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

[A-wal]

No one said that coordinate systems are "meaningless", they are well-defined entities which are very useful for making calculations that try to figure out the answer to questions about coordinate-invariant physical facts, like the proper time along a given worldline between two events on that worldline. However, coordinate systems are quite arbitrary, i.e. there is no physical reason why you must use one as opposed to any other, and coordinate-dependent claims like the claim that a clock's rate of ticking approaches zero as it approaches the horizon (true in Schwarzschild coordinates but not in Kruskal-Szekeres coordinates) don't really "describe physical reality", they just describe properties of the coordinate system.

If they didn't describe reality then they really would be meaningless. They do describe reality, just from a very limited perspective. The Schwarzschild coordinates don't define a moment in time when any object crosses any event horizon. I understand what you're getting at. From the point of view of the free-faller time has to behave normally. But that's just how their wrist watch behaves. They would surely view the watch of the distant observer moving faster. If the free-faller were actually able to reach the horizon then it would view the distant observers watch moving infinitely quickly because time dilation would be infinite because they're frozen in time, so they wouldn't really be able to observe anything, just like approaching c.

In Schwarzschild coordinates this is true, but not in any coordinate-independent sense. Indeed the question of how fast an object is "moving through time" is an inherently coordinate-dependent one, just as much so as the question of how quickly an object's x-coordinate is changing.

Of course, but that doesn't contradict what I'm saying.

If you're talking about observation, this can be defined in a coordinate-independent way, since it only concerns pairs of events which occur at the same localized point in spacetime (for example, if the event of my clock reading T=20 seconds coincides with the event of my being hit by the light from your clock reading T=15 seconds, this is something all coordinate systems agree on). But the fact that the hovering observer sees the image of the falling observer slow down as the falling observer approaches the horizon doesn't imply that time is "really" going slower for the falling observer.

Yes it does. In the same way that time is "really" going slower for someone accelerating in flat space-time. If they were to meet up then more time would have passed for the distant observer.

To see this, just consider the case of the Rindler horizon in flat SR spacetime. If you have a family of accelerating observers who are at rest in Rindler coordinates, and then graph their worldlines from the perspective of an ordinary inertial frame, you get something like this:

Note that none of these curved paths will ever cross the diagonal dotted line which bounds them, which is the path of a light beam and which also represents the "Rindler horizon"; and since all light paths are diagonal in a diagram of an inertial frame, you can see why no event on the left side of the Rindler horizon can ever send a signal that will intersect the worldlines of any of the accelerating observers, thus they will never see anything that happens in the region of spacetime that's on the left side of the horizon in the diagram.

Now consider an ordinary inertial observer at rest in this inertial frame, whose worldline would just be a vertical line. Any such observer who starts out on the right side of the Rindler horizon will eventually cross to the left side. Just as with an observer hovering outside of a black hole, the accelerating observers will visually see the inertial observer's clock slow down more and more as he approaches the horizon (try drawing diagonal lines representing light signals emanating from points on the inertial observer's worldline just before crossing the horizon, and you'll see that they take longer and longer to intersect the worldline of an accelerating observer), and will see it take an infinite time for him to actually cross. But from the point of view of the inertial frame this diagram is drawn in, you can also see that nothing special is happening to the inertial observer as he approaches and crosses the horizon, and since he's at rest in this frame his clock always runs at the same rate in this frame.

I understand and agree, and don't see how any of that is relevant to the question.

No, again consider drawing an inertial observer with a vertical worldline in the above diagram. If you consider the point he crosses the Rindler horizon, and then draw a diagonal line sloping downward from that point and see where it intersects with one of the accelerating worldlines, then whatever point on the accelerating worldline it intersects with, that will be the event on the accelerating observer's worldine that the inertial observer is seeing as he crosses the horizon.

There is a very close analogy between lines of constant Rindler coordinate drawn in an inertial frame (as in the diagram above) and lines of constant Schwarzschild radius drawn in a Kruskal-Szekeres diagram--see here. So, everything I'm saying about observers outside the Rindler horizon vs. observers who cross it can be translated into statements about the black hole event horizon in the context of a Kruskal-Szekeres diagram.

Accelerate those observers up to the speed of light on that diagram. You can't. You'd get a singularity.

 

[Dale]

They would surely view the watch of the distant observer moving faster. If the free-faller were actually able to reach the horizon then it would view the distant observers watch moving infinitely quickly because time dilation would be infinite because they're frozen in time, so they wouldn't really be able to observe anything, just like approaching c.

Repeating an incorrect statement multiple times does not make it correct. In fact, given a stationary observer at any finite distance in Rindler coordinates a free-falling observer would actually see the stationary observer's clock slow down due to the relativistic Doppler effect. The free-falling observer doesn't see anything unusual about the universe as they cross the event horizon, which happens in a finite amount of proper time.

 

[A-wal]

Repeating an incorrect statement multiple times does not make it correct. In fact, given a stationary observer at any finite distance in Rindler coordinates a free-falling observer would actually see the stationary observer's clock slow down due to the relativistic Doppler effect.

Okay, but again irrelevant. Doppler shift isn't really time dilation and it's a thought experiment that assumes no Doppler shift, or an anti Doppler device if you prefer.

The free-falling observer doesn't see anything unusual about the universe as they cross the event horizon, which happens in a finite amount of proper time.

I still don't see how it can happen at all.

Let's keep bringing the free-falling observer and the distant observer together, then separating them again putting the free-falling observer back where they were each time. After every 5 mins on the clock of the distant observer they meet up and compare watches. There'll be more time dilation the closer the free-falling observer gets to the event horizon.

On the free-fallers watch there will never be a time when it's too late for them to meet up again. The time difference between the two will just keep on increasing until there isn't a black hole there anymore.

 

[Austin0]

Okay, but again irrelevant. Doppler shift isn't really time dilation and it's a thought experiment that assumes no Doppler shift, or an anti Doppler device if you prefer.

I still don't see how it can happen at all.

Let's keep bringing the free-falling observer and the distant observer together, then separating them again putting the free-falling observer back where they were each time. After every 5 mins on the clock of the distant observer they meet up and compare watches. There'll be more time dilation the closer the free-falling observer gets to the event horizon.

On the free-fallers watch there will never be a time when it's too late for them to meet up again. The time difference between the two will just keep on increasing until there isn't a black hole there anymore.

What you seem to be describing is Zeno's infinite divisability paradox in a modern form. But Achilles (the rabbit) does catch up and cross the line
in the real world

 

[Dmitry67]

As I understand, A-wal has mental block because many people believe that there is an objective way to map time of person A to time of person B. Even in classical twin paradox, when traveling twin returns, the one who was on Earth could say - you see, YOUR clock was ticking slower.

However, it is valid only because traveling twin returns. Or, if he does not return but dies on distant planet, it is possible to make a chain of observers (relatively at rest) and this way synchronize him back. Map his day of death into some "common galaxy time".

However, it is not possible in case of BH: spacetime is being ripped apart, A and B will never even meet, nor one can build a sequence of observers passing information from A to B. Literally, flow of time is ripped into 2 separate flows.

Finally the statement is incorrect:

"Let's keep bringing the free-falling observer and the distant observer together, then separating them again putting the free-falling observer back where they were each time. After every 5 mins on the clock of the distant observer they meet up and compare watches. There'll be more time dilation the closer the free-falling observer gets to the event horizon."

