Bell's Spaceships Paradox explained.

[Hurkyl]

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?

 

[Fredrik]

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.

 

[Wizardsblade]

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. =)

 

[Fredrik]

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?

 

[Fredrik]

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.

 

[Wizardsblade]

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.

 

[Fredrik]

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).

 

[Fredrik]

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).

 

[Fredrik]

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.

 

[Hurkyl]

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, x_a(t) - x_b(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 f_a = f_b.

(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.

 

[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 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.

 

[yuiop]

Follow on to post #65

[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?

 

[mysearch]

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.

 

[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.

 

[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.

 

[Hurkyl]

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.

That's because the nose and tail do not undergo the same acceleration, so the situation is quite different. If the effect is large, you have to program the tail engine to give more thrust than the rear engine. If the effect is small, you can simply let the tension in the hull take care of things.

 

[Fredrik]

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.

The fact that f(t) is the same for both follows immediately from translation invariance and the fact that the rockets are identical. And so does the fact that the rocks are thrown at the same time in the original rest frame.

The rockets can't accelerate at different rates in the original rest frame because that would imply that they aren't identical, or that the laws of physics are different at different positions in space.

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.

OK, I understand that this can be confusing. The reason is that the single ship would be approximately Born rigid. I will explain what that means.

Imagine a rocket, originally at rest in an inertial frame, that accelerates extremely fast from 0 to a high velocity and then stops accelerating. Imagine that the acceleration is so high that it should look instantaneous in a space-time diagram. Now, how would you draw the world lines of the endpoints of the rockets? Should you choose option A or B (defined below)?

Option A: Draw them so that the endpoints get accelerated at the same time in the frame that's co-moving with the rocket before the boost.
Option B: Draw them so that the endpoints get accelerated at the same time in the frame that's co-moving with the rocket afterthe boost.

Neither of those choices would be close to what actually happens when a "rigid" object gets accelerated. I'm putting "rigid" in quotes, because there aren't any truly rigid bodies in SR, and I'm about to explain why. Option A would imply that the rocket gets forcefully stretched so that it can remain the same length in the frame that's co-moving with it before the boost. (The forceful stretching compensates for the Lorentz contraction). Option B would impy that the rocket gets forcefully compressed so that it can remain the same length in the frame that's co-moving with it after the boost.

It should be clear from this example that solid objects can't be accelerated without some forceful stretching or compressing of the material. Hence no rigid bodies in SR.

The closest you can get to "actually rigid" is something called "Born rigid". It means that at any time in the original rest frame, the distance between any two nearby points on the rocket will have changed from their original length by a factor of gamma. This is what's expected to happen to solid objects that are accelerated slowly.

This is why the two rockets in Bell's scenario aren't equivalent to a single object. Internal forces in the "single object" would make it approximately Born rigid so that the front accelerates a bit slower than it would have if it had been a separate object with it's own engine.

 

[peter0302]

Option A: Draw them so that the endpoints get accelerated at the same time in the frame that's co-moving with the rocket before the boost.
Option B: Draw them so that the endpoints get accelerated at the same time in the frame that's co-moving with the rocket afterthe boost.

What about:
Option C: Draw them so that the endpoints get accelerated at the same time in the frame that's co-moving with the rockets during the boost.

Option B would impy that the rocket gets forcefully compressed so that it can remain the same length in the frame that's co-moving with it after the boost.

The rocket only appears to be compressed to the inertial observer - exactly as SR says it will.

Sorry, I promised I would shut up.

 

[MeJennifer]

What about:
Option C: Draw them so that the endpoints get accelerated at the same time in the frame that's co-moving with the rockets during the boost.

There is no single comoving frame for the two rockets but instead there are two different comoving frames, one for each rocket.

 

[Fredrik]

What about:
Option C: Draw them so that the endpoints get accelerated at the same time in the frame that's co-moving with the rockets during the boost.

That doesn't make sense since we're talking about an instantaneous boost.

The rocket only appears to be compressed to the inertial observer - exactly as SR says it will.

Yes, in the frame that's co-moving with the rocket before the boost, option B implies that after the boost, the rocket's length is 1/gamma times its rest length. You got that right, but you missed something else:

In the frame that's co-moving with the rocket after the boost, option B implies that the length of the rocket is not changed by the boost, and that contradicts the Lorentz contraction formula.

So option A causes a problem in the "before" frame, and option B causes a problem in the "after" frame. Option A is the only option that avoids the problem in the "after" frame, and option B is the only option that avoids the problem in the "before" frame, so any option that's different from A and B would cause problems in both frames.

Sorry, I promised I would shut up.

No one has asked you to shut up. I don't mind a discussion as long as it's going somewhere.

 

[peter0302]

In the frame that's co-moving with the rocket after the boost, option B implies that the length of the rocket is not changed by the boost, and that contradicts the Lorentz contraction formula.

Why is that a problem? The object that's moving never believes its own length is contracting - it believes everything else is contracting. So in the frame that's co-moving with the rocket after the boost, the length of the rocket indeed should not be changed.

No one has asked you to shut up. I don't mind a discussion as long as it's going somewhere.

I feel we're going in circles for one of two reasons: either my understanding of SR is so off that my points are nonsense or no one is understanding my points. Either way I don't think the discussion is making any progress and I'm sorry if it's my fault.

 

[Fredrik]

The force of gravity due to an infinite flat plate would be independant of the distance of the test particle from the plate (constant).

If someone has calculated that, then they must have calculated the metric of that space-time. I'll probably try to find it later. I'm curious what it would look like.

 

[Fredrik]

Why is that a problem? The object that's moving never believes its own length is contracting - it believes everything else is contracting. So in the frame that's co-moving with the rocket after the boost, the length of the rocket indeed should not be changed.

You seem to be confusing the inertial frame that's co-moving with the rocket after the boost with a non-inertial, accelerating frame.

There's no such thing as "its own frame" here. There's one object, two velocities, and two inertial frames. In one of the frames, the rocket's velocity changes from 0 to v and its length changes by a factor of gamma. In the other frame, the rocket's velocity changes from -v to 0 and its length doesn't change.

I don't know how to make it more clear. Are the laws of physics different in the two inertial frames? Does Lorentz contraction only occur in one of them?

In that second frame, the rocket was Lorentz contracted when it moved with velocity -v, and it isn't Lorentz contracted when it's at rest, but its length at rest is the same as its Lorentz contracted length. This is only possible if the rocket was forcefully compressed at the time of the boost.

I feel we're going in circles for one of two reasons: either my understanding of SR is so off that my points are nonsense or no one is understanding my points.

I don't mean to be rude, but I think you are pretty far off. That's not exactly uncommon though. This stuff is pretty tricky. I do however feel that I understand what you're saying. The things you say are often wrong, but they're not nonsense.

 

[peter0302]

In that second frame, the rocket was Lorentz contracted when it moved with velocity -v, and it isn't Lorentz contracted when it's at rest, but its length at rest is the same as its Lorentz contracted length. This is only possible if the rocket was forcefully compressed at the time of the boost.

That's where you lost me. Why is the rocket's length at rest (when it's finished accelerating) the same as its Lorentz-contracted length when its velocity was -v (before it started accelerating)? Does not an object always have its proper length when it's at rest? What's the basis for arguing that the rocket will be forcefully compressed at the time of the boost? Is this a feature of relativistic acceleration or classical acceleration? In classical acceleration, the system (two rockets plus string) should remain the same size because the extra tension caused by the pull from the front rocket is balanced out by the release of tension from the push of the rear rocket. So the only way your argument works is if there's some feature of acceleration in SR that I'm just not getting.

In short, what's causing this physical compression?

 

[Fredrik]

That's where you lost me. Why is the rocket's length at rest (when it's finished accelerating) the same as its Lorentz-contracted length when its velocity was -v (before it started accelerating)?

Because that's how I defined "option B". It's not implied by the physics. I'm just saying that if we draw the world lines of the endpoints as described by option B, then the physical interpretation of that diagram must be that the rocket was forcefully compressed at the time of the boost.

