Lorentz Contraction: Exploring Standard Relativity & Bell's Paradox

In summary, the conversation discusses the controversy surrounding Bell's standard spaceship paradox and whether or not the string connecting the spaceships would break in different formulations of relativity. The conclusion is that the string would indeed break due to the changing electromagnetic forces between atoms in the string. However, there is still debate over the details of the calculation and the role of the launch frame's perspective.
  • #246
cfrogue said:
Agreed, but this v is relative to the instantaneous rest frame which is not the launch frame.
This instantaneous rest frame is auxiliary in order to solve the problem.

The paper calls the frame S'.

v is relative to S.
 
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  • #247
atyy said:
v is relative to S.

Let me look at the paper again.
 
  • #248
atyy said:
v does not become 0 in the launch frame unless the ships decelerate.

Geez, I am wrong, you are right.

That is, as the velocity in S increases, the distance between the spaceships in their rest system S′ increases.

http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

This is not good then.

This implies this new distance remains after the acceleration ends.
Do you read it this way?
 
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  • #249
atyy said:
v does not become 0 in the launch frame unless the ships decelerate.

There is an ambiguity.

Although the spaceships are accelerating, the system S′ is a Lorentz system moving at constant velocity. Since each ship is instantaneously at rest in this system, the length d′ = λd is the rest frame distance between the ships. As such, it is the physical distance between the ships. If there were an inextensible cable between the ships, it would snap at the start of motion of the ships.


http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf
 
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  • #250
cfrogue said:
Geez, I am wrong, you are right.

That is, as the velocity in S increases, the distance between the spaceships in their rest system S′ increases.

http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

This is not good then.

This implies this new distance remains after the acceleration ends.
Do you read it this way?

The distance d in S remains constant, as part of the specification of this scenario.

The distance d' in S' (S' is not really a single frame, it is the instantaneous rest frame, which changes with v) remains the same if the acceleration stops and the ships continue to move with constant velocity relative to S.
 
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  • #251
cfrogue said:
Universal generalization of "distance is a coordinate-dependent" is not logically true.
Yes, it is logically universally true, however I think you may be misunderstanding some of the terminology based on your comments.
cfrogue said:
1) The distance between objects in an inertial frame is constant and it is not the case that distance is a coordinate-dependent.
Whether or not something is constant is not at all the same thing as whether or not something is coordinate independent. For a quantity A to be constant simply means that dA/dt = 0. For a quantity to be coordinate-independent means that A=A' where A' is the coordinate transform of A. They are two completely independent concepts.
cfrogue said:
2) If there exists relative motion, then the stationary frame will calculate length contraction for the moving frame metrics.
Yes, you are correct. Mathematically, A≠A' where A is the distance in the stationary frame and A' is the distance in the moving frame. Therefore, as I said above, distance is coordinate-dependent (or "relative").
cfrogue said:
3) If there is an accelerating frame, then the frame will experience metric expansion in the direction of acceleration. However, the launch frame will calculate a constant distance for objects in the accelerating frame.
Again, A≠A', the distance is coordinate-dependent.
 
  • #252
DaleSpam said:
Yes, it is logically universally true, however I think you may be misunderstanding some of the terminology based on your comments.Whether or not something is constant is not at all the same thing as whether or not something is coordinate independent. For a quantity A to be constant simply means that dA/dt = 0. For a quantity to be coordinate-independent means that A=A' where A' is the coordinate transform of A. They are two completely independent concepts.Yes, you are correct. Mathematically, A≠A' where A is the distance in the stationary frame and A' is the distance in the moving frame. Therefore, as I said above, distance is coordinate-dependent (or "relative").Again, A≠A', the distance is coordinate-dependent.


Sorry, I do not see your terminology as descriptive enough.

And, strictly from a logical point of view, it does not satisfy condition 1. For me, distance is coordinate-dependent means just that. Because the coordinates can be the identity coords in an inertial frame, this does not meet condition 1.

