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

[schaefera]

As I understand, we should treat length contraction as if all space in the moving frame is contracted-- that is, the distance between two objects as well as the objects themselves are cotracted by a factor of gamma.

If this is correct, wouldn't the string in Bell's paradox NOT break- because a stationary observer does see the string contract, but only just as much as the distance between the ships shrinks?

 

[Nugatory]

As I understand, we should treat length contraction as if all space in the moving frame is contracted-- that is, the distance between two objects as well as the objects themselves are cotracted by a factor of gamma.

If this is correct, wouldn't the string in Bell's paradox NOT break- because a stationary observer does see the string contract, but only just as much as the distance between the ships shrinks?

Here's a simpler version of Bell's spaceship paradox. The simplification loses Bell's more important and interesting point about how accelerations at diffent locations transform, but it does explain why the string breaks.

Suppose the spaceships don't start at rest; instead they're zooming by the ground-based observer at a constant velocity and separated by a distance D in the ground-based frame. Furthermore, the two spaceships aren't yet connected by the string. Instead, two daredevil acrobats are standing on the ground, separated by the same distance D and holding the ends of a string of that length.

As seen by the acrobats and the ground frame observer, the spaceships are separated by a distance D and so are the acrobats. So each acrobat will see a spaceship pass overhead at the same time in their ground frame. At this exact moment, both acrobats leap into the air and grab hold of the spaceship passing overhead.

Being strong and capable daredevil acrobats, they manage to grab their spaceship with one hand while holding onto their end of the string with the other. They each experience an enormous jerk as their spaceship drags them off, but once they've recovered from that shock, we have our two acrobats and the string now moving at the speed of the two spaceships.

The string was of length D when it was rest relative to the ground-based observer, but now it's moving relative to that observer, so is length-contracted. The distance between the spaceships is still D relative to that observer, so the string breaks.

For an observer on either spaceship (both moving at the same constant speed, so in the same frame) relativity of simultaneity means that the two acrobats grab the spaceships at different times. The lead acrobat goes first, while the trailing acrobat is still standing on the ground. The string stretches and breaks as one end is being pulled by the lead ship while the other end is still anchored.

 

[schaefera]

That makes a lot of sense, so thank you for the explanation!

To transfer from this example to the actual Bell's paradox, does the acceleration in Bell's paradox kind of "smear out" the acceleration in what you describe? So while the stretching out happens in your example essentially instantaneously (well, in the time it takes the attach the string), in Bell's version it happens over the team it takes to reach the speed we already assume in this other example.

 

[Nugatory]

To transfer from this example to the actual Bell's paradox, does the acceleration in Bell's paradox kind of "smear out" the acceleration in what you describe? So while the stretching out happens in your example essentially instantaneously (well, in the time it takes the attach the string), in Bell's version it happens over the team it takes to reach the speed we already assume in this other example.

I'd rather say that my version (I did say that it loses the important part of Bell's paradox) just skips over the analysis of the acceleration by considering only the starting and ending conditions while ignoring the way we got from one to the other. The "essentially instantaneous" bit is only a way of making this trick a bit more palatable; you could get the same result from the original thought experiment by considering the situation after both ships have shut off their engines so are no longer accelerating but instead coasting at the same constant velocity.

 

[schaefera]

Ok, makes sense! My one last question about it is this: why wouldn't the acceleration of the ships in Bell's version mean that, as seen by an outside observer, the distance between them is contracted at the same rate that the string itself contracts? This would cancel out the string's contraction.

You see, in your version the fact that the rope starts at D and is suddenly moving (making its length less than D) is affected by the fact that the ships remain distance D apart. But if the ships accelerate WITH the string, wouldn't they become contracted as the string does?

 

[someGorilla]

The space between the ships doesn't shrink because they move in such a way that it doesn't shrink. By assumption. That's the whole point of Bell's paradox!

 

[schaefera]

But how don't they? I'm thinking of applying the Lorentz transform to their positions are random points in time-- because, at a single instant we can treat them as moving at constant velocity (yes?) and I think the distance between them in the non-moving frame will always have to be shrunk just like any object is contracted.

