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Stupider-er Twins Question

  1. Jun 6, 2008 #1
    In the standard "twins paradox", we have a turnaround 8 ly from earth, v = 0.8c. Omitting the math, the ship's twin observes earth's clock to "jump time" from 3.6 yrs to 16.4 yrs (adjusting for the Doppler effect). OK, this makes sense, since the ship's twin (instantaneously) switched frames from one in which he left earth 6 yrs ago to one in which he left earth 27.33 yrs ago, as determined from his (current) frame.

    If the ship's twin can "see" the earth clock "jump time" during the instantaneous turnaround, why wouldn't he "see" his own clock "jump time" during the (instantaneous) turnaround?

    Thanks,
    Al
     
  2. jcsd
  3. Jun 6, 2008 #2
    Hello Al68

    Quote:-

    ----why wouldn't he "see" his own clock "jump time" during the (instantaneous) turnaround?
    ----

    I haven't thought about the first part of your question. But in answer to your last part, nobody ever sees anything different happen to their own clock. It just carries on as normal.

    Matheinste.
     
  4. Jun 6, 2008 #3
    well he never really 'sees' anything jump. due to the shift in simultaneity his 'calculation' of the earth twins time will jump.
     
    Last edited: Jun 6, 2008
  5. Jun 6, 2008 #4
    Yeah, he "calculates" that since (in his return frame) he left earth 27.33 yrs ago, that earth's clock should read 16.4 yrs. (27.33 * 0.6).

    But also, he could watch earth's clock through a telescope and adjust for the Doppler effect. According to the textbook resolution, it would read 16.4 yrs just after the turnaround(in ship's frame).

    Thanks,
    Al
     
  6. Jun 6, 2008 #5
    Would it be any more abnormal to see his own clock "jump" ahead than to see earth's clock jump ahead?
    After all, if it were real (non-instantaneous) acceleration, the ship's twin would calculate that less time passed on earth than for him during the turnaround, as measured in the ship's frame.
     
  7. Jun 6, 2008 #6
    But also, he could watch earth's clock through a telescope and adjust for the Doppler effect.

    that is what i was referring to.
     
  8. Jun 6, 2008 #7
    OK, sure.
    In most of the resolutions I've seen, they use the word "see" the earth's clock "jump" time, or something similar. Of course, the jump isn't actually seen, but the clock (or signals) is actually seen, and, assuming the ship's twin is aware of the Doppler effect, he will know that what he saw was the earth's clock instantaneously advance ahead during the (instantaneous) turnaround.

    Thanks,
    Al
     
  9. Jun 6, 2008 #8
    I know that some consider this nitpicking but I think this is essential in understanding relativity. We cannot directly observe the clock. The only thing we can observe are the light signals from the clock, and light signals cause "tricky" effects when they are exchanged between two objects which move with respect to each other or between two objects in a curved spacetime.

    The only directly observable effect in relativity are the proper accumulated time differentials between objects that follow different paths in spacetime from one event to another.
     
    Last edited: Jun 6, 2008
  10. Jun 6, 2008 #9
    I know that some consider this nitpicking but I think this is essential in understanding relativivty. We cannot directly observe the clock. The only thing we can observe are the light signals from the clock, and light signals cause "tricky" effects when they are exchanged between two objects which move with respect to each other or which have a different gravitational potential.

    The only directly observable effect apart from light signals are the proper accumulated time differentials between objects that follow different paths in spacetime from one event to another.
     
  11. Jun 7, 2008 #10

    Fredrik

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    That's a pretty strange question. You're asking why one clock, at one event in space-time, isn't showing two different times in two different frames. How could it? It's just one event.

    Consider a car that crashes into another car. That's one event. What you're asking is like asking why the cars didn't miss each other in another frame.
     
    Last edited: Jun 7, 2008
  12. Jun 7, 2008 #11
    Yes, I know it's a strange question. Absurd, in fact. But we're assuming that the ship can just switch frames without accelerating. And the actual time since he left earth(in his new frame) has changed according to SR.
    And the ship's twin "sees" the earth clock show two different times corresponding to the two different times that the two different ship frames would judge as the time elapsed since the ship left earth. In the ship's new frame, at the time of the turnaround, the ship's twin left earth 27.33 years ago, and the earth's clock should read 16.4 yrs, accordingly. He can observe earth's clock instantaneously "jump time", which is also strange, even though this fact is somewhat hidden by the Doppler effect.
     
