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

[Buckleymanor]

That is not what you said above. Above you said nothing about the time prior to when the ball was rising and decelerating, you simply claimed that a ball rising and decelerating is an example of push gravity, which is wrong as I demonstrated.

Look, this is easy to show mathematically. For a ball falling in a gravitational field the law is:
x''=-g
Which, if we are given initial conditions x(0)=x_0 and x'(0)=v_0 leads to the equation of motion:
x(t)=-\frac{1}{2}g t^2 + v_0 t + x_0

Now, if we time reverse the equation by making the substitution t=-T we get:
x(T)=-\frac{1}{2}g T^2 - v T + x
Note, that this expression is the same as the above equation of motion except with the initial condition x'(0)=-v_0. Taking the second derivative of this expression (wrt T) we recover the same law
x''=-g
where gravity is still "pull" gravity. This is what is meant by the time reverse symmetry of gravity.

Peter Donis is correct in his statements about the time reversal of gravity. Your comments about it being inconsistent are wrong as demonstrated both by the counter example and by the math above. In fact, as the math shows, for any scenario of an object falling under pull gravity, the time reverse is also an object falling under pull gravity only with the opposite initial velocity.

Opposite initial velocity not possible and inconsistent the math might show it is, well then let's take a view of your bouncing ball from the ground upwards. Small bounce then bigger then larger still.Why don't you ask from the view of this where is this extra energy is coming from.It don't look the same, it might have reverse symmetry but that ain't the same as being symmetrical.

 

[Dale]

Opposite initial velocity not possible and inconsistent the math might show it is, well then let's take a view of your bouncing ball from the ground upwards. Small bounce then bigger then larger still.Why don't you ask from the view of this where is this extra energy is coming from.It don't look the same, it might have reverse symmetry but that ain't the same as being symmetrical.

For the third time, I carefully specified no inelasticity or friction. I.e. All of the asymmetry you are thinking about comes from thermodynamics, not gravity.

However, the important point is that a ball going upwards and decelerating is NOT push gravity as you asserted in post 315. Do you understand that now?

 

[A-wal]

On re-reading this and the question it was in response to, I realized I should clarify a couple of things:

(1) The family of observers that are "hovering" outside the black hole horizon, at rest relative to each other, are *not* inertial observers; they are accelerated and feel weight. No two inertial observers in a black hole spacetime can be at rest relative to each other for more than a single instant unless they are at rest relative to each other at the same radial coordinate r (which r coordinate that is will change with time, as the observers free fall towards the hole, but if they both start free falling at the same r, they will always be at the same r, so they will always be at rest relative to each other--at least, as long as they are close enough together that the tangential tidal gravity is negligible).

(2) Even if we allow the family of "hovering" observers, at rest relative to each other, to define a "background" coordinate system for the spacetime, that family of observers does not cover the entire spacetime around the black hole. Inside the horizon, there are *no* "hovering" observers *at all*--*all* observers, no matter how they move or how hard they accelerate, *must* decrease their radial coordinate r with time. So inside the horizon, there are *no* observers, inertial *or* accelerated, that are at rest even for an instant relative to the family of "hovering" observers outside the horizon.

Item (2) is what I was referring to in the quote above: there is a region of spacetime around a black hole (the region inside the horizon) where it is impossible for any observer, even for an instant, to be at rest relative to what A-wal calls the "river bed" (or "pebbles" or "background" or whatever term you want to use). Such observers can only exist outside the horizon. But as item (1) shows, even outside the horizon, observers at rest relative to the "river bed" are *not* inertial observers. That's what I wanted to clarify.

They cover the entire space-time outside the black hole, which is the entire space-time if I'm right. Take a line of them going from our starting position. They start off evenly spaced then spread out as we move into a higher gravitational field. When we reach the second one from the end we see the biggest gap, with the last one almost at the horizon. This observer is very time-dilated, so it could be traveling towards the horizon at just under c and it still wouldn't have time to reach it. When we reach the last observer we see that the black hole’s gone. "Did you see the horizon?" "No, I just got here and it was already gone."

While they're in the same place, yes, the effect of gravity is the same--but they only remain in the same place for an instant because their velocities are different. Then they move apart and the curvature of spacetime is different at their two different locations, so they experience different things.

You're using relative velocity then. You were talking about light rays and there at rest relative to each other if they occupy the same point in space-time.

Still doesn't make any difference. Your claim that an object would have to "break the light barrier" relative to the hovering objects implicitly assumes that the "frame" defined by the hovering objects covers the entire spacetime. Once again, that assumption is false in the standard GR model. (See below for more details on this.)

My non-standard version is disproved by an assumption of the standard one? That hardly seems fair.

Ok, at least I understand why you don't see a difference. I don't agree, but at least I understand how this particular claim of yours ties in with the rest of your claims.

Super.

It's not that it "can't form", it's that the mathematical solution describes an object that has existed for an infinite time in the past. If you have good physical reasons to believe that the entire universe has only existed for a finite time in the past (which we do), then obviously any mathematical solution describing an object that would have had to exist for an infinite time in the past is not a good candidate for describing an actual, physical object. But it's still a perfectly valid solution mathematically.

Only if you start the equation part-way through.

I've mentioned making correct predictions before. What does your nice, simple model predict for the following:

(1) The precession of the perihelion of Mercury's orbit?

(2) The bending of light by the Sun?

(3) The changes in the orbits of binary pulsars due to the emission of gravitational waves?

(4) The precession of gyroscopes orbiting the Earth due to gravitomagnetism?

The standard GR model that predicts all these phenomena correctly *also* predicts that black holes will behave as I've been describing. That's why physicists believe in the standard GR model of black holes that I've been describing. If you can show how your model reproduces all these correct predictions *without* requiring black holes to behave as I've been describing, please do so. But you can't just wave your hands and say, "well, obviously my model looks just like GR outside the horizon", because the way GR arrives at all the above predictions is inseparably linked, mathematically, to the way it describes black holes and their horizons. So you have to start from scratch, and work through how your model would deal with the above phenomena, *without* making use of any of the machinery or results of GR.

(And no, you can't get any of the above results just by applying non-relativistic Newtonian gravitational theory. That's why I chose these examples.)

What could possibly make you think that I'm saying gravity is non-relativistic and shouldn't predict those things? Just the opposite. I've already started from scratch so I don't need to again. I think you might though.

Because you can't directly assign any physical meaning to the "relative velocity" of two objects at different places. Suppose I have observer A, well outside the horizon and hovering at a constant r. Then I have observer B, who has just freely fallen through the horizon. In order to make sense of the "relative velocity" of A and B, I have to implicitly assume a third observer, C, who is at the same radial coordinate r as B (i.e., r a little less than the radius of the horizon), but who is at rest relative to A, so that I can say that the relative velocity of A and B is equal to the relative velocity of C and B (which I can assign a direct physical meaning to because C and B are at the same place). But in the standard GR model, there can't be any such observer C; *no* observer inside the horizon can "hover" at a constant radius, not even for an instant. So the only way of physically assigning a meaning to the concept "relative velocity of A and B" breaks down if A is outside the horizon and B is inside.

Yes exactly, it breaks down if an object is allowed to cross the horizon.

(One clarification: by "physical meaning of relative velocity" I mean a meaning that would justify the requirement that the "relative velocity" of two objects can't be faster than light. If two observers are at the same place, then I can apply special relativity locally and impose that requirement. But I can't do it for observers that are separated, if the curvature of spacetime is significantly changed from one to the other. Of course, I can arbitrarily define the "relative velocity" of A and B by simply using, for example, dr/dt, the derivative of the radial coordinate r with respect to the "time" coordinate t. But this meaning of "relative velocity" does *not* require that the relative velocity can't be faster than light, because it's just an arbitrary number; it doesn't correspond to anything that any possible physical observer could ever observe.)

Of course it can't be faster than light. The reason it doesn't correspond to anything that any possible physical observer could ever observe is because it never happens. That's like saying it's perfectly okay to break the light barrier because if you did you wouldn't be able to measure you're relative velocity anyway. You have to reach c first before you can claim that it would work if you did (which it doesn't). You have to reach the horizon before you can talk about how that could work (which doesn't either).

Why do you think there's a push? There's no push anywhere. Objects emerge from the white hole singuarity, but that's not a "push" because it's not due to any "force" from the singularity; the objects just emerge. As soon as they emerge, they start decelerating, so the only "force" observed is a pull.

Yea like the ball jumping off the ground by itself if you run time backwards. What makes it a white hole then? The fact that nothing can reach the horizon despite the fact that they pull? They're all white holes then. What happens on the approach to a white hole? Presumably you don't just get stuck when you hit the horizon. It happens in the same way I've been describing doesn't it?

What makes you think this? And don't say "I heard it somewhere" (see next comment). Give me some sort of logical argument, based on premises we all accept, that makes this seem reasonable to you. (To me, as should be obvious from my previous posts, it's just wrong as it stands.)

I was thinking black holes loose all of their energy very quickly and then they're gone. Their energy gets released as gravity waves and the same should be true of matter. It would just take a lot longer. The energy it's losing to curve space-time can't last forever.

You keep on saying "I heard somewhere" something, and give a vague description of it, but can't give any actual reference or explain what you mean beyond the vague description. That's not very helpful in understanding what you're talking about. Even the OP in this thread suffers from this problem.

If you are thinking of things like normal matter "quantum tunnelling" into other states (as described, for example, in the page linked to below), yes, according to QM that will eventually happen if nothing else does, but that process doesn't require any energy, and normal matter certainly doesn't do any work or "build up" energy while it's "waiting" for this to happen.

Yea that was probably what I was thinking of. I wasn't thinking it built up energy, I was thinking it lost it until it collapsed or whatever. That's not what I thought quantum tunnelling was. I don't want to get into an in depth conversation about quantum mechanics but I thought it was when two solid objects are able to pass through each other because neither has an exact position, so they can't possibly hit each other?

I have never said any such thing. I have said repeatedly that only one "thing" curves spacetime, and that's the stress-energy tensor (which includes what you are calling "matter", and "pressure", and also includes "energy" as standard physics uses the term, but you don't always use that term correctly).

I *have* said that what you sometimes refer to as "energy" (meaning something like firing a rocket engine to accelerate, and therefore feeling weight) *is* different from the stress-energy tensor curving spacetime, because acceleration (in the sense of feeling weight) curves your worldline, not spacetime, and you can have a curved worldline in a flat spacetime, so the two concepts are distinct. (See further comment below on this.)

The fact that you view them differently doesn't make them distinct and neither does the fact that you can have a curved world-line in flat space-time. A curved world-line is just another way of saying your area of space-time is curved because of acceleration. You can even feel tidal force. I believe you call it g-force.

I'm quite sure what I think. I'm also quite sure that you don't understand it, and that the reason you don't understand it is that your thinking is based on assumptions that you think are obviously true, whereas I have a consistent model in which they're false. I keep on asking you to give actual arguments for your assumptions, instead of just assuming they're true even though I've repeatedly said I don't accept them, but you never do, you just keep asserting them. It's like trying to explain how matrix multiplication works to a person who keeps insisting that multiplication *has* to be commutative, even though it keeps being pointed out that in fact, matrix multiplication is *not* commutative, just as a matter of mathematical fact.

You keep saying this and I'm not sure what you're after. I'm explaining in multiple ways how I see whatever particular aspect it is I'm talking about. I'm not really sure what you're asking for. Equations? That's not going to happen. You want to know why it would work the way I think rather than your way? I don't see how the version you're telling me about works. I have another version and I can see exactly how this one works. There is of course an outside chance that I don't get something about the standard version and I'm wrong, not everyone else. That's why I thought it would be a good idea to start this thread and see where I might be missing something. I'm aware that going into this conversation with a preconceived model isn't ideal, but that just makes us even. I have no idea what matrix multiplication is.

You are ignoring one key difference between what I've been saying and what you've been saying. Every time you have made a statement I disagree with, I have given a physical reason why I disagree, whereas you have just kept on asserting your statements without ever responding to the physical reason for my disagreement. Take the statement in *bold* above. You have asserted it repeatedly, without ever responding to the physical reason I've given for why the two cases *are* different: because the curvature of an object's path (which is determined by whether or not it feels weight--objects on curved paths feel weight, objects on straight paths do not) is completely independent of the curvature of spacetime itself (which is determined by whether tidal gravity exists, or in more explicit terms, by whether two objects, both in free fall, both starting out close together and at rest with respect to each other at a given time, continue to remain at rest with respect to each other for all time or not--if they do, spacetime is flat, if they don't, spacetime is curved).

"...because the curvature of an object's path (which is determined by whether or not it feels weight--objects on curved paths feel weight, objects on straight paths do not)..." That could just as easily describe both. "is completely independent of the curvature of spacetime itself (which is determined by whether tidal gravity exists, or in more explicit terms, by whether two objects, both in free fall, both starting out close together and at rest with respect to each other at a given time, continue to remain at rest with respect to each other for all time or not--if they do, spacetime is flat, if they don't, spacetime is curved)." Which could just as easily apply to two objects close together with a constant source of energy close by. The difference is that it works backwards (pushes rather than pulls) so the object behind would catch the closer object and overtake then pull away from it because of it's increased velocity relative to the energy source.

In the quote above, in the part that's _underlined_, you at least have given some sort of amplification of your statement, but it's still wrong, and the reason why it's wrong has been brought up repeatedly in this thread: a family of accelerated "Rindler observers" in flat spacetime has precisely the set of properties you describe, but the spacetime is still flat, as is easily seen by the test I gave above (*inertial* observers that start at rest with respect to each other remain at rest with respect to each other for all time). So your argument is still wrong. (And don't say that you've looked at Rindler observers and it doesn't change your mind, because every time you've described how you think Rindler observers and a Rindler horizon work, you've gotten it wrong, as I've pointed out repeatedly.)

How does the fact that an object can accelerate to the point where it can't ever be caught by other objects from certain frames show that the curvature from matter is distinct from the curvature from energy, or what you would call a curved world-line in flat space-time?

So the reason why you have not been "getting through" to me is that, as I've said before, I already understand the mistakes you're making, so seeing you continue to make them does not change my mind.

If you understood what mistakes I was making you would have no problem convincing me that they are mistakes. Explain to me why GR is more fundamental than SR. Explain to me why energy curves relatively while gravity does it absolutely. It's curved either way.


You said you wanted an "amplification"/application of what I'm describing. Okay, time for a new coordinate system. We're going to follow the horizon. Remember when I said a black hole is a four-dimensional sphere? That means it would be the same from any angle of space-time that you viewed it from. When it forms it rushes outwards at the speed of light. Nothing can come into contact during this phase because any proper time/length between you and the horizon is being dilated/contracted as it moves towards you. You wouldn't see it coming (even if you could) because it's moving at c, so by the time you see it it's there. Phase two (past the apex of the sphere). It recedes at the speed of light (so good luck chasing it). Nothing special has happened, it's just got past the point where G>C. The gravity wave continues to spread outwards and anything it touches would “see” the black hole for the first time. It would look extremely time-dilated and length contracted though, so it appears as though it's much bigger (really it's just the singularity) and exists for much more time than it does in its own frame. If you reversed the arrow of time then the collapse becomes the initial expansion and the initial expansion becomes the collapse. It looks exactly the same in reverse. It's the same from all sides.

 

[PeterDonis]

Take a line of them going from our starting position. They start off evenly spaced then spread out as we move into a higher gravitational field.

You're not specifying the scenario precisely enough, which is leading you into confusion. If you specify that the line of observers is separated by equal increments of the Schwarzschild exterior radial coordinate r, then as you get closer to the horizon, equal increments of r translate into larger and larger increments of actual proper distance, so yes, they would get more spread out. But that's because you specified the scenario as equal increments of r.

If you instead specify the scenario such that each observer in the line of observers fires his rockets in such a way as to maintain equal proper distance from the next observer "above" him, then that is also a perfectly consistent and realizable scenario, and in that scenario the observers do *not* "spread out" as you get closer to the horizon. They are able to maintain constant proper distance from each other, and they are able to do so using a constant acceleration (i.e., they feel a constant, unchanging weight--the weight gets larger for observers closer to the horizon).

You're using relative velocity then. You were talking about light rays and there at rest relative to each other if they occupy the same point in space-time.

