Does Bell's Paradox Suggest String Shouldn't Break Due to Length Contraction?

[Eli Botkin]

yuiop:
First, you are not addressing (in your scenario 1) the issues that are being discussed in the Bell Paradox scenario. I’m certain that there are countless scenarios of two accelerating vehicles, connected by a string, wherein the string must break. Whether or not your selected scenarios do indeed make breakage certain, is something I would have to check mathematically, and that takes time.

At this point I’m not sure that it holds my interest since, as I said above, there are many scenarios that ensure that outcome.

Second, a note about your scenario 1:
Instructing the rockets “to stay 1 km apart at all times as measured in their own instantaneous rest frame” means that before they start they need to know what each of their accelerations, as function of time, needs to be. Those accelerations won’t be constants as in the Bell scenario. And there is more than one such set of acceleration histories that could suit the 1 km requirement. A calculation headache ;-)

Now your scenario 2:
“…the rocket pilots are instructed to stay 1 km apart as measured in the ground reference frame.” This is what happens in the ground frame when the acceleration histories are identically the same for both rockets. This is the Bell scenario.

But you need to tell me why the “…the un-tensioned length of the string should be 1/2 km…” in the ground frame. If you think it is because “…the rockets are moving at 0.866c relative to the ground,…”, then why is the rocket separation still 1 km, though the rocket frame (which is the string’s frame) is also moving at 0.866c relative to the ground?

Ultimately the question comes down to this:
1. Why is the string's length, as transformed between inertial frames, being treated differently than the rockets' separation length.
2. Arguments for breakage always seem to hinge on scenarios as viewed by observers that never see the rockets approaching each other, when in fact, there are such observers.

 

[Eli Botkin]

yuiop:
First, you are not addressing (in your scenario 1) the issues that are being discussed in the Bell Paradox scenario. I’m certain that there are countless scenarios of two accelerating vehicles, connected by a string, wherein the string must break. Whether or not your selected scenarios do indeed make breakage certain, is something I would have to check mathematically, and that takes time.

At this point I’m not sure that it holds my interest since, as I said above, there are many scenarios that ensure that outcome.

Second, a note about your scenario 1:
Instructing the rockets “to stay 1 km apart at all times as measured in their own instantaneous rest frame” means that before they start they need to know what each of their accelerations, as function of time, needs to be. Those accelerations won’t be constants as in the Bell scenario. And there is more than one such set of acceleration histories that could suit the 1 km requirement. A calculation headache ;-)

Now your scenario 2:
“…the rocket pilots are instructed to stay 1 km apart as measured in the ground reference frame.” This is what happens in the ground frame when the acceleration histories are identically the same for both rockets. This is the Bell scenario.

But you need to tell me why the “…the un-tensioned length of the string should be 1/2 km…” in the ground frame. If you think it is because “…the rockets are moving at 0.866c relative to the ground,…”, then why is the rocket separation still 1 km, though the rocket frame (which is the string’s frame) is also moving at 0.866c relative to the ground?

Ultimately the question comes down to this:
1. Why is the string's length, as transformed between inertial frames, being treated differently than the rockets' separation length.
2. Arguments for breakage always seem to hinge on scenarios as viewed by observers that never see the rockets approaching each other, when in fact, there are such observers.

 

[A.T.]

1. Why is the string's length, as transformed between inertial frames, being treated differently than the rockets' separation length.

What? The string's length and the rockets' separation are the same thing.

2. Arguments for breakage always seem to hinge on scenarios as viewed by observers that never see the rockets approaching each other

I gave you the reason for breakage in the frame where they don't approach each other.

 

[Austin0]

Austi0:
you say " Yet we also have to assume that it will never actually reach the lead ship."

That need not be an assumption. The Minkowski diagram shows that the two ship hyperbolic worldlines are the same shapes, laterally displaced from one another and, therefore, never intersecting.

Hi , you are of course quite right that the two worldlines would be identical in shape but they are not simply displaced laterally but also vertically (temporally).
So with a certain magnitude of lead time for the trailing ship, the lines could intersect even with identical curvature. So the assumption part is; that the maximum possible time difference due to relative simultaneity for the distance between them, in any frame, is always going to be less than this threshold magnitude.

 

[Dale]

Remember the Pole/Barn Paradox? Did you ponder whether or not the pole was stiff or elastic before applying the SR transformations ?

The stiffness of the pole is irrelevant in the barn/pole paradox. The pole is not under tension and is moving inertially, so whether it is made of rubber or steel the calculations are the same.

So again, what is wrong with assuming a stiff string in the Bell's spaceship scenario? Do you object because you think the assumption is non-standard or because you think it is wrong? Please answer these questions directly instead of with an evasion.

 

[Dale]

yes i noticed this myself also. And found it somewhat mysterious. The trailing ship by starting before the lead ship develops a velocity relative to that ship, and given equal proper acceleration there is no reason to assume that this velocity will diminish over time. Yet we also have to assume that it will never actually reach the lead ship.
The only explanation I could come up with is the diminishing coordinate acceleration, in the frame in which it starts first, results in the velocity differential asymptotically approaching zero.
SO it never reaches the lead ship.
What do you think, does this sound right?

This is correct. As they both asymptotically approach c in such a frame their relative coordinate velocity clearly goes to zero, meaning that the distance approaches some asymptotic value. It happens that this asymptotic value is non-zero, and so regardless of the length of the rope and the initial separation, eventually it breaks when it contracts below that finite distance.

