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

[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.

Ha...llelujah
Ha...llelujah
Hallelujah
Hallelujah
Halle...lujah

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.

 

[A-wal]

I take it the circle works then?

I know that the formula for finding length contraction/time dilation shows that it becomes more pronounced at higher speeds. If you draw four straight lines joining up the time and space lines to make a diamond shape then the four triangles around the edge would be the difference between a smooth motion (ie .5c = .5 time dilation/length contraction) and the actual equation. In the circle if their proper time stays equal then the curve takes it more up than left at lower speeds then catches up at the end.

This isn't just a graphical representation. This is literally what's happening. All I'm doing is substituting time for another spatial dimension.

 

[A-wal]

How come as soon as you lot think I've made a mistake you can't wait to post "No! You're wrong! You don't understand!" but when I'm asking about something that might work you've all gone away? It's not as even as if it's vague and undefined. Does the circle work or not?

 

[Dale]

Sorry, in our last exchange you basically told me to butt out so I was trying to do so.

It looks like you are doing a spacetime diagram. Traditionally they are drawn with time vertical, but that is just a convention and there is no reason not to do it the way you have done. You are correct that an inertial object follows a straight line and that an accelerating object follows a curved line.

I don't understand why you have a "c" on the space axis (things on the space axis should have units of distance, not speed), and I don't know what significance (if any) you are attaching to the circle.

 

[A-wal]

I wasn't telling you to butt out. It's just that you said twice that I didn't get something to do with simultaneity but wouldn't tell me what it was. I don't think I was missing anything. It just seems that you're much more interested in trying to make others look stupid rather than helping them understand something. I'm going by things I've read in other threads as well. I'm still grateful that you've taken the time to reply. Some of it's been very helpful.

Anythings speed through space-time is always c from any frame. The speed through time and space individually depends which frame you're measuring it from. From you're own perspective you're always moving through space at zero, which makes sense because you can't run away from yourself. So you're always moving through time at c. The circle is just because I originally thought of it as a simple way to compare time dilation and length contraction between two frames and you need to use the same length of proper time to do that, so it always goes to the edge of the circle without acceleration and you then just draw a straight line down to see how much it's shortened by. It also works for gravity/acceleration with the c line being the event horizon.

What about the matter entering a black hole being converted into energy and released at the singularity? It seems to me to be simple and the only way it could work. It answers two questions at once and it's true that no time has passed at the singularity no matter how long the black hole's been there from the outside, and that's where matter that's crossed the horizon has to end up.

I have another question though. I've had a chance to let what I've been told sink in and I've figured out the problem I was having. I was told that if an object is moving in at a faster rate behind another slower object then the faster one will undergo more time dilation from the perspective of a third distant observer, and that there will be a point when it's too late for anything including a signal sent at light speed to reach the closest observer from the closest observers perspective. But what about from the second faster moving observers perspective? It's following the first observer in and it can't possibly enter the horizon after the first observer from its own perspective. But it can't possibly enter the horizon before the first observer from the first observers perspective. We have a contradiction. Which one crosses the horizon first? The only conclusion is that all observers cross at the same time. This makes sense from the third distant observers perspective because they can't see anything cross the horizon until it's gone, which is what I meant before by a black hole being the same as the singularity. Length contraction and time dilation force anything that does cross the horizon to do it when the black hole exists at a single point in time and space. The fact that the second observer is slowing at a faster rate than the first from the third observers perspective just explains how the second observer may never be able to overtake the first observer before reaching the horizon from the third observers perspective.

Let's assume that black holes do loose mass during their lives for whatever reason. The first observer measures the size of the horizon just before it crosses and sees that the second observer hasn't set off yet. The second observer catches up to the first at the event horizon. It should never be too late because it can never witness the first one crossing. Now they both cross together but the second observer measures the size of the horizon as well and says it's smaller than the first one made it, yet they're now in the same place at the same time. If they do all cross at the same time then it makes sense as long as black holes have a life span. The first observer measures the size of the horizon just before it crosses and sees that the second observer hasn't set off yet, and knows that it can't reach the horizon before the black hole's gone. It still works even if the second observer moves in slower than the first because time dilation would slow the first one down so the second one catches up.

This is the only way I can get my head round it without breaking the law that nothing can be destroyed and without contradicting itself. If I've made a mistake somewhere along the line please point out where exactly it went wrong.

 

[Dmitry67]

It's following the first observer in and it can't possibly enter the horizon after the first observer from its own perspective. But it can't possibly enter the horizon before the first observer from the first observers perspective

Begin from here.
There are 2 types of horizons, apparent and absolute. For an observer at infinity, apparent and absolute horizons are at the same position (almost).

But it is not true for a falling observer: when it falls, apparent horizon recedes in front of him. So, if you 'jump' inside the BH to 'resque' a ship 'frozen near the horizon', then you would see how horizons moves deeper and deeper inside the BH, ship 'unfreezes' and falls down too; so you can't catch up with him and can't resque.

And for obvious reasons, no observer can ever pass thru his own apparent horizon, because it would mean that infinite tidal forces had torn his body apart. Fortunaly, in GR freely-falling observer can always assume that his part of space is locally flat, so apparent horizon is always at some distance from him

 

[JesseM]

I have another question though. I've had a chance to let what I've been told sink in and I've figured out the problem I was having. I was told that if an object is moving in at a faster rate behind another slower object then the faster one will undergo more time dilation from the perspective of a third distant observer

I didn't say that, I just said that its coordinate velocity would decrease at a faster rate.

But what about from the second faster moving observers perspective? It's following the first observer in and it can't possibly enter the horizon after the first observer from its own perspective.

Why not? And what do you mean by "its own perspective"? Are you just talking about local coordinate-independent facts like whether the event of the second observer crossing the horizon happens at a later or earlier proper time than the event of the second observer passing the first observer? (if so, the answer is that crossing the horizon happens at an earlier proper time than passing the first observer, if he manages to pass the first observer at all before hitting the singularity) Or are you talking about a "frame" for the second observer, so it can make judgments about things like the velocity of the first observer as a function of time and the time at which the first observer crosses the horizon? If the latter, what coordinate system would this "frame" correspond to?

As always, any statement you can make about observers falling into a black hole has an analogous statement involving observers crossing the Rindler horizon. If you have two inertial observers moving towards the horizon, it may be that the second departed too late to catch up with the first before the first crossed the horizon, even if the second has a higher velocity. In Rindler coordinates, neither would reach the horizon in finite coordinate time, but the second one's velocity would decrease faster than the first so that the second would never pass the horizon at any finite coordinate time. Would you say in this case that the second observer with a higher velocity is "following the first observer in and it can't possibly enter the horizon after the first observer from its own perspective"? If you wouldn't say that in the case of two inertial observers crossing the Rindler horizon, why do you think the case of the black hole is any different?

Length contraction and time dilation force anything that does cross the horizon to do it when the black hole exists at a single point in time and space.

Once and for all, do you agree that "length contraction and time dilation" can only be defined relative to a particular choice of coordinate system, that the fact that length contraction and time dilation approach infinity as you approach the event horizon in Schwarzschild coordinates has no more coordinate-independent "reality" than the fact that length contraction and time dilation approach infinity as you approach the Rindler horizon in Rindler coordinates?

The fact that the second observer is slowing at a faster rate than the first from the third observers perspective just explains how the second observer may never be able to overtake the first observer before reaching the horizon from the third observers perspective.

No, it's a statement about how things work in a parcticular coordinate system, namely Schwarzschild coordinates. What do you mean when you talk about a given observer's "perspective"?? It's not uncommon in SR to talk about an inertial observer's "perspective" as a shorthand for their inertial rest frame, because every inertial observer has a unique inertial rest frame. However, non-inertial observers don't have unique non-inertial rest frames, there are an infinite number of different ways to construct a non-inertial coordinate system where a particular non-inertial observer is at rest. Physicists just don't talk about the "perspectives" of observers in GR, because it would be totally unclear what coordinate system they were referring to! And I suspect that you don't have any well-defined coordinate system in mind when you talk about the "perspectives" of different observers falling into the black hole--if you don't, your comments along these lines would seem to be totally meaningless.

Let's assume that black holes do loose mass during their lives for whatever reason. The first observer measures the size of the horizon just before it crosses and sees that the second observer hasn't set off yet.

"Measures the size" is another meaningless phrase unless you are referring to a particular coordinate system. The only way to define the "size" of an extended object in GR is to have a coordinate system in which you can calculate things like coordinate length or coordinate volume.

The second observer catches up to the first at the event horizon.

Why "at the event horizon"? Depending on how late the second observer departs, it may be impossible for the second observer to catch up with the first until after they are both inside the horizon, or it may even be impossible for the second observer to ever catch up to the first observer before hitting the singularity.

It should never be too late because it can never witness the first one crossing.

If by "witness" you just mean "see the light from the event of the crossing", the second observer will witness the first observer crossing at the moment the second observer crosses the horizon himself (the event of the first crossing the horizon may look like it happened a great distance away at the moment of the second crosses, though). Exactly the same is true if the two observers are crossing the Rindler horizon in flat spacetime, and the second observer isn't able to catch up with the first observer before the first crosses the horizon. Every argument you make I am going to meet with an exact analogy involving the Rindler horizon, so it really is important that you understand the analogy and explain where you think it breaks down (you seemed confused about how the analogy worked before, please reread my clarification in post #148, starting from the paragraph that begins "But I'm not talking about..." right above the second graphic I included in that post)

 

[A-wal]

Begin from here.
There are 2 types of horizons, apparent and absolute. For an observer at infinity, apparent and absolute horizons are at the same position (almost).

But it is not true for a falling observer: when it falls, apparent horizon recedes in front of him. So, if you 'jump' inside the BH to 'resque' a ship 'frozen near the horizon', then you would see how horizons moves deeper and deeper inside the BH, ship 'unfreezes' and falls down too; so you can't catch up with him and can't resque.

And for obvious reasons, no observer can ever pass thru his own apparent horizon, because it would mean that infinite tidal forces had torn his body apart. Fortunaly, in GR freely-falling observer can always assume that his part of space is locally flat, so apparent horizon is always at some distance from him

Yea, that's pretty much exactly what I was describing earlier with the receding horizon but it was shot down.

I have another question though. I've had a chance to let what I've been told sink in and I've figured out the problem I was having. I was told that if an object is moving in at a faster rate behind another slower object then the faster one will undergo more time dilation from the perspective of a third distant observer

I didn't say that, I just said that its coordinate velocity would decrease at a faster rate.

Okay, that was misleading. Loses more velocity through time dilation, not undergo more time dilation.

Originally Posted by A-wal
But what about from the second faster moving observers perspective? It's following the first observer in and it can't possibly enter the horizon after the first observer from its own perspective.

Why not? And what do you mean by "its own perspective"? Are you just talking about local coordinate-independent facts like whether the event of the second observer crossing the horizon happens at a later or earlier proper time than the event of the second observer passing the first observer? (if so, the answer is that crossing the horizon happens at an earlier proper time than passing the first observer, if he manages to pass the first observer at all before hitting the singularity) Or are you talking about a "frame" for the second observer, so it can make judgments about things like the velocity of the first observer as a function of time and the time at which the first observer crosses the horizon? If the latter, what coordinate system would this "frame" correspond to?

I meant the former. How can crossing the horizon happen before the second observer catches the first one? The second observer can't witness the first one crossing the horizon.

As always, any statement you can make about observers falling into a black hole has an analogous statement involving observers crossing the Rindler horizon. If you have two inertial observers moving towards the horizon, it may be that the second departed too late to catch up with the first before the first crossed the horizon, even if the second has a higher velocity. In Rindler coordinates, neither would reach the horizon in finite coordinate time, but the second one's velocity would decrease faster than the first so that the second would never pass the horizon at any finite coordinate time. Would you say in this case that the second observer with a higher velocity is "following the first observer in and it can't possibly enter the horizon after the first observer from its own perspective"? If you wouldn't say that in the case of two inertial observers crossing the Rindler horizon, why do you think the case of the black hole is any different?

Either the second observer can witness the first crossing the horizon before they themselves get there or they can't. You can't have it both ways.

Originally Posted by A-wal
Length contraction and time dilation force anything that does cross the horizon to do it when the black hole exists at a single point in time and space.

Once and for all, do you agree that "length contraction and time dilation" can only be defined relative to a particular choice of coordinate system, that the fact that length contraction and time dilation approach infinity as you approach the event horizon in Schwarzschild coordinates has no more coordinate-independent "reality" than the fact that length contraction and time dilation approach infinity as you approach the Rindler horizon in Rindler coordinates?

Once and for all you say? Does that mean you're not going to ask me this same question ever again? If the first and second ships meet up again then less time would have passed for the first ship, so there is a coordinate-independent reality.

Originally Posted by A-wal
The fact that the second observer is slowing at a faster rate than the first from the third observers perspective just explains how the second observer may never be able to overtake the first observer before reaching the horizon from the third observers perspective.

No, it's a statement about how things work in a parcticular coordinate system, namely Schwarzschild coordinates. What do you mean when you talk about a given observer's "perspective"?? It's not uncommon in SR to talk about an inertial observer's "perspective" as a shorthand for their inertial rest frame, because every inertial observer has a unique inertial rest frame. However, non-inertial observers don't have unique non-inertial rest frames, there are an infinite number of different ways to construct a non-inertial coordinate system where a particular non-inertial observer is at rest. Physicists just don't talk about the "perspectives" of observers in GR, because it would be totally unclear what coordinate system they were referring to! And I suspect that you don't have any well-defined coordinate system in mind when you talk about the "perspectives" of different observers falling into the black hole--if you don't, your comments along these lines would seem to be totally meaningless.

I meant the third observer is sitting there watching the other two so we see what's happing from the perspective of someone who's maintaining a constant distance from the black hole. I spose you'd call it Schwarzschild coordinates. Let me tell you what's meaningless. Those coordinate systems your so found of! I'm not saying they can't be useful, but they're meaningless in themselves. It's what they're describing that has meaning (my one's different because it's not describing it, it's literal), so let's just stick to describing things in real terms. What happens if they do this? What would they see if this happens? Otherwise we're translating it into a graph then translating it back into conceptual reality. It's very limiting to use graphs to form an understanding of something. I think it's meant to be the other way round.

Originally Posted by A-wal
Let's assume that black holes do loose mass during their lives for whatever reason. The first observer measures the size of the horizon just before it crosses and sees that the second observer hasn't set off yet.

"Measures the size" is another meaningless phrase unless you are referring to a particular coordinate system. The only way to define the "size" of an extended object in GR is to have a coordinate system in which you can calculate things like coordinate length or coordinate volume.

I meant that if the first were to measure the size of the black hole just before crossing the horizon and the second one does the same using the same method when reaching the horizon then if they got different results it would mean that they crossed at different times in the black holes life despite crossing the horizon at the same time.

Originally Posted by A-wal
The second observer catches up to the first at the event horizon.

Why "at the event horizon"? Depending on how late the second observer departs, it may be impossible for the second observer to catch up with the first until after they are both inside the horizon, or it may even be impossible for the second observer to ever catch up to the first observer before hitting the singularity.

Because the second observer can't witness the first crossing the horizon until they cross it themselves.

Originally Posted by A-wal
It should never be too late because it can never witness the first one crossing.

If by "witness" you just mean "see the light from the event of the crossing", the second observer will witness the first observer crossing at the moment the second observer crosses the horizon himself (the event of the first crossing the horizon may look like it happened a great distance away at the moment of the second crosses, though).

"See the light from the event of the crossing" is exactly the same as "the event of the crossing". The light's moving slowly because time is. The event of the first one crossing may happen at a great distance away at the moment the second one crosses? The first one can't cross at all from the perspective of the second one at least until the second one reaches the horizon, at which point the first object jumps to a great distance away? WTF?

 

[JesseM]

Okay, that was misleading. Loses more velocity through time dilation, not undergo more time dilation.

I don't see that the faster loss of velocity has anything to do with time dilation, it's just a consequence of the coordinate transformation.