In the stationary frame, Einstein said,
Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the ``stationary system.''
http://www.fourmilab.ch/etexts/einstein/specrel/www/
 
  • #253
cfrogue said:
I mean that the posted paper shows the nature of SR's acceleration that the accelerating frame sees an expanding metric in the direction of acceleration when compared to the instantaneous rest frame.
Again, I don't know what the phrase "expanding metric" is supposed to mean.
JesseM said:
What does it mean to see a constant metric? Certainly the spacetime metric is the same in all frames, and in relativity the spacetime metric is the fundamental one. I suppose you can talk about a spatial metric in any given coordinate system, but what does it mean to say it's expanding? Are you just saying that the distance between the rockets expands? But of course even in classical Newtonian mechanics the distance between rockets will expand if they have different coordinate accelerations, yet in this situation I doubt you would talk about an "expanding metric". Do you have any exact definition for this phrase?
cfrogue said:
The metric I describe is only the one in the frame.
That doesn't really answer my questions above. Again, normally in relativity the metric is a spacetime metric which gives an invariant measure of spacetime "distance" along any path (along timelike worldlines, this is equivalent to the proper time), so the metric does not in any way depend on what frame you choose. I was asking if you were instead talking about a spatial metric, which I suppose would be defined in terms of the coordinate distance between points at a given moment in a given frame. Or does neither of these describe what you mean? A "metric" is always defining some measure of distance on a manifold, so what measure are you using?
 
  • #254
cfrogue said:
I would be interested in your calculation.

This seems to be an area that has been left off from a rigorous analysis.

In the prior post, there are also some interesting possibilities after the acceleration is done.

Ok, I don't think I can do better than Bell, or Lorentz for that matter!

Let's try to set up electrostatic equilibrium of 3 charges in row (Q ... -q ... Q), and the distance between neighbouring charges is r. For the rightmost charge to be in equilibrium require Q/((2r)^2)=(q/r^2). If r is finite, then q=Q/4, but in which case r can be any finite value, so there is no unique equilibrium. In the case of relativistic quantum mechanics, I believe the ground state is unique, then one can do the calculation in the launch frame without appealing to the Lorentz symmetry of the laws. But that's more pain than I'd like to go through at the moment.

So basically Lorentz's "calculation in the launch frame" holds if we grant him quantum mechanics and a unique ground state."Consequently, if, neglecting the effects of molecular motion, we suppose each particle of a solid body to be in equilibrium under the action of the attractions and repulsions exerted be its neighbours, and if we take for granted the there is but one configuration of equilibrium, we may draw the conclusion that the system Σ' , if the velocity w is imparted to it, will of itself change into the system Σ. [emphasis mine]" http://en.wikisource.org/wiki/Electromagnetic_phenomena
 
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  • #255
cfrogue said:
And, strictly from a logical point of view, it does not satisfy condition 1. For me, distance is coordinate-dependent means just that. Because the coordinates can be the identity coords in an inertial frame, this does not meet condition 1.
Frankly, this response leaves me questioning your honesty. By this absurd line of reasoning the volume of your stereo does not depend on the amplifier because you can set the gain to 0 dB, or more directly that the linear system y=A.x is independent of A because A may be the identity matrix. This level and quality of argument is characteristic of trolls and crackpots.

The term "coordinate independent" means that a quantity does not change under any arbitrary coordinate transform; if something is not coordinate independent then it is "coordinate dependent". The term "Lorentz invariant" means that a quantity does not change under any arbitrary Lorentz transform; if something is not "Lorentz invariant" then it is "Lorentz variant" or "relative". Distance is coordinate dependent. I hope that is clear enough for you.
 
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  • #256
cfrogue said:
Universal generalization of "distance is a coordinate-dependent" is not logically true.

It is existentially quantified by the following.

1) The distance between objects in an inertial frame is constant and it is not the case that distance is a coordinate-dependent.

2) If there exists relative motion, then the stationary frame will calculate length contraction for the moving frame metrics.

3) If there is an accelerating frame, then the frame will experience metric expansion in the direction of acceleration. However, the launch frame will calculate a constant distance for objects in the accelerating frame.

These are all different behaviors based on the particular SR motion and thus the phrase "distance is a coordinate-dependent" is not a universally descriptive phrase.

1) and 2) above are not generally true. The distance between the ships in this case being constant in the launch frame is specific to the scenario, just to make the point simple.

