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

[A-wal]

If those two questions have a different answer as I'm assuming they do (no to the first and yes to the second) then I don't see how they can be consistent with each other, unless they're describing different horizons. I also don't see how changing coordinate system can make the same object behave differently with respect to the black hole (or anything for that matter). Surely all coordinate systems that are accurate have to tell the exact same story!

The event horizon of a black hole is something that has a definite radius around the singularity providing an object has no inertial velocity relative to it. If the object doesn't expend energy to counteract the gravity from the black hole then it will accelerate towards it. The closer the object gets to the event horizon, the slower time moves from the perspective of that object, making it perceive everything else to be speeding up of course. If it were to reach the horizon it would be frozen in time, but it can't because time moves slower the closer it gets, just like an accelerating object comparing itself to an object in its original frame in flat space-time.

Besides, length contraction would make the black hole smaller from the objects perspective in the dimension of a straight line between the two, giving it an oblong event horizon. But time dilation would mean the black hole would evaporate before anything could cross the horizon even if it wasn't for length contraction. I think to tell if an object did cross the horizon you would have to see the black hole die. Any objects that disappear with it obviously did cross the horizon, but not until then.

I don't see how anything changes if you use a different coordinate system?

 

[Dale]

Surely all coordinate systems that are accurate have to tell the exact same story!

Actually, no. Some coordinates do not cover all of spacetime. The Rindler coordinates and Schwarzschild coordinates share the fact that there are some events in their respective spacetimes which are outside of the coordinate system. For example, what is the Schwarzschild coordinate of something halfway between the event horizon and the singularity?

 

[A-wal]

I'm liking the Schwarzschild one more and more. It says an object can't cross the horizon judging by what you just said, correct? How can an object reach the horizon when it would be infinitely time dilated from any other frame?

 

[Dale]

I'm liking the Schwarzschild one more and more. It says an object can't cross the horizon judging by what you just said, correct? How can an object reach the horizon when it would be infinitely time dilated from any other frame?

It would only be infinite time dilated for an object at rest in the Schwarzschild coordinates. For a free-falling observer the time dilation is finite.

 

[A-wal]

That's paradoxical! In one it does and in the other it doesn't. How can two different coordinate systems that give two different outcomes both be right? What if someone at rest in Schwarzschild coordinates waited till the end of the black holes life? What would someone who was falling in see when looking at a more distant object? You're saying the situation is non equivalent to looking at a closer object? That doesn't make sense!

I'm not giving up until I've got this straight in my head.

 

[Dale]

That's paradoxical! In one it does and in the other it doesn't. How can two different coordinate systems that give two different outcomes both be right?

There is no outcome which is disagreed-upon by different coordinate systems. The difference is not the coordinate systems, it is the motion of the observer. A free falling observer detects a finite redshift in any coordinate system and an accelerating observer (stationary in Schwarzschild coordinates) detects an infinite redshift in any coordinate system.

What if someone at rest in Schwarzschild coordinates waited till the end of the black holes life?

Then by definition it is not Schwarzschild coordinates. The Schwarzschild metric is stationary.

You really would be better off looking into Rindler coordinates. You can learn everything you need to know about the event horizon without the confusion of spacetime curvature.

 

[A-wal]

Apparently the Rindler coordinates are to show the relativistic effects of an object under acceleration. Like the fact that the back of the object has to accelerate faster than the front from the perspective of a non-accelerating observer because of length contraction. And apparently this can also be used to show the effects of curved space-time instead of acceleration, which makes sense because they're the same thing. An event horizon appears that the accelerating object can't reach for an observer using these coordinates, so an event horizon that can't be reached is created when applied to curved space-time as well.

I can't see any self-consistent way of resolving this without the event horizon receding as it's approached. A distant observer would see the in-falling object undergo length contraction and time dilation as it approaches and it would reach infinity at the horizon, so it can never get there, and conversely the object itself would observe the rest of the universe speeding up as it got closer, becoming infinitely reverse time dilated at the event horizon, again meaning it can never actually get there.

I'm assuming that I'm going wrong when I switch it round and look at it from the perspective of the in-falling observer. So I need to understand how the situation can be both non-symmetric and self-consistent at the same time (without moving the goal posts).

 

[JesseM]

That's paradoxical! In one it does and in the other it doesn't.

Schwarzschild coordinates don't specifically predict it doesn't cross the horizon, it's just that no finite time coordinate can be assigned to the event of crossing the horizon. And you can actually assign a finite set of coordinates to events on the object's worldline once it's inside the horizon, see the left diagram here (from Gravitation by Misner/Thorne/Wheeler):

 

[Dale]

Apparently the Rindler coordinates are to show the relativistic effects of an object under acceleration. Like the fact that the back of the object has to accelerate faster than the front from the perspective of a non-accelerating observer because of length contraction. And apparently this can also be used to show the effects of curved space-time instead of acceleration, which makes sense because they're the same thing. An event horizon appears that the accelerating object can't reach for an observer using these coordinates, so an event horizon that can't be reached is created when applied to curved space-time as well.