Why not? And what do you mean by "its own perspective"? Are you just talking about local coordinate-independent facts like whether the event of the second observer crossing the horizon happens at a later or earlier proper time than the event of the second observer passing the first observer? (if so, the answer is that crossing the horizon happens at an earlier proper time than passing the first observer, if he manages to pass the first observer at all before hitting the singularity) Or are you talking about a "frame" for the second observer, so it can make judgments about things like the velocity of the first observer as a function of time and the time at which the first observer crosses the horizon? If the latter, what coordinate system would this "frame" correspond to?

I meant the former. How can crossing the horizon happen before the second observer catches the first one? The second observer can't witness the first one crossing the horizon.

Sure he can, the second observer can witness the first crossing the horizon at the moment the moment he himself crosses the horizon. Again just consider the case of two observers that cross the Rindler horizon, where the second one won't see the first crossing it until the moment the second crosses the horizon. Of course that doesn't mean the second observer is right on top of the first observer as they cross, the second observer may see the first crossing the horizon at a great distance away (remember that it is only in Rindler coordinates that the Rindler horizon has an unchanging coordinate position, in any inertial coordinate system the Rindler horizon is expanding outward at the speed of light...similarly, for a black hole it is true in Schwarzschild coordinates that the event horizon has a fixed position, but in Kruskal-Szekeres coordinates it is also expanding outward at the speed of light, and likewise the event horizon is moving outward with a speed of c in the local inertial frame of a freefalling observer at the moment she crosses the horizon). Do you think this is a problem?

Either the second observer can witness the first crossing the horizon before they themselves get there or they can't. You can't have it both ways.

The second observer cannot witness the the first crossing before the second crosses the horizon, the second will only see this at the moment the second crosses the horizon.

Length contraction and time dilation force anything that does cross the horizon to do it when the black hole exists at a single point in time and space.

Once and for all, do you agree that "length contraction and time dilation" can only be defined relative to a particular choice of coordinate system, that the fact that length contraction and time dilation approach infinity as you approach the event horizon in Schwarzschild coordinates has no more coordinate-independent "reality" than the fact that length contraction and time dilation approach infinity as you approach the Rindler horizon in Rindler coordinates?

Once and for all you say? Does that mean you're not going to ask me this same question ever again? If the first and second ships meet up again then less time would have passed for the first ship, so there is a coordinate-independent reality.

Sure, overall elapsed proper time for two observers who cross paths twice is coordinate-independent. But the very fact that it is coordinate-independent means all coordinate systems will agree on this, so you cannot possibly use this fact to support the claim that time dilation and length contraction go to infinity on the horizon, since even a coordinate system (like Kruskal-Szekeres coordinates) where they don't go to infinity at the horizon will make the same prediction about elapsed proper time for any two observers who cross paths twice.

So, do you agree that the statement that time dilation and length contraction go to infinity at the horizon is a purely-coordinate dependent one that's true in Schwarzschild coordinates but has no coordinate-independent reality? If you do agree with that, then I don't understand what you meant by the statement that "Length contraction and time dilation force anything that does cross the horizon to do it when the black hole exists at a single point in time and space".

The fact that the second observer is slowing at a faster rate than the first from the third observers perspective just explains how the second observer may never be able to overtake the first observer before reaching the horizon from the third observers perspective.

No, it's a statement about how things work in a parcticular coordinate system, namely Schwarzschild coordinates. What do you mean when you talk about a given observer's "perspective"?? It's not uncommon in SR to talk about an inertial observer's "perspective" as a shorthand for their inertial rest frame, because every inertial observer has a unique inertial rest frame. However, non-inertial observers don't have unique non-inertial rest frames, there are an infinite number of different ways to construct a non-inertial coordinate system where a particular non-inertial observer is at rest. Physicists just don't talk about the "perspectives" of observers in GR, because it would be totally unclear what coordinate system they were referring to! And I suspect that you don't have any well-defined coordinate system in mind when you talk about the "perspectives" of different observers falling into the black hole--if you don't, your comments along these lines would seem to be totally meaningless.

I meant the third observer is sitting there watching the other two so we see what's happing from the perspective of someone who's maintaining a constant distance from the black hole. I spose you'd call it Schwarzschild coordinates.

No, I most certainly would not. In my view, the only meaningful statements you can make about the "perspectives" of different observers in GR are coordinate-independent statements about when light beams from various faraway events hit their own worldlines, in terms of their own proper time. An observer in GR does not have a "frame" or any other notion of a "perspective" besides this simple visual one, and coordinate systems are just coordinate systems, they don't represent the perspective of any particular observer. From your 'I meant the former' comment in response to my own 'Why not? And what do you mean by "its own perspective"?' above, perhaps you would actually agree with this, and I was misunderstanding you by thinking you were invoking some notion of frames--can you clarify this? Whenever you talk about an observer's "perspective", are you always just talking about coordinate-independent facts about what their clocks read when light rays from various events reach their eyes? If so, then do you agree that such facts will be the same regardless of what coordinate system we use, so it's fine to figure them out using Kruskal-Szekeres coordinates (where it takes only a finite coordinate time to reach the event horizon, just like with observers crossing the Rindler horizon when viewed in inertial coordinates) rather than Schwarzschild coordinates (where it takes an infinite coordinate time, just like with Rindler coordinates)?

Let me tell you what's meaningless. Those coordinate systems your so found of! I'm not saying they can't be useful, but they're meaningless in themselves. It's what they're describing that has meaning (my one's different because it's not describing it, it's literal), so let's just stick to describing things in real terms.

In general relativity, all calculations about coordinate-independent facts depend on having a coordinate system and a metric expressed in terms of those coordinates. Of course once we have figured out the coordinate-independent facts themselves we don't have to talk about the coordinate systems any more, but you seem unsatisfied when I just tell you what the coordinate-independent facts are, like the fact that both observers pass the horizon in finite time according to their own clocks, and that the second observer sees the image of the first observer crossing the horizon at a significant distance at the moment the second observer himself crosses the horizon.

I meant that if the first were to measure the size of the black hole just before crossing the horizon

Again, how do you "measure the size" of anything without using a coordinate system? Are you just talking about the apparent visual size, i.e. what percentage of your field of vision is occupied by the black hole just before crossing the horizon? (if you're interested in visual appearance, see the animations and images on this page)

Because the second observer can't witness the first crossing the horizon until they cross it themselves.

Yes, but they see it happening far away if they waited a significant time before pursuing the first observer.

It should never be too late because it can never witness the first one crossing.

If by "witness" you just mean "see the light from the event of the crossing", the second observer will witness the first observer crossing at the moment the second observer crosses the horizon himself (the event of the first crossing the horizon may look like it happened a great distance away at the moment of the second crosses, though).

"See the light from the event of the crossing" is exactly the same as "the event of the crossing". The light's moving slowly because time is. The event of the first one crossing may happen at a great distance away at the moment the second one crosses? The first one can't cross at all from the perspective of the second one at least until the second one reaches the horizon, at which point the first object jumps to a great distance away? WTF?

No, there is no "jump", the visual position of the horizon (which could be seen as a black region against a background filled with stars) continues to look far-off as you approach it, and even at the moment you cross it yourself. Again see this page, particularly the section labeled "Through the horizon". Also, by the same token if you're following another observer in, the moment you see the event on the first observer's worldline that occurred when they crossed the horizon, their apparent visual distance may be different than the apparent visual distance of the black sphere you see below you, i.e. it may not look like they're crossing the horizon at that moment.

 

[A-wal]

I don't see that the faster loss of velocity has anything to do with time dilation, it's just a consequence of the coordinate transformation.

Um, I was specifically referring to the effect of gravitational time dilation on relative velocity. NOTHING is a consequence of coordinates.

Sure, overall elapsed proper time for two observers who cross paths twice is coordinate-independent. But the very fact that it is coordinate-independent means all coordinate systems will agree on this, so you cannot possibly use this fact to support the claim that time dilation and length contraction go to infinity on the horizon, since even a coordinate system (like Kruskal-Szekeres coordinates) where they don't go to infinity at the horizon will make the same prediction about elapsed proper time for any two observers who cross paths twice.

So, do you agree that the statement that time dilation and length contraction go to infinity at the horizon is a purely-coordinate dependent one that's true in Schwarzschild coordinates but has no coordinate-independent reality? If you do agree with that, then I don't understand what you meant by the statement that "Length contraction and time dilation force anything that does cross the horizon to do it when the black hole exists at a single point in time and space".

I would have thought it was obvious. Any external observer can never witness an object crossing the horizon.

No, I most certainly would not. In my view, the only meaningful statements you can make about the "perspectives" of different observers in GR are coordinate-independent statements about when light beams from various faraway events hit their own worldlines, in terms of their own proper time. An observer in GR does not have a "frame" or any other notion of a "perspective" besides this simple visual one, and coordinate systems are just coordinate systems, they don't represent the perspective of any particular observer. From your 'I meant the former' comment in response to my own 'Why not? And what do you mean by "its own perspective"?' above, perhaps you would actually agree with this, and I was misunderstanding you by thinking you were invoking some notion of frames--can you clarify this? Whenever you talk about an observer's "perspective", are you always just talking about coordinate-independent facts about what their clocks read when light rays from various events reach their eyes? If so, then do you agree that such facts will be the same regardless of what coordinate system we use, so it's fine to figure them out using Kruskal-Szekeres coordinates (where it takes only a finite coordinate time to reach the event horizon, just like with observers crossing the Rindler horizon when viewed in inertial coordinates) rather than Schwarzschild coordinates (where it takes an infinite coordinate time, just like with Rindler coordinates)?

Yes, that is what I meant and I don't want to use any coordinate systems apart from what's actually observed. You seem to desperately want to keep using them though.

Again, how do you "measure the size" of anything without using a coordinate system? Are you just talking about the apparent visual size, i.e. what percentage of your field of vision is occupied by the black hole just before crossing the horizon? (if you're interested in visual appearance, see the animations and images on this page)

Any, that will do. As long as they use the same method so they can compare.

Sure he can, the second observer can witness the first crossing the horizon at the moment the moment he himself crosses the horizon. Again just consider the case of two observers that cross the Rindler horizon, where the second one won't see the first crossing it until the moment the second crosses the horizon. Of course that doesn't mean the second observer is right on top of the first observer as they cross, the second observer may see the first crossing the horizon at a great distance away (remember that it is only in Rindler coordinates that the Rindler horizon has an unchanging coordinate position, in any inertial coordinate system the Rindler horizon is expanding outward at the speed of light...similarly, for a black hole it is true in Schwarzschild coordinates that the event horizon has a fixed position, but in Kruskal-Szekeres coordinates it is also expanding outward at the speed of light, and likewise the event horizon is moving outward with a speed of c in the local inertial frame of a freefalling observer at the moment she crosses the horizon). Do you think this is a problem?

The second observer cannot witness the the first crossing before the second crosses the horizon, the second will only see this at the moment the second crosses the horizon.

In general relativity, all calculations about coordinate-independent facts depend on having a coordinate system and a metric expressed in terms of those coordinates. Of course once we have figured out the coordinate-independent facts themselves we don't have to talk about the coordinate systems any more, but you seem unsatisfied when I just tell you what the coordinate-independent facts are, like the fact that both observers pass the horizon in finite time according to their own clocks, and that the second observer sees the image of the first observer crossing the horizon at a significant distance at the moment the second observer himself crosses the horizon.

Yes, but they see it happening far away if they waited a significant time before pursuing the first observer.

No, there is no "jump", the visual position of the horizon (which could be seen as a black region against a background filled with stars) continues to look far-off as you approach it, and even at the moment you cross it yourself. Again see this page, particularly the section labeled "Through the horizon". Also, by the same token if you're following another observer in, the moment you see the event on the first observer's worldline that occurred when they crossed the horizon, their apparent visual distance may be different than the apparent visual distance of the black sphere you see below you, i.e. it may not look like they're crossing the horizon at that moment.

Not a single mention of the event horizon seemingly being in two places at once and now FIVE! That info would have been very handy before, especially when I was talking about the receding horizon. It could have saved us both a lot of time. Right, how does that happen? If you're saying they're seeing a time delayed image of them crossing where the horizon used to be then the black hole has to have grown in the time between the two objects crossing. The event horizon by definition is the point at which an object can no longer possibly find the energy to escape, so how can it be in two places at the same time? It should be (and appear to be) in the same place for all objects. Also if the event horizon continues to look far off as you approach it then light is escaping from the “actual” horizon.

 

[JesseM]

Um, I was specifically referring to the effect of gravitational time dilation on relative velocity. NOTHING is a consequence of coordinates.

There is no notion of "relative velocity" (or gravitational time dilation) outside of coordinate systems, they are frame-dependent concepts in GR.

So, do you agree that the statement that time dilation and length contraction go to infinity at the horizon is a purely-coordinate dependent one that's true in Schwarzschild coordinates but has no coordinate-independent reality? If you do agree with that, then I don't understand what you meant by the statement that "Length contraction and time dilation force anything that does cross the horizon to do it when the black hole exists at a single point in time and space".

I would have thought it was obvious. Any external observer can never witness an object crossing the horizon.

No, it's not obvious at all, so you need to explain it. "An external observer can never witness an object crossing the horizon" has no clear connection to the statement "Length contraction and time dilation force anything that does cross the horizon to do it when the black hole exists at a single point in time and space". How does not seeing things cross the horizon have anything to do with length contraction or time dilation or the notion that "the black hole exists at a single point in time and space"?

Yes, that is what I meant and I don't want to use any coordinate systems apart from what's actually observed. You seem to desperately want to keep using them though.

I'm fine with not using them as long as you don't try to sneak in inherently coordinate-dependent notions like "relative velocity" or "length contraction" (and we can only talk about 'time dilation' in a coordinate-independent way if we're talking about a twin-paradox-like situation where two observers meet twice and compare the elapsed time on their clocks between meetings, any notion of 'time dilation' other than this is also inherently coordinate-dependent).

Again, how do you "measure the size" of anything without using a coordinate system? Are you just talking about the apparent visual size, i.e. what percentage of your field of vision is occupied by the black hole just before crossing the horizon? (if you're interested in visual appearance, see the animations and images on this page)

Any, that will do. As long as they use the same method so they can compare.

OK, if you just want to talk about apparent visual size, then for a Schwarzschild black hole both will see the BH as being the same apparent size at the moment they each cross the horizon.

Not a single mention of the event horizon seemingly being in two places at once and now FIVE! That info would have been very handy before, especially when I was talking about the receding horizon.

I thought you were talking about what "really" happens in some frame, you didn't specify that you wanted to talk about visual appearances.

It could have saved us both a lot of time. Right, how does that happen?

I don't know of any way to "explain" visual phenomena like this except by calculating which events have light beams that all converge on the observer at the moment he crosses the horizon, which would require using a coordinate system to work out the timing of events and the light travel times (and in any cases the calculations for the apparent size and the horizon would be above my head I think, though the appendix of this paper discusses angular size of spheres in GR, and http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000078000002000204000001&idtype=cvips&gifs=yes&ref=no probably goes into more detail but isn't available for free online).

If you're saying they're seeing a time delayed image of them crossing where the horizon used to be then the black hole has to have grown in the time between the two objects crossing.

"Grown" in what sense? Remember, the only notion of size you wanted to talk about was apparent visual size--do you just mean the apparent visual size grows as the falling observer approaches it? Well naturally, that would be true of any body!

The event horizon by definition is the point at which an object can no longer possibly find the energy to escape, so how can it be in two places at the same time? It should be (and appear to be) in the same place for all objects.