No. To talk about time dilation the falling observer MUST RETURN. Of pass somehow information to the distant observer. If falling observer is sending signals every second, the following would happen:

Falling obsserver time / Distant observer received (numbers are my guess):
0s 0s
1s 1.1s
2s 2.5s
3s 5s
4s 30s
5s 1year
6s NEVER
7s NEVER

 

[Demystifier]

In this context, it is useful to think about gravity NOT as geometry, but as just another force in flat spacetime. (Yes, general relativity can be formulated in this way as well. See, e.g., Feynman's Lectures on Gravitation.) In this view, all what black hole does is that it does not allow light or anything else to escape from it. So, from the point of view of a static observer outside the black hole, it takes a FINITE time for a freely falling observer to cross the horizon. However, the static observer just cannot observe it, because the black hole slows down the light (or any other information) sent by the freely falling observer. In fact, the horizon can be thought of as a kind of a curtain. The static observer cannot see what happens behind the curtain, but it doesn't make the events behind the curtain less real.

In other words, "to be" and "to be observed" are different things. Physicists often consider these two things to be the same (which causes a lot of confusion, especially in quantum mechanics and black hole physics), but they are not.

Thus, one should NOT say: "For me (the static observer), the freely falling observer will never cross the horizon." Instead, one should say: "The freely falling observer will cross the horizon in a finite time, but I (the static observer) will never be able to see it." When you put it this way, all weirdness of horizons suddenly disappears.

 

[Dale]

Okay, but again irrelevant. Doppler shift isn't really time dilation and it's a thought experiment that assumes no Doppler shift, or an anti Doppler device if you prefer.

The Doppler effect is certainly relevant to how he would "view the distant observers watch". In any case, time dilation is part of the relativistic Doppler shift, so even after he accounts for the increasing distance the free-falling observer will still determine that the Rindler observer's watch is running slow (in the free-falling frame).

Let's keep bringing the free-falling observer and the distant observer together, then separating them again putting the free-falling observer back where they were each time.

This doesn't make any sense. If you do that then either the free-falling observer is no longer free-falling or the Rindler observer is no longer uniformly accelerating.

 

[JesseM]

If they didn't describe reality then they really would be meaningless.

Not sure what you mean by "describe reality". Of course once you have chosen a well-defined coordinate system, there is a single correct answer to any question about physical events in the context of this coordinate system. But the choice of how you label points in spacetime in order to construct the coordinate system is totally arbitrary, the laws of nature don't say that one choice of labeling scheme is better than any other (at least not in the context of general relativity where the Einstein field equations work in any smooth coordinate system). Do you disagree?

From the point of view of the free-faller time has to behave normally. But that's just how their wrist watch behaves. They would surely view the watch of the distant observer moving faster. If the free-faller were actually able to reach the horizon then it would view the distant observers watch moving infinitely quickly

Why do you believe that? Everything here is perfectly analogous to the SR example of an accelerating observer with a Rindler horizon vs. an inertial observer who crosses the horizon (here the accelerating observer is analogous to the observer hovering outside the black hole, and the inertial observer is analogous to the observer who falls through the event horizon). Do you agree that the accelerating observer will see the inertial observer's clock slow down as the inertial observer approaches the Rindler horizon, and will in fact see it take an infinite time for the inertial observer to reach the Rindler horizon? Do you agree that, nevertheless, the inertial observer does not see the accelerating observer moving infinitely quickly as he crosses the Rindler horizon, instead he sees the accelerating observer's clock reading some finite time at the moment he crosses the Rindler horizon? If you agree this is perfectly self-consistent in the case of a Rindler horizon in SR, why do you think the same can't be true for a black hole horizon in GR, especially given that the Kruskal-Szekeres diagram of the two observers in GR looks identical to the Minkowski diagram of the two observers in SR?

because time dilation would be infinite because they're frozen in time

"Frozen in time" visually, or in some other sense? It's true that visually the outside observer sees their rate of ticking approach zero as they approach the black hole horizon, but exactly the same is true for the accelerating observer watching the inertial observer approaching the Rindler horizon. Do you think the inertial observer crossing the Rindler horizon is experiencing infinite time dilation?

Note that none of these curved paths will ever cross the diagonal dotted line which bounds them, which is the path of a light beam and which also represents the "Rindler horizon"; and since all light paths are diagonal in a diagram of an inertial frame, you can see why no event on the left side of the Rindler horizon can ever send a signal that will intersect the worldlines of any of the accelerating observers, thus they will never see anything that happens in the region of spacetime that's on the left side of the horizon in the diagram.

Now consider an ordinary inertial observer at rest in this inertial frame, whose worldline would just be a vertical line. Any such observer who starts out on the right side of the Rindler horizon will eventually cross to the left side. Just as with an observer hovering outside of a black hole, the accelerating observers will visually see the inertial observer's clock slow down more and more as he approaches the horizon (try drawing diagonal lines representing light signals emanating from points on the inertial observer's worldline just before crossing the horizon, and you'll see that they take longer and longer to intersect the worldline of an accelerating observer), and will see it take an infinite time for him to actually cross. But from the point of view of the inertial frame this diagram is drawn in, you can also see that nothing special is happening to the inertial observer as he approaches and crosses the horizon, and since he's at rest in this frame his clock always runs at the same rate in this frame.

Yes it does. In the same way that time is "really" going slower for someone accelerating in flat space-time.

No it isn't, not if you're talking about the rate of ticking at a given instant on their worldline. The instantaneous rate of ticking depends only on velocity, so you can always find a frame where at this moment their velocity is smaller than the velocity of the inertial twin, so the inertial twin's clock is ticking slower.

If they were to meet up then more time would have passed for the distant observer.

Yes, but here you are comparing elapsed time over an entire trip (the proper time each experiences between the moment they depart and the moment they reunite). All frames agree on the total elapsed time, but they disagree on whose clock was ticking slower during any brief phase of the trip.

I understand and agree, and don't see how any of that is relevant to the question.

If your argument is that seeing a clock's rate of ticking approach zero implies it's approaching infinite time dilation, and therefore that an observer traveling along with that clock would see the rate of your clock approaching infinity, then if you followed what I was saying you can see that this plainly isn't true in the case of the Rindler horizon. If you think there is some important difference between the case of the inertial observer crossing the Rindler horizon (being watched by an accelerating observer) and the case of the falling observer crossing the black hole horizon (being watched by the observer hovering outside), and that this difference explains why it is reasonable to believe "infinite time dilation" applies to the second case but not the first, please explain the precise nature of this difference.

No, again consider drawing an inertial observer with a vertical worldline in the above diagram. If you consider the point he crosses the Rindler horizon, and then draw a diagonal line sloping downward from that point and see where it intersects with one of the accelerating worldlines, then whatever point on the accelerating worldline it intersects with, that will be the event on the accelerating observer's worldine that the inertial observer is seeing as he crosses the horizon.

There is a very close analogy between lines of constant Rindler coordinate drawn in an inertial frame (as in the diagram above) and lines of constant Schwarzschild radius drawn in a Kruskal-Szekeres diagram--see here. So, everything I'm saying about observers outside the Rindler horizon vs. observers who cross it can be translated into statements about the black hole event horizon in the context of a Kruskal-Szekeres diagram.

Accelerate those observers up to the speed of light on that diagram. You can't. You'd get a singularity.

What does this statement have to do with anything? No observer gets accelerated up to the speed of light in either the case of an observer crossing the Rindler horizon or the case of an observer crossing the black hole event horizon. In the Kruskal-Szekeres diagram all light-like worldlines are diagonals at 45 degrees just like in a Minkowski diagram, and the worldline of an observer falling into the black hole always remains closer to vertical than 45 degrees.

 

[A-wal]

The part where they keep meeting up was a bad example. I knew that would get ripped apart. It just complicates things. That's why I said instantly meet up, but let's stick to what's possible, if not practical.

I don't think the rules of self-consistency need to be broken in the case of black holes and I think it is analogous to the twin paradox in the sense that there is an objective answer to the difference between how much time has passed between the two observers.