Essentially correct. I wouldn't say that the horizon cannot be reached, I would say that it cannot be reached in a finite amount of coordinate time. You can reach it in a finite amount of proper time simply by falling.

conversely the object itself would observe the rest of the universe speeding up as it got closer, becoming infinitely reverse time dilated at the event horizon, again meaning it can never actually get there.

The infalling observer doesn't see any of these effects. They pass through the event horizon without incident and without any observable effect other than they can no longer send a signal to the accelerating observer.

So I need to understand how the situation can be both non-symmetric and self-consistent at the same time (without moving the goal posts).

The two observres are physically non-symmetric, so you expect non-symmetric results and observations.

 

[A-wal]

If no finite time coordinate can be assigned to the event of crossing the horizon then it's meaningless to describe what happens after it has crossed. In that diagram what happen to the observers view of the rest of the universe? As they undergo more and more time dilation won't their view of the rest of the universe speed up? It would reach infinate speed at the horizon but black holes con't last for an infinate amount of time, so if you try to cross the event horizon af a black hole then you'll never be able to get there in time.

The infalling observer doesn't see any of these effects. They pass through the event horizon without incident and without any observable effect other than they can no longer send a signal to the accelerating observer.

This is a contradiction to me, sorry.

The two observres are physically non-symmetric, so you expect non-symmetric results and observations.

Same again.

 

[JesseM]

If no finite time coordinate can be assigned to the event of crossing the horizon then it's meaningless to describe what happens after it has crossed.

Why? Suppose you have an ordinary inertial coordinate system to describe your movements today, and at t=10350 seconds you step outside of your house. Now, let's transform your movements into a different non-inertial coordinate system:

x' = x
y'=y
z'=z
t' = t^2/(t-10350)

In this coordinate system, no finite time can be assigned to the event of your stepping outside the house. But that doesn't mean it can't be used to label events that happen to you after you step outside, say at t=10400 seconds in the original inertial coordinate system.

In that diagram what happen to the observers view of the rest of the universe? As they undergo more and more time dilation won't their view of the rest of the universe speed up?

No, not for a nonrotating black hole--if they are watching some clock hovering outside as they fall in, they will only see it reading some finite time as they cross the horizon (see the 'Will you see the universe end?' section of this page from the Usenet Physics FAQ). On the other hand, a rotating black hole has both an inner and an outer event horizon, and an observer crossing the inner horizon would see the entire infinite future of the outside universe, which also means that light falling into the black hole becomes blueshifted to infinity at the inner horizon and the energy density goes to infinity there, which is why general relativity's predictions about the inside of rotating black holes are treated as suspect and physicists think a theory of quantum gravity is needed to understand what happens at the inner horizon.

It would reach infinate speed at the horizon but black holes con't last for an infinate amount of time,

In pure general relativity a black hole would last an infinite time, at least unless the universe collapsed in a big crunch. Quantum gravity will probably give them a finite lifetime due to Hawking radiation though.

so if you try to cross the event horizon af a black hole then you'll never be able to get there in time.

There is no physical significance to the fact that it takes an infinite Schwarzschild coordinate time to reach the horizon, that's just a bug in how the coordinate system is defined, not really any different than the coordinate system I defined at the start of the post which cannot assign a finite time to the event of your stepping out your door. Always remember, coordinate systems are just arbitrary ways of labeling events, they don't have any deep physical significance.

 

[Dale]

This is a contradiction to me, sorry.

What is contradictory about it? Remember I was speaking about the event horizon in Rindler coordinates on flat spacetime, which I think will be a better starting point.

 

[A-wal]

First I’m told that my question has no basis unless it’s defined within a coordinate system and I’m not even wrong unless I do that, then coordinate systems are meaningless and don’t describe physical reality.


If an object is in a stronger gravitational field then it will be moving through time slowly relative to someone in a lower gravitational field. As an object approaches the event horizon it will become infinitely time dilated so no amount of time will be enough to observe an object crossing it.

So if we reverse that then the person falling in will see the distant observer as moving quickly through time, and as no amount of time is enough to see the observer cross the horizon; any given amount of time will pass for the observer approaching the horizon before they actually reach it, surley?

 

[Dale]

If an object is at rest in a stronger gravitational [STRIKE]field[/STRIKE] potential then it will be moving through time slowly relative to someone at rest in a lower gravitational [STRIKE]field[/STRIKE] potential. As an object approaches the event horizon it will become infinitely time dilated so no amount of time will be enough to observe an object crossing it.

So if we reverse that then the person [STRIKE]falling in[/STRIKE] at rest in a stronger gravitational potential will see the distant observer as moving quickly through time

I have corrected your statement to make it correct.

 

[JesseM]

First I’m told that my question has no basis unless it’s defined within a coordinate system and I’m not even wrong unless I do that, then coordinate systems are meaningless and don’t describe physical reality.

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 an object is in a stronger gravitational field then it will be moving through time slowly relative to someone in a lower gravitational field.

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.

As an object approaches the event horizon it will become infinitely time dilated so no amount of time will be enough to observe an object crossing it.

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. 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.

So if we reverse that then the person falling in will see the distant observer as moving quickly through time, and as no amount of time is enough to see the observer cross the horizon; any given amount of time will pass for the observer approaching the horizon before they actually reach it, surley?

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.

 

[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.