Why should it "appear to be" in the same place for all objects? On the contrary, if a row of observers are falling in one after the other, then when an observer in the rear crosses the horizon the light he's seeing from the other observers at that moment is the light they emitted at the moment each of them crossed the horizon, but this observer will still see the other observer in a row in front of him, at a range of different apparent visual distances. It's really no different from the idea in SR that if you have a row of observers who enter the future light cone of some event E in succession (which would also mean they were crossing the Rindler horizon of some possible observer accelerating at a constant rate), then at the moment the observer in the rear first enters the future light cone, the light he's seeing from other observers at that moment is the light each of them emitted at the moment they entered the future light cone, and at this moment they still appear to be in a row at different distances. (And of course just as the Rindler horizon looks like the edge of a light cone when plotted in Minkowski coordinates, so the black hole event horizon looks like the edge of a light cone when plotted in Kruskal-Szekeres coordinates, which is a coordinate system where all light worldlines are straight lines at 45 degrees from the time axis, just like in Minkowski coordinates. But I'm putting this in parentheses because you may not want to think about the representation of the horizon and of light signals in some coordinate system--the paragraph above outside of the parentheses deals only with visual observations without explaining to explain why this is what the falling observer sees, so if you don't want to deal with coordinate systems you pretty much have to accept it without explanation).

Also if the event horizon continues to look far off as you approach it then light is escaping from the “actual” horizon.

No, it isn't. But again, if you just make wrong statements without providing your reasoning, I can't really help you to see where your error lies. It may help to remember that the apparent size of the horizon is based on the set of angles where no light from distant stars can reach your eyes, if there was a solid shell hovering just an infinitesimal fraction of a nanometer above the event horizon, I'm pretty sure the visual size of the solid shell would be different from the apparent size of the region where no distant stars can be seen (it seems like it would have to, since you would see the shell rush up to meet you and you'd punch through it a moment before you crossed the horizon, but at the moment you cross the horizon the black sphere where stars are blocked still looks like it's far away from you).

edit: actually, looking at the "through the horizon" section here, they say:

As you fall through the horizon, at 1 Schwarzschild radius, something quite unexpected happens. You thought you were going to fall through the red grid that supposedly marks the horizon. But no. The red grid still stands off ahead of you.
Instead, the horizon splits into two as you pass through it. Click on Penrose diagrams to understand more about why the horizon splits in two.

And the Penrose diagram, which is basically just like a compacted version of a Kruskal-Szekeres diagram, shows why the horizon appears to split in two as you cross it in this animation. If there was a spherical shell eternally hovering just above the event horizon, in this graph it would look like a hyperbola that hugs extremely close to both the antihorizon (really the white hole event horizon) on the bottom and the regular horizon on top, so I think up until just before the moment you reached it, the shell's apparent size would look about the same size as the black region (since on the page the red grid marking the antihorizon is the same size as the black region obscuring stars), and then suddenly it would split off from the black sphere and rush up to meet you (or perhaps just part of it would rush up to meet you, like a spike or bump rising out of the sphere) like the white grid representing the horizon (see the third animation on the main page, or the white 'Schwarzschild bubble' in the similar animations on http://casa.colorado.edu/~ajsh/singularity.html).

 

[A-wal]

There is no notion of "relative velocity" (or gravitational time dilation) outside of coordinate systems, they are frame-dependent concepts in GR.

Relative velocity still applies in curved space-time.

No, it's not obvious at all, so you need to explain it. "An external observer can never witness an object crossing the horizon" has no clear connection to the statement "Length contraction and time dilation force anything that does cross the horizon to do it when the black hole exists at a single point in time and space". How does not seeing things cross the horizon have anything to do with length contraction or time dilation or the notion that "the black hole exists at a single point in time and space"?

The reason an external observer can never witness an object crossing the horizon is length contraction/time dilation. But they will have to see everything cross the horizon eventually if the black hole has a limited life span. From this (any external) perspective it would have to mean that it happens when the black hole exists at a single point in time and space at the very end of its life.

I'm fine with not using them as long as you don't try to sneak in inherently coordinate-dependent notions like "relative velocity" or "length contraction" (and we can only talk about 'time dilation' in a coordinate-independent way if we're talking about a twin-paradox-like situation where two observers meet twice and compare the elapsed time on their clocks between meetings, any notion of 'time dilation' other than this is also inherently coordinate-dependent).

If gravitational time dilation applies when two objects with different world lines meet up then it always applies. If not, then at what distance would it suddenly not apply?

OK, if you just want to talk about apparent visual size, then for a Schwarzschild black hole both will see the BH as being the same apparent size at the moment they each cross the horizon.

I take it this assumes an ever-lasting black hole?

I thought you were talking about what "really" happens in some frame, you didn't specify that you wanted to talk about visual appearances.

Same thing isn't it?

I don't know of any way to "explain" visual phenomena like this except by calculating which events have light beams that all converge on the observer at the moment he crosses the horizon, which would require using a coordinate system to work out the timing of events and the light travel times (and in any cases the calculations for the apparent size and the horizon would be above my head I think, though the appendix of this paper discusses angular size of spheres in GR, and http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000078000002000204000001&idtype=cvips&gifs=yes&ref=no probably goes into more detail but isn't available for free online).

Okay so there is the actual horizon and an apparent one when viewing an object that crossed at an earlier time, and it appears to be beyond the actual horizon. That would mean that there's always an actual horizon and an apparent one because things like that never just start happen, it's always a matter of degree. That's why I questioned it earlier when it seemed you were implying there was a jump. So the two horizons appear closer the further you are away? Presumably if you were to move away from the black hole then the object that appears beyond your event horizon would appear to move back outwards?

"Grown" in what sense? Remember, the only notion of size you wanted to talk about was apparent visual size--do you just mean the apparent visual size grows as the falling observer approaches it? Well naturally, that would be true of any body!

I meant that object that crossed earlier had a larger size relative to the black hole than you do now, meaning the black hole has grown since (assuming both objects are the same size of course).

Why should it "appear to be" in the same place for all objects? On the contrary, if a row of observers are falling in one after the other, then when an observer in the rear crosses the horizon the light he's seeing from the other observers at that moment is the light they emitted at the moment each of them crossed the horizon, but this observer will still see the other observer in a row in front of him, at a range of different apparent visual distances. It's really no different from the idea in SR that if you have a row of observers who enter the future light cone of some event E in succession (which would also mean they were crossing the Rindler horizon of some possible observer accelerating at a constant rate), then at the moment the observer in the rear first enters the future light cone, the light he's seeing from other observers at that moment is the light each of them emitted at the moment they entered the future light cone, and at this moment they still appear to be in a row at different distances. (And of course just as the Rindler horizon looks like the edge of a light cone when plotted in Minkowski coordinates, so the black hole event horizon looks like the edge of a light cone when plotted in Kruskal-Szekeres coordinates, which is a coordinate system where all light worldlines are straight lines at 45 degrees from the time axis, just like in Minkowski coordinates. But I'm putting this in parentheses because you may not want to think about the representation of the horizon and of light signals in some coordinate system--the paragraph above outside of the parentheses deals only with visual observations without explaining to explain why this is what the falling observer sees, so if you don't want to deal with coordinate systems you pretty much have to accept it without explanation).

You cannot see an object cross the horizon until you do so the closer object could still find the energy to escape. From your perspective though it shouldn't be able to escape because it's beyond your horizon. Let's assume it can and does escape. It starts to head towards you. What then? If it pulls right up and sits along side you then your horizon is now its horizon, so it's still just outside it. It's moved to you but it's still the same distance away from the horizon?

No, it isn't. But again, if you just make wrong statements without providing your reasoning, I can't really help you to see where your error lies. It may help to remember that the apparent size of the horizon is based on the set of angles where no light from distant stars can reach your eyes, if there was a solid shell hovering just an infinitesimal fraction of a nanometer above the event horizon, I'm pretty sure the visual size of the solid shell would be different from the apparent size of the region where no distant stars can be seen (it seems like it would have to, since you would see the shell rush up to meet you and you'd punch through it a moment before you crossed the horizon, but at the moment you cross the horizon the black sphere where stars are blocked still looks like it's far away from you).

The horizon is just in front of you but you can see an object beyond it, meaning light is escaping the horizon. And what does the earlier in falling object see when looking back at the later one? Does that mean the apparent horizon can be in front or behind you, or would an object behind have to actually catch up to you to reach the horizon. That means their two views of each other would be contradictory. One says there's space between them and one doesn't.

edit: actually, looking at the "through the horizon" section here, they say:

And the Penrose diagram, which is basically just like a compacted version of a Kruskal-Szekeres diagram, shows why the horizon appears to split in two as you cross it in this animation. If there was a spherical shell eternally hovering just above the event horizon, in this graph it would look like a hyperbola that hugs extremely close to both the antihorizon (really the white hole event horizon) on the bottom and the regular horizon on top, so I think up until just before the moment you reached it, the shell's apparent size would look about the same size as the black region (since on the page the red grid marking the antihorizon is the same size as the black region obscuring stars), and then suddenly it would split off from the black sphere and rush up to meet you (or perhaps just part of it would rush up to meet you, like a spike or bump rising out of the sphere) like the white grid representing the horizon (see the third animation on the main page, or the white 'Schwarzschild bubble' in the similar animations on http://casa.colorado.edu/~ajsh/singularity.html).

Okay, but that doesn't apply to your view of an object that falls in at an earlier time because there can't be a jump, like we said. So this notion of a split horizon must be describing something else, yes?

 

[JesseM]

Relative velocity still applies in curved space-time.

No, there is no frame-independent notion of "relative velocity" in curved spacetime. Take a look at all the posters explaining this to Anamitra in this thread, for example.

The reason an external observer can never witness an object crossing the horizon is length contraction/time dilation.

It is true that visually, a clock falling towards a black hole will appear to run slower and slower as it approaches, and will also appear increasingly compressed in the radial direction. But usually "time dilation" is understood in terms of a slowdown in the rate a clock is ticking relative to coordinate time in some reference frame--visual slowdown is not the same as "time dilation". Consider the simple example of Doppler shift in SR, which causes a clock moving towards you to appear to be ticking faster than your clock despite the fact that it's ticking slow in your frame, and which also causes a clock moving away from you to appear to be ticking even slower than the time dilation factor. Likewise with "length contraction"--the visual apparent length of an object may be different than its length-contracted length in your frame due to the fact that the light you see at any given moment was actually emitted by different parts of the object at different times in your frame (the Penrose-Terrell effect). So if you're referring to visual appearances, you should use some other term besides "length contraction/time dilation" as people (like me in earlier posts) are liable to misinterpret your meaning.

But they will have to see everything cross the horizon eventually if the black hole has a limited life span. From this (any external) perspective it would have to mean that it happens when the black hole exists at a single point in time and space at the very end of its life.

Yes, in the case of an evaporating block hole it's apparently true that theoretically (assuming you could detect light of arbitrary redshift, and assuming light was emitted continuously rather than in discrete photons), you would see light from objects crossing the horizon at the moment the black hole finally evaporated to zero radius (see the section 'What about Hawking radiation? Won't the black hole evaporate before you get there?' on this page). But that's just a statement about when the light emitted by those observers as they crossed the horizon finally reaches the outside observer, it doesn't imply that the black hole actually was point-sized for the infalling observers when they crossed the horizon.

I'm fine with not using them as long as you don't try to sneak in inherently coordinate-dependent notions like "relative velocity" or "length contraction" (and we can only talk about 'time dilation' in a coordinate-independent way if we're talking about a twin-paradox-like situation where two observers meet twice and compare the elapsed time on their clocks between meetings, any notion of 'time dilation' other than this is also inherently coordinate-dependent).

If gravitational time dilation applies when two objects with different world lines meet up then it always applies. If not, then at what distance would it suddenly not apply?

Your language is overly vague, I don't know what it means for it to "apply" or "not apply". To be precise about what I mean, suppose you have two observers who compare clock readings at a single point, fly apart, and then reunite and compare clock readings again. Pick any point A on the worldline of one observer, and any point B on the worldline of the second observer. Then it can be true that different coordinate systems disagree on whether the clock at point A was ticking slower than the clock at point B or vice versa, and likewise it can be true that other observers with different paths through space time disagree about the exact ration between the visual rate of ticking of the clock at point A vs. the clock at point B. Nevertheless, all frames/observers agree about the total elapsed time on each clock when they finally reunite at a single point in spacetime. Would you say this means time dilation "applies" or "doesn't apply" at points A and B on each clock's worldline?

OK, if you just want to talk about apparent visual size, then for a Schwarzschild black hole both will see the BH as being the same apparent size at the moment they each cross the horizon.

I take it this assumes an ever-lasting black hole?

Right, I'm talking about the Schwarzschild solution. For a black hole that was shrinking due to Hawking radiation or growing due to infalling matter, I'm pretty confident the apparent visual size would be different for observers who cross the horizon at different times (i.e. in terms of visual appearances, neither sees the other as right next to themselves at the moment they're crossing the horizon, even though they fall in along the same radial direction)

I thought you were talking about what "really" happens in some frame, you didn't specify that you wanted to talk about visual appearances.

Same thing isn't it?

No, but my way of talking may be confusing, when I said "really" I wasn't talking about frame-invariant physical facts, but rather the "real" coordinates of events in a given frame, as distinguished from visual appearances. As I mentioned above, even in SR things like visual clock rates and visual lengths can be different from the "real" amount of time dilation and length contraction in the observer's inertial rest frame.

Okay so there is the actual horizon and an apparent one when viewing an object that crossed at an earlier time, and it appears to be beyond the actual horizon. That would mean that there's always an actual horizon and an apparent one because things like that never just start happen, it's always a matter of degree. That's why I questioned it earlier when it seemed you were implying there was a jump.

Well, normally the horizon isn't something you can "see" directly, but if we imagine that somehow every event on the horizon generated photons which traveled in all directions of the future light cone of that event, so that the horizon was visible in its own right (as opposed to our just being able to see a black area where light from distant stars is blocked), then you would only begin to "see" the horizon at the moment you crossed it, since none of the photons generated at the horizon would make it outside the horizon, but they could reach observers falling into the horizon. Meanwhile for a Schwarzschild black hole, there is also an "antihorizon" which consists of both a white hole event horizon in our universe and a black hole event horizon leading to a different universe (see the maximally extended Kruskal-Szekeres diagram for a discussion). So if we imagine all events on the antihorizon were also generating photons of a different color (say, red) which traveled in all directions in the future light cone of the event that generated them, then as you traveled towards the black hole you would see a red sphere representing the white hole's event horizon, and after you crossed the horizon you'd continue to see the red sphere, but now you'd be seeing light from the black hole horizon in the other universe. But after you crossed the horizon you'd also suddenly start seeing light from black hole horizon in your own universe--if you imagine that this light is white instead of red, then you should be able to follow what's depicted in the third movie from the top on this page, where we approach a black hole covered with red grid lines and then suddenly a new set of white grid lines appear (forming a sort of dome shape above us) at the moment we cross the horizon. If you read about the maximally extended Kruskal-Szekeres diagram, you should be able to follow this similar Penrose diagram that I linked to earlier, which shows which events on the horizon and antihorizon emitted the red and white light reaching you at each point in your fall. And I also linked to http://casa.colorado.edu/~ajsh/singularity.html which shows somewhat cruder animations of what a falling observer would see (again with the antihorizon in red and the white 'Schwarzschild bubble' that you see after crossing the horizon in white), with somewhat more detailed accompanying explanation (for example, he mentions that at the moment you cross the horizon the white Schwarzschild bubble would first appear as a straight line reaching from you to the red horizon, only later expanding into a bubble, and that at that moment other objects which fell into the black hole along the same axis would be arrayed along this line, all appearing as they did the moment they crossed the horizon).

So the two horizons appear closer the further you are away?

No, why do you say that?

"Grown" in what sense? Remember, the only notion of size you wanted to talk about was apparent visual size--do you just mean the apparent visual size grows as the falling observer approaches it? Well naturally, that would be true of any body!

I meant that object that crossed earlier had a larger size relative to the black hole than you do now

And are you using "size" to refer only to apparent visual size, i.e. how much of the visual field is taken up by the black region devoid of stars (or the red antihorizon which apparently would coincide with it)? If so then I think as long as the black hole wasn't significantly changing its coordinate size (in some appropriate coordinate system like Eddington-Finkelstein coordinates) due to evaporation or significant influxes of matter, than observers who fell in at different times also shouldn't see a difference in apparent visual size at the moment they cross the horizon. If you want "size" to refer to something more than apparent visual size, I don't see how you can define any other notion of size without reference to a coordinate system.

You cannot see an object cross the horizon until you do so the closer object could still find the energy to escape. From your perspective though it shouldn't be able to escape because it's beyond your horizon.

What do you mean "beyond your horizon"? If the object escapes, then you will always see it as outside the red antihorizon, and if you ended up crossing the horizon yourself you would presumably see it remain outside the white "Schwarzschild bubble" (which consists of light from events on the black hole event horizon) which you are underneath.

Let's assume it can and does escape. It starts to head towards you.

And what are "you" doing? Remaining outside the horizon too, or crossing through the black hole event horizon? If the latter, then you will see it pass by you and move up to a greater radius before you pass the event horizon and begin to see the "Schwarzschild bubble".

The horizon is just in front of you but you can see an object beyond it, meaning light is escaping the horizon.

If other objects fell into the black hole horizon from your universe, you will always see them above the red antihorizon, not "beyond" it (though you could potentially see objects beyond it that were in regions IV or III of the Kruskal diagram, in either the white hole interior region or the "other universe" respectively. Once you cross the horizon yourself and see yourself under the white Schwarzschild bubble, every other object that fell in before you will be under the bubble too, and you could look above you and see other objects entering the bubble from above.

And what does the earlier in falling object see when looking back at the later one?

The earlier object would also see a white Schwarzschild bubble that appeared at the moment it crossed the horizon (again assuming for the sake of visualization that each event on the event horizon emits white light in all possible directions, in reality of course the bubble wouldn't actually be a visible shape), and assuming light from your crossing the horizon had time to reach the earlier object before it hit the singularity, it would at some point see you cross through the top of this expanding bubble.

 

[Austin0]

.

Well, normally the horizon isn't something you can "see" directly, but if we imagine that somehow every event on the horizon generated photons which traveled in all directions of the future light cone of that event, so that the horizon was visible in its own right (as opposed to our just being able to see a black area where light from distant stars is blocked), then you would only begin to "see" the horizon at the moment you crossed it,

.

Since this thread seems to be winding down I hope it is alright to ask a somewhat off topic question.
With the Rindler horizon; is there any visual image that can be associated with it as viewed from inside the accelerating frame?
Or is it more a question of light that will never reach you but you were not aware of anyway?
SO a diminished field of relative obscurity?
Thanks

 

[JesseM]

Since this thread seems to be winding down I hope it is alright to ask a somewhat off topic question.
With the Rindler horizon; is there any visual image that can be associated with it as viewed from inside the accelerating frame?
Or is it more a question of light that will never reach you but you were not aware of anyway?
SO a diminished field of relative obscurity?
Thanks

Well, note that the Rindler horizon has an upper and lower component (or maybe it would be better to say there are two horizons?), as seen in this diagram from the wikipedia article which plots both the horizons and lines of constant Rindler position and time in a Minkowski diagram. The upper component behaves like a black hole horizon in that the Rindler observers can't see anything that crosses it (though an observer on the "Rindler wedge" can cross it herself if she doesn't remain at rest in Rindler coordinates), but the lower component behaves like a white hole horizon in that Rindler observers can see light from events "below" it (the light rays would be traveling parallel to the upper horizon in the diagram), and it's impossible for any observer on the Rindler wedge to cross it (in Minkowski coordinates it's moving away from anyone on the Rindler wedge with a speed of c). So, in terms of Minkowski coordinates this means an observer on the Rindler wedge can see light from events that occurred arbitrarily far in either direction along the x-axis, including events much further than the current distance of the Rindler horizon. So certainly an inertial observer on the Rindler wedge would see nothing unusual, and at any given moment an accelerating Rindler observer should see the same thing as an inertial observer at the same point in spacetime who is instantaneously at rest relative to themselves.

 

[A-wal]

No, there is no frame-independent notion of "relative velocity" in curved spacetime. Take a look at all the posters explaining this to Anamitra in this thread, for example.

That's not what I meant. If something has a relative velocity of .5c for example then that velocity doesn't disappear in curved space-time. All space-time is at least a little bit curved so there'd be no such thing as relative verlocity at all.

It is true that visually, a clock falling towards a black hole will appear to run slower and slower as it approaches, and will also appear increasingly compressed in the radial direction. But usually "time dilation" is understood in terms of a slowdown in the rate a clock is ticking relative to coordinate time in some reference frame--visual slowdown is not the same as "time dilation". Consider the simple example of Doppler shift in SR, which causes a clock moving towards you to appear to be ticking faster than your clock despite the fact that it's ticking slow in your frame, and which also causes a clock moving away from you to appear to be ticking even slower than the time dilation factor. Likewise with "length contraction"--the visual apparent length of an object may be different than its length-contracted length in your frame due to the fact that the light you see at any given moment was actually emitted by different parts of the object at different times in your frame (the Penrose-Terrell effect). So if you're referring to visual appearances, you should use some other term besides "length contraction/time dilation" as people (like me in earlier posts) are liable to misinterpret your meaning.

I stand by my claim that the reason an external observer can never witness an object crossing the horizon is length contraction/time dilation. Gravitational time dilation/length contraction doesn't make it any less real.

Yes, in the case of an evaporating block hole it's apparently true that theoretically (assuming you could detect light of arbitrary redshift, and assuming light was emitted continuously rather than in discrete photons), you would see light from objects crossing the horizon at the moment the black hole finally evaporated to zero radius (see the section 'What about Hawking radiation? Won't the black hole evaporate before you get there?' on this page). But that's just a statement about when the light emitted by those observers as they crossed the horizon finally reaches the outside observer, it doesn't imply that the black hole actually was point-sized for the infalling observers when they crossed the horizon.

You're still sure about that then. I think it does.

Your language is overly vague, I don't know what it means for it to "apply" or "not apply". To be precise about what I mean, suppose you have two observers who compare clock readings at a single point, fly apart, and then reunite and compare clock readings again. Pick any point A on the worldline of one observer, and any point B on the worldline of the second observer. Then it can be true that different coordinate systems disagree on whether the clock at point A was ticking slower than the clock at point B or vice versa, and likewise it can be true that other observers with different paths through space time disagree about the exact ration between the visual rate of ticking of the clock at point A vs. the clock at point B. Nevertheless, all frames/observers agree about the total elapsed time on each clock when they finally reunite at a single point in spacetime. Would you say this means time dilation "applies" or "doesn't apply" at points A and B on each clock's worldline?

It's just easier to compare them when there's no distance between them because "all frames/observers agree about the total elapsed time on each clock when they finally reunite at a single point in spacetime" like you said. Use common sense instead of thinking in graphs. They can never ocupy the same exact space, but that doesn't mean they can't compare watches. If they can do that then how far away from each other do they have to be before they can't?

No, but my way of talking may be confusing, when I said "really" I wasn't talking about frame-invariant physical facts, but rather the "real" coordinates of events in a given frame, as distinguished from visual appearances. As I mentioned above, even in SR things like visual clock rates and visual lengths can be different from the "real" amount of time dilation and length contraction in the observer's inertial rest frame.

Yes, okay Doppler shift is a good example of visual appearance being different to what's actually happening, but we're ignoring Doppler shift.

So the two horizons appear closer the further you are away?

No, why do you say that?

Just because when would the horizon split? I just assumed it was always split and it's only noticable when you get close.

Well, normally the horizon isn't something you can "see" directly, but if we imagine that somehow every event on the horizon generated photons which traveled in all directions of the future light cone of that event, so that the horizon was visible in its own right (as opposed to our just being able to see a black area where light from distant stars is blocked), then you would only begin to "see" the horizon at the moment you crossed it, since none of the photons generated at the horizon would make it outside the horizon, but they could reach observers falling into the horizon. Meanwhile for a Schwarzschild black hole, there is also an "antihorizon" which consists of both a white hole event horizon in our universe and a black hole event horizon leading to a different universe (see the maximally extended Kruskal-Szekeres diagram for a discussion). So if we imagine all events on the antihorizon were also generating photons of a different color (say, red) which traveled in all directions in the future light cone of the event that generated them, then as you traveled towards the black hole you would see a red sphere representing the white hole's event horizon, and after you crossed the horizon you'd continue to see the red sphere, but now you'd be seeing light from the black hole horizon in the other universe. But after you crossed the horizon you'd also suddenly start seeing light from black hole horizon in your own universe--if you imagine that this light is white instead of red, then you should be able to follow what's depicted in the third movie from the top on this page, where we approach a black hole covered with red grid lines and then suddenly a new set of white grid lines appear (forming a sort of dome shape above us) at the moment we cross the horizon. If you read about the maximally extended Kruskal-Szekeres diagram, you should be able to follow this similar Penrose diagram that I linked to earlier, which shows which events on the horizon and antihorizon emitted the red and white light reaching you at each point in your fall. And I also linked to http://casa.colorado.edu/~ajsh/singularity.html which shows somewhat cruder animations of what a falling observer would see (again with the antihorizon in red and the white 'Schwarzschild bubble' that you see after crossing the horizon in white), with somewhat more detailed accompanying explanation (for example, he mentions that at the moment you cross the horizon the white Schwarzschild bubble would first appear as a straight line reaching from you to the red horizon, only later expanding into a bubble, and that at that moment other objects which fell into the black hole along the same axis would be arrayed along this line, all appearing as they did the moment they crossed the horizon).

And are you using "size" to refer only to apparent visual size, i.e. how much of the visual field is taken up by the black region devoid of stars (or the red antihorizon which apparently would coincide with it)? If so then I think as long as the black hole wasn't significantly changing its coordinate size (in some appropriate coordinate system like Eddington-Finkelstein coordinates) due to evaporation or significant influxes of matter, than observers who fell in at different times also shouldn't see a difference in apparent visual size at the moment they cross the horizon. If you want "size" to refer to something more than apparent visual size, I don't see how you can define any other notion of size without reference to a coordinate system.

What do you mean "beyond your horizon"? If the object escapes, then you will always see it as outside the red antihorizon, and if you ended up crossing the horizon yourself you would presumably see it remain outside the white "Schwarzschild bubble" (which consists of light from events on the black hole event horizon) which you are underneath.

And what are "you" doing? Remaining outside the horizon too, or crossing through the black hole event horizon? If the latter, then you will see it pass by you and move up to a greater radius before you pass the event horizon and begin to see the "Schwarzschild bubble".

If other objects fell into the black hole horizon from your universe, you will always see them above the red antihorizon, not "beyond" it (though you could potentially see objects beyond it that were in regions IV or III of the Kruskal diagram, in either the white hole interior region or the "other universe" respectively. Once you cross the horizon yourself and see yourself under the white Schwarzschild bubble, every other object that fell in before you will be under the bubble too, and you could look above you and see other objects entering the bubble from above.

The earlier object would also see a white Schwarzschild bubble that appeared at the moment it crossed the horizon (again assuming for the sake of visualization that each event on the event horizon emits white light in all possible directions, in reality of course the bubble wouldn't actually be a visible shape), and assuming light from your crossing the horizon had time to reach the earlier object before it hit the singularity, it would at some point see you cross through the top of this expanding bubble.

Wft are you on about. Keep it simple.

Let's go back to the two horizons. One just in front of you and one some distance away just in front of another object that very proberbly crossed the horizon earlier. How could it have if it's always possible for it to escape. This isn't just visual, it's real. If it can escape then in what sense is the horizon in front of you (which the other object is inside) real? What happens if you mantain your distance and the other object moves back out to your possition? It would mean it's moved but the event horizon has followed it. Just ignore the absolute horizon. Why would there even be one? It would be subject to gravitational length contraction which would mean its size would decrease the closer you got, going all the way up to infinity at the horizon. If it's infinite in one "frame" then it's infinite in all of them. You can never reach it in exactly the same way you can never reach c. The black hole becomes the singularity at 0 range, like the whole universe becomes a singularity at c.

 

[JesseM]

That's not what I meant. If something has a relative velocity of .5c for example then that velocity doesn't disappear in curved space-time.

But in fact it does "disappear" in the sense of there no longer being a unique objective truth about "relative velocity". You can define a notion of relative velocity that's coordinate-dependent, or a notion of relative velocity based on parallel transport over a particular path through spacetime, but there are always multiple possible coordinate systems and multiple possible paths, so there's no "objective" physical notion of relative velocity for objects at different points in spacetime in GR. For example, take a look at p. 167 of this textbook which discusses the notion of the relative velocity 4-vector in SR, and then in a footnote says "The concept of relative four-vector cannot be extended to theories of physics formulated over a non-linear space, i.e., curved spaces (e.g., General Relativity)" Likewise see this page from the site of physicist John Baez, who says:

In special relativity, we cannot talk about absolute velocities, but only relative velocities. For example, we cannot sensibly ask if a particle is at rest, only whether it is at rest relative to another. The reason is that in this theory, velocities are described as vectors in 4-dimensional spacetime. Switching to a different inertial coordinate system can change which way these vectors point relative to our coordinate axes, but not whether two of them point the same way.

In general relativity, we cannot even talk about relative velocities, except for two particles at the same point of spacetime -- that is, at the same place at the same instant. The reason is that in general relativity, we take very seriously the notion that a vector is a little arrow sitting at a particular point in spacetime. To compare vectors at different points of spacetime, we must carry one over to the other. The process of carrying a vector along a path without turning or stretching it is called `parallel transport'. When spacetime is curved, the result of parallel transport from one point to another depends on the path taken! In fact, this is the very definition of what it means for spacetime to be curved. Thus it is ambiguous to ask whether two particles have the same velocity vector unless they are at the same point of spacetime.

If you still don't believe that there is no coordinate- and path-independent notion of "relative velocity" for objects at different points in spacetime in GR, feel free to start a new thread asking the resident GR experts in the forum about this.

All space-time is at least a little bit curved so there'd be no such thing as relative verlocity at all.

No such thing in GR, at least if you want a single uniquely correct objective "relative velocity" as opposed to a coordinate-dependent or path-dependent one.

I stand by my claim that the reason an external observer can never witness an object crossing the horizon is length contraction/time dilation. Gravitational time dilation/length contraction doesn't make it any less real.

OK, but you completely failed to address my question about whether you were talking about visuals or something else. I can think of only 3 senses in which we can talk about clocks ticking at different rates in GR:

1. Visual appearances--how fast the an observer sees the image of another clock ticking relative to his own clock
2. Local comparisons of elapsed times, as in the twin paradox where each twin looks at how much time each has aged between two meetings
3. Coordinate-dependent notions of how fast each clock is ticking relative to coordinate time at a particular moment (which depends on the definition of simultaneity in your chosen coordinate system)

If you are confident there is some other sense in which we can compare the rates of different clocks, please spell it out with some reference to the technical definition you are thinking of in GR. If you have some kind of hunch or intuition that there should be some "real truth" about relative clock rates but can't back it up with any technical details, please consider that hunches and intuitions are often untrustworthy in modern physics, and thinking intuitions can take precedence over precise mathematical definitions is a common feature of physics crackpots (see items 12-14 on Are you a quack? from physicist Warren Siegel). Finally, if you agree that those three are the the only ways of comparing clock rates that make sense in GR, please tell me which you are referring to when you talk about "time dilation" near the event horizon being the explanation for why an external observer can never witness anything crossing it.

Either way, please give me a clear answer to this question, don't just keep talking about "time dilation" without explaining what you mean by that phrase.

But that's just a statement about when the light emitted by those observers as they crossed the horizon finally reaches the outside observer, it doesn't imply that the black hole actually was point-sized for the infalling observers when they crossed the horizon.

You're still sure about that then. I think it does.

Unless this is just a vague hunch you need to explain why you think it does with some sort of detailed non-handwavey argument.

It's just easier to compare them when there's no distance between them because "all frames/observers agree about the total elapsed time on each clock when they finally reunite at a single point in spacetime" like you said. Use common sense instead of thinking in graphs.

Again, trusting "common sense" over precise mathematical definitions is a sure path to becoming a crackpot. A little thought shows that believing there is a "real truth" about the relative rate of ticking of clocks in different regions of spacetime is equivalent to believing there must be some "real truth" about simultaneity--for example, if I say clock A is running half as fast as clock B, that's equivalent to saying that if clock A showing a time T is simultaneous with clock B showing a time T', then clock A showing a time T + delta-t must be simultaneous with clock B showing a time T' + 2*delta-t. But as long as there is no objective truth about simultaneity, what's to stop you from picking a different definition of simultaneity where clock A showing T is still simultaneous with clock B showing T', but clock A showing T + delta-t is now simultaneous with B showing T + 0.5*delta-t or T + 3*delta-t?

They can never ocupy the same exact space, but that doesn't mean they can't compare watches. If they can do that then how far away from each other do they have to be before they can't?

Any distance greater than zero means there is no basis for comparing rates other than the 3 I mentioned earlier.

Yes, okay Doppler shift is a good example of visual appearance being different to what's actually happening, but we're ignoring Doppler shift.

If you're not talking about visual appearances I say your only remaining options are #2 and #3 in my list above. If you disagree, please explain exactly what your own fourth option would be, or if you just have a hunch there should be a fourth option but can't think of any technical way to define it in GR.

Just because when would the horizon split? I just assumed it was always split and it's only noticable when you get close.

No, the second (white) horizon doesn't appear until the moment you cross the horizon. Please read carefully the section "At the horizon, the Schwarzschild surface" of http://casa.colorado.edu/~ajsh/singularity.html which I directed you to earlier, particularly this bit:

The small white dot indicates our point of entry through the horizon. Remarkably, the Schwarzschild surface, the red grid, still appears to stand off at some distance ahead of us. The white dot is actually a line which extends from us to the Schwarzschild surface still ahead, though we only ever see it as a dot, not as a line. The dot-line marks the formation of the Schwarzschild bubble (see below), and our entry into that bubble. Persons who fell through the Schwarzschild surface at this precise point before us would lie arrayed along this dot-line. At this instant, as we pass through the horizon into the Schwarzschild bubble, we see all the other persons who passed through this location before us also pass through the horizon into the bubble.

Wft are you on about. Keep it simple.

If you're going to get hostile, and refuse to think about details in the name of some illusory "simplicity", I won't continue this conversation.

Let's go back to the two horizons.

"Get back to"? Defining the detailed meaning of the "two horizons" was exactly what I was doing in the sections you quoted and responded to with the dismissive "wtf" comment.

One just in front of you and one some distance away just in front of another object that very proberbly crossed the horizon earlier.

No, the reason I gave such detailed descriptions was in hopes that you'd follow along and not just jump back into relying on your own vague intuitions and hunches. Read what I said again (which is just a summary of the section of the page I quoted above, which I recommend reading in full):

he mentions that at the moment you cross the horizon the white Schwarzschild bubble would first appear as a straight line reaching from you to the red horizon, only later expanding into a bubble, and that at that moment other objects which fell into the black hole along the same axis would be arrayed along this line, all appearing as they did the moment they crossed the horizon

And I also said:

If the object escapes, then you will always see it as outside the red antihorizon, and if you ended up crossing the horizon yourself you would presumably see it remain outside the white "Schwarzschild bubble" (which consists of light from events on the black hole event horizon) which you are underneath.

So you would never see the second (white in the diagrams) visual horizon as "in front" of you as you suggested, instead it first appears as a straight line extending from you to the first (red in the diagrams) visual horizon at the moment you actually cross the horizon, and then it immediately expands into a bubble which you are underneath, as is any other object whose light you are seeing from a moment after that object crossed the horizon.

How could it have if it's always possible for it to escape. This isn't just visual, it's real. If it can escape then in what sense is the horizon in front of you (which the other object is inside) real?

For an eternal black hole, the red horizon is actually a physically separate horizon, the "antihorizon" one that borders the bottom of "our" exterior region I and the top of the alternate exterior region III in the maximally extended Kruskal-Szekeres diagram. The falling object genuinely never crosses this horizon, it's a white hole horizon in our universe and a black hole horizon in another exterior universe inaccessible from our own.

For a more realistic black hole that formed at some finite time from a collapsing star, you wouldn't actually be able to "see" any horizon from the outside, in the sense that light emitted from events on an event horizon would never reach anyone outside, at least not unless the black hole evaporated away. However, this section of the other site on falling into a black hole I linked to earlier also seems to say that if you could see the highly redshifted image of the collapsing star long after the black hole had formed, it would occupy almost exactly the same visual position as the red antihorizon of an eternal black hole:

The Penrose diagram shows that the horizon is really two distinct entities, the Horizon, and the Antihorizon. The Horizon is sometimes called the true horizon. It's the horizon you actually fall through if you fall into a black hole. The Antihorizon might reasonably called the illusory horizon. In a real black hole formed from the collapse of the core of a star, the illusory horizon is replaced by an exponentially redshifting image of the collapsing star. As the collapsing star settles towards its final no-hair state, its appearance tends to that of a no-hair black hole.

What happens if you mantain your distance and the other object moves back out to your possition? It would mean it's moved but the event horizon has followed it. Just ignore the absolute horizon.

I don't know what you mean by "absolute horizon"--are you talking about the white horizon in the diagram (the Schwarzschild bubble which can only be seen once you cross the horizon yourself), the red horizon in the diagram (the antihorizon), or something else? If either of those, your statement that "the event horizon has followed it" doesn't make sense, as long as you remain outside the horizon you'll never see the white horizon, and if you maintain a constant radius the red horizon should maintain a constant visual size.

Why would there even be one?

One what?

It would be subject to gravitational length contraction which would mean its size would decrease the closer you got, going all the way up to infinity at the horizon.

What would be subject to gravitational length contraction, the object or the horizon? And just as with "time dilation", please specify whether by "length contraction" you mean visual appearances, or frame-dependent length, or something else.

If it's infinite in one "frame" then it's infinite in all of them.

Nope, that's just flat-out wrong. I already told you many times that time dilation and length contraction don't go to infinity at the horizon in Kruskal-Szekeres coordinates, and also that in ordinary Minkowski spacetime you do have infinite time dilation and length contraction at the Rindler horizon if you use Rindler coordinates, but obviously this is a purely coordinate-based effect which disappears if you use ordinary inertial coordinates in the same spacetime.

You can never reach it in exactly the same way you can never reach c.

Obviously you have a powerful intuition that this is true, but you continue to fail to provide any detailed argument as to why anyone else should believe this, and you seem unwilling to consider that your own intuitions might be wrong (beware the Dunning-Kruger effect, another major cause of crackpotism IMO). Do you think you can "never reach" the Rindler horizon just because time dilation and length contraction go to infinity as you approach the horizon in Rindler coordinates, and a Rindler observer can never see anything reach the horizon? If not please explain what makes the black hole event horizon different.

The black hole becomes the singularity at 0 range, like the whole universe becomes a singularity at c.

Neither of these claims makes any sense in relativity, if you relied more on math and less on intuitions you might figure out why.

 

[A-wal]

But in fact it does "disappear" in the sense of there no longer being a unique objective truth about "relative velocity". You can define a notion of relative velocity that's coordinate-dependent, or a notion of relative velocity based on parallel transport over a particular path through spacetime, but there are always multiple possible coordinate systems and multiple possible paths, so there's no "objective" physical notion of relative velocity for objects at different points in spacetime in GR. For example, take a look at p. 167 of this textbook which discusses the notion of the relative velocity 4-vector in SR, and then in a footnote says "The concept of relative four-vector cannot be extended to theories of physics formulated over a non-linear space, i.e., curved spaces (e.g., General Relativity)" Likewise see this page from the site of physicist John Baez, who says:

If you still don't believe that there is no coordinate- and path-independent notion of "relative velocity" for objects at different points in spacetime in GR, feel free to start a new thread asking the resident GR experts in the forum about this.

That's still not what I meant. If something has a relative velocity of .5c for example then that velocity definitely doesn't disappear in curved space-time. Objects still move relative to each other in curved space-time is all I meant.

No such thing in GR, at least if you want a single uniquely correct objective "relative velocity" as opposed to a coordinate-dependent or path-dependent one.

I didn't say there was.

OK, but you completely failed to address my question about whether you were talking about visuals or something else. I can think of only 3 senses in which we can talk about clocks ticking at different rates in GR:

1. Visual appearances--how fast the an observer sees the image of another clock ticking relative to his own clock
2. Local comparisons of elapsed times, as in the twin paradox where each twin looks at how much time each has aged between two meetings
3. Coordinate-dependent notions of how fast each clock is ticking relative to coordinate time at a particular moment (which depends on the definition of simultaneity in your chosen coordinate system)

If you are confident there is some other sense in which we can compare the rates of different clocks, please spell it out with some reference to the technical definition you are thinking of in GR. If you have some kind of hunch or intuition that there should be some "real truth" about relative clock rates but can't back it up with any technical details, please consider that hunches and intuitions are often untrustworthy in modern physics, and thinking intuitions can take precedence over precise mathematical definitions is a common feature of physics crackpots

Funny, I thought letting precise mathematical definitions take precedence over intuitions is a common feature of physics crackpots. Maths only describes, it can't explain anything.

(see items 12-14 on Are you a quack? from physicist Warren Siegel).

I'm a quack?

Finally, if you agree that those three are the the only ways of comparing clock rates that make sense in GR, please tell me which you are referring to when you talk about "time dilation" near the event horizon being the explanation for why an external observer can never witness anything crossing it.

Either way, please give me a clear answer to this question, don't just keep talking about "time dilation" without explaining what you mean by that phrase.

The mass of the singularity determines the diameter of the event horizon. Say 100 s-units at my current range of say 100 d-units. I now halve the distance to 50 d-units. Make sense? It shouldn't. If I measure where half the distance is from my starting position (we'll say there's a marker keeping a constant distance from the event horizon from it's own perspective) and then travel to that position and measure my new current distance to the event horizon then it wouldn't be 50 d-units. It would be more because I'm now in space-time that was contracted from my previous perspective. I'd have to get closer to be 50 d-units away, but then I'd have the same problem. The closer you get the further you need to go, but this obviously can't go on for ever. It stops at the singularity. What happens to the s-units?

Unless this is just a vague hunch you need to explain why you think it does with some sort of detailed non-handwavey argument.

That's exactly what I've been doing! Non-technical does not = handwavey!

Again, trusting "common sense" over precise mathematical definitions is a sure path to becoming a crackpot. A little thought shows that believing there is a "real truth" about the relative rate of ticking of clocks in different regions of spacetime is equivalent to believing there must be some "real truth" about simultaneity--for example, if I say clock A is running half as fast as clock B, that's equivalent to saying that if clock A showing a time T is simultaneous with clock B showing a time T', then clock A showing a time T + delta-t must be simultaneous with clock B showing a time T' + 2*delta-t. But as long as there is no objective truth about simultaneity, what's to stop you from picking a different definition of simultaneity where clock A showing T is still simultaneous with clock B showing T', but clock A showing T + delta-t is now simultaneous with B showing T + 0.5*delta-t or T + 3*delta-t?

Nothing!

Any distance greater than zero means there is no basis for comparing rates other than the 3 I mentioned earlier.

I never said there was.

If you're not talking about visual appearances I say your only remaining options are #2 and #3 in my list above. If you disagree, please explain exactly what your own fourth option would be, or if you just have a hunch there should be a fourth option but can't think of any technical way to define it in GR.

#2.

No, the second (white) horizon doesn't appear until the moment you cross the horizon. Please read carefully the section "At the horizon, the Schwarzschild surface" of http://casa.colorado.edu/~ajsh/singularity.html which I directed you to earlier, particularly this bit:

So there is a jump? This is getting silly again.

If you're going to get hostile, and refuse to think about details in the name of some illusory "simplicity", I won't continue this conversation.

Hostile? You've lead a very sheltered life my friend. Please don't end the conversation. If I'm wrong then I'd like to understand why.

"Get back to"? Defining the detailed meaning of the "two horizons" was exactly what I was doing in the sections you quoted and responded to with the dismissive "wtf" comment.

You were starting to sound like a crackpot.

No, the reason I gave such detailed descriptions was in hopes that you'd follow along and not just jump back into relying on your own vague intuitions and hunches. Read what I said again (which is just a summary of the section of the page I quoted above, which I recommend reading in full):

I'm confused. I'm approaching the horizon and I can't see any object cross the horizon from the outside. I now cross the horizon and I see those same objects suddenly jump to some point along a line and the time since they crossed determines how far along the line they jump to?

And I also said:

So you would never see the second (white in the diagrams) visual horizon as "in front" of you as you suggested, instead it first appears as a straight line extending from you to the first (red in the diagrams) visual horizon at the moment you actually cross the horizon, and then it immediately expands into a bubble which you are underneath, as is any other object whose light you are seeing from a moment after that object crossed the horizon.

Hmm, I'm still not buying it.

For an eternal black hole, the red horizon is actually a physically separate horizon, the "antihorizon" one that borders the bottom of "our" exterior region I and the top of the alternate exterior region III in the maximally extended Kruskal-Szekeres diagram. The falling object genuinely never crosses this horizon, it's a white hole horizon in our universe and a black hole horizon in another exterior universe inaccessible from our own.

For a more realistic black hole that formed at some finite time from a collapsing star, you wouldn't actually be able to "see" any horizon from the outside, in the sense that light emitted from events on an event horizon would never reach anyone outside, at least not unless the black hole evaporated away. However, this section of the other site on falling into a black hole I linked to earlier also seems to say that if you could see the highly redshifted image of the collapsing star long after the black hole had formed, it would occupy almost exactly the same visual position as the red antihorizon of an eternal black hole:

I can honestly see no need for the "true" horizon.

I don't know what you mean by "absolute horizon"--are you talking about the white horizon in the diagram (the Schwarzschild bubble which can only be seen once you cross the horizon yourself), the red horizon in the diagram (the antihorizon), or something else? If either of those, your statement that "the event horizon has followed it" doesn't make sense, as long as you remain outside the horizon you'll never see the white horizon, and if you maintain a constant radius the red horizon should maintain a constant visual size.

I'm just having trouble with the transition from what's observed before reaching the horizon and how it maintains continuity if it can be crossed.

One what?

Absolute horizon. As in the one that doesn't recede but maintains a constant distance from the singularity despite the fact that distance is meant to be relative.

What would be subject to gravitational length contraction, the object or the horizon? And just as with "time dilation", please specify whether by "length contraction" you mean visual appearances, or frame-dependent length, or something else.

I meant the distance between the event horizon and singularity depends on distance it's viewed from.

Nope, that's just flat-out wrong. I already told you many times that time dilation and length contraction don't go to infinity at the horizon in Kruskal-Szekeres coordinates, and also that in ordinary Minkowski spacetime you do have infinite time dilation and length contraction at the Rindler horizon if you use Rindler coordinates, but obviously this is a purely coordinate-based effect which disappears if you use ordinary inertial coordinates in the same spacetime.

In the same space-time? When comparing objects at different distance from an event horizon they can't possibly be in the same space-time.

Obviously you have a powerful intuition that this is true, but you continue to fail to provide any detailed argument as to why anyone else should believe this, and you seem unwilling to consider that your own intuitions might be wrong (beware the Dunning-Kruger effect, another major cause of crackpotism IMO). Do you think you can "never reach" the Rindler horizon just because time dilation and length contraction go to infinity as you approach the horizon in Rindler coordinates, and a Rindler observer can never see anything reach the horizon? If not please explain what makes the black hole event horizon different.

Stop calling me names! It's mean.

Neither of these claims makes any sense in relativity, if you relied more on math and less on intuitions you might figure out why.

And if you relied more on intuitions and less on maths then you might question the things you're told a bit more. I thought the universe did become a singularity at c (as far as light, all energy for that matter is concerned).