 

[bcrowell]

Ok, makes sense! My one last question about it is this: why wouldn't the acceleration of the ships in Bell's version mean that, as seen by an outside observer, the distance between them is contracted at the same rate that the string itself contracts? This would cancel out the string's contraction.

I've written up an analysis here that may help: http://www.lightandmatter.com/html_books/genrel/ch02/ch02.html#Section2.3 (Scroll down to example 10.) The crucial point is that due to the relativity of simultaneity, different observers disagree about whether the two spaceships' velocities are matched.

The space between the ships doesn't shrink because they move in such a way that it doesn't shrink. By assumption. That's the whole point of Bell's paradox!

This isn't my understanding of how the paradox is normally stated.

 

[schaefera]

This also makes sense-- but why doesn't the distance between the ships shrink along with the string? Shouldn't it all be length contracted?

 

[Dale]

If this is correct, wouldn't the string in Bell's paradox NOT break- because a stationary observer does see the string contract, but only just as much as the distance between the ships shrinks?

But the distance between the ships doesn't shrink in the stationary observer's frame.

 

[someGorilla]

This isn't my understanding of how the paradox is normally stated.

Then I'm probably mistaken. I thought it's normally stated as a scenario with two ships having equal velocity at a given moment and equal constant acceleration as seen by a third (inertial) observer. Which implies the distance between them is constant for the same observer. Am I wrong?

 

[Nugatory]

Ok, makes sense! My one last question about it is this: why wouldn't the acceleration of the ships in Bell's version mean that, as seen by an outside observer, the distance between them is contracted at the same rate that the string itself contracts? This would cancel out the string's contraction.

You see, in your version the fact that the rope starts at D and is suddenly moving (making its length less than D) is affected by the fact that the ships remain distance D apart. But if the ships accelerate WITH the string, wouldn't they become contracted as the string does?

Ooops, you're right.

 

[A.T.]

Then I'm probably mistaken. I thought it's normally stated as a scenario with two ships having equal velocity at a given moment and equal constant acceleration as seen by a third (inertial) observer. Which implies the distance between them is constant for the same observer. Am I wrong?

You're right. I think it was just a misunderstanding which frame is meant.

 

[A.T.]

This also makes sense-- but why doesn't the distance between the ships shrink along with the string? Shouldn't it all be length contracted?

Why should the distance shrink if the rockets have always the same speed in the inertial frame?

 

[harrylin]

As I understand, we should treat length contraction as if all space in the moving frame is contracted [..]

Bell explained in his book that his example was rejected by those who understood length contraction like, it seems, you understood it: as a magical "space contraction" between unconnected objects. However, length contraction (Lorentz contraction) should be treated as a physical effect, as both Lorentz and Einstein described it. As a matter of fact, it was the purpose of Bell to highlight that point with his spaceship example.
-http://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox

One may better understand length contraction as a physical contraction of bodies. This cannot affect the distance between accelerating rockets. However, in combination with a different synchronization of clocks, the result is that for a reference system that accelerated from rest to a certain speed, after re-synchronization all space in the stationary system appears to be contracted.

 

[Eli Botkin]

Schaefera:
Bell’s Spaceship Paradox is indeed paradoxical if you apply only SRT in seeking a solution. One must always keep in mind that the physics must be coordinate-free. Every observer must conclude the same result for the string, regardless of the observer’s motion, if the solution is to be valid.

For Bell’s “stationary” observer the string endpoints remain eternally separated by the same distance and so that observer would conclude that the string, except for other possible reasons, would remain unbroken.

An inertial observer (at some constant speed V) traveling in the same direction as the spaceships would say that the leading spaceship initiated its acceleration first. This (for this observer) results in an increasing separation between ships and an expectation of string breakage.

But what about a third inertial observer moving in the opposite direction with speed -V? This observer says that the rear ship accelerated first and so is approaching the leading ship and there is no reason to expect that the string will break.

So, if the string does break it needs to be because it is being accelerated, which is a feature common to all observers. My best guess is that GRT (and maybe QM) needs to be invoked to answer this “paradox”.

 

[Dale]

Bell’s Spaceship Paradox is indeed paradoxical if you apply only SRT in seeking a solution.

No, it isn't. It is merely an apparent paradox, not a real one. I.e. It is nothing more than a homework-style problem that tends to stump novices.

