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I Twin Paradox - Observation at Turnaround Moment

  1. Jul 11, 2017 #1

    Andy_K

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    Dear All,

    I am doing the Special Relativity course at World Science U, and came across this question which I got wrong:

    When 100 nanoseconds have elapsed on traveling Gracie's watch, she immediately stops, turns around, and heads back toward George at the same speed of her outbound journey, 3/5 light speed. Just as she starts the return journey, how much time will she say has elapsed on George's watch since the journey began?

    The answer is 170 ns. But I'm confused, shouldn't she observe 80 ns on George's earthbound watch (based on time dilation) at the turnaround moment? Based on the illustration from the course, it clearly shows that the light ray reaching the turnaround point is still from an older moment in time. Or am I misunderstanding the question?

    Below is a diagram I'm referring to from his lecture video. Wouldn't Gracie only start seeing later moments of George along the journey back towards him?

    77ceM6c.png


    The lecture video of the above screenshot is as follows (starting at 7th minute):




    Can someone please enlighten me? Thank you.
     
  2. jcsd
  3. Jul 11, 2017 #2

    Ibix

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    Your reasoning is correct for the outbound leg, but the question is about the inbound leg. What time will George's watch read when Gracie's friend returns? What must it read just after turnaround, then?

    Incidentally, this is not a good way to handle "what the traveler says about the stay-at-home's watch" precisely because you get these wildly inconsistent answers near turnaround.
     
  4. Jul 11, 2017 #3

    Andy_K

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    Thank you for helping, Ibix. From what I understand, wouldn't Gracie (the traveler) be seeing sped up images of George (and his watch) as she travels back to earth?

    If so, wouldn't George's watch read 80ns (or perhaps a little more) right at turnaround point, and then go faster along her journey back? What I don't understand is how come there's a sudden huge jump right at the turnaround moment to 170ns.

    Shouldn't she only see George's watch display 170ns when she is more than halfway back to earth, based on the above diagram which shows George's yellow lines? 170ns means George already reached 85% of the entire journey duration, and his clock should only show that time when Gracie is almost back and not while she's still turning around?



     
  5. Jul 11, 2017 #4

    Orodruin

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    You need to differentiate between what an observer acutually will see and what is happening in a particular inertial frame. Time dilation is related to how events are described in an inertial frame. The signal diagram and what signals arrive when is more related to what an observer would actually see.
     
  6. Jul 11, 2017 #5

    Andy_K

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    Thank you, Orodruin. Do you mean that 170ns is what is actually "happening" to George when Gracie is turning around, although she will "see" 80ns?

    Can you please guide me on how to calculate and arrive at the 170ns answer?


     
  7. Jul 11, 2017 #6

    Ibix

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    There's a difference between what you see and how you interpret it. For example, the still picture in your OP shows you that, at turnaround, the traveller will have just received the 10ns pulse from the stay-at-home. So the traveller sees the stay-at-home's watch reading 10ns at turnaround, then racing through the next 240ns over the return journey.

    But the traveler can just subtract out the travel time of the light pulses to work out "what time it is now" at home. But you'll get different answers using the outbound and inbound inertial frames - that's where your 80ns and 170ns come from.
     
  8. Jul 11, 2017 #7

    Andy_K

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    I see. So if Gracie is "seeing" 80ns on George's watch at the turnaround point, George's watch is actually already 170ns? Can you please guide me on how to calculate and arrive at that figure.


     
  9. Jul 11, 2017 #8

    Ibix

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    Use the Lorentz transforms. You are just going to confuse yourself mucking around with the time dilation formula. Time dilation alone cannot resolve the twin paradox - that's kind of the point.
     
  10. Jul 11, 2017 #9
    You have to subtract that 80 ns from the total time that elapses on the staying twin's clock. Gracie spends 200 ns travelling, dilating George's time to ##\frac{5}{4}(200\ \mathrm{ns})## or 250 ns.
     
  11. Jul 11, 2017 #10

    jbriggs444

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    Gracie is "seeing" 10 ns on George's watch just prior to the turnaround point.

    Gracie is "observing" 80 ns on George's watch just prior the turnaround point (i.e. +70 ns to account for the not-yet-seen light from George that is already in flight).

    Gracie is "observing" 170 ns on George's watch just after the turnaround point. There is a discontinuity in what is "observed" because Gracie has changed her inertial frame and, accordingly, her idea of what time is "now" over where George's watch is. The hyperplane of simultaneity has swept forward along George's timeline to a point where George's watch has "now" ticked all the way up to 170 ns.

    Grace is still "seeing" 10 ns on George's watch just after the turnaround point. There is no discontinuity in what is "seen".

    Gracie has spent 100 ns outbound and has "observed" George's watch tick from 0 to 80 ns (net 80 ms).
    Gracie will spend 100 ns inbound and will "observe" George's watch tick from 170 to 250 ns (net 80 ms).

    Gracie has spent 100 ns outbound and has "seen" George's watch tick from 0 to 10 ns (net 10 ns).
    Gracie will spend 100 ns inbound and will "see" George's watch tick from 10 to 250 ns (net 240 ns).

    [All of us learn to wrap our heads around the twin's paradox differently. My intuition handles it with rotating planes of simultaneity]
     
  12. Jul 14, 2017 #11

    Andy_K

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    Thank you for your guidance!