Maybe I am being a cock, because I'm getting frustrated. First the name calling isn't nice. Secondly I told you that I don't like all the coordinate stuff but you continue to ask question like: "Do you think you can "never reach" the Rindler horizon just because time dilation and length contraction go to infinity as you approach the horizon in Rindler coordinates, and a Rindler observer can never see anything reach the horizon? If not please explain what makes the black hole event horizon different." I have no idea and honestly couldn't care less. I'd never even heard of Rindler coordinates or Schwarzschild coordinates or half the stuff that's been mentioned here until this conversation so it's not a reason for me thinking anything. All I care about is understanding what happens and why. And finally, I don't think all of your recent explanations are an understanding in any meaningful sense of the word, though I do appreciate the time and effort you've put in. I think they're what happens when you let maths lead the "understanding" instead of doing it the other way round. There, I said it.


P.S. Sorry about the delays between posts. I've written all these posts while at work and I've been off for three nights.

 

[YLW]

Hi, I seldom post so I'm really, really so, so sorry if I've posted in the wrong thread. But I've just been reading through a Discovery magazine article Back From The Future and here are excerpts. (http://discovermagazine.com/2010/apr/01-back-from-the-future/article_view?b_start:int=1&-C=)
________
"A series of quantum experiments shows that measurements performed in the future can influence the present. Does that mean the universe has a destiny—and the laws of physics pull us inexorably toward our prewritten fate?..."

"The boat trip has been organized as part of a conference sponsored by the Foundational Questions Institute to highlight some of the most controversial areas in physics. Tollaksen’s idea certainly meets that criterion. And yet, as crazy as it sounds, this notion of reverse causality is gaining ground. A succession of quantum experiments confirm its predictions—showing, bafflingly, that measurements performed in the future can influence results that happened before those measurements were ever made.

As the waves pound, it’s tough to decide what is more unsettling: the boat’s incessant rocking or the mounting evidence that the arrow of time—the flow that defines the essential narrative of our lives—may be not just an illusion but a lie."

"Just last year, physicist John Howell and his team from the University of Rochester reported success. In the Rochester setup, laser light was measured and then shunted through a beam splitter. Part of the beam passed right through the mechanism, and part bounced off a mirror that moved ever so slightly, due to a motor to which it was attached. The team used weak measurements to detect the deflection of the reflected laser light and thus to determine how much the motorized mirror had moved.

That is the straightforward part. Searching for backward causality required looking at the impact of the final measurement and adding the time twist. In the Rochester experiment, after the laser beams left the mirrors, they passed through one of two gates, where they could be measured again—or not. If the experimenters chose not to carry out that final measurement, then the deflected angles measured in the intermediate phase were boringly tiny. But if they performed the final, postselection step, the results were dramatically different. When the physicists chose to record the laser light emerging from one of the gates, then the light traversing that route, alone, ended up with deflection angles amplified by a factor of more than 100 in the intermediate measurement step. Somehow the later decision appeared to affect the outcome of the weak, intermediate measurements, even though they were made at an earlier time.

This amazing result confirmed a similar finding reported a year earlier by physicists Onur Hosten and Paul Kwiat at the University of Illinois at Urbana-Champaign. They had achieved an even larger laser amplification, by a factor of 10,000, when using weak measurements to detect a shift in a beam of polarized light moving between air and glass."
____________
So, asking as a layman, do the experiments confirm backward or retrocausality and that the arrow of time can move backwards as well?

Once again, so so sorry if I've posted wrongly.

 

[Austin0]

Well, note that the Rindler horizon has an upper and lower component (or maybe it would be better to say there are two horizons?), as seen in this diagram from the wikipedia article which plots both the horizons and lines of constant Rindler position and time in a Minkowski diagram. The upper component behaves like a black hole horizon in that the Rindler observers can't see anything that crosses it (though an observer on the "Rindler wedge" can cross it herself if she doesn't remain at rest in Rindler coordinates), but the lower component behaves like a white hole horizon in that Rindler observers can see light from events "below" it (the light rays would be traveling parallel to the upper horizon in the diagram), and it's impossible for any observer on the Rindler wedge to cross it (in Minkowski coordinates it's moving away from anyone on the Rindler wedge with a speed of c). So, in terms of Minkowski coordinates this means an observer on the Rindler wedge can see light from events that occurred arbitrarily far in either direction along the x-axis, including events much further than the current distance of the Rindler horizon. So certainly an inertial observer on the Rindler wedge would see nothing unusual, and at any given moment an accelerating Rindler observer should see the same thing as an inertial observer at the same point in spacetime who is instantaneously at rest relative to themselves.

Hi When you refer to the upper and lower components are you talking about the lower deceleration part of the diagram and the upper or acceleration half of the diagram??

When you say light from events arbitrarily far , I assume this also means distant in time also ,,is this correct??

That the events from much further than the horizon would then have to have occurred further in the past, yes??
If so this is more or less what I thought.

So if I am understanding you correctly, both Rindler and CMIRF observers would visually see both Doppler shift and aberration but nothing identifiable beyond that. yes?

Thanks for your explication

 

[JesseM]

(response to post #167, part 1)

That's still not what I meant. If something has a relative velocity of .5c for example then that velocity definitely doesn't disappear in curved space-time. Objects still move relative to each other in curved space-time is all I meant.

No such thing in GR, at least if you want a single uniquely correct objective "relative velocity" as opposed to a coordinate-dependent or path-dependent one.

I didn't say there was.

If you're talking about a purely coordinate-dependent or path-dependent notion of "relative velocity", then you must agree that just because two objects at distant locations have a relative velocity of 0.5c in some coordinate system or with velocities parallel-transported along some path, there may be some other coordinate system or path where the relative velocity is 0.9c or 0.1c or even 0. So it seems to me the relative velocity of 0.5c does disappear in curved spacetime, in the sense that it there is no objective sense in which two objects can be said to have a relative velocity of 0.5c as opposed to 0.9c or 0.

Funny, I thought letting precise mathematical definitions take precedence over intuitions is a common feature of physics crackpots. Maths only describes, it can't explain anything.

No physics theory has ever "explained" why matter/energy/particles behave in the way they do, it just gives equations describing their behavior. And anyone with experience dealing with physics crackpots will recognize that quite a lot of them are motivated by being unsatisfied with such abstract mathematical models, and mistakenly think that the equations should be understood in terms of some intuitive concrete model. Consider for example a few items from physicist John Baez's famous crackpot index:

15. 10 points for each statement along the lines of "I'm not good at math, but my theory is conceptually right, so all I need is for someone to express it in terms of equations".

...17. 10 points for arguing that while a current well-established theory predicts phenomena correctly, it doesn't explain "why" they occur, or fails to provide a "mechanism".

Or consider some of the characteristics of crackpot arguments in A brief field guide to scientific crackpots:

6. Criticisms of existing theories which rely on “common sense”. This particular branch of’ ‘crackpottery’ reminds me of a personal anecdote. Years ago I was wandering through the lamp department of a Service Merchandise store, when I noticed bright red signs prominently displayed: “CAUTION! Light bulbs are hot! Do not touch!”

Common sense is evidently a terribly inaccurate method of understanding the world! I’ve mentioned in previous posts numerous modern “common sense” ideas which were anything but when first proposed, among them: Newton’s laws, heliocentrism, the germ theory of disease, the brain as the center of intelligence, the theory of atoms.

For relativity denialists, the idea that space and time are intermingled violates common sense. They have no strong quantitative criticism of the theory (which they probably don’t understand anyway), only a mushy notion that it “feels funny”. Relativity felt funny to a lot of people, who raised objections to the theory in the form of apparent paradoxes — all of which were resolved successfully. The lesson physicists learned from this is that “common sense” is an artifact of our limited perceptions and place in the universe. In fact, modern science was really born when people realized that the universe might work differently in circumstances different from those experienced in daily life.

7. A complete absence of quantitative analysis. In a 165-page “primer” on geocentricity (discussed in another post), the long-disavowed notion that the Earth is the center of the universe, a crank author uses no calculations to back up his extravagant alternative theory of celestial motion. He claims the theory works just fine, but does it? Without calculations which can be checked for errors and compared to experiment, there’s no immediate way to tell. The relativity denialists mentioned earlier also present no quantitative results to back up their ideas, only appeals to “common sense”. The Templeton Foundation, a group dedicated to promoting the links of science and religion, once asked for proponents of intelligent design (dressed-up creationism) to submit research proposals. “They never came in,” said the Foundation’s senior vice president.
These are some of the characteristics and arguments of crackpot scientists, and are things which should raise a red flag if you come across them. Anybody have any other characteristics to add to the list?

(see items 12-14 on Are you a quack? from physicist Warren Siegel).

I'm a quack?

To me words like "crackpot" and "quack" don't really refer to characteristics of a person but rather common styles of argument used by people who think they have found conceptual flaws in mainstream scientific theories that they have never studied in any great detail--and having had experience arguing with denialists of a bunch of different mainstream theories on the internet, it does ring true that there are quite lot of commonalities to their style of argument regardless of what theory it is they're attacking, with an over-emphasis on intuitions and "common sense" being one prominent one (think of global warming denialists who have arguments like 'if the Earth is getting warmer, why did we have record snowstorms last year' or evolution deniers who say 'if man evolved from monkeys, how can monkeys still be around today'?) Warren Siegel attempts to compile a bunch of these commonalities, and I directed you to 12-14 above because they also dealt with the favoring of common sense over math:

12. "My theory doesn't need any complicated math."

Then how do you calculate anything? Science is not just knowing "what goes up must come down", but when and where it comes down.

Note: Quacks come in slightly different levels of sophistication in math. Some use only words, and no numbers whatsoever, but lots of pictures. The worst one I ever corresponded with claimed that dimensions did not physically exist, but were just abstract mathematical concepts, and you could never prove the existence of anything unless you could do it without equations. After giving him the examples of directions, he claimed that "up" and "down" did not physically exist.

Better ones actually know arithmetic, but no algebra, so even E=mc2 is usually beyond them. They will quote lots of numbers, which they "predicted" by some numerology, but never functions (like cross sections). They don't understand units, or conventions, and will not appreciate that some constants of nature may be more natural with extra factors of 2π or so, or that some are actually not constants (like running couplings).

13. "Numbers aren't important in science."
I guess you can throw out your clock.

14. "How you explain something is more important than the numbers."
Try that the next time you pay a bill.

For a more detailed discussion of the necessity to abandon common-sense intuitions about what a good model of physics should look like, and accept that physics is ultimately just about coming up with elegant mathematical models which accurately predict observation, consider this section from Richard Feynman's book The Character of Physical Law:

On the other hand, take Newton's law for gravitation, which has the aspects I discussed last time. I gave you the equation:

F=Gmm'/r^2

just to impress you with the speed with which mathematical symbols can convey information. I said that the force was proportional to the product of the masses of two objects, and inversely as the square of the distance between them, and also that bodies react to forces by changing their speeds, or changing their motions, in the direction of the force by amounts proportional to the force and inversely proportional to their masses. Those are words all right, and I did not necessarily have to write the equation. Nevertheless it is kind of mathematical, and we wonder how this can be a fundamental law. What does the planet do? Does it look at the sun, see how far away it is, and decide to calculate on its internal adding machine the inverse of the square of the distance, which tells it how much to move? This is certainly no explanation of the machinery of gravitation! You might want to look further, and various people have tried to look further. Newton was originally asked about his theory--'But it doesn't mean anything--it doesn't tell us anything'. He said, 'It tells you how it moves. That should be enough. I have told you how it moves, not why.' But people are often unsatisfied without a mechanism, and I would like to describe one theory which has been invented, among others, of the type you migh want. This theory suggests that this effect is the result of large numbers of actions, which would explain why it is mathematical.

Suppose that in the world everywhere there are a lot of particles, flying through us at very high speed. They come equally in all directions--just shooting by--and once in a while they hit us in a bombardment. We, and the sun, are practically transparent for them, practically but not completely, and some of them hit. ... If the sun were not there, particles would be bombarding the Earth from all sides, giving little impuleses by the rattle, bang, bang of the few that hit. This will not shake the Earth in any particular direction, because there are as many coming from one side as from the other, from top as from bottom. However, when the sun is there the particles which are coming from that direction are partially absorbed by the sun, because some of them hit the sun and do not go through. Therefore the number coming from the sun's direction towards the Earth is less than the number coming from the other sides, because they meet an obstacle, the sun. It is easy to see that the farther the sun is away, of all the possible directions in which particles can come, a smaller proportion of the particles are being taken out. The sun will appear smaller--in fact inversely as the square of the distance. Therefore there will be an impulse on the Earth towards the sun that varies inversely as the square of the distance. And this will be the result of a large number of very simple operations, just hits, one after the other, from all directions. Therefore the strangeness of the mathematical relation will be very much reduced, because the fundamental operation is much simpler than calculating the inverse of the square of the distance. This design, with the particles bouncing, does the calculation.

The only trouble with this scheme is that it does not work, for other reasons. Every theory that you make up has to be analysed against all possible consequences, to see if it predicts anything else. And this does predict something else. If the Earth is moving, more particles will hit it from in front than from behind. (If you are running in the rain, more rain hits you in the front of the face than in the back of the head, because you are running into the rain.) So, if the Earth is moving it is running into the particles coming towards it and away from the ones that are chasing it from behind. So more particles will hit it from the front than from the back, and there will be a force opposing any motion. This force would slow the Earth up in its orbit, and it certainly would not have lasted the three of four billion years (at least) that it has been going around the sun. So that is the end of that theory. 'Well,' you say, 'it was a good one, and I got rid of the mathematics for a while. Maybe I could invent a better one.' Maybe you can, because nobody knows the ultimate. But up to today, from the time of Newton, no one has invented another theoretical description of the mathematical machinery behind this law which does not either say the same thing over again, or make the mathematics harder, or predict some wrong phenomena. So there is no model of the theory of gravity today, other than the mathematical form.

If this were the only law of this character it would be interesting and rather annoying. But what turns out to be true is that the more we investigate, the more laws we find, and the deeper we penetrate nature, the more this disease persists. Every one of our laws is a purely mathematical statement in rather complex and abstruse mathematics.

...[A] question is whether, when trying to guess new laws, we should use seat-of-the-pants feelings and philosophical principles--'I don't like the minimum principle', or 'I do like the minimum principle', 'I don't like action at a distance', or 'I do like action at a distance'. To what extent do models help? It is interesting that very often models do help, and most physics teachers try to teach how to use models and to get a good physical feel for how things are going to work out. But it always turns out that the greatest discoveries abstract away from the model and the model never does any good. Maxwell's discovery of electrodynamics was made with a lot of imaginary wheels and idlers in space. But when you get rid of all the idlers and things in space the thing is O.K. Dirac discovered the correct laws for relativity quantum mechanics simply by guessing the equation. The method of guessing the equation seems to be a pretty effective way of guessing new laws. This shows again that mathematics is a deep way of expressing nature, and any attempt to express nature in philosophical principles, or in seat-of-the-pants mechanical feelings, is not an efficient way.

It always bothers me that, according to the laws as we understand them today, it takes a computing machine an infinite number of logical operations to figure out what goes on in no matter how tiny a region of space, and no matter how tiny a region of time. How can all that be going on in that tiny space? Why should it take an infinite amount of logic to figure out what one tiny piece of space/time is going to do? So I have often made the hypothesis that ultimately physics will not require a mathematical statement, that in the end the machinery will be revealed, and the laws will turn out to be simple, like the chequer board with all its apparent complexities. But this speculation is of the same nature as those other people make--'I like it', 'I don't like it',--and it is not good to be too prejudiced about these things.

The mass of the singularity determines the diameter of the event horizon.

Only if you are talking about some coordinate-dependent notion of "diameter", like the diameter in Schwarzschild coordianates or Eddington-Finkelstein coordinates. Please, if you are going to talk about "distance" then please tell me if you are talking about distances in some coordinate system, or apparent visual distances and sizes, or if you think there is some third option.