 

[A.T.]

Bell’s Spaceship Paradox is indeed paradoxical if you apply only SRT in seeking a solution.
...
For Bell’s “stationary” observer the string endpoints remain eternally separated by the same distance and so that observer would conclude that the string, except for other possible reasons, would remain unbroken.

If you apply only SRT, as you suggest, the “stationary” observer will still conclude that all the atoms in the string are contracting during the acceleration, and therefore cannot span the constant length anymore. So the “stationary” observer will conclude that the string breaks, if he applies SRT correctly.

 

[bcrowell]

No, it isn't. It is merely an apparent paradox, not a real one. I.e. It is nothing more than a homework-style problem that tends to stump novices.

Not just novices. The story is that Bell went around the CERN cafeteria posing the question, and virtually every physicist he asked got it wrong.

 

[Eli Botkin]

DaleSpam: All paradoxes are apparent. I meant only that it remains an (apparent) paradox if you seek its explanation only through SRT.

A.T. : I disagree with your interpretation of SRT. You are misinterpreting the usual dictum “moving objects contract their length in the direction of their motion.” In the inertial frame in which the two spaceships simultaneously start their journey it is a given that they maintain a constant separation. 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. SRT makes no such distinction. The string ends must maintain the same constant separation. In this inertial frame the moving string is NOT contracted. But its length is less than its length would be when measured by an observer in a frame co-moving with the string (which we call its “proper” length). Don’t interpret the words “proper length’” to mean “true length.” In SRT we are dealing only with comparisons of measurements (actually coordinates) between different inertial frames. Coordinates are labels that are assigned to events and SRT tells you how to change labels as you change observers. That label changing procedure (Lorentz-Fitzgerald transformations) is the “truth” claimed by SRT.

If you doubt any of this I suggest that you set up a calculation as follows:
In frame S place a rod of length L (as measured when it was at rest in S). Set the rod to be moving at speed V in S. Measure it now by emitting a light signal that is reflected off both ends. You will note that the length is still L in this frame.

You can also note a Minkowski diagram for a rod of length L that is at rest for t<0. Apply an impulsive acceleration at t=0 so that the rod is then moving at speed V when t >0. Note that in S the rod’s length is still L.

bcrowell: Though Bell opted for breakage of the string, it should be noted that he did not arrive at that conclusion through (nor base his argument on) the application of SRT spacetime transformations. His conclusion was centered on the expectation of a modification of the EM field between the strings atomic components.

 

[Dale]

DaleSpam: All paradoxes are apparent. I meant only that it remains an (apparent) paradox if you seek its explanation only through SRT.

No, it doesn't, it can be fully explained with SR. The string breaks in any frame if the distance between the ships is greater than the length of the string in that frame. You can determine the length of the string and the distance between the ships in any frame using only SR.

 

[Doc Al]

If you doubt any of this I suggest that you set up a calculation as follows:
In frame S place a rod of length L (as measured when it was at rest in S). Set the rod to be moving at speed V in S. Measure it now by emitting a light signal that is reflected off both ends. You will note that the length is still L in this frame.

How are you accelerating that rod? You cannot keep its length at L without keeping it under great tension as it moves.

 

[schaefera]

Bell explained in his book that his example was rejected by those who understood length contraction like, it seems, you understood it: as a magical "space contraction" between unconnected objects. However, length contraction (Lorentz contraction) should be treated as a physical effect, as both Lorentz and Einstein described it. As a matter of fact, it was the purpose of Bell to highlight that point with his spaceship example.
-http://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox

One may better understand length contraction as a physical contraction of bodies. This cannot affect the distance between accelerating rockets. However, in combination with a different synchronization of clocks, the result is that for a reference system that accelerated from rest to a certain speed, after re-synchronization all space in the stationary system appears to be contracted.

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

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.

 

[Eli Botkin]

DaleSpam:
It would please me no end to see a math solution (from any discipline: SRT, GRT, QM, etc.) that proves that the string must break. A math solution does not consist of a repetition of the statement that the string breaks because a moving string contracts. There are some observers for whom the string does not contract (the spaceship separation decreases).