  13. Jun 7, 2008 #12


    why would he 'calculate' that? he has only to look at his clock to see how much time has passed.
     
    Last edited: Jun 7, 2008
  14. Jun 7, 2008 #13
    the simple answer is because the shift in simultaneity that occurs during acceleration depends on how far away the object is. up close nothing much changes but farther and farther away it changes more and more.
     
  15. Jun 8, 2008 #14

    Fredrik

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    I don't think you do. What happens at an event has absolutely nothing to do with coordinates. Space-time is completely coordinate independent.

    It has to accelerate, but there's no upper bound on how fast it can accelerate, so we simplify by assuming that the acceleration takes (almost) no time at all. It's obvious from the space-time diagram that this doesn't change anything relevant.

    Not the "actual time". The thing that changes is the difference between the time coordinates of the event where he left Earth and the event where he turned around. That difference obviously depends on what coordinate system we're using.

    No. The clock on Earth will show different times at the two events on Earth that are simultaneous with the turnaround event in the two frames.

    Not really. The clock must show different times at different events along its world line. If if didn't, it wouldn't be a clock. So the fact that it does can certainly not be as strange as it would be to have a clock show two different times at one event, which is impossible by definition.
     
  16. Jun 8, 2008 #15
    OK, let's say that a buoy passes earth at 0.8c (at rest with ship) at t = 10 yrs earth time, t' = 6 yrs ship time. How far apart are the ship and buoy before the turnaround, as measured in each frame?

    Thanks,
    Al
     
  17. Jun 8, 2008 #16
    The Earth observer will say the bouy and ship are 8 light years apart while the ship observer will say the bouy and ship are 13.333 light years apart. Note that at the same time the ship observer will say Earth and ship are 4.2 light years apart.
     
  18. Jun 8, 2008 #17

    Fredrik

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    If I assume that you meant that "Earth at t=10 years" is an event on the buoy's world line, then I'm getting the same answers as kev: 8 light-years in Earth's frame and 13.333... light-years in the ship's frame.

    Note that "Earth at t=10 years" is not the same as "Earth at t'=6 years". The latter is actually the same as "Earth at t=3.6 years".

    Some details: The event on the buoy's world line that's simultaneous in the ship's frame with the turnaround event is located at (t,x)=(-7.777...,-14.222...). (Solve t=1/0.8x+10=0.8x+3.6 for t and x). A Lorentz transformation tells us that the coordinates in the ship's frame of this event are (t',x')=(6,-13.333...). The turnaround event has coordinates (t',x')=(6,0), so the distance is -13.333... light-years.

    Kev also said that "at the same time the ship observer will say Earth and ship are 4.2 light years apart". I'm getting a different result. At the turnaround event, the coordinates of Earth in the ship's frame are (t',x')=(6,6*(-0.8))=(6,-4.8), so the distance between Earth and the buoy is -4.8-(-13.333...)=8.5333... light-years.
     
  19. Jun 8, 2008 #18
    I think Kev meant 4.8 ly for the distance between the ship and earth in the ship's frame. I have to assume, though, that since the buoy is at rest with the ship, the ship's twin and an observer on the buoy will consider the turnaround event to be simultaneous with the buoy passing earth at t'=6 yrs. And, since the ship and buoy are at rest, their distance should be the same at any time in earth's frame prior to the turnaround (in earth's frame).
    So how far apart are the ship and buoy in earth's frame prior to the turnaround?

    Thanks,
    Al
     
    Last edited by a moderator: Jun 8, 2008
  20. Jun 9, 2008 #19

    Fredrik

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    Wait a minute. Does the buoy pass Earth at t=10 or t'=6? It's impossible to answer without knowing that. If you meant t'=6, kev and I both gave you the answer to the wrong question.

    "Earth" represents a vertical line in the space-time diagram. "t=10" represents a horizontal line. "t'=6" represents a line with slope v (like one of the blue lines in my diagram). The Earth line intersects both the t=10 line and the t'=6 line, but not at the same event.

