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

[A-wal]

Still, no object in front of you can reach the horizon before you do, meaning all objects reach the horizon at the exact moment that you do. This clearly doesn't work. It's like trying to reach c in the other example. If they were able to reach the object they're accelerating towards then they'd all do it at the same time.

Hi A-wal, welcome back!

Thanks. I seem to be able to think clearer when I'm in pain.

 

[PAllen]

Still, no object in front of you can reach the horizon before you do

Wrong. For a chain of infalling observers, each would see the lower object crossing the horizon 'a little' before they cross, all in very finite time on their own 'watch'. Assuming an ancient, stable, super-massive black hole, nothing would be distinguishable about the horizon, locally. Starting from well before the horizon, distant stars would be blue shifted and optically distorted, but there would be no further sudden change reflecting passage of the horizon.

 

[Passionflower]

Wrong. For a chain of infalling observers, each would see the lower object crossing the horizon 'a little' before they cross, all in very finite time on their own 'watch'. Assuming an ancient, stable, super-massive black hole, nothing would be distinguishable about the horizon, locally. Starting from well before the horizon, distant stars would be blue shifted and optically distorted, but there would be no further sudden change reflecting passage of the horizon.

You mean red shifted.

 

[PAllen]

You mean red shifted.

Yes, I was thinking of static observers near the horizon, though discussing in-falling observers. For in-falling observers, the reality is redshift/blue shift of distant stars is direction dependent. If falling straight in, maximum red shift for those behind you, maximum blue shift for those in front of you, others in between.

 

[Passionflower]

Yes, I was thinking of static observers near the horizon, though discussing in-falling observers. For in-falling observers, the reality is redshift/blue shift of distant stars is direction dependent. If falling straight in, maximum red shift for those behind you, maximum blue shift for those in front of you, others in between.

By the way radially free falling observers can observe far away stars blue shifted, but the closer they get to the EH the less of a possibility that is as their proper velocity based Doppler factor overruns the gravitationally based Doppler factor.

One could plot this out taking several radially free falling observers with different energies. For all the proper velocity at the EH will be c but their proper velocities on their way to the EH will differ and thus the red/blue shift factor wrt to far away stars will also.

 

[PAllen]

By the way radially free falling observers can observe far away stars blue shifted, but the closer they get to the EH the less of a possibility that is as their proper velocity based Doppler factor overruns the gravitationally based Doppler factor.

One could plot this out taking several radially free falling observers with different energies. For all the proper velocity at the EH will be c but their proper velocities on their way to the EH will differ and thus the red/blue shift factor wrt to far away stars will also.

Right. If instead of in-falling from far, one imagined the radial in-fall of static, near horizon, observer whose rocket ran out of fuel, they would initially see all stars blue-shifted.

 

[Passionflower]

Right. If instead of in-falling from far, one imagined the radial in-fall of static, near horizon, observer whose rocket ran out of fuel, they would initially see all stars blue-shifted.

Exactly.

 

[PeterDonis]

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. An objects velocity is always c / the distance squared to the destination object. None of these objects would be able to reach the 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.

I don't understand what you're describing here. You are saying that the accelerating objects are supposed to be accelerating *towards* the distant object? If so, how are you coming up with the conclusion that they can never reach it?

Also, what accelerations are the objects feeling? Are they all feeling the same acceleration? Or does the acceleration they feel vary with their starting distance from the distant object? If so, how?

 

[Dale]

Still, no object in front of you can reach the horizon before you do, meaning all objects reach the horizon at the exact moment that you do.

You have to define "in front" and "before". If you use coordinate independent definitions (e.g. Based on light cones) then you will get coordinate independent results. If you simply use the Schwarzschild coordinates then obviously you get coordinate-dependent results. There is nothing particularly meaningful about that.

 

[Dale]

So they all have to cross the horizon at the same time.

Actually, now that I think about it I am not sure this is true even in Schwarzschild coordinates. Infinity minus infinity isn't 0, it is undefined.

I think that you would have to find the difference in coordinate time between when the two observers cross a given radius and then take the limit of that as the radius approaches the Schwarzschild radius. That limit might be zero, but it isn't obvious. In any case it is coordinate dependent

 

[A-wal]

Wrong. For a chain of infalling observers, each would see the lower object crossing the horizon 'a little' before they cross, all in very finite time on their own 'watch'. Assuming an ancient, stable, super-massive black hole, nothing would be distinguishable about the horizon, locally. Starting from well before the horizon, distant stars would be blue shifted and optically distorted, but there would be no further sudden change reflecting passage of the horizon.

That can’t be right. If they see them fall past the horizon before they reach it themselves then the object that fell in would have to come back out if the other one pulled away.

I don't understand what you're describing here. You are saying that the accelerating objects are supposed to be accelerating *towards* the distant object? If so, how are you coming up with the conclusion that they can never reach it?

Because if they reached it then they would be moving at c relative to it. They become more and more ‘distance shortened’ as they get closer.

Also, what accelerations are the objects feeling? Are they all feeling the same acceleration? Or does the acceleration they feel vary with their starting distance from the distant object? If so, how?

They’re accelerating completely smoothly so that each individual atom is accelerating independently based on its distance from the object. The only acceleration they feel is the difference between the different parts of those objects, which doesn’t become noticeable until they get fairly close.

You have to define "in front" and "before". If you use coordinate independent definitions (e.g. Based on light cones) then you will get coordinate independent results. If you simply use the Schwarzschild coordinates then obviously you get coordinate-dependent results. There is nothing particularly meaningful about that.

