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Clocks Within Each Ship in Bell Spaceship Paradox

  1. Mar 22, 2015 #1
    I believe length contraction always makes more sense when integrated with reminders of relativity of Simultaneity.

    Let's say the engines are at the back end of each rocket. For the viewer "A" in the initial frame, they begin moving and continue accelerating simultaneously, and clocks next to the engines are seen in synch.

    However right away, "A" will measure that for each rocket, a clock by the engine is running faster than the clock by the head of that same rocket. I have never seen this pointed out in a discussion of this paradox, and I think that this may be one reason people so often are mistaken about this situation. The idea that "both spaceships accelerate and keep their clocks synchronized" to "A" distracts the reader and puts one in the mind of a situation without SR. The reminder that within each rocket, the clocks appear out of synch to "A" might snap the view back to SR.

    We might attempt to apply this same logic to the rope, but right away we run into a problem, since the back end of the rope is connected to the front of the back rocket, and the front of the rope is connected to the back of the front rocket. If the back end of the rope was accelerating in synch with the backs of the rockets, the rope back clock could not be connected to the back rocket front clock, since they are out of synch. And the similar applies to the front of the rope being attached to the front ship back clock. We are asking of a suggested nonstressed rope that it fulfill contradictory requirements.
     
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  3. Mar 22, 2015 #2

    Dale

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    And everybody knows that fulfilling contradictory requirements is highly stressful.
     
  4. Mar 24, 2015 #3
    Rearranging:
    Relativity of simultaneity must certainly be dealt with when using less convenient inertial reference systems, and I agree that it is useful. When this is properly done it emerges that although length contraction is "relative", it plays a role in any inertial frame.
    I'm not sure if I simply misunderstand what you mean, but it sounds like a wrong argument. "A" would measure if it were technically feasible that for each rocket, a clock by the engine is running extremely slightly slower than the clock by the head of that same rocket due to the rocket's length contraction. However that's not spectacular at all.
    Or perhaps you mean that an observer "B" inside one rocket, when verifying clock synchronization, will discover that the rocket has accelerated from the fact that the synchronization of the instantaneously co-moving inertial frame at approximately that time does not correspond anymore with that of "A"'s reference frame (a clock by the engine will now appear to be behind compared to the one near the head).
    Adding such useless detail is more than likely to distract from Bell's striking example of a breaking string.
    Certainly not! It's key to Bell's clear and straightforward argument about the physical consequences of SR's length contraction. His case is perfectly SR and it's a simple scenario with identical rockets.
    Clocks that are out of sync can certainly be connected.. I'm afraid that I can't follow your logic at all.

    Note that "Bell's spaceship paradox" was only paradoxical for his colleagues who misunderstood SR; it was his example to drive home their misconception. How do you think that your discussion better clarifies Bell's example than he did himself?
     
    Last edited: Mar 24, 2015
  5. Mar 24, 2015 #4
    I'm comparing this scenario to the more familiar one of the single train/ship which travels by, and which was accelerated at some unknown time in the past, all of its clocks now in synch. There are no stresses there, and there is no "real" length contraction which can be divorced from RoS. In a frame where the clocks at both ends are in synch, there is no length contraction. In a different frame, the clocks at both ends are now out of synch, and there is measured a length contraction.

    The ladder paradox, too, shows that very starkly.

    Would the rope break in Bell's example if the ships did not contract (in the initial frame)? No. Would the ships contract (in the initial frame) without the clocks at their front and back ends being out of synch (in the initial frame)? No.

    The alternative is to add a detail to Bell's example whereby each ship has an engine at the front and at the back, and so all 4 engines fire on the same program, in synch from the initial frame. In that example, the ships themselves would stretch and break apart.

