• yuiop
yuiop
It seems odd that Fredrik and Pete0302 seem to disagree on this topic and yet they both manage to draw wrong conclusions.

...
I saw that you talked about two people jumping from different floors of the same building earlier in this thread. Your argument seems to be that since the distance between those guys (as measured by one of them) won't increase, even if we pretend that the pull of gravity doesn't vary with altitude, the distance between the rockets (as measured by one of the rockets) won't either. I would reverse that argument and say that since my posts above prove in a very simple way that the string must break, we are forced to conclude that the distance between the two guys in free-fall must also increase.

The distance between the two guys in free-fall in a uniform gravitational field will not increase.

The reason is this:

2 observers (A and B) that are in free fall in a uniform free fall (after jumping off a tall building is equivalent to 2 inertial observers (C and D) far out in space far away from any large gravitational body that are at rest with respect to each other and watching a tall but massless building being accelerated past them by a rocket.

Observers (C and D) do not see the distance between them as changing with time and any string joining them will not be stretched and the same is true for observers A and B. It is is also true that observers A and B and observers C and D do not feel any acceleration. They only see it when they look at the building.

2 observers (E and F) that are accelerating with constant equal acceleration according to an inertial observer are not equivalent to the situation of observers A and B and nor are they equivalent to the situation of observers C and D and this is made clear when it is noted that observers E and F feel acceleration and would know the difference even with their eyes closed.

2 observers (G and H) that are on separate floors of a tall building located in a gravitational field (and at rest with their respective floors) are in an equivalent situation to observers (J and K) that are in a massless building far away from any significant massive body that is being accelerated artificially. An inertial observer watching the building being accelerated in space sees the building as length contracting as it accelerates. Observers G and H in the gravitational field obviously do not see the distance between floors of their building as increasing over time. Because the situations of observers G and H and J and K are equivalent it is also obvious that observers J and K do not not see their separation as increasing over time. Neither the situation of observers G and H and J and K are are equivalent to that of observers E and F even though all thee pairs of observers actually feel acceleration.

The nearest equivalent to the situation of observers E and F is that of two observers in a building being artificially accelerated and being artificially expanded at the same time, or two observers on different floors of a tall building in a gravitational field where the building is getting rapidly taller over time. Observers E and F are equivalent to the classic situation described in Bell's paradox and Bell's paradox can not be compared to the situation of observers (A and B) or (C and D) or (G and H) or (J and K) as described above because none of those situations are equivalent.

...

I solved the original problem completely in my first two posts in this thread, so I encourage you to take a look at them again and try to find something wrong with my argument. l also recommend that you draw a space-time diagram.

You could always refer to the diagram I posted in post#1 of this thread which has the paths and points in spacetime accurately drawn using geometrical software with coordinates tranformed using the the Lorentz transformations.

What if the Earth accelerated right along with them as soon as they launched? Would the string still break?

The original Bell's paradox does not include the Earth as a gravitational body but just as a point of reference. As in the twins paradox the Earth is not meant to represent a source of acceleration and is loosely used as inertial reference frame even though it is not in reality. In the though classic thought experiments, the Earth is imagined to be an ideal massless point of reference with no significant gravitational field. As such it would make no difference if the Earth accelerated right along with accelerating rockets. The rockets are only required to maintain constant proper acceleration which they can measure without even looking out of a window by using onboard accelerometers. If the Earth is replaced by a small spacestation it should be clear that the spacestation accelerating after the rockets have accelerated would make little difference to the proper acceleration measured by the onboard rocket accelerometers.

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Would this not also apply to the two rockets when they toss their first rocks in Earths frame?
It is a co-moving frame w/r/t the rockets, but this implies that the two rocket's clocks are not synced.
You can consider four different frames here. The frames can be co-moving with a rocket either before or after the toss, and they can be co-moving with either rocket A or rocket B.

The two frames that are co-moving with the rockets before the toss have the same velocity as the original rest frame. (I prefer not to think of it as the Earth frame, because that suggests gravitational effects that we aren't going to consider anyway). The only difference between them is a translation. At this time, in these frames, the clocks are both showing 0.

If we consider the frame that's co-moving with rocket B (which is behind rocket A) after the toss, then the clock on B still shows 0 immediately after the toss (assuming that the toss was instantaneous). The event on rocket A that's simultaneous with the event where rocket B tossed its first rock, occurs some time after rocket A tossed its first rock. So at this time, in this frame, clock A shows 0 and clock B shows a positive time.