Actually, now that I think about it I am not sure this is true even in Schwarzschild coordinates. Infinity minus infinity isn't 0, it is undefined.

I think that you would have to find the difference in coordinate time between when the two observers cross a given radius and then take the limit of that as the radius approaches the Schwarzschild radius. That limit might be zero, but it isn't obvious. In any case it is coordinate dependent

From the perspective of an in-falling observer; no observer in front of you can possibly reach the horizon before you do!

 

[PeterDonis]

Because if they reached it then they would be moving at c relative to it. They become more and more ‘distance shortened’ as they get closer.

I still don't understand. Are you trying to describe (1) what you think GR actually says, or (2) what your non-GR model says, or (3) what happens in a scenario in flat spacetime (no gravity) that you think is analogous to the scenario of a black hole?

If it's #1, your description is simply wrong. GR does not say what you are claiming it says.

If it's #2, how does your model make the predictions you are describing?

If it's #3, then I don't see how the following fits in:

They’re accelerating completely smoothly so that each individual atom is accelerating independently based on its distance from the object. The only acceleration they feel is the difference between the different parts of those objects, which doesn’t become noticeable until they get fairly close.

This can't happen in flat spacetime; you're basically saying the only "acceleration" felt by the objects is tidal acceleration, but there is no tidal acceleration in flat spacetime. Which brings me back to either #1 or #2 above, but see my questions about those.

 

[PeterDonis]

Actually, now that I think about it I am not sure this is true even in Schwarzschild coordinates. Infinity minus infinity isn't 0, it is undefined.

I think this is a better way of putting it, because the important point, to me, is that Schwarzschild coordinates are singular at the horizon. That means you can't use them to even describe relationships like "before", "after", "in front of", "in back of" at the horizon in the first place. A-wal is assuming that t = infinity at the horizon in Schwarzschild coordinates implies that all infalling objects reach the horizon "at the same time", when what it actually means is that Schwarzschild coordinates can't be used to describe the horizon at all.

 

[PAllen]

That can’t be right. If they see them fall past the horizon before they reach it themselves then the object that fell in would have to come back out if the other one pulled away.

No, what I said is correct. The answer to your objection is simply that if A and B are a pair of infallers, with B at a little higher r value, then B sees (absorbs) light from A crossing the horizon at the moment B crosses the horizon. So, indeed, no light leaves the horizon. This is no different from any normal situation where if I see light from you now, and you are moving in a known direction, I interpret that 'now' you are further in the direction you were moving. That is, when B crosses the horizon, it looks like A crossed before (and A is, indeed, inside already), and is now a bit inside. Note that light emitted directly away from the singularity at the horizon remains frozen at the horizon - waiting, if you will, for the next infaller.

If B pulls away any time before crossing the horizon, then, indeed, they never see light from A at or inside the horizon. This would look a lot like a Rindler horizon. B suddenly accelerates madly to escape, A gets red shifted and frozen, just like a Rindler horizon.

One final observation is that B never sees A reach the singularity. At the moment B reaches the singularity, the last light they see from A is from a little before A reached the singularity.

 

[A-wal]

I still don't understand. Are you trying to describe (1) what you think GR actually says, or (2) what your non-GR model says, or (3) what happens in a scenario in flat spacetime (no gravity) that you think is analogous to the scenario of a black hole?

If it's #1, your description is simply wrong. GR does not say what you are claiming it says.

If it's #2, how does your model make the predictions you are describing?

If it's #3, then I don't see how the following fits in:

It’s 3.

This can't happen in flat spacetime; you're basically saying the only "acceleration" felt by the objects is tidal acceleration, but there is no tidal acceleration in flat spacetime. Which brings me back to either #1 or #2 above, but see my questions about those.

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. In terms of actual acceleration it’s slightly different because the individual parts of an object are held together by an electro-magnetic field which creates resistance, which is felt as tidal force in the first scenario and proper acceleration in the second one.

I think this is a better way of putting it, because the important point, to me, is that Schwarzschild coordinates are singular at the horizon. That means you can't use them to even describe relationships like "before", "after", "in front of", "in back of" at the horizon in the first place. A-wal is assuming that t = infinity at the horizon in Schwarzschild coordinates implies that all infalling objects reach the horizon "at the same time", when what it actually means is that Schwarzschild coordinates can't be used to describe the horizon at all.

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.

No, what I said is correct. The answer to your objection is simply that if A and B are a pair of infallers, with B at a little higher r value, then B sees (absorbs) light from A crossing the horizon at the moment B crosses the horizon. So, indeed, no light leaves the horizon. This is no different from any normal situation where if I see light from you now, and you are moving in a known direction, I interpret that 'now' you are further in the direction you were moving. That is, when B crosses the horizon, it looks like A crossed before (and A is, indeed, inside already), and is now a bit inside. Note that light emitted directly away from the singularity at the horizon remains frozen at the horizon - waiting, if you will, for the next infaller.

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 B pulls away any time before crossing the horizon, then, indeed, they never see light from A at or inside the horizon. This would look a lot like a Rindler horizon. B suddenly accelerates madly to escape, A gets red shifted and frozen, just like a Rindler horizon.

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

One final observation is that B never sees A reach the singularity. At the moment B reaches the singularity, the last light they see from A is from a little before A reached the singularity.

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.

 

[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: