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

[Dale]

Yes, the Lorentz contraction formula IS derived from the Lorentz-Fitzgerald transformation equations. My point was that they stated the contraction formula as the first equation in their paper, never discussing its applicability in the problem they were presenting, just assuming it was applicable because there was motion.

The usual practice in the scientific literature is to simply handle such background proofs etc. by reference, otherwise all of the unnecessary historical background detracts from the main point of the paper, especially for knowledgeable readers. So the practice of "jumping in" like that is well accepted, particularly with common formulas like that, but they should have included a specific reference.

In any case, regardless of any critiques of the style of the paper, it should be clear that Bells paradox can be resolved with SR alone.

 

[Eli Botkin]

DaleSpam:
I have no problem with using well known formulae without beginning from scratch to derive them. But these authors, without further ado, claim that each spaceship must undergo Lorentz contraction (in the at-rest frame) from its initial at-rest length simply because they were now in motion in that frame. I'm saying that applying the contraction formula this way is incorrect. Though the ship's length (in the at-rest frame) maintains its at-rest length, it is already Lorentz-contracted from the ship's length that would be measured by a co-moving observer.

You maintain that "Bell's paradox can be resolved with SR alone." I'ld be pleased to see how. Can you describe to me the Minkowski diagram that does this? As you know, a Minkowski diagram is well suited for demonstrating issues of Lorentz-contraction.

 

[Nugatory]

It should be noted too that SRT makes no distinction between how to compute the space interval between two events on two different bodies or on the same body. So its hard to see why they chose to.

When you're dealing with points on the same body, all accelerations take place at the same time in the reference frame in which that body is at rest. However, Bell very carefully constructed his thought experiment with two spaceships so that all accelerations take place at the same time in the ground observer's frame. Thanks to the relativity of simultaneity, these are not the same thing, so the worldlines of the two spaceships are not the worldlines of the two ends of a rod lying between the spaceships at the moment of takeoff and accelerated to the final velocity of the spaceships. That's the basis for the different treatment of the two situations.

 

[Nugatory]

But despite all this, why doesn't the space between two objects shrink as per post #23, just as an extended object between those two locations would?

The paper you link to tries to explain this, but I don't understand why applying Lorentz transformations says this should be true.

Consider the situation at the moment that the ships take off, as viewed from the reference frame that is moving at the final speed. Because both ships take off at the same time in the ground observer's frame, relativity of simultaneity means that they don't take off at the same time in that moving frame. In fact, in that frame the lead ship takes off first, lengthening the distance between itself and the trailing ship (which breaks the string between them). Lorentz-contract that increasing distance and you'll get the constant distance that the ground observer sees - but of course the ground observer also sees the string Lorentz-contracting so that it can no longer span that constant distance, so again the string breaks.

 

[harrylin]

Do the Lorentz transformations not require that to a stationary observer the distance between two points is smaller than the distances measured by somebody moving with that other frame?

Yes indeed. Now, as explained by Dewan and Beran (see the elaboration in Wikipedia) we have the situation that the distance between the spaceships as measured in in the launch pad frame is the same as it was before they took off, since they took of simultaneously and with the same acceleration. Consequently, in perfect agreement with what you say, the distance as measured in the moving frame must be greater (increased).
As Nugatory pointed out, according to observations in a co-moving frame this is so because the first spaceship took off earlier. Note that also from that point of view length contraction plays a role; however in this case the spaceships are seen as slowing down, so that now the string would hang loose (due to length de-contraction) if according to that frame the ships had taken of simultaneously.

Wouldn't that require that it is all of space shrinking in the direction of motion, not just objects?

I hope that it's now clear that that does not follow. It even wouldn't make any sense, if you consider the following:

A...B...C...D

A and D are reference towers of the launch pad frame. B and C are the space ships. Imagine that the spaceships take off horizontally and quickly reach very high speed, before closely passing (or hitting) D. According to you we then have:

A...B...C.D

Which means that, following your reasoning, the space ships would shrink the distance between the towers, pulling them together!
There is no law of physics corresponding to such an idea; it would be magic.

As in, x1-x2=(x1'-x2')/(gamma), if we measure at the same time in S'... And this would imply that frame S sees the distance between points in S' as larger than what S measures.