 

[Nugatory]

2. Arguments for breakage always seem to hinge on scenarios as viewed by observers that never see the rockets approaching each other, when in fact, there are such observers.

Indeed there are such observers, and an observer moving, as viewed by the ground observer, in the opposite direction from the two spaceships is one. Let's call that observer the "left-mover", and then take on your question #1.

1. Why is the string's length, as transformed between inertial frames, being treated differently than the rockets' separation length.

They are being transformed from the frame in which they are constant and equal to L in the same way; the trick is that these frames are different. In one case we're transforming a distance that is constant in the front-ship frame (and therefore contracted by different amounts in all other frames including the ground frame). In the other case we are transforming a distance that is constant in the ground observer's frame (and therefore contracted by different amounts in all other frames including the front-ship frame). When I transform them both into the frame of your left-moving observer, I will get different contractions because the relative velocity between the "from" frame and the left-moving frame is different for the two transformations.

Thus, the left-mover sees the distance between the ships contract, but sees the length of the string contract even more (his speed is greater relative to the frame in which the string's length is L then it is relative to the frame in which the ship separation is L). Ground observer happens to be using the only frame in which the ship separation is constant, but in that frame the string contacts because it is moving. Lead-ship observer occupies a frame in which the length of the string is constant, but in that frame the distance between the ships is increasing (because the two ships are maneuvering to maintain a constant separation in ground observer's frame, so the separation distance is not constant in the front-ship frame).

Lemme know if this isn't clear enough... I have a space-time diagram that shows how the three frames (ground, front ship, and left mover) are related, along with the Lorentz transformation calculations.

(as an aside, I find that when time dilation or length contraction are giving me trouble, it often helps to go back to the Lorentz transformations themselves. Length contraction and time dilation can be derived by applying the Lorentz transforms to particular thought experiments, and if you're working with a scenario that doesn't match this those thought experiments, it's easy to misapply contraction and dilation).

 

[Eli Botkin]

A.T.:
You say "What? The string's length and the rockets' separation are the same thing."
Why the puzzlement? Isn't that what I'm implying when I'm asking (with tongue in cheek) "Why is the string's length, as transformed between inertial frames, being treated differently than the rockets' separation length? What I'm asking is why the string and the separation aren't being treated in the same SR way if, as you say, "they are the same" and so should be. Do we agree on this?

Please direct me to the reply number where you've answered my question about frames where the ships' separation reduces with time

 

[Eli Botkin]

Nugatory:
Indeed this isn't clear enough. To me it sounds like a comparison of apples and oranges. I would expect the comparison in frame A to be transformed to one in frame B , thereby assuring that the time coordinate is the same for all events in frame A and again in frame B. Why the need to transform one distance from frame A into frame B, then the other distance from frame C into frame A. Certainly both distances exist in frame A.

Maybe your spacetime diagram would clear this up for me. Thanks

 

[Eli Botkin]

DaleSpan:
Here is one thing that is wrong with assuming a stiff string.

Inertial observer A is moving opposite to the ships' acceleration, all of them with respect to the ground frame. As you now agree, for A the rear ship is closing in on the front ship. Or at least it should be except for its prevention by that stiff string whose compression forces acting on the ships plays havoc with the Bell Paradox scenario;-)

Unless, of course, there are no compression forces and the scenario continues to evolve without string compression (nor breakage as viewed in other frames).

 

[A.T.]

What I'm asking is why the string and the separation aren't being treated in the same SR way

No idea what "treated in the same SR way" means. And no idea why you even use two differnt terms for the same thing: string length = separation distance in any frame.

Please direct me to the reply number where you've answered my question about frames where the ships' separation reduces with time

The reason for breakage valid for all frames:

The atoms of the string cannot span the separation distance anymore

In some frames this is because the separation distance increased, in others because the atoms are contracted.

 

[Eli Botkin]

Nugatory:
I'm rereading your reply #107 and have some questions:

1. You say, "They are being transformed from the frame in which they are constant and equal to L in the same way;...". I read "the frame" to mean a single frame, and I presume it to be the ground frame in which they (both the string length and the separation) have, and maintain, the same length L. Is that what you meant?

But then you add, "...the trick is that these frames are different." You've switched to the plural "these," which I presume is more than one frame. Please clear that up for me.

2. I think you are saying that the string is of constant length L in the front-ship's co-moving frames. Would that then also be true for the rear-ship's co-moving frames? And if so, could that be a problem since there exists a relative velocity between the two ships in any co-moving frame of either ship? Of course, this constant string length in the front-ship co-moving frames is an assumption on your part which may not be assured of realism any more than would be an assumption of constant separation in all co-moving frames ;-)

 

[Nugatory]

Nugatory:
Indeed this isn't clear enough. To me it sounds like a comparison of apples and oranges. I would expect the comparison in frame A to be transformed to one in frame B , thereby assuring that the time coordinate is the same for all events in frame A and again in frame B. Why the need to transform one distance from frame A into frame B, then the other distance from frame C into frame A. Certainly both distances exist in frame A.

Maybe your spacetime diagram would clear this up for me. Thanks

[Edit - fixed a transposed A and B]
I'm busy cleaning up the picture now... But while I'm doing that, could I ask you to read the below, then read my previous post again more carefully?