This stuff is hard to explain. Let's go back to something very basic: What's a rigid body in pre-relativistic physics? It's an object with the property that the distance between any two component parts will remain the same, no matter what you do to it. I'm trying to show you that no such rigid bodies can exist in SR, even as an approximation.

I'm doing that by considering what a space-time diagram of an accelerating rigid body would look like. Option A describes an object that's "rigid" (according to the pre-relativistic definition) in the frame that's co-moving with the rocket before the boost. When we think about the physical interpretation of option A, we see that it represents an object that's being forcefully stretched.

Option B describes an object that's "rigid" (according to the pre-relativistic definition) in the frame that's co-moving with the rocket after the boost. When we think about the physical interpretation of option B, we see that it represents an object that's being forcefully compressed.

What we can learn from this is that if an object is truly rigid in one frame, it isn't rigid in another. This is why we say that there are no rigid objects in SR. Because of that, physicists (I'm guessing Born) have defined something called "Born rigid". An object that goes through Born rigid acceleration has the property that the distance between two nearby points on the object in a frame that's co-moving with one of them is approximately equal to the original proper distance between the points. (The approximation becomes exact in the limit where the proper distance between the points goes to zero).

A rocket is approximately Born rigid if it's accelerated slowly.

Does not an object always have its proper length when it's at rest?

It does, assuming that you have given it enough time to settle down. If you e.g. give something a boost by hitting it with a hammer, it will first compress a bit, and a shock wave will bounce back and forth between the end points for a while, but if you wait a while, it will get its original proper length back.

In short, what's causing this physical compression?

In the unrealistic cases (my options A and B), the answer is that nothing can cause it, except something like a little rocket attached to each atom, forcing each component part to accelerate the way we have chosen. That's why those cases are unrealistic.

In the realistic (Born rigid) case, the only stretching and compressing that's going on is caused by internal interactions (between molecules) that transmit the force from the engine located in the rear to other parts of the rocket, and then strive to restore every part of the rocket to its original rest length in the locally co-moving frames.

 

[Fredrik]

By the way, your description of what should happen in Bell's spaceship scenario (rocks being thrown at the same time in co-moving frames) is a description of Born rigid acceleration of two points that are infinitesimally close. (I'm actually not 100% sure that they need to be infinitesimally close. I should probably give that some thought).

 

[yuiop]

By the way, your description of what should happen in Bell's spaceship scenario (rocks being thrown at the same time in co-moving frames) is a description of Born rigid acceleration of two points that are infinitesimally close. (I'm actually not 100% sure that they need to be infinitesimally close. I should probably give that some thought).

As far as I can recall, observers at the tail and nose of a rocket undergoing perfect and constant Born rigid acceleration will measure their separation distance to remain constant over time no matter how long the rocket is.

 

[Hurkyl]

As far as I can recall, observers at the tail and nose of a rocket undergoing perfect and constant Born rigid acceleration will measure their separation distance to remain constant over time no matter how long the rocket is.

If their method of measurement is proper distance along their line of simultaneity. At least in the case of a constant acceleration (the motion is such that a line of simultaneity for one observer will also be a line of simultaneity for the other one); I am suspicious of this claim for a variable acceleration.

 

[yuiop]

...

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.

If in case 1 a value of r=100Rs as in case 2 ,then both cases will obtain the same value of [\gamma_g], but that same gamma factor is obtained from two very different values of gravitational acceleration, as you have noted. I think gravitational time dilation is more a function of gravitational potential (gradient) than of acceleration.

It can also be noted that the gravitational gamma factor \gamma_g = \frac{1}{\sqrt{1-2GM/rc^2}} at height R from a body can be equated to \gamma_v = \frac{1}{\sqrt{1-v^2/c^2}} where v is the velocity of a hypothetical particle at height R that has fallen from infinity in a classical Newtonian gravitational field.