    But that is not the situation in Bell's standard paradox. I didn't think about this until I was looking at the wiki page, and the contracted ships.

    https://en.wikipedia.org/wiki/Bell's_spaceship_paradox
     
  6. Mar 24, 2015 #5
    The ships could be made (artificially) to not contract in the initial frame; the string would still break.
    I think that I already clarified that the clocks inside the rocket are only very slightly out of synch according to "A", and that this is caused by the rocket's contraction (due to the contraction, the front clock's linear speed is all the time very slightly less than that of the rear clock). The inverse is incorrect: the rate at which clocks tick or the way they are synchronized has no effect on the length of the rocket - that is unphysical!
    And then, I'm afraid that someone else will propose to put clocks in the 4 engines in order to clarify that the engines contract. :wink:
    It's easy to make the string much longer than the rockets, so that their contraction can be ignored; no need to complicate it with 4 engines or breaking rockets. And see next!
    That drawing nicely illustrates that the contraction of the rockets can be made totally irrelevant. If the rockets as drawn there are made with some technical means to have a "proper stretching" such that they do not length contract in the original rest frame, it will make no difference at all for the breaking of the string.
     
    Last edited: Mar 24, 2015
  7. Mar 24, 2015 #6
    Agreed. The string would break and the ships would break.

    All of these effects are "only slight" depending upon the actual velocities in play. Also by out of synch I should clarify that I don't mean simply not ticking at the same *rate* but more importantly in the ladder/barn way, i.e. the back clock of each ship is presumably made set earlier due to being at the back. Surely this applies? I'm not up to doing the math for the accelerating case, but in a non-accelerating case, a moving ship is seen to be shortened, and with the back clock set later than the front - by viewers in the "platform" frame. This is the "out of synch" I refer to. The RoS out-of-synch which accompanies length contraction.

    Of course. I was grasping for an inference, a comparison with the non-stressed familiar 2nd-frame-train, not implying direct cause and effect.

    Not at all. I would think pointing out that the clocks within the rockets are not in synch would shatter the sleight-of-hand of the way the "two simultaneously accelerating" rockets scenario is set up.

    It doesn't matter how long the rockets are compared to the rope/space between them. If they contract, it will put stress on the rope. Why does this not happen with the 2nd-frame-train? Everything contracts together. The scenario here has been designed so that they whole getup can't contract *together*.

    If we set the string up for "proper stretching" as well, it won't break. My real point is that for platform observers, an unaccelerated moving flotilla of ships is expected to be shortened and with rear clock set ahead of front clock. I'm trying to find the differences here. I realize that on an unaccelerated moving ship, clocks can be synchronized, such that they appear unsynchronized for platform viewer. I'm not quite sure what can be said analogously for the case of a single accelerating ship, its clocks, and comparison with platform viewer.
     
  8. Mar 24, 2015 #7
    Certainly not! Once more, your statement in your first post, that "right away, "A" will measure that for each rocket, a clock by the engine is running faster than the clock by the head of that same rocket", is totally wrong.
    The back clock of each ship is just like the clocks of the two ships set in synch to the original rest frame before take-off, and if their travel histories are identical, that cannot change.
    Let's take the example of your 4 clocks that are made to accelerate identically in "A"'s rest frame thanks to a slight artificial stretching of the rockets. For that case they will remain in synch with each other according to the original frame, the rest frame of "A".

    That is - happily - not required to understand Bell's spaceship example. :smile:
    It accompanies length contraction if the operator performed a so-called Einstein synchronization at that velocity.
    Not at all: "simultaneously" refers to the launch pad reference system. It's as much a "sleight-of-hand" as synchronous clocks in the GPS system that you may use in your car.
    :bugeye: Please look again carefully at the Wikipedia sketch. Once more: as pictured there, Bell's example is insensitive to length contraction by the rockets.
    It sounds as if you are bugged by the bug that bugged Bell's colleagues. Material objects length contract with their increase of speed, but their speed increase cannot affect the space between them.
    :bugeye: The point of Bell's Spaceship example is that the string undergoes "proper stretching" so that it breaks...
    Here you are getting to the essential point: shortened as compared to a measurement with instruments of the flotilla, if those have first been synchronized according to convention. :oldeyes:
    The main difference here is that Bell discusses physical effects due to a change in velocity, while the other examples merely examine differences in measurements between two independent inertial reference systems.