Yes, that's correct. As described from the launch pad frame: If the occupant of the space ships would put a ruler between them, they would measure a greater distance because their ruler has shortened. More practical, instead of a mechanical ruler they could use a laser with mirror and a fast time laps detector to detect the reflected laser pulse: due to their speed they will measure that the return time of the laser pulse is increased which they may interpret as an increased distance.

 

[A.T.]

A.T. : You wish to apply a different rule to the separation of the two hooks that hold the string ends than you would apply to the same string ends.

I have no idea what you mean.

1) The distance between the hooks and the length of the string is the same, unless the string breaks. That distance is given to stay constant as measured in the inertial frame. That's simply the scenario, I'm not applying any rules here yet.

2) SRT says that the atoms of the string will contract as they accelerate. Here I apply SRT.

Thus, at some point those contracted atoms cannot fill the still unchanged distance anymore. The string breaks.

 

[Doc Al]

Doc Al:
The easiest way to see this is via a Minkowski diagram for an inertial frame. Draw 2 orthogonal axes, x horizontal, t vertical. Select the x-interval 0 to 1 to represent a log length on the x-axis. If the log is at rest in this frame, then the worldlines of the log’s end-points are vertical lines parallel to the t-axis.

If the log is at rest for t < 0, then the end-point worldlines there are as stated above.

Imagine an impulsive acceleration of the log at t = 0 where the log and its end-points very suddenly acquires speed V. Now, for t > 0, the end-points’ worldlines are again parallel but sloping upward to the right.

In this frame the log length is always measured in the x-direction. The separation of the end-point worldlines in the x-direction is still 1. In this frame the log length remains 1.

If you still disagree then you will have to show why other end-point worldlines are appropriate for t > 0.

It should be noted that there is a "length contraction" in the picture. The log-length measurement by an observer in an inertial frame co-moving with the rod will yield a rod length > 1. So its in this sense that the moving rod is "contracted" from its "proper length."

Note that in this problem what is referred to as "proper length" is a value that depends on V. So we should not think of "proper length" as an absolute true length. After all, this is relativity theory :-)

Again, the only way you can have the now moving rod retain its original length when measured from the original frame is if you stretch it massively, most likely destroying it. This is not an unstressed rod! (And this is precisely why the string breaks in Bell's example.)

An example where the rod is not stressed, where its length is merely measured from a moving frame, would always have the measured length being shorter than its proper length by the usual factor of gamma. That's a simple frame change as described by the Lorentz transformations.

 

[Dale]

You maintain that "Bell's paradox can be resolved with SR alone." I'ld be pleased to see how.

Sure.

Problem: Given two spaceships, A and B, initially at instantaneous rest separated by distance d in an inertial frame in flat spacetime, the "launch frame". The two spaceships follow identical coordinate acceleration profiles in the launch frame with constant proper acceleration, a, in the positive x direction. The two ships are tethered by a massless stiff string of proper length L. The string breaks if the distance between the ships in the momentarily co-moving inertial frame (MCIF) is greater than L. Does the string break?

Solution: Use units of time and length such that c=1 and a=1. Then without loss of generality the y and z directions can be neglected and the worldline of the two ships can be written as:$$r_A=(t,x)=\left( \sinh(t_A) , \cosh(t_A) \right)$$$$r_B=\left( \sinh(t_B) , \cosh(t_B) + d \right)$$

The four-velocity of A is given by:$$u_A=\frac{\frac{d}{dt_A}r_A}{\left| \frac{d}{dt_A}r_A \right|} = \left( \cosh(t_A) , \sinh(t_A) \right)$$

The three velocity is therefore given by:$$v=\frac{\sinh(t_A)}{\cosh(t_A)} = \tanh(t_A)$$

So the Lorentz transformation to the (primed) MCIF of A is:$$\Lambda = \left( \begin{array}{cc} \frac{1}{\sqrt{1-v^2}} & -\frac{v}{\sqrt{1-v^2}} \\ -\frac{v}{\sqrt{1-v^2}} & \frac{1}{\sqrt{1-v^2}} \end{array} \right) = \left( \begin{array}{cc} \cosh (t_A) & -\sinh (t_A) \\ -\sinh (t_A) & \cosh (t_A) \end{array} \right)$$$$r'_A=\Lambda \cdot r_A=(0,1)$$$$r'_B=\Lambda \cdot r_B= \left( \cosh (t_A) \sinh (t_B)-\sinh (t_A) (\cosh (t_B)+d),\cosh (t_A-t_B)+d \cosh (t_A) \right)$$

