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Twin Paradox- a quick(ish) question

  1. May 5, 2009 #1
    Hi all, I'm new here, I was on another forum asking this question but there was a lack of a response, so I hope you guys can help me out!

    Ok, so after a few weeks of grappling with the twin paradox, I finally accept that the twin that travels on the rocket and back is the one that ages less.

    (however)

    If both twins had a stopwatch, the stay at home twin's watch would register a longer time when the twins meet again. My question is: what would the twins observe about the other twins stopwatch (imagine they have extremely good vision :P )?
    The travelling twin would have to, at some point, see the stay at home twin's watch go faster, other wise how would it reach a larger value than his own?

    Hope that makes sense, please help!!
     
  2. jcsd
  3. May 5, 2009 #2

    atyy

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  4. May 5, 2009 #3

    George Jones

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    I wrote a(n unreadably) terse quantitative analysis of what atty posted,

    https://www.physicsforums.com/showthread.php?p=1384776#post1384776.

    A simpler version:

    Two twins, Alfred and Betty, are together on the planet Omicron 7. After synchronizing their watches, Betty sets off on a return trip from Omicron 7 to Earth and back to Omicron 7, and Alfred remains the whole time on Omicron 7. The distance between Omicron 7 and Earth is 3.75 lightyears in the (approximately) inertial reference frame of Omicron 7 and the Earth. Betty takes the most direct route and moves at a constant speed of 3/5 lightspeed during both the outgoing and incoming segments of the trip. Both the outgoing an incoming legs of Betty's trip take 5 years according to her watch.

    As Betty travels, Betty uses a telescope to watch Alfred's wristwatch. During Betty's outgoing leg, she sees Alfred's second hand spin slower than hers by a factor of 2. Thus, Betty sees Alfred's watch advance by 5/2 = 2.5 years during the outgoing leg. During Betty's incoming leg, she sees Alfred's second hand spin faster than hers by a factor of 2. Thus, Betty sees Alfred's watch advance by 5*2 = 10 years during the incoming leg.

    During the whole trip:

    Betty's watch advances by 5 + 5 = 10 years;

    Alfred's watch advances by 2.5 + 10 = 12.5 years.
     
  5. May 5, 2009 #4
    Thanks! Makes sense now :)

    That's cleared up alot for me so thanks again!
     
  6. May 5, 2009 #5

    DrGreg

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    George's argument is one the simplest explanations of the twins paradox and it follows directly from Einstein's postulates of relativity. The fact the same factor of 2 applies to the outward and inward journeys is no coincidence. Consider a third person Charlie who is on Earth all the time watching Alfred and Betty through a telescope. On the outward journey, of course he sees Alfred's clock hand spin at half the speed of Betty's hand, because that is what Betty sees (and it is the same light travelling from Alfred to Betty that continues to Charlie).(1) But Charlie and Alfred are stationary relative to each other, so they see each other's clock hands spin at the same rate. Therefore Charlie sees his own clock hand spin at half the speed of Betty's hand. Or in other words, as Betty travels towards Charlie, Charlie sees Betty's hand spin at twice the speed as his own.

    On the return journey, Betty sees Alfred moving towards her at exactly the same speed that Charlie saw Betty moving towards him in the outward journey. By the principle of relativity,(2) both situations are equivalent, so Betty sees Alfred's hand spin at twice the speed as her own.

    This argument is an example of k-calculus. (So called because you can use the same argument with any other number instead of 2, e.g. k.)


    (1) This is an application of Einstein's 2nd postulate: light emitted by Alfred towards Betty and Charlie travels at the same speed as light emitted by Betty towards Charlie.

    (2) i.e. Einstein's 1st postulate
     
    Last edited: May 5, 2009
  7. May 6, 2009 #6

    NWH

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    You're assuming the gravity on Earth and Omicron 7 are the same though, right?
     
  8. May 6, 2009 #7

    George Jones

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    I assumed that gravity is negligible, i.e., special relativity.
     
  9. May 6, 2009 #8

    NWH

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    Got you!
     
  10. May 7, 2009 #9

    otg

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    I wonder, how is it [that Alfred's watch is speeding from Betty's point of view] consistent with the fact (?) that a watch relative to which one is moving, slows down?

    I see the argument with Doppler shift as clear as a day, but I don't find a "pure" time dilation argument to why the symmetry is broken. The gamma factor does not take into account whether one is approaching or moving away.

    If one does not use Doppler shift as an argument, I don't find a way to prove that the situation is not possible to reverse [as in: Betty is at rest and Alfred is moving away and Alfred is younger at the reunion]

    Any ideas?
     