The curtain you refer to isn't because of slowed light. It's because of slowed time.

I am aware of the difference between "to be" and "to be observed". I'm trying to stick to "to be" (which is why there's no need to bring in Doppler shift).

I meant coordiante systems can acurately describe reality from a limited perspective.

The difference with an accelerating observer in flat space-time is that they can't accelerate up to c. In fact that's not a difference as far is I can see because reaching the event horizon of a black hole is equivalent to reaching c to my mind.

I think of the horizon and c to be 45 degrees.

What you seem to be describing is Zeno's infinite divisability paradox in a modern form. But Achilles (the rabbit) does catch up and cross the line
in the real world

I was talking about that in another thread here. The way I think of that paradox is that you can keep dividing the distance but the time it takes to move that far gets divided by the same amount so it cancels out. So the problem really just breaks down to the fact that if something is infinitely divisible than how can it ever be finite. That's why I don't think anything is infinitely divisible. I don't think it applies to this situation though. In this thought experiment we have a definite limit; the length of time the black hole lasts for.

If there were a series of observers at various distances from it then the life span of the black hole would be longer the further away it was measured from. At the event horizon it would be exactly 0, so how could anything cross it?

 

[Dmitry67]

If there were a series of observers at various distances from it then the life span of the black hole would be longer the further away it was measured from. At the event horizon it would be exactly 0, so how could anything cross it?

Correct, they will never OBSERVE falling spaceship crossing the horizon
How it affects an ability of the spaceship to cross the horizon?

As simplification, it is easier to think that when you see 'frozen' spaceship near the horizon, the spaceship is inside BH long time ago, it just took too much time for the light from it to come back.

There is anothe misconception about BH based on the 'time dilation': people tend to think, that as observer is 'frozen' near the horizon, you can fly much later to BH and to 'save' him. No. Too late. If you try to approach him, he 'unfreezes' and goes deeper and deeper so you can't catch him. As you fall in BH, the hoziron recedes in front of you so you will never be able to cross it (position of an apparent horizon is observer dependent)

 

[Dmitry67]

If there were a series of observers at various distances from it then the life span of the black hole would be longer the further away it was measured from. At the event horizon it would be exactly 0, so how could anything cross it?

I re-read your question, I was thinking you were talking about falling observers, not BH at whole. I had partly answered in my previous post: as soon as you approach BH, horizon moves deeper (for you) and co-falling on different distances observers 'unfreeze'.

The is an interesting question: does BH exist?
Our notion of exist, existed, will exist is applicable to flat time.
In curved spacetime it depends on how you define it (in some cases it can be observer depenedent. We really don't want the 'EXISTANCE' to be observer dependent, so we tend to say that if something exists for one observer, if exists for all observers)
Finally, in Black World there is no difference between existed, exists, will exist, so BH definitely exists as spacetime structure.

 

[Dale]

I think it is analogous to the twin paradox in the sense that there is an objective answer to the difference between how much time has passed between the two observers.

You are correct, it is analogous. There is only an objective answer if the two twins meet up again. At any point where they are separated by some distance there is no objective answer to which is older and how much time has elapsed. This is known as the relativity of simultaneity and is one of the most difficult concepts to learn.

 

[JesseM]

The difference with an accelerating observer in flat space-time is that they can't accelerate up to c. In fact that's not a difference as far is I can see because reaching the event horizon of a black hole is equivalent to reaching c to my mind.

Why do you think they are equivalent? From the point of view of a locally inertial frame in the neighborhood of a free-falling observer near the horizon, the horizon itself is moving outward at c, which means if they just stay where they are it'll sweep past them at light speed. Exactly the same is true for the "Rindler horizon", which is at a fixed position in Rindler coordinates but from the perspective of an inertial frame, it's moving outwards at c.

I think of the horizon and c to be 45 degrees.

That's true, in a Kruskal-Szekeres diagram. In KS coordinates the horizon is at 45 degrees, and all worldlines of light beams are at 45 degrees, with worldlines of slower-than-light observers being closer to vertical than 45 degrees. So, in a KS diagram you can illustrate how a slower-than-light observer can cross the horizon, and once he does no path in his future light cone will allow him to escape the horizon or avoid hitting the singularity. Look at the right-hand diagram on p. 835 of Gravitation, where the black hole event horizon is the 45-degree line that goes through the origin and is marked "r = 2M, t=+infinity", and the worldline that passes through the three events A, A' and A'' represents a slower-than-light-observer who falls through the event horizon:

If there were a series of observers at various distances from it then the life span of the black hole would be longer the further away it was measured from. At the event horizon it would be exactly 0, so how could anything cross it?

If you consider a series of hovering observers at constant radius, then the time for them to see some large amount of time to pass (say, 1 billion years) for objects at very large radii (say, 100 light-years from the black hole) would approach zero in the limit as the hovering observers approach the radius of the event horizon. But of course, it's impossible to actually hover at the event horizon, just like it's impossible for a massive object to travel at the speed of light! And for a falling observer, they don't see events in the external universe get arbitrarily fast as they approach the horizon in this way, the above only applies to hovering observers.

The same would be true for the accelerating observers in flat spacetime who are at constant distance from the Rindler horizon, but whose worldlines look like hyperbolas in the inertial frame diagram:

If you consider hyperbolas closer and closer to the diagonal dotted line representing the Rindler horizon, then since their velocity in the inertial frame is closer and closer to 45 degrees (light speed) the time dilation they experience relative to the inertial frame is greater and greater, which means that if they're getting light from on object much further away (whose worldline is close to vertical in the inertial frame, so it experiences little time dilation), they'll see it extremely sped up. If you consider the limit of the set of hyperbolas as the distance from the Rindler horizon approaches zero, the degree to which they see distant objects sped-up approaches infinity as well. But at the same time, you can draw a vertical line in this diagram representing an inertial observer who crosses the horizon without seeing distant objects speed up at all as he approaches the horizon.

If you can understand this in the case of the Rindler diagram (and be sure to ask questions if it isn't clear), why do you think the same situation couldn't hold in the case of a black hole?

 

[A-wal]

You are correct, it is analogous. There is only an objective answer if the two twins meet up again. At any point where they are separated by some distance there is no objective answer to which is older and how much time has elapsed. This is known as the relativity of simultaneity and is one of the most difficult concepts to learn.

Yea I know that simultaneous events doesn't even make sense without of point of reference.

That's true, in a Kruskal-Szekeres diagram. In KS coordinates the horizon is at 45 degrees, and all worldlines of light beams are at 45 degrees, with worldlines of slower-than-light observers being closer to vertical than 45 degrees.

Oops. I meant 90 degrees, sorry. Ninety's much simpler.

If you consider a series of hovering observers at constant radius, then the time for them to see some large amount of time to pass (say, 1 billion years) for objects at very large radii (say, 100 light-years from the black hole) would approach zero in the limit as the hovering observers approach the radius of the event horizon. But of course, it's impossible to actually hover at the event horizon, just like it's impossible for a massive object to travel at the speed of light! And for a falling observer, they don't see events in the external universe get arbitrarily fast as they approach the horizon in this way, the above only applies to hovering observers.

Right, so basically you're saying that an observer who fights against the inward pull of gravity by accelerating away from it experiences time dilation, but an observer who let's themselves just fall doesn't? So a series of observers at various distances from the black hole but not maintaining their distance and just drifting towards it won't experience any time dilation relative to each other and if they meet up in flat space-time after the black hole has gone then they'll all agree on how long the black hole was there for and their watches will all show the same time? Are you sure?

 

[Dale]

Yea I know that simultaneous events doesn't even make sense without of point of reference.

If you understand that then do you also understand how the fact that the stationary observer sees the free-falling clock slow down does not imply that the falling observer sees the stationary clock speed up?