Setting the timelike component of $r'_B$ to 0 and solving gives:$$t_A=\cosh ^{-1}\left(\frac{\cosh (t_B)+d}{\sqrt{2 d \cosh (t_B)+d^2+1}}\right)$$

Substituting back gives:$$r'_B=\left( 0, \sqrt{1 + d^2 + 2 d \cosh(t_B) } \right)$$

The distance between the two ships in the MCIF is then given by the difference between the spacelike components of $r'_A$ and $r'_B$:$$d' = x'_B-x'_A = \sqrt{1 + d^2 + 2 d \cosh(t_B) } - 1$$

Note that d' is monotonically increasing and that:$$\lim_{t_B\to \infty } \, d' = \infty$$

Therefore, regardless of L, the string breaks. Specifically, the string breaks at:$$t_B = \cosh ^{-1}\left(\frac{-d^2+L^2+2 L}{2 d}\right)$$

Which is 0 for the special case of d=L.

 

[schaefera]

Yes indeed. Now, as explained by Dewan and Beran (see the elaboration in Wikipedia) we have the situation that the distance between the spaceships as measured in in the launch pad frame is the same as it was before they took off, since they took of simultaneously and with the same acceleration. Consequently, in perfect agreement with what you say, the distance as measured in the moving frame must be greater (increased).
As Nugatory pointed out, according to observations in a co-moving frame this is so because the first spaceship took off earlier. Note that also from that point of view length contraction plays a role; however in this case the spaceships are seen as slowing down, so that now the string would hang loose (due to length de-contraction) if according to that frame the ships had taken of simultaneously.

I hope that it's now clear that that does not follow. It even wouldn't make any sense, if you consider the following:

A...B...C...D

A and D are reference towers of the launch pad frame. B and C are the space ships. Imagine that the spaceships take off horizontally and quickly reach very high speed, before closely passing (or hitting) D. According to you we then have:

A...B...C.D

Which means that, following your reasoning, the space ships would shrink the distance between the towers, pulling them together!
There is no law of physics corresponding to such an idea; it would be magic.


Yes, that's correct. As described from the launch pad frame: If the occupant of the space ships would put a ruler between them, they would measure a greater distance because their ruler has shortened. More practical, instead of a mechanical ruler they could use a laser with mirror and a fast time laps detector to detect the reflected laser pulse: due to their speed they will measure that the return time of the laser pulse is increased which they may interpret as an increased distance.

With respect to the tower example, though, let's say that in their rest frame (S) they are separated by D meters. Doesn't the Lorentz transformation require that in S', moving with the ships, the distance is D/gamma?

 

[harrylin]

With respect to the tower example, though, let's say that in their rest frame (S) they are separated by D meters. Doesn't the Lorentz transformation require that in S', moving with the ships, the distance is D/gamma?

Sure. My point was that the distance between the towers must remain D as measured with S.

Once more: If according to measurements in S, the space between the rockets contracts with the rockets (as you proposed), then we obtain that the towers will be pulled towards each other by the motion of the rockets in-between the towers. There is no physical mechanism to predict or explain such a weird effect.

 

[schaefera]

Ah, I understand! I was mixing up what would shrink in which frame. It's in the moving frame that they have to contract.

 

[harrylin]

Ah, I understand! I was mixing up what would shrink in which frame. It's in the moving frame that they have to contract.

Right. Here we have one of many examples to illustrate that acceleration breaks observational symmetry.

Elaborating on my post #15: while the distance between these accelerating rockets appears constant in S, the distance between the towers appears to be contracted in S'. In S' even the distance between stars will appear to be contracted, as the physical cause is fully ascribed to changes of measurement by the accelerating system - nothing happens to the stationary system.

 

[schaefera]

By breaking symmetry, you mean that given the fact that S' is accelerating, we must consider S as stationary while if S' were moving at a constant speed they'd both be on equal footing? This might have been my issue, because I tried to treat each on equally initially.