If we consider the frame that's co-moving with rocket A after the toss, then the clock on A still shows 0 immediately after the toss. The event on rocket B that's simultaneous with the event where rocket A tossed its first rock, occurs some time before rocket B tossed its first rock. So at this time, in this frame, clock B shows 0 and clock A shows a negative time.

yuiop
Let Z be a spacestation of isgnificant mass far out in space.

A and B are 2 rockets initially at rest with Z and spatially separated along the x-axis on which observers A,B and C all lie on.

Rockets A and B accelerate simultaneously according to observer Z to reach a new velocity v relative to Z.

Rockets A and B accelerate simultaneously again as observered by observer Z.

Since observer Z sees the the second burst of acceleration happen simultaneously, it follows that the the rocket observers in the frame with velocity v relative to Z would not see the the second burst of acceleration as happening simultaneously.

This is because if two spatially separated events are simultaneous in one reference frame (Z) it is impossible to find another inertial reference frame (Z') that has velocity relative to Z such that frame Z' will also consider the events to be simultaneous.

Therefore the assumption the two rockets share an (instantaneous) inertial reference frame is incorrect or the assumption that they also see the acceleration events as simultaneous is incorrect.

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It seems odd that Fredrik and Pete0302 seem to disagree on this topic and yet they both manage to draw wrong conclusions.

The distance between the two guys in free-fall in a uniform gravitational field will not increase.
It's not that strange, considering that we're talking about two different problems. It's certainly possible to be right about one and wrong about the other.

I know I understand the spaceship problem, because I've been thinking about that a lot, but it seems I didn't give the other problem enough thought.

You could always refer to the diagram I posted in post#1 of this thread which has the paths and points in spacetime accurately drawn using geometrical software with coordinates tranformed using the the Lorentz transformations.
Oooh, nice. I actually didn't read #1 until now.

One thing that would be nice to see in the diagram on the left is this: Two dots, somewhere in the middle of each world line, marking two events where the two clocks show the same time. (They would be on the same horizontal line in the diagram). Also, simultaneity lines, showing what events co-moving observers consider simultaneous with those two events.

2 observers (E and F) that are accelerating with constant equal acceleration according to an inertial observer are not equivalent to the situation of observers A and B and nor are they equivalent to the situation of observers C and D and this is made clear when it is noted that observers E and F feel acceleration and would know the difference even with their eyes closed.

2 observers (G and H) that are on separate floors of a tall building located in a gravitational field (and at rest with their respective floors) are in an equivalent situation to observers (J and K) that are in a massless building far away from any significant massive body that is being accelerated artificially. An inertial observer watching the building being accelerated in space sees the building as length contracting as it accelerates. Observers G and H in the gravitational field obviously do not see the distance between floors of their building as increasing over time. Because the situations of observers G and H and J and K are equivalent it is also obvious that observers J and K do not not see their separation as increasing over time. Neither the situation of observers G and H and J and K are are equivalent to that of observers E and F even though all thee pairs of observers actually feel acceleration.

This seems to disagree with...

Acceleration does not cause time dilation. This is known as the http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Clock_Hypothesis" and has been experimentally verified up to about 10^18 g. Consider also muons created from cosmic rays in the upper atmosphere. They do not accelerate but instead are created at their high relative velocity. They are a textbook example of time dilation without acceleration.

You can either say that velocity causes time dilation or that time dilation is just what happens when a clock takes a shorter path through spacetime. I prefer the second approach, which is the spacetime geometric explanation.

I take this to mean that E and F is equivalent to G and H and J and K.

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Sorry I just realized my mistake with what Kev was saying. But I tend to think that this paradox is talking about the G,H / J,K pairs not the E,F pair. What makes you think its the E,f pair and not the others?

yuiop
...

Oooh, nice. I actually didn't read #1 until now.

One thing that would be nice to see in the diagram on the left is this: Two dots, somewhere in the middle of each world line, marking two events where the two clocks show the same time. (They would be on the same horizontal line in the diagram). Also, simultaneity lines, showing what events co-moving observers consider simultaneous with those two events.

I have uploaded two more diagrams.

The first has solid magenta lines added that show the lines of simultaneity of an inertial observer that remains in the initial reference frame of the green and yellow Bell's rockets.

The second has solid blue lines added that shows the lines of simultaneity of the blue rocket that represents an inertial observer that stays in the final reference frame of the Bells rockets.