    If you meant t=10, then we already gave you the answer. It's 8 light-years. This is very obvious. Earth is always at x=0, by definition of the unprimed coordinates, and the ship turns around at t=10, x=0.8*10=8, so the distance is 8-0=8.
     
  21. Jun 10, 2008 #20
    I meant t'=6. The buoy passes earth simultaneously (in the outbound ship/buoy frame) with the ship turnaround.

    Thanks,
    Al
     
  22. Jun 10, 2008 #21

    Fredrik

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    OK, so the buoy passes Earth at t=3.6 years. At this time in Earth's frame, the ship is 3.6*0.8=2.88 light-years from Earth. Since the ship and the buoy move at the same velocity, the distance (in all frames) between them will be constant. So it's still 2.88 light-years in Earth's frame when the rocket turns around at t=10.

    The distance in the ship's frame is 2.88/gamma=4.8 light-years.
     
  23. Jun 11, 2008 #22
    OK, after the turnaround, in the ship's new frame, where is the buoy?

    Thanks,
    Al
     
  24. Jun 12, 2008 #23
    OK, I'll try an answer. The ship turnaround is at t'=6 yrs in ship/buoy frame, t = 3.6 yrs in earth frame. Since the buoy is at earth and can observe earth's clock locally, the turnaround event occurs at 3.6 yrs in earth frame, 2.88 ly away from earth. After the turnaround, buoy is still at earth moving toward ship. Ship arrives at earth at t' = 12 yrs on ship's clock, t = 7.2 yrs on earth's clock. Earth twin is younger when they reunite. Earth twin will observe ship's clock to run slower the whole time, but "jump" ahead during the turnaround, since it was not simultaneous (earth's velocity relative to the ship changed). I won't bother drawing a spacetime diagram, just switch the ship and earth around on the diagrams all over the net.

    Oh, yeah, I didn't mention the space station in my "resolution". Oh, well, the textbook resolution doesn't mention a buoy, so that's OK.

    And I didn't follow the rule that says we have to consider who "felt" acceleration during the turnaround. I guess that's OK, too, since that rule isn't in Einstein's original SR.

    So, did I resolve this version of the "twins paradox", or am I missing something?

    Thanks,
    Al
     
  25. Jun 12, 2008 #24

    Fredrik

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    You seem to have misunderstood how to do this (and maybe also what an event is). The turnaround is an event that has coordinates (10,8) in Earth's frame. The event (3.6,0) is the event where the buoy passes Earth. These two events are simultaneous in the ship's frame before the turnaround, but it's absolutely not true that...
    The event (3.6,2.88) in Earth's frame is the event on the ship where the ship's clock shows 2.16 years.

    There is only one turnaround event and it's at (10,8) in Earth's frame, at (6,0) in the ship's frame before the turnaround, and at (0,4.8) in the buoy's frame (if the buoy's clock is set to 0 when it passes Earth).

    The event on the buoy's world line that's simultaneous in the ship's new frame with the turnaround event is at (11.40,6.24) in Earth's frame. Let's call this event Z. If we take the turnaround event to be the origin of the ship's new coordinates, then event Z has coordinates (0,-1.05) in the ship's new frame.

    It will be a good exercise for you to try to obtain these results, but don't start with the calculations. The first thing you should do is draw a space-time diagram (from Earth's point of view) so you can see that these results are reasonable. You should think really hard about how to draw simultaneity lines.
     
    Last edited: Jun 12, 2008
  26. Jun 13, 2008 #25
    Hi Fredrik,

    Thanks for your response.

    I only presented my "resolution" to show that the math will work my way just by re-defining the event at (6,4.8) in the ship's frame and (3.6,2.88) in earth's frame, instead of "at the space station". This is defining a different event, which was my point. It's like saying the earth switched frames instead of the ship. I only did it to show that the math will work either way, not because I don't understand how to do it the way you did. I understand how, I'm just not sure why to do it that way instead of the way in my "resolution". Also in my "resolution", there is no space station. The distance traveled to the turnaround point is defined as the distance between the ship and buoy.

    In your results, the twin on the ship will never observe the buoy pass the ship, if I read it right. Will the buoy observe the ship pass it?

    Thanks,
    Al
     
    Last edited by a moderator: Jun 13, 2008
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