Not if they're moving in different directions. An ingoing light ray is not at rest relative to an outgoing one, even if they both pass through the same event (same point in spacetime), any more than a light ray moving to the left is at rest relative to a light ray moving to the right.

My non-standard version is disproved by an assumption of the standard one? That hardly seems fair.

It's not a matter of "disproof". You are asserting that your version is the only *possible* version. To do that, it's not enough to assert your assumptions; you have to prove that they are logically necessary.

I've already started from scratch so I don't need to again.

Great, then please answer the questions as I posed them (feel free to ignore what I said about Newtonian gravity if that doesn't fit with your model).

What happens on the approach to a white hole? Presumably you don't just get stuck when you hit the horizon.

You can't reach a white hole horizon from the outside because it's an ingoing null surface; it's moving radially inward at the speed of light. So no matter how fast you move inward, you can't catch it.

I don't want to get into an in depth conversation about quantum mechanics but I thought it was when two solid objects are able to pass through each other because neither has an exact position, so they can't possibly hit each other?

That's one possible manifestation of quantum tunnelling, but quantum tunnelling itself is a much more general concept; it basically covers any of a multitude of cases where some kind of transition that is impossible classically has a non-zero probability (even if the probability is very, very small) of happening when quantum mechanics is taken into account. Another example would be a particle quantum tunnelling through a potential barrier that is classically not passable; it's possible to model radioactivity this way.

In the particular case I was talking about, the idea is that, since iron (more precisely, a particular isotope of iron, which I believe is Fe-56 but I'm not certain going just from memory) has a lower energy than any other nucleus, there is a non-zero probability, quantum mechanically, that a nucleus of any other atom can quantum tunnel to become an iron nucleus. Classically, this can't happen because there is a huge potential barrier; it takes a lot of energy to rearrange the nucleons in a nucleus, even if the final energy of the rearranged state is lower. But quantum mechanically, because of the uncertainty principle, there is a (very, very small) probability that a quantum fluctuation could give any nucleus the (temporary) energy it needs to rearrange itself into an iron nucleus.

You can even feel tidal force. I believe you call it g-force.

In general I'm not responding to simple repeated assertions of your assumptions, but you keep on saying this particular one so pardon my shouting: "TIDAL FORCE" IS *NOT* THE SAME AS "G FORCE"! The reason I focus in on this is that it's not just your assumption; it's a false statement of fact. Tidal gravity and "g force" (or "acceleration due to gravity") are two *different* physical phenomena. This is true even in Newtonian gravity.

You keep saying this and I'm not sure what you're after.

The particular question I keep asking (can you prove that your assumptions, which are false in standard GR, are logically necessary) is kind of rhetorical, because I don't think you can do it; your assumptions are *not* logically necessary, which is why there can be a consistent model (standard GR) in which they are false.

However, there is something else you can do: tell me how your model would account for the four experimental observations I gave.

"...because the curvature of an object's path (which is determined by whether or not it feels weight--objects on curved paths feel weight, objects on straight paths do not)..." That could just as easily describe both.

No, feeling weight vs. not feeling weight is a genuine physical distinction--we can observe it and measure it. The fact that you don't think it makes a difference in your model does not mean it's not a genuine physical, observable distinction. In standard GR, it is a crucial distinction in the theory as well. If it isn't in your model, fine; explain how your model accounts for the four experimental observations I listed.

"is completely independent of the curvature of spacetime itself (which is determined by whether tidal gravity exists, or in more explicit terms, by whether two objects, both in free fall, both starting out close together and at rest with respect to each other at a given time, continue to remain at rest with respect to each other for all time or not--if they do, spacetime is flat, if they don't, spacetime is curved)." Which could just as easily apply to two objects close together with a constant source of energy close by.

What does "a constant source of energy close by" mean? Does it mean the two objects are being pushed by the source of energy, and therefore are accelerated and feel weight? Then see my previous comment above. Again, in standard GR, tidal gravity is exactly as I've defined it in the sentence you quoted. If it means something different in your model, fine; show me how your model accounts for the four experimental observations I listed.

If you understood what mistakes I was making you would have no problem convincing me that they are mistakes.

Oh, if only that were true.

I'll have to digest the last part of your post some more before I can respond; I'm not sure I understand yet what you're trying to describe.

 

[PeterDonis]

When we reach the second one from the end we see the biggest gap, with the last one almost at the horizon. This observer is very time-dilated, so it could be traveling towards the horizon at just under c and it still wouldn't have time to reach it. When we reach the last observer we see that the black hole's gone. "Did you see the horizon?" "No, I just got here and it was already gone."

On re-reading, I realized that I should respond to this point as well. First, if the observer is traveling towards the horizon at just under c, the observer is not "hovering" at a constant r, which is what you specified at the start of the scenario. Please make up your mind what scenario you're talking about.

Second, considering the observer falling towards the horizon at just under c, waaay back in post #230, I posted the calculation showing that the proper time to reach the horizon for an infalling observer is finite, as well as the proper distance to the horizon, for both a "hovering" observer and an infalling observer--with the infalling observer's distance being length-contracted relative to the "hovering" observer's distance, as expected. As those calculations show, the "length contraction" and "time dilation" as the horizon is approached are *not* sufficient to prevent an infalling observer from reaching the horizon.

I even explained the reason back then, but I only did it explicitly in the case of time, so I'll restate it again for both time and distance. As the horizon is approached, "time dilation" means that the increment of proper time per unit of coordinate time t goes to zero while coordinate time t goes to infinity, while "length contraction" means that the increment of proper distance per unit of coordinate radius r goes to infinity while the increment of radius left to reach the horizon goes to zero. But in the case of time, the proper time per unit of coordinate time t goes to zero faster than t goes to infinity, while in the case of distance, the increment of radius left to the horizon goes to zero faster than the proper distance per unit of coordinate radius r goes to infinity. In both cases, the result is a finite sum, meaning a finite proper time and proper distance to the horizon.

 

[PeterDonis]

You said you wanted an "amplification"/application of what I'm describing. Okay, time for a new coordinate system.

But I thought you objected to switching coordinate systems? You've certainly objected to me and others using anything other than Schwarzschild coordinates.

We're going to follow the horizon. Remember when I said a black hole is a four-dimensional sphere? That means it would be the same from any angle of space-time that you viewed it from.

I can make sense of this statement in isolation, but I don't see how it leads to the description you go on to give of what the black hole looks like to someone "moving in time". It seems to me that your model of the black hole as a 4-D sphere would lead to the following description. I'll reduce it to just two dimensions (one of time and one of space) so it's easier to visualize.

Let R be the "spacetime radius" of the black hole. We'll use coordinates where the units of time and space are the same, so R is the radius of the black hole in both dimensions. Then, on a spacetime diagram, the hole just looks like a circle with radius R around the origin. The "bottom" of the circle (which would be the "South Pole" of the 4-D sphere if we added the other two dimensions back in) is at t = -R, x = 0. The "top" of the circle is then the event where the black hole evaporates and ceases to exist, and it's at t = +R, x = 0. The "maximum expansion" point of the hole is at t = 0, and the hole's "edges" are at x = -R and x = +R at t = 0.

What this describes, to me, is the following: the hole comes into existence and is expanding, at that instant, at infinite speed. It gradually slows down its speed of expansion until it halts for an instant at "maximum expansion"; then it begins to contract again, at first very slowly, then faster and faster, until at the final instant it is imploding at infinite speed and then evaporates and disappears.

I don't know if this is what you intended to describe, but I see at least one major difference with what you seemed to be describing: the hole's expansion and contraction are not limited to the speed of light. I don't see how you could get around this without destroying the symmetry of the whole thing (i.e., without making the time dimension "special"), which kind of defeats your apparent purpose in constructing this model in the first place (to put the time dimension on the same footing as the others).

 

[Jake Minneman]

Physics do not distinguish past and future in terms of quantum theory.

 

[A-wal]

You're not specifying the scenario precisely enough, which is leading you into confusion. If you specify that the line of observers is separated by equal increments of the Schwarzschild exterior radial coordinate r, then as you get closer to the horizon, equal increments of r translate into larger and larger increments of actual proper distance, so yes, they would get more spread out. But that's because you specified the scenario as equal increments of r.

If you instead specify the scenario such that each observer in the line of observers fires his rockets in such a way as to maintain equal proper distance from the next observer "above" him, then that is also a perfectly consistent and realizable scenario, and in that scenario the observers do *not* "spread out" as you get closer to the horizon. They are able to maintain constant proper distance from each other, and they are able to do so using a constant acceleration (i.e., they feel a constant, unchanging weight--the weight gets larger for observers closer to the horizon).

I was describing the first one. They’re equally spread out from the perspective of the starting frame of the free-faller, and the first hoverer. They get further apart as you get pulled in. To reach the horizon you would have to move faster than light relative to the hoverers/riverbed.

Not if they're moving in different directions. An ingoing light ray is not at rest relative to an outgoing one, even if they both pass through the same event (same point in spacetime), any more than a light ray moving to the left is at rest relative to a light ray moving to the right.

That’s not how light rays move. They spread out radially, at c locally.

It's not a matter of "disproof". You are asserting that your version is the only *possible* version. To do that, it's not enough to assert your assumptions; you have to prove that they are logically necessary.

You’re using a finding of the standard model to suggest a finding of the model I’m using is wrong. If the model I’m using is right then the assumption of the standard model doesn’t apply anyway.

Great, then please answer the questions as I posed them (feel free to ignore what I said about Newtonian gravity if that doesn't fit with your model).

What questions? If I ignore the four about Newtonian gravity you haven’t asked me anything.

You can't reach a white hole horizon from the outside because it's an ingoing null surface; it's moving radially inward at the speed of light. So no matter how fast you move inward, you can't catch it.

That sounds familiar. That’s what a black hole does from the horizons point of view. Could you please list just the differences between a black hole and a white hole?

That's one possible manifestation of quantum tunnelling, but quantum tunnelling itself is a much more general concept; it basically covers any of a multitude of cases where some kind of transition that is impossible classically has a non-zero probability (even if the probability is very, very small) of happening when quantum mechanics is taken into account. Another example would be a particle quantum tunnelling through a potential barrier that is classically not passable; it's possible to model radioactivity this way.

In the particular case I was talking about, the idea is that, since iron (more precisely, a particular isotope of iron, which I believe is Fe-56 but I'm not certain going just from memory) has a lower energy than any other nucleus, there is a non-zero probability, quantum mechanically, that a nucleus of any other atom can quantum tunnel to become an iron nucleus. Classically, this can't happen because there is a huge potential barrier; it takes a lot of energy to rearrange the nucleons in a nucleus, even if the final energy of the rearranged state is lower. But quantum mechanically, because of the uncertainty principle, there is a (very, very small) probability that a quantum fluctuation could give any nucleus the (temporary) energy it needs to rearrange itself into an iron nucleus.

Thanks.

In general I'm not responding to simple repeated assertions of your assumptions, but you keep on saying this particular one so pardon my shouting: "TIDAL FORCE" IS *NOT* THE SAME AS "G FORCE"! The reason I focus in on this is that it's not just your assumption; it's a false statement of fact. Tidal gravity and "g force" (or "acceleration due to gravity") are two *different* physical phenomena. This is true even in Newtonian gravity.

Different how?

The particular question I keep asking (can you prove that your assumptions, which are false in standard GR, are logically necessary) is kind of rhetorical, because I don't think you can do it; your assumptions are *not* logically necessary, which is why there can be a consistent model (standard GR) in which they are false.

I’m not sure GR is entirely self-consistent.

However, there is something else you can do: tell me how your model would account for the four experimental observations I gave.

GR accounts for them. Are you saying that my model would give different predictions to GR that wouldn’t be consistent with those experimental observations? How so? I’m not saying space-time isn’t curved. I’m saying that’s one way of looking at it, and energy curves space-time at lot more in the opposite direction if you want to take that view.

No, feeling weight vs. not feeling weight is a genuine physical distinction--we can observe it and measure it. The fact that you don't think it makes a difference in your model does not mean it's not a genuine physical, observable distinction. In standard GR, it is a crucial distinction in the theory as well. If it isn't in your model, fine; explain how your model accounts for the four experimental observations I listed.

If you’re accelerating in flat space-time you could easily use a coordinate system in which you are at rest while the universe curves around you to move you into a new relative position.

What does "a constant source of energy close by" mean? Does it mean the two objects are being pushed by the source of energy, and therefore are accelerated and feel weight? Then see my previous comment above. Again, in standard GR, tidal gravity is exactly as I've defined it in the sentence you quoted. If it means something different in your model, fine; show me how your model accounts for the four experimental observations I listed.

If you’re not using energy to accelerate while in curved space-time you could easily use a coordinate system in which you and everything else are accelerating in flat space-time. Curvature, gravity and acceleration are exactly same thing.

Oh, if only that were true.

On re-reading, I realized that I should respond to this point as well. First, if the observer is traveling towards the horizon at just under c, the observer is not "hovering" at a constant r, which is what you specified at the start of the scenario. Please make up your mind what scenario you're talking about.

I haven't changed my mind. All observers except the in-faller start off equally spaced and maintain constant proper acceleration in order to hover.

Second, considering the observer falling towards the horizon at just under c, waaay back in post #230, I posted the calculation showing that the proper time to reach the horizon for an infalling observer is finite, as well as the proper distance to the horizon, for both a "hovering" observer and an infalling observer--with the infalling observer's distance being length-contracted relative to the "hovering" observer's distance, as expected. As those calculations show, the "length contraction" and "time dilation" as the horizon is approached are *not* sufficient to prevent an infalling observer from reaching the horizon.

They would have to be! Are you adding them together?

I even explained the reason back then, but I only did it explicitly in the case of time, so I'll restate it again for both time and distance. As the horizon is approached, "time dilation" means that the increment of proper time per unit of coordinate time t goes to zero while coordinate time t goes to infinity, while "length contraction" means that the increment of proper distance per unit of coordinate radius r goes to infinity while the increment of radius left to reach the horizon goes to zero. But in the case of time, the proper time per unit of coordinate time t goes to zero faster than t goes to infinity, while in the case of distance, the increment of radius left to the horizon goes to zero faster than the proper distance per unit of coordinate radius r goes to infinity. In both cases, the result is a finite sum, meaning a finite proper time and proper distance to the horizon.

Why would the proper time go to zero faster than t goes to infinity and the proper distance go to zero faster than the radius r goes to infinity?

But I thought you objected to switching coordinate systems? You've certainly objected to me and others using anything other than Schwarzschild coordinates.

No. I object to having to use multiple coordinate systems to describe the same thing. This is Schwarzschild coordinates. It’s the perspective of the horizon from the outside. It’s an impossible frame, like the perspective of light. The horizon itself doesn’t experience anything because it’s moving at c, so it only exists for one infinitely small moment in an infinitely small space, the singularity.

I can make sense of this statement in isolation, but I don't see how it leads to the description you go on to give of what the black hole looks like to someone "moving in time". It seems to me that your model of the black hole as a 4-D sphere would lead to the following description. I'll reduce it to just two dimensions (one of time and one of space) so it's easier to visualize.

Let R be the "spacetime radius" of the black hole. We'll use coordinates where the units of time and space are the same, so R is the radius of the black hole in both dimensions. Then, on a spacetime diagram, the hole just looks like a circle with radius R around the origin. The "bottom" of the circle (which would be the "South Pole" of the 4-D sphere if we added the other two dimensions back in) is at t = -R, x = 0. The "top" of the circle is then the event where the black hole evaporates and ceases to exist, and it's at t = +R, x = 0. The "maximum expansion" point of the hole is at t = 0, and the hole's "edges" are at x = -R and x = +R at t = 0.

What this describes, to me, is the following: the hole comes into existence and is expanding, at that instant, at infinite speed. It gradually slows down its speed of expansion until it halts for an instant at "maximum expansion"; then it begins to contract again, at first very slowly, then faster and faster, until at the final instant it is imploding at infinite speed and then evaporates and disappears.

I don't know if this is what you intended to describe, but I see at least one major difference with what you seemed to be describing: the hole's expansion and contraction are not limited to the speed of light. I don't see how you could get around this without destroying the symmetry of the whole thing (i.e., without making the time dimension "special"), which kind of defeats your apparent purpose in constructing this model in the first place (to put the time dimension on the same footing as the others).