Not exactly what you asked for but the rest can easily be extrapolated.

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yuiop
Sorry I just realized my mistake with what Kev was saying. But I tend to think that this paradox is talking about the G,H / J,K pairs not the E,F pair. What makes you think its the E,f pair and not the others?

Have a look at this link to a drawing of the classic Minkoski diagram of an accelerating rocket. http://www.mathpages.com/home/kmath422/Image5456.gif it has the worldlines converging and represents Born rigid acceleration ,where the accelerating observers measure their separation to be constant while the inertial observer measures the separation of the accelerating observers to be length contracting. This represents the classic equivalence of an artificaly accelerated rocket to gravitational acceleration. The equivalence princple is only valid over a short range so if you wanted to nit pick you could point out that this represents acceleration proportional to GM/R rather the normal Newtonian GM/R^2. You could also point out that this classic equivalence diagram shows the event horizon being at the origin rather than at 2GM/Rc^2 but we won't go there for now. However, the point is that the acceleration is inversely proportional to the distance from the origin as is normal for gravitational type acceleration. This represents the G/H and J/K pairs.

The rockets in Bell's paradox experience constant and equal acceleration. Their wordlines in the Minkowski diagram would be parallel (translated) rather than converging as shown in the Born rigid case. The bell's rockets have acceleration that is independent of the displacement along the x axis. That is the E/F type acceleration described earlier.

peter0302
I believe I finally found the flaw in the logic and, I believe this proves the string will not break.

The very first assumption they make is that the following equations are true for all times "t":
x_a(t) = a0 + f(t)
x_b(t) = b0 + f(t)

This assumes that the observer's coordinate (let's call this O) is zero. But in reality the equations are:
x_a(t) = (a0 - O) + f(t)
x_b(t) = (b0 - O) + f(t)

Now here's where I believe the error is. I believe (a0 - O) and (b0 - O) need to be divided by gamma because those lengths, corresponding to the respective starting points of the rockets, should be length contracted just as surely as the length of the rockets themselves, no different than if the rockets had tails that extended all the way back to the Observer at the start of the experiment. In other words, once the rockets are moving at relativistic velocities, you can no longer simply add their original starting points relative to the inertial observer when the rockets were at rest without accounting for relativistic effects - length contraction - on that displacement caused by high velocity of the rocket. Otherwise you're adding apples and oranges.

If that's right, then the equation becomes:

x_a(t) = (a0 - O)/gamma + f(t)
x_b(t) = (b0 - O)/gamma + f(t)

x_a(t) - x_b(t) = (a0-O-b0+O)/gamma

x_a(t) - x_b(t) = (a0-b0)/gamma

So the apparent distance between the two rockets DOES length contract, exactly as we would expect it to if they were treated as one object, they do not appear to accelerate at the same rate in the inertial observer's frame, although they do in their own frame, and therefore the string does not break.

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This assumes that the observer's coordinate (let's call this O) is zero. But in reality the equations are:
x_a(t) = (a0 - O) + f(t)
x_b(t) = (b0 - O) + f(t)

Now here's where I believe the error is. I believe (a0 - O) and (b0 - O) need to be divided by gamma because those lengths, corresponding to the respective starting points of the rockets, should be length contracted just as surely as the length of the rockets themselves,
Those lengths are Lorentz contracted in a frame that's co-moving with one of the rockets, but the equations above are in the original rest frame.

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I believe I finally found the flaw in the logic and, I believe this proves the string will not break.

The very first assumption they make is that the following equations are true for all times "t":
x_a(t) = a0 + f(t)
x_b(t) = b0 + f(t)

This assumes that the observer's coordinate (let's call this O) is zero. But in reality the equations are:

x_a(t) = (a0 - O) + f(t)
x_b(t) = (b0 - O) + f(t)
There is no third observer in Wikipedia's presentation of the experiment. And if there was a third observer stationary in the (t, x)-coordinate chart, a0 was defined to be the initial x coordinate of spaceship A, not the initial coordinate distance between the third observer and the third observer, so you'd still be wrong.