You've mixed up the common destination frame when you ask "Why the need to transform one distance from frame A into frame B, then the other distance from frame C into frame A?"

We're trying to transform a distance known in frame A into C and another distance known in frame B into C, and we should expect that the A->C transformation is different than the B->C transformation because A and B are different.

Frame A: Ground observer.
Frame B: Front-ship observer
Frame C: Left-moving observer
We know the length of the string as measured in frame B.
We know the separation between the ships as measured in frame A.

What is the length of the string as measured in frame C? We use the B->C transformation on the known length in B to find what the frame C observer measures.

What is the separation between the ships as measured in frame C? We use the A->C transformation on the known separation in A to find what the frame C observer measures.

And once we know the distances as measured in frame C... We compare them to see if the string breaks.

 

[Dale]

DaleSpan:
Here is one thing that is wrong with assuming a stiff string.

Inertial observer A is moving opposite to the ships' acceleration, all of them with respect to the ground frame. As you now agree, for A the rear ship is closing in on the front ship. Or at least it should be except for its prevention by that stiff string whose compression forces acting on the ships plays havoc with the Bell Paradox scenario;-)

Unless, of course, there are no compression forces and the scenario continues to evolve without string compression (nor breakage as viewed in other frames).

What compression? Strings don't sustain compression. They are slack or in tension.

You seem to be unaware of the standard behavior of an idealized massless stiff string. This is basic first-semester Newtonian physics. An idealized massless stiff string is slack if the distance between its attachment points is less than its length and it cannot be stretched without breaking regardless of the tension applied. Do you understand that behavior of idealized massless stiff strings from Newtonian physics? In many first semester problems it is simply called an ideal string.

 

[Eli Botkin]

A.T.:
In this thread all other contributors have elected to distinguish between string length and the separation distance (length) between the ships. The string's ends have been fastened to the two ships. The differentiation is being made because the issue in the Bell Paradox is whether or not these two different lengths change in a way that requires the string to break.

By "treated in the same SR way" I mean only: dealt with through application of of the SR transformation equations.

Hope that clears up both points for you.

You say "The atoms of the string cannot span the separation distance anymore.
In some frames this is because the separation distance increased, in others because the atoms are contracted."

That sounds very nice, but I fear it needs SR's (or any other accepted theory like GR and QM) mathematical verification. If you feel up to that, I'ld appreciate that contribution.

 

[Eli Botkin]

DaleSpan:
Oops, of course you're right that " Strings don't sustain compression. They are loose or in tension."

Except, maybe, when there is only one space dimension, x, in the problem and there is no y or z to hang into ;-)

Better yet, would your solution have a problem if we replaced your "fixed" string with a "fixed" wooden pole so that we could deal with both tension and compression?

 

[Eli Botkin]

Nugatory:
I think that you've mixed up your frames:
"We know the length of the string as measured in frame A."
"We know the separation between the ships as measured in frame B."

"What is the length of the string as measured in frame C? We use the B->C transformation on the known length in B..."
"What is the separation between the ships as measured in frame C? We use the A->C transformation on the known separation in A..."

But I understand what you're doing. I'll keep A for string and B for separation.

But we also know the separation between the ships as measured in frame A? Why go to B?
Is there something special about the observer in B as opposed to the observer in A? or the observer D in the aft-ship? Will B and D agree on separation distance at the same time on their respective clocks? And if not then, then when? Oh..., so many questions ;-)

 

[Nugatory]

Nugatory:
I'm rereading your reply #107 and have some questions:

1. You say, "They are being transformed from the frame in which they are constant and equal to L in the same way;...". I read "the frame" to mean a single frame, and I presume it to be the ground frame in which they (both the string length and the separation) have, and maintain, the same length L. Is that what you meant?

But then you add, "...the trick is that these frames are different." You've switched to the plural "these," which I presume is more than one frame. Please clear that up for me.

The two frames I am referring to are:
1) The ground observer's frame, in which the separation of the ships is always L and the length of the string is something less than L as soon as the ships start moving.
2) The lead-ship frame in which the length of the string is always L and the separation between the ships is steadily increasing as long as the ships are accelerating.

The key to understanding this is to recognize that the path through spacetime of the trailing ship is not the same as the path of the trailing end of the string (except that we've tied the trailing end of the string to the trailing spaceship so the string breaks when its trailing end cannot follow its natural path).

2. I think you are saying that the string is of constant length L in the front-ship's co-moving frames. Would that then also be true for the rear-ship's co-moving frames?

Not the same while the two ships are accelerating, but the same once the engines are cut off and they're drifting at the same constant speed (zero relative to each other, so no relativistic effects). Note that they are not comoving while they're accelerating; there is relative velocity between them as they're accelerating because the acceleration has been specified to provide a constant separation in the ground observer's frame so the must be a non-constant separation in all other frames, including their own.

We're starting a red herring discussion below - read it if you want, but don't respond until you're SURE that you understand what I've said above.

And if so, could that be a problem since there exists a relative velocity between the two ships in any co-moving frame of either ship? Of course, this constant string length in the front-ship co-moving frames is an assumption on your part which may not be assured of realism any more than would be an assumption of constant separation in all co-moving frames ;-)

Strictly speaking, if the ships are comoving (that is, zero relative velocity) then they are in the same frame, except perhaps for a linear transformation of the origin. They will agree about all measured lengths, the rate of all observed clocks, and in general won't see any relativistic weirdness when they compare measurements.