The distance between the ships being length contracted (constant instead of increasing as in the co-moving frame) in the launch frame is due to relative velocity, which increases with time. Greater relative velocity equals greater contraction, which is why, since it is stipulated that the distance between the ships remains constant in the launch frame, it must increase with time in the co-moving frame.
 
  • #257
JesseM said:
Again, I don't know what the phrase "expanding metric" is supposed to mean.

Yea, normally one uses a metric space to determine a metric I agree.

What I mean is that the distance within the accelerating frame between objects, the two ships in this example, expands as the acceleration continues.

Once the acceleration stops, the normal at rest frame measurements snap back.

JesseM said:
That doesn't really answer my questions above. Again, normally in relativity the metric is a spacetime metric which gives an invariant measure of spacetime "distance" along any path (along timelike worldlines, this is equivalent to the proper time), so the metric does not in any way depend on what frame you choose. I was asking if you were instead talking about a spatial metric, which I suppose would be defined in terms of the coordinate distance between points at a given moment in a given frame. Or does neither of these describe what you mean? A "metric" is always defining some measure of distance on a manifold, so what measure are you using?

Yes, I am only talking about a spatial metric without time.

I have not thought about what it looks like because I assume it is handled by GR.

But, what seems to be clear is that the distance bewteen the two ships increases as the acceleration continues.

Once the accelerations stops, the distance between the ships returns back to the original d.
Is this what you calculate?
 
  • #258
atyy said:
Ok, I don't think I can do better than Bell, or Lorentz for that matter!

Let's try to set up electrostatic equilibrium of 3 charges in row (Q ... -q ... Q), and the distance between neighbouring charges is r. For the rightmost charge to be in equilibrium require Q/((2r)^2)=(q/r^2). If r is finite, then q=Q/4, but in which case r can be any finite value, so there is no unique equilibrium. In the case of relativistic quantum mechanics, I believe the ground state is unique, then one can do the calculation in the launch frame without appealing to the Lorentz symmetry of the laws. But that's more pain than I'd like to go through at the moment.

So basically Lorentz's "calculation in the launch frame" holds if we grant him quantum mechanics and a unique ground state."Consequently, if, neglecting the effects of molecular motion, we suppose each particle of a solid body to be in equilibrium under the action of the attractions and repulsions exerted be its neighbours, and if we take for granted the there is but one configuration of equilibrium, we may draw the conclusion that the system Σ' , if the velocity w is imparted to it, will of itself change into the system Σ. [emphasis mine]" http://en.wikisource.org/wiki/Electromagnetic_phenomena

This analysis does not account for the "push" from the back ship.
If one ship were pulling a string without the 2nd ship, this would be complete above.

However, the back ship is accelerating into the string.

Thus, I would assume from the launch frame, there exists an equilibrium at the center of the string.

I am not sure if I am thinking this through correctly.
 
  • #259
DaleSpam said:
Frankly, this response leaves me questioning your honesty. By this absurd line of reasoning the volume of your stereo does not depend on the amplifier because you can set the gain to 0 dB, or more directly that the linear system y=A.x is independent of A because A may be the identity matrix. This level and quality of argument is characteristic of trolls and crackpots.

The term "coordinate independent" means that a quantity does not change under any arbitrary coordinate transform; if something is not coordinate independent then it is "coordinate dependent". The term "Lorentz invariant" means that a quantity does not change under any arbitrary Lorentz transform; if something is not "Lorentz invariant" then it is "Lorentz variant" or "relative". Distance is coordinate dependent. I hope that is clear enough for you.

I am perfectly fine with your views.

I simply would not use your terminology for my stated reasons.

If you want to call me names for that, so be it.
 
  • #260
cfrogue said:
This analysis does not account for the "push" from the back ship.
If one ship were pulling a string without the 2nd ship, this would be complete above.

However, the back ship is accelerating into the string.

Thus, I would assume from the launch frame, there exists an equilibrium at the center of the string.

I am not sure if I am thinking this through correctly.

Yes, Lorentz's and Bell's arguments only say that the moving equilibrium length in the launch frame of the string should be shorter than its stationary equilibrium length in the launch frame. What we know from the specification of the ships' acceleration that if the string does not break when it is moving, it must occupy a greater length in the launch frame than its moving equilibrium length in the launch frame, so it must be stressed at some point - but it does not tell us which at which point it is stressed, nor where the string eventually breaks.
 