Edit:Corrected a typo

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There is no third observer in Wikipedia's presentation of the experiment. And if there was a third observer stationary in the (t, x)-coordinate chart, a0 was defined to be the initial x coordinate of spaceship A, not the initial coordinate distance between the third observer and the third observer, so you'd still be wrong.
FYP

peter0302
Neither of you addressed my point, that you cannot use a measurement taken in an inertial frame and add it to a distance calculated using the proper acceleration of a relativistically moving object. You've got to include the gamme factor in the original distance measurement (a0, b0), OR adjust the displacement function for one of the two ships, which means f(t) for the first ship is not the same as f(t) for the second ship in the inertial observer's frame. Either way, you get the same result: x(a)-x(b) length contracts as the ships move faster.

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Neither of you addressed my point,
Your argument contains errors, and is therefore invalid.

And both of our responses touch upon what I think is your underlying problem -- in your (misguided) attempts to apply relativistic ideas, you've apparently forgotten how to use coordinates to analyze a problem.

peter0302
Since you can't give me any specific criticisms, your critiques are not helpful.

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Since you can't give me any specific criticisms, your critiques are not helpful.
There is no third observer in Wikipedia's presentation of the experiment. And if there was a third observer stationary in the (t, x)-coordinate chart, a0 was defined to be the initial x coordinate of spaceship A, not the initial coordinate distance between the third observer and the third observer, so you'd still be wrong.

Wait a minute, didn't I already give this specific criticism of your agument?

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Neither of you addressed my point
Since you can't give me any specific criticisms, your critiques are not helpful.
We both pointed out where your argument goes wrong. I'll try again.

You seem to have misunderstood the meaning of the formula $x_A(t)=a_0+f(t)$. The left-hand side is the position of rocket A, in the original rest frame, at time t. The first term on the right is the position of rocket A, in the original rest frame, at time 0. The second term on the right is the distance that rocket A has moved, in the original rest frame, at time t.

It's all in the original rest frame, so it doesn't add apples to oranges. That's what you're doing when you're using a modified version of this formula with one term expressed in a different frame.

To say that a0 is Lorentz contracted in the original rest frame is equivalent to saying that the distance between my eyes gets Lorentz contracted in my rest frame when someone on my right starts walking away from me.

I guess the thing that bothers me about this is the whole simulatanity idea and what exactly it mean when. The way I see it is that anytime you view an object with a velocity (I'm leaving gravity out of this for simplicity) you lose simualtanity. I.e. you get the ladder through the barn. But when I think about this I look at it as instances. For example:
Instant 1) Everything is at rest (no contraction, velocities, and everything is simulataious.
Instant 2) The 2 rockets toss out a rock (for simplicity I like rocks because its instantanious change in velocity) and have the a change in velocity. At this time the spacestation no longer shares a common frame and it losses simualtanialty w/r/t the rockets. (I do not see how this is any differnt than if the space station tossed a rock in the other direction and left the 2 rockets in place.)
Instant 3) The station now views the back rocket tossing the rock before the front rocket does. This tells me that it would appear as though the back rocket is accelerating faster and catching up to the back rocket. (all the while both rockets are still seeing each other sitting still, again just as the spacestation was launching rocks.)
Instant 4) Instant 3 would continue until from the spacestations view the back rockets stops tossing rocks and then the front rocket would finish tossing rockets leaving the distance between the 2 rockets shortented (contacted) and constant because they would both now have the same velocity according to the space station. (From the view of the rockets they have never moved in relation to each other.

Insite as to this flaw would be wonderful. =)

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2 observers (A and B) that are in free fall in a uniform free fall (after jumping off a tall building is equivalent to 2 inertial observers (C and D) far out in space far away from any large gravitational body that are at rest with respect to each other and watching a tall but massless building being accelerated past them by a rocket.
I'd like to continue the discussion about the two guys jumping from different floors. I would like to see a geometric explanation too. It's not that I don't accept the simple solution by direct reference to the equivalence principle. I'm just interested in what a an explanation in terms of space-time geometry would look like.

What are we really saying here? That the spatial distance between two parallel geodesics is the same everywhere? Isn't that only true when there's no curvature? (It certainly isn't true on a sphere). Is there curvature in this case? Is a "homogeneous gravitational field" a curved space-time, or is it a flat space-time with some other funny property?

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Instant 3) The station now views the back rocket tossing the rock before the front rocket does.
This one is wrong. In the station's frame, the two rockets always toss their rocks at the same time. You may have confused simultaneity with the different times it takes information about what happened on the rockets to reach the space station.

peter0302
There is no third observer in Wikipedia's presentation of the experiment. And if there was a third observer stationary in the (t, x)-coordinate chart, a0 was defined to be the initial x coordinate of spaceship A, not the initial coordinate distance between the third observer and the third observer, so you'd still be wrong.