It’s much simpler than that. It’s the same horizon that spreads outwards at c in flat space-time. It works the opposite way when gravity > c, until it moves far enough away that c > gravity again. Then it corrects itself by behaving normally and rushing away at the speed of light. The upshot is a black hole. I think what you call a white hole is the arse end of a black hole, when it rushes back after it’s reached the point when it’s no longer greater than c.

 

[PeterDonis]

I was describing the first one. They're equally spread out from the perspective of the starting frame of the free-faller, and the first hoverer. They get further apart as you get pulled in.

Suppose each hoverer sends out radar pings to measure his distance from the next hoverer above him. He satisfies himself that the distance he measures remains constant. Then he sends a radio message way out to a "bookkeeper" who is very, very far away from the hole, giving the distance he measures and the fact that it's holding constant. When the bookkeeper compares the distance measurements he receives from all the hoverers, will he find that (1) all those distance measurements are the same, or that (2) they increase as the hoverers get closer to the hole?

It seems to me that (1) is implied by saying that the hoverers are equally spread out. But if (1) is the case, then the hoverers *cannot* be separated by equal increments of the radial coordinate r, which is what "the first one" was that you said you were describing. If the hoverers are separated by equal increments of the radial coordinate r, then the bookkeeper must find (2) when he compares the measurements. But if (2) is the case, then there is *no* observer who will see the hoverers as "equally spread out" in the sense of their distance measurements all being the same (and I can't make sense of the phrase "equally spread out" any other way).

That's not how light rays move. They spread out radially, at c locally.

Are you being deliberately obtuse? Say I'm hovering high above the Earth, directly between the Earth and the Moon. Someone on the Moon sends a laser beam towards the Earth, and someone on the Earth sends a laser beam towards the Moon, in such a way that the two beams meet at my location. Therefore the two beams pass through the same event in spacetime, but they are moving in different directions, so their worldlines are different; the Moon to Earth beam is radially ingoing, and the Earth to Moon beam is radially outgoing. That sort of thing is what I was describing.

What questions? If I ignore the four about Newtonian gravity you haven't asked me anything.

The four questions I asked in that same post are the ones I would like you to answer. They were not "about" Newtonian gravity; I just remarked that you can't get the correct answer to any of them using just Newtonian gravity. You appear to agree with that, which is fine. Now please answer them, using your model.

That sounds familiar. That's what a black hole does from the horizons point of view. Could you please list just the differences between a black hole and a white hole?

A black hole horizon is an *outgoing* null surface. There is *no* "point of view" from which a black hole's horizon is moving inward.

A white hole horizon is an *ingoing* null surface. There is no point of view from which a white hole's horizon is moving outward.

Different how?

I've already answered this many, many times. Go back and read my previous posts.

I'm not sure GR is entirely self-consistent.

I understand that. I also understand that you have admitted you don't know enough math to be able to follow the proofs that it is self-consistent, which is why I haven't bothered much with this point.

GR accounts for them. Are you saying that my model would give different predictions to GR that wouldn't be consistent with those experimental observations?

We've already established that there is at least one experiment (the rope experiment we discussed earlier in the thread) for which your model gives a different prediction than GR. Also your model's picture of tidal gravity and your claim that "energy curves spacetime" don't appear consistent with standard GR. So I want you to explain those observations *without* using the machinery of standard GR, since you don't appear to understand that machinery. I want you to use *your* machinery, the machinery that leads you to say the things you keep saying about tidal gravity and energy curving spacetime, which are not consistent with standard GR. That means you can't use any of the results of standard GR; you have to derive your predictions using just your machinery.

I'm not saying space-time isn't curved. I'm saying that's one way of looking at it, and energy curves space-time at lot more in the opposite direction if you want to take that view.

And I'm saying that your claim that "energy curves spacetime...in the opposite direction" is not just a different "view" of what standard GR says; it's inconsistent with what standard GR says. I've already explained why several times.

If you're accelerating in flat space-time you could easily use a coordinate system in which you are at rest while the universe curves around you to move you into a new relative position.

And changing the coordinate system has no effect on the physics, as you yourself have said multiple times. If spacetime was flat in the original coordinates, it's still flat in the new coordinates.

If you're not using energy to accelerate while in curved space-time you could easily use a coordinate system in which you and everything else are accelerating in flat space-time. Curvature, gravity and acceleration are exactly same thing.

No, they're not. See previous comment; changing the coordinate system doesn't change the physics. If spacetime was curved in the original coordinates, it's still curved in the new ones.

Why would the proper time go to zero faster than t goes to infinity and the proper distance go to zero faster than the radius r goes to infinity?

Go back and read the computation I posted quite a while back. It's obvious from the integrals I wrote there, but to repeat briefly: the proper time goes to zero quadratically in t while t goes to infinity only linearly; the increment of radius left to the horizon goes to zero linearly in r while the increment of proper distance per unit of radius goes to infinity as the square root of r.

No. I object to having to use multiple coordinate systems to describe the same thing.

And yet in two of the earlier quotes that I commented on, you were arguing that you could do just that in order to show that curvature of spacetime and "curvature due to energy" were somehow the same.

It's much simpler than that. It's the same horizon that spreads outwards at c in flat space-time. It works the opposite way when gravity > c, until it moves far enough away that c > gravity again. Then it corrects itself by behaving normally and rushing away at the speed of light. The upshot is a black hole. I think what you call a white hole is the arse end of a black hole, when it rushes back after it's reached the point when it's no longer greater than c.

How does any of this correspond to your statement that the black hole is a 4-D sphere? I don't get it.

 

[Dale]

IMO, A-wal will never answer the questions that PeterDonis has asked because he does not have a working model in his mind, but he believes his preconceptions so strongly that he cannot learn about anything. He has had several people explain the same concept in many different ways, but his mental state is so rigidly inflexible that he has not made one bit of progress in over 300 posts.

A-wal, if you really want to learn, then you should stick with flat spacetime and try to understand the Rindler event horizon. Once you understand that then you may be able to add curvature and make some progress, but you will have to let go of a lot of your preconceptions first to even understand that much.

 

[A-wal]

Suppose each hoverer sends out radar pings to measure his distance from the next hoverer above him. He satisfies himself that the distance he measures remains constant. Then he sends a radio message way out to a "bookkeeper" who is very, very far away from the hole, giving the distance he measures and the fact that it's holding constant. When the bookkeeper compares the distance measurements he receives from all the hoverers, will he find that (1) all those distance measurements are the same, or that (2) they increase as the hoverers get closer to the hole?

An observer who is very, very far away would measure the distances as less than the starting frame of the in-faller and furthest hoverer, so let's have the book keeper in this frame. They see an equal distance between the hoverers, like I said. The first hoverer can be the bookkeeper.

It seems to me that (1) is implied by saying that the hoverers are equally spread out. But if (1) is the case, then the hoverers *cannot* be separated by equal increments of the radial coordinate r, which is what "the first one" was that you said you were describing. If the hoverers are separated by equal increments of the radial coordinate r, then the bookkeeper must find (2) when he compares the measurements. But if (2) is the case, then there is *no* observer who will see the hoverers as "equally spread out" in the sense of their distance measurements all being the same (and I can't make sense of the phrase "equally spread out" any other way).

It's 1. The hoverers have to expend different amounts of energy to maintain a constant separation from the perspective of the bookkeeper.

Are you being deliberately obtuse? Say I'm hovering high above the Earth, directly between the Earth and the Moon. Someone on the Moon sends a laser beam towards the Earth, and someone on the Earth sends a laser beam towards the Moon, in such a way that the two beams meet at my location. Therefore the two beams pass through the same event in spacetime, but they are moving in different directions, so their worldlines are different; the Moon to Earth beam is radially ingoing, and the Earth to Moon beam is radially outgoing. That sort of thing is what I was describing.

Not in this instance. The gravity and therefore the time dilation/length contraction due to gravity, or curvature if you prefer, and therefore the local speed of light would be the same for both, which was my original point.

The four questions I asked in that same post are the ones I would like you to answer. They were not "about" Newtonian gravity; I just remarked that you can't get the correct answer to any of them using just Newtonian gravity. You appear to agree with that, which is fine. Now please answer them, using your model.

I've mentioned making correct predictions before. What does your nice, simple model predict for the following:

(1) The precession of the perihelion of Mercury's orbit?

(2) The bending of light by the Sun?

(3) The changes in the orbits of binary pulsars due to the emission of gravitational waves?

(4) The precession of gyroscopes orbiting the Earth due to gravitomagnetism?

The standard GR model that predicts all these phenomena correctly *also* predicts that black holes will behave as I've been describing. That's why physicists believe in the standard GR model of black holes that I've been describing. If you can show how your model reproduces all these correct predictions *without* requiring black holes to behave as I've been describing, please do so. But you can't just wave your hands and say, "well, obviously my model looks just like GR outside the horizon", because the way GR arrives at all the above predictions is inseparably linked, mathematically, to the way it describes black holes and their horizons. So you have to start from scratch, and work through how your model would deal with the above phenomena, *without* making use of any of the machinery or results of GR.

(And no, you can't get any of the above results just by applying non-relativistic Newtonian gravitational theory. That's why I chose these examples.)

(1) I have no idea what precession of the perihelion means. I take it you mean explain why the observations are different from Newtonian predictions using my model. Newtonian calculations don’t account for length contraction/time dilation.
(2) It’s bent by gravity.
(3) Loss of energy in the system.
(4) What is gravitomagnetism?

I’m sorry if these answers seem deliberately obtuse. It’s because I’m having trouble working out what you’re getting at.

A black hole horizon is an *outgoing* null surface. There is *no* "point of view" from which a black hole's horizon is moving inward.

A white hole horizon is an *ingoing* null surface. There is no point of view from which a white hole's horizon is moving outward.

I thought so. The horizon rushes outwards at c. That's what you call a black hole. Then it rushes back at c, because it's a sphere. That's what you call a white hole. It can't hit you though. The first you'd see of it would be when it's rushing back at c (c if you account for length contraction and time dilation).

I've already answered this many, many times. Go back and read my previous posts.

I don’t see a single difference other than in appearance.

We've already established that there is at least one experiment (the rope experiment we discussed earlier in the thread) for which your model gives a different prediction than GR. Also your model's picture of tidal gravity and your claim that "energy curves spacetime" don't appear consistent with standard GR. So I want you to explain those observations *without* using the machinery of standard GR, since you don't appear to understand that machinery. I want you to use *your* machinery, the machinery that leads you to say the things you keep saying about tidal gravity and energy curving spacetime, which are not consistent with standard GR. That means you can't use any of the results of standard GR; you have to derive your predictions using just your machinery.

Why can't I use GR? I don't think GR is completely wrong. I think it's no more fundamental than SR and the two should be on an equal footing. So no greater than c travel or curvature that can't be curved back by energy, or any crap like that that may make sense to you mathematically but doesn't work conceptually. Maybe it's just that I can't conceptualise it properly, but if it's right then it can be done.

And I'm saying that your claim that "energy curves spacetime...in the opposite direction" is not just a different "view" of what standard GR says; it's inconsistent with what standard GR says. I've already explained why several times.

Of course it's inconsistent with what GR says. That's why it says that objects can’t cross an event horizon or break the light barrier. In fact it says they’re the same thing.

And changing the coordinate system has no effect on the physics, as you yourself have said multiple times. If spacetime was flat in the original coordinates, it's still flat in the new coordinates.

Flat is a matter of perspective.

No, they're not. See previous comment; changing the coordinate system doesn't change the physics. If spacetime was curved in the original coordinates, it's still curved in the new ones.

Curved is a matter of perspective.

Go back and read the computation I posted quite a while back. It's obvious from the integrals I wrote there, but to repeat briefly: the proper time goes to zero quadratically in t while t goes to infinity only linearly; the increment of radius left to the horizon goes to zero linearly in r while the increment of proper distance per unit of radius goes to infinity as the square root of r.

I don't understand it though. I understand what you’re saying but I don't understand why that description of what happens is the right one because I don't understand how to get to that equation.

And yet in two of the earlier quotes that I commented on, you were arguing that you could do just that in order to show that curvature of spacetime and "curvature due to energy" were somehow the same.

I don’t object to using multiple coordinate systems that are consistent with each other, because then it’s a different view of the exact same thing. What I do object to is having to use multiple coordinate systems to describe something completely. It means they contradict each other. The entire space time isn't covered? Where did it go?

How does any of this correspond to your statement that the black hole is a 4-D sphere? I don't get it.

It's description of what a four-dimensional bubble looks like from the perspective of something that perceives a moving time-line. It doesn't matter what angle you approach it from, including any angle of the arrow of time. It's the same from all sides.

IMO, A-wal will never answer the questions that PeterDonis has asked because he does not have a working model in his mind, but he believes his preconceptions so strongly that he cannot learn about anything. He has had several people explain the same concept in many different ways, but his mental state is so rigidly inflexible that he has not made one bit of progress in over 300 posts.

A-wal, if you really want to learn, then you should stick with flat spacetime and try to understand the Rindler event horizon. Once you understand that then you may be able to add curvature and make some progress, but you will have to let go of a lot of your preconceptions first to even understand that much.

I'm still not talking to you. You're too mean.

 

[Dale]

I'm still not talking to you. You're too mean.

I am too mean and you are too closed-minded. I think there is ample proof of both assertions.

Being mean doesn't prevent me from learning new things, but unfortunately for you being closed-minded does prevent you. If your goal is to learn, then you need to fix your attitude regardless of how bad my attitude is.

 

[PeterDonis]

A-wal, I'm not sure we're going to make much more progress, because you keep on saying things that are simply false from the standpoint of GR, yet you continue to think that you are somehow using GR, just in a different way. You're not. Whatever model you think you have in your head, it isn't GR. I'll illustrate with a few specific statements from your last post:

I don’t see a single difference other than in appearance.

You were saying this in response to my statement that tidal gravity and "acceleration due to gravity" were two different, distinct physical phenomena. I'm not going to repeat my explanation of that again; if you didn't get it the first half dozen times, you're not going to get it now. I only mention it to remark that, in GR (and even in Newtonian gravity), these two things *are* two different, distinct physical phenomena. So if you don't see a difference, then you don't understand GR (or even Newtonian gravity, in this instance).

Flat is a matter of perspective.

Curved is a matter of perspective.

This is about as close to a "money quote" as I think it's likely to get. In standard GR, both of these statements are false. Egregiously false. As in, "two plus two equals five" false. If you really believe these two statements, it's no wonder you don't understand GR. In GR, whether or not spacetime is flat or curved is a fundamental, invariant physical fact, and it is capable of direct physical measurement; I described how in several previous posts where I explained how to tell if tidal gravity is present--in GR, tidal gravity and spacetime curvature are the same thing, so detecting one is the same as detecting the other.

Once again: I'm not arguing, right now, that the statements I've made above, about what GR says, are "true". I think they are, but I'm not arguing that right now. Right now, I'm simply saying that the statements I've made above, about what GR says, *are* what GR says, and yet you don't believe them; you hold beliefs that are contrary to these basic statements of what GR says. (If you want confirmation that I'm right about what GR says, read Kip Thorne's Black Holes and Time Warps, or any of a number of relativity textbooks or other references.) So again, whatever model you think you have in your head, it isn't GR. It just isn't. Period.

***

I also have some comments on your last post that aren't really related to the above, but are just general clarifications or corrections:

(1) I have no idea what precession of the perihelion means. I take it you mean explain why the observations are different from Newtonian predictions using my model.

Yes. See the Wikipedia page on tests of GR for starters:

http://en.wikipedia.org/wiki/Tests_of_general_relativity

(2) It’s bent by gravity.

How much does gravity bend the light? How do I figure that out so I can compare the predicted amount of bending with the observed amount? An actual calculation would be nice, but I'll settle for now for a description of how such a calculation could be done.

(3) Loss of energy in the system.

By what mechanism? How does the energy get out of the system, and why is energy lost that way?

(4) What is gravitomagnetism?

It's what Gravity Probe B is trying to detect:

http://www.nasa.gov/vision/universe/solarsystem/19apr_gravitomagnetism.html

The Wikipedia page above on tests of GR also talks about it; another term that is used is "frame dragging".

