1. Jul 29, 2012

### 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?

2. Jul 29, 2012

### Staff: Mentor

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.

3. Jul 29, 2012

### 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.

4. Jul 29, 2012

### Staff: Mentor

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.

5. Jul 29, 2012

### 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?

6. Jul 30, 2012

### 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!

7. Jul 30, 2012

### 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.

8. Jul 30, 2012

### bcrowell

Staff Emeritus
I've written up an analysis here that may help: http://www.lightandmatter.com/html_books/genrel/ch02/ch02.html#Section2.3 [Broken] (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.

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

Last edited by a moderator: May 6, 2017
9. Jul 30, 2012

### 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?

10. Jul 30, 2012

### Staff: Mentor

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

11. Jul 30, 2012

### someGorilla

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?

12. Jul 30, 2012

### Staff: Mentor

Ooops, you're right.

13. Jul 30, 2012

### A.T.

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

14. Jul 30, 2012

### A.T.

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

15. Jul 31, 2012

### harrylin

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.

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.

16. Jul 31, 2012

### 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”.

17. Jul 31, 2012

### Staff: Mentor

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.

18. Aug 1, 2012

### A.T.

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.

19. Aug 1, 2012

### bcrowell

Staff Emeritus
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.

20. Aug 1, 2012

### 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.

21. Aug 1, 2012

### Staff: Mentor

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.

22. Aug 1, 2012

### Staff: Mentor

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

Last edited: Aug 1, 2012
23. Aug 1, 2012

### schaefera

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.

Last edited: Aug 1, 2012
24. Aug 1, 2012

### 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).

25. Aug 1, 2012

### 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).