 

[JesseM]

Oops. I meant 90 degrees, sorry. Ninety's much simpler.

Well, the horizon is a vertical line in a Schwarzschild diagram, if that's what you mean by 90 degrees. Similarly, the Rindler horizon is a vertical line in a Rindler coordinate diagram. But Kruskal-Szekers diagrams and Minkowski diagrams have the advantage of always showing light paths at the same slope (45 degrees), and of not showing all objects taking an infinite coordinate time to reach the horizon. Both KS diagrams and Minkowski diagrams show the horizons at 45 degrees just like light worldlines.

Right, so basically you're saying that an observer who fights against the inward pull of gravity by accelerating away from it experiences time dilation, but an observer who let's themselves just fall doesn't?

Time dilation isn't absolute, it's always relative to your coordinate system. But you can say that an observer who accelerates to stay at constant Schwarzschild radius will see distant clocks (which are also at constant Schwarzschild radius, but very far from the black hole) as running very quickly, with the visual rate of the distant clocks approaching infinity in the limit as the hovering observer's radius approaches the Schwarzschild radius. Whereas a falling observer might see distant clocks running somewhat faster depending on how he was falling, but he wouldn't see the rate of distant clocks approach infinity as he approached the horizon, he would continue to see the distant clocks ticking at a finite rate as he crossed the horizon.

So a series of observers at various distances from the black hole but not maintaining their distance and just drifting towards it won't experience any time dilation relative to each other and if they meet up in flat space-time after the black hole has gone then they'll all agree on how long the black hole was there for and their watches will all show the same time? Are you sure?

No, why would what I said previously imply I believed in such an exact agreement? In most cases, observers who take different paths through curved spacetime and later reunite will find they have aged differently.

 

[A-wal]

If you understand that then do you also understand how the fact that the stationary observer sees the free-falling clock slow down does not imply that the falling observer sees the stationary clock speed up?

No, not if we're talking about actual time dilation and ignoring Doppler shift. You can ignore what is observed in flat space-time and just use acceleration to work work what's really happening. There's only one true answer to how much time difference there is between the two, despite what they actually see happening.

Time dilation isn't absolute, it's always relative to your coordinate system. But you can say that an observer who accelerates to stay at constant Schwarzschild radius will see distant clocks (which are also at constant Schwarzschild radius, but very far from the black hole) as running very quickly, with the visual rate of the distant clocks approaching infinity in the limit as the hovering observer's radius approaches the Schwarzschild radius. Whereas a falling observer might see distant clocks running somewhat faster depending on how he was falling, but he wouldn't see the rate of distant clocks approach infinity as he approached the horizon, he would continue to see the distant clocks ticking at a finite rate as he crossed the horizon.

Okay then, let's go back to that example. They are all accelerating against gravity and staying stationary relative to each other and the black hole. They measure the life span of the black hole. An infinitely distant observer would presumably say that the black hole has always been there and it always will be. Not possible though, just like accelerating to c. An infinitely close observer who reaches the horizon would presumably say that it lasted for no time at all. Not possible again though, just like reaching c. Are you saying that the black hole would have a minimum life span or are you saying that its life span would vary depending on your velocity as well as your position relative to it?

No, why would what I said previously imply I believed in such an exact agreement? In most cases, observers who take different paths through curved spacetime and later reunite will find they have aged differently.

It sounded like that's what you were implying. I'm trying to understand exactly how acceleration reduces the effects of time dilation.

 

[JesseM]

Okay then, let's go back to that example. They are all accelerating against gravity and staying stationary relative to each other and the black hole. They measure the life span of the black hole.

In pure general relativity a black hole can never disappear, so unless the universe collapses, the black hole's life span is infinite. Quantum gravity would almost certainly allow black holes to evaporate but GR can't model this situation.

Also, the Schwarzschild solution describes a black hole of constant size that not only lasts infinitely long into the future, but has been there infinitely far in the past. GR has other solutions corresponding to a black hole that forms at some finite time from a collapsing star, though.

An infinitely distant observer would presumably say that the black hole has always been there and it always will be.

If you're talking about pure GR, all observers outside the horizon would say that. And if you're talking about a black hole that evaporates, why do you think the distant observer would say it lasts forever?

Not possible though, just like accelerating to c.

Why is it not possible? An infinitely long-lived black hole is a valid GR solution.

An infinitely close observer who reaches the horizon would presumably say that it lasted for no time at all.

Not if the observer reaches the horizon by falling in, such an observer would continue to see it "last" after falling through the horizon. If you're talking about a hovering observer at the horizon, that's just impossible in GR, so GR doesn't tell you what such an impossible observer would see. You can consider the limit of a series of hovering observers as their radius approaches the horizon, and maybe if the black hole evaporated then in this limit they would see it evaporate instantly, but for a Schwarzschild black hole the lifespan wouldn't be well-defined in this limit.

Are you saying that the black hole would have a minimum life span or are you saying that its life span would vary depending on your velocity as well as your position relative to it?

Again, in pure GR all external observers say the lifespan is infinite.

I'm trying to understand exactly how acceleration reduces the effects of time dilation.

"Time dilation" has no coordinate-independent meaning, do you understand that? If you want to talk about something objective you should talk about what different observers see in a visual sense.

 

[DrGreg]

There's only one true answer to how much time difference there is between the two, despite what they actually see happening.

That is only true for two observers who are a zero distance apart at both the start and end of their journeys. Whilst they are separated by a distance there is no "one true answer", only a number of different coordinate-dependent answers.

 

[Dale]

No, not if we're talking about actual time dilation and ignoring Doppler shift. You can ignore what is observed in flat space-time and just use acceleration to work work what's really happening. There's only one true answer to how much time difference there is between the two, despite what they actually see happening.

Hi A-wal, this is not correct, you do not understand the relativity of simultaneity. It is the single most difficult concept of relativity.

I would strongly recommend that you postpone further GR-related questions about black holes and event horizons until after you have learned SR, and understand the relativity of simultaneity.

 

[A-wal]

Steady. I should have said there's only one true answer from any given perspective. I'll try to be clearer. From the perspective of an observer in the original frame of the two inertial observers there is an absolute truth as to which one is more time dilated and the order that things happen in. I don't see why there shouldn't be in the case of a black hole. I understand that there's no preferred frame, or is that not what you meant?

If you're talking about pure GR, all observers outside the horizon would say that. And if you're talking about a black hole that evaporates, why do you think the distant observer would say it lasts forever?

Infinitely far away = -infinite time dilation / infinitely close = infinite time dilation.

Why is it not possible? An infinitely long-lived black hole is a valid GR solution.

Infinitely far away is impossible. I said I was going to stick to what's possible didn't I? I can't even follow my own rules.

I'm trying to understand exactly how acceleration reduces the effects of time dilation.

I was being thick when I said that. It's acceleration that causes time dilation and when you're in free-fall you're not accelerating. Does that mean that the only time dilation in the case of a free-fall observer is in fact only Doppler shift? So free-fall is the equivalent of a different inertial frame?

I know Hawking radiation is unproven but it's not really important to this situation though. I'm really just using the finite life span as an easy way of getting my head round the existence of the black hole as a whole. I don't think it really changes anything I've asked. The workings of the event horizon wouldn't change. My questions and examples assume a finite life span just for simplicity and it's handy to have flat space-time at the end to compare watches.

What I'm really having trouble with how the differences in perspective are resolved when all is said and done. If an external observer not only can't ever see an in-falling observer cross the horizon, but also could always possibly see them escape then from their perspective the absolute truth is that nothing can cross the horizon. Is it always possible or is it too late when they get close to the horizon. One post says yes, one says no. Is it their light that you can never see cross or is it them?

