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The age old twin paradox

  1. Jan 19, 2014 #1
    Two twins A and (evil) B. Twin A remains "stationary" while evil B goes at near light speed to-and-fro. Ignoring General Relativity (acceleration/deceleration) evil B experiences time dilation and as result, when they meet up again, B's clock is behind A's clock.

    Look at it from B's point of view, B is stationary and A goes to-and-fro. Why won't A's clock be behind B? Why doesn't A experience time dilation?
     
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  3. Jan 19, 2014 #2

    PeterDonis

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    Before posing any questions about the twin paradox, I highly recommend reading the Usenet Physics FAQ article on it:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

    The short answer to your question is that A and B are not symmetric; A is at rest in the same inertial frame the whole time but B is not. However, there are *lots* of subtleties, which are addressed in the article.
     
  4. Jan 19, 2014 #3

    phinds

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    No such thing. No one ever "experiences" time dilation, it is something observed by a different reference frame. For example, you, right now as you read this, are traveling at almost the speed of light from some reference frame. Do you fell dilated?
     
  5. Jan 19, 2014 #4

    ghwellsjr

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    A's clock will be behind B.
     
  6. Jan 19, 2014 #5

    PeterDonis

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    If I'm understanding the scenario correctly, A remains at rest in the same inertial frame the whole time while B does not. That means A's clock will show more elapsed time than B's when they meet up again.
     
  7. Jan 19, 2014 #6

    Dale

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    You can certainly ignore General relativity, but you cannot ignore acceleration.
     
  8. Jan 19, 2014 #7

    ghwellsjr

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    I understood him to be describing two different scenarios. In the first, he said A is stationary and B travels to and fro. In the second he said B is stationary and A travels to and fro. I took him at his word. But, until he clarifies, we won't know. We get ill-defined scenarios like this all the time, don't we?
     
  9. Jan 19, 2014 #8

    DrGreg

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    It seems pretty clear to me that he was asserting that these are two different ways of describing the same scenario. Of course, they are not.
     
  10. Jan 19, 2014 #9

    ghwellsjr

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    And, of course, that's the way I answered his question.
     
  11. Jan 19, 2014 #10

    Nugatory

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    You don't need GR to deal with acceleration and deceleration. Special Relativity handles them just fine, as long as there's no gravity.
     
  12. Jan 20, 2014 #11

    tom.stoer

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    You can't look at it from B's point of view (in SRT) b/c he does not define an inertial frame ;-)

    It's better to introduce an inertial frame with coordinate time t and define the proper times τA and τB. A and B follow two worldlines CA and CB. For their proper times you find

    [tex]\tau[C_{A,B}] = \int_{C_{A,B}} d\tau = \int_{t_1}^{t_2} dt \sqrt{1-\vec{v}_{A,B}^2(t)}[/tex]

    The two curves are defines such that the they intersect at t1 and t2. At t2 A and B can compare their proper times τA and τB.

    Iff A remains at rest in the inertial frame (with coordinate time t), then his proper time and the coordinate time t coincide, i.e.

    [tex]\tau[C_{A}] = {t_2}-{t_1}[/tex]

    And for B you may define a circular path CB with constant speed |vB| along CB. Then you find

    [tex]\tau[C_{B}] = \sqrt{1-v_{B}^2}\cdot({t_2}-{t_1}) = \sqrt{1-v_{B}^2}\cdot\tau[C_{A}][/tex]

    Now the situation is not symmetric for A and B b/c their curves aren't (in any inertial frame).

    But if the curves (other curves, of course) are symmetric (in some inertial frame) then the proper times are identical.
     
    Last edited: Jan 20, 2014
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