In this context an analogy can be made. (Like all analogies it should not be pushed too far) . Imagine some fluid fills all space. far from any gravitational bodies the fluid is stationary and uniformly distributed. A physical body moving relative to this imaginary fluid length contracts and time dilates as per the gamma factor of Special Relativity. Now you could imagine this fluid flowing towards gravititational bodies with a flow velocity that is equal to the hypothetical particle free falling from infinity, at any given distance from the gravitational body. Now a physical body that is stationary with respect to the gravitational body would experience this magical imaginary fluid passing through it and time dilate and length contract by a factor that is indentical to the time dilation and length contraction of a body moving with the same velocity relative to the static fluid far out in flat space.

 

[Wizardsblade]

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 do not bleive that translational invariance and "the laws of physics are differnt at differnt positions in space" are the same thing. I say this because I agree that in any inertial frame the laws of physics are the same at differnt positions in space, but for translation to happed an object must undergo a velocity and hence reletivistic effects. Translational invariance implies that rigid bodies exsist.

This stuff is hard to explain. Let's go back to something very basic: What's a rigid body in pre-relativistic physics? It's an object with the property that the distance between any two component parts will remain the same, no matter what you do to it. I'm trying to show you that no such rigid bodies can exist in SR, even as an approximation.

A simple change in position will contract the length, for at least a moment, even if it comes back to rest again. Therefore, because this it is relativistic paradox, translational invariance can not be used as an argument.

 

[Wizardsblade]

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.

Instance 1 - I do not believe anyone has any problems with this =).

Instance 2 - I believe most people accept this. To disprove this one would have to show that there is a relativistic difference between the station tossing a rock and the rockets simualtanialty tossing rocks, ie the view from one frame to another is not identical in length contraction, time dilation etc. There are of course non relativistic differences, who tossed rocks, mass changes etc.

Instance 3 - Assuming one accepts instance 2 and its consequences, one needs only look at the "ladder through the barn" paradox to see this must be the logical outcome from instance 2. In this paradox there are 2 frames that both see each other contracted length and time dilated. The analogy to Bell's paradox, via instance 2, is simply that the ladder is the spacestation, the 2 barn doors are the rockets, and the act of opening/closing the doors simualtanialty is tossing rockets. So that if one agree with instance 2 and that the lack of simualtanialty solves the "ladder through the barn" paradox. Then one must agree with the outcome of instance 3.

Instance 4 - This is just the logical extension and ending of instance 3 that shows the final conclusion. If one has come to an agreement with instance 1-3 then this is the logical physical conclusion.

 

[Fredrik]

I do not bleive that translational invariance and "the laws of physics are differnt at differnt positions in space" are the same thing.

What translation invariance means in SR is that a translation is an isometry of Minkowski space. That implies that if you start with Minkowski space and apply a translation, what you get is still Minkowski space. Translation invariance is a stronger statement than the somewhat ill-defined "the laws of physics are the same at different positions".

If you are unfamiliar with the term "isometry", then think about a linear bijective map between two vector spaces (as an analogy). Such a function is an isomorphism between the vector spaces, and the existence of an isomorphism makes the second vector space equivalent to the first in every way that can possibly matter. The same goes for isometries. Isometries are the isomorphisms of manifolds with metrics.

I say this because I agree that in any inertial frame the laws of physics are the same at differnt positions in space, but for translation to happed an object must undergo a velocity and hence reletivistic effects. Translational invariance implies that rigid bodies exsist.

A translation is just a mathematical function of the form f(x)=x+a, so there are no velocities involved. You're thinking of a boost, which is also a kind of isometry.

Translational invariance implies that rigid bodies exsist.

It certainly doesn't.

A simple change in position will contract the length, for at least a moment, even if it comes back to rest again. Therefore, because this it is relativistic paradox, translational invariance can not be used as an argument.

I don't understand what you're trying to say here. Yes, if I change the velocity of an object, it will contract, and when I restore it to its original velocity it will uncontract. There's no paradox here (or anywhere else in SR).

I used translational variance for one thing only: To argue that the world lines of the two rockets will be identical, as in this picture.

I don't understand how anyone can disagree with that. The rockets have the same properties. The laws of physics are the same at every event they will pass through. The rockets don't interact in any way. It's obvious that the world lines must look like in the picture.

