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

[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.

 

[jtbell]

Thread closed pending final mentor action.