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Twins/triplets again, but *truly* symmetric

  1. Dec 28, 2004 #1
    Guys, I'm trying (just for fun) to map out quantitatively from each traveller's perspective what happens in the following situation. Imagine the classic twins paradox, with triplets instead of twins, but not for the purposes of avoiding the turn-around. In my question, Triplet A stays on Earth, Triplet B travels in +X, turns around and returns to Earth, and Triplet C travels in -X a path exactly symmetric to the one travelled by Triplet B, turning around at the same time as measure on Earth. Eventually the travelling triplets reunite on earth, at the same time measured in the Earth frame.

    When the triplets reunite, I'm hapy that B and C are the same age as eachother and both younger than A, BUT: how does the clock of C look from B's perspective? This must be the same as how the clock of B looks from C's perspective, even though B and C are moving relative to eachother and have truly symmetrical worldlines. What stops B and C arguing about who is younger? I was hoping to do some calculations that showed the opposing clock doing strange (slow) things and finally catching up with 'mine' where I am either B or C, and then showing that the same happens if I am C or B.............
  2. jcsd
  3. Dec 28, 2004 #2
    The problem has been analysed on the internet - Seems it was set up to show SR incorrect - but if you go through all the transforms, symmetrical voyages result in symmetrical answers. In actuality, what one traveling triplet does can have no influence upon the proper clock rate of the other - both journeys being equal - the ages of the travelers upon return must also be equal under any theory that is internally consistent
  4. Dec 28, 2004 #3
    Thanks yogi, I have trawled the web but all I can find are justifications of why there is no paradox. This I accept anyway. What I want to do is work things through myself, creating an Excel spreadsheet with time & location of salient events in all reference frames, using Lorentz. I am halfway there, but I'm having trouble interpreting what I have done. I am hoping someone can say something that unlocks my understanding.

    I've managed to use the Lorentz equations to measure the intervals between clock ticks on the Earth clock in the earth frame and a moving frame, and get the same answer (as expected) with more time and more space between successive ticks when measured in the moving frame. I don't have difficulty with the equations, I think I'm just battling with a mindset that insists on an external overview of the situation (probably represented by the successive rows of my spreadsheet). I'll happily put the spreadsheet on my website with a link if anyone wants to see it.......

    What I'm trying to work towards is a quantitative understanding of how triplet B's clock looks (both 'as seen' and 'as deduced') to triplet C, and vice versa. I'm not really concerned with triplet A, he's just there for completeness.
  5. Dec 28, 2004 #4


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    In the usual way of defining the noninertial rest frame for C:

    During the outbound trip, B's clock will run slower than the time coordinate of the frame.
    During the time when C accelerates, B's clock will run much faster than the time coordinate.
    During the inbound trip, B's clock will run slower than the time coordinate.


    If you want to see something really weird, try this scenario:

    There are two stationary planets, each with a triplet A and B. The triplet C starts at the first planet, goes halfway towards the other planet, turns around, and goes back to the first planet.
    Last edited: Dec 28, 2004
  6. Dec 28, 2004 #5
    Thanks Hurkyl. How would I calculate what happens during the acceleration - can this be done in SR with the acceleration being a step change in velocity?
    Last edited: Dec 28, 2004
  7. Dec 28, 2004 #6
    iq2150- Here is an internet article that deals with your version of the triplet problem ==down near the end of the page - the analysis is not complete and its conclusionary - but you might find it interesting:
    http://www.wcug.wwu.edu/~erikba/ziegler/1-2.html [Broken]

    If that link doesn't work, you can search google under Gordon Ziegler
    Last edited by a moderator: May 1, 2017
  8. Dec 28, 2004 #7


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    Kinda -- acceleration "is" an infinitessimal change in velocity over an infinitessimal change in time... wait, that sounds like differential calculus!
  9. Jan 10, 2005 #8
    This is very similar to thread

    Note you have not defined your problem to be truly “symmetric” yet.
    That would require the out bound ref frame for B to be the same frame as the inbound frame for C. Which would of course means the outbound frame for C would be the same as frame as for inbound B. And since they both spend exactly the same amount of time in exactly the same reference frames of course they would expect to age the same as each other.

