Instantaneous Age Changes in the Twin "paradox"

Discussion in 'Physics & Math' started by Mike_Fontenot, Sep 2, 2018.

  1. Mike_Fontenot Registered Member

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    The apparent paradox in the twin "paradox" scenario arises because it would seem that each twin should conclude that the other twin is ageing more slowly, due to the well-known "time-dilation" result, during the entire trip, except for the single instant at the instantaneous turnaround ... "surely" nothing could happen to peoples' ages during a single instant. But that assumption is wrong: the traveling twin must conclude that the home twin's age instantaneously increases during the instantaneous turnaround. There are various ways to obtain this result, but by far the easiest and quickest way is to use the equation described in this posting.

    The change in the home-twin's (her) age, before and after an instantaneous velocity change at some instant t in the traveler's (his) life, is given by the very simple "delta_CADO_T equation":

    delta_CADO_T(t) = - L(t) * delta_v(t),

    where

    delta_v(t) = v(t+) - v(t-),

    and where t- and t+ are the instants of his life immediately before and immediately after his instantaneous velocity change at t. The quantities v(t+) and v(t-) are their relative speeds at the instants t+ and t-, according to her. v is positive when the twins are moving apart, and negative when they are moving toward each other. The quantity L(t) is their distance apart when he is age t, according to her.

    So, getting the change in her age during an instantaneous velocity change by him is very simple: you just multiply the negative of their distance apart (according to her) by the change in his velocity. Couldn't be simpler.

    For example, take a case where their relative velocity right before his turnaround is v = 0.9 ly/y (they are moving apart), and right after his instantaneous velocity change their relative velocity is v = -0.8 ly/y (they are moving toward one another). Then

    delta_v = (-0.8) - (0.9) = -1.7 ly/y.

    Suppose that their distance apart at the turnaround is 20 ly. Then

    delta_CADO_T = - 20 * (-1.7) = 34.0 years,

    so he says that she instantaneously got 34 years older during his instantaneous turnaround. Couldn't be simpler.

    Now, suppose that at some later instant t in his life, he decides to instantaneously change his velocity again, this time from -0.8 ly/y to 0.7 ly/y. So this time, he is instantaneously changing from going toward her to going away from her. In this case, we have

    delta_v = (0.7) - (-0.8) = 1.5 ly/y.

    Suppose their distance apart now 18 ly. Then

    delta_CADO_T = - 18 * (1.5) = -27.0 years,

    so he says that she instantaneously got 27 years younger during his instantaneous turnaround. Couldn't be simpler.

    The above information was intentionally designed to be as concise and "narrowly-focused" as possible. Much more complete and wide-ranging information about the traveler's perspective in the twin "paradox" is contained in my webpage:



    https://sites.google.com/site/cadoequation/cado-reference-frame
     
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  3. Gawdzilla Sama Valued Senior Member

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    Why would anybody conclude the ageing was instantaneous? Time was passing for both, but at different rates.
     
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  5. mathman Valued Senior Member

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    For both twins time is passing at the same rate in his own coordinate system. The different rates result from coordinate system changes.
     
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  7. Neddy Bate Valued Senior Member

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    Yes, in SR, idealized instantaneous velocity changes can result in instantaneous age changes. The person whose velocity changes instantaneously is not the one whose age changes, but rather it is other people who are located far away.

    Another way to visualize this is to imagine a very fast train speeding past a station. Let there be clocks on the train, all Einstein-synchronized to each other. Someone standing on the train would say that all the clocks on the train are simultaneously displaying the same time as one another, (after correcting for the time it takes for light to travel from a clock to his eyes). But using that same method, someone standing at the station would say that the clocks on the train are not displaying the same time as one another, but rather a clock at the rear of the train would be "ahead" of a clock at the front of the train, in terms of the times the clocks are displaying simultaneously.

    Then, if the person standing at the station jumps from the station onto the moving train, he would suddenly say that all of the clocks on the train display the same time as one another. That means either the clock at the rear must have jumped back in time, or the clock at the front must have jumped ahead in time, or both. But only according to the person who jumped from the station to the train, because someone who was always standing on the train would say that all the clocks on the train were always displaying the same time as one another.
     
    Last edited: Sep 2, 2018
  8. Dr_Toad It's green! Valued Senior Member

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    Or perhaps you could choose not to use discrete maths, and adopt calculus, rather than relying on Zeno or Houdini.

    This topic was dead out the gate.
     
  9. Gawdzilla Sama Valued Senior Member

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    2,790
    I didn't know that ... before 1964 anyway. Good old Bell films.
     
  10. Mike_Fontenot Registered Member

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    54
    Good example!
     
  11. Neddy Bate Valued Senior Member

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    1,666
    Apologies in advance if I have misunderstood you. But of course instantaneous acceleration is an idealized notion, used for the sake of simplicity.

    In my train example, the person jumping from the station onto the high-speed moving train would die from the impact, but that does not change the fact that before he jumped he determined that the clocks on the train all displayed different times simultaneously, and it also does not change the fact that the people on the train determined that the clocks on the train all displayed the same time simultaneously.
     
  12. Dr_Toad It's green! Valued Senior Member

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    See, that's the problem. As far as I know, nothing on a macroscopic level does anything instantaneously. Only in the strange realm of QM does this occur, or in the minds of folks who don't quite get it yet.
     
  13. Gawdzilla Sama Valued Senior Member

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    2,790
    How many Gs are we talking about here?
     
  14. Neddy Bate Valued Senior Member

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    1,666
    I think we are talking about infinite gees, lol.
     
  15. Dr_Toad It's green! Valued Senior Member

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    At the risk of repeating myself, that's the problem.

    Lol. Christ, I said that in sarcasm, but it still tastes bad.
     
  16. Neddy Bate Valued Senior Member

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    1,666
    But it's not really a problem, because the velocity change won't really be instantaneous. So there won't be infinite Gs, and the people at the front and rear of the train will not instantaneously change in age. It would all happen over an arbitrarily small period of time, in reality.
     
    Last edited: Sep 3, 2018
  17. Dr_Toad It's green! Valued Senior Member

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    I kenned that ye had a wee problem there, Neddy, but I never knew it ran so deep.
     
  18. Gawdzilla Sama Valued Senior Member

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    2,790
    Sorry, I'm on a cell phone while working with 75% function in only one eye.
     
  19. Gawdzilla Sama Valued Senior Member

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    2,790
    So, instead of infinite Gs we only get a few billions. Got it.
     
  20. Neddy Bate Valued Senior Member

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    1,666
    Yes, but Special Relativity doesn't cease to exist just because it is undetectable at slow speeds. So my guy can jump on a slower moving train instead, if you prefer. Like a hobo, train hopping.
     
  21. Dr_Toad It's green! Valued Senior Member

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    Are you channeling Farshite?
     
  22. Neddy Bate Valued Senior Member

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    1,666
    No, I actually do understand SR, unlike Mr Shite. Sorry if something I've written has made you think otherwise.
     
  23. Neddy Bate Valued Senior Member

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    1,666
    For the guy standing in the train station, the time shown on any train-clock can be found using t' = γ(t - (vx / c²)) where γ = 1 / √(1 - (v²/c²)) and t' is the time on the train clock, t is the time on the station-clock, and x is the location of the train-clock along the axis of the train's motion. So you can see that at any given time t, he gets different values for t' depending on x. That is because the train-clocks are not all showing the same time simultaneously, at least not according to the station reference frame.
     
    Last edited: Sep 3, 2018

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