# Bell's Spaceships Paradox explained.

• yuiop
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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.

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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.

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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.

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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?

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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.

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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.

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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.

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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.

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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.

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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.

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• Rocket A.PNG
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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.

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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.

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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.

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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.

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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?

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

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.

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

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

Do you guys not see that what I said is equivalent to what Fredrik said? Make T=1 second so that -T = -1 second then it is clear that at x = 1 light second t=0 simultaneously with x=0, t=0. Ergo what I am currently seeing with my eyes at a distance x=1 light second is what happened when my T=-1 second. I just do not have the finesse that Fredrik has =).

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OK, I'm glad we agree. I misunderstood you then.

MeJennifer
Ergo what I am currently seeing with my eyes at a distance x=1 light second is what happened when my T=-1 second. I just do not have the finesse that Fredrik has =).
We cannot see at distances, that is just popular speak.

What you are currently seeing with your eyes are photons hitting your retina with a distance of 0 and it is happening now. Yes these photons have an origin but so does a tennis ball hitting one's head. Would you say that you feel a tennis ball currently hitting your head from a distance that happened in the past?

peter0302
Ok, now that's nitpicky.

Wizardsblade
We cannot see at distances, that is just popular speak.

What you are currently seeing with your eyes are photons hitting your retina with a distance of 0 and it is happening now. Yes these photons have an origin but so does a tennis ball hitting one's head. Would you say that you feel a tennis ball currently hitting your head from a distance that happened in the past?

This is off topic, but our eyes (plural) can distinguish distance by triangulation. Of course we only see what is currently hitting our eyes, hence the definition of simultaneity is not what we see but rather what has already been stated above.