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Twins Paradox clarification

  1. Apr 16, 2014 #1
    Hi, from what I understand about the twin paradox, Is the resolution essentially that since the traveling twin undergoes acceleration when the ship reverses direction, so since the velocity of the twin is no longer constant, the inertial reference frames are no longer valid? Then does that also means that the twins are aging at the same rate only until the brother undergoes acceleration?

    And I was also thinking about spacetime. Since time is a dimension, one brother is lying in a relatively constant distance through time, but the other brother is traveling much farther distance, so time must have slowed for him since he is traveling through two dimensions (space and time) ?
     
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  3. Apr 16, 2014 #2

    PeterDonis

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  4. Apr 16, 2014 #3
    Yes, the acceleration resolves the paradox.

    But be careful! One cannot say that the twins are "aging at the same rate", since "same rate" implies they can share a common definition of time. Since they're moving relative to one another, they can't.

    You can pick a frame, and compare the rates, but the comparison will be frame-dependent (and hence not very meaningful). In either twin's rest frame (indeed, in most other frames too), they are aging at different rates.

    TBH, I don't quite understand what you're saying here.
    Sorry! :-)
     
  5. Apr 16, 2014 #4
    It looks like he is getting at the idea that the space time interval must be constant for all observers. Since one has an increase in changing distance a decrease in time is needed for the space time interval to remain constant for all.

    OP, if you havent heard of this jargon, check it out.
     
  6. Apr 16, 2014 #5

    russ_watters

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    The brother undergoes an acceleration at the very start of the scenario: that's how they come to be traveling at different speeds!
     
  7. Apr 17, 2014 #6
    Thanks, that's what I mean and I need to learn more about spacetime.

    So if that's true, why bother explaining that the traveling brother undergoes acceleration and that there are actually 3 reference frames because his velocity isn't constant? The spacetime continuum explanation seems much more simpler.
     
  8. Apr 17, 2014 #7

    PeterDonis

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    Many people (including me) agree with you. But there are still plenty of people who want an explanation in terms of reference frames and acceleration, which is why the Usenet Physics FAQ article I linked to gives multiple ways of understanding what's going on. (It does say that the spacetime explanation is the most general, since by drawing in appropriate curves on the spacetime diagram you can illustrate how all the other explanations work.)
     
  9. Apr 17, 2014 #8

    ghwellsjr

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    I'm confused. You say that you prefer the spacetime explanation over "an explanation in terms of reference frames and acceleration" but the link you provided in post #2 shows just that. What's the difference?
     
  10. Apr 17, 2014 #9

    Dale

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    I agree. Also, the spacetime explanation works in General Relativity, where most of the other explanations fail.
     
  11. Apr 17, 2014 #10

    PeterDonis

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    The link I provided in post #2 shows both kinds of explanation, the spacetime explanation (they call it the "Spacetime Diagram Analysis") and the explanation in terms of reference frames and acceleration (that's in the Introduction). They also say, as I noted, that the spacetime diagram analysis is the most general; as the FAQ says (on the "Too Many Analyses" page):

    This is the point of view I was agreeing with.
     
  12. Apr 17, 2014 #11
    Let me ask for some clarification about what it means to be in the same inertial frame of reference -- Does this mean that two observers are experiencing the same magnitude of acceleration or does it mean accel + velocity + direction?
     
  13. Apr 18, 2014 #12

    ghwellsjr

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    Saying two observers are "in the same inertial frame of reference" is very sloppy terminology because everything and everybody is in every inertial frame of reference and there is no preferred frame. But the terminology is quite common and usually means that two observers are at rest according to a single inertial frame of reference, which, of course means, no acceleration, no velocity, and whatever distance in whatever direction they are from each other, it never changes. Next time someone uses that terminology, why don't you ask them what they mean?
     
  14. Apr 18, 2014 #13

    ghwellsjr

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    I think when you say "reference frames and acceleration", you mean one reference frame before acceleration and another frame after acceleration, also known as "frame jumping", correct?

    But as long as one frame is use throughout, then that is the same as the "Spacetime Diagram Analysis", correct?

    Unfortunately though, the Spacetime Diagram Analysis is limited to in-line scenarios which means that it is no where near universal, wouldn't you say?
     
  15. Apr 18, 2014 #14

    PeterDonis

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    I'm trying to capture the options as described by the OP, for example in post #6.

    It depends on how you're using the frame. The key point of the Spacetime Diagram Analysis is that different curves between the same two points in spacetime can have different lengths (proper times), simply as a matter of geometry. Picking a single convenient frame in which to do the analysis makes it easy to compute the proper time along each worldline, and to draw the worldlines in a single spacetime diagram so that the relationships between them are visually evident. But that's not the only thing people use frames for.


    Why do you think that? You can draw a spacetime diagram of any scenario you like; there's no requirement that the time axis of the diagram has to correspond to any of the worldlines of interest.
     
  16. Apr 18, 2014 #15

    Dale

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    For me, the key part of the spacetime diagram analysis posted in the FAQ is the use of the spacetime interval. That is the metric in an inertial frame, and that is the key part that makes it generalizable to general relativity. The OP seems to be (correctly) focusing on the spacetime interval part of the explanation, which is commendable and is the point that I was reacting to.
     
  17. Apr 18, 2014 #16

    morrobay

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  18. Apr 19, 2014 #17

    ghwellsjr

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    Here's Dr Greg's graph:

    attachment.php?attachmentid=14191&d=1212060478.png

    I can see the accelerations (in red) and the velocities (in blue) but where do you see any clarification on time dilation?
     
    Last edited: Apr 19, 2014
  19. Apr 19, 2014 #18

    morrobay

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    With explanation. As Yuiop says in Dr Greg's graph: With identical acceleration events the difference in time dilation is accounted for only by time in relative higher velocities. And in also answering the OP question :
    Then does that also mean the twins are aging at the same rate only until the brother undergoes
    acceleration ?
     
  20. Apr 19, 2014 #19

    ghwellsjr

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    Here's that graph again:

    attachment.php?attachmentid=14191&d=1212060478.png

    Yuiop misspoke. When he said "time dilation", he meant and should have said "aging". The point of the graph and yuiop's explanation is that since A and B both experienced identical acceleration events and therefore attained identical velocities (and identical time dilations or aging rates), the difference in their aging is accounted for only because A spends more time at the higher velocity than B does.

    Neither the diagram nor its explanation addresses that question by the OP. Chogg answered that question in post #3 by saying that the aging rate (or the time dilation) is frame-dependent. So even before the turn-around acceleration, they are not aging at the same rate in the diagram above.
     
  21. Apr 19, 2014 #20

    morrobay

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    Time dilation and aging are related by Δt' = γΔt. For .6c, γ=1.25 So "time between ticks" of
    travelers clock is longer. Hence less aging.

    Regarding your second paragraph : I stated exactly the same thing : Even before turnaround point
    they are not aging at same rate. The overall picture is that the traveling twins aging is dependent
    on velocity before turnaround point and acceleration / frame change at turnaround point.
    See this numerical example I did in post #3 www.physicsforums.com/showthread.php?t=734695
     
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