Wait a minute, didn't I already give this specific criticism of your agument?
Wow, I knew the egos would start making an appearance before long.

It is irrelevant whether there is an observer in Wikipedia's version. a0 and b0 are with respect to some origin, which I am calling "O". It is the distance between "O" and "a0" and "b0", respectively, which must contract, and therefore, x_a(t) - x_b(t) must also contract.

It's also (or alternatively, depending on how you want to look at it) incorrect to assume that f_a(t) = f_b(t). They are accelerating in THEIR reference frames - it is not an external force in the intertial frame that is being applied. Their velocities are therefore the same at all times. They are in the same frame at all times. Things are simultaneous for them at all times. Therefore, things must be non-simultaneous for the inertial observer, which is why he sees their separation decrease.

One of the things about being a lawyer is you know how to spot when someone's not addressing your argument, and this is one of those cases. I'm tired of belaboring the point. You disagree, fine. I'll have to talk to the folks at CERN, I guess, who are the only ones who apparently know what's going on.

This one is wrong. In the station's frame, the two rockets always toss their rocks at the same time. You may have confused simultaneity with the different times it takes information about what happened on the rockets to reach the space station.

I believe if you look closely at the ladder through the barn thought expiroment and call the doors rockets and the ladder a spacestation and the opening/closing of the doors tossing rocks you will find that is correct.

yuiop
...
The distance between the two guys in free-fall in a uniform gravitational field will not increase.

The reason is this:

2 observers (A and B) that are in free fall in a uniform free fall (after jumping off a tall building is equivalent to 2 inertial observers (C and D) far out in space far away from any large gravitational body that are at rest with respect to each other and watching a tall but massless building being accelerated past them by a rocket.

I'd like to continue the discussion about the two guys jumping from different floors. I would like to see a geometric explanation too. It's not that I don't accept the simple solution by direct reference to the equivalence principle. I'm just interested in what a an explanation in terms of space-time geometry would look like.

What are we really saying here? That the spatial distance between two parallel geodesics is the same everywhere? Isn't that only true when there's no curvature? (It certainly isn't true on a sphere). Is there curvature in this case? Is a "homogeneous gravitational field" a curved space-time, or is it a flat space-time with some other funny property?

For what it worth, while I said the distance between two observers jumping from different floors remains constant that is only true in a uniform gravitational field where the acceleration of gravity is inversely proportional to gravitational radius. If two observers jump from different floors of a building in a normal gravitational field like that of the Earth with acceleration proportional to 1/R^2 then the gap will gradually increase because the lower victim will experience greater acceleration because she is nearer the centre of the massive body they are falling towards. This is a tidal effect and (as far as I know curvature is synonomous with tidal effects - but don't quote me) That in itself in also a simplification because the coordinate velocity of a falling object starts to de-accelerate at a certain point due to time dilation. The coordinate speed of light deep in a gravitational well is slower than higher up. That means that at certain altitudes in a strong gravitational field it is entirely possible that if a stationary object that is released it will accelerate while an object dropped from higher up will be slowing down (in coordinate terms) at the same point. Things get complicated at this point because it becomes apparent that the gravitational acceleration of a falling body depends not only on the distance of the falling body from the massive body it is falling towards but also on the instantaneous velocity of the falling object. That is something I would like to explore in more detail in the future.

The equivalence principle only applies where the two observers are so close to each other that the difference in acceleration is negligable. It is a bit like the assumptions made in the aproximation of mgh for potential energy where g is considered to be constant over height h.

yuiop
... You disagree, fine. I'll have to talk to the folks at CERN, I guess, who are the only ones who apparently know what's going on.

If I recall correctly, I think Bell said the folks at CERN eventually agreed with Bell after a period of reflection on the problem. It was just their initial instincts that disagreed with Bell's conclusion.

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It is irrelevant whether there is an observer in Wikipedia's version. a0 and b0 are with respect to some origin, which I am calling "O". It is the distance between "O" and "a0" and "b0", respectively, which must contract, and therefore, x_a(t) - x_b(t) must also contract.
It's also irrelevant if you take the spatial position of O to be zero or not. It's not wrong to introduce a new frame that has the same velocity as the original rest frame and a different spatial origin. It's just unnecessary.

You are however doing something that's very wrong: You're not paying attention to what frames you're using.