 

[harrylin]

By breaking symmetry, you mean that given the fact that S' is accelerating, we must consider S as stationary while if S' were moving at a constant speed they'd both be on equal footing? This might have been my issue, because I tried to treat each on equally initially.

Right! More precisely, commonly S' is another inertial system, for example the final inertial system in which the rockets are in rest when they switch of the engines. Up to then the rockets continuously switch inertial rest systems. S and S' are on equal footing, but the accelerating rockets are not on equal footing with the towers or the (I'm simplifying here) "stationary" stars.

 

[schaefera]

Thank you so much for the help! I see where my confusion was initially-- I was trying to reconcile views that shouldn't have been considered.

One final question, concerning the spacetime diagrams: when we draw the diagram to show the acceleration (not assuming it instantaneous), the spacetime diagrams seem to imply that while if we measure in a stationary frame the distance remains constant, in the moving frame the distance between the ships actually approaches infinite [this is from Post #8's picture]. Must this break down at some point, or would this really happen?

 

[Eli Botkin]

DaleSpam:
I thank you for your prompt analytical reply. I’m familiar with this math since I too have done this. However, you have based your conclusion that the string breaks on the sole result that the observers on the ships see the separation between ships to be monotonically increasing. And I do agree that they do observe this.

But are they the only arbiters, don’t other observers count? What about the “stationary” observer who notes that the ship separation is unchanging? Or better yet, what about the observer moving in the opposite direction who notes that the ships are reducing their separation? To what does he/she attribute the breakage? Should one attribute a physical event (breakage) to a computation in some particular coordinate frame?

I guess that is why there are analysts who seek to distinguish (as SRT does not) the relationship between two events on the same string versus two events not on the string. And there is also always the fall-back to claiming that the string is shortening because it went from not-moving to moving, an often misconstrued repeat of “moving bodies contract.”

I would suggest that you look at the Minkowski diagrams (or do the math) for observers other than those co-moving with the two ships. Thanks again.

 

[Nugatory]

DaleSpam:
I would suggest that you look at the Minkowski diagrams (or do the math) for observers other than those co-moving with the two ships.

Ahhhh... the Minkowski diagram is the same for all observers. The way you introduce another observer into a Minkowski diagram is by drawing that a new set of x and t axes on the diagram to represent the new observer's coordinate system.

But I suspect that what you're really asking for is a Minkowski diagram in which that other observer's x and t axes make a 90-degree angle on the sheet of paper? That's a fair request, as it's lot easier to visualize.

 

[A.T.]

What about the “stationary” observer who notes that the ship separation is unchanging?

See post #41

And there is also always the fall-back to claiming that the string is shortening because it went from not-moving to moving, an often misconstrued repeat of “moving bodies contract.”

The string is not shortening in the “stationary” frame. It can only break. It breaks if its atoms cannot span the separation distance anymore.

If contracting atoms are too abstract, then replace the string with a chain. The accelerating chain links will get shorter and shorter, so at some point they cannot span the constant separation distance, and the chain breaks somwhere.

 

[Eli Botkin]

Nugatory:
Not so. One diagram can contain any number of observer coordinates. The x,t axes for a particular observer need not be at a right angle to each other. The scale on any axis is determined by the hyperbolae that are asymptotic to the light ray at 45 degrees. Such a single diagram is usefull in showing how things appear to all observers. There is nothing special about the observer assigned to the right angle coordinates.

 

[Nugatory]

Nugatory:
Not so. One diagram can contain any number of observer coordinates. The x,t axes for a particular observer need not be at a right angle to each other. The scale on any axis is determined by the hyperbolae that are asymptotic to the light ray at 45 degrees. Such a single diagram is usefull in showing how things appear to all observers. There is nothing special about the observer assigned to the right angle coordinates.

I certainly agree with everything you say after the words "not so"... but it is precisely because one diagram can contain any number of observer coordinates and there is nothing special about the 90-degree pair that I didn't understand what you meant when you asked for "the Minkowski diagrams ... for observers other than those co-moving with the two ships'.

 

[Dale]

I thank you for your prompt analytical reply.