The assumption of constant string length in the front-moving ship's frame comes the fact that the string is at rest relative to the front-moving ship... Always has been, always will be, as long as it is tied to the front-moving ship so moves with it. This is assuming that the string has the decency to break at the knot that attaches it to the trailing ship - but if you don't like that assumption, we can break the string anywhere else and what I said about the the constant length will still apply to whatever part of the string remains attached to the front-moving ship.

 

[Nugatory]

Nugatory:
I think that you've mixed up your frames:
"We know the length of the string as measured in frame A."
"We know the separation between the ships as measured in frame B."

"What is the length of the string as measured in frame C? We use the B->C transformation on the known length in B..."
"What is the separation between the ships as measured in frame C? We use the A->C transformation on the known separation in A..."

But I understand what you're doing. I'll keep A for string and B for separation.

You're right, I did. I just corrected it.

 

[Eli Botkin]

Nugatory:
I’ve been thinking some more about your solution. It occurs to me that there are many close similarities with DaleSpan’s solution.

DaleSpan assumed that the string was, what he termed, “fixed.” That is, the string was always of length L. He then correctly computed the ships’ separation distance in a ship’s co-moving frame and that, of course, was always > L. This, he believes allows him to claim that the string always breaks. It would, of course, have to break at T = 0.

In your own solution you haven’t assumed the string length to be “fixed” but instead selected frame A for its pre-transformation frame, and in frame A the string length is always L. Your co-moving frame B, as in DaleSpan’s case, yields a separation length > L. Then, transforming these two lengths from their respective frames to any frame C will result in the transposed string length being less than the transposed separation length. So you too can claim that the string will break.

Since the separation length is also available in A you could transform both lengths from any selected time in frame A to frame C and then compare them. That seems the more direct approach, avoiding what might be thought by some to be putting a thumb on the scale ;-)

 

[yuiop]

yuiop:
First, you are not addressing (in your scenario 1) the issues that are being discussed in the Bell Paradox scenario. I’m certain that there are countless scenarios of two accelerating vehicles, connected by a string, wherein the string must break.

Yes there are countless acceleration schemes where the connecting string must break, but in Scenario 1 I am discussing a unique acceleration scheme where the string tension remains constant (and negligible) and the string does not break. This unique acceleration scheme is know as Born rigid acceleration.

At this point I’m not sure that it holds my interest since, as I said above, there are many scenarios that ensure that outcome.

The whole point of Scenario 1 is to illustrate a method that does not break the string and any other acceleration scheme that allows the separation to be greater than the Born rigid separation (e.g. the Bell's rocket paradox acceleration scheme) will break the string, eventually.

Now your scenario 2:
“…the rocket pilots are instructed to stay 1 km apart as measured in the ground reference frame.” This is what happens in the ground frame when the acceleration histories are identically the same for both rockets. This is the Bell scenario.

But you need to tell me why the “…the un-tensioned length of the string should be 1/2 km…” in the ground frame. If you think it is because “…the rockets are moving at 0.866c relative to the ground,…”, then why is the rocket separation still 1 km, though the rocket frame (which is the string’s frame) is also moving at 0.866c relative to the ground?

The rockets are 1 km apart, because the pilots have been instructed to stay 1 km apart ;) Assuming the pilots are obedient, the only reason they would not be 1 km apart as measured in the ground frame is because the string is so tough that the rocket engines are not strong enough to stretch the string. Unlike Dalespam, I am assuming a fairly elastic string that can easily stretch to twice its own rest length before breaking. Think of it as an elastic band if you prefer.

Ultimately the question comes down to this:
1. Why is the string's length, as transformed between inertial frames, being treated differently than the rockets' separation length.

The string is subject to the laws of nature and its untensioned length depends on its velocity relative to the observer. The gap between the rockets is whatever the rocket pilots want it to be.
Mathematically they are treated (transformed) the same way. If the rocket pilots measure the gap and the (stretched) connecting string to be 2km when they are moving at 0.866 relative to the ground, then the ground based observers will measure the gap and the (stretched) connecting string to both be 1 km so the transformation factor for both the gap and the (stretched) connecting string will be 0.5. No discrimination there.

2. Arguments for breakage always seem to hinge on scenarios as viewed by observers that never see the rockets approaching each other, when in fact, there are such observers.

Yes, there are such observers, but for any observer the gap will always be larger than the (un-tensioned) proper length of the string, when the rockets and the string are moving relative to the observer in Bell's paradox.

The idea of the two scenarios is so that you can see why the string does not break in the first scenario and why it must break in the second scenario, by examining the differences between the two scenarios.

Here is a 3rd scenario, which I hope will settle the matter. It is still based on the acceleration scheme outlined in Bell's rocket paradox as in Scenario 2.

Scenario 3

This time there is a very rigid titanium pole attached to the back of the leading rocket, but not attached to the rear rocket. It is 1km long and simply trails behind the leading rocket and acts as a ruler. As before, the rockets are connected by a 1 km long elastic string , which is designed to break when stretched to twice the rest length of the titanium pole. The rockets accelerate gently to 0.866c relative to the ground.

Measurements in the ground based frame:
At this point the length of the pole is 1/2 km and the separation between the rockets (and the length of the stretched string) is 1 km so in this frame the string is stretched to twice the length of the pole and so must snap.