  • #261
atyy said:
Yes, Lorentz's and Bell's arguments only say that the moving equilibrium length in the launch frame of the string should be shorter than its stationary equilibrium length in the launch frame. What we know from the specification of the ships' acceleration that if the string does not break when it is moving, it must occupy a greater length in the launch frame than its moving equilibrium length in the launch frame, so it must be stressed at some point - but it does not tell us which at which point it is stressed, nor where the string eventually breaks.

The latest paper on the subject says:

Bell’s paradox was that his intuition told him the cable would
break, yet there was no change in the distance between the ships in system S.
He suggested resolving the paradox by stating that a cable between the ships
would shorten due to the contraction of a physical object proposed by Fitzgerald
and Lorentz, while the distance between the ships would not change. This
resolution however contradicts special relativity which allows no such difference
in any measurement of these two equal lengths
.

http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf
 
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  • #262
cfrogue said:
The latest paper on the subject says:

Bell’s paradox was that his intuition told him the cable would
break, yet there was no change in the distance between the ships in system S.
He suggested resolving the paradox by stating that a cable between the ships
would shorten due to the contraction of a physical object proposed by Fitzgerald
and Lorentz, while the distance between the ships would not change. This
resolution however contradicts special relativity which allows no such difference
in any measurement of these two equal lengths
.

http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

I believe that paper's own analysis is right, but I think it's assessment of Bell's analysis is wrong - ie. I think Bell was essentially right (but I also don't think the paper understands Bell's analysis, and so misrepresents it).
 
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  • #263
cfrogue said:
Yea, normally one uses a metric space to determine a metric I agree.

What I mean is that the distance within the accelerating frame between objects, the two ships in this example, expands as the acceleration continues.

Once the acceleration stops, the normal at rest frame measurements snap back.
What kind of accelerating frame are you thinking of? The simplest way to define a non-inertial rest frame for an object which accelerates is to have the frame's time coordinate match up with the object's proper time along its worldline, and then say that at any given event on the object's worldline, the non-inertial frame's definition of simultaneity and distance at the time of that event should match up with the object's instantaneous inertial rest frame at that event on its worldline. In this case, since the distance to the back ship increases in the instantaneous inertial rest frame of the front ship, if we define the front ship's non-inertial rest frame in this way, the distance will increase in the non-inertial frame. Then once the two ships stop accelerating, the non-inertial frame's definition of distance will match up with that of their inertial rest frame. Is that something close to what you meant?
cfrogue said:
But, what seems to be clear is that the distance bewteen the two ships increases as the acceleration continues.

Once the accelerations stops, the distance between the ships returns back to the original d.
Is this what you calculate?
What do you mean by "the original d"? If you define the non-inertial frame of one of the ships in the way I suggested, it won't return to the d seen in the launch frame, once they stop accelerating the distance between them in the non-inertial frame will be equal to the distance between them in their rest frame. At no point will the distance between them in the non-inertial frame contract, it'll just stop expanding once they've both stopped accelerating. If you want to use a different type of non-inertial frame you have to explain how it defines distance, time, and simultaneity.
 
  • #264
1) The distance between objects in an inertial frame is constant and it is not the case that distance is a coordinate-dependent.

2) If there exists relative motion, then the stationary frame will calculate length contraction for the moving frame metrics.

Al68 said:
1) and 2) above are not generally true. The distance between the ships in this case being constant in the launch frame is specific to the scenario, just to make the point simple.

The accelerating ships are not an inertial frame and so the above logic does not apply.

Al68 said:
The distance between the ships being length contracted (constant instead of increasing as in the co-moving frame) in the launch frame is due to relative velocity, which increases with time. Greater relative velocity equals greater contraction, which is why, since it is stipulated that the distance between the ships remains constant in the launch frame, it must increase with time in the co-moving frame.


I do not agree with this.

The integral for the solution only involves the accelerating frame and the instantaneous at rest frame, S'.

The launch frame is not part of the decision process in this paper.
http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

The launch frame exists as a separate entity that only concludes the distance does not change between the ships but does not participate in the integral for the solution.
 
