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Proper Time : Constant Velocity Clock vs Constant Acceleration Clock

  1. Apr 11, 2012 #1

    morrobay

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    1. The problem statement, all variables and given/known data
    3 clocks at to
    A clock on Earth
    B clock above earth 4.66 * 1014 meters
    C clock above earth 3.30 * 1013 meters
    B accelerates to .6 c and arrives at earth when earth clock reads 2.6 *106 sec
    = 30 days With velocity 1.8 *10^8 m/sec with gamma = .8 B clock reads 2.07 * 10^6 sec = 24 days
    C clock travels to earth with constant acceleration in respect to earth frame of 9.81 m/sec2
    and arrives at the same time as B . C clock velocity 2.5 * 10^7 m/sec
    What is the proper time on Clock C ?
    2. Relevant equations
    Integral to to 2.6 * 106 sec. sqrt [ 1-v(t)2/c2] dt
    So Int to to t1 sqrt [ 1-1.01*10-15t2] dt



    3. The attempt at a solution I put Sqrt{1-ax2] in The Integrator
    and plugged in values and got 15 days proper time on C clock ?
     
    Last edited: Apr 12, 2012
  2. jcsd
  3. Apr 12, 2012 #2

    morrobay

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    The evaluation of the above integral =
    1/2x sqrt [ 1-ax2] + sin-1 ( sqrt a) x/2 sqrt a

    Note: Im posting this problem because there are endless discussions on proper time
    but not many numerical answers
     
    Last edited: Apr 12, 2012
  4. Apr 13, 2012 #3

    morrobay

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    Last edited: Apr 14, 2012
  5. Apr 14, 2012 #4

    Dale

    Staff: Mentor

    All your work is good. But it seems like you are getting bad values out of your numerical integration routine. Perhaps it is a numerical precision problem.

    I plug the same integral in to Mathematica and get 2.588E6 s = 29.95 days.
     
  6. Apr 14, 2012 #5

    morrobay

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    Would you expect that the acellerating clock C in the original problem would have
    essentially the same proper time as clock A ( 30 days proper time ) ?
    Also clock B had 24 days proper time with .6c

    note correction : a = 1.07 * 10 ^-15 from v(t^2)/c^2 = (9.81)^2 m/s^2 / 9*10^16
    but does not change values too much.
    Yes Im having a few numerical precision problems.
     
    Last edited: Apr 14, 2012
  7. Apr 14, 2012 #6

    Dale

    Staff: Mentor

    Yes. 3.3E13 m / 2.6E6 s is only an average speed of .04 c which corresponds to an average time dilation factor less than 1.001, so I would expect the clock to not be significantly time dilated overall.
     
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