Measurements in the rocket based frame:
At this point the length of the pole is 1 km and the separation between the rockets (and the length of the stretched string) is 2 km so in this frame the string is stretched to twice the length of the pole and so must snap.

Measurements in any refrence frame:
The string is stretched to twice the length of the pole and so must snap. This includes reference frames where the rockets appear to getting closer together.

 

[A.T.]

A.T.:
In this thread all other contributors have elected to distinguish between string length and the separation distance (length) between the ships. The string's ends have been fastened to the two ships. The differentiation is being made because the issue in the Bell Paradox is whether or not these two different lengths change in a way that requires the string to break.

You are confusing string's length and string's proper length.

string's length = separation distance in all frames, so there is no need to differentiate between the two.

The condition for breakage is:

separation distance in strings rest frame ≠ string's proper length

You say "The atoms of the string cannot span the separation distance anymore.
In some frames this is because the separation distance increased, in others because the atoms are contracted."

That sounds very nice, but I fear it needs SR's (or any other accepted theory like GR and QM) mathematical verification.

The length contraction applies to the atoms, like it does to any moving physical object:
http://en.wikipedia.org/wiki/Length_contraction

If you have fears that SR might predict different outcomes from different frames, then it's up to you to calculate the scenario from all possible frames, so you can sleep well again.

 

[harrylin]

[..] I'm asking (with tongue in cheek) "Why is the string's length, as transformed between inertial frames, being treated differently than the rockets' separation length? What I'm asking is why the string and the separation aren't being treated in the same SR way if, as you say, "they are the same" and so should be. [..]

Are you asking about the mathematical explanation or the physical explanation?
The mathematical explanation was already given by Dewan et al (the illustration imposes a constant separation as measured in the launch pad frame; next Lorentz transform to any frame you like).
As for a physical explanation, see my replies #15 and #47 to the OP.

 

[Eli Botkin]

yuiop:
I appreciate your replies and will review them with care. A quick read, however, indicates an incorrect statement that appears often in many of this thread’s replies, leading to erroneous conclusions.
That statement is:
“Natural length contraction of the pole as the rockets accelerate…”

This is a misinterpretation of the often stated SR edict that “Moving bodies contract their length”.

Follow this Minkowski image for a moment. A rigid rod of length L lies on the x-axis in inertial frame A (its left end is at x=0, its right end is at x=L). The rod is initially at rest in frame A.
At time T=0 in A the rod undergoes an impulsive acceleration in the +x direction bringing the rod to a velocity V relative to frame A. The two rod endpoints of course have worldlines. When T<0 in A (rod at rest in A) those worldlines are vertical and parallel, separated by distance L. When T>0 in A (rod moving in A) the worldlines are still parallel, but now sloped to the right. Their separation, at any value of T>0, in the x direction is still L. The now moving rod’s length as measured in frame A has not contracted from the length measured before motion in frame A.

Can we measure the length of the moving rod in frame A? Yes, by sending and timing a light signal which is then reflected from both rod ends to the observer and applying a simple computation. Whether moving in A or not, the measurement by the A observer will yield the same length L.
Message: No length contraction due to motion within a frame takes place when length is measured in that frame.

So where does length contraction come into this picture?
After acceleration the rod, no longer at rest in A, is at rest in a frame B whose T’ axis is parallel to the final worldlines. An observer at rest in B will measure the rod and find its length to be L’ >L. [To see this draw the B-frame’s x’ axis.] Contraction has occurred in this sense: Measurement L in frame A, which is in motion relative to the rod’s rest frame B, is a contraction of the rod’s measurement L’ that is made in its rest frame B. Moving rods are contracted ! But only in the sense outlined above.

 

[Eli Botkin]

A.T.:
Surprise! A string (or pole or anythig) can have more than one "proper" length. It depends when and by whom it is measured. Start with a pole at rest in frame A and therefore with proper length L. Now accelerate this pole to velocity V in frame A. In its new rest frame B its "proper" length will be L' > L. Even proper length is relative, not a universal truth.

 

[A.T.]

This is a misinterpretation of the often stated SR edict that “Moving bodies contract their length”.

“Moving bodies contract their length” is already a misinterpretation of SR. The correct description of length contraction is: "bodies are shorter in a frame where they move, than in their rest frame". Length contraction relates lengths in different frames, not lengths at different time points during an acceleration.

Whether an accelerating object contracts, depends on the scenario. If it is prevented from contracting by external forces, like the string in Bell's Paradox, then it doesn't contract, but deforms internally and eventually breaks.

 

[harrylin]

A.T.:
Surprise! A string (or pole or anythig) can have more than one "proper" length. It depends when and by whom it is measured. Start with a pole at rest in frame A and therefore with proper length L. Now accelerate this pole to velocity V in frame A. In its new rest frame B its "proper" length will be L' > L. Even proper length is relative, not a universal truth.

"Proper length": while the length of a "moving" 1m ruler will be determined as less than 1 m in the "stationary" frame, it will still be determined as 1m in a co-moving frame - in shorthand, its "proper length" is still 1m. Proper length does not depend on by whom it's measured, because it is defined as measured in a co-moving inertial system.

 

[A.T.]

Surprise! A string (or pole or anythig) can have more than one "proper" length. It depends when and by whom it is measured.

