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Acceleration and twins.

  1. May 28, 2008 #1
    In the original impure twins paradox thread I tried to introduce a scheme that tried to put the travelling twin under the same effective acceleration as the Earth twin to try and elliminate the effect of gravity from the classic twins paradox. How successeful that was, is debatable.


    This new version puts both twins under identical acceleration histories, while still producing differential ageing. Hopefuly, this will refute all claims that General Relativity is required to resolve the classic twins paradox.

    Earth in the classic twins paradox is replaced by a small space station so that we can ignore gravity. Twin A launches from the spacestation at the start of year 2000. At the start of year 2010 (according to the spacestation calender) Twin A turns around and starts to head back and at the same time twin B launches with exactly the same acceleration and reaches exactly the same cruise velocity as twin A did 10 years earlier. In the year 2015 twin B passes twin A on his way back and turns around so that they both arrive back at the space station in the year 2020.

    Now both twins experienced one launch acceleration event, one turnaround acceleration event and one de-acceleration event and reasonable care was taken to ensure they accelerated at identical rates and cruised at identical velocities. They only differ in launch and turnaround dates and total journey times. Twin A will now be younger than twin B and acceleration is NOT the explanation of the differential ageing because they both experienced IDENTICAL acceleration.
     
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  3. May 28, 2008 #2

    Janus

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    It is not just the magnitude and duration of the acceleration that counts, it is also the distance between the twins and the space station when they accelerate. Also, the acceleration does not effect the clock of the accelerating twin, but effects what the accelerating twin determines is happening to the other twin's and the station's clocks.

    Even In GR it is not the difference in local g-force/acceleration that counts but the difference in gravitational potential. For instance, if you have a uniform gravitational field and placed two clocks at different heights in this field, the clock higher in the field will run faster, even though it experiences exactly the same g-force as the lower clock.

    For like reasons, a clock in the nose of an accelerating rocket will run faster than one at the tail, even though they experience the same acceleration.

    A secondary problem is that you say that twin B launches at the same time as A turns around but fail to say according to who? A or B?
     
  4. May 28, 2008 #3
    Well I probably should of made it clear that all times were according to a third observer that remained as a witness on the unaccelerated space station.
     
  5. May 29, 2008 #4

    DrGreg

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    kev, that is an excellent example you gave to show that acceleration does not cause dilation. To help other readers I enclose a spacetime diagram. Inertial motion is blue and acceleration is red.

    Considering this as a 2D Euclidean drawing, route A is longest and route C is shortest. In the topsy-turvy geometry of space time, this means A experiences the shortest proper time and C the longest. (C is a third observer who stays on the space station.)

    A and B both experience the same acceleration for the same time, but A's total elapsed time is shorter than B's.
     

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    Last edited: May 29, 2008
  6. May 29, 2008 #5

    Thanks Dr Greg and the spactime diagram you produced is exactly what I had in mind. :smile:

    However, in my experience things are never so clean cut in relativity and there are always alternative interpretations. For example you could come up with an elaborate equation that is a function of acceleration and distance from the observer to account for time dilation and I think that is what Janus was getting at, though I am not sure such an equation has ever been published.

    I think one way the issue can resolved is to look at the emperical evidence from short half life particles that are routinely stored in cyclotrons. If particles are injected into a cyclotron at say 0.8 so that they follow a circular path will their half life be extended by a factor of 1.6666 as this they travelling in a straight line at 0.8c or will there be additional time dilation due to the centripetal acceleration? Or is it possible to calculate a time dilation factor of 1.6666 for the particles entirely due to the centripetal acceleration and completely ignore the linear velocity?
     
  7. May 29, 2008 #6
    The example shows that acceleration by itself is not the cause of the difference in the ageing rates of the twins in this case, because any time dilation due to acceleration is cancelled out. It does not prove that acceleration by itself has no effect on time dilation. That is what I was trying to get at in my last post.

    There is also a minor typo in the second last sentance of my last post. It should have read:

    "If particles are injected into a cyclotron at say 0.8c, so that they follow a circular path, will their half life be extended by a factor of 1.6666 as if they travelling in a straight line at 0.8c or will there be additional time dilation due to the centripetal acceleration?"
     