    A similar issue arises with time dilation: the discussion of effects of change in velocity ("twin paradox") is quite different from the discussion about how two systems in inertial motion measure each other (mutual time dilation).

    Observational symmetry if broken with a change of velocity.

    Does that help?
     
    Last edited: Mar 24, 2015
  9. Mar 24, 2015 #8
    I have to start again. :) One rocket, with acceleration = 0, moving with positive velocity V wrt ref frame S, is measured by observers in S to be shortened (compared to an identical rocket sitting next to the platform) in the direction of motion, and also to have a clock at the back which is set later than the clock in the front (assuming that the denizens have Einstein sync'd them). Now, give that rocket a slight acceleration, and all of those effects still hold, and their extent is slightly changing with time. For observers in S, the clock at the back is getting increasingly ahead compared with a clock at the nose, and the rocket is getting progressively shorter. Any problem there? (The denizens won't be able to do exact Einstein sync if accelerating though, right?)
     
    Last edited: Mar 24, 2015
  10. Mar 25, 2015 #9
    I immediately noticed two problems with that:

    1. Once more: according to S the clock at the back is not getting increasingly ahead compared with a clock at the nose (I explained why it's even getting slightly less ahead).
    By what physical means, do you think, would S propose that the clock at the back should tick faster than the clock at the nose??

    2. You seem to try to show something completely different from what Bell's spaceship example shows!
    How does your example demonstrate that one should not confound the length contraction of material objects with the space between them? It was apparently the mistaken notion of space contraction that made Bell's example a "paradox" for many of Bell's colleagues (this is also referred to in the intro in Wikipedia).
     
  11. Mar 25, 2015 #10

    stevendaryl

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    What you're saying isn't correct.

    Let's assume that the rocket is "Born-rigid". What that means is that it has the same length [itex]L[/itex] in any frame in which it is (momentarily) at rest. Then it follows, by the mathematics of relativity, that as the rocket accelerates, it keeps getting shorter and shorter, as viewed in the original rest frame S.

    Now, think about what "getting shorter" means. It means that the rear end of the rocket is getting closer to the front end of the rocket. Which means that the rear end is traveling (slightly) faster than the front end. Which means (by time dilation) that the rear clock is running slightly slower than the front clock.

    So what you're saying is exactly backwards. The rear clock gets farther and farther behind the front clock.
     
  12. Mar 25, 2015 #11

    stevendaryl

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    I think a point of confusion is clock synchronization in different frames.

    According to the Lorentz transformations, if (1) a rocket is moving at constant velocity v relative to frame S, and (2) the clocks at the front and the rear are synchronized, according to the rocket's reference frame (call it S'), then according to frame S the front clock will be behind the rear clock by an amount

    [itex]\delta t' = \frac{v L'}{c^2}[/itex]

    where [itex]L'[/itex] is the length of the rocket in its own rest frame (by length contraction, [itex]L = \frac{L'}{\gamma}[/itex] is the length in frame S).

    Note the phrase: if the clocks are synchronized in frame S'. That's not going to happen naturally; the people on board the rocket have to adjust the clocks to make that happen. They have to SET the front clock so that it's synchronized with the back clock.

    So you can imagine a constantly accelerating rocket to be approximated by the following discrete process:
    1. The rocket is initially at rest in some frame [itex]S_0[/itex]. The clocks at the front and rear are synchronized in that frame.
    2. At time [itex]t=0[/itex], the rocket accelerates instantaneously to speed [itex]\delta v[/itex] relative to [itex]S_0[/itex]. So it's at rest in a new frame, [itex]S_1[/itex].
    3. The front clock must be set back by an amount [itex]\delta t' = \frac{\delta v L}{c^2}[/itex] in order for the two clocks to be in synch in frame [itex]S_1[/itex].
    4. At time [itex]t = t_1[/itex], the rocket again accelerates to speed [itex]\delta v[/itex] relative to [itex]S_1[/itex].
    5. Again, the front clock must be set back.
    6. etc.
    Every time the rocket accelerates, the front clock must be set back. If you didn't continually adjust the front clock, then the front clock would get farther and farther ahead of the rear clock.
     
