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Position Matters?

  1. Feb 2, 2010 #1
    Twins Jack and Jill are initially at rest at x=L/2 in IRF K. Jill travels to the Origin at speed v, and Jack travels to x=L at the same speed. They are still the same age. At some pre-arranged time each twin ignites a rocket and begins accelerating in the positive x direction. Each twin shuts the thruster down at a later, pre-arranged time. They simultaneously come to rest in IRF K’, traveling in the +x direction of K at speed v>0. They decide to come together again, traveling toward one another at the same speed v relative to K’. Relative to K, Jill must travel faster than v and Jack must travel slower than v. When they get together again, Jack has aged less than Jill. Comments?
     
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  3. Feb 2, 2010 #2
    I don't know what i'm talking about, and i certainly dont know the mathematics to back it up, but i'd like to take a stab to see how my intuition holds up against the knowledgable guys here.

    relative to K',
    jack and jill did not start accelerating from their respective positions simultaneously, so theyre not the same age when they initially come to rest in K'.

    also, wouldn't the final accellerations relative to K play some part on their age difference in K?
     
  4. Feb 2, 2010 #3

    DrGreg

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    Just to clarify, at this point have they both come to rest again relative to K?
     
  5. Feb 2, 2010 #4
    Assuming they both come to rest in K.
    Assuming they accelerate at the same rate as measured in frame K or by onboard accelerometers.
    Assuming at the same time, as measured in frame K or the same time as measured by their onboard clocks before resynchronisation.
    The above assumptions imply they simultaneously (as measured in frame K) come to rest in IRF K’ . It turns out that it is not important whether you mean simultaneous in frame K or K' but it is important to aware of the difference. Whenever you say "simultaneously" you should specify according to what observer because simultaneity is relative.
    Again, when they decide to come together again, did you mean they start the final phase simultaneously as measured in K, or simultaneously as measured in K' after they have resynchronised their clocks in K'?
    I am sure you meant to say that Jack will have aged more than Jill, because as you have already stated, Jill has travelled faster (and further) than Jack in frame K when they get back together again.

    Simply put, Jack had a head start of L in the x direction and if they both take off simultaneously as measured in K and travel in the positive x direction and Jill catches up with Jack, then Jill has travelled further than Jack over the same time period (as measured in K) so Jill has travelled faster and aged less.
     
    Last edited: Feb 3, 2010
  6. Feb 2, 2010 #5
    In the trival case where L = the Origin, Jack and Jill don't part company. All frames will agree on the simultaneity of their actions, because they happen at the same place as each other. Their worldlines will be identical, and so will their ages.

    Otherwise... I assume the first pre-arranged time is according to K. And when you say they're the same age as Jill reaches the Origin, and Jack reaches x, I take it their age is measured in K. (In another frame, these events may happen at different times.)

    When you say simultaneously come to rest in IRF K', do you mean that this event is simultaneous in K', as well as that they come to rest in K', or do they stop simultaneously in K?

    If the simultaneity of their stopping in K' is according to K, then Jill ages less than Jack, as her spacetime trajectory between their parting and meeting is longer.

    If the simultaneity of their stopping in K' is according to K', then it's still Jill who ages less. But according to K, Jill stops first, so--if I've got this right--she has to accelerate even more, in comparison to Jack, than in the previous case, Therefore Jill would age even less, compared to Jack, than she did if they stopped in K' at a moment that was simutaneous in K.
     
  7. Feb 2, 2010 #6

    Mentz114

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    How do you know this is true ? Can we see your calculation, please ?

    Without specifying the acceleration phase it's not possible to work out the proper times.

    Draw a space-time diagram, then we'll see what's going on.
     
  8. Feb 3, 2010 #7
    They are at equal conditions at start.
    Use Jack and Jill own referentials.

    As long as they apply the some amount of thrust(energy), during the same time (in their own referential) they will stay equivalent in all aspects.
    I think that this can be used as a principle (same action imply the same consequences).

    pre-arranged time: It is not impossible the use of an 'instant observer', independent of light travel to do the analisys, where 'pre-arranged time' makes sense. I bet that this step makes everything simpler.
    I know that it is not the usual Einstein way of syncronization.
     