You are welcome.

I’m familiar with this math since I too have done this. However, you have based your conclusion that the string breaks on the sole result that the observers on the ships see the separation between ships to be monotonically increasing.

That and the fact that the limit is infinite, therefore it will eventually break regardless of the actual values of L or d. It is possible for a function to be monotonically increasing to a finite limit, in which case the breakage would depend on the details of how big L was compared to the limiting value of d'.

And I do agree that they do observe this.

Excellent, therefore the string breaks and Bell's paradox is resolved using only SR.

But are they the only arbiters, don’t other observers count? What about the “stationary” observer who notes that the ship separation is unchanging? Or better yet, what about the observer moving in the opposite direction who notes that the ships are reducing their separation? To what does he/she attribute the breakage? Should one attribute a physical event (breakage) to a computation in some particular coordinate frame?
...
I would suggest that you look at the Minkowski diagrams (or do the math) for observers other than those co-moving with the two ships.

The laws of physics are frame invariant, therefore if it breaks in one frame then it breaks in all frames.

I leave the details of any other frame as an exercise for the interested reader (you). I would be glad to look over your efforts if you get stuck somewhere.

 

[Eli Botkin]

A.T.:
Let’s try this. Have the string attached to the two ships only loosely. Say each end is firmly in a sleeve such that when you tug the string it will slide in the sleeve.

With the ships always maintaining the same separation why will the string slide out of a sleeve? And which sleeve, fore or aft? And why is the space between the string atoms being reduced but not the space between the ships?

 

[schaefera]

I think you may be confused on the same point as I was! Try considering the two frames: S which is stationary and S' which move with the ships after they finish their acceleration.

In S, the distance between the ships remains constant because this is the frame in which they accelerate the same, so the statement of the paradox requires this. But the string is length contracted and therefore breaks. It's not that the space between the ships remains constant but the space between the string atoms is shrunk-- it's that the atoms themselves are shrunk, but not the space they are in.

In S', the ships do not accelerate the same way, and the distance between the two is increasing. Since the front ship begins accelerating first, I'd imagine that the string has to slip from the front ship's sheath. This agrees with the fact that (I'm assuming the sheaths to be frictionless) since the ships accelerate forwards the string would slip backwards in the sheaths.

 

[Eli Botkin]

DaleSpam:
It must surely also follow that if it doesn't break in one frame then it doesn't break in all frames ;-)

The main point of frame invariance is that if you can show that it breaks in one frame then you should be able to show that it breaks in other frames.

 

[Eli Botkin]

schaefera:

Please tell me why you think "...that the atoms themselves are shrunk, but not the space they are in."

 

[Dale]

It must surely also follow that if it doesn't break in one frame then it doesn't break in all frames ;-)

Indeed. If you think that you can show that it doesn't break in any frame, then please post your work and I am sure we can figure out where you made the mistake.

The main point of frame invariance is that if you can show that it breaks in one frame then you should be able to show that it breaks in other frames.

The main point of frame invariance is that you can always choose to work the problem in the easiest frame.

 

[Eli Botkin]

DaleSpam:
I agree "...that you can always choose to work the problem in the easiest frame." However, with this type of problem, where there are questions of interpretation, it can't hurt (and may help) to seek that same solution, from other directions, in other frames.

An even "easier" frame than the ship's co-moving frame would have the observer moving at some constant speed, V, in the direction opposed to the ship's motion. Can your solution be found there?

Bear in mind that I've never claimed that the string doesn't break. My claim is that string breakage has not been proven, not even by Bell himself. Bell did opine that it would break because of the internal EM field distortion (which I find strange since , through V, the distortion is also frame dependent). But that doesn't make a proof.

I must admit, though I have great respect and admiration for Bell's contributions to the understanding of QM, I am puzzled by his claim that breakage is due to Fitzgerald contraction though he knows there is no such contraction of the separation between ships.

If the correct solution is that breakage must take place, then I think the culprit is more likely to be the ships' acceleration, a component of the problem that exists in all frames.

Thanks for "listening."

 

[Austin0]

You are welcome.

That and the fact that the limit is infinite, therefore it will eventually break regardless of the actual values of L or d. It is possible for a function to be monotonically increasing to a finite limit, in which case the breakage would depend on the details of how big L was compared to the limiting value of d'.