What would happen when and if the black hole's gone? Would the light from all the objects that fell in simply vanish? They can't get length contracted out of existence because the curvature in the area is being reduced as the black hole looses mass.

 

[Dale]

I should have said there's only one true answer from any given perspective. I'll try to be clearer. From the perspective of an observer in the original frame of the two inertial observers there is an absolute truth as to which one is more time dilated and the order that things happen in.

This is a really bad misuse of the word "absolute". In relativity absolute means coordinate independent or frame invariant. For something to be an "absolute truth" from "the perspective of an observer" is a contradiction in terms.

I don't see why there shouldn't be in the case of a black hole.

There is no difference here. Time dilation is coordinate dependent in both cases.

 

[JesseM]

Steady. I should have said there's only one true answer from any given perspective. I'll try to be clearer. From the perspective of an observer in the original frame of the two inertial observers there is an absolute truth as to which one is more time dilated and the order that things happen in. I don't see why there shouldn't be in the case of a black hole.

Sure, relative to any given coordinate system in a black hole spacetime, there is a unique answer (as DaleSpam said, 'absolute' is confusing here, since it normally refers to something frame-independent) "as two which one is more time dilated and the order that things happen in." But you have to specify the precise coordinate system you want to use, you can't just talk about this or that observer's "frame" since unlike with inertial frames in SR there is no unique way to define a "rest frame" for a given observer in GR.

If you're talking about pure GR, all observers outside the horizon would say that. And if you're talking about a black hole that evaporates, why do you think the distant observer would say it lasts forever?

Infinitely far away = -infinite time dilation / infinitely close = infinite time dilation.

This wouldn't be true in Schwarzschild coordinates, for example. In Schwarzschild coordinates, if two observers are hovering at different radii r₁ and r₂ outside the event horizon of a black hole with an event horizon at the Schwarzschild radius r_0 = 2GM/c^2, the ratio of the rates their clocks tick would be:

\frac{\sqrt{1 - \frac{r_0}{r_1}}}{\sqrt{1 - \frac{r_0}{r_2}}}

So if we let r₁ be some distance slightly outside the horizon like r₁ = 1.3333...*r₀, and consider the limit as r₂ approaches infinity, in this limit \frac{r_0}{r_2} would just approach 0, so the whole fraction would approach:

\frac{\sqrt{1 - \frac{1}{1.333...}}}{\sqrt{1 - 0}} = \sqrt{1 - 0.75} = 0.5

So, in this case the near clock (at 4/3 of the radius of the event horizon) is just ticking at half the rate of the clock infinitely far away.

I was being thick when I said that. It's acceleration that causes time dilation and when you're in free-fall you're not accelerating. Does that mean that the only time dilation in the case of a free-fall observer is in fact only Doppler shift? So free-fall is the equivalent of a different inertial frame?

In curved spacetime you have gravitational time dilation too, so if two observers are both in free-fall, they can cross paths twice and one observer will have aged more than the other between meetings.

What I'm really having trouble with how the differences in perspective are resolved when all is said and done. If an external observer not only can't ever see an in-falling observer cross the horizon, but also could always possibly see them escape then from their perspective the absolute truth is that nothing can cross the horizon. Is it always possible or is it too late when they get close to the horizon. One post says yes, one says no. Is it their light that you can never see cross or is it them?

If the falling observer is sufficiently close to the horizon, then unless the falling observer decides to activate his rockets and avoid the horizon himself, past a certain point there's no way the distant observer can stop the falling observer from reaching the horizon, because any signal the distant observer sent would not be able to catch up with the falling observer before the falling observer crossed it.

Anyway, do you agree that exactly the same issue applies to the Rindler horizon? An inertial observer approaching the Rindler horizon (which is moving outward at the speed of light in an inertial frame) can always activate his rockets to accelerate and avoid crossing it at any moment before he actually does cross it. And any of the accelerating observers at constant Rindler coordinate will never see the inertial observer cross the Rindler horizon. Do you think this constitutes some genuine paradox in SR?

What would happen when and if the black hole's gone? Would the light from all the objects that fell in simply vanish? They can't get length contracted out of existence because the curvature in the area is being reduced as the black hole looses mass.

Any light they emitted before crossing the horizon would escape, and light they emitted outward at the exact moment they were crossing the horizon would also escape at the moment the black hole evaporated completely (see the 'What about Hawking radiation?" section here). Any light they emitted inside the horizon would probably never escape, though we'd need a theory of quantum gravity to be sure. If that doesn't answer your question, could you clarify what you're asking here?

 

[A-wal]

I did say my terminology was off and I only know how to speak English. Okay, absolute means frame independent, gotcha.

Anyway, do you agree that exactly the same issue applies to the Rindler horizon? An inertial observer approaching the Rindler horizon (which is moving outward at the speed of light in an inertial frame) can always activate his rockets to accelerate and avoid crossing it at any moment before he actually does cross it. And any of the accelerating observers at constant Rindler coordinate will never see the inertial observer cross the Rindler horizon. Do you think this constitutes some genuine paradox in SR?

No I don't. I think it's different because in the case SR the distance between them is changing and with a black hole it isn't. It's all happening in the same place. I don't see how in the case of GR you can accelerate to a frame that crosses the horizon. If they're free falling then they're just in a different inertial frame and the event horizon would always be in the direction they're traveling in.

If the falling observer is sufficiently close to the horizon, then unless the falling observer decides to activate his rockets and avoid the horizon himself, past a certain point there's no way the distant observer can stop the falling observer from reaching the horizon, because any signal the distant observer sent would not be able to catch up with the falling observer before the falling observer crossed it.

But if you were to observe your signal heading towards the nearer ship then you would actually see it reach them if you never see them cross the horizon. If from one perspective an object crosses the horizon and yet from another it never does then surely they can't both be right. From the perspective of the distant observer it takes less and less time for them to accelerate against gravity in order to hover the closer they get to the event horizon, reduced to zero at the horizon (just like it takes infinate energy to reach c).

This is a story about some Nerbleonians from the planet Zorcreb who live for a purploid years. I ran out of appropriate made up words so I started using colours. That's how big a number it is. Lawa can't understand what the mean Mesapald and Sejmes were on about. They seem to be under the impression that matter can mysteriously pass through an area so time dilated and length contracted that it would disappear out of existence. Lawa wasn't buying it and was sure that the same principle would apply as when you try to accelerate up to c in flat space-time. Lawa convinces them to follow him to the nearest black hole and watch what happens when he talks to them as he's supposedly falling into the mystical never ever realm. Mesapald and Sejmes pass the time by playing scrabble. A l’squillion years pass before they receive a message from the ridiculously time dilated Lawa. He asks them not to go anywhere and just wait for his next message. "He must have sent that just before he crossed over to the never ever realm." A b’jillion years later they receive another message "I'm still waiting." "Okay we give up. We've been through every word in every language in the universe at least six times, and we're running low on beers. You can come back now".

If a distant observer can't see light ever reach the horizon because of time dilation then presumably no information at all can reach them. Yet in your scenario information is not only passing through infinitely time dilated and length contracted space but also making the return journey. If they watch the black hole and matter can fall in then they could measure the increase in the black holes mass. They would then know if it fell or not. That's a paradox.

 

[Dale]

I think it's different because in the case SR the distance between them is changing and with a black hole it isn't. It's all happening in the same place.

What would make you believe that? In both SR and GR and for both the accelerated and inertial observers the distance is always increasing. In both SR and GR for the accelerated observer the velocity decreases although it is always positive (asymptotically approaches 0 from above). In both SR and GR for the inertial observer the velocity continually increases. The situation is very analogous.