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  • #265
atyy said:
I believe that paper's own analysis is right, but I think it's assessment of Bell's analysis is wrong - ie. I think Bell was essentially right (but I also don't think the paper understands Bell's analysis, and so misrepresents it).

This is why discussion occurs.

So, based on this paper, can you explain the Bell position and how it is consistent with this paper?

Or, are they logically inconsistent?
 
  • #266
JesseM said:
What kind of accelerating frame are you thinking of? The simplest way to define a non-inertial rest frame for an object which accelerates is to have the frame's time coordinate match up with the object's proper time along its worldline, and then say that at any given event on the object's worldline, the non-inertial frame's definition of simultaneity and distance at the time of that event should match up with the object's instantaneous inertial rest frame at that event on its worldline. In this case, since the distance to the back ship increases in the instantaneous inertial rest frame of the front ship, if we define the front ship's non-inertial rest frame in this way, the distance will increase in the non-inertial frame. Then once the two ships stop accelerating, the non-inertial frame's definition of distance will match up with that of their inertial rest frame. Is that something close to what you meant?.

This is what I meant.

JesseM said:
What do you mean by "the original d"? If you define the non-inertial frame of one of the ships in the way I suggested, it won't return to the d seen in the launch frame, once they stop accelerating the distance between them in the non-inertial frame will be equal to the distance between them in their rest frame. At no point will the distance between them in the non-inertial frame contract, it'll just stop expanding once they've both stopped accelerating. If you want to use a different type of non-inertial frame you have to explain how it defines distance, time, and simultaneity.


d for me is the distance between the ships from the POV of the ships before accelerating.

Now, I have a question.

Will the ships return to their original distance between them before acceleration once the acceleration stips?
 
  • #267
cfrogue said:
d for me is the distance between the ships from the POV of the ships before accelerating.

Now, I have a question.

Will the ships return to their original distance between them before acceleration once the acceleration stips?
Not if we define the ship's non-inertial frame in the way I did above. After all, the definition says that for an event on the ship's worldline after that ship has stopped accelerating, the ship's definition of distance at that moment will match the definition of distance in its current inertial rest frame, not the definition of distance in the original inertial frame they were at rest in before accelerating (the launch frame).
 
  • #268
cfrogue said:
The accelerating ships are not an inertial frame and so the above logic does not apply.
You can analyze the ships from the perspective of an inertial frame. As I told you on the GPS thread, you can use any frame you like to analyze any physical situation you like, you could use an accelerating frame to analyze a collection of inertial objects or an inertial frame to analyze a collection of accelerating objects, the choice of what frame to use is never forced on you by the motion of any physical objects.
cfrogue said:
I do not agree with this.

The integral for the solution only involves the accelerating frame and the instantaneous at rest frame, S'.

The launch frame is not part of the decision process in this paper.
http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf
Do any of the equations in that paper make use of a non-inertial frame? Skimming the paper, it seemed like everything was analyzed from the perspective of different inertial frames.
 
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  • #269
JesseM said:
Not if we define the ship's non-inertial frame in the way I did above. After all, the definition says that for an event on the ship's worldline after that ship has stopped accelerating, the ship's definition of distance at that moment will match the definition of distance in its current inertial rest frame, not the definition of distance in the original inertial frame they were at rest in before accelerating (the launch frame).

Are you arguing that the ships will be at a distance d' > d of the original launch frame distance after the acceleration ceases?

If so, you have produced a paradox.
 
  • #270
JesseM said:
You can analyze the ships from the perspective of an inertial frame. As I told you on the GPS thread, you can use any frame you like to analyze any physical situation you like, you could use an accelerating frame to analyze a collection of inertial objects or an inertial frame to analyze a collection of accelerating objects, the choice of what frame to use is never forced on you by the motion of any physical objects.

Do any of the equations in that paper make use of a non-inertial frame? Skimming the paper, it seemed like everything was analyzed from the perspective of different inertial frames.

The paper focuses on the accelerating frame and an instanteneous at rest frame.

The launch frame is incidental.
 