Why can't I use GR? I don't think GR is completely wrong.

You don't even agree with the basic assumptions of GR; in fact, I'm not sure you understand them, for the reasons I gave at the start of this post. So whatever you think is not completely wrong, it isn't GR. In any case, whatever model you have in your head is *not* GR, as I showed above, so no, you can't use GR. You have to use your model, all by itself.

Not in this instance. The gravity and therefore the time dilation/length contraction due to gravity, or curvature if you prefer, and therefore the local speed of light would be the same for both, which was my original point.

Let me correct what you just said: *at the event where the ingoing and outgoing light beams meet*, the strength of gravity is the same for both. But they are traveling in different directions, so the strength of gravity is only the same for both beams at that one event; and since they are traveling in different directions, they will move into different regions of spacetime, where the strength of gravity is different.

I don't understand it though. I understand what you’re saying but I don't understand why that description of what happens is the right one because I don't understand how to get to that equation.

Yes, I know you don't. But you asked the question and I gave you the answer. I don't think there's any quick way to explain to you "how to get to that equation", because you don't even agree with the basic assumptions of GR to start with, as I've shown above. But for the record, what I said is in fact the answer, according to standard GR.

I don’t object to using multiple coordinate systems that are consistent with each other, because then it’s a different view of the exact same thing. What I do object to is having to use multiple coordinate systems to describe something completely. It means they contradict each other. The entire space time isn't covered? Where did it go?

You don't have to use multiple coordinate systems to describe the spacetime around a black hole completely. Kruskal coordinates do it all by themselves. The point I and others have been making is that Schwarzschild coordinates don't.

It's description of what a four-dimensional bubble looks like from the perspective of something that perceives a moving time-line. It doesn't matter what angle you approach it from, including any angle of the arrow of time. It's the same from all sides.

Then you'd better define what you mean by "four-dimensional bubble", because I don't see how what you're describing is at all related to the obvious definition of that term, which is what I gave in a previous post. A description of the bubble as it would appear on a space-time diagram, or even just a two-dimensional projection of it, with one time and one space coordinate, like the one I gave in a previous post, would help.

 

[A-wal]

A-wal, I'm not sure we're going to make much more progress, because you keep on saying things that are simply false from the standpoint of GR, yet you continue to think that you are somehow using GR, just in a different way. You're not. Whatever model you think you have in your head, it isn't GR. I'll illustrate with a few specific statements from your last post:

I've promoted Special Relativity to the level of General Relativity because I think they're equivalent. I call it General Special Relativity. More specifically I think acceleration and gravity, and the event horizon of c in flat space-time and the event horizon of a black hole are equivalent. The model I'm using is General Special Relativity in tandem with General Relativity. I shall call it Special General Special Relativity.

You were saying this in response to my statement that tidal gravity and "acceleration due to gravity" were two different, distinct physical phenomena. I'm not going to repeat my explanation of that again; if you didn't get it the first half dozen times, you're not going to get it now. I only mention it to remark that, in GR (and even in Newtonian gravity), these two things *are* two different, distinct physical phenomena. So if you don't see a difference, then you don't understand GR (or even Newtonian gravity, in this instance).

No, I think tidal gravity and acceleration in flat space-time are equivalent. Acceleration due to gravity is a change in relative velocity because of gravity.

This is about as close to a "money quote" as I think it's likely to get. In standard GR, both of these statements are false. Egregiously false. As in, "two plus two equals five" false. If you really believe these two statements, it's no wonder you don't understand GR.

You do it all the time, with world lines. Objects don't follow curved lines from their own perspective. They follow straight lines through curved space-time. That's making curvature disappear. Do that in reverse so that they become curved lines in flat space-time.

In GR, whether or not spacetime is flat or curved is a fundamental, invariant physical fact, and it is capable of direct physical measurement; I described how in several previous posts where I explained how to tell if tidal gravity is present--in GR, tidal gravity and spacetime curvature are the same thing, so detecting one is the same as detecting the other.

Yes tidal gravity and curvature are the same thing, and so is acceleration. It's just like when you accelerate in flat space-time, there's no difference. That's what I meant when I said you could construct a coordinate system where the curvature disappears and view it purely as acceleration in flat space-time. In flat space-time the horizon of the universe and a black/white hole becomes a Rindler horizon, and no it can't be crossed.

Once again: I'm not arguing, right now, that the statements I've made above, about what GR says, are "true". I think they are, but I'm not arguing that right now. Right now, I'm simply saying that the statements I've made above, about what GR says, *are* what GR says, and yet you don't believe them; you hold beliefs that are contrary to these basic statements of what GR says. (If you want confirmation that I'm right about what GR says, read Kip Thorne's Black Holes and Time Warps, or any of a number of relativity textbooks or other references.) So again, whatever model you think you have in your head, it isn't GR. It just isn't. Period.

Just because I think it can be looked at in more than one way doesn't mean I think it's wrong.

Yes. See the Wikipedia page on tests of GR for starters:

http://en.wikipedia.org/wiki/Tests_o...ral_relativity

Homework? Seriously? Do you know how many links I've followed or things I've looked up during this thread? One. The Rindler horizon. I've already said that the Newtonian model doesn't use time dilation and length contraction. Please tell me what you're getting at.

How much does gravity bend the light? How do I figure that out so I can compare the predicted amount of bending with the observed amount? An actual calculation would be nice, but I'll settle for now for a description of how such a calculation could be done.

Use acceleration in flat space-time. When an object accelerates it alters the speed of light so that it's no longer c from the perspective of the accelerator. This is the equivalent to viewing it as curved. The sun is a constant source of energy that accelerates anything in the system by an amount depending on how much mass/energy the sun has and on the diffusion of the strength of the original field, which is an inverse square. That's because the energy is being spread evenly throughout the spatial dimensions. In zero dimensions it would be infinite, in one it would never decrease, in two it would be directly proportionate to the distance, and in three it's an inverse square.

By what mechanism? How does the energy get out of the system, and why is energy lost that way?

A gravity wave is just a change in the strength of the field, which moves at c. If the field has weakened it means there's less mass in the system so matter is being converted into energy.

It's what Gravity Probe B is trying to detect:

http://www.nasa.gov/vision/universe/...magnetism.html

The Wikipedia page above on tests of GR also talks about it; another term that is used is "frame dragging".

I thought fame dragging was just spin. The object has a relative velocity without moving anywhere. It's still undergoing time-dilation and length contraction as if it was linear velocity, but because it's angular velocity instead and the object isn't actually going anywhere it act like time-dilation and length contraction from gravity, and that's tidal force. You could probably use that stress-energy tensor or whatever you called it to view the spin as energy and explain it as real gravity rather than spin.

You don't even agree with the basic assumptions of GR; in fact, I'm not sure you understand them, for the reasons I gave at the start of this post. So whatever you think is not completely wrong, it isn't GR. In any case, whatever model you have in your head is *not* GR, as I showed above, so no, you can't use GR. You have to use your model, all by itself.

No I don't because I'm not saying the GR description is wrong. I'm saying you could construct a coordinate system where the curvature disappears. I'll pick on SR instead if you want. You could easily construct a coordinate system where there is no acceleration and the universe is curved through energy. Do that but the opposite for gravity, so that it's just energy that accelerates things rather than curvature.

Let me correct what you just said: *at the event where the ingoing and outgoing light beams meet*, the strength of gravity is the same for both. But they are traveling in different directions, so the strength of gravity is only the same for both beams at that one event; and since they are traveling in different directions, they will move into different regions of spacetime, where the strength of gravity is different.

Yes. That's what I was saying. Correct what I said how? You just reworded it. The direction they're moving in has no impact on the properties of the space-time they occupy so they can't behave differently. It's not as if they carry inertia. If light can't get out, it can't get in.

Yes, I know you don't. But you asked the question and I gave you the answer. I don't think there's any quick way to explain to you "how to get to that equation", because you don't even agree with the basic assumptions of GR to start with, as I've shown above. But for the record, what I said is in fact the answer, according to standard GR.

I do agree with the basic assumptions of GR. More than you even. Like there being no difference between tidal gravity and acceleration in space-time for example. I also think it can be looked at in more than one way. It's relative.

You don't have to use multiple coordinate systems to describe the spacetime around a black hole completely. Kruskal coordinates do it all by themselves. The point I and others have been making is that Schwarzschild coordinates don't.

Then they contradict each other, simple as that. And don't say the entire space-time isn't covered in one of them. That doesn't actually mean anything other than either the two coordinate systems contradict each other and therefore one of them is wrong, or one of them is incomplete. But it should be possible to put the entire space-time into the coordinate system. You can't because it doesn't exist and any coordinate system that does include a traversable event horizon is just plain wrong, imo.

Then you'd better define what you mean by "four-dimensional bubble", because I don't see how what you're describing is at all related to the obvious definition of that term, which is what I gave in a previous post. A description of the bubble as it would appear on a space-time diagram, or even just a two-dimensional projection of it, with one time and one space coordinate, like the one I gave in a previous post, would help.

I can make sense of this statement in isolation, but I don't see how it leads to the description you go on to give of what the black hole looks like to someone "moving in time". It seems to me that your model of the black hole as a 4-D sphere would lead to the following description. I'll reduce it to just two dimensions (one of time and one of space) so it's easier to visualize.

Let R be the "spacetime radius" of the black hole. We'll use coordinates where the units of time and space are the same, so R is the radius of the black hole in both dimensions. Then, on a spacetime diagram, the hole just looks like a circle with radius R around the origin. The "bottom" of the circle (which would be the "South Pole" of the 4-D sphere if we added the other two dimensions back in) is at t = -R, x = 0. The "top" of the circle is then the event where the black hole evaporates and ceases to exist, and it's at t = +R, x = 0. The "maximum expansion" point of the hole is at t = 0, and the hole's "edges" are at x = -R and x = +R at t = 0.

Okay.

What this describes, to me, is the following: the hole comes into existence and is expanding, at that instant, at infinite speed. It gradually slows down its speed of expansion until it halts for an instant at "maximum expansion"; then it begins to contract again, at first very slowly, then faster and faster, until at the final instant it is imploding at infinite speed and then evaporates and disappears.

Why would it slow down? It's a sphere, not some weird four-dimensional oblong type shape.

I don't know if this is what you intended to describe, but I see at least one major difference with what you seemed to be describing: the hole's expansion and contraction are not limited to the speed of light. I don't see how you could get around this without destroying the symmetry of the whole thing (i.e., without making the time dimension "special"), which kind of defeats your apparent purpose in constructing this model in the first place (to put the time dimension on the same footing as the others).

Not limited to the speed of light, why?

I am too mean and you are too closed-minded. I think there is ample proof of both assertions.

Being mean doesn't prevent me from learning new things, but unfortunately for you being closed-minded does prevent you. If your goal is to learn, then you need to fix your attitude regardless of how bad my attitude is.

You serious? I've tried to get my head round the standard descriptions time and again. If there's something I'm not getting then I would have thought it would have been shown to me by now because there are people here more interested in helping others understand than they are in trying to feel good about themselves. It seems like whenever I come up with anything that refutes your scripture you and the others like you do one of three things: 1). Ignore it completely. 2). Make some hand-wavy comment in an attempt to marginalise the problem. "The entire space-time isn't covered"/"We need to be careful when drawing conclusions about when things happen when they refute the way we say it works" 3). Lash out like a spoilt child. And you've got the nerve to call me close-minded. I bet you only come here to try to make yourself feel important. Grow up and get some self-esteem. Until then, cough see you next Tuesday.

 

[Dale]

If there's something I'm not getting then I would have thought it would have been shown to me by now

You have been shown over and over and over and over by multiple different people each making a good faith effort to explain, and explain again, and try again in a different way. It is pointless.

It seems like whenever I come up with anything that refutes your scripture you and the others like you do one of three things: 1). Ignore it completely. 2). Make some hand-wavy comment in an attempt to marginalise the problem. "The entire space-time isn't covered"/"We need to be careful when drawing conclusions about when things happen when they refute the way we say it works" 3). Lash out like a spoilt child.

2) and 3) I agree with, but 2) is required by you since you cannot understand the math and have completely and immediately dismissed any math which has been presented. 1) is simply not correct, what comment do you feel has been ignored completely? Please go through the thread and find an example of an important (i.e. "refutes [my] scripture") comment that was ignored completely.

FYI, that "refutes your scripture" comment is a very typical crackpot statement

 

[PeterDonis]

A-wal, once again, I don't think we're going to make much more progress, because you keep on claiming you are somehow using GR, but you don't even agree with the basic principles of GR. Whatever model you think you have in your head, it is not GR and doesn't use any part of GR. So I won't bother responding, for the most part, except to point out the specific points where you contradict GR.

No, I think tidal gravity and acceleration in flat space-time are equivalent.

And in GR, they are *not*. If you believe they are equivalent, you are contradicting one of the basic principles of GR.

Objects don't follow curved lines from their own perspective. They follow straight lines through curved space-time. That's making curvature disappear. Do that in reverse so that they become curved lines in flat space-time.

Nope, none of this is consistent with GR. It may well be part of your model, but it isn't consistent with GR.

Yes tidal gravity and curvature are the same thing, and so is acceleration.

Nope, in GR acceleration is *not* the same as tidal gravity/curvature of spacetime.

In flat space-time the horizon of the universe and a black/white hole becomes a Rindler horizon, and no it can't be crossed.

Wrong two ways. The "horizon of the universe" is *not* analogous to a Rindler horizon, and a Rindler horizon *can* be crossed. Again, what you say may be part of your model, but it isn't GR (or even SR, in this case, since a Rindler horizon and the fact that it can be crossed can be analyzed purely in SR, since it's in flat spacetime).

Homework? Seriously? Do you know how many links I've followed or things I've looked up during this thread?

And I should care about this why?

When an object accelerates it alters the speed of light so that it's no longer c from the perspective of the accelerator. This is the equivalent to viewing it as curved. The sun is a constant source of energy that accelerates anything in the system by an amount depending on how much mass/energy the sun has and on the diffusion of the strength of the original field, which is an inverse square.

As far as I can make sense of this at all, it basically tells me that you think the bending of light should be proportional to the strength of gravity (the "amount of acceleration") at the closest point of approach of the light beam to the surface of the sun. Is that what you're thinking? Also, how does the speed of the light changing affect the amount of bending?

A gravity wave is just a change in the strength of the field, which moves at c. If the field has weakened it means there's less mass in the system so matter is being converted into energy.

What causes the field to weaken?

You could probably use that stress-energy tensor or whatever you called it to view the spin as energy and explain it as real gravity rather than spin.

But the stress-energy tensor is part of GR, and your model contradicts GR, so you can't use it.

No I don't because I'm not saying the GR description is wrong. I'm saying you could construct a coordinate system where the curvature disappears.

And, as I keep on telling you, you *can't do that* in GR, because in GR, curvature is an invariant; it doesn't depend on the coordinate system you use. You *can't* make it "disappear" by choosing coordinates. Again, your statement here contradicts a basic principle of GR, so you *are* saying GR is wrong.

The direction they're moving in has no impact on the properties of the space-time they occupy so they can't behave differently.

Even if they move into different *regions* of spacetime, where the properties of spacetime *are* different?

It's not as if they carry inertia.

If you mean the light beams, you're wrong, they do carry inertia. Everything that has energy has inertia in GR.

I do agree with the basic assumptions of GR.

Then you've been doing a great job of hiding it, by continuing to make statements that contradict basic principles of GR, even after I've pointed out the contradictions multiple times. The statements you keep making are not just "alternate views" that you can somehow finesse; they are flat out contradictions of GR, pure and simple.

I understand that this stuff is hard for you to grasp. But I have to call things the way I see them, and what I see is that you have a model in your head that contradicts GR, no matter how you slice it. You don't *think* it does, but that's because the model in your head is muddled, not because GR somehow admits the interpretations you're trying to put on it. That's the way I see it.

Why would it slow down? It's a sphere, not some weird four-dimensional oblong type shape.

It's a sphere in *spacetime* (or a circle in the two-dimensional projection I described). That means its "shape" determines how it moves, not just how it looks in space. You apparently don't understand the implications of your own model, let alone GR.

 

[PeterDonis]

And I should care about this why?