Excellent, therefore the string breaks and Bell's paradox is resolved using only SR.

The laws of physics are frame invariant, therefore if it breaks in one frame then it breaks in all frames.

I leave the details of any other frame as an exercise for the interested reader (you). I would be glad to look over your efforts if you get stuck somewhere.

Hi First I will say that I personally have the view that the light speed electromagnetic interactions involved in maintaining physical structure do result in a contraction through the tensile forces. I.e. ; the string will break.
Even so I feel impelled to play Devil's advocate here.
Just looking at the scenario from a target frame of say 0.8c it is clear that the leading ship initiates acceleration before the trailing ship. This not only means increasing the separation but also creating a velocity differential that will persist even after the trailing ship is also accelerating.
Therefore relative to that frame, the distance between the ships must continue to increase without bounds as you calculated with your analysis.
This itself seems problematic. How do you reconcile frame agreement of time and position observations of the ships when the distance remains constant in one frame and is indefinitely expanding in the other frame??
Or conversely: Assume the ships started accelerating from signals that were simultaneous
in the 0.8c frame. Now in that frame the distance remains constant, but in the launch frame the trailing ship fires up first and must eventually intercept and overtake the lead ship.
How do we reconcile these contradictory expectations?

Adopting for the moment a purely kinematic interpretation of SR which has been expressed by many knowledgeable people on this forum in the past.
Assuming no contraction of either the distance or the string. At a velocity of 0.8 c what would be the measurement of that distance?
According to an 0.8c inertial frame the clock in the launch frame would be running ahead at the instantaneous location of the lead ship.Therefore the trailing ship and string would move forward an additional distance before reaching a clock with the same proper time. I.e.; Would be measured as being contracted relative to the initial separation.

As I said I don't necessarily accept this interpretation but I can't dismiss it out of hand.

I might point out that if the physical reality of contraction as implied by the Maxwell maths is actually valid this leads strongly to the logical conclusion that all motion is actual and in a sense absolute, even if it is not determinable or measurable in non-relative quantitative terms. Would you agree?? ;-)

 

[Dale]

I agree "...that you can always choose to work the problem in the easiest frame."

Then you must agree that proving an outcome in anyone frame is sufficient to prove the outcome.

Bear in mind that I've never claimed that the string doesn't break. My claim is that string breakage has not been proven

I proved it above. If you disagree then please point out any mistake I made and I will correct it. If I made no mistakes, then I have proven it.

 

[A.T.]

A.T.:
Let’s try this. Have the string attached to the two ships only loosely. Say each end is firmly in a sleeve such that when you tug the string it will slide in the sleeve.

With the ships always maintaining the same separation why will the string slide out of a sleeve?

Beacuse it will contract, so it cannot span the distance between the sleeves.

And which sleeve, fore or aft?

An ideal mass-less string? From both I guess.

And why is the space between the string atoms being reduced

Because the EM-fields that hold the atoms together are contracting, and are pulling the atoms closer together. If the atoms are forced to span a constant length (like in the original scenario) the bounds between the atoms will break somewhere.

but not the space between the ships?

That is given in the scenario. They accelerate such that the distance between the string attachments stays constant.

 

[Dale]

Therefore relative to that frame, the distance between the ships must continue to increase without bounds as you calculated with your analysis.
This itself seems problematic. How do you reconcile frame agreement of time and position observations of the ships when the distance remains constant in one frame and is indefinitely expanding in the other frame??

Why should this be problematic or need reconciliation? The worldlines of the spaceships are curved. This same effect happens with curved lines in Euclidean geometry.

Take a piece of paper and draw two copies of the same curved shape separated by some fixed horizontal distance. Then draw a set of parallel oblique lines and see how the distances are not fixed on the oblique lines even though they are fixed on the horizontal lines.

Adopting for the moment a purely kinematic interpretation of SR which has been expressed by many knowledgeable people on this forum in the past.
Assuming no contraction of either the distance or the string. At a velocity of 0.8 c what would be the measurement of that distance?

I don't know what you are asking here, nor why you would assume no contraction.