No, it doesn't depend on "by whom" it is measured. It is always measured in the rest frame. As for the "when": Sure if you stretch a rubber band you change its proper length, so it is different than before.

Why is this a "surprise"?

 

[Nugatory]

A.T.:
Surprise! A string (or pole or anything) can have more than one "proper" length. It depends when and by whom it is measured. Start with a pole at rest in frame A and therefore with proper length L. Now accelerate this pole to velocity V in frame A. In its new rest frame B its "proper" length will be L' > L. Even proper length is relative, not a universal truth.

I'm sorry, but that's just plain wrong, and may be be contributing greatly to some of the confusion further up in this thread.

The proper length is defined as a frame-invariant quantity, namely the length of the pole as measured by an observer at rest relative to that pole (and back in post 81 DaleSpam explained how the proper length may be defined when such an observer is not available).

The pole will always have the same length when measured in its rest frame (that's a frame in which the pole is not moving, so if the pole is moving at speed v relative to you then so is its rest frame) and that length is its proper length. Everyone moving relative to the pole measures an apparent length L'<L which depends on the their velocity relative to the pole.

 

[Eli Botkin]

Nurgatory:
We agree and we disagree. We agree that "The proper length is defined as the length of the pole as measured by an observer at rest relative to that pole."

We disagree when you add "The proper length is defined as a frame-invariant quantity." Defining it as such is counter to what a Minkowski diagram shows: The rest frame before acceleration and the rest frame after acceleration show different "proper" lengths.

 

[Eli Botkin]

A.T.:
See my reply 130 (where no stretching has taken place.)

 

[A.T.]

We disagree when you add "The proper length is defined as a frame-invariant quantity." Defining it as such is counter to what a Minkowski diagram shows: The rest frame before acceleration and the rest frame after acceleration show different "proper" lengths.

If you think that there is a contradiction between "frame invariant" and "different before and after acceleration", then you simply don't understand what "frame invariant" means. Frame invariance has nothing to do with staying constant over time. And SR's length contraction has nothing to do with changing over time. See my post #126.

See my reply 130

See above.

(where no stretching has taken place.)

Define "stretching". I mean by "stretching" : changing the proper length of an elastic body, by applying forces. By my definition there is stretching in your scenario.

 

[Nugatory]

We disagree when you add "The proper length is defined as a frame-invariant quantity." Defining it as such is counter to what a Minkowski diagram shows: The rest frame before acceleration and the rest frame after acceleration show different "proper" lengths.

It does no such thing.

First, let's double-check a piece of terminology because that phrase "the rest frame" is dangerous (which, BTW, is why in my previous posts I've been so careful to say "the frame of the ground observer" instead of "the rest frame" and "the frame in which the object is at rest" instead of "the rest frame"):

Consider two frames, A and B. Frame B is moving at a velocity v relative to frame A (so of course I could just as correctly say that frame A is moving at a velocity -v relative to frame B). I have a pole that is at rest in frame A. Because the pole is at rest in frame A, that is its "rest frame". When I accelerate the pole to speed v, it is now at rest in frame B and moving with speed v relative to frame A. Frame B is now the "rest frame" of the pole and frame A is moving at a velocity -v relative to the pole so no longer is the rest frame of the pole.

Do you agree with this definition, that is that the "rest frame" of an object is whatever frame the object is at rest in, so changes when the speed of the object changes?

If the answer is "no", you're looking at the wrong place in your Minkowski diagram because youve misunderstood what the definition of a rest frame is.
If the answer is "yes", then you're misreading the scale on the axes in the Minkowski diagram. In the rest frame, the spatial separation between the two ends of the pole will always be L, the proper length of the pole although it may not look that way on a piece of paper.

Pick a "yes" or a "no" answer for followup, please? I don't want to spend a lot of time writing up answer to the wrong question.

 

[A.T.]

We disagree when you add "The proper length is defined as a frame-invariant quantity." Defining it as such is counter to what a Minkowski diagram shows: The rest frame before acceleration and the rest frame after acceleration show different "proper" lengths.

It does no such thing.

He might be talking about a scenario where the proper length indeed changes over time (like a elastic string in Bell's Paradox). But that of course in no way contradicts the fact that proper length is frame invariant. He is apparently confused about different frames vs. different time points.

 

[Eli Botkin]

Nugatory:
Yes, your definition of rest frame is also my definition of rest frame. There should be no confusion when adopting the short phrase “rest frame” to stand in for the complete phrase “the frame in which the object is at rest. ”And no, I’m not, as A.T. suspects, considering an elastic string.

If you exam the Minkowski diagram, which reflects the SR transformation equations, you will note that a rod which is of “proper” length L in its rest frame A, will, after acceleration to velocity V, be at rest in a different frame B (as I’m certain you already know). In that new rest frame B the rod will have a different, larger, “proper” length (based on the transformed distance scale to the new frame).

I do know what the term “frame invariance” means and I admit that when I wrote that it failed for proper length I was using it in a different sense than its intended use. I was using it as short-hand for saying that what we define as “proper” length is not a universal, all-time property of a rigid body. The rigid body’s “proper” length can be altered by intermediate accelerations.

I hope I was clearer this time.

 

[ghwellsjr]

The rigid body’s “proper” length can be altered by intermediate accelerations.

Of course you can alter the length of a rigid body by applying forces to it at more than one point. That's what Bell's Paradox is all about.