Instance 1 - I do not believe anyone has any problems with this =).

Instance 2 - I believe most people accept this.

1-2 are OK.

Instance 3 - Assuming one accepts instance 2 and its consequences, one needs only look at the "ladder through the barn" paradox to see this must be the logical outcome from instance 2. In this paradox there are 2 frames that both see each other contracted length and time dilated. The analogy to Bell's paradox, via instance 2, is simply that the ladder is the spacestation, the 2 barn doors are the rockets, and the act of opening/closing the doors simualtanialty is tossing rockets. So that if one agree with instance 2 and that the lack of simualtanialty solves the "ladder through the barn" paradox. Then one must agree with the outcome of instance 3.

Not at all. Let's say that rocket A (the one behind rocket B (yes I changed my naming convention from my previous posts to agree with the Wikipedia article)) tosses its second rocket at t=10 seconds. In a frame that's co-moving with rocket A just before the second toss, the second toss is simultaneous with an event on rocket B where the clock on rocket B shows a later time, say t=11 seconds. Why would rocket B wait until its clock shows 11 seconds before it tosses the second rock? It can't do that if the rockets are identical. The rockets being identical means among other things that the computers on the rockets that control the tosses are programmed the same way. What you're saying implies that the computer on rocket B is programmed differently than the computer on rocket A.

Instance 4

...is irrelevant since instance 3 is wrong.

 

[peter0302]

I'm right on the same page as Wizardsblade.

In a frame that's co-moving with rocket A just before the second toss, the second toss is simultaneous with an event on rocket B where the clock on rocket B shows a later time, say t=11 seconds.

I don't think that's right. After the first toss and just before the second toss, rocket A and rocket B are going at the same speed relative to the inertial observer. They're in the same frame. Why would they not be?

Therefore the inertial observer does indeed see the closer rocket accelerate sooner and edge closer to the other.

 

[Fredrik]

I don't think that's right. After the first toss and just before the second toss, rocket A and rocket B are going at the same speed relative to the inertial observer. They're in the same frame. Why would they not be?

It's a pretty simple consequence of how simultaneity works in SR.

Look at the really ugly space-time diagram I have uploaded. It shows the world line of rocket A in the original rest frame, as it tosses a rock and then another. The events that are simultaneous with the event where rocket A throws the second rock, in the frame that's co-moving with rocket A just before the second throw, are the ones on the red line. You're saying that rocket B throws the second rock at the event where the world line of rocket B (not drawn in the diagram) intersects the red line. The clock on rocket B must definitely show a later time then, simply because of the slope of the red line. (I didn't bother to try to get th slope exactly right. It's supposed to make the same angle with the x-axis as the world line makes with the time axis).

Why? Because the two clocks show the same times at events on the blue line. That follows immediately from the fact that the rockets are identical.

 

[Wizardsblade]

1-2 are OK.

I am glade we agree up to instance 2. Then about instance 3 you said...

Not at all. Let's say that rocket A (the one behind rocket B (yes I changed my naming convention from my previous posts to agree with the Wikipedia article)) tosses its second rocket at t=10 seconds. In a frame that's co-moving with rocket A just before the second toss, the second toss is simultaneous with an event on rocket B where the clock on rocket B shows a later time, say t=11 seconds. Why would rocket B wait until its clock shows 11 seconds before it tosses the second rock? It can't do that if the rockets are identical. The rockets being identical means among other things that the computers on the rockets that control the tosses are programmed the same way. What you're saying implies that the computer on rocket B is programmed differently than the computer on rocket A.

It is important to think back to instance 2 where we agreed that there is no relativistic difference between the space station tossing a rock and the two rockets tossing rocks simultaneously in their starting frame. From here say that the simultaneity of the rockets is still in tacked because if we looked at if from the spaceship tossing a rock then it is quite clear that the rockets are completely unaffected in their still stationary frame. Now we can looks at instance 3 the same way as we looked at instance 2 if instance 2 was viewed by a third rocket (rocket C) traveling in the same direction the space station will when it tosses a rock and rocket C has the speed that the space station will achieve (after it has tossed it's rock).