I might point out that if the physical reality of contraction as implied by the Maxwell maths is actually valid this leads strongly to the logical conclusion that all motion is actual and in a sense absolute, even if it is not determinable or measurable in non-relative quantitative terms. Would you agree?? ;-)

I don't know what you mean by this either.

 

[Austin0]

Or conversely: Assume the ships started accelerating from signals that were simultaneous
in the 0.8c frame. Now in that frame the distance remains constant, but in the launch frame the trailing ship fires up first and must eventually intercept and overtake the lead ship.
How do we reconcile these contradictory expectations?

You missed this one.

Adopting for the moment a purely kinematic interpretation of SR which has been expressed by many knowledgeable people on this forum in the past.
Assuming no contraction of either the distance or the string. At a velocity of 0.8 c what would be the measurement of that distance?
According to an 0.8c inertial frame the clock in the launch frame would be running ahead at the instantaneous location of the lead ship.Therefore the trailing ship and string would move forward an additional distance before reaching a clock with the same proper time. I.e.; Would be measured as being contracted relative to the initial separation.

I don't know what you are asking here, nor why you would assume no contraction.

The assumption of no contraction was the kinematic interpretation of SR. That all effects were purely the result of relative motion. Coordinate evaluations without necessary physical imp0lications. As I said this is not my assumption but an assumption made by others who adopt this interpretation.
How many times over the years has someone asked , "is contraction real or what causes it" and the expert response has been that it is not necessarily physical but merely an effect of relative motion?
I merely noted that employing this assumption to clock desynchronization could produce a coordinate contraction without an assumption of actual physical contraction.

I might point out that if the physical reality of contraction as implied by the Maxwell maths is actually valid this leads strongly to the logical conclusion that all motion is actual and in a sense absolute, even if it is not determinable or measurable in non-relative quantitative terms. Would you agree?? ;-)

I don't know what you mean by this either.

Likewise, this is a perennial question, "Is motion real " and the majority response has been a resounding NO , it is purely relative with no physical actuality.
But if one accepts the validity of the physical contraction implied by Maxwell then logically it should be said that, yes it is real, with actual physical consequences but is simply undetectable and unquantifiable. So I will ask you , in your view is motion real or purely relative??

 

[Dale]

You missed this one.

It appears to be the same question as the other one. Similarly with the question below.

The assumption of no contraction was the kinematic interpretation of SR. That all effects were purely the result of relative motion. Coordinate evaluations without necessary physical imp0lications. As I said this is not my assumption but an assumption made by others who adopt this interpretation.
How many times over the years has someone asked , "is contraction real or what causes it" and the expert response has been that it is not necessarily physical but merely an effect of relative motion?
I merely noted that employing this assumption to clock desynchronization could produce a coordinate contraction without an assumption of actual physical contraction.

My usual response to such questions, and to this one, is to ask the person to define "real" and "actual" and "physical". Those terms are quite ambiguous and without a solid definition of them the question cannot be answered.

If you have a scenario which is described completely in one inertial frame, then you can use the Lorentz transform to obtain the equivalent scenario in any other inertial frame and be assured that the same laws of physics explain the scenario in both frames.

 

[A.T.]

How many times over the years has someone asked , "is contraction real or what causes it" and the expert response has been that it is not necessarily physical but merely an effect of relative motion?

Is death from rifle bullets physical? Nah, merely an effect of relative motion.

 

[Eli Botkin]

DaleSpam:
I've reviewed your math (your earlier reply #43) and find no error. What you've shown for certain is that for a sequence of instantaneously co-moving observers the ships' separation is some d' > d. I'm familiar with the math, having done this myself. And of course your conclusion favoring string breakage is pre-ordained since you've also stipulated that the string must remain L (= d) at any d'.

But why does every co-moving observer "see" d expand to a value d' > d but not see L undergo a proportionate expansion to an L' > L? I think that would be an essential point to address so others couldn't claim that you have, in effect, proved what you assumed.

By selecting the frames of the co-moving observers (who always deal with a d' > d) you have avoided the problem of finding a solution for frames of observers for whom d' < d (and, as you know, there are many such frames.)

I note that you don't claim string contraction, as many others (including Bell) do. Is that because you've selected a massless string (no atoms to contract)? ;-)