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I Lines of simultaneity in Twin Paradox spacetime diagram

  1. Jun 30, 2016 #1
    I set up a Twin Paradox scenario and accompanying spacetime diagram to help better understand the resolution, but I had a question about the diagram I was hoping someone here could help answer. Please excuse the hastily drawn diagram!
    314wcyd.jpg

    (Note: the ' frame corresponds to the outbound trip, the '' frame to the return trip. I mistakenly jotted down tD as 6 yr when it should actually be 16yr. My bad!)

    A traveler is making a 6 lightyear voyage to a distant star at a constant speed of 0.6c. His twin remains on Earth. In the traveler's frame, the 6 ly distance is contracted to 4.8 ly (γ = 1.25). I have drawn lines of simultaneity on the diagram, as well as incoming and outgoing information at the point at which the traveler reaches his destination. Once he does, he immediately turns around and goes home at the same speed. I know that if the traveler and his twin are exchanging timestamp information, that the twin on Earth sees the traveler's clock read 8 years upon reaching the star, while his own reads 16 years. The traveler sees his own clock read 8 years upon reaching the star, while his twin's reads only 4 years. I have tA = 4 yr in the diagram and tD = 16 yr (note errata).

    I understand that the magic of the twin paradox scenario happens when the traveler changes his reference frame and heads home. The geometry of the spacetime diagram suggests that there is an immediate shift in the lines of simultaneity between the traveler frame and the twin frame. This feels pretty reminiscent of the Doppler effect. What I don't know is how to calculate tB and tC, that is, the time on Earth the moment before and the moment after acceleration of the traveler back toward Earth, respectively, and how to interpret that result. Am I to understand that the traveler's acceleration corresponds with an IMMEDIATE jump in the simultaneous time on Earth? How big is this jump (tc-tB)? Why does this occur, and why is it independent of any details of the acceleration (except initial and final velocities)?

    Thanks for any insight!
     
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  3. Jun 30, 2016 #2

    robphy

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  4. Jun 30, 2016 #3

    Ibix

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    Lines of simultaneity get complicated when you accelerate. For example, if you take your lines of simultaneity and extend them to the right, you will find that the ones associated with the primed frames cross the ones from the double primed frame. So, the picture that you constructed would imply that clocks jump forwards on Earth and actually jump backwards for points to the right - things happen twice. That doesn't really make sense.

    If you want to construct a coherent set of lines of simultaneity, you need to look up ways to construct coordinate systems for accelerated frames of reference. One way of doing it is Dolby and Gull's radar coordinates: https://arxiv.org/abs/gr-qc/0104077. I'd recommend not worrying about it for now, however.

    The thing you seem to have forgotten is the travel time of light. The twins cannot see each other "now". They can only see each other in the past. If you add a series of light pulses to your diagram, you'll be able to work out what the twins see. I'll just toot my own horn here and point you to my interactive Minkowski diagrams: www.ibises.org.uk/Minkowski.html. If you scroll down to the botton there's a button to draw that for you...
     
  5. Jun 30, 2016 #4

    robphy

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    In your diagram, OB is the hypotenuse of a Minkowski-right triangle.
    That's calculated like a time-dilation problem.

    Let Q be the turn-around event, let P be the midpoint-event of [or better the event simultaneous with Q for] the stay at home twin, and Z be the reunion event [as in my diagram].
    So, ##OP=\gamma OQ##... since OQ is the hypotenuse and OP and PQ are Minkowski-perpendicular.
    Similarly, ##OQ=\gamma OB## since OB is the hypotenuse and OB and BQ are Minkowski-perpendicular.
    (Note OPQ and OQB are Minkowski-similar triangles.)
    For OC, calculate CZ analogous to OB [but using the return trip]... then OC=OZ-CZ.
     
    Last edited: Jun 30, 2016
  6. Jun 30, 2016 #5
    Ibix,

    I actually had drawn another diagram where I drew in light pulses, and those definitely helped me make sense of the effect as it was occurring in both frames. I have two such pulses in the diagram above going to and from the point of turnaround of the traveler. Where they intersect with the Earth frame at A and D, I understand to be A: the pulse that the traveler sees upon reaching his destination and D: the point at which the twin on Earth sees the traveler reach his destination. Interestingly, both observers agree what the traveler's clock says at the destination (8 yr) but they do not agree on what the Earth clock says when the ship reaches the destination (4 yr vs 16yr) highlighting the relativity of simultaneity.