Say 100 s-units at my current range of say 100 d-units. I now halve the distance to 50 d-units. Make sense? It shouldn't. If I measure where half the distance is from my starting position (we'll say there's a marker keeping a constant distance from the event horizon from it's own perspective) and then travel to that position and measure my new current distance to the event horizon then it wouldn't be 50 d-units. It would be more because I'm now in space-time that was contracted from my previous perspective.

I don't know what you mean by "perspective". Are you talking about visual appearances at 100 d-units vs. 50 d-units, or are you imagining that the observer will use different coordinate systems depending on his distance, or what?

Unless this is just a vague hunch you need to explain why you think it does with some sort of detailed non-handwavey argument.

That's exactly what I've been doing! Non-technical does not = handwavey!

If it wouldn't be obvious to a physicist well-versed in the mathematics how to translate your verbal argument into a detailed mathematical one, then hell yes it's "handwavey". In physics verbal arguments are only meaningful insofar as they can be understood as shorthand for a technical argument, where the meaning of the shorthand is clear enough that you don't have to bother laboriously spelling everything out in technical terms. This is not to say that physicists don't appeal to physical intuitions in situations where they're groping for the correct answer to some problem they don't know how to solve in a rigorous mathematical way, but I think they always do so with an eye towards developing a technical mathematical argument, I think you'd be hard-pressed to find a physicist who believes that some verbal argument is sufficient in itself to demonstrate some claim even if they don't know how to translate it into technical terms (and don't think other physicists would be able to either).

 

[JesseM]

(response to post #167, part 2)

A little thought shows that believing there is a "real truth" about the relative rate of ticking of clocks in different regions of spacetime is equivalent to believing there must be some "real truth" about simultaneity--for example, if I say clock A is running half as fast as clock B, that's equivalent to saying that if clock A showing a time T is simultaneous with clock B showing a time T', then clock A showing a time T + delta-t must be simultaneous with clock B showing a time T' + 2*delta-t. But as long as there is no objective truth about simultaneity, what's to stop you from picking a different definition of simultaneity where clock A showing T is still simultaneous with clock B showing T', but clock A showing T + delta-t is now simultaneous with B showing T + 0.5*delta-t or T + 3*delta-t?

Nothing!

So does that mean you are retracting all your earlier statements suggesting there is some objective sense in which we can talk about which clock experiences more time dilation at different distances from the horizon, apart from the question of which clock experiences more total elapsed time between two meetings which each occur at a single point in space time? For example, this comment from post #161:

(and we can only talk about 'time dilation' in a coordinate-independent way if we're talking about a twin-paradox-like situation where two observers meet twice and compare the elapsed time on their clocks between meetings, any notion of 'time dilation' other than this is also inherently coordinate-dependent).

If gravitational time dilation applies when two objects with different world lines meet up then it always applies. If not, then at what distance would it suddenly not apply?

Or this one from post #165:

To be precise about what I mean, suppose you have two observers who compare clock readings at a single point, fly apart, and then reunite and compare clock readings again. Pick any point A on the worldline of one observer, and any point B on the worldline of the second observer. Then it can be true that different coordinate systems disagree on whether the clock at point A was ticking slower than the clock at point B or vice versa, and likewise it can be true that other observers with different paths through space time disagree about the exact ration between the visual rate of ticking of the clock at point A vs. the clock at point B. Nevertheless, all frames/observers agree about the total elapsed time on each clock when they finally reunite at a single point in spacetime. Would you say this means time dilation "applies" or "doesn't apply" at points A and B on each clock's worldline?

It's just easier to compare them when there's no distance between them because "all frames/observers agree about the total elapsed time on each clock when they finally reunite at a single point in spacetime" like you said. Use common sense instead of thinking in graphs. They can never ocupy the same exact space, but that doesn't mean they can't compare watches. If they can do that then how far away from each other do they have to be before they can't?

If you don't retract these comments, then how does this square with your agreement that depending on our choice of simultaneity convention we can reach different conclusions about whether clock B was ticking faster or slower than clock A between the events of clock A reading a time T and clock A reading a time T + delta-t?

If you're not talking about visual appearances I say your only remaining options are #2 and #3 in my list above. If you disagree, please explain exactly what your own fourth option would be, or if you just have a hunch there should be a fourth option but can't think of any technical way to define it in GR.

#2.

OK, #2 was "Local comparisons of elapsed times, as in the twin paradox where each twin looks at how much time each has aged between two meetings". But in this case you can't say anything about time dilation except when talking about total elapsed time between two local comparisons, in particular you can't say one clock was ticking slower when it was closer to the horizon. If you agree with that, it seems to me you are changing your tune from your comments in post #161 and #165 when the statements of mine you were disagreeing with were just statements that we can only talk about differences in total elapsed time and nothing else.

No, the second (white) horizon doesn't appear until the moment you cross the horizon. Please read carefully the section "At the horizon, the Schwarzschild surface" of the page which I directed you to earlier, particularly this bit:

So there is a jump? This is getting silly again.

In visual appearances there is a "jump" in the sense that the visual horizon which is shown in white in the diagrams is completely absent for any observer who hasn't yet crossed the event horizon, then at the moment the observer crosses it they will suddenly see it appear as a straight line reaching from their position down to the antihorizon depicted in red, after which it expands into a bubble enclosing them. I already explained this in previous posts, like the long explanation in post #162 which you gave the dismissive "wtf" response to:

But after you crossed the horizon you'd also suddenly start seeing light from black hole horizon in your own universe--if you imagine that this light is white instead of red, then you should be able to follow what's depicted in the third movie from the top on this page, where we approach a black hole covered with red grid lines and then suddenly a new set of white grid lines appear (forming a sort of dome shape above us) at the moment we cross the horizon.

...

And I also linked to http://casa.colorado.edu/~ajsh/singularity.html which shows somewhat cruder animations of what a falling observer would see (again with the antihorizon in red and the white 'Schwarzschild bubble' that you see after crossing the horizon in white), with somewhat more detailed accompanying explanation (for example, he mentions that at the moment you cross the horizon the white Schwarzschild bubble would first appear as a straight line reaching from you to the red horizon, only later expanding into a bubble, and that at that moment other objects which fell into the black hole along the same axis would be arrayed along this line, all appearing as they did the moment they crossed the horizon).

Hostile? You've lead a very sheltered life my friend. Please don't end the conversation. If I'm wrong then I'd like to understand why.

Of course I am not particularly offended by the phrase "wtf", it's the dismissive attitude it suggests that bothers me. The rules of this forum don't allow people to use it as a platform to try to make arguments against mainstream ideas in relativity, though they do allow for people who are confused by some aspect of relativity to ask questions about how some apparent problem or contradiction can be avoided. The crucial difference here is one of attitude--in the second case the person is willing to take as a default hypothesis that there almost certainly is some way around the problem they see and they genuinely want to learn what it is, in the first case they are confident that they really have discovered a genuine problem which shows that the mainstream view is wrong. Your dismissive comment about my detailed explanation, along with other comments suggesting you probably hadn't read it carefully and tried to visualize what I was talking about, makes it seem like your attitude is closer to that of an anti-relativity type who is trying to prove there is a flaw, as opposed to someone who trusts that there is most likely an error in their thinking and is making an honest effort to listen to others who are knowledgeable about the subject and trying to explain where the error might lie. If you do want to learn, then please pay attention to what I say, ask questions about the aspects of my explanation you don't understand, and don't make dismissive comments like tossing aside long explanations with "wtf", or comments like this:

You were starting to sound like a crackpot.

"Crackpot" is not just a generic insult for any argument that sounds weird to you, again it refers to a specific style of trying to argue against mainstream views. If you think anything I was telling you doesn't match what mainstream physicists (like the one who wrote up the webpages I was basing my comments on) would say about visual appearances for an observer falling into a black hole, please point out where, as far as I know I was just summarizing the explanations on those pages.

I'm confused. I'm approaching the horizon and I can't see any object cross the horizon from the outside. I now cross the horizon and I see those same objects suddenly jump to some point along a line and the time since they crossed determines how far along the line they jump to?

No, there is no jump in the apparent position of the objects who appear to be arrayed at various distances below you both before and after you cross the horizon, and likewise there is no jump in the position of the red antihorizon which all the objects appear to remain above, it's just that there is a sudden appearance of the white horizon itself, which first appears as a line going from your position, through all the objects below you falling on the same radial path, down the red antihorizon (at the moment it appears, the light from all the objects below you that lie along it is the light they emitted at the moment they crossed the horizon themselves). Then it immediately expands into a bubble which encloses both you and these objects below you, with the top rising above you so that objects may cross into the bubble above you (and at the moment you see any object above you reach the boundary of the white bubble, you are seeing light from that object at the moment it crossed the event horizon). The reason it all works this way would be a lot easier to understand if you familiarized yourself with the details of the Kruskal-Szekeres diagram and then looked carefully at this diagram, imagining adding some worldlines of objects which fell through the horizon at earlier and later times than the object with the blue worldline. Again if there's some aspect of these diagrams you don't understand, or you don't see the relevance to my explanation of the visual appearances above, please ask questions about this stuff rather than just dismissing it.

So you would never see the second (white in the diagrams) visual horizon as "in front" of you as you suggested, instead it first appears as a straight line extending from you to the first (red in the diagrams) visual horizon at the moment you actually cross the horizon, and then it immediately expands into a bubble which you are underneath, as is any other object whose light you are seeing from a moment after that object crossed the horizon.

Hmm, I'm still not buying it.

Not a substantive response. Do you have some specific reason to doubt that this is what GR would predict we'd see when falling into a Schwarzschild black hole if light was being emitted by events on the antihorizon and also from events on the horizon? If not, do you think there is some inconsistency or other problem with GR's predictions?

For an eternal black hole, the red horizon is actually a physically separate horizon, the "antihorizon" one that borders the bottom of "our" exterior region I and the top of the alternate exterior region III in the maximally extended Kruskal-Szekeres diagram. The falling object genuinely never crosses this horizon, it's a white hole horizon in our universe and a black hole horizon in another exterior universe inaccessible from our own.

For a more realistic black hole that formed at some finite time from a collapsing star, you wouldn't actually be able to "see" any horizon from the outside, in the sense that light emitted from events on an event horizon would never reach anyone outside, at least not unless the black hole evaporated away. However, this section of the other site on falling into a black hole I linked to earlier also seems to say that if you could see the highly redshifted image of the collapsing star long after the black hole had formed, it would occupy almost exactly the same visual position as the red antihorizon of an eternal black hole:

I can honestly see no need for the "true" horizon.

I didn't use the words "true horizon", what part of my above explanation are you referring to? Do you mean the white horizon as opposed to the red antihorizon? Do you doubt that there are solutions to the Einstein field equations involving both eternal Schwarzschild black holes as well as black holes that form from collapsing stars, and that when you consider the "maximally extended" version of these spacetimes (the meaning of 'maximally extended' is discussed in the Kruskal-Szekeres article, ask questions if you don't understand something about it), they include a region of spacetime where anything inside the region will inevitably hit the singularity, and a region where it's possible to avoid the singularity, with the boundary between the two regions defined as the "event horizon"?

I'm just having trouble with the transition from what's observed before reaching the horizon and how it maintains continuity if it can be crossed.

All physical objects behave in a continuous way visually, but the white horizon behaves in a discontinuous way visually as it goes from completely absent before you cross the horizon to suddenly appearing as a straight line which expands to a bubble after you cross the horizon. Again the reason for the sudden appearance should be fairly understandable if you follow how light behaves in Kruskal-Szekeres coordinates and also follow what's going on in this diagram. And remember that this sudden appearance is just based on the assumption (made for the sake of visualization) that every point in spacetime which lies on the event horizon emits white light, it would similarly be true in Minkowski spacetime that if you had some event E and assumed that every event on its future light cone emitted white light, you wouldn't actually see any of this white light until the moment you yourself entered the future light cone of E, at which moment you would see a straight white line reaching from you to the event E itself (with objects that entered the future light cone earlier along the same radial path to the position of E appearing arrayed along this line), which would then expand into an ellipsoid enclosing yourself and the objects that entered the future light cone earlier (and expanding to enclose objects which entered the future light cone of E after you). If you're having trouble understanding why GR would predict what I told you about the appearance of the black hole horizon, see if you can at least understand why this would be true for the appearance of a future light cone in SR, and ask questions if you don't.

Absolute horizon. As in the one that doesn't recede but maintains a constant distance from the singularity despite the fact that distance is meant to be relative.

Do you mean "constant distance from the singularity" in terms of some coordinate system like Schwarzschild coordinates, or in terms of visual appearances (in which case it's not correct to say the visual appearance of the horizon maintains a constant distance from the singularity, both visual horizons appear to change in apparent size as you approach and anyway the singularity has no apparent visual position, it can't be 'seen' any more than the Big Crunch singularity in a contracting universe could be 'seen' in advance), or do you imagine there is some other notion of "distance" besides coordinate distance and apparent visual distance?

What would be subject to gravitational length contraction, the object or the horizon? And just as with "time dilation", please specify whether by "length contraction" you mean visual appearances, or frame-dependent length, or something else.

I meant the distance between the event horizon and singularity depends on distance it's viewed from.

Any comment about "distance" is meaningless unless you specify what you mean by that word. Please answer my question: are you referring to apparent visual distance, or to distance in some coordinate system, or do you claim there is some third notion of "distance" aside from these? (I suppose you could also talk about the integral of ds^2 along some specific spacelike path, like the worldline of a hypothetical tachyon, which would have a coordinate-independent value just like proper time along a timelike worldline).

Nope, that's just flat-out wrong. I already told you many times that time dilation and length contraction don't go to infinity at the horizon in Kruskal-Szekeres coordinates, and also that in ordinary Minkowski spacetime you do have infinite time dilation and length contraction at the Rindler horizon if you use Rindler coordinates, but obviously this is a purely coordinate-based effect which disappears if you use ordinary inertial coordinates in the same spacetime.

In the same space-time? When comparing objects at different distance from an event horizon they can't possibly be in the same space-time.

You seem to have some confused idea about what "spacetime" means, it just refers to the continuous curved 4D manifold consisting of every possible point in space and time where a physical event could occur, including events at different distance from the horizon, along with a definite geometry (curvature at every point, defined by the metric) assigned to this manifold. "Same spacetime" just means we are talking about the same geometry--events which both occur on the same solution to the Einstein field equations (say, the Schwarzschild metric as opposed to the Minkowsk metric or a FLRW metric)

 

[JesseM]

(response to post #167, part 3)

Stop calling me names! It's mean.

Again terms like "crackpot" are meant to refer to a style of argument which you are skirting dangerously close to, not a comment on your personality or intellect.

And if you relied more on intuitions and less on maths then you might question the things you're told a bit more. I thought the universe did become a singularity at c (as far as light, all energy for that matter is concerned).

No, see the various threads on how it's meaningless to talk about the "perspective" of an observer moving at c, like this one. If you consider what some inertial landmarks look like for an observer moving at v relative to them in the limit as v approaches c, some quantities do approach infinity in this limit, but in any case this would only be approaching a coordinate singularity as opposed to a genuine physical singularity (where some quantity approaches infinity at a given point in all coordinate systems which approach arbitrarily close to that point, like the curvature singularity at the center of a black hole.)

Maybe I am being a cock, because I'm getting frustrated. First the name calling isn't nice. Secondly I told you that I don't like all the coordinate stuff but you continue to ask question like: "Do you think you can "never reach" the Rindler horizon just because time dilation and length contraction go to infinity as you approach the horizon in Rindler coordinates, and a Rindler observer can never see anything reach the horizon? If not please explain what makes the black hole event horizon different."

