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Twin paradox

  1. Jul 23, 2017 #1
    1. The problem statement, all variables and given/known data
    In the twins paradox, suppose that Florence begins at rest beside Methuselah, then accelerates in Methuselah’s x-direction with an acceleration a equal to one Earth gravity, “1g”, for a time ##T_F/4## as measured by her, then accelerates in the −x-direction at 1g for a time ##T_F/2## thereby reversing her motion, and then accelerates in the +x-direction at 1g for a time ##T_F/4## thereby returning to rest beside Methuselah. Show that the total time lapse as measured by Methuselah is:
    $$T_M=\frac{4}{g}\sinh{(\frac{gT_F}{4})}$$

    2. Relevant equations
    (1) ##T_F=\int d\tau = \int\sqrt{dt^2-\delta_{ij}dx^idx^j}=\int_0^{T_M}\sqrt{1-v^2}dt## where ##d\tau## is proper time of Florence

    3. The attempt at a solution
    I tried to integrate (1) with ##v=gt## but I have got some crazy result with ArcSin and square root. It is impossible to extract ##T_M##.
    Is my procedure generally wrong? Or it exists some better way how get result for this case?
     
  2. jcsd
  3. Jul 23, 2017 #2

    TSny

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    This is not the correct equation for the velocity ##v## of Florence relative to Methuselah. According to Methuselah, Florence does not have a constant acceleration during the first quarter of the trip. (Note that your equation would make ##v## become greater than the speed of light for large enough ##t##.)

    Florence "feels" a constant acceleration of g, but g is not the acceleration of Florence relative to Methuselah's frame. g is the acceleration of Florence relative to an inertial reference frame that is instantaneously co-moving with Florence.

    Have you studied how accelerations transform between inertial frames?
     
  4. Jul 24, 2017 #3
    Ou I forgot it. Transformation of acceleration is ##a=\frac{a'}{\gamma^3(1+v'\beta)^3}##. And velocity of Florence in her own frame is ##v'=0##. Right?
    And ##v=gt##, may I use only becouse ##gt<<c## is it true?
    But still there is ##\beta## in transformation of ##a##, but now ##\beta## is not constant. May I write ##\beta=gt##?
     
    Last edited: Jul 24, 2017
  5. Jul 24, 2017 #4

    TSny

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    Yes, that's right.
    No, there is nothing in the problem statement that would allow you to make that approximation.
    With ##v'=0##, you have

    ##a=\frac{a'}{\gamma^3}##.

    Write this out explicitly in terms of ##g##, ##v##, and ##dv/dt##.
     
  6. Jul 24, 2017 #5
    Hmm... I thought that for accelerated motion ##u=\frac{at}{\sqrt{1-a^2t^2/c^2}}##.
     
  7. Jul 24, 2017 #6

    TSny

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    Where did you get this formula? I believe there is a sign error in the denominator.
    What does ##a## represent here, the acceleration of Florence according to Florence, or the acceleration of Florence according to Methuselah? What does ##u## represent?
     
  8. Jul 24, 2017 #7
    ##a=\frac{dv}{dt}=\frac{a'}{\gamma^3}=(1-\beta^2)^{3/2}\frac{dv'}{dt'}=(1-g^2t^2)^{3/2}g## Right?
     
  9. Jul 24, 2017 #8
    Yes...there should be + sign. Sorry.

    How would it looks like? I can't imagine this situation. How can be Florence accelerated with respect to herself? In her frame she will be in rest, no?

    So I mean ##a## and ##u## are acceleration and velocity of Florence measurred by Methuselah.
     
  10. Jul 24, 2017 #9

    TSny

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    You cannot write ##\beta## as ##gt## because Florence does not have a constant acceleration ##g## relative to Methuselah.

    Write ##\beta## as ##v## (since you are apparently letting ##c = 1##). You then have

    ##\frac{dv}{dt}=(1-v^2)^{3/2}g##. This is a differential equation whose solution will give you ##v(t)##.

    It is helpful to rearrange it as ##\frac{1}{(1-v^2)^{3/2}}\frac{dv}{dt} = g##. The trick is to note that

    ##\frac{1}{(1-v^2)^{3/2}}\frac{dv}{dt} = \frac{d}{dt} \left( \frac{v}{(1-v^2)^{1/2}} \right)##
     
  11. Jul 24, 2017 #10

    TSny

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    So, that would change your equation to

    ##u=\frac{at}{\sqrt{1+ a^2t^2/c^2}}##

    So, ##u## is the same as ##v## that you used in your first post. OK.
    But the equation ##u=\frac{at}{\sqrt{1+ a^2t^2/c^2}}## is still not correct if you are saying that ##a## is the acceleration of Florence relative to Methuselah. The correct formula is ##u=\frac{a't}{\sqrt{1+ a'^2t^2/c^2}}## where ##a'## is the acceleration that Florence feels.

    Good point. By "the acceleration of Florence according to Florence" I meant "the acceleration of Florence relative to an instantaneously co-moving inertial frame". That is, it's the acceleration ##g## that Florence feels.
     
  12. Jul 24, 2017 #11
    Oooook :-) Now I understand, I have to be careful from what perspective are all variables taken.
    Thank you for a hint...it is helpful. So than I have ##v=\frac{gt}{\sqrt{1+g^2t^2}}##...nice.
    Now I perfectly understand also the previous point, thanks this particular example. And also YOU ;-)
    Now I put it in formula from #1 post and integrate...
     
  13. Jul 24, 2017 #12

    TSny

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    Yes. Good.
     
  14. Jul 24, 2017 #13
    Unfortunately I have another problem, with limits of integral in #1. How to interpret that it stops after some part of time and go back.
    My result of integration is$$T_F=\frac{1}{g}(Arcsinh(gT2_M)-Arcsinh(gT1_M))$$

    I suppose it doesn't matter of change of direction of velocity and also acceleration (it is seen from first formula where is ##v^2##. And from result, if ##g## changes sign doesn't matter, because of sinh is an odd function.) So a time lapse in first part and third part is same and limits of the integral are from ##0## to ##TM1## resp. ##TM3## and it is ##T_F/4##, but in second part it's different it travels half of Florence time ##T_F/2##. Where ##T_M=TM1+TM2+TM3##.
    What am I missing?
     
  15. Jul 24, 2017 #14

    TSny

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    Break it up into four intervals ##T_M=TM1+TM2+TM3 + TM4## where each interval lasts ##T_F/4## of proper time. Argue that each quarter contributes the same.
     
  16. Jul 24, 2017 #15
    I understand, I thought about that, but I wasn't sure, wheather TM3 is same. Every of TMx starts with ##v=0## but TM3 starts with ##v\neq 0##. Actually I would say it isn't same, in #1 there is ##\int\sqrt{1-v^2}##.
     
  17. Jul 24, 2017 #16

    TSny

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    ##TM1## and ##TM3## start with ##v = 0## and end with ##v = v_{max}##.

    ##TM2## and ##TM4## start with ##v = v_{max}## and end with ##v = 0##.
     
  18. Jul 24, 2017 #17
    I'm sorry I can't read :-) I've read it again and now I understand what the motion is.
    Now it is clear, THANK YOU very much.
     
  19. Jul 24, 2017 #18

    TSny

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    OK. Good.
     
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