The spatial distance between the points that have spatial coordinates a0 and O in the original rest frame, is not Lorentz contracted in that frame. It is Lorentz contracted in frames that are co-moving with one of the rockets, but that's irrelevant. It's irrelevant because the equation you want to change is expressed in the coordinates of the original rest frame.

If you want to use a version of $x_A(t)=a_0+f(t)$ that has a gamma factor on a0, you must express all three terms in a co-moving frame, not just one of them.

One of the things about being a lawyer is you know how to spot when someone's not addressing your argument, and this is one of those cases.
It really isn't. You made a huge error in the first step. I explained what your mistake was, and Hurkyl did the same. Then you said that we weren't specific enough to be helpful. I honestly have no idea what can be more specific and more helpful than telling you what your mistake was, and explaining why it was a mistake.

After that, I explained it again, and you still claim that we haven't addressed your argument. What a bizarre thing to say. It's been addressed three times, by two different people. I'm explaining it again in this post, so now it's four times.

By the way, why haven't you adressed my arguments? (My first two posts in this thread).

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I believe if you look closely at the ladder through the barn thought expiroment and call the doors rockets and the ladder a spacestation and the opening/closing of the doors tossing rocks you will find that is correct.
No, this is something I'm sure of. The rockets would always toss their rocks at the same time in the original rest frame. This is a consequence of translation invariance and the fact that the rockets are identical. (If the world lines of the two rockets aren't exactly the same in the original rest frame, then either the rockets aren't identical or the laws of physics are different at different positions in space).

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For what it worth, while I said the distance between two observers jumping from different floors remains constant that is only true in a uniform gravitational field where the acceleration of gravity is inversely proportional to gravitational radius.
I assume you meant "where the acceleration of gravity is constant", or "isn't inversely proportional..."

Yes, I understand that the "gravitational field" must be constant in a large enough region for your previous argument to hold. I'm just curious what the geometry of space-time is in such a region, and also what the solution to this problem is in terms of geometry.

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Wow, I knew the egos would start making an appearance before long.
I didn't realize that sarcasticly remarking that you have ignored my post was an 'ego'.

It is irrelevant whether there is an observer in Wikipedia's version. a0 and b0 are with respect to some origin, which I am calling "O". It is the distance between "O" and "a0" and "b0", respectively, which must contract,
a0 and b0 are simply numbers. The phrase "distance between "O" and "a0" is nonsensical.

Just to make the point, I will remind you that this is all I need to say: I have identified a particular flaw in your argument, and that is sufficient to invalidate the it. What comes next is purely for your benefit, in hopes that it will help you learn something.

You can locate an object "O" at the origin in the original coordinate system. You can also place another object "S" at rest relative to "O", separated from it by a proper distance a0. Then, if you were to consider other coordinate frames, the coordinate distance between the objects S and O would indeed appear to be length contracted... but that doesn't change a0, which is merely a number.

and therefore, x_a(t) - x_b(t) must also contract.
Again, xa(t) - xb(t) is simply a number, and it is nonsense to talk about it contracting.

It's also (or alternatively, depending on how you want to look at it) incorrect to assume that f_a(t) = f_b(t). They are accelerating in THEIR reference frames - it is not an external force in the intertial frame that is being applied.
It's not an assumption. If you actually grind through the calculus, you will find the explicit formula (for some constant K):
$$f_a(t) = f_b(t) = \sqrt{K^2 + c^2 t^2} - K$$
It's far simpler, however, to apply translation invariance: if you translate rocket A to the right by b0-a0, it would follow the same path as rocket B, from which it's clear that fa = fb.

(the given formula is only for the interval of t's in which the rocket is accelerating)

Their velocities are therefore the same at all times.
By what measure of simultaneity are you making that assertion? This statement is true if simultaneity is determined by the original coordinate system, but it looks like that's not the measure you're using.

They are in the same frame at all times.
By what measure of simultaneity are you making that assertion? And I assume what you mean to say is that they are "stationary relative to each other at all times"... (using the measure of simultaneity you specify)

Things are simultaneous for them at all times. Therefore, things must be non-simultaneous for the inertial observer,
What "things"? How is 'simultaneous for them' measured?

One of the things about being a lawyer is you know how to spot when someone's not addressing your argument, and this is one of those cases.
Mathematics has somewhat stricter standards for the word 'argument' than a courtroom.