  13. Mar 25, 2015 #12
    Instead of acceleration, let us consider a sequence of rockets, each moving with constant velocity, only progressively faster. One way to visualize the clocks being out of synch to the platform viewer is that there is a light beacon in the center of the rocket. For observers in that vehicle's frame, the ping guarantees that the end clocks are in synch. For a viewer in initial frame S, the light signal rushes backward to the rear of the vehicle, triggering the clock to tick forward, before the corresponding signal from the beacon reaches the (receeding) front clock. Therefore the clock at the rear is set to a later time than the one in the front. Now let's move on to a faster moving vehicle. The effect is more pronounced. Observers in S now measure that the rear of the 2nd vehicle is set even later with regard to the front clock than in the initial vehicle's case. etc. No?
     
  14. Mar 25, 2015 #13

    stevendaryl

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    When people say that the front clock on a rocket runs faster than the rear clock, they are NOT talking about synchronization. Forget about synchronization of front and rear clocks, and instead, consider the following thought experiment:

    Take two clocks in the rear of the rocket. Set them both to [itex]t=0[/itex]. Leave one clock in the rear, and bring the other clock to the front. Let both clocks continue to run for one year. Then bring the front clock back to the rear. Then the clock that had been in the front of the rocket will show more elapsed time than the clock that had been in the rear of the rocket the whole time.
     
  15. Mar 25, 2015 #14

    stevendaryl

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    You cannot keep the clocks in the front and rear synchronized in an accelerating rocket, if they are identical clocks.
     
  16. Mar 25, 2015 #15
    Yes, this is precisely the effect I wish to ignore. I am attempting to find a way to bring the insights of the ladder/barn paradox - in which there is no "real" shortening apart from the one related to relativity of simultaneity - into the situation of the Bell's Spaceship paradox. I'm not quite there yet, and am hoping to get there, a step at a time.
     
  17. Mar 25, 2015 #16
    Perhaps if someone can confirm this for me - this other (very similar to other cases being discussed here) situation. Just like Bell's, except that the programmed identical acceleration goes on for a finite amount of time, then the engines shut off (the times at which they do so only being the same in the platform frame) and both ships coast after that. The platform observers now find that even though the backs of the ships are the same distance from one another as they began, and the fronts of the ships are the same distance from one another as they began, that both now coasting ships are now both shortened. This is correct?
     
  18. Mar 25, 2015 #17

    stevendaryl

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    Hmm. What I described is a real effect. It's not a coordinate effect.
     
  19. Mar 25, 2015 #18

    stevendaryl

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    That is correct, as I understand it.
     
  20. Mar 25, 2015 #19
    I'm trying to figure out if this is similar enough to Bell's Spaceship paradox to be illustrative. The ships are shortened, and while the distance between the backs of the ships remains the same to original observer, the ships are shortened to him, and so the distance from the back of the front ship to the front of the back ship is shortened >>although not by as great a percentage as either ship<<. Observers on the ships agree that their ships are now farther apart than they were before their acceleration phase. They will not be surprised to see the rope has broken. Is there something extra, something mysterious, about the "always accelerating" detail of Bell's Spaceship paradox which negates or calls into question the relevance of this particular narrative?
     
  21. Mar 25, 2015 #20

    A.T.

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    The breaking of the rope has nothing to with the contraction of the ships in the original rest frame. The rope is attached to points on the ships which have a constant distance in the original rest frame, but it still will break.

    640px-Dewan-Beran-Bell-Paradox.svg.png
     
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