On thinking over this particular comment, I realized I should probably amplify it. I know it seems snarky as it stands, but it's actually a serious question, which I'll amplify as follows:

How do you think I learned about this stuff? Do you think I had somebody just drop a ready-made explanation of GR in my lap? Do you think I had somebody sit down and patiently go through all the key experimental results? I've never even taken a formal course in GR; I had to learn it all on my own. (I have taken one formal course, if you want to call it that--it was a summer course, not for credit--in SR, and I've taken a number of courses that used SR indirectly, but even those courses expected me to do a lot of the work.) I got interested in GR when I was in graduate school; one of my office mates had a copy of Misner, Thorne, & Wheeler's Gravitation on his desk, and one day we got talking about it, and I thought, this seems like a neat subject, I think I'll look into it. So I borrowed the book and started reading it and trying to work the problems.

I'm not saying I recommend this particular procedure as the best way of learning about GR, you understand; but that's how I got started. Nor am I trying to say that I got anywhere close to working even a substantial fraction of all the problems in MTW--*that* I would only have done under the pressure of an actual class. ;-)

Nowadays it's a lot easier, because there are good resources on the web (many of which have been linked to in this thread), and there is at least one very good layman's book, Kip Thorne's Black Holes and Time Warps, which I've referred to before. There's also an updated version out now of Taylor & Wheeler's Spacetime Physics (which is the book, in its original edition, that I first learned SR from, but it covers the basics of GR as well). There are also plenty of sites (such as Living Reviews in Relativity or the Usenet site on Experimental Tests of Relativity, which have been linked to in this thread) that discuss the key experimental results. Heck, even Wikipedia has decent discussions of many of them (though you do have to be careful with Wikipedia).

The point I'm trying to make is that if you come to a forum like this and make claims like those you've been making, and say things that make it obvious that you don't understand even the basics of relativity, or know about experimental facts that are talked about all over the place when relativity is discussed, don't be surprised if we start referring you to the material that's already out there instead of explaining everything from scratch here. If you read the material and have something specific in it that you can't understand, then by all means come here and post a specific question with a specific reference, so we know what you're asking about. But don't expect us to do all the hard work for you. Often there's no substitute for slogging through it yourself.

 

[A-wal]

A-wal, once again, I don't think we're going to make much more progress, because you keep on claiming you are somehow using GR, but you don't even agree with the basic principles of GR. Whatever model you think you have in your head, it is not GR and doesn't use any part of GR. So I won't bother responding, for the most part, except to point out the specific points where you contradict GR.

You're going to a lot of trouble to explain this stuff to someone who you think is a lost cause. You've started making claims that simply aren't true. I'm not sure if you're trying to make me look bad, but it's not me that looks bad when you do that. Doesn't use any part of GR? Yes it does.

And in GR, they are *not*. If you believe they are equivalent, you are contradicting one of the basic principles of GR.

I thought that was one of the basic principles of GR. Close enough anyway. I modified it a bit.

Nope, none of this is consistent with GR. It may well be part of your model, but it isn't consistent with GR.

Now I'm confused. I thought the GR view was of straight lines (geodesics?) in curved space-time. From a distance the path of an object looks curved, but to the object it's they're traveling in a straight line and space-time is curved.

Nope, in GR acceleration is *not* the same as tidal gravity/curvature of spacetime.

Very odd. I screams it to me.

Wrong two ways. The "horizon of the universe" is *not* analogous to a Rindler horizon, and a Rindler horizon *can* be crossed. Again, what you say may be part of your model, but it isn't GR (or even SR, in this case, since a Rindler horizon and the fact that it can be crossed can be analyzed purely in SR, since it's in flat spacetime).

I'll explain what I was thinking. The Rindler can't be crossed from the perspective of the inertial observer, but the accelerator can move normally. That doesn't mean they can actually cross the horizon though because time dilation and length contraction mean that even though the accelerator isn't restricted in any way from their own perspective, time dilation and length contraction mean that they are restricted from the perspective of the inertial observer. So the edge of the universe is an event horizon.

And I should care about this why?

I didn't say you should. I was wrong anyway. I did look at a couple of things that were suggested come to think of it. Can't remember what they were though. I'm just saying that's no way to learn. Not for me anyway.

As far as I can make sense of this at all, it basically tells me that you think the bending of light should be proportional to the strength of gravity (the "amount of acceleration") at the closest point of approach of the light beam to the surface of the sun. Is that what you're thinking? Also, how does the speed of the light changing affect the amount of bending?

The speed of light changing in flat space-time is the equivalent of bending in curved space-time. I think the bending of light should be proportional to the amount of energy/acceleration in that area, which can be worked out by the mass and distance of the source. Gravity as a force in flat space-time still uses time dilation and length contraction, but through acceleration, not curved space-time. I doubt if Newtons equations took into account the fact that the speed of light/time effectively changes through acceleration relative to an inertial observer. I'm not saying GR is wrong and space-time isn't curved. I'm not even offering an alternative. I think the two are the same because whether you view it as acceleration in flat space-time or curvature makes no difference. If the space between two objects changes over time does that mean that the objects themselves moved, or does it mean the space-time between them is curved? It's exactly the same.

What causes the field to weaken?

Loss of mass. Matter being converted into energy.

But the stress-energy tensor is part of GR, and your model contradicts GR, so you can't use it.

What do you mean I can't use it? Why not? And that wasn't my explanation, it was just a comment that you could do it that way. Why are you being so awkward?

And, as I keep on telling you, you *can't do that* in GR, because in GR, curvature is an invariant; it doesn't depend on the coordinate system you use. You *can't* make it "disappear" by choosing coordinates. Again, your statement here contradicts a basic principle of GR, so you *are* saying GR is wrong.

NO I'M NOT! I'm saying it's not the whole truth. It's true within it's own context, but GR is basically a description of gravity using the mechanism of curved space-time so of course anything that deviates from this is not GR. The point I've been trying to make is that I don't think that's the only way of doing it. But a different way of doing it would have to amount to the exact same thing in practice or they'd contradict each other. If I'm right about not being able to reach the event horizon then GR with it's curved space-time should describe this. It doesn't and I'm trying to work out why.

Even if they move into different *regions* of spacetime, where the properties of spacetime *are* different?

Then they no longer occupy the same area of space-time.

If you mean the light beams, you're wrong, they do carry inertia. Everything that has energy has inertia in GR.

But they don't move relatively. They're constant, so how do arrive at a value for inertia? I suppose that's constant as well.

Then you've been doing a great job of hiding it, by continuing to make statements that contradict basic principles of GR, even after I've pointed out the contradictions multiple times. The statements you keep making are not just "alternate views" that you can somehow finesse; they are flat out contradictions of GR, pure and simple.

If you really believe that I'm flat out contradicting relativity then it only proves that you either don't really understand the concepts properly or you're not listening, or I'm not explaining myself properly.

I understand that this stuff is hard for you to grasp. But I have to call things the way I see them, and what I see is that you have a model in your head that contradicts GR, no matter how you slice it. You don't *think* it does, but that's because the model in your head is muddled, not because GR somehow admits the interpretations you're trying to put on it. That's the way I see it.

I don't agree, so I don't understand? It's actually quite easy to understand, mostly. Galilean Relativity: There's no such thing as absolute movement. Only relative movement makes sense. Special Relativity: The speed of light is constant so the only way everything can remain consistent for all observers is if either space or time are just as relative as movement. It's split evenly between both. Time and one spatial dimension because any two objects are separated spatially by a one-dimensional line. General Relativity: Includes constant inwards acceleration from gravity, expressed in the form of curved space-time (distances between objects changing rather than movement of the objects themselves) because gravity tends to stick around for a bit. Could also be expressed in flat space-time, although you would obviously still have length contraction and time dilation. If that's what you class as curvature then it is curved whichever way you choose to look at it. There's certain things I've highlighted that I can't get my head around, but I believe that's because it can't be done (although maybe it just can't be done by me yet). It's very hard to agree with when it keeps contradicting itself. I think GR is muddled. That's the way I see it.

It's a sphere in *spacetime* (or a circle in the two-dimensional projection I described). That means its "shape" determines how it moves, not just how it looks in space. You apparently don't understand the implications of your own model, let alone GR.

THAT'S EXACTLY WHAT I MEANT! If it's a sphere then it's "movement" should remain constant. What you're describing is an oblong because it's speed changes with time. That's not a sphere.

How do you think I learned about this stuff? Do you think I had somebody just drop a ready-made explanation of GR in my lap? Do you think I had somebody sit down and patiently go through all the key experimental results? I've never even taken a formal course in GR; I had to learn it all on my own. (I have taken one formal course, if you want to call it that--it was a summer course, not for credit--in SR, and I've taken a number of courses that used SR indirectly, but even those courses expected me to do a lot of the work.) I got interested in GR when I was in graduate school; one of my office mates had a copy of Misner, Thorne, & Wheeler's Gravitation on his desk, and one day we got talking about it, and I thought, this seems like a neat subject, I think I'll look into it. So I borrowed the book and started reading it and trying to work the problems.

I could learn it that why, but I don't see the point. Plus I like the back and forth of a debate. It helps me think. I'd switch off and not take it in if just read about it. I have read a couple of laymen books but I'm not going to learn much more that way. I could read something slightly more advanced but I don't want to start thinking of it like that because I think it takes you away from understanding something and into memorising it. I could even learn the equations if I wanted but I simply get turned off by them. If you can't explain it using words then you don't understand the equations. Or at least you don't fully understand what they represent.

I'm not saying I recommend this particular procedure as the best way of learning about GR, you understand; but that's how I got started. Nor am I trying to say that I got anywhere close to working even a substantial fraction of all the problems in MTW--*that* I would only have done under the pressure of an actual class. ;-)

Memorising and learning aren't the same things. Proving you can work through any equation known to Man doesn't prove that you understand anything.

Nowadays it's a lot easier, because there are good resources on the web (many of which have been linked to in this thread), and there is at least one very good layman's book, Kip Thorne's Black Holes and Time Warps, which I've referred to before. There's also an updated version out now of Taylor & Wheeler's Spacetime Physics (which is the book, in its original edition, that I first learned SR from, but it covers the basics of GR as well). There are also plenty of sites (such as Living Reviews in Relativity or the Usenet site on Experimental Tests of Relativity, which have been linked to in this thread) that discuss the key experimental results. Heck, even Wikipedia has decent discussions of many of them (though you do have to be careful with Wikipedia).

I don't think I'd find anything that would answer my questions. If I thought it would help I'd do it.

The point I'm trying to make is that if you come to a forum like this and make claims like those you've been making, and say things that make it obvious that you don't understand even the basics of relativity, or know about experimental facts that are talked about all over the place when relativity is discussed, don't be surprised if we start referring you to the material that's already out there instead of explaining everything from scratch here. If you read the material and have something specific in it that you can't understand, then by all means come here and post a specific question with a specific reference, so we know what you're asking about. But don't expect us to do all the hard work for you. Often there's no substitute for slogging through it yourself.

When I say I don't understand it doesn't mean I don't get it. It means I don't understand how it could work that way. Don't even understand the basic concepts? Do you think I've been blagging it this whole time? I'm going to say this again because I think it's a very important point: Just because someone doesn't agree with you doesn't mean they don't understand what you're saying. It does sometimes and it does a lot of the time with relativity because some people don't understand the concepts. They're a bit trippy at first. But I think I've proved I understand the concepts. One of the reasons I posted those blogs was to show that I do at least understand the basic concepts of relativity. Here's the problem. I think it can be expanded. You're interpreting this as a contradiction of GR. In a way it is, but it's not as simple as that. It's true within its own context, but GR is a description using curved space-time, so of course anything that doesn't look at it as curved isn't GR. That doesn't mean I'm saying GR is wrong. I think it's right except that if you want to look at gravity as curved space-time then you have to do exactly the same for acceleration as well. I think of it more as an expansion of GR rather than a contradiction of it. I'm not trying to get anyone to do any work for me, other than trying to help me see if there really is something I've been missing over and over again.

You have been shown over and over and over and over by multiple different people each making a good faith effort to explain, and explain again, and try again in a different way. It is pointless.

Because I haven't had satisfactory answers yet. It maybe because I can't get my head round it. I don't think so though because I understand what I'm being told, I just don't agree. It could be because it hasn't been explained in a way that I understand or it could be that what I'm being told doesn't add up.

2) and 3) I agree with, but 2) is required by you since you cannot understand the math and have completely and immediately dismissed any math which has been presented. 1) is simply not correct, what comment do you feel has been ignored completely? Please go through the thread and find an example of an important (i.e. "refutes [my] scripture") comment that was ignored completely.

I will. I was going to do that anyway at some point. I'm going to go through it from the beginning and pick up some of the things I let go because I wanted to move on and didn't want to stick the knife in at the time.

FYI, that "refutes your scripture" comment is a very typical crackpot statement

If you say so. It was a bit of a cheap shot.


I'm going to try something else because this is starting to not work. It's mainly because I'm not taking any time on my posts. I'm too impatient and I do it at work when I've got other stuff to do and I keep getting interrupted. I'm just writing it then posting it. You're getting the first thing that comes to mind. I'll take my time from now on and won't post every half-thought that enters my head. I'll try to make future posts more focused and be clearer about what I mean. I'll also try to be a bit less cocky and petulant when I disagree with something, although I can't make any promises. Conflict does breed creativity but so does peace and quiet. I'll try asking questions instead of making assertions. Hopefully me and Dalespam can be friends again. Group hug.

 

[PeterDonis]

You're going to a lot of trouble to explain this stuff to someone who you think is a lost cause.

I didn't say I thought you were a lost cause; I said I didn't think we could make much further progress when we're not even using words the same way. You think you're saying the same things that GR is saying, but just "modified a bit". You think you're just taking GR and tweaking it some to make it fit better with the model you have in your head. *I* think you are flat out contradicting GR; you are building a model in your head that's based on premises that are simply false in GR, and so naturally what comes out looks like nonsense to me.

Right now I'm not even trying to argue which one of us is right. I'm saying that we can't even communicate if we can't agree on the meanings of words. You're using them one way and I'm using them another. Unless we can agree on what the words we're using mean, we're stuck.

The key words that seem to me to be causing confusion are "flat" and "curved" with respect to spacetime itself, and "straight" and "curved" with respect to individual worldlines. Again, please understand: I'm not trying to argue, right now, whether the standard model of GR is right or wrong. I'm simply saying that these are the standard definitions of these words in GR; they are used to refer to the precise physical concepts/measurements I'm going to describe. If you use the words any other way, you're not using them the way standard GR uses them.

Spacetime is flat if no tidal gravity is present. It is curved if tidal gravity is present. Tidal gravity is physically measured as follows: take two nearby, point-like (i.e., no internal structure), freely falling objects which are at rest with respect to each other at some instant. If they remain at rest with respect to each other, spacetime is flat; if they do not (i.e., if they either get closer together or farther apart as time passes), spacetime is curved.

A worldline is straight if an observer traveling along the worldline is freely falling--i.e., feels no weight. A worldline is curved if an observer traveling along the worldline feels weight--i.e., is not freely falling. The word "accelerated" may also be used to refer to a worldline that is curved, or an observer traveling on such a worldline. However, it is important to remember that this "acceleration" (which is more precisely called "proper acceleration") is *not* the same as "coordinate acceleration", which can be present for freely falling bodies if you adopt non-inertial coordinates. For example, a freely falling rock has "coordinate acceleration" with respect to an observer standing at rest on the surface of the Earth, even though the rock's worldline is straight and the observer's worldline is curved, in the sense of the above definitions of those terms.

If you're going to claim to be using GR, you have to use these words with the above definitions. If you don't--if, for example, you say that tidal gravity and acceleration, in the sense of "proper acceleration", look the same to you--then you're simply saying something that's false (obviously false, from the definitions above), and I don't see the point of responding. If you want to argue for some other model where tidal gravity and "acceleration" (with some other definition you will need to supply for that term) *are* the same, then go ahead, but please first give me a precise definition of what you mean by "acceleration" and how I can measure it physically, so I can see if I agree that it can be "the same" as tidal gravity. (Or for that matter, if you are thinking of something different than what I defined above when you use the term "tidal gravity", then please define it precisely and use some other term, so we can be clear what we're talking about.)