So it seems to me that if you accept instance 2 you must accept the consequence of instance 3. (I can try to draw diagrams if it will help, let me know It will take a day or so though.)

 

[Wizardsblade]

Fredrik,
I think that I see where our difference is coming from. What I call simultaneous takes the travel time of light into account, ei something I see 1 light second away happened 1 second ago, it is not happening right now. The other way is to say what you see now is what is happening now. I believe you use the earlier definition for simultaneous in instance 1-2, but use the latter in instance 3. And this change in convention is causing the confusion.

 

[Fredrik]

It is important to think back to instance 2 where we agreed that there is no relativistic difference between the space station tossing a rock and the two rockets tossing rocks simultaneously in their starting frame. From here say that the simultaneity of the rockets is still in tacked because if we looked at if from the spaceship tossing a rock then it is quite clear that the rockets are completely unaffected in their still stationary frame.

OK, I get what you're saying. You're not making a mistake here. The mistake was in instance 2. I was wrong to agree with it.

If the space station tosses a rock, that can't be equivalent with the rockets tossing one rock each. I have uploaded another ugly space-time diagram to show why.

The diagram shows the world lines of both rockets, in a coordinate system where they are initially at rest, as they give themselves a boost by tossing a rock each. This happens where the world lines intersect the red line. Events on the red line are simultaneous in this frame. Events on the blue line are simultaneous in any frame that's co-moving with one of the rockets at any event on that rocket's world line after that rocket tosses its first rock and before it tosses its second rock. In particular, events on the blue line are simultaneous in the frame that's co-moving with rocket B (the world line on the right) immediately after its first toss. So in rocket B's frame, right after its first toss, rocket A hasn't tossed its first rock yet.

This proves that the situations are not equivalent.

(I hope it's not confusing that I accidentally swapped the meaning of red and blue from my last diagram).

I believe you use the earlier definition for simultaneous in instance 1-2, but use the latter in instance 3. And this change in convention is causing the confusion.

That is absolutely not true. I'm always using the standard definition of simultaneity.

What I call simultaneous takes the travel time of light into account, ei something I see 1 light second away happened 1 second ago, it is not happening right now. The other way is to say what you see now is what is happening now.

That definition of simultaneity would be very confusing and probably also useless. Open any book on SR and you will see that that's not how they do it. This is how they do it:

Assume that both space and time have the same properties at every event, and that the speed of light is c in every inertial frame. Now pick one of those inertial frames and suppose that there's a mirror at some point along the x axis. Suppose also that you emit light from x=0, in the positive x direction, at t=-T, and that it's reflected by the mirror and returns to x=0 at t=T. Now the reflection event must have been simultaneous with the event t=0,x=0. That implies that we must assign time coordinate 0 to the reflection event, and the fact that the speed of light is c implies that we must assign the x coordinate cT to the reflection event.

You could, alternatively, start with the definition of Minkowski space and use that to define the inertial frames mathematically.

Either way, this simple statement will always hold: Two events are simultaneous in an inertial frame if they have the same time coordinate in that frame.

 

[MeJennifer]

something I see 1 light second away happened 1 second ago, it is not happening right now.

What you see is nothing but photons hitting your retina, it is not something away, it is a local event.

 

[peter0302]

So the bottom line is when two objects are separated in space and begin acclerating at the same time with the same proper acceleration, their clocks will not stay in sync. So if they both instantaneously accelerate to .5c, the front rocket will believe the rear rocket hasn't even started moving yet, and the rear rocket will think the front rocket had a huge head start.

Is that right?? That's the only way the Bell argument can work.

 

[yuiop]

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. =)

...
Instance 2 - I believe most people accept this. To disprove this one would have to show that there is a relativistic difference between the station tossing a rock and the rockets simualtanialty tossing rocks, ie the view from one frame to another is not identical in length contraction, time dilation etc. There are of course non relativistic differences, who tossed rocks, mass changes etc.