Please try to keep track of the context. That comment of mine was in response to your own statement from the end of post #165 where you were talking about "frames" (coordinate systems) too:

Let's go back to the two horizons. One just in front of you and one some distance away just in front of another object that very proberbly crossed the horizon earlier. How could it have if it's always possible for it to escape. This isn't just visual, it's real. If it can escape then in what sense is the horizon in front of you (which the other object is inside) real? What happens if you mantain your distance and the other object moves back out to your possition? It would mean it's moved but the event horizon has followed it. Just ignore the absolute horizon. Why would there even be one? It would be subject to gravitational length contraction which would mean its size would decrease the closer you got, going all the way up to infinity at the horizon. If it's infinite in one "frame" then it's infinite in all of them. You can never reach it in exactly the same way you can never reach c. The black hole becomes the singularity at 0 range, like the whole universe becomes a singularity at c.

So that's why I responded with the comment about Rindler coordinates, because presumably you don't think in that case that if time dilation is "infinite in one frame then it's infinite in all of them".

I'd never even heard of Rindler coordinates or Schwarzschild coordinates or half the stuff that's been mentioned here until this conversation so it's not a reason for me thinking anything.

Uh, I never said it was a "reason for you thinking anything", your thinking seems to be based on vague intuitions about things like "distance" and "time dilation" that are based on statements you've read by physicists (or physics popularizers) in various places, without realizing that physicists generally use such terms in the context of some specific coordinate system. So your failure to realize this means you try to imitate the way physicists would use such terms but without sufficient understanding of how they are just verbal shorthand for coordinate-dependent statements, and this causes a lot of your arguments to be not even wrong (also see Feynman's essay about http://www.lhup.edu/~DSIMANEK/cargocul.htm, where people imitate some of the external forms of scientific explanation but without the underlying technical substance). I'm trying to teach you to think more clearly about the terms you're throwing around--that's why I ask these questions offering you lists of different possible meanings that might be assigned to terms like "distance" and "time dilation" (like coordinate distance, apparent visual distance, etc.) and emphasize that your claims don't really make sense under any of these specific meanings.

I think they're what happens when you let maths lead the "understanding" instead of doing it the other way round. There, I said it.

That, fundamentally, is what modern physics is all about, and the sooner you realize that the sooner you'll be on track to thinking like physicists do. Any self-consistent mathematical model is potentially an accurate model of the universe, there are no real criteria that should cause you to reject any mathematical model besides the fact that it doesn't give accurate predictions. Think of it this way, if you had access to a computer of the Gods that could simulate any arbitrarily large collection of fundamental particles or points in spacetime or other basic entities, all constrained to obey some general mathematical law that could be programmed into the simulation, then any possible rules that could be programmed into the computer that would lead to the evolution of simulated beings whose empirical observations match our own should be considered a viable candidate for the "program" governing the behavior of our own universe. To argue otherwise would be to argue that we have some totally non-empirical reasons for believing the laws of physics must take one type of form (one that has some kind of appeal to our 'common sense' intuitions) as opposed to other types of forms that might still produce correct predictions about all empirical observations.

 

[JesseM]

Hi When you refer to the upper and lower components are you talking about the lower deceleration part of the diagram and the upper or acceleration half of the diagram??

I said "the Rindler horizon has an upper and lower component"--I was talking only about the horizon, which would be the two straight red lines at 45 degrees in the diagram which form the upper and lower boundary of the "Rindler wedge". These are analogous to the black hole event horizon and white hole event horizon that form the upper and lower boundaries of the exterior region I in the Kruskal-Szekeres diagram for a Schwarzschild black hole.

When you say light from events arbitrarily far , I assume this also means distant in time also ,,is this correct??

Right, for example if at some point an accelerating Rindler observer receives light from an event that happened 1000000 light years away in some inertial frame (perhaps far beyond the Rindler horizon), then this event would have happened 1000000 years earlier in that inertial frame. Note that if this event happened outside the Rindler horizon, it must have happened somewhere in the past light cone of the event at the center of that diagram, the one at the "tip" of the Rindler wedge where the two components of the horizon meet. In the Kruskal-Szekeres diagram this past light cone of the event at the center would correspond to the white hole interior region.

That the events from much further than the horizon would then have to have occurred further in the past, yes??
If so this is more or less what I thought.

Right.

So if I am understanding you correctly, both Rindler and CMIRF observers would visually see both Doppler shift and aberration but nothing identifiable beyond that. yes?

Right again.

 

[Anamitra]

Two gravitating objects may be observed at some point of time,$${t=t_{0}}$$
A few seconds ago they were at a greater distance and were definitely attracting each other[and not repelling]. Another few seconds ago they were still attracting each other[and the force was not of repulsive nature].
Let's see it mathematically:
$${F_{G}}{=}{\frac{dp}{dt}}$$
For t-->-t
We seem to have,
$${F_{G}}{=}{-}{\frac{dp}{dt}}$$
But
$${p}{=}{m}{\frac{dx}{dt}}$$, for a constant mass particle
Here also we need to change the sign of t
Finally we have[after time reversal],
$${F_{G}}{=}{\frac{dp}{dt}}$$

For a particle with variable mass,

$${F}{=}{m}{\frac{dv}{dt}}{+}{v}{\frac{dm}{dt}{=}{m}{\frac{dv}{dt}}{+}{\frac{dx}{dt}}{\frac{dm}{dt}$$

In the last term to the right we need to bring two changes [$${t-->}{-}{t}$$] in time and so it does not change sign[assuming mass to increase in the forward direction of time].And the first term also does not change sign. So ,attraction remains attraction.
[Relativistically mass should increase with time with increasing speed.]

 

[A-wal]

If you're talking about a purely coordinate-dependent or path-dependent notion of "relative velocity", then you must agree that just because two objects at distant locations have a relative velocity of 0.5c in some coordinate system or with velocities parallel-transported along some path, there may be some other coordinate system or path where the relative velocity is 0.9c or 0.1c or even 0. So it seems to me the relative velocity of 0.5c does disappear in curved spacetime, in the sense that it there is no objective sense in which two objects can be said to have a relative velocity of 0.5c as opposed to 0.9c or 0.

Only if you are talking about some coordinate-dependent notion of "diameter", like the diameter in Schwarzschild coordianates or Eddington-Finkelstein coordinates. Please, if you are going to talk about "distance" then please tell me if you are talking about distances in some coordinate system, or apparent visual distances and sizes, or if you think there is some third option.

I don't know what you mean by "perspective". Are you talking about visual appearances at 100 d-units vs. 50 d-units, or are you imagining that the observer will use different coordinate systems depending on his distance, or what?

Do you mean "constant distance from the singularity" in terms of some coordinate system like Schwarzschild coordinates, or in terms of visual appearances (in which case it's not correct to say the visual appearance of the horizon maintains a constant distance from the singularity, both visual horizons appear to change in apparent size as you approach and anyway the singularity has no apparent visual position, it can't be 'seen' any more than the Big Crunch singularity in a contracting universe could be 'seen' in advance), or do you imagine there is some other notion of "distance" besides coordinate distance and apparent visual distance?

Any comment about "distance" is meaningless unless you specify what you mean by that word. Please answer my question: are you referring to apparent visual distance, or to distance in some coordinate system, or do you claim there is some third notion of "distance" aside from these? (I suppose you could also talk about the integral of ds^2 along some specific spacelike path, like the worldline of a hypothetical tachyon, which would have a coordinate-independent value just like proper time along a timelike worldline).

All I ever meant (as I keep telling you) is that objects can obviously move relative to each other in curved space-time, and that the velocity can be measured. Obviously if you measure it differently you’ll get a different result. But as long as we keep assuming they stick to the same method then what’s the problem? Besides can’t we just assume the shortest path? In fact, from now on I’ll always use the shortest path between any specified objects or phenomena unless I expressly specify otherwise. Now you’ll never have to ask me that again.

No physics theory has ever "explained" why matter/energy/particles behave in the way they do, it just gives equations describing their behavior.

Plenty of theories explain why matter/energy/particles behave in the way they do. In fact that's what every physics theory attempts to do. General relativity explains Newtons laws for example.

If it wouldn't be obvious to a physicist well-versed in the mathematics how to translate your verbal argument into a detailed mathematical one, then hell yes it's "handwavey". In physics verbal arguments are only meaningful insofar as they can be understood as shorthand for a technical argument, where the meaning of the shorthand is clear enough that you don't have to bother laboriously spelling everything out in technical terms. This is not to say that physicists don't appeal to physical intuitions in situations where they're groping for the correct answer to some problem they don't know how to solve in a rigorous mathematical way, but I think they always do so with an eye towards developing a technical mathematical argument, I think you'd be hard-pressed to find a physicist who believes that some verbal argument is sufficient in itself to demonstrate some claim even if they don't know how to translate it into technical terms (and don't think other physicists would be able to either).

I am not a physicist! It’s not fair for you to expect me to know how to put it in technical terms. It doesn't mean I don't get it. Equasions are the shorthand for whatever it is they represent.

So does that mean you are retracting all your earlier statements suggesting there is some objective sense in which we can talk about which clock experiences more time dilation at different distances from the horizon, apart from the question of which clock experiences more total elapsed time between two meetings which each occur at a single point in space time? For example, this comment from post #161:

Or this one from post #165:

If you don't retract these comments, then how does this square with your agreement that depending on our choice of simultaneity convention we can reach different conclusions about whether clock B was ticking faster or slower than clock A between the events of clock A reading a time T and clock A reading a time T + delta-t?

When I said use your common sense I was just saying that people can still compare watches in curved space-time. You disagree?

OK, #2 was "Local comparisons of elapsed times, as in the twin paradox where each twin looks at how much time each has aged between two meetings". But in this case you can't say anything about time dilation except when talking about total elapsed time between two local comparisons, in particular you can't say one clock was ticking slower when it was closer to the horizon. If you agree with that, it seems to me you are changing your tune from your comments in post #161 and #165 when the statements of mine you were disagreeing with were just statements that we can only talk about differences in total elapsed time and nothing else.

Of course they can say one clock was ticking slower when it was closer to the horizon. One clock was ticking slower when it was closer to the horizon if one spent all that time closer to the horizon and less time has passed for it.

I didn't use the words "true horizon", what part of my above explanation are you referring to?

It was from something you either quoted or linked.

Do you mean the white horizon as opposed to the red antihorizon? Do you doubt that there are solutions to the Einstein field equations involving both eternal Schwarzschild black holes as well as black holes that form from collapsing stars, and that when you consider the "maximally extended" version of these spacetimes (the meaning of 'maximally extended' is discussed in the Kruskal-Szekeres article, ask questions if you don't understand something about it), they include a region of spacetime where anything inside the region will inevitably hit the singularity, and a region where it's possible to avoid the singularity, with the boundary between the two regions defined as the "event horizon"?

I believe there is a region of space-time where anything inside would inevitably hit the singularity. I also believe it’s analogous to saying that there is a velocity that exists that is greater than c. I also believe this velocity can’t ever be reached.

You seem to have some confused idea about what "spacetime" means, it just refers to the continuous curved 4D manifold consisting of every possible point in space and time where a physical event could occur, including events at different distance from the horizon, along with a definite geometry (curvature at every point, defined by the metric) assigned to this manifold.

I just see it as the distance between objects, which is relative and the difference is the curve. To our linier perspective it means that everything with relative velocity moves in straight lines but through curved space-time – gravity. If you want to create your own curve you accelerate.

"Same spacetime" just means we are talking about the same geometry--events which both occur on the same solution to the Einstein field equations (say, the Schwarzschild metric as opposed to the Minkowsk metric or a FLRW metric)

Oh okay. My point was that one is in space-time that’s more length contracted/time dilated than the other. I don’t see how infinite time dilation/ length contraction can disappear if you change coordinate systems. You can’t change reality by measuring differently (yes I know you can in quantum mechanics). Change the value of something yes, but not get it to or from infinity. That still seems contradictory.

No, see the various threads on how it's meaningless to talk about the "perspective" of an observer moving at c, like this one. If you consider what some inertial landmarks look like for an observer moving at v relative to them in the limit as v approaches c, some quantities do approach infinity in this limit, but in any case this would only be approaching a coordinate singularity as opposed to a genuine physical singularity (where some quantity approaches infinity at a given point in all coordinate systems which approach arbitrarily close to that point, like the curvature singularity at the center of a black hole.)

I though it’s meaningless because the universe would be a singularity at c. I thought photons have a very short life span but it doesn’t matter because they’re infinitely time dilated meaning there frozen in time. It’s that’s true than surely the universe would be perceived as a singularity at c, if you reach c, but you can’t.

Uh, I never said it was a "reason for you thinking anything", your thinking seems to be based on vague intuitions about things like "distance" and "time dilation" that are based on statements you've read by physicists (or physics popularizers) in various places, without realizing that physicists generally use such terms in the context of some specific coordinate system. So your failure to realize this means you try to imitate the way physicists would use such terms but without sufficient understanding of how they are just verbal shorthand for coordinate-dependent statements, and this causes a lot of your arguments to be not even wrong (also see Feynman's essay about http://www.lhup.edu/~DSIMANEK/cargocul.htm, where people imitate some of the external forms of scientific explanation but without the underlying technical substance). I'm trying to teach you to think more clearly about the terms you're throwing around--that's why I ask these questions offering you lists of different possible meanings that might be assigned to terms like "distance" and "time dilation" (like coordinate distance, apparent visual distance, etc.) and emphasize that your claims don't really make sense under any of these specific meanings.

Read the blogs I wrote ages ago and tell me if I’m under the wrong impression about anything. You think I don’t understand the concepts just because I don’t know how to speak your language. If I am wrong about anything then by all means tell me but stop just telling me I don’t get it. Don’t get what? The fact that length contraction and time dilation have to be expressed in specific coordinate systems to make sense? No they don’t. Maybe they do on paper.

That, fundamentally, is what modern physics is all about, and the sooner you realize that the sooner you'll be on track to thinking like physicists do.

I really don’t want to think like physicists do. I want to get this straight in my head and still think the way I do. I don’t know how you can do it like that. Would you watch your favorite dvd in binary code? The code is necessary but no one cares what it looks like.

Any self-consistent mathematical model is potentially an accurate model of the universe, there are no real criteria that should cause you to reject any mathematical model besides the fact that it doesn't give accurate predictions. Think of it this way, if you had access to a computer of the Gods that could simulate any arbitrarily large collection of fundamental particles or points in spacetime or other basic entities, all constrained to obey some general mathematical law that could be programmed into the simulation, then any possible rules that could be programmed into the computer that would lead to the evolution of simulated beings whose empirical observations match our own should be considered a viable candidate for the "program" governing the behavior of our own universe. To argue otherwise would be to argue that we have some totally non-empirical reasons for believing the laws of physics must take one type of form (one that has some kind of appeal to our 'common sense' intuitions) as opposed to other types of forms that might still produce correct predictions about all empirical observations.

Of course you can't just assume intuition is right. That's stupid. But it always starts and ends with that. It decides what you test for and what conclusions you draw from the results. Without it you're trying to paint in the dark.


I don’t mind if I’m wrong. My ego isn’t tied up in this and I have nothing to prove. I find it difficult to accept what I don’t understand and I’m not convinced by what I’ve been told. How the hell can an object that can never reach the horizon from any external perspective ever cross the horizon from its own perspective? Is not just the light from those objects that’s frozen. How could it be if they could always escape? They’re moving slower and slower through time relative to you because time in that region is moving slower and slower relative to you. If the time dilation/length contraction go up to infinity then no given time can ever long enough and no distance can ever be short enough locally if it’s infinitely length contracted from a distance! Are those inertial coordinates you use to describe an object crossing the event horizon even relative? Does it take into account the fact that you’re constantly heading into an ever increasingly sharpening curve?

ObserverA measures the distance between ObserverB (who is much closer) and the horizon using some coordinate system or other. Then after moving right next to ObserverB measures it to be more than it seemed before in the same coordinate system. ObserverA: "You've moved!" ObserverB: "No I haven't!" (Because it's not length contracted when you're actually there). You have to un-length contract the space, making the distance longer, right? But if it’s infinitely length contract at the horizon then there’s going to be an infinite amount of space (and time) between you and it before you get there.