    Although stated as ‘symmetric’ you found a need to detail your problem with the turn around time in Earth time was the same for both??? If that was not assumed already, I cannot assume the speeds are the same ether. Thus you could use .8c for C to go out and back. While having B go out and back at 1km/hr or .1c (age about the same as A) or any other speed maybe .5c .6c or .8c.
    Using .8c will give you the completely “symmetric” view. But solving all four will give you the best understanding of why the “PARADOX” is not a paradox to you.

    And yes when working SR assume your ‘accelerations’ are at a maximum rate for an infinitesimal period of time so the affect on each clock is Zero during the acceleration or turn around.
  10. Jan 25, 2005 #9
  11. Jan 26, 2005 #10


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    Let's say that the travelling triplets are travelling at 0.6c relative to the earth, on both legs of the trip. In the earth-triplet's frame, here are the coordinates of the four significant events:

    both travelling triplets leave earth: x=0 light years, t=0 years
    triplet moving in -x direction turns around: x=-0.6 light years, t=1 year
    triplet moving in +x direction turns around: x=0.6 light years, t=1 year
    triplets reunite: x=0 light years, t=2 years

    Now, we have two other frames here--the frame whose origin is moving in the +x direction (let's say this frame uses coordinates x' and t'), and the frame whose origin is moving in the -x direction (let's say this frame uses coordinates x'' and t''). During both triplet's outbound trip, their clocks will match the coordinate time in these frames. The Lorentz transform from the earth frame to the +x frame looks like this:

    [tex]x' = 1.25(x - 0.6t)[/tex]
    [tex]t' = 1.25(t - 0.6x/c^2)[/tex] [in these units [tex]c^2[/tex] has the value 1]

    And the Lorentz transform from the earth frame to the -x frame looks like this:

    [tex]x'' = 1.25(x + 0.6t)[/tex]
    [tex]t'' = 1.25(t + 0.6x/c^2)[/tex]

    So, in the +x frame, the coordinates of those four events are:

    both travelling triplets leave earth: x'=0 light years, t'=0 years
    triplet moving in -x direction turns around: x'=-1.5 light years, t'=1.7 years
    triplet moving in +x direction turns around: x'=0 light years, t'=0.8 years
    triplets reunite: x'=-1.5 light years, t'=2.5 years

    And in the -x frame, the coordinates of those four events are:

    both travelling triplets leave earth: x''=0 light years, t''=0 years
    triplet moving in -x direction turns around: x''=0 light years, t''=0.8 years
    triplet moving in +x direction turns around: x''=1.5 light years, t''=1.7 years
    triplets reunite: x''=1.5 light years, t''=2.5 years

    The triplet that starts out moving in the +x direction will agree with that frame up until he turns around, but after that he will be in the -x frame, except that his clock time will be 0.9 years behind the coordinate time of the -x frame (because his clock read 0.8 years at the moment he turned around, but the coordinate time of that event in the -x frame is t''=1.7 years). So from his point of view, up until the moment he turned around the other travelling triplet hadn't turned around yet (since in the +x frame he turns at t'=0.8 years, but the other triplet doesn't turn until t''=1.7 years), but as soon as he turns around, he switches to the -x frame and his definition of simultaneity changes, so now he will say the other triplet has already turned around and is returning to earth (I can calculate the exact position of the other triplet from his point of view, immediately before and immediately after he turns around, if you want to see the numbers). The fact that he suddenly skips over part of the other triplet's trip when he turns around instantaneously explains how it makes sense that the other triplet will be the same age as him when they reuinite, despite the fact that during both the inbound and outbound leg he saw the other triplet aging more slowly than himself. And of course, the situation is completely symmetric from the other triplet's point of view.
    Last edited: Jan 26, 2005
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