the horizon). Less well known, perhaps, is that two infallers at close radial separation can continue two way communication inside the horizon for a little while (until the true singularity interferes; all assuming the ideal Schwarzschild geometry, which presumably doesn't actually exist in nature).

Where’s 3 Mr Maths? Work out the point when no signal sent from a more distant observer will reach a free-faller and how this horizon moves in relation to the free-faller and you’ll see it’s exactly equivalent to a Rindler horizon. Are you saying a Rindler horizon has more in common with an event horizon than it does with the horizon I just described?

LOL.

That’s exactly what I do every time I see the words ‘science advisor’ by the name of nearly everyone in here.

So basically I'm dividing c by the ratio of the current distance d to the starting distance D? That would mean if d = D/2 then v = c/4, as you say. But by that formula, if d = D, then v= c; in other words, v = c at the *starting* point. Which already violates SR. But then you say:

I have to simplify everything you say and you go out of your way to complicate everything I say. How the hell does it violate SR? It doesn’t mean anything actually travels at c.

So does "zero distance" mean "zero distance from the starting point" or "zero distance from the target object"? If it's distance from the starting point, then you're postulating an object that starts out moving at c, which violates SR, as I just said. (You're also requiring the object to accelerate *away* from the target, but you've repeatedly said that you are postulating an object accelerating *towards* the target.)

If it's distance from the target object, then what is the velocity at the starting point, a distance D from the target object? Is it zero? You can't get that out of your formula; to get a zero by dividing c by a distance squared, the distance would have to be infinite.

Yes it’s distance from the target. I’ll use energy instead because you seem to struggle with relative velocity as well as basic maths when you have to think for yourself. Then it doesn’t matter what their distance or velocity is.

I think it was Bertrand Russell who said: "The whole problem with the world is that fools are so sure of themselves and wise men so full of doubts."

Speak for yourself. You so used to parroting others work and words that it’s become second nature hasn’t it? People who question things more, including themselves are less likely to be wrong. That’s common sense. The problem is a stubborn refusal to accept that what you’ve been taught is wrong. I’m sorry but I can’t help you with that. All I can do is explain why I think it works this way and not your way, and see if at least one of you has a level of understanding that justifies your religiously held beliefs. It’s not looking promising. And you’ve just proved your own point. You have no doubts about gr. You think it must be right if everyone else believes it. Trouble is everyone thinks that.

That's for sure.

If that’s an indication of your level of wit then at least I wouldn’t have to worry about you making me look silly even if I was wrong.

Object A and object B are falling into the black hole. A is a little bit ahead of B. A emits three light signals directly outward, towards B: #1 just before A crosses the horizon, when he is at a radial distance d above the horizon; #2 exactly when A crosses the horizon; and #3 just after A crosses the horizon, when he is at a radial distance d below the horizon.

B will receive all three signals: he will receive #1 just before he himself crosses the horizon, at a radial distance above the horizon slightly *larger* than d (because the outgoing light beam is able to move outward slightly to meet up with B); he will receive #2 exactly as he crosses the horizon; and he will receive #3 after he crosses the horizon, at a radial distance *below* the horizon slightly larger than d (because even outgoing light rays inside the horizon fall inward, so B has to fall slightly farther to catch the light than A did when he emitted it).

So B will indeed see A cross the horizon; but the light signals that show him that will only reach him when he himself crosses the horizon.

What? When B reaches the horizon they see A cross the horizon? But then the horizon is in two places at once, or all the in-falling objects are in the same place at the same time. Neither makes any sense and the stupid thing is there’s no reason to think that they should because there’s absolutely no reason to think an object can reach an event horizon. There’s no reason I can think of to suspect that it works any differently than it does for an observer accelerating towards c in flat space-time.

Just for completeness, consider a third observer, C, who falls right along with B until B receives the first light signal from A (signal #1, emitted when A is a distance d above the horizon). At that point, C fires his rockets very, very hard, so that he "hovers" a distance d above the horizon. Then C, unlike B, will *not* see A cross the horizon, because the light signals #2 and #3 that would show him that never get outside the horizon, where he is.