The speed of light changing in flat space-time is the equivalent of bending in curved space-time.

If you took out the phrase "in flat spacetime", referring to the speed of light changing, and just used this as a description of something that happens in curved spacetime, I would buy it as acceptable (though I still think it's a confusing way of describing how curved spacetime works). But by including the phrase "in flat spacetime", you ruined it. In flat spacetime, the speed of light is the same everywhere. The light cones line up with each other everywhere. There is no tidal gravity, no spacetime curvature. If you think you can somehow finesse "curved" out of that, then you are *not* using the word "curved" the way GR defines it. If you insist on arguing for some model where somehow the statement above makes sense to you, then please give precise definitions of what you mean by the terms, since you obviously can't be using them in the standard senses I defined above.

I'm not saying GR is wrong and space-time isn't curved. I'm not even offering an alternative. I think the two are the same because whether you view it as acceleration in flat space-time or curvature makes no difference.

If you meant "curvature" as in "a curved worldline", then this would be OK, since an accelerated worldline (in the sense of "proper acceleration") is a curved worldline (by the definition I gave above), whether spacetime itself is flat or curved. But from the context (and from all the times you've said things like this before), it's evident to me that you meant "curvature" as in "curved spacetime", and that is *not* the same, or analogous to, or "another view of" acceleration of a worldline, as the terms are used in GR (and as should be obvious from my definitions above).

THAT'S EXACTLY WHAT I MEANT! If it's a sphere then it's "movement" should remain constant. What you're describing is an oblong because it's speed changes with time. That's not a sphere.

Draw a spacetime diagram, as I suggested. Use just one dimension of space and one of time. Draw a circle (which is the projection of a 4-D sphere into the two dimensions of the diagram), with its center at the origin, and equal radius in all "directions" in the diagram. That means, for example, that the circle passes through the points t = -R, x = 0; t = 0, x = -R; t = 0, x = R; and t = R, x = 0 (where R is the radius of the circle). That is what a "sphere" in spacetime would look like.

Now read up from the bottom of the diagram up (I'm viewing the diagram so that the t axis is vertical and the x-axis is horizontal) and tell me what the circle, which is the "worldline" of the edges of the hole, indicates, physically, as time advances. In particular:

(1) The circle is horizontal at t = -R, x = 0; what does this indicate about the speed at which the two edges of the hole are moving apart at that instant?

(2) The two sides of the circle are both vertical at t = 0, x = -R and t = 0, x = R. What does this indicate about what the two opposite edges of the hole are doing at t = 0?

(3) The circle is horizontal again at t = R, x = 0; what does this indicate about the speed at which the two edges of the hole are moving together at that instant?

When I say I don't understand it doesn't mean I don't get it. It means I don't understand how it could work that way.

I know that's what you mean. But at the same time:

But I think I've proved I understand the concepts.

Not to me you haven't. All you've shown me is that you are either unable or unwilling to use the standard terms used in GR in the standard way they are used, to refer to the standard concepts they are supposed to refer to. If you want to bring in other concepts, fine; then define other terms to refer to them. But when you use the terms "flat", "straight", "curved", etc. in a way that is obviously inconsistent with their standard usage, that tells me that you're either unable or unwilling to communicate clearly.

That doesn't mean I'm saying GR is wrong. I think it's right except that if you want to look at gravity as curved space-time then you have to do exactly the same for acceleration as well.

Read the definitions above again. Notice that I defined "curved spacetime" and "acceleration of a worldline" (in the sense of "proper acceleration") entirely in terms of physical observations--and they're completely different and independent physical observations. So in terms of those definitions, which are the standard ones in GR, the statement of yours that I just quoted is simply false; there is simply no way to "look at" gravity as somehow the same as acceleration. If you think you can, then you're not just trying to extend or supplement GR; you're contradicting it, and as the computer geeks say, "garbage in, garbage out". If you're thinking of something else as "acceleration", then please define it precisely so we can talk about "A-wal's acceleration" without confusing it with something else.

Edit: For clarity, once again, "gravity" in the sense of "curved spacetime" means "tidal gravity" as I defined it above, and *that* is the sense of "gravity" for which there is no way to "look at" it as the same as acceleration (in the sense of "proper acceleration"). If you're using "gravity" in some other sense, then "gravity" in your sense isn't the same as curved spacetime.

 

[Dale]

Because I haven't had satisfactory answers yet. It maybe because I can't get my head round it. I don't think so though because I understand what I'm being told, I just don't agree. It could be because it hasn't been explained in a way that I understand or it could be that what I'm being told doesn't add up.

But not being satisfied with the answers is a very different statement than that you haven't been shown any answer. You can (and do) disagree with or not understand the answers, but they have been provided multiple times.

I will. I was going to do that anyway at some point. I'm going to go through it from the beginning and pick up some of the things I let go because I wanted to move on and didn't want to stick the knife in at the time.

That would be good, I may indeed have never responded to some point. If I am responding to a long post I tend to only respond to the one or two most important points. I figure that the other poster will repeat one of the remaining points if they think it is more important than I deemed it. It is not that I am avoiding a topic, just prioritizing my response, particularly when the secondary topics build on the important point.

I'll also try to be a bit less cocky and petulant when I disagree with something, although I can't make any promises. Conflict does breed creativity but so does peace and quiet. I'll try asking questions instead of making assertions. Hopefully me and Dalespam can be friends again. Group hug.

I can't make any promises either, but I can try to be less dismissive and caustic too.

However, if I do get back to responding substantively, I think I will do so using much more math and much less english. My firm opinion from the previous several hundred posts is that you in particular need to learn the math if you are going to understand GR. Way too much is getting "lost in translation" between the math and the english. I would also recommend that you focus on the Rindler horizon in flat spacetime first, in order to understand the nature of an event horizon and coordinate charts on subsets of the manifold.

 

[A-wal]

Three questions:

1). Using the river model; what speed would a free-faller have accelerated to relative to the hovering observers of the river bed when they hit the horizon?

2). Let’s do this the other way round. New four-dimensional shape. This one does expand and contract at a constant speed. What would that shape look like?

3). Why is the length contraction and time dilation from gravity considered as curvature when the length contraction and time dilation from acceleration isn’t? Seems a bit hypocritical.

 

[Dale]

2). Let’s do this the other way round. New four-dimensional shape. This one does expand and contract at a constant speed. What would that shape look like?

A cone.

3). Why is the length contraction and time dilation from gravity considered as curvature when the length contraction and time dilation from acceleration isn’t? Seems a bit hypocritical.

The Riemann curvature tensor is defined as:

{R^\ell}_{ijk}=<br /> \frac{\partial}{\partial x^j} \Gamma_{ik}^\ell-\frac{\partial}{\partial x^k}\Gamma_{ij}^\ell<br /> +\sum^{n}_{s=1}(\Gamma_{js}^\ell\Gamma_{ik}^s-\Gamma_{ks}^\ell\Gamma_{ij}^s)

http://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Curvature_tensors

Where the Christoffel symbols are given by:

\Gamma^m_{ij}=\frac12 g^{km} \left(<br /> \frac{\partial}{\partial x^i} g_{kj}<br /> +\frac{\partial}{\partial x^j} g_{ik}<br /> -\frac{\partial}{\partial x^k} g_{ij}<br /> \right)

The curvature tensor evaluates to 0 for a metric like the Rindler metric where there is acceleration. So non-gravitational acceleration is not due to curvature of the manifold.

There is a different concept, called the covariant derivative, which measures the proper acceleration felt by a particle on a given worldline. I.e. this is the acceleration which would be measured by an accelerometer.

\nabla_j v^i=\frac{\partial v^i}{\partial x^j}+\Gamma^i_{jk}v^k

http://en.wikipedia.org/wiki/List_o...y#Christoffel_symbols.2C_covariant_derivative

In the Rindler metric, a particle with a constant position coordinate has a constant and non-zero covariant derivative, despite the fact that the curvature tensor is 0. So it is clear that acceleration and curvature are different things.

 

[cosmik debris]

Three questions:

1). Using the river model; what speed would a free-faller have accelerated to relative to the hovering observers of the river bed when they hit the horizon?

2). Let’s do this the other way round. New four-dimensional shape. This one does expand and contract at a constant speed. What would that shape look like?

3). Why is the length contraction and time dilation from gravity considered as curvature when the length contraction and time dilation from acceleration isn’t? Seems a bit hypocritical.

1). The local speed of the event horizon is c.
2). I don't know what this means.
3). They are considered different because they are. GR and the acceleration thing are only equivalent to first order.

I really shouldn't wade in at this late stage but I have if only to reinforce what Dale and Peter have told you. You have so many misunderstandings about GR and SR and you keep repeating them. You don't seem to have learned anything in hundreds of posts. These are mathematical subjects and maths is used because it is the only unambiguous way of expressing the relationships. But not understanding the maths is only a small part of your problem as I see it, your main problem is your refusal to move forward.

Listen to Dale and Peter, I don't know how they have remained so patient.

 

[PeterDonis]

I posted a response earlier but somehow it seems to have vanished, at least when I view the site, so I'm posting again. Apologies if this ends up being a double post.

1). Using the river model; what speed would a free-faller have accelerated to relative to the hovering observers of the river bed when they hit the horizon?

There are two possible answers, depending on how the term "relative speed" is interpreted:

(1a) If "relative speed" means "the speed at which a hypothetical observer hovering at a constant radius r would see the free-faller falling, even if such an observer can't actually exist", then the answer at the horizon is the speed of light, c (and inside the horizon it's some speed greater than c). But if "relative speed" is interpreted this way, then the river model does *not* require that relative speed be less than c (for objects with nonzero rest mass). "Relative speed" interpreted this way is just a ratio of coordinates, and has no direct physical meaning. Sometimes it so happens that you can equate it to something that does have direct physical meaning, but there's no requirement that you must always be able to do so.

(1b) If "relative speed" means "the speed at which an actual hovering observer would observe the free-faller to be falling", then the question has no meaning at or inside the horizon since there are no such hovering observers in that region of spacetime. This is the only definition of "relative speed" that requires relative speed to be less than that of light (for objects with nonzero rest mass); at or inside the horizon, it's still true that actual physical objects with mass must move at less than the speed of light relative to each other; but it's also true that they must all be moving inward (i.e., decreasing radius with time), so none of them can be "hovering".

None of this contradicts the river model--at least not the "standard" river model, the one described in this paper:

http://arxiv.org/abs/gr-qc/0411060

If you have some other model in mind as the "river model", please give a reference.

3). Why is the length contraction and time dilation from gravity considered as curvature when the length contraction and time dilation from acceleration isn’t? Seems a bit hypocritical.

DaleSpam gave the mathematical answer to this; I just want to supplement a little bit what he said.

I gave the standard GR definitions of curvature of spacetime and curvature of a worldline (i.e., proper acceleration) in non-mathematical terms in an earlier post. (DaleSpam's math is a more precise way of saying what I said.) So it's not true that acceleration (if you mean proper acceleration) is not considered as curvature: it's just curvature of a worldline instead of curvature of spacetime. Those two types of curvature are different, independent fundamental concepts in GR. Length contraction and time dilation are not fundamental concepts in GR; they are just "side effects" of curvature, and the fact that both types of curvature happen to give rise to them does not indicate that both types of curvature are somehow "the same". They just both happen to have similar side effects in that particular way.

 

[Dale]

So it's not true that acceleration (if you mean proper acceleration) is not considered as curvature: it's just curvature of a worldline instead of curvature of spacetime. Those two types of curvature are different, independent fundamental concepts in GR.

good way of saying it.

 

[A-wal]

I'll explain this purely in the context of Schwarzschild coordinates. Apparently this system is incomplete and the entire space-time isn't covered in these coordinates, but I haven't got a clue what that means. The entire space-time is covered but it's hidden. Not very well hidden, but apparently well enough. A black hole is an area of space-time where time dilation and length contraction have gone beyond the point that they would if you had accelerated to c, beyond infinity. But the Schwarzschild coordinates don't actually allow this to happen. Instead, while time dilation and length contraction would shrink the size and lifespan of the black hole down to a single point in time and space (the singularity) if you get close enough, from a distance we "see" curvature. If we look at the inside we see an area where objects would be accelerated by gravity beyond the point of c. Objects obviously can't do that, but c becomes a "visible" place. In other words there is now a specific area of space-time that no amount of acceleration could ever take you to. This is because the time dilation and length contraction from the acceleration of gravity would be infinite at the horizon, in the same way they are at c. Everything outside is moving slower than c and everything inside would be moving at over c, which is why you're getting a black hole in the first place, but the horizon isn't always in the same place. It moves. It's relative, like c.

The space-time’s compressed from a distance so that objects closer to the singularity appear shorter and slower than the objects are in their own frames, just like an accelerator does in flat space-time. The horizon represents infinite time dilation and length contraction, just like you get at c. So there's a potentially infinite amount of space-time between you and the horizon (just like you get a potentially infinite amount of space-time between you and the c in flat space-time) which is determined by the life-span of the black hole. This means that it's not possible for an object to reach the horizon from the perspective of anything on the outside, and it's not possible for an object to reach the horizon from the perspective of anything on the inside either because of more than infinite curvature. There are apparently other coordinate systems that do allow an in-faller to reach and cross the horizon but they directly contradict Schwarzschild coordinates because they can't cross from the perspective of the outside, even from the perspective of the parts of the in-falling object that haven't crossed yet. How can you move smoothly between the Schwarzschild coordinates, where an object can't cross the horizon, to a coordinate system where it can? At what point does the changeover happen? It should be smooth, but it can't be. In your version an object can't ever cross one second, and then it does.

Anything at the horizon would be moving at c relative to everything in the universe outside the horizon, even the atom next to it. The rope isn't needed, it just makes it easier to visualise. There would be no way of observing an in-faller and knowing whether or not they made it to the horizon, which doesn't make sense. If an object hasn't crossed from the perspective of an observer then it hasn't crossed (ignoring the delay of the time the light takes to reach you of course). There’s no reason for this sudden jump to infinity at the horizon if you treat it the same way as c, and everything’s simpler and makes a lot more sense. It takes infinite energy to accelerate an object to c. You think you've got it with a black hole but it doesn't work.

A cone.

In three dimensions it’s a cone, but only because there's a spatial dimension missing. Extend that cone into the horizontal direction you're using for time and you have half a sphere. Now I'm going to describe what I think happens during the formation (and whole life span come to think of it) of a black hole. A gravity wave (a change in the strength of gravity which moves outwards at c) is released from the singularity. This is the horizon, so the sphere is limited to c. It can't "hit" anything as it expands because the space-time that makes up the black (and white) hole is being created behind it, like length contracting from the back in Rindler coordinates when an object accelerates in flat space-time. The gravity wave carries on moving outwards at c and does hit objects in its path but that's not the event horizon. This is the first you would notice that something has changed. When the horizon has reached its maximum size (when gravity has no longer curved space-time beyond c) it rushes inwards at c because it’s determined by how close an object could get to it in that amount of time. There's the other half of the sphere, and your white hole. We now have a four-dimensional bubble that nothing can ever reach from any direction (including of time) formed around the singularity, from a distance anyway. Because this is all happening at c it isn't actually experienced by the black/white hole at all, in the same way that light doesn't experience time. It's infinitely small from its own perspective but it’s bigger the more distance there is between you and it.

The Riemann curvature tensor is defined as:

{R^\ell}_{ijk}=<br /> \frac{\partial}{\partial x^j} \Gamma_{ik}^\ell-\frac{\partial}{\partial x^k}\Gamma_{ij}^\ell<br /> +\sum^{n}_{s=1}(\Gamma_{js}^\ell\Gamma_{ik}^s-\Gamma_{ks}^\ell\Gamma_{ij}^s)

http://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Curvature_tensors

Where the Christoffel symbols are given by:

\Gamma^m_{ij}=\frac12 g^{km} \left(<br /> \frac{\partial}{\partial x^i} g_{kj}<br /> +\frac{\partial}{\partial x^j} g_{ik}<br /> -\frac{\partial}{\partial x^k} g_{ij}<br /> \right)

The curvature tensor evaluates to 0 for a metric like the Rindler metric where there is acceleration. So non-gravitational acceleration is not due to curvature of the manifold.