     
  13. Jul 14, 2017 #12

    Andy_K

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    Thank you for taking the time to illustrate the answer with detailed explanation, Jbriggs444! I am much clearer now. I have been confused between what is "seen" and "observed", especially at the turnaround point.

     
  14. Jul 14, 2017 #13

    robphy

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    Here is a spacetime diagram on rotated graph paper
    that can help visualize the comments that people made in this thread.

    The (3/5)c case is on the left.

    Using the language in the comments...
    • What you "see [visually... determined by receiving light-signals]" is based on light-signals joining source events and receiver events.
      • In my diagram, these light-signals are the dotted-lines, parallel to the grid of the rotated graph paper
    • What you "observe [happening simultaneously-"now" according to you.. determined (say) by radar]" is based on
      your instantaneous space-axis ("your spaceline"), which is Minkowski-perpendicular to your worldline.
      • In my diagram, I show the ticks of each observer's light-clock as light-clock diamonds. (All light-clock diamonds have the same area.)
      • The diagonals encode perpendicularity.
        Thus, the spacelines (lines of simultaneity) are the solid-lines, parallel to the spacelike-diagonal of the light-clock diamond.
        The set of spacelines for Bob change at the turn around event Z.
    • Because of the nice choice of numbers (exploiting pythagorean triples [here 3,4,5].. which leads to rational doppler factors [here 2]), you can count diamonds to obtain your answers.
    (3/5)c for 10 ticks each way
    RRGP-twin-3_5.png RRGP-twin-12_13.png

    For the situation in the Brian Greene video [which was done well... I think I'll look at more that set of videos],
    it turns out (presumably derived in a later video as he promised) that the outgoing and incoming speeds are (12/13)c for 5 ticks each way.
    Here again is a pythagorean triple [5,12,13] , which corresponds to the rational doppler factor of 5.
    You can construct all of the light rays that Greene draws in his animation.

    Side note:
    on the diagram, the doppler factor suggests how to stretch out Alice's diamond along the light cone in the forward direction, while squeezing down by the same factor in the other direction (to maintain a fixed enclosed area)... all while keeping the edges parallel to the grid. (This is a Lorentz-Transformation.)
     
  15. Jul 15, 2017 #14
    What I most like about the doppler explanation is the unassailable geometry of the presentation (I first saw it in the Baez articles long ago). Most newcomers should be able to see "where all the time goes" in a maths-free environment, even if it takes them a while to really get to grips with it. I think this is the best we can do when asked for an "intuitive" description of the twin paradox.

    This bedrock is a potential refuge when studying the more formal mathematics of the Lorentz Transform (and puts the meaning of time dilation and length contraction in a more theoretical rather than actual perspective - offline processing as Green says).

    Yes, still banging the same drum, but I think is is worth it . . . ;)
     
  16. Jul 16, 2017 #15

    Andy_K

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    Hi Robphy,

    Thank you very much for taking the time and effort to explain the confusion, and even supplementing it with detailed graphs! It is certainly very helpful indeed.

    May I know if you have any suggested software or tool that I can use to easily model spacetime diagrams, perhaps a ready-made tool that I can just input figures i.e. coordinates, velocity etc and study the results in a graphical format?

    Yup, this lecture on Special Relativity by Brian Greene is excellent! I recently shared on this course here too :)
    https://www.physicsforums.com/threa...ld-science-university-by-brian-greene.919398/


     
  17. Jul 16, 2017 #16

    Andy_K

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    Thank you m4r35n357, the reference with multiple graphs is very nice and a good maths-free way to help visualize the twin paradox from various perspectives.

    The subject is simply mind-boggling (and addictive, I would say), and fortunately there are so many helpful friends like you in this forum :)


     
  18. Jul 16, 2017 #17
    Sorry for spelling your name wrongly Professor! I can't edit my earlier post, but if a mod is prepared to edit it and delete this one, that would be great.
     
  19. Jul 16, 2017 #18

    robphy

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    I'm glad my comments and graphs were useful.
    My diagrams are created with GeoGebra ( https://www.geogebra.org/ ), a general-purpose dynamical-geometry software package.
    I used a variant of https://www.geogebra.org/m/HYD7hB9v#material/VrQgQq9R and https://www.geogebra.org/m/Ny2s7H7N ...
    While they may be usable to an enduser, it will take some effort to customize it for your own situation.

    With nice numbers (related to pythagorean triples), [with practice] it is easy to sketch your own diagrams on rotated graph paper.
    See my insight for details: https://www.physicsforums.com/insights/relativity-rotated-graph-paper/
     
  20. Jul 17, 2017 #19

    Andy_K

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    Thank you for your nice article / insight, it's simply awesome! May I know if you are a physics professor?

    The GeoGebra software looks interesting and useful, I've downloaded it to experiment. :)


     
  21. Jul 17, 2017 #20

    robphy

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    Thanks!
    In my search for an apparently-elusive permanent position, my academic title has changed over the years.
    Right now, I am an instructor of physics.

    Yes, GeoGebra is great!
    I really appreciate that constructions can be done geometrically (e.g. define this point as the intersection of these two lines... or define this line as the tangent line to that curve). Otherwise, I would have to do it algebraically then write it as code or commands.
    In addition, it's free, it runs on various platforms, it can be shared via the web (so end users don't have to actively install anything), and it has an active community.
     
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