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yuiop
No, this is something I'm sure of. The rockets would always toss their rocks at the same time in the original rest frame. This is a consequence of translation invariance and the fact that the rockets are identical. (If the world lines of the two rockets aren't exactly the same in the original rest frame, then either the rockets aren't identical or the laws of physics are different at different positions in space).

I am in total agreement with Fredrik on this point.

yuiop
I assume you meant "where the acceleration of gravity is constant", or "isn't inversely proportional..."

Yes, I understand that the "gravitational field" must be constant in a large enough region for your previous argument to hold. I'm just curious what the geometry of space-time is in such a region, and also what the solution to this problem is in terms of geometry.

Yes, there is some ambivalance in my statement and well done for spotting it ;) It is a point I am not absolutely clear on. In another thread a long discussion was had about the difference between flat and curved space and whether you can have a gravitational field and still call it flat space. As usual, as with most of the threads the issue was never really settled to the satifaction of all parties and put down to differences of semantics etc.

Now the classic equivalence of an accelerating rocket and a gravitational field is usually illustrated with a minkowski diagram like the one attached to this post and
using acceleration equations as defined here http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken] with acceleration a=1/R where R is the distance from the orgin.

Refering to the diagram, the two parallel vertical green lines represent two inertial observers in "non-gravitational" space. A notices that the part of the rocket goingpast him at t=0 has an acceleration of 1/2 while B notices that the part of the rocket going past him is accelerating with an acceleration of 1/3. In other words they are not measuring the same acceleration. This is equivalent to observers A and B free falling in a gravitational field that varies according to 1/R (inversely proportional to radial displacement) rather than the usual 1/R^2 (inversely proportional to the square of the radial displacement) of Newtonian gravity due to a spherical mass. So the conclusion is that in order for the free falling observers to measure there separation as constant while falling the field has to vary according to 1/R rather than the constant acceleration that is usually assumed when people talk of a uniform gravitational field. The uniform part is talking about the acceleration being uniform horizontally which is not the case in the curved gravitational field of a spherical body. For example, a very long straight rod placed on the surface of spherical body will not experience the same acceleration at the extremities of the rod than at the part of the rod that is touching the surface of the sphere due to the curvature of the surface and gravitational field of the spherical body. That is my interpretation. I hope it makes some sort of sense :P

Going back to the flat/curved debate the following canbe noted. The force of gravity due to an infinite flat plate would be independant of the distance of the test particle from the plate (constant). The force of gravity due to an infinite cylinder varies according to the inverse of the distance (F=GMm/R). The field is flat parallel to the long axis of the cylinedr while it is curved transverse to the long axis of the cylinder. The force of gravity due to a spherical body varies according to the inverse of the square of the radial distance (F=GMm/R^2) and the field is curved whatever the orientation.

The equivalence principle requires that observers inside a closed lab that is being artificially accelerated would be unable to determine that they are not in a gravitational field. With accurate enough measurements (in a fairly large artificially accelerated lab) they could detect that the acceleration varies according to 1/R vertically and is constant in all directions, transverse to the acceleration.

The equivalence principle therefore requires that there is at least a theoretical/hypothetical gravitational body that has a field that varies according to 1/R and is flat in any orientation or the artificially accelerated observers could claim they know for certain that they are not in a gravitational field. Fortunately there is such a hypothetical gravitational body. An infinitely long rod with a square cross section would have the desired properies.

[EDIT] None of the above is "text book" and I present the above thoughts for discussion and clarification.

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yuiop

[EDIT] There is still the problem that particles falling in a lab in the field of the square sectioned infinite rod would not fall parallel to each other so we are back to the equivalence principle being a local principle where the lab is small enough that the fact that falling paths are not exactly parallel is aproximated out. As far as I can tell then, there is no one hypothetical body that can satisfy both the requirements of parallel falling paths and gravity proportional to 1/R to avoid the equivalence principle being an aproximation that is only accurate within an infinitesimal local region.

It can be shown that if a circle is aproximated by a polygon, the perimeter of the polygon gradually gets closer to the value 2piR as the number of sides is increased, but no matter how many sides the polygon has it never quite exactly matches 2piR. Can it be proven that the aproximations made in the equivalence principle will not ultimately fail in accuracy in the same way, when GR is pushed to extremes?

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General response to #65

I am still reading through all the arguments presented throughout this thread, as well as the references #1. As I understand the Bell paradox so far, the problem seems to involve both length contraction and lines of simultaneity, both of which are effects of relativity. However, the presence of acceleration in the paradox seems to have led to much discussion about the equivalence of gravity and acceleration and their relativistic effects. I also noted a question raised in #65, which I would like to try and get some initial clarification.