 

[Dale]

A string (or pole or anythig) can have more than one "proper" length. It depends when and by whom it is measured.

No, proper length is invariant, its value does not depend on the reference frame in which it is measured. Your ignorance of basic SR definitions does not invalidate my proof.

 

[Eli Botkin]

Ghwellsjr:
Yes, bodies can be deformed, stretched, compressed, etc by forces. But that’s irrelevant here. This rigid body was accelerated by forces that were distributed and applied uniformly at every point of the body so as to get all points to accelerate in an equivalent manner. Clearly this is an idealism, not a realism.

The outcome here is that regardless of how complicated the acceleration and deceleration history might be, the change in proper length depends solely on the final velocity change. This is a relativistic effect independent of the acceleration history and the force levels applied.

 

[Dale]

Eli Botkin, you really need to learn some basic definitions. I suggest you google "proper length", "proper time", and "Born rigid motion".

A) there is nothing physically wrong with assuming the string to be of a stiff material that breaks rather than stretches.

B) there is also nothing wrong with performing calculations in any desired frame, including the momentarily comoving inertial frame.

C) in the momentarily comoving inertial frame the failure condition is that the distance between the ships is greater than the proper length

D) my proof follows

 

[Eli Botkin]

DaleSpam:
I presume that by invariant you mean that unless the body is deformed by applied forces and torques, every inertial observer who finds the body at rest in his/her frame will measure it to have the same "proper" length (assuming the same metrics in each frame). And that this is so regardless of the accelerations the body undergoes.

Or, in other words, once ten feet long at rest in frame A, then also ten feet long when at rest in frame B, even if it had to accelerrate to come to rest in B.

If you believe that then that's ok. But a careful check of the transformation equations and the help of a Minkowski diagram might be useful to you.

 

[ghwellsjr]

Ghwellsjr:
Yes, bodies can be deformed, stretched, compressed, etc by forces. But that’s irrelevant here. This rigid body was accelerated by forces that were distributed and applied uniformly at every point of the body so as to get all points to accelerate in an equivalent manner. Clearly this is an idealism, not a realism.

If you accelerate a rigid body identically at more than one point along the direction of acceleration, then its length will change. If it is truly rigid, then it will break. Otherwise it will be stretched.

The outcome here is that regardless of how complicated the acceleration and deceleration history might be, the change in proper length depends solely on the final velocity change. This is a relativistic effect independent of the acceleration history and the force levels applied.

If you accelerate a rigid body at only one point, then it will end up with the same Proper Length no matter what speed it ends up at.

 

[Eli Botkin]

DaleSpan:
In answer to 140:
I find no serious fault with your A,B,C
What I find unsettling is using the string's proper length in the co-moving frame when that length applies only to the ground frame. Note that the ships' proper separation changes by the SR transformation equations in either ship's co-moving frame (which is a momentary rest frame). Why is not the string length subject to that transformation?

I guess its your rejection of my view on proper length that I expounded recently ;-)

 

[Eli Botkin]

ghwellsjr:
1) Sounds like you've settled the Bell Paradox issue: The string will break because it is rigid and is being accelerated at more than one point identically. Thank you.

2) Question: If you accelerate a body at one point,then do you not leave the other points behind? ;-)

 

[ghwellsjr]

ghwellsjr:
1) Sounds like you've settled the Bell Paradox issue: The string will break because it is rigid and is being accelerated at more than one point identically. Thank you.

You're welcome.

2) Question: If you accelerate a body at one point,then do you not leave the other points behind? ;-)

If it is a rigid body and you don't accelerate it beyond its ability to endure the stress, then the points behind will get dragged along and the points ahead will get pushed along until the acceleration ends and the body assumes the same shape and size at its new velocity (relative to its velocity prior to acceleration) as it had before.

 

[Austin0]

So again, what is wrong with assuming a stiff string in the Bell's spaceship scenario? Do you object because you think the assumption is non-standard or because you think it is wrong? Please answer these questions directly instead of with an evasion.

A question has occurred to me:
If we assume a realistic rod connecting the ships so there is some degree of flex without breaking. The ships are spaced prior to acceleration such that there is a small degree of arc,
Say a drop of 5 cm. in the middle
In the launch frame, after an initial extremely short period as the momentum propagates through the rod, there should be no decrease in the deviation of the middle of the rod until eventually enough velocity is achieved to cause measurable contraction.
But in a frame moving in the same direction as the acceleration, the lead ship begins accelerating/moving first.
This would seem to indicate that the slack in the rod must instantly diminish to some extent. In a frame with a high velocity and therefore a greater interval between initiation of the front and rear ships, this seems like it would be significant.
Of course any changes in measurement of the difference in deviation would be transverse to motion, so the relative velocities would not affect this measurement in any frame.

Besides an actual coordinate displacement of the front ship relative to the rear , there would also be the resulting velocity away from the rear ship which would immediately continue taking up the slack and reducing the arc deviation from straight.
Without setting numbers, it still seems safe to say that a very small reduction of distance in the launch frame from the separation that would draw the rod taut would result in the small degree of sag I am talking about.
So it seems reasonable to suppose that a very small increase in the coordinate separation would then remove the deviation and render the rod straight between the ships.
Any thoughts??

 

[Eli Botkin]

A note to all the Bell Paradox contributors with whom I’ve interacted.

I’ve enjoyed the time spent but it is now at a point where everyone is just *sticking to their guns*.