    It sounds like where I got confused was in trying to apply lines of simultaneity in an overly simplistic manner to make sense of that relativity. I see what you mean about the lines crossing to the right of the traveler path. There is definitely a discontinuity there on the graph which confused me, since I saw other resources demonstrate those lines in the same way. I will definitely check out your radar coordinates paper, thank you.
     
  7. Jun 30, 2016 #6
    robphy, thank you for the resource, but I don't think that gets at the questions I had with this Paradox. I do fully understand the resolution and how to apply spacetime contraction and Minkowski geometry in order to make sense of the result, I was just confused as to what happens on Earth from Bob's pespective upon reaching Q and turning around, but I think Ibix's response has shed some light on that for me.

    The "clocks jumping forward" thing on Earth is what got to me as odd, I didn't think to look at the extension for frames to the right of ct' and see that something obviously was wrong with my interpretation. Looks like I didn't apply those lines in a correct way.
     
  8. Jun 30, 2016 #7
    Ibix, I think that radar coordinates paper perfectly addresses my issue, thank you for the find! Highly interesting result from the looks of it.
     
  9. Jun 30, 2016 #8

    pervect

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    Let's add some additional labels to this diagram. Let O is below A, the point where the trip stars. E is above D, the point where the trip ends. And let T be the point where the turnaround occurs, the rightmost point on the diagram

    That's a good thought, hold onto it. Better yet, calculate the value of the dopopler shift.

    The doppler shift factor is ##k = \sqrt{\frac{1+\beta}{1-\beta}}##
    See https://en.wikipedia.org/wiki/Relativistic_Doppler_effect
    Plugging the numbers in, we get k=2.

    The time dilation factor is ##\sqrt{\frac{1}{1-\beta^2}}##.
    As previously mentioned, this is 1.25 when we plug the numbers in.

    6 light years out and back takes a round trip time of twenty years, so OE is 20 years.
    4.8 light years out and back takes 16 years, so OT is 8 years and TE is 8 years.

    Check: OT+TE = 16 years, mutliplied by gamma gives us 20 years. Check.

    Because of the doppler shift factor being 2:1 (as we calculated above) we can say that OT is twice OB. This is really the key point. If the observer "at rest" sends pulses out every second, they'll be received by the moving observer every two seconds. If the observer at rest sends out a pulse every year, it will be receive by the moving observer every two years.

    So if the observer in motion receives a pulse at 8 years on the moving observer's clock, it must have be sent at year four on the stationary observer's clock. Thus OB is 4 years.

    The same arguments (and symmetry) gives us that CE is 4 years.

    As far as acceleration goes. In the diagram above, the velocity changes instantaneously, so the definition of simultaneity changes instantly. Perhaps you're thinking that simultaneity has some significance outside the conventions of a particular observer? It doesn't. Similarly, considering acceleration makes the problem harder and serves as a distraction, it's not important to the explanation of the twin paradox.
     
  10. Jun 30, 2016 #9
    I think my diagram was kind of confusing and poorly drawn. AT is supposed to follow that dotted line and represents the light pulse from Earth that reaches the traveler at their destination. OA thus should be 4. BT is the solid line, and is supposed to be a line of constant x' (simultaneity). CT represents a line of constant x'', both diverge from the same point given the instantaneous acceleration, and I wondered what the physical significance of that was. Ibix pointed out that those lines are bunk, and that I would need a different coordinate system to accommodate hypersurfaces from an accelerated frame.

    In hindsight, I may very well have been caught in that classic line of thinking. In my head, I was wondering, "If the traveler has a telescope that can view Earth and his twin's clock, what does he see as he accelerates from his destination back the other direction?" and the answer is of course, as you have pointed out, he sees Earth's clock speed up fourfold, likewise, the twin on Earth looking backward at the traveler's clock would see the same speed-up.
     
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