 

[Eli Botkin]

Doc Al:
Of course the external forces applied to accelerate a distributed body will set up internal forces which will either retain the bodies integrity or destroy it. But, assuming that it survives the acceleration's time interval, we are dealing with a then unstressed body moving at constant speed V. It is the length before and after acceleration that is the same in that inertial frame (the one in which it was initially at rest).

 

[Nugatory]

DaleSpam:
It would please me no end to see a math solution (from any discipline: SRT, GRT, QM, etc.) that proves that the string must break. A math solution does not consist of a repetition of the statement that the string breaks because a moving string contracts. There are some observers for whom the string does not contract (the spaceship separation decreases).

There's a pretty good one at http://skfiz.wdfiles.com/local--files/materialy/space_ships.pdf - a pure SR explanation, though there's a small digression through GR at the end. And it's not surprising that there's a pure SR explanation, because the thought experiment is happening in flat space-time so SR works, despite the acceleration. And of course there's Bell's own resolution of the problem.

As with just about all SR paradoxes, you'll find that there's a carefully concealed "at the same time" assumption hidden in the problem statement. In this case, it is in the statement about the acceleration profiles that the two spaceships follow. If they both follow the same acceleration profile in the ground observer's frame (that is, both vary their thrust at the exact same time in the ground observer's frame) they will maintain a constant separation in that frame - but then because of relativity of simultaneity they will not both be on the same acceleration profile in their own frame so the separation in their own frame will change.

 

[Doc Al]

Doc Al:
Of course the external forces applied to accelerate a distributed body will set up internal forces which will either retain the bodies integrity or destroy it. But, assuming that it survives the acceleration's time interval, we are dealing with a then unstressed body moving at constant speed V. It is the length before and after acceleration that is the same in that inertial frame (the one in which it was initially at rest).

Why would you think the length of the moving rod would be the same? It would be Lorentz contracted, per the usual SR rules.

 

[Eli Botkin]

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 :-)

 

[schaefera]

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.

 

[Eli Botkin]

Nugatory:
Thanks for the link to the Matsuda and Kinogarbagea’ paper. As you know, they conclude that the separation in inertial frame S between the two space spaceships remains constant, whereas each spaceship is individually Lorentz-contracted. They are correct about the separation constancy (yes it is self-evident). However, their conclusion about each ship’s contraction is based on blindly using (assuming!) the Lorentz contraction formula as applicable in this problem. In other words, they’ve assumed what they set out to prove.

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.

 

[Eli Botkin]

schaefera:
You raise a valid question. The answer is that both would do the same. If one contracts, the other does too. If one is unchanged, then the other is unchanged. Check my earlier replies.

 

[Dale]

However, their conclusion about each ship’s contraction is based on blindly using (assuming!) the Lorentz contraction formula as applicable in this problem.

The Lorentz contraction formula can be derived directly from the Lorentz transform:

http://en.wikipedia.org/wiki/Length_contraction#Derivation

 

[schaefera]

So what it boils down to is that if in one frame we require the distance of separation to remain constant, it must change in another-- and in this instance we require the distance in a stationary frame to remain the same?

 

[Eli Botkin]

DaleSpam:
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.

 

[Eli Botkin]

schaefera:
I'm not sure what you mean by using the word "require." In the Bell Spaceship Paradox problem the distance between spaceships remains constant while the ships accelerate (as viewed by the "stationary" observer.) Therefore any taught string connecting the ships would (in that frame) also be of that same constant length.

 

[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(<br /> \begin{array}{cc}<br /> \frac{1}{\sqrt{1-v^2}} &amp; -\frac{v}{\sqrt{1-v^2}} \\<br /> -\frac{v}{\sqrt{1-v^2}} &amp; \frac{1}{\sqrt{1-v^2}} \end{array}<br /> \right) = \left(<br /> \begin{array}{cc}<br /> \cosh (t_A) &amp; -\sinh (t_A) \\<br /> -\sinh (t_A) &amp; \cosh (t_A) \end{array}<br /> \right)r&#039;_A=\Lambda \cdot r_A=(0,1)r&#039;_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&#039;_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&#039;_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&#039;_A and r&#039;_B:d&#039; = x&#039;_B-x&#039;_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&#039; = \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?