Right so B and C fall together and C brakes and hovers before reaching the horizon. Let’s give this a little nudge to show just how fragile it is. A through to Y are falling into the black hole. Z hovers at a very short distance away from the horizon. Z must either see all those letters of the alphabet in a line in front of them between them and the horizon, or see them all in the same place just in front of them. If Z sees them all in a line in front of him and none of them have crossed the horizon then there will always be space in between Z and the horizon. If they’re all on top of each other then the conveyer belt gets jammed.

They observe that the object crosses the horizon before they themselves do. They make the observation that leads to that conclusion when they themselves cross the horizon.

And what are the specific reasons why you think it makes more sense to use those coordinates instead of the same coordinates you’d use for an accelerating observer in flat space-time?

I.e. they are separated by some finite distance, so when the observer crosses the horizon they receive the light from the objects crossing, account for the finite speed of light, and conclude that the objects crossing happened earlier.

Your explanation doesn’t make sense. Objects can cross the event horizon but they can’t be seen doing it until the observing object reaches the horizon? So what happens when an object hovers just above the horizon? It will never see any of the other objects crossing the horizon no matter how close it stops to the horizon. It’s either that or they can see objects crossing the horizon and I can rip both to shreds.

The same thing happens with observers falling across a Rindler horizon in flat spacetime.

The accelerator observes them crossing the Rindler horizon when they stop accelerating but a free-faller doesn’t see an object in front of them crossing the event horizon if they start to hover!


Scenario 1).
You claim that objects can reach even horizons in a finite amount of their own proper time. If a line of objects were continuously falling into a black hole like a conveyer belt then this would cause a few problems. None of the objects in front can reach the horizon before they themselves do. So they all have to cross the horizon at the same time. If everything that will have to reach a horizon has to do it at a specific time then at what point in the black holes life will that be? A through to Y are falling into the black hole. Z hovers at a very short distance away from the horizon. Z must either see all those letters of the alphabet in a line in front of them between them and the horizon, or see them all in the same place just in front of them. If Z sees them all in a line in front of him and none of them have crossed the horizon then there will always be space in between Z and the horizon. If they’re all on top of each other then the conveyer belt gets jammed.

Scenario 2).
If we do the same thing with a row of accelerating objects that accelerate harder the closer they are to a distant object and started them off equally spaced then they would start to separate, and at a quicker rate the closer they are to the object that they're heading towards. The amount of energy the individual parts of the objects feel is always infinity / the distance squared to the destination object. The actual amount of energy can be anything you like (just as a singularity can have any mass), as long as that amount of energy quadruples when you halve the distance and is divided by four when you halve the distance then it will work with any amount of energy at any distance with any starting velocity relative to the destination. None of these objects would be able to reach the relatively stationary object that they're heading towards. They'd just get more and more time dilated and length contracted as they get closer at a slower and slower rate from any distance.


Clearly you can’t see when you’re screwed, even if it’s staring you right in the face. It’s checkmate. We can keep doing this if you want to but it’s over. In case you hadn’t realized, refused to accept or simply hadn’t noticed, I’ve got GR by the bollocks and I’m not letting go until I’ve killed it. It’s a lie. I can make this as nasty as you want it to be. I’m just getting warmed up. I’m so glad I don’t have to fumble around in the dark with equations and coordinate systems. You'd better starting making sense soon or you’re all going to start looking a bit stupid. I actually care about this stuff and I’m not going to let a bunch of parrots who have memorised some words and numbers but have very little understanding of what they mean when they’re put together (as demonstrated by PeterDonis’s recent comment that they’ll always get there in the end even they have to reach c to do it, as well as about a dozen others) carry on pretending to have some kind of deep understanding when in reality they don’t have a clue. Seems to me like none of you are interested in the truth. All you care about is defending something you don’t even fully understand. It really is just like talking to a bunch of god worshipers. You can’t even give me a self-consistent alternative!

 

[Passionflower]

I’ve got GR by the bollocks and I’m not letting go until I’ve killed it. It’s a lie. I can make this as nasty as you want it to be. I’m just getting warmed up. I’m so glad I don’t have to fumble around in the dark with equations and coordinate systems. You'd better starting making sense soon or you’re all going to start looking a bit stupid.

Oh dear, perhaps it is time to close this topic before A-wal embarrasses himself even more.