Again, I think you misunderstand SR, particularly the relativity of simultaneity. I think you should leave GR until you get a good foundation in SR. You should understand Rindler coordinates before you tackle Schwarzschild coordinates. It seems to me that all of your confusion about black holes would be resolved.

 

[JesseM]

Anyway, do you agree that exactly the same issue applies to the Rindler horizon? An inertial observer approaching the Rindler horizon (which is moving outward at the speed of light in an inertial frame) can always activate his rockets to accelerate and avoid crossing it at any moment before he actually does cross it. And any of the accelerating observers at constant Rindler coordinate will never see the inertial observer cross the Rindler horizon. Do you think this constitutes some genuine paradox in SR?

No I don't. I think it's different because in the case SR the distance between them is changing and with a black hole it isn't.

Distance between who and who, and in what coordinate system? In Schwarzschild coordinates the distance between the freefalling observer and the observer at constant Schwarzschild radius is changing, though the rate of change approaches zero in the limit as the distance of the freefalling observer to the horizon approaches zero. But then it's also true that in Rindler coordinates, if we consider an inertial observer approaching the horizon with an observer at constant Rindler distance, the rate at which the distance between them is changing approaches zero in the limit as the distance of the inertial observer from the horizon approaches zero. So, I don't see a difference.

I don't see how in the case of GR you can accelerate to a frame that crosses the horizon.

Are you talking about a particular coordinate system, or are you talking about what's true in coordinate-independent terms? In Schwarzschild coordinates it's true that any object takes an infinite coordinate time to reach the horizon, but you can still show that in the limit as coordinate time approaches infinity, the falling observer's proper time (time as measured by their own clock) only approaches some finite value, the same value that would correspond to the proper time the observer actually crosses the horizon as calculated in a coordinate system where this happens at finite coordinate time.

If the falling observer is sufficiently close to the horizon, then unless the falling observer decides to activate his rockets and avoid the horizon himself, past a certain point there's no way the distant observer can stop the falling observer from reaching the horizon, because any signal the distant observer sent would not be able to catch up with the falling observer before the falling observer crossed it.

But if you were to observe your signal heading towards the nearer ship then you would actually see it reach them if you never see them cross the horizon.

No you wouldn't. At every moment in Schwarzschild coordinates they are moving towards the horizon, but at a slower and slower and slower rate; the same is true of your signal. The way the speed of the falling observer and the speed of the signal slow down with time, it works out that there is always some finite distance between the signal and the observer in Schwarzschild coordinates (again assuming the signal was emitted at some time when the falling observer was already close enough to the horizon that all coordinate systems agree the signal can't catch up to the observer at any time before he reaches the horizon)

If from one perspective an object crosses the horizon and yet from another it never does then surely they can't both be right.

Would you say "surely they can't both be right" about the two perspectives on an inertial observer approaching the Rindler horizon? If not, what's the difference?

From the perspective of the distant observer it takes less and less time for them to accelerate against gravity in order to hover the closer they get to the event horizon

By definition an observer approaching the horizon is not "hovering" at constant Schwarzschild radius, so I don't understand what you mean here.

This is a story about some Nerbleonians from the planet Zorcreb who live for a purploid years. I ran out of appropriate made up words so I started using colours. That's how big a number it is. Lawa can't understand what the mean Mesapald and Sejmes were on about. They seem to be under the impression that matter can mysteriously pass through an area so time dilated and length contracted that it would disappear out of existence.

"Time dilated and length contracted" relative to what coordinate system?

Lawa convinces them to follow him to the nearest black hole and watch what happens when he talks to them as he's supposedly falling into the mystical never ever realm. Mesapald and Sejmes pass the time by playing scrabble. A l’squillion years pass before they receive a message from the ridiculously time dilated Lawa. He asks them not to go anywhere and just wait for his next message. "He must have sent that just before he crossed over to the never ever realm." A b’jillion years later they receive another message "I'm still waiting." "Okay we give up. We've been through every word in every language in the universe at least six times, and we're running low on beers. You can come back now".

If they wait too long to send the signal, then as I said above it's impossible for their message to reach Lawa at any finite Schwarzschild time (or at any time they can see--if the signal is a physical object they can watch as it falls, they will always see it behind Lawa). Meanwhile if a calculation of Lawa's free-fall trajectory in some more useful coordinate system shows that he should cross the horizon at a proper time of ten minutes after he left Mesapald and Sejmes, then if he started a stopwatch at the moment he left them and if he never turns on his rockets to stop free-falling, then Mesapald and Sejmes will see his stopwatch getting closer and closer to reading 10 minutes but never quite reaching that value.

And of course, exactly the same would be true if Lawa was simply moving inertially towards the Rindler horizon while Mesapald and Sejmes were accelerating in such a way to maintain a constant distance from the horizon in Rindler coordinates.

Yet in your scenario information is not only passing through infinitely time dilated and length contracted space but also making the return journey.

What do you mean "return journey"? If someone falls all the way to the horizon they can never return! And if someone falls for a while but accelerates back before reaching the horizon, then although their time dilation factor (in Schwarzschild coordinates) may have been very large at the closest point to the horizon, it was never infinite.

If they watch the black hole and matter can fall in then they could measure the increase in the black holes mass.

How would they measure that? They can measure the curvature of spacetime at some distance away from the black hole, but that changes as soon as the falling object falls to a point closer to the horizon than that distance (and if we make the simplifying assumption that the falling object consist of a spherically symmetric shell of matter falling inward, then once it's fallen past your radius, the curvature of spacetime in your neighborhood will be exactly the same as if its mass had already been added to the center of the black hole). For more on this point see How does the gravity get out of the black hole? from the Usenet Physics FAQ:

Purely in terms of general relativity, there is no problem here. The gravity doesn't have to get out of the black hole. General relativity is a local theory, which means that the field at a certain point in spacetime is determined entirely by things going on at places that can communicate with it at speeds less than or equal to c. If a star collapses into a black hole, the gravitational field outside the black hole may be calculated entirely from the properties of the star and its external gravitational field before it becomes a black hole. Just as the light registering late stages in my fall takes longer and longer to get out to you at a large distance, the gravitational consequences of events late in the star's collapse take longer and longer to ripple out to the world at large. In this sense the black hole is a kind of "frozen star": the gravitational field is a fossil field. The same is true of the electromagnetic field that a black hole may possess.

 

[A-wal]

Again, I think you misunderstand SR, particularly the relativity of simultaneity. I think you should leave GR until you get a good foundation in SR. You should understand Rindler coordinates before you tackle Schwarzschild coordinates. It seems to me that all of your confusion about black holes would be resolved.

You keep saying that. As I understand it the relativity of simultaneity is simply says that if you accelerate to a new inertial frame then length contraction and time dilation change the distances and time between objects, so there's no frame independent truth about any measurements of distance or time. But if something happens or doesn't happen in one frame then that's true in all frames so I don't see how it applies here. If there's something else that you think I missing and it does apply to this situation could you tell me what it is rather than just keep telling me I don't get it.

No you wouldn't. At every moment in Schwarzschild coordinates they are moving towards the horizon, but at a slower and slower and slower rate; the same is true of your signal. The way the speed of the falling observer and the speed of the signal slow down with time, it works out that there is always some finite distance between the signal and the observer in Schwarzschild coordinates (again assuming the signal was emitted at some time when the falling observer was already close enough to the horizon that all coordinate systems agree the signal can't catch up to the observer at any time before he reaches the horizon)

That doesn't make sense because time dilation and length become more pronounced, not less over relatively shorter distances. The signal would never reach a frame where the object it's heading towards crosses the horizon. You can never see an object reach the horizon no matter how close you get in the same way as in SR one observers view of another's time never actually freezes, because c can't be reached. Actually the signal example doesn't work because it would always travel away at c locally, so you couldn't follow it. We'll just use another observer heading in behind but with more inertia. The one behind will have to catch up before the horizon is reached and that's regardless of the distance between them to start with because there is no external frame in which it the closer observer crosses the horizon, so it can't ever be too late.