  8. May 29, 2008 #7

    tiny-tim

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    Hi kev! :smile:

    No, DrGreg is right …

    If there is no gravity, then time dilation relative to an inertial observer depends entirely on velocity and nothing else. :smile:

    Acceleration is only ever mentioned when people try (and fail) to measure time dilation relative to a non-inertial observer. :cry:
     
  9. May 29, 2008 #8

    Dale

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    This exact experiment has been done by Bailey et al. as referenced in the Clock Hypothesis section of the FAQ. The centripetal acceleration was ~10^18 g and no additional time dilation was detected beyond that predicted by accounting for velocity alone.
     
  10. May 31, 2008 #9
    isn't the earth twin accelerating at the same rate as the other?if so, shouldn't HE age less? Time does not dilate.only the distance increases.
     
  11. May 31, 2008 #10

    Dale

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    Actually, the spacetime interval decreases for the non-inertial observer. This is the measurement of distance that has absolute meaning.
     
  12. Jun 1, 2008 #11
    Isn't this statement the equivalent (according to Mach's Principle) of saying that there is a "preferred" rest frame?

    Al
     
  13. Jun 1, 2008 #12

    tiny-tim

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    Hi Al! :smile:

    No, because only inertial observers qualify as having rest-frames.

    So inertial observers are preferred over non-inertial observers because they have rest-frames! :smile:
     
  14. Jun 1, 2008 #13
    Well, according to Mach's principle, the only difference between the two is a non-inertial observer is accelerating relative to the "fixed stars", not just relative to an inertial frame.

    Wouldn't that make the "fixed stars" a preferred frame, if we define "non-inertial" as acceleration relative to them? Otherwise, what is the acceleration relative to?

    I personally don't believe acceleration is relevant to the twins paradox at all, since the asymmetry is obvious even if if there were no acceleration involved. It just seems like saying that whichever observer is not accelerating relative to the "fixed stars" is "preferred" is equivalent to saying that the "fixed stars" constitute a preferred frame.

    Al
     
  15. Jun 1, 2008 #14

    tiny-tim

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    Hi Al! :smile:

    But the stay-at-home twin keeps the same rest-frame, while the prodigal twin has at least two different rest-frames.

    I'm not sure what position Mach's principle has in modern relativity, but on any view it can't apply to an observer who changes his frame, can it? :smile:
     
  16. Jun 1, 2008 #15
    "Also, the acceleration does not effect the clock of the accelerating twin,"


    isnt that exactly the point they are making??
     
  17. Jun 1, 2008 #16

    Dale

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    No, the spacetime interval can be calculated in any frame, including non-inertial ones. All you need is the metric tensor or the line element in your chosen frame.
     
  18. Jun 1, 2008 #17
    My understanding is that's exactly who it applies to.

    Al
     
  19. Jun 1, 2008 #18
    so as i see it, special relativity is perfectly capable of solving the twins paradox but you must choose an inertial frame from which to view the events and you must stick with it throughout.

    if you want to view it from the point of view of the accelerating twin (or any other non-inertial frame of reference) then you must use general relativity.

    so the only question is whether one can take it for granted that one can determine whether a given frame is inertial or not. can one determine which twin is being accelerated and which one is not?
     
  20. Jun 1, 2008 #19
    It can be very easily determined if a given frame is inertial or not. Simply use an accelerometer. If the accelerometer reads zero you are in an inertial reference frame. If another observers velocity is changing over time relative to you then they are not in an intertial reference frame. If you jump from a tall building you are in a inertial reference frame because your accelerometer will read zero while you are in free fall. The observer that remains on the tall building watching you accelerate away from him is not an inertial observer because his accelerometer is not reading zero.
     
    Last edited: Jun 1, 2008
  21. Jun 1, 2008 #20

    Hurkyl

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    Well, you see it wrongly. The whole point of learning about the Lorentz transformations is so that you can easily transfer between two different sets of coordinates. The method of generalized coordinates is perfectly applicable to special relativity, thus you don't need to choose orthonormal coordinates (i.e. an 'inertial frame').

    Finally, you don't even have to use coordinates at all. For example, the relevant math for most twin scenarios is nothing more than elementary (Minkowski) plane geometry, and can be solved with (hyperbolic) trigonometry.
     
    Last edited: Jun 1, 2008
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