  • #271
cfrogue said:
Are you arguing that the ships will be at a distance d' > d of the original launch frame distance after the acceleration ceases?
In the non-inertial frame I defined, yes.
cfrogue said:
If so, you have produced a paradox.
How?
cfrogue said:
The paper focuses on the accelerating frame and an instanteneous at rest frame.
Again, I didn't see anywhere in the paper where they referred to an accelerating frame. Can you provide a quote, or point to an equation that you think involves quantities measured in an accelerating frame?
 
  • #272
cfrogue said:
This is why discussion occurs.

So, based on this paper, can you explain the Bell position and how it is consistent with this paper?

Or, are they logically inconsistent?

Bell's argument is logically consistent with the argument in the paper.

Let's start in the launch frame. In that frame we know that Maxwell's equations hold. Maxwell's equations tell us the electric field for an electric charge that is stationary in the launch frame, as well as one that is moving in the launch frame.

The field of a moving charge is "flattened" compared with the field of a stationary charge. This can be seen from http://farside.ph.utexas.edu/teaching/jk1/lectures/node26.html, Eq 265. It can also be obviously seen pictorially from http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html by comparing the "linear" case (ie. constant velocity) with v=0 and with v=0.9.

Because of this flattening, if we imagine that electrical forces are one of the things that hold particles in a string together, we may imagine that Maxwell's equations predict that the equilibrium length of a moving string is shorter than that of a stationary string. If this is correct, then all we have made use of is Maxwell's equations in the launch frame. This is not totally correct because classical electrostatics does not predict a unique equilibrium configuration, and also because a string is held together not just by classical electrical forces, but also by quantum mechanics (eg. Pauli exclusion principle, the uniqueness of the ground state) and not just classical electrostatics. With these additional assumptions, then we can predict that a string's moving equilibrium length is shorter than its stationary equilibrium length.

An essential part of these heuristics was the flattening of the electric field of a moving charge using only Maxwell's equations in the launch frame. So a crucial question in seeing if the calculations in the launch frame and in other frames match up is: is this flattening predicted by using Maxwell's equations in the rest frame of the moving charge, then Lorentz transforming to the launch frame? The answer is yes.
 
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  • #273
cfrogue said:
Al68 said:
1) and 2) above are not generally true. The distance between the ships in this case being constant in the launch frame is specific to the scenario, just to make the point simple.
The accelerating ships are not an inertial frame and so the above logic does not apply.
Huh? That makes no sense whatsoever. I was referring to your statement that "1) The distance between objects in an inertial frame is constant and it is not the case that distance is a coordinate-dependent." and "3) If there is an accelerating frame, then the frame will experience metric expansion in the direction of acceleration. However, the launch frame will calculate a constant distance for objects in the accelerating frame." None of those statements are true generally for obvious reasons.
Al68 said:
The distance between the ships being length contracted (constant instead of increasing as in the co-moving frame) in the launch frame is due to relative velocity, which increases with time. Greater relative velocity equals greater contraction, which is why, since it is stipulated that the distance between the ships remains constant in the launch frame, it must increase with time in the co-moving frame.
I do not agree with this.

The integral for the solution only involves the accelerating frame and the instantaneous at rest frame, S'.

The launch frame is not part of the decision process in this paper.
http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

The launch frame exists as a separate entity that only concludes the distance does not change between the ships but does not participate in the integral for the solution.
Seriously? You are now objecting to anyone referring to the launch frame? None of your response even remotely explains why you "do not agree with this".
cfrogue said:
Are you arguing that the ships will be at a distance d' > d of the original launch frame distance after the acceleration ceases?
Of course the ships will be farther apart in a co-moving inertial frame than in the launch frame. Why on Earth would they "snap back" to their original distance in any frame just because they kill their engines? Magic?
 
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  • #274
JesseM said:
In the non-inertial frame I defined, yes.

How?
Because, the distance between the accelerating frame will exceed that of the launch frame.


JesseM said:
Again, I didn't see anywhere in the paper where they referred to an accelerating frame. Can you provide a quote, or point to an equation that you think involves quantities measured in an accelerating frame?
Yea, it does not directly refer to it but does infer it.

But, what do you think this means?