Would you say "surely they can't both be right" about the two perspectives on an inertial observer approaching the Rindler horizon? If not, what's the difference?

It's different with SR because there's no contradiction with inertial frames. In the case of a black hole you've got an event horizon that can't be crossed unless you yourself cross it. In the case of acceleration in flat space-time an observer in an inertial frame sees the accelerator as moving slowly through time and length contracted. The accelerator would perceive the inertial observer as moving through time quicker than themselves. I know that's not literally true because of Doppler shift but if that was taken away they would. There's an equivalence that doesn't apply to the way you're describing black holes, but it should because gravity and acceleration are the same thing.

By definition an observer approaching the horizon is not "hovering" at constant Schwarzschild radius, so I don't understand what you mean here.

Bad wording again. I should have said escape. I meant you wouldn't need to accelerate for any length of time to avoid crossing the horizon because time dilation would mean it doesn't exist for any length of time at the horizon in the same way that length contraction would mean it would have no size.

What do you mean "return journey"? If someone falls all the way to the horizon they can never return! And if someone falls for a while but accelerates back before reaching the horizon, then although their time dilation factor (in Schwarzschild coordinates) may have been very large at the closest point to the horizon, it was never infinite.

I just meant information is coming back to you that they did cross the horizon.

How would they measure that? They can measure the curvature of spacetime at some distance away from the black hole, but that changes as soon as the falling object falls to a point closer to the horizon than that distance

Oh yea. Bugger.


Rather than thinking of it as a singularity in the middle surrounded by the black hole I think it makes a lot more sense to see it as one thing. The singularity appears to get stretched in time and space the further away you are from it.


Figure 1: This is what I meant by using right angles for c and the event horizon. I haven't checked if this works but there's no reason why it shouldn't.

Figure 2: You can see that at half the speed of light (45 degrees) time dilation and length contraction are between a quarter and a third, ish. It's frame independent because from the other ones perspective you get this.

Figure 3: To represent acceleration you use a curved line.


To see it from the accelerators point of view you would need a sphere. An inertial observer would be moving round the edge while an accelerator could take a short cut through the sphere.

 

[JesseM]

No you wouldn't. At every moment in Schwarzschild coordinates they are moving towards the horizon, but at a slower and slower and slower rate; the same is true of your signal. The way the speed of the falling observer and the speed of the signal slow down with time, it works out that there is always some finite distance between the signal and the observer in Schwarzschild coordinates (again assuming the signal was emitted at some time when the falling observer was already close enough to the horizon that all coordinate systems agree the signal can't catch up to the observer at any time before he reaches the horizon)

That doesn't make sense

I assure you that this is what would happen in Schwarzschild coordinates if you sent a signal too late to catch up with the falling object; both would travel more and more slowly as they approached the horizon, but the distance between them would never reach zero.

because time dilation and length become more pronounced, not less over relatively shorter distances.

How do you think that's incompatible with what I just said? Time dilation in Schwarzschild coordinates does become more pronounced as you approach the horizon, which is why the signal travels slower and slower and never manages to catch up with the falling observer.

The signal would never reach a frame where the object it's heading towards crosses the horizon.

I don't know what you mean by "reach a frame". We are analyzing the problem in a single frame, Schwarzschild coordinates--if you want to do something different please specify what frame or frames you want to use. In Schwarzschild coordinates, it's true that the object never crosses the horizon, but its speed also never reaches exactly zero so it's always getting slightly closer, and meanwhile the speed of the signal is continually decreasing too, in such a way that it never quite catches up to the object at any time in Schwarzschild coordinates (again assuming the signal was sent out too late).

Are you familiar with the idea that the infinite geometric series 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... 1/2^N + ... will never quite reach a sum of 1, although it gets ever closer to it? You can get an idea of why this is the case from looking at this image:

Anyway, the point is that in Schwarzschild coordinates, both the falling observer and the signal approach the horizon in a way that resembles this--each second the distance they travel is smaller than the distance they traveled in the previous second, in such a way that their position never quite reaches the position of the horizon, no matter how many seconds we count. And for a signal sent out too late to catch up to the falling observer, it would also be true that the distance between them is continually closing but it never quite reaches zero--if the gap between the signal and the falling observer is 1 at some time (in whatever units you want to use), then at some later time the distance is only 1 - 1/2, at a later time it's 1 - (1/2 + 1/4), at a later time it's 1 - (1/2 + 1/4 + 1/8), and so forth.

You can never see an object reach the horizon no matter how close you get in the same way as in SR one observers view of another's time never actually freezes, because c can't be reached.

That's true.

Actually the signal example doesn't work because it would always travel away at c locally, so you couldn't follow it.

It doesn't "always travel at c locally" in Schwarzschild coordinates! In Schwarzschild coordinates the speed of the signal continually decreases as it approaches the horizon. Light only travels at c in a locally inertial coordinate system, but such coordinate systems have an infinitesimal size in GR so they can't be used to define the distance between two things at different points in curved spacetime, like the falling object and the signal.

We'll just use another observer heading in behind but with more inertia. The one behind will have to catch up before the horizon is reached and that's regardless of the distance between them to start with because there is no external frame in which it the closer observer crosses the horizon, so it can't ever be too late.

Nope, doesn't work that way. If both observers are sent from the same radius, then whichever had a larger initial speed in Schwarzschild coordinates will necessarily be decelerating at a greater rate in Schwarzschild coordinates, i.e. its speed in these coordinates is dropping more quickly (and the same is true for light, its speed drops more quickly as it approaches the horizon than massive falling objects). This may help understand why the one sent out later can avoid ever catching up to the one sent out earlier even if the one sent out later had a greater initial speed.

Anyway, remember anything you say about what happens to two objects (or an object and a light signal) approaching the black hole horizon in Schwarzschild coordinates can also be said of two objects approaching the Rindler horizon in Rindler coordinates. Do you understand that in Rindler coordinates, observers moving towards the horizon who have constant speed in inertial coordinates will instead have constantly decreasing speed in Rindler coordinates, and will never quite reach the Rindler horizon at any finite Rindler time? Do you think this means it is always possible in Rindler coordinates for a second object to catch up with one sent towards the horizon earlier, regardless of how much later the second is sent? After all you can always give the second object a greater initial speed in Rindler coordinates...

It's different with SR because there's no contradiction with inertial frames.

But I'm not talking about comparing inertial frames, I'm talking about comparing the Rindler coordinate frame (a non-inertial coordinate system) with inertial frames. This diagram was showing what lines of constant position and time in Rindler coordinates look like when graphed in an inertial frame:

The black hyperbolas labeled s=1, s=2 etc. represent lines of constant position in Rindler coordinates, the gray angled lines labeled q=0.25, q=0.5 etc. represent lines of constant time in Rindler coordinates. So if you have a series of events at known coordinates in the inertial frame, you could use this diagram to figure out their approximate s and q value (or use the coordinate transformation to get more accurate values), then a diagram from the perspective of Rindler coordinates would be one where all the s-lines of constant position are just vertical lines, while all the q-lines of constant time are just horizontal lines, forming a nice grid. Then if you plot a series of events on the worldline of an object which crossed the Rindler horizon at some finite time in the inertial frame, in the Rindler graph the object would seem to move slower and slower as it approached the horizon, never quite reaching it at any finite Rindler time-coordinate.

In the case of a black hole you've got an event horizon that can't be crossed unless you yourself cross it.