There is a different concept, called the covariant derivative, which measures the proper acceleration felt by a particle on a given worldline. I.e. this is the acceleration which would be measured by an accelerometer.

\nabla_j v^i=\frac{\partial v^i}{\partial x^j}+\Gamma^i_{jk}v^k

http://en.wikipedia.org/wiki/List_o...y#Christoffel_symbols.2C_covariant_derivative

In the Rindler metric, a particle with a constant position coordinate has a constant and non-zero covariant derivative, despite the fact that the curvature tensor is 0. So it is clear that acceleration and curvature are different things.

Well it's not clear to me. I know you said you'd be using equations, and I shouldn't moan that I'm finally getting the kind of response that may actually answer my questions, but it'd be lovely to get it in English as well. I take it "In the Rindler metric" means using Rindler coordinates, and "constant and non-zero covariant derivative" means constantly accelerating at the same rate. So an object that is undergoing zero proper acceleration can still be accelerating in Rindler coordinates despite the fact that there’s no curvature? There’s no curvature because it’s being expresses as acceleration instead. You also said "there is a different concept, called the covariant derivative, which measures the proper acceleration felt by a particle on a given worldline." What's stopping that from being viewed as curvature? Whatever energy’s accelerating that particle will also accelerate every other object in the universe, a bit, so how’s that different from curvature? I'm still failing to see the distinction. Maybe I misunderstood what you said?

1). The local speed of the event horizon is c.
2). I don't know what this means.
3). They are considered different because they are. GR and the acceleration thing are only equivalent to first order.

1). Yes it is, and that's okay for an event horizon but not for matter approaching it.

2).That's okay. Apparently shapes can be hard.

3).They are considered different because they are? Why are GR and the acceleration thing only equivalent to first order?

I really shouldn't wade in at this late stage but I have if only to reinforce what Dale and Peter have told you. You have so many misunderstandings about GR and SR and you keep repeating them. You don't seem to have learned anything in hundreds of posts. These are mathematical subjects and maths is used because it is the only unambiguous way of expressing the relationships. But not understanding the maths is only a small part of your problem as I see it, your main problem is your refusal to move forward.

Listen to Dale and Peter, I don't know how they have remained so patient.

They're not misunderstandings about GR and SR. I know how they say it works. Don't assume I have misunderstandings from a few posts that you've read where I've used the concepts in a way you're not used to. Read the blogs I posted if you want a more conventional description, then you can tell me I have so many misunderstandings if you think there's some things I’ve gotten wrong. I have been listening to Dale and Peter, intently. I still haven’t had my questions properly answered. It may be that there's something I don't get but I don't think so. After this long I don't think I’m jumping to conclusions. They've probably stayed patient because we've only recently starting talking about it in this context. I was building up to it. I wanted to learn the official stance and see if I was missing something. It has been a while though. It would have been quicker to go to collage. I wouldn't be able to earn money at the same time though. These are not mathematical subjects. There’s only one mathematical subject and that’s mathematics. If that’s all you’re using then that’s all you’re doing because you have no conceptual understanding of what it is you're describing. In other words you don’t know what you’re talking about. But I do agree that maths is the only unambiguous way of expressing the relationships. It's not open to interpretation. It's a tool that should be based on understanding, not the other way round. I'll move on when someone explains to me why it doesn't work this way.

There are two possible answers, depending on how the term "relative speed" is interpreted:

(1a) If "relative speed" means "the speed at which a hypothetical observer hovering at a constant radius r would see the free-faller falling, even if such an observer can't actually exist", then the answer at the horizon is the speed of light, c (and inside the horizon it's some speed greater than c). But if "relative speed" is interpreted this way, then the river model does *not* require that relative speed be less than c (for objects with nonzero rest mass). "Relative speed" interpreted this way is just a ratio of coordinates, and has no direct physical meaning. Sometimes it so happens that you can equate it to something that does have direct physical meaning, but there's no requirement that you must always be able to do so.

(1b) If "relative speed" means "the speed at which an actual hovering observer would observe the free-faller to be falling", then the question has no meaning at or inside the horizon since there are no such hovering observers in that region of spacetime. This is the only definition of "relative speed" that requires relative speed to be less than that of light (for objects with nonzero rest mass); at or inside the horizon, it's still true that actual physical objects with mass must move at less than the speed of light relative to each other; but it's also true that they must all be moving inward (i.e., decreasing radius with time), so none of them can be "hovering".

None of them can be hovering on the other side of the event horizon, but you've got to reach it first. Remember the riverbed covers the entire "exterior" of the black hole, which is allowed. What speed would the free-faller be moving relative to the riverbed, or any object outside the horizon for that matter?

None of this contradicts the river model--at least not the "standard" river model, the one described in this paper:

http://arxiv.org/abs/gr-qc/0411060

If you have some other model in mind as the "river model", please give a reference.

The one I'm using might be a bit different, seeing as the first and only time I've heard of it was when somebody mentioned it in this thread and I thought it was a nice way of looking at it. Still do.

Use the river model, but add a river bed to act as a kind of ether. Not a real ether, just an imaginary one to compare our starting position with the horizon. We start far enough away for the gravitational pull of the black hole to be negligible enough to ignore. Our river bed provides us with an uncurvable surface that’s at rest relative to our starting position and the black hole. We wait a while and eventually notice we’re moving towards it as the river slowly moves us along relative to our starting position. We don't feel as though we're accelerating because we're still at rest relative to the river. That river moves relative to the river bed in exactly the same way any object using energy to accelerate would. It can’t reach c. The river changes in relation to the river bed as it moves faster, and the black hole changes with it. Time dilation and length contraction relative to the river bed mean that the its life span and size decrease exponentially as we approach the horizon. We would be traveling at the speed of light at the horizon if it made sense for anything to move that fast. You could try to use energy to accelerate you through the river to get you there faster, but that would just have the effect of adding very high velocities.

An observer who is very, very far away would measure the distances as less than the starting frame of the in-faller and furthest hoverer, so let's have the book keeper in this frame. They see an equal distance between the hoverers, like I said. The first hoverer can be the bookkeeper.

The hoverers have to expend different amounts of energy to maintain a constant separation from the perspective of the bookkeeper.

Anything else you need clarified?

DaleSpam gave the mathematical answer to this; I just want to supplement a little bit what he said.

I gave the standard GR definitions of curvature of spacetime and curvature of a worldline (i.e., proper acceleration) in non-mathematical terms in an earlier post. (DaleSpam's math is a more precise way of saying what I said.) So it's not true that acceleration (if you mean proper acceleration) is not considered as curvature: it's just curvature of a worldline instead of curvature of spacetime. Those two types of curvature are different, independent fundamental concepts in GR. Length contraction and time dilation are not fundamental concepts in GR; they are just "side effects" of curvature, and the fact that both types of curvature happen to give rise to them does not indicate that both types of curvature are somehow "the same". They just both happen to have similar side effects in that particular way.

What a coincidence. They both have the exact same two consequences and they're both caused by curvature. What are you defining as curvature anyway? It just means you're viewing movement as a change in the space-time between objects rather than actual movement of the objects themselves. Tomato, tomato. You’re saying it’s okay that matter can reach the equivalent of c if it’s because of curved space-time, but it doesn’t make any difference. What makes them distinct concepts? The results shouldn’t be any different from viewing it as accelerating towards c in flat space-time. You could view acceleration as curved space-time as well to completely switch it round and it still wouldn’t make any difference. Energy doesn't just affect the object that's producing it. It spreads out to affect every object in the universe to some extent, but it's an inverse square so most objects are hardly affected at all. I can't see any reason why this can't be viewed as curvature of space-time, or why time dilation and length contraction aren’t fundamental concepts and curvature as just one way of looking at it in either case.

 

[PeterDonis]

A-wal, I see that you went to considerable trouble to "explain" your viewpoint, but your explanation (which is the first part of your post, before you responded to others) doesn't change my viewpoint at all. I still think you're making the same mistake I've thought you were making all along. So I still don't see much point in responding to it, but I do want to comment on a couple of things:

Anything at the horizon would be moving at c relative to everything in the universe outside the horizon, even the atom next to it.

By "anything at the horizon", do you mean an object free-falling inwards that is just crossing the horizon, relative to an object hovering just outside the horizon? If so, then there is an interpretation of "moving at c" that makes the above statement true, but that interpretation is *not* one that requires objects to move slower than c, as I explained in a previous post. See further comments below on the "river bed".

There would be no way of observing an in-faller and knowing whether or not they made it to the horizon, which doesn't make sense.

Why not? Why must the laws of physics ensure that an observer anywhere in the universe can always receive information from everywhere else in the universe? Why must it be impossible that there are some portions of the universe that simply can't send signals to other portions? I realize it's counterintuitive, but what iron law of physics or logic do you think makes it impossible? Just saying it doesn't make sense to you isn't enough.

Read the blogs I posted if you want a more conventional description, then you can tell me I have so many misunderstandings if you think there's some things I’ve gotten wrong.

Which blogs are you referring to? I don't remember the links, but it's been a long thread...

None of them can be hovering on the other side of the event horizon, but you've got to reach it first. Remember the riverbed covers the entire "exterior" of the black hole, which is allowed. What speed would the free-faller be moving relative to the riverbed, or any object outside the horizon for that matter?

You keep on assuming, without proof or argument, that there is some genuine physical restriction on the speed of objects relative to the "river bed". Why must there be such a restriction? The river bed itself is imaginary; it doesn't exist. Why must the speed of anything be restricted relative to something that doesn't exist?

The one I'm using might be a bit different

It is if you think objects are restricted to moving slower than c relative to the river bed. Read the paper I linked to; it makes perfectly clear that the "river model" works just fine inside the horizon, even though objects are moving "faster than light" relative to the "river bed" there, because the river itself is moving "faster than light" relative to the "river bed" there. But that's fine because the river bed doesn't actually exist.

What a coincidence. They both have the exact same two consequences

No. They both have two particular consequences in common, length contraction and time dilation. I did *not* say that they have *all* consequences in common. They don't. One very important consequence they do *not* have in common is that objects moving in curved spacetime do not have to feel weight; objects can free-fall in curved spacetime. But an object moving on a curved worldline *has* to feel weight; that's what "curved worldline" means.

You’re saying it’s okay that matter can reach the equivalent of c if it’s because of curved space-time

No. I'm saying that in curved spacetime, the "equivalent of c" is moving within the light cones at any given event. It does *not* require that objects have to move "slower than c" relative to objects that are in parts of spacetime where the light cones are tilted differently.

 

[Dale]

How can you move smoothly between the Schwarzschild coordinates, where an object can't cross the horizon, to a coordinate system where it can? At what point does the changeover happen? It should be smooth, but it can't be. In your version an object can't ever cross one second, and then it does.

I highly recommend Sean Carroll's lecture notes located here:
http://lanl.arxiv.org/abs/gr-qc/9712019v1

Your questions above are addressed in chapter 2 entitled "Manifolds". In particular, be sure to carefully study from 33-35 on maps and diffeomorphisms and 37-39 on coordinate charts. Note in particular the explanation about coordinate systems being defined on open subsets of the manifold and that some manifolds require multiple charts to span the entire manifold.

In three dimensions it’s a cone, but only because there's a spatial dimension missing.

You are correct. This is a good example of why math is preferable to English. A cone has the equation:
x^2+y^2-z^2=0

A sphere has the equation:
x^2+y^2+z^2=r^2

A hypercone has the equation:
w^2+x^2+y^2-z^2=0

A 4D hypersphere has the equation:
w^2+x^2+y^2+z^2=r^2

The figure you asked about is technically a hypercone. It is most definitely not a hypersphere.

Now I'm going to describe what I think happens during the formation (and whole life span come to think of it) of a black hole.

The math of black hole formation is much more complicated than the math for the static Schwarzschild spacetime. I think you need to learn to walk before you try to run. I would approach the topics in the following order:
1) Rindler spacetime - acceleration in flat spacetime, coordinate systems, event horizons
2) Schwarzschild spacetime - curvature, singularities, static spacetimes
3) black hole formation - non-static spacetimes, stress-energy, stellar structure

You simply aren't ready for it now, but when you are here is a good reference:
http://www.phys.uu.nl/~prokopec/MichielBouwhuis_bh2.pdf

Well it's not clear to me. I know you said you'd be using equations, and I shouldn't moan that I'm finally getting the kind of response that may actually answer my questions, but it'd be lovely to get it in English as well.

You have gotten it in English for more than 350 posts; I'm not saying anything new here with the math that hasn’t already been said multiple times before in English. The English approach didn't work and I don't want to repeat that; all that leads to is confusion on your part and nastiness on my part. I think you really need to tackle the math head-on. Otherwise you won't be able to learn GR.

I take it "In the Rindler metric" means using Rindler coordinates, and "constant and non-zero covariant derivative" means constantly accelerating at the same rate. So an object that is undergoing zero proper acceleration can still be accelerating in Rindler coordinates despite the fact that there’s no curvature?

All correct, yes. Let's go ahead and work this out in detail.

In units where c=1 the Rindler metric is given by:
ds^2 = -x^2 dt^2 + dx^2 + dy^2 + dz^2
or
g=\left(<br /> \begin{array}{cccc}<br /> -x^2 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)

Using the expression that I gave above for the Christoffel symbols, we find that there are three which are non-zero:
\Gamma^{t}_{tx}=\Gamma^{t}_{xt}=\frac{1}{x}
\Gamma^{x}_{tt}=x

We can then use the expression above to determine the curvature tensor. For example, we have
{R^x}_{txt}=<br /> \frac{\partial}{\partial x} \Gamma_{tt}^x-\frac{\partial}{\partial t}\Gamma_{tx}^x<br /> +\sum^{n}_{s=1}(\Gamma_{xs}^x\Gamma_{tt}^s-\Gamma_{ts}^x\Gamma_{tx}^s)=<br /> 1-0+(0-1) = 0

Similarly, all the other 64 elements of the curvature tensor also evaluate 0. So the curvature is 0, meaning that the Rindler metric describes a flat spacetime.

Now, on the other hand, if we have a particle at a constant x position in the Rindler metric then its tangent vector is given by:
v=\left(\frac{1}{\sqrt{x^2}}, 0, 0, 0 \right)

Plugging that into the equation above for the covariant derivative gives:
\nabla v = \left(0, \frac{1}{x} ,0 ,0 \right)

So the covariant derivative is non-zero, meaning that an accelerometer on a stationary particle (coordinate acceleration = 0) in the Rindler metric will measure a non-zero proper acceleration in the positive x direction whose magnitude is inversely proportional to x. This example shows the difference between curvature which is a rank 4 tensor, the covariant derivative of the tangent vector (proper acceleration) which is a rank 1 tensor, and the coordinate acceleration which is not a tensor.

 

[A-wal]

A-wal, I see that you went to considerable trouble to "explain" your viewpoint, but your explanation (which is the first part of your post, before you responded to others) doesn't change my viewpoint at all. I still think you're making the same mistake I've thought you were making all along.

I really just wanted to clarify the way I see it as much as possible. It wasn't really any trouble. I just waited until I knew what I wanted to say before I started writing it instead of forcing it. I was actually less effort than usual. It just takes a lot longer. What mistake do you think I'm still making? Thinking the entire space-time can be contained within just one coordinate system?

By "anything at the horizon", do you mean an object free-falling inwards that is just crossing the horizon, relative to an object hovering just outside the horizon? If so, then there is an interpretation of "moving at c" that makes the above statement true, but that interpretation is *not* one that requires objects to move slower than c, as I explained in a previous post. See further comments below on the "river bed".

I’m not talking about coordinate acceleration. I'm using proper acceleration, otherwise know as tidal force. Anything crossing the horizon (if that were possible) would have to be moving at c relative to every single other object in the universe that's not inside the horizon, including moving at c relative to the rest of the object that hasn't crossed yet. You said the rope would always break, well so would the object crossing the horizon. It would also be feeling infinite proper acceleration, so anything that gets pulled past the horizon would be ripped apart. That jump shouldn't be there, and why can objects survive it in GR?

Why not? Why must the laws of physics ensure that an observer anywhere in the universe can always receive information from everywhere else in the universe? Why must it be impossible that there are some portions of the universe that simply can't send signals to other portions? I realize it's counterintuitive, but what iron law of physics or logic do you think makes it impossible? Just saying it doesn't make sense to you isn't enough.