  9. Feb 3, 2010 #8
    A good point and a legitimate point of view.
     
  10. Feb 3, 2010 #9
    Yes, that was the idea. They are both at rest in K when they fire up their thrusters.
     
  11. Feb 3, 2010 #10
    They shut their thrusters down simultaneously according to the clocks of K.
     
  12. Feb 3, 2010 #11
    You're right. My apologies for the mental short circuit.
     
  13. Feb 3, 2010 #12
    I meant that they come to rest in simultaneously in K', as determined by the clocks in K.
     
  14. Feb 3, 2010 #13
    Right. I fouled up here. Jack will be older than Jill once they join up again.
     
  15. Feb 3, 2010 #14

    Mentz114

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    What a pathetic thread.:cry: It's like arguing about the size of sqrt(2) when you could just calculate the value. Instead of handwaving and guesswork, why not draw the worldlines for the scenario in question and calculate the proper-times ? It's called 'science', some of you may have heard of it.:biggrin:
     
  16. Feb 3, 2010 #15

    Mentz114

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    After my disrepectful remarks :redface:, I have no choice but to draw the diagram myself. I think each traveller has 4 phases of which I've labelled A 1, B1 etc. The small triangle and text show how to calculate the proper-time for each leg. Maybe GRDixon can fix the pic to fit his scenario and work out the proper-times. I had some trouble understanding when the changeovers happen after phase B. The purple line is the worldline of someone at L/2, from whose viewpoint the diagram is drawn. ( whoops, the origin is at L/2, makes no difference).
     

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  17. Feb 3, 2010 #16
    I'll have a go, but must confess that I'm just now reading "Relativity Simply Explained" by Martin Gardner. In brief, I'm out of my depth, at times, with some of the bright young minds who express their views on Physics Forums, particularly in GRT. I had a look at your diagram and like it! I'm going to learn how to do this stuff. At 71, I expect it's not going to be as easy as things were 50 years ago (ouch!) at MIT. But I know that it will be as much fun as ever. Speaking of GRT, I was anticipating more comments to my doggerel thread a la the Equivalence Principle. Thanks for commenting. Yours in Physics, GRD.
     
  18. Feb 3, 2010 #17
    "WHEELER'S FIRST MORAL PRINCIPLE. Never make a calculation until you know the answer" (Spacetime Physics, p. 60).
     
  19. Feb 3, 2010 #18

    DrGreg

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    The wording is still a little ambiguous, so I've drawn two possible diagrams. They both start the same.

    (A) Jack and Jill begin moving apart, each at v relative to K in opposite directions

    (B) They come to rest in K, simultaneously in K

    (C) They start to accelerate identically, simultaneously in K.

    (D) When they each reach a speed of v relative to K, they stop accelerating. This will occur simultaneously in K. At this point, both are at rest in K'.

    Up to this point, by considering the symmetry relative to K, the proper time (i.e. time experienced by Jack or Jill) for each stage (A-B, B-C, C-D) is equal for Jack and Jill. So at the events D1 and D2, Jack and Jill are the same age.

    This is where I now have to make an assumption.


    Assumption 1 (left hand diagram)

    (E) Jack and Jill begin moving towards each other, each at v relative to K' in opposite directions simultaneously in K'

    (F) Jack and Jill meet and come to rest relative to K'.

    Because K and K' have different definitions of simultaneity, Jill ages less from D1 to E1 than Jack does from D2 to E2. Now considering the symmetry relative to K', Jack and Jill age the same amount from E to F. Over the whole journey the only segment where there's an age difference is D to E, where Jack ages the most.


    Assumption 2 (right hand diagram)

    (E) Jack and Jill begin moving towards each other, each at v relative to K' in opposite directions simultaneously in K

    (F) Jack and Jill meet and come to rest relative to K'.

    Jack and Jill age the same amount from D to E. From E to F, Jack is moving slower than Jill relative to K, and so ages more. Over the whole journey the only segment where there's an age difference is E to F, where Jack ages the most. So, same outcome as assumption 1.


    I suspect the questioner intended assumption 2, and may also have intended that the B-C and D-E segments are all zero duration.
     

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  20. Feb 3, 2010 #19
    That is so fine. I did have Assumption 2 in mind, but both are interesting. Now if I could only tie this in with the Equivalence Principle ... :) Muchas Gracias.
     
  21. Feb 3, 2010 #20
    I'm glad my intuitive hunch that jack ages the most with either assumption was correct (this time). :smile:
    I suspect we might not have addressed the crux of what you were asking in this thread because the title says "position matters?" and we have not really discussed what you meant by that. Also, could you post a link to the EP thread that you asked about earlier?
     
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