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Twin Paradox (Again)

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Anne and Joe are twins, happily living in an inertial frame. On their 20th birthday Joe decides
    to take a rocket.

    (a) According to Anne the rocket moves with constant speed [tex]v = \frac{3c}{5}[/tex]. For 6 months it moves away from earth and then returns in time for Anne's 21st birthday. How old is Joe on Anne's 21st birthday?

    (b) According to Anne the rocket is initially at rest and then accelerates such that its velocity is always proportional to [tex]\sqrt{t}[/tex], where t is the time elapsed since the start of the trip. After two years the rocket reaches a velocity of 3c/5. How old is Joe on Anne's 21st birthday?

    (c) Joe is on his trip, according to (b). How old is Anne on Joe's 21st birthday?

    2. Relevant equations

    [tex]d\tau = \int^{t_1}_{t_0} \frac{dt}{\gamma\left(v\left(t\right)\right)}[/tex]

    3. The attempt at a solution

    For part a, I have,

    [tex]d\tau = \frac{dt}{\gamma}[/tex]

    [tex]\gamma = 5/4[/tex]

    [tex]dt = 1\text{yr}[/tex]

    [tex]d\tau = 4/5 \text{yrs}[/tex]

    So Anne has aged one year, and Joe has aged 4/5 = 0.8 years. So Joe is about 20 yrs and 292 days old.

    For part b I'm having trouble getting my head around how to incorporate the time dependence of the velocity into the integral (which is what we have been led into doing). I'm not sure whether to use the proper acceleration or to try to evaluate the integral as it is.

    Any pointers would be greatly appreciated.
     
    Last edited: Feb 14, 2010
  2. jcsd
  3. Feb 14, 2010 #2

    vela

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    Just integrate the integral as it is. You have that [itex]v(t)=k\sqrt{t}[/itex] where k is a constant, so you can find [itex]\gamma[/itex] as a function of time t.
     
  4. Feb 14, 2010 #3
    I've given that a go and I've ended up with an expression in terms of the speed of light and some constant. My best guess from here is to use the information in the question to try and determine what this constant is in terms of how it relates to the acceleration of the rocket?

    EDIT:

    Ok, for part b,

    [tex] v = k\sqrt{2}[/tex]
    [tex] d\tau = \int^{t_1}_0 \sqrt{1 - \frac{k^2 t}{c^2}}\, dt[/tex]
    [tex] d\tau = \frac{2}{3k^2}\left( c^2 + \left(k^2 - c^2\right) \sqrt{1 - \frac{k^2}{c^2}}\right)[/tex]

    So, now to determine what k is. I had a bit of a think about the relationship between acceleration and velocity, and used the usual relationships from calculus to do this, (?!)

    [tex] \int^t_0 a \, dt = v +c[/tex]

    Using a bit of reverse math,

    [tex] a = \frac{3c}{10\sqrt{2t}}[/tex]

    Since then,

    [tex] \int^2_0 \frac{3c}{10\sqrt{2t}} \, dt = \frac{3c}{5}[/tex]

    Which is the velocity as defined in the problem statement. From here we integrate a to determine the velocity, complete with constant of proportionality,

    [tex] \int \frac{3c}{10\sqrt{2t}} \, dt = \frac{3c\sqrt{t}}{5\sqrt{2}}[/tex]

    [tex] k = \frac{3c}{5\sqrt{2}}[/tex]

    Now sub back into the original expression above to find,

    [tex] d\tau = 0.953 [/tex]

    Quite different to the answer to part a, but what we would expect, no? Lower average velocity, so therefore the difference in time as perceived by the twins is less than before?
     
    Last edited: Feb 14, 2010
  5. Feb 14, 2010 #4

    vela

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    You also made solving for k way harder than you had to. You have an expression for v(t), so set v(2)=3/5 c and solve for k. You did get the right answer for k, though.

    You seem to have made a mistake in the first integral. For one thing, t1 doesn't appear in your result. I think if you figure that out, you'll get the right answer.
     
  6. Feb 14, 2010 #5
    Sorry, I've taken t1 to be the elapsed time interval that we are interested in, and that happens to be 1 year, so t1 is just 1.
     
  7. Feb 14, 2010 #6

    vela

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    Ah, okay. I see you also got a new numerical answer, which agrees with what I found. Your reasoning is correct. Joe is moving more slowly, so he ages more during the trip than when moving at a constant speed of 3/5 c.
     
  8. Feb 14, 2010 #7
    Ahh great, nice to know that I'm getting the hang of this course - just about! Thanks very much for your help!
     
  9. Feb 15, 2010 #8
    In regards to part c, can you just consider in joes frame that anne is moving and thus anne will have aged more slowly, giving t=0.953 as in part b?
     
  10. Feb 15, 2010 #9

    vela

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    I don't think so because the rocket's accelerating messes things up. I'll have to think about this a bit.
     
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