Because time can't be infinitely time dilated because nothing with mass can reach c. Whether c is an event horizon in curved space-time or light speed in flat space-time makes no difference because they're exactly the same thing. The information would reach you eventually. I know that the Rindler horizon means information wouldn’t reach a constantly accelerating observer, but you can’t accelerate forever because that would take infinite energy. It doesn’t make sense because you can see that there's enough time dilation and length contraction to stop anything reaching the horizon if you constantly compare the in-faller with a more distant observer. If the in-faller is moving through time at half speed relative to the distant observer then the distant observer is moving at double speed relative to the in-faller. If the black hole is about to die from the perspective of the distant observer and the in-faller hasn’t crossed yet then it’s about to die from the perspective of the in-faller and the in-faller hasn’t crossed yet (after taking into account the time it takes for the light to reach you).

You can see that there’s a contradiction if you measure proper time using the age/size of the black hole as a ruler. An in faller measures the size of the black hole when they supposedly cross the horizon and finds it’s about half way though its life. A distant observer sees that the black hole is near the end of its life and the in-faller still hasn’t crossed. An in-faller doesn't reach the event horizon in a finite amount of proper time using Schwarzschild coordinates, so why is it okay to switch to a coordinate system that contradicts this and assume this one is the right one for crossing the horizon, and if it is the right one then why doesn't it invalidate the Schwarzschild coordinates?

Which blogs are you referring to? I don't remember the links, but it's been a long thread...

Special Relativity made simple and General Relativity made simple. The links are by my name.

You keep on assuming, without proof or argument, that there is some genuine physical restriction on the speed of objects relative to the "river bed". Why must there be such a restriction? The river bed itself is imaginary; it doesn't exist. Why must the speed of anything be restricted relative to something that doesn't exist?

I forgot to quote this bit.

Don't think of it as one entity. Think of all the pebbles on the riverbed as individual 'hovering' observers. Now you can't move at c relative to them.

So it does exist. In fact it's everything that exists. Anything reaching the horizon would reach c relative to anything outside the horizon.

It is if you think objects are restricted to moving slower than c relative to the river bed. Read the paper I linked to; it makes perfectly clear that the "river model" works just fine inside the horizon, even though objects are moving "faster than light" relative to the "river bed" there, because the river itself is moving "faster than light" relative to the "river bed" there. But that's fine because the river bed doesn't actually exist.

It's not fine. It means that anything inside the horizon is moving faster than c relative to everything outside it. The fact that you wouldn't be able to see them doing it doesn't make it okay. I used the evenly spaced accelerating hoverers for the riverbed to remove the curvature and show the proper acceleration going up towards infinity and the relative velocity of the in-faller approaching c towards the horizon. The hoverers get further apart as you accelerate to keep you BELOW c. You could use the in-fallers velocity relative to the hoverers to work out the time dilation and length contraction just as you would in flat space-time, so it makes no sense to say there would be enough time dilation and length contraction to stop you from reaching the horizon.

No. They both have two particular consequences in common, length contraction and time dilation. I did *not* say that they have *all* consequences in common. They don't. One very important consequence they do *not* have in common is that objects moving in curved spacetime do not have to feel weight; objects can free-fall in curved spacetime. But an object moving on a curved worldline *has* to feel weight; that's what "curved worldline" means.

An object free-falling in curved space-time does always feel weight, but it’s called tidal force instead. That’s what “curved space-time” means. Free-falling is the equivalent to a different but constant relative velocity except that you’re always accelerating. You’re always moving into a higher gravitational field because you’re always getting closer to the source of the gravity, so you’re always feeling tidal force. Tidal force in curved space-time is the equivalent to acceleration in flat space-time. What else don’t they have in common?

No. I'm saying that in curved spacetime, the "equivalent of c" is moving within the light cones at any given event. It does *not* require that objects have to move "slower than c" relative to distant objects that have light cones tilted differently.

I would have thought you could view the light the cones as tilted when accelerating in flat space-time in exactly the same way as you can with curved space-time. The Rindler horizon would be when they tilt to 90 degrees?

I highly recommend Sean Carroll's lecture notes located here:
http://lanl.arxiv.org/abs/gr-qc/9712019v1

Your questions above are addressed in chapter 2 entitled "Manifolds". In particular, be sure to carefully study from 33-35 on maps and diffeomorphisms and 37-39 on coordinate charts. Note in particular the explanation about coordinate systems being defined on open subsets of the manifold and that some manifolds require multiple charts to span the entire manifold.

Thanks, but I’m still not sure why you would need multiple charts to span the entire manifold. It just says that you do. I still think you've gone off the edge of the map and said here there be monsters. Space-time curves round past 90 degrees at the horizon. That means all the space-time in the universe could fit in the gap between wherever you are and the horizon, just like all the velocity in the universe would fit between you and c.

You are correct. This is a good example of why math is preferable to English. A cone has the equation:
x^2+y^2-z^2=0

A sphere has the equation:
x^2+y^2+z^2=r^2

A hypercone has the equation:
w^2+x^2+y^2-z^2=0

A 4D hypersphere has the equation:
w^2+x^2+y^2+z^2=r^2

The figure you asked about is technically a hypercone. It is most definitely not a hypersphere.

I thought I was describing a four-dimensional circle. If a hypersphere is something else then how would that expand and contract? Why would it be a hypercone? It takes up the same amount of space in each dimension and the equivalent in time as well, so it should be a hypersphere. If it expands and then contracts at a constant rate so that the time it exists for is equivalent to how big it gets (which would happen if it always moves at c) then it's the same size in every dimension, so it's a sphere.

The math of black hole formation is much more complicated than the math for the static Schwarzschild spacetime. I think you need to learn to walk before you try to run. I would approach the topics in the following order:
1) Rindler spacetime - acceleration in flat spacetime, coordinate systems, event horizons
2) Schwarzschild spacetime - curvature, singularities, static spacetimes
3) black hole formation - non-static spacetimes, stress-energy, stellar structure

You simply aren't ready for it now, but when you are here is a good reference:
http://www.phys.uu.nl/~prokopec/MichielBouwhuis_bh2.pdf

But different coordinate systems should be just different ways of looking at the same thing, so it should be as simple as the static Schwarzschild space-time. I just don’t see how it could work any other way. A black hole would have no way of holding itself up. If you treat the event horizon in the same way as you treat c then you have an area that’s length contracted and time dilated beyond the point where it’s reachable and it would expand at c, and then recede at c because that’s how far you could reach with c as a speed limit. That way it doesn’t need to stay for any length of time in its own frame and doesn’t have to worry about holding itself up. c is a limit on how far you can go in certain amount of time. When G>C there's an area where you can't go because they'll never be enough time. It's nothing but an effect of time dilation and length contraction over the area of space time that they exceed c. It only looks like something from a distance. There's no space-time between the EV and the singularity, just like there's no space-time outside of the universe or before the big bang.

You have gotten it in English for more than 350 posts; I'm not saying anything new here with the math that hasn’t already been said multiple times before in English. The English approach didn't work and I don't want to repeat that; all that leads to is confusion on your part and nastiness on my part. I think you really need to tackle the math head-on. Otherwise you won't be able to learn GR.

I don’t think the equations will help me understand it because learning the equations won’t show me why those equations are the right ones. Besides, I don’t think in equations so learning all the maths in the world won’t help me to understand why it works in a way that seems to me to be completely self-contradictory and unnecessary. They would only help me to describe something that I don’t think can happen. I don’t think learning the equations are essential for learning GR, or anything other than the equations themselves.

All correct, yes. Let's go ahead and work this out in detail.

In units where c=1 the Rindler metric is given by:
ds^2 = -x^2 dt^2 + dx^2 + dy^2 + dz^2
or
g=\left(<br /> \begin{array}{cccc}<br /> -x^2 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)

Using the expression that I gave above for the Christoffel symbols, we find that there are three which are non-zero:
\Gamma^{t}_{tx}=\Gamma^{t}_{xt}=\frac{1}{x}
\Gamma^{x}_{tt}=x

We can then use the expression above to determine the curvature tensor. For example, we have
{R^x}_{txt}=<br /> \frac{\partial}{\partial x} \Gamma_{tt}^x-\frac{\partial}{\partial t}\Gamma_{tx}^x<br /> +\sum^{n}_{s=1}(\Gamma_{xs}^x\Gamma_{tt}^s-\Gamma_{ts}^x\Gamma_{tx}^s)=<br /> 1-0+(0-1) = 0

Similarly, all the other 64 elements of the curvature tensor also evaluate 0. So the curvature is 0, meaning that the Rindler metric describes a flat spacetime.

Now, on the other hand, if we have a particle at a constant x position in the Rindler metric then its tangent vector is given by:
v=\left(\frac{1}{\sqrt{x^2}}, 0, 0, 0 \right)

Plugging that into the equation above for the covariant derivative gives:
\nabla v = \left(0, \frac{1}{x} ,0 ,0 \right)

So the covariant derivative is non-zero, meaning that an accelerometer on a stationary particle (coordinate acceleration = 0) in the Rindler metric will measure a non-zero proper acceleration in the positive x direction whose magnitude is inversely proportional to x. This example shows the difference between curvature which is a rank 4 tensor, the covariant derivative of the tangent vector (proper acceleration) which is a rank 1 tensor, and the coordinate acceleration which is not a tensor.

But if there’s no curvature that would mean there’s no proper acceleration if they’re the same thing, so it’s fine. The fact that there is coordinate acceleration but no proper acceleration is irrelevant. In my riverbed example they are at rest despite the fact that the proper acceleration in different for each one. You're saying that you use a different process to work out the two types of curvature, but I don’t know why they would be different. Is it a different kind of tensor for curved space-time because you're describing what would happen in four dimensions rather than two? After you've plotted the path of the free-faller you could then uncurve that path and you should be left with acceleration in flat space-time. You could do the same thing the other way round using energy for mass, making acceleration a source of curvature.

This is a good coordinate system. You could view time as a spatial dimension if you look at space as expanding everywhere at c. In this view the past is inwards and the future is outwards. Everything is at the centre of its own bubble. Energy has no mass so it’s carried outwards as space expands making light expand at c in a sphere shape. You can use energy to accelerate making the light bubble warp and causing the expansion to pass through you more slowly in one spatial dimension so you get time dilation and length contraction. The expansion is also slowed through mass. Matter doesn’t expand outwards like energy but that movement has to go somewhere. The electrons are held in place by the strong nuclear force and are forced to orbit instead of expanding. Whether you could look at the space as curved by the strong nuclear force and the electrons moving in straight lines through curved space-time on a tiny scale I don't know but I suspect you probably could. The orbit of the electrons causes length contraction and time dilation because they have a different relative velocity, which radiates outwards from the source. So in these coordinates gravity is caused by the frame dragging of atoms. A black hole happens when the time dilation and length contraction are greater than the expansion over a certain area. You can’t reach that area because it’s obvious using these coordinates that you would always run out of time.

That coordinate system is only good for viewing one moment at a time though. If you want to view more than one time at once you can remove one spatial dimension to turn it into the ball analogy so that the surface of the ball represents all three spatial dimensions in the present with the past towards the centre and the future further out. The bubbles are then cones instead of spheres.

 

[PeterDonis]

What mistake do you think I'm still making? Thinking the entire space-time can be contained within just one coordinate system?

Not precisely; it is thinking that there is a flat background to the spacetime (what you call the "river bed") that constrains how things can move (or, equivalently, constrains how much the spacetime can curve). Any time you reason based on an analogy between the horizon and reaching light speed in flat spacetime, you're making this mistake. It shows itself in other ways as well, some of which I'll comment on below.

Anything crossing the horizon (if that were possible) would have to be moving at c relative to every single other object in the universe that's not inside the horizon, including moving at c relative to the rest of the object that hasn't crossed yet. You said the rope would always break, well so would the object crossing the horizon. It would also be feeling infinite proper acceleration, so anything that gets pulled past the horizon would be ripped apart.

There are at least two misstatements here:

(1) The freely falling object feels *no* proper acceleration. Zero. You keep on using the term "proper acceleration" incorrectly. If you're talking about freely falling objects, then just stop yourself whenever you want to use the term "proper acceleration", because it's always zero for a freely falling object.

(2) You are mixing up two scenarios. One (the "rope" scenario) involves a rope that is not freely falling--at least, one end of it is not (because it's attached to an observer that is hovering at a constant radius above the horizon). Such a rope will indeed always break when it crosses the horizon (actually, as I noted before, a real rope would break well before that, but in the absolute limiting case where the rope's tensile strength is the maximum allowed by relativity, it would break as it crossed the horizon). But that is because the top end of the rope is constrained; it is not freely falling.

The second scenario, which you are wrongly conflating with the "rope" scenario, is that of an object freely falling through the horizon--i.e., not attached to any other object, just falling by itself. Such an object *might* break as it crossed the horizon, if its extension in the radial direction was long enough compared to the strength of tidal gravity at the horizon. But if the freely falling object is small enough that tidal effects can be neglected over its length (by "neglected" I mean "compared to the tensile strength of the object", so obviously the material the object is made of can make a difference; an object made of styrofoam will have to be smaller than one made of steel for tidal effects to be negligible), then it will *not* break. I can set up a local inertial frame around the object as it crosses the horizon, and within that frame, the object is simply floating at rest, the force on it is negligible, and it holds together just fine.

(In that local inertial frame, by the way, the horizon is an outgoing light ray; for example, if I set up the local coordinates such that the origin is the point where the center of mass of the freely falling object crosses the horizon, and the positive x direction is radially outgoing, then the horizon is the line t = x. Since the object is at rest in this frame, the worldline of its center of mass is just the t axis; and the worldlines of other parts of the object are just vertical lines parallel to the t axis. So all the parts of the object are at rest relative to each other; even those that have not yet crossed the horizon--they are below the line t = x--are at rest relative to those that have crossed the horizon--they are above the line t = x. This should be obvious since the object is freely falling, and if it is short enough that tidal effects are negligible, all parts of the object are falling at the same rate.)

I know that the Rindler horizon means information wouldn’t reach a constantly accelerating observer, but you can’t accelerate forever because that would take infinite energy.

And if an accelerating observer in flat spacetime stops accelerating, they fall through the Rindler horizon and can now see the information beyond it. Similarly, if an observer hovering above a black hole stops accelerating, they are no longer hovering; they fall into the hole and can now see the information inside the horizon that was hidden from them before. None of this changes the fact that observers above the horizon can't receive information from below it *while they remain above the horizon*.

You can see that there’s a contradiction if you measure proper time using the age/size of the black hole as a ruler.

You keep on bringing in a finite lifetime for the black hole, when we've already agreed that's irrelevant; you said many posts ago that the horizon of even an "eternal" black hole, which lasts for an infinite time in the past and future, could not be reached. So reasoning based on a finite black hole lifetime is irrelevant. Please re-state your argument in a way that applies to an eternal black hole, so we can get that case nailed down first.

Special Relativity made simple and General Relativity made simple. The links are by my name.

Ok, found them. I'll take a look.

I forgot to quote this bit.So it does exist. In fact it's everything that exists. Anything reaching the horizon would reach c relative to anything outside the horizon.

You're assuming that the entire spacetime can be covered by a family of hovering observers, all at rest relative to each other. That's another form of the mistake I referred to at the start of this post.

It's not fine. It means that anything inside the horizon is moving faster than c relative to everything outside it.

And your continued refusal to accept this possibility in a curved spacetime is another form of the same mistake.

An object free-falling in curved space-time does always feel weight, but it’s called tidal force instead. That’s what “curved space-time” means.

No, it isn't. If you really think you're saying something relevant here, you will need to give a detailed scenario that illustrates why you think objects moving solely under the influence of tidal gravity feel weight. In standard GR (and in standard Newtonian gravity, for that matter), they don't.

I would have thought you could view the light the cones as tilted when accelerating in flat space-time in exactly the same way as you can with curved space-time. The Rindler horizon would be when they tilt to 90 degrees?

No. In flat spacetime the light cones are never tilted at all. All of the light cones throughout the spacetime line up exactly with each other. That's part of what "flat spacetime" means.

Also, the motion of an observer doesn't affect the light cones, in either flat or curved spacetime.