We denote the spaceships as L and R, each having acceleration a(t) in the positive x direction in system S
http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf
 
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  • #275
Al68 said:
Huh? That makes no sense whatsoever. I was referring to your statement that "1) The distance between objects in an inertial frame is constant and it is not the case that distance is a coordinate-dependent." and "3) If there is an accelerating frame, then the frame will experience metric expansion in the direction of acceleration. However, the launch frame will calculate a constant distance for objects in the accelerating frame." None of those statements are true generally for obvious reasons.
You and I are not communicating.

You cannot refute 1 and 2, if so do that please.



Al68 said:
Seriously? You are now objecting to anyone referring to the launch frame? None of your response even remotely explains why you "do not agree with this". Of course the ships will be farther apart in a co-moving inertial frame than in the launch frame. Why on Earth would they "snap back" to their original distance in any frame just because they kill their engines? Magic?

I showed the math, you show yours.
 
  • #276
Al68 said:
None of your response even remotely explains why you "do not agree with this". Of course the ships will be farther apart in a co-moving inertial frame than in the launch frame. Why on Earth would they "snap back" to their original distance in any frame just because they kill their engines? Magic?

More directly, if you support the proposition that the accelerating ships will retain their > d positions after the acceleration stops, you will find that the launch frame will completely disagree with your assessment since the ships maintained their distance.

So, where is your math to reconcile this?
 
  • #277
cfrogue said:
Because, the distance between the accelerating frame will exceed that of the launch frame.
What's paradoxical about that? The distance in the non-inertial frame began to grow larger than d from the moment they began accelerating. Anyway, you don't even need to consider a non-inertial frame here; just using the Lorentz transformation and their x(t) functions in the launch frame, you can easily show that the distance between them in other inertial frames can be larger than d. Different inertial frames always disagree on the distance between a pair of objects, just like they disagree on the time between a pair of events, that's just a feature of how the Lorentz transformation works.
cfrogue said:
Yea, it does not directly refer to it but does infer it.
How so? Do you think any of the variables in the equations they use refer to quantities in a non-inertial frame?
cfrogue said:
But, what do you think this means?

We denote the spaceships as L and R, each having acceleration a(t) in the positive x direction in system S
http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf
That quote is from p. 4, and if you look at the context it's clear that S is what we have been calling the "launch frame". Do you disagree?
 
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  • #278
cfrogue said:
More directly, if you support the proposition that the accelerating ships will retain their > d positions after the acceleration stops, you will find that the launch frame will completely disagree with your assessment since the ships maintained their distance.

So, where is your math to reconcile this?

Try reading the proposed scenario. The constancy of the distance between the ships in the launch frame is not a natural occurrence but a stipulation in the set up of the problem.

Apart from that there is a logical inconsistency in what you say. You say the ships will "snap back" to the original distance. This, apart from being ridiculous, assumes that they acquired a greater distance through the acceleration, which although true, refutes your argument for constancy.

Matheinste.
 
  • #279
matheinste said:
Try reading the proposed scenario. The constancy of the distance between the ships in the launch frame is not a natural occurrence but a stipulation in the set up of the problem.

Apart from that there is a logical inconsistency in what you say. You say the ships will "snap back" to the original distance. This, apart from being ridiculous, assumes that they acquired a greater distance through the acceleration, which although true, refutes your argument for constancy.

Matheinste.

Fine, I will argue my point from SR here.

Any logic that assumes the ships will have a greater distance after acceleration stops from the accelerating frame will contradict the launch frame that sees the distance as constant and then contracted by the normal LT calculations. You must have logic and equations to support this position.

I want to see all this.
 
  • #280
cfrogue said:
Fine, I will argue my point from SR here.

Any logic that assumes the ships will have a greater distance after acceleration stops from the accelerating frame will contradict the launch frame that sees the distance as constant and then contracted by the normal LT calculations. You must have logic and equations to support this position.

I want to see all this.
What do you mean when you say the launch frame "sees the distance as constant and then contracted by the normal LT calculations"? The launch frame sees the distance as constant at all times, even after the acceleration stops, so why do you say "then contracted"? Do you mean the distance seen in the launch frame once the ships stop accelerating is contracted relative to the distance between them in the ships' new inertial rest frame? Of course that's true, it's why the distance is greater in the ships' rest frame!
 
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