Same is true of the Rindler horizon--regardless of whether you plot it in inertial coordinates or Rindler coordinates, you can see that no one who remains outside the horizon (like the accelerating observers who have a constant position in Rindler coordinates) will ever get a signal from any event on or beyond the horizon, the only way to see these events is to cross the horizon yourself.

So again--what's the difference? Maybe with my additional explanations, you can finally see that there is no difference, that the way the same events are viewed in Rindler coordinates vs. inertial coordinates exactly mirrors the way the same events are viewed in Schwarzschild coordinates vs. Kruskal-Szkeres coordinates?

In the case of acceleration in flat space-time an observer in an inertial frame sees the accelerator as moving slowly through time and length contracted. The accelerator would perceive the inertial observer as moving through time quicker than themselves.

When you talk about what each "sees" or "perceives", do you mean what happens in an inertial coordinate system where the inertial observer is at rest vs. what happens in a non-inertial coordinate system the accelerating observer is at rest (like Rindler coordinates)? In this case you're incorrect to say that the accelerating observer "would perceive the inertial observer as moving through time quicker than themselves", at least if we are talking about the inertial observer vs. the Rindler observer. In the Rindler coordinates used by the accelerating observer, the inertial observer falling towards the Rindler horizon is becoming increasingly time-dilated, his clock running slower and slower relative to Rindler coordinate time as he approaches the horizon, approaching infinite time dilation in the limit as his distance from the horizon approaches zero. And visually the accelerating observer will also see the inertial observer's clock running slower and slower in a visual sense.

So again, how is this different from an observer at constant Schwarzschild radius using Schwarzschild coordinates to define the time dilation for a falling observer? Your statement above suggests you might be confused about the analogy, the falling observer in the black hole spacetime is supposed to be analogous to the inertial observer in flat spacetime, while the observer hovering at constant Schwarzschild radius is supposed to be analogous to the accelerating observer at constant Rindler position coordinate...it's not the other way around!

Bad wording again. I should have said escape. I meant you wouldn't need to accelerate for any length of time to avoid crossing the horizon because time dilation would mean it doesn't exist for any length of time at the horizon in the same way that length contraction would mean it would have no size.

Statements about time dilation and length contraction are meaningless unless you specify a coordinate system--do you disagree? In Schwarzschild coordinates the horizon exists at every value of the coordinate time and it's always at a finite Schwarzschild radius, so I don't see how the above statement makes sense in the context of Schwarzschild coordinates if you're talking about the horizon itself, although it's true that any object falling towards the horizon gets more and more time-dilated and shrinks to a smaller and smaller length in these coordinates (exactly the same is true about anything approaching the Rindler horizon in Rindler coordinates, so would you say 'you wouldn't need to accelerate for any length of time to avoid crossing the Rindler horizon'?)

I just meant information is coming back to you that they did cross the horizon.

How is it doing that? No matter what coordinate system we use, observers outside the horizon can only receive signals from events that also occurred some finite distance outside the horizon. And no matter how close some object is to the horizon, their time dilation never quite reaches zero in Schwarzschild coordinates, so you can never be sure they didn't accelerate away during the last gazillionth of a second on their clock.

Rather than thinking of it as a singularity in the middle surrounded by the black hole I think it makes a lot more sense to see it as one thing. The singularity appears to get stretched in time and space the further away you are from it.

No. the separation between distinct physical events is defined by the metric (which defines 'separation' in a coordinate-independent way), and regardless of what coordinate system you use, there are events inside the horizon which do have some separation from the singularity, so it's just not correct to view the entire interior + horizon as part of the singularity.

Figure 1: This is what I meant by using right angles for c and the event horizon. I haven't checked if this works but there's no reason why it shouldn't.

Figure 2: You can see that at half the speed of light (45 degrees) time dilation and length contraction are between a quarter and a third, ish. It's frame independent because from the other ones perspective you get this.

Figure 3: To represent acceleration you use a curved line.

I don't understand these diagrams at all, you need to give more explanation. Where is the event horizon? What does the circle represent? What does the thick black line parallel to the time axis represent? What does the second thick black line at an angle relative to the first (or the thick black curve in the third diagram) represent? Do you think this is how events near a black hole would work in some coordinate system defined in the right way, or are the diagrams supposed to show something frame-independent?

 

[A-wal]

Nope, doesn't work that way. If both observers are sent from the same radius, then whichever had a larger initial speed in Schwarzschild coordinates will necessarily be decelerating at a greater rate in Schwarzschild coordinates, i.e. its speed in these coordinates is dropping more quickly (and the same is true for light, its speed drops more quickly as it approaches the horizon than massive falling objects). This may help understand why the one sent out later can avoid ever catching up to the one sent out earlier even if the one sent out later had a greater initial speed.

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That's all I needed. I was thinking that it didn't make sense because you could always catch an object closer to the horizon because it's always more time dilated. But of course it's percentage based, so an object that's moving faster would loose more speed. I don't know where the hell that mental block came from. I wonder how long it would have taken for that to dawn on me. It's always so bloody simple. Poor Lawa took his ignorance to his grave.

Thanks a lot for your time and patience, much appreciated. I hope I didn't frustrate you too much. It's been a good mental exercise and I've learned some of the more technical ways of defining things. Hopefully you can understand where I was coming from now and why I asked the questions that I did.

Now, about those diagrams. I came up with this when it occurred to me that all objects are only separated from each other by two dimensions because you can always draw a straight one dimensional line between any two objects. The thick black horizontal line represents an inertial observer. You're always horizontal because you're always static in space and moving through time at c. The length of the lines represent proper time. The thick diagonal line represents a second observer in a different inertial frame, but from the perspective of the first observer. The angle of the diagonal line comes from its speed relative to c, so .5c is 45 degrees. Draw a line down from the end of the diagonal line to the horizontal line to see how time dilated/length contracted the second observer is from your perspective. It's frame-independent without acceleration because you can move the diagonal line so that it's horizontal and the other observers world line is now diagonal (diagram 2). The third diagram represent an inertial observers perspective of an accelerating observer.

Forget what I said about using a sphere because there is no speed limit from the perspective of an accelerator. It doesn't work because you could reach any object in the universe as quickly as you like in proper time if you could stand the g force because length contraction/time dilation would bring the object to you. Maybe warping the circle into a oblong would do the trick.

Anyway this wasn't why I originally posted. There was a reason I named this thread The Arrow Of Time. I had an idea about what happens inside the event horizon that ties into those diagrams. I just didn't understand how an object could reach the horizon in the first place. This is extremely simple and to me it makes perfect sense. Inside the horizon time moves backwards. At the singularity no time has passed at all since the formation of the black hole. So any object entering the horizon would be pulled back through time to the exact moment in time and space that the black hole formed where it pass through the eye of a metaphorical needle and be converted into the simplest form of energy, which is why the formation of black holes is always proceeded by a gamma ray burst. The only information to survive the black hole is matters value, so no paradoxes. Obviously the value of the gamma rays would have to match the value of all the matter that falls in over the black holes life span. I tend to think of time as coming from the expansion of the universe with matter slowing it down. In the case of a black hole the universe is locally contracting making time move backwards. I may be getting very carried away with this bit but maybe the black hole actually reduces in mass when matter enters it until the same value of matter has gone in as the value of the energy that originally came out (as it's the same stuff), would would kind of go along with the whole working in reverse thing.

 

[A-wal]

Forget what I said about using a sphere because there is no speed limit from the perspective of an accelerator. It doesn't work because you could reach any object in the universe as quickly as you like in proper time if you could stand the g force because length contraction/time dilation would bring the object to you. Maybe warping the circle into a oblong would do the trick.

Got it! To do it from the accelerators point of view you would need to make the circle smaller or bigger, as in less than or more than (overlap) 360 degrees, depending on whether you were accelerating or decelerating relative to the other one.