...The mistake was in instance 2. I was wrong to agree with it...

Hi Wizardblade,
Fredrik was right to disagree (eventually) with instance 2 and everything that follows from that conclusion. There IS a relativistic difference between the station tossing a rock and the rockets simultaneousy tossing rocks. When the space station toses rocks he sees the the rockets getting progressively closer together while the the rockets see themselves as remaining a constant distance apart. That is NOT equivalent to the situation described in the classic Bell's rockets paradox. When the rockets toss rocks simultaneously according to the space station they ramain a constant distance apart as measured by the spacestation, while the rocket observers will measure there separation distance to be increasing over time.

 

[yuiop]

By the way, your description of what should happen in Bell's spaceship scenario (rocks being thrown at the same time in co-moving frames) is a description of Born rigid acceleration of two points that are infinitesimally close. (I'm actually not 100% sure that they need to be infinitesimally close. I should probably give that some thought).

As far as I can recall, observers at the tail and nose of a rocket undergoing perfect and constant Born rigid acceleration will measure their separation distance to remain constant over time no matter how long the rocket is.

This wikipedia article on Rindler Coordinates supports my assertion above. http://en.wikipedia.org/wiki/Rindler_space#Notions_of_distance

"There are other notions of distance, but the main point is clear: while the values of these various notions will in general disagree for a given pair of Rindler observers, they all agree that every pair of Rindler observers maintains constant distance. The fact that very nearby Rindler observers are mutually stationary follows from the fact, noted above, that the expansion tensor of the Rindler congruence vanishes identically. However, we have shown here that in various senses, this rigidity property holds at larger scales. "

The "rigidity property" they talk of here is the property of the mutual separation of rindler observers remaining constant over time and not the usual meaning of a rigid body being infinitely incompressable.

 

[yuiop]

So the bottom line is when two objects are separated in space and begin acclerating at the same time with the same proper acceleration, their clocks will not stay in sync. So if they both instantaneously accelerate to .5c, the front rocket will believe the rear rocket hasn't even started moving yet, and the rear rocket will think the front rocket had a huge head start.

Is that right?? That's the only way the Bell argument can work.

That is the way I visualise the situation.

 

[peter0302]

Ok so now another question - if that's right - WHEN does the string break? It clearly cannot break right away - because then it would be breaking in the PAST of the S frame.

 

[Wizardsblade]

Can't we test this? Isn't this equivalent to a charged particle (the space station) and a wire without current (the electrons are the rockets and at rest). When we apply current we give the “rockets” velocity and we can do the math to see if they contract or stay the same distance apart in the “space stations” frame.

 

[Fredrik]

So the bottom line is when two objects are separated in space and begin acclerating at the same time with the same proper acceleration, their clocks will not stay in sync.

Yes, that's right. The clocks show the same times at events that are simultaneous in the original rest frame, but they don't show the same time at events that are simultaneous in a frame that's co-moving with one of the rockets.

So if they both instantaneously accelerate to .5c, the front rocket will believe the rear rocket hasn't even started moving yet, and the rear rocket will think the front rocket had a huge head start.

Is that right?? That's the only way the Bell argument can work.

Yes, that's exactly how it works.

Ok so now another question - if that's right - WHEN does the string break? It clearly cannot break right away - because then it would be breaking in the PAST of the S frame.

It does break right away. Let's say that it breaks by disconnecting itself from rocket B. This event is located where the blue line intersects the world line of rocket B (the one on the right in the diagram). This event has a higher time coordinate (in the original rest frame) than the event where rocket B tossed its first rock, so it doesn't occur in the past.

Both rockets agree that the string broke because rocket B took off first.

It's not a problem that this event is simultaneous in B's frame with events that are in the past in S's frame. What matters is that there is no inconsistency when you describe all events from one frame. All the "paradoxes" of SR are the result of incorrectly describing different parts of the story in different frames.

 

[Hurkyl]

What I call simultaneous takes the travel time of light into account, ei something I see 1 light second away happened 1 second ago, it is not happening right now.

How do you measure that distance, without first having a notion of simultaneity?