There is only repetition of positions held, both on your parts and on mine. I think none of us has the full answer to this riddle, though some may think they do; but it’s good that we keep on trying. I’ll chime in on other issues if I think I can make a useful contribution. My thanks to all.

Eli Botkin

 

[ghwellsjr]

A question has occurred to me:
If we assume a realistic rod connecting the ships so there is some degree of flex without breaking. The ships are spaced prior to acceleration such that there is a small degree of arc,
Say a drop of 5 cm. in the middle
In the launch frame, after an initial extremely short period as the momentum propagates through the rod, there should be no decrease in the deviation of the middle of the rod until eventually enough velocity is achieved to cause measurable contraction.
But in a frame moving in the same direction as the acceleration, the lead ship begins accelerating/moving first.
This would seem to indicate that the slack in the rod must instantly diminish to some extent. In a frame with a high velocity and therefore a greater interval between initiation of the front and rear ships, this seems like it would be significant.
Of course any changes in measurement of the difference in deviation would be transverse to motion, so the relative velocities would not affect this measurement in any frame.

Besides an actual coordinate displacement of the front ship relative to the rear , there would also be the resulting velocity away from the rear ship which would immediately continue taking up the slack and reducing the arc deviation from straight.
Without setting numbers, it still seems safe to say that a very small reduction of distance in the launch frame from the separation that would draw the rod taut would result in the small degree of sag I am talking about.
So it seems reasonable to suppose that a very small increase in the coordinate separation would then remove the deviation and render the rod straight between the ships.
Any thoughts??

Nobody is ever going to carry out this experiment with actual spaceships and a connecting rod or string. You seem to think that a more complicated experiment would provide a more convincing demonstration. What's wrong with a simple string?

 

[ghwellsjr]

ghwellsjr:
1) Sounds like you've settled the Bell Paradox issue: The string will break because it is rigid and is being accelerated at more than one point identically. Thank you.

You're welcome.

A note to all the Bell Paradox contributors with whom I’ve interacted.

I’ve enjoyed the time spent but it is now at a point where everyone is just *sticking to their guns*.

There is only repetition of positions held, both on your parts and on mine. I think none of us has the full answer to this riddle, though some may think they do; but it’s good that we keep on trying. I’ll chime in on other issues if I think I can make a useful contribution. My thanks to all.

Eli Botkin

Are you saying you weren't sincere in your thanks to me?

If you think that none of us has the full answer to this riddle, then you are admitting that you don't think you have the full answer. But don't extrapolate your own confusion on to the rest of us.

And it's not a riddle or a paradox. It's a simple problem with a simple answer. It's a shame you have given up learning and understanding.

 

[yuiop]

A note to all the Bell Paradox contributors with whom I’ve interacted.

I’ve enjoyed the time spent but it is now at a point where everyone is just *sticking to their guns*.

There is only repetition of positions held, both on your parts and on mine. I think none of us has the full answer to this riddle, though some may think they do; but it’s good that we keep on trying. I’ll chime in on other issues if I think I can make a useful contribution. My thanks to all.

Eli Botkin

You said you would take a closer look at post #121 https://www.physicsforums.com/showpost.php?p=4041097&postcount=121 so it seems a shame you appear to be bowing out without keeping your promise. If you take a closer look at scenario 3 in that post it should make it crystal clear to you that the string stretches and eventually breaks in Bel's rocket paradox.

If you are still not convincd, then if you stick around a little while, I will post another scenario (4) which should make it even clearer.

Or is it just that you have already realized the string must break and just do not know how to gracefully accept your position was wrong?

 

[yuiop]

yuiop:
I appreciate your replies and will review them with care. A quick read, however, indicates an incorrect statement that appears often in many of this thread’s replies, leading to erroneous conclusions.
That statement is:
“Natural length contraction of the pole as the rockets accelerate…”

This is a misinterpretation of the often stated SR edict that “Moving bodies contract their length”.

I deleted that statement to avoid introducing a distraction, but it appears that you were replyiing to an old version of the post perhaps from an email notification. Anyway, it appears I was too late and you did allow the statement to distract you from the main points of post #121.

Bell introduced his rocket paradox to make an important point about the nature of length contraction and many so called experts at the time did not really understand length contraction and so came to wrong conclusion (similar to you) that the string does not break. If you take a different approach and accept that the string breaks (as most experts today would agree) then you work backwards and figure out why that must be the case and learn about the true nature of length contraction. In your earlier lecture to me on length contraction you said:

The two rod endpoints of course have worldlines. When T<0 in A (rod at rest in A) those worldlines are vertical and parallel, separated by distance L. When T>0 in A (rod moving in A) the worldlines are still parallel, but now sloped to the right. Their separation, at any value of T>0, in the x direction is still L. The now moving rod’s length as measured in frame A has not contracted from the length measured before motion in frame A.

... Can we measure the length of the moving rod in frame A? Yes, by sending and timing a light signal which is then reflected from both rod ends to the observer and applying a simple computation. Whether moving in A or not, the measurement by the A observer will yield the same length L.

Clearly you do not understand length contraction or spacetime diagrams. If the length of the rod is L when at rest in A, then when the rod is moving in A and the worldlines of the rod slope to the right, the length of the rod is not still L as measured by A. Maybe you do understand that measurements in A are made using clocks and rulers at rest in A's inertial reference frame?