  11. May 7, 2009 #10

    JesseM

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    The time dilation equation only works in inertial frames, it can't be applied in a non-inertial one where Betty is at rest throughout the trip. It is true that if you take the inertial frame where Betty is at rest during the outbound phase before the turnaround (assuming Betty moves inertially prior to turning around), then in this frame Alfred ages more slowly than Betty during the outbound phase; but after the turnaround, in this frame Betty has an even higher velocity than Alfred, and therefore ages more slowly than him. Likewise, in the inertial frame where Betty is at rest during the inbound phase after the turnaround, in this frame Alfred ages more slowly than Betty during the inbound phase, but in this frame Betty had a higher velocity than Alfred during the outbound phase and was aging slower than him during that phase. And you can't combine results from the two frames for the half of the trip where Betty was at rest in each frame, because the two frames disagree about simultaneity, and therefore have a large disagreement about Alfred's age at the moment of the turnaround. For example, suppose Betty moves at 0.6c for 25 years in Alfred's frame, then returns at 0.6c for another 25 years, so Alfred ages 50 years by the time she returns but Betty only ages 25*sqrt(1 - 0.6^2) = 25*0.8 = 20 years during each phase, so she is only 40 years older when she returns. In the first of Betty's two different rest frames, the frame where Betty was at rest during the outbound phase, at the moment of the turnaround Betty is 20 years older than when she departed Alfred, while Alfred is only 20*0.8 = 16 years older than when Betty departed. But then in the frame where Betty was at rest during the inbound phase, Betty is still 20 years older at the moment of the turnaround but Alfred is already 34 years older, And she'll age another 20 years from then until when she reunites with Alfred while he'll only age 16 more years, but because he was already 34 at the moment of the turnaround in this frame he'll be 34 + 16 = 50 years older when they reunite.
     
  12. May 7, 2009 #11

    otg

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    thank you, I knew I had seen it somewhere, it's just that thing about "Alfred har aged 16 years and instantaneosly he has aged 34" that bothers me... but I guess I have to live with that :)
     
  13. May 7, 2009 #12

    JesseM

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    But it isn't "instantaneous" in that frame, it's only "instantaneous" if you try to imagine constructing a non-inertial frame where Betty is at rest throughout the trip, a frame whose measurements match those of the first inertial frame during the outbound phase and those of the second inertial frame during the inbound phase. In the inertial frame where Betty is at rest during the inbound phase, prior to the turnaround Betty had been traveling even faster than Alfred--using the velocity addition formula, we can see that in this frame Betty must have had a velocity of (0.6c + 0.6c)/(1 + 0.6*0.6) = 1.2c/1.36 = 0.88235c, so in this frame her clock must have been slowed by sqrt(1 - 0.88235^2) = 0.4706. So if she aged 20 years during the outbound phase in this frame, the "actual" time in this frame before turning around must have been 20/0.4706 = 42.5 years. And in this frame Alfred is always moving at 0.6c so he always has a time dilation factor of sqrt(1 - 0.6^2) = 0.8, so during those 42.5 years he must have aged 42.5*0.8 = 34 years in this frame. So you see in this frame nothing special happened to Alfred at the moment of the turnaround itself, their relative ages at the moment of the turnaround are just a consequence of their velocities in this frame since Betty departed from Alfred.
     
    Last edited: May 7, 2009
  14. May 7, 2009 #13

    Ich

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    Yes, it should bother you. Nobody claimed that.
    See it as a change in perspective, like in this analogon: if you look at Rigel, 800 ly away, and then turn your head 90 degrees to the right, Rigel is suddenly 1100 ly further to the left. It would certainly bother you if someone claimed that Rigel moved 1100 ly in less than a second. You don't have to live with that.
     
  15. May 7, 2009 #14

    otg

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    Yes, and I realized my error in formulation [that's why I'm back, but I see that I don't have to rephrase myself :)].

    thank you very much for the comments
     
  16. May 13, 2009 #15

    otg

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    hmm... gave it some more thought, and I can't really convince myself that it would be any different to make the same argument in the inertial frame where Alfred is at rest during the inbound leg, for instance... I find that Alfred is younger than Betty at the reunion...
    Inbound Alfred has a velocity relative to the outbound Alfred which is higher than the velocity relative to Betty at all times...

    Where's my glitch?

    Darn my head is turning inside out :)
     
  17. May 13, 2009 #16

    otg

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    perhaps it was in that I forgot to shrink the distance earth-star in the spaceship-outbound-rest-frame... yikes this is not to be done late at night...
     
  18. May 13, 2009 #17

    otg

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    nope... didn't help me...
     
  19. May 13, 2009 #18

    JesseM

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    I thought Alfred was the one moving inertially--if you pick the inertial frame where he's at rest during the inbound leg, then he's also at rest during the outbound leg, no?
     
  20. May 13, 2009 #19

    Mentz114

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    Somewhere in this forum ( can't find it ) is a descrption of what a traveller sees as he accelerates away from the earth, turns round and comes back. On the outward trip the traveller sees the earth apparently turning more and more slowly as he moves away, and on the return trip sees the earth spinning faster and faster ( due to separation and light delay) until, when she touches down the number of rotations seen is equal to the number experienced by those left behind. Anything else would be a contradiction. But the travellers clock will still be behind those left on earth. So the sight of planet earth turning more or less slowly or quickly means nothing since they'll agree at the end on the same number. But the travellers clock will still be showing less elapsed time than those left behind.

    It doesn't matter a swat which frames are inertial or not, the elapsed times on clocks travelling a worldline ( according to SR and GR) is the sum of dt2-dx2. End of story. You don't need to know much else to resolve the twins non-paradox.
     
  21. May 13, 2009 #20

    sylas

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    George does it in [post=2186296]msg #3[/post] of this thread.
     
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