In another thread a long discussion was had about the difference between flat and curved space and whether you can have a gravitational field and still call it flat space.

I will state my initial question and then provide some background as to why it is being raised:

Are acceleration and gravity relativistic effects rather than the cause?

As a generalisation, relativistic effects on spacetime are often described in terms of an associated value of $$[\gamma]$$. Normally, the value of $$[\gamma]$$ is defined in terms of either velocity and/or gravity, i.e.

[1] $$\gamma_v = \frac{1}{\sqrt{1-v^2/c^2}}$$

[2] $$\gamma_g = \frac{1}{\sqrt{1-Rs/r}}$$

Where $$[Rs=2GM/c^2]$$ corresponds to the Schwarzschild radius, which if substituted into [2] gives:

[3] $$\gamma_g = \frac{1}{\sqrt{1-2GM/rc^2}}$$

However, [3] can be transposed further in terms of gravitational acceleration [g] via the classical equation $$F = ma = GMm/r^2}$$, such that $$[a=g=GM/r^2]$$, which from [3] seems to lead to:

[4] $$\gamma_g = \frac{1}{\sqrt{1-2gr/c^2}}$$

Now while [g] is acceleration due to gravity and there is the general acceptance of the equivalence of gravity and acceleration, equation [4] does not directly relate the value of $$[\gamma]$$ to [g], but rather the product [gr]. I believe this is best illustrated by 2 examples:

Case-1:
A super-massive black hole (M=1.5E12) solar masses has an event horizon [Rs=4.55E15m], but a relatively small value of [g=9.82], i.e. directly comparable to Earth’s gravity. However, the product [gr], where [r=Rs] leads to an infinite value of $$[\gamma_g]$$.

Case-2:
In contrast, another black hole (M=3.84) solar masses has an event horizon [Rs=1E4m], but with an enormous value of [g=4.47E7]. However, with [r=100Rs], the product [gr] leads to a value of $$[\gamma_g=1.01]$$.

So the implication seems to be that gravitational acceleration itself does not affect the geometry of spacetime, rather the product [gr] defines a position in spacetime, which is subject to curvature due to mass [M] that then leads to a given value of [g].

If the assumptions forwarded are valid, does this mean that acceleration [a], in isolation, has no effect on spacetime, other than leading to a variable velocity, which affect $$[\gamma_v]$$ not $$[\gamma_g]$$ ?

In part, the reason for raising these issues was to determine whether there was any consensus that the Bell paradox could be resolved in terms of special relativity only. However, would appreciate any other thoughts on the issues raised.

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peter0302
If you want to use a version of $x_A(t)=a_0+f(t)$ that has a gamma factor on a0, you must express all three terms in a co-moving frame, not just one of them.
Ok! I agree with you! That is why f(t) cannot be equal for both rockets in the inertial frame. If the two rockets are self-propelling themselves at a constant 1g, then the "Rocks" so to speak are being thrown at the same time in their co-moving frame, and therefore not simultaneously in the inertial frame. If the rockets were being pushed by a force originating in the _inertial_ frame, then you'd be right.

I was just saying that length contracting the original distance between them accomplished the same thing.

By what measure of simultaneity are you making that assertion? And I assume what you mean to say is that they are "stationary relative to each other at all times"... (using the measure of simultaneity you specify)
Yes, for the umpteenth time, the ships are propelling themselves at a constant rate. The laws of physics must hold for them in their frames. Therefore, by any measure of displacement vs. acceleration, they each measure their velocity wrt the Earth to be the same at all times in their frame. So when ship A sees himself at X distance from the searth, ship B sees himself at X + (b0-a0) from the earth.

You know, it's possible I'm a moron. I'll grant you that. So I'm going to shut up after this. But this the last thing I want to say. No one has given a good reason why you don't treat the ships + string exactly the same as you would a single ship with two engines connected by a titanium hull.

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MeJennifer
If the assumptions forwarded are valid, does this mean that acceleration [a], in isolation, has no effect on spacetime, other than leading to a variable velocity, which affect $$[\gamma_v]$$ not $$[\gamma_g]$$ ?
If an object of mass is accelerated it means that energy carriers are exchanged, and, as a consequence, the EM distribution must change, and this implies a change in curvature.

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