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Relativity Question (Twin Paradox)

  1. Sep 17, 2006 #1
    Hi all,

    I posted this question in the homework section yesterday but has gone unanswered so far.

    Most of you are probably familiar with the twin paradox of relativity, which is the basis of this problem. I think I understand it fairly well but am having trouble with one specific detail. Here is the setup:

    We have two planets, the earth and canopus, that are separated by a distance of 99 light years in the earth frame. A rocket (R1) traveling from the earth to canopus travels at such a speed that it will arrive at canopus in 101 years. Upon arriving at canopus, the occupant in the rocket will instantaneously jump from the first rocket to a second rocket (R2) traveling the same speed but now pointed towards the earth.

    For the earth, there is a string a clocks that infinitely stretch out towards canopus (the positive direction) and away from the direction of canopus (the negative direction). These clocks are synchronized to the clock on earth. A similar set of clocks are attached to the rockets and are synchronized to the clock on the rocket.

    There are six events that occur in the problem.

    Event 1: R1 leaves earth
    Event 2: R1 reaches canopus
    Event 3: Rocket station S1 passes earth
    (S1 is the station that reaches earth at the same time that R1 reaches canopus in the R1 frame)
    Event 4: R2 leaves canopus
    Event 5: Rocket station S2 passes earth
    (S2 is the station that reaches earth at the same time that R2 leaves canopus in the R2 frame)

    There are six clocks

    Clock 1: on earth
    Clock 2: on canopus
    Clock 3: on R1
    Clock 4: on S1
    Clock 5: on R2
    Clock 6: on S2

    I am told to find the times of all six clocks at each of the six events in the earth frame.

    Attempted Solution
    Well, I found the times of all the clocks of, except clock 4 at event 1 and clock 6 at event 6. If I can find out how to obtain the time from clock 4, I can probably find the time for clock 6 as well.

    I started by using the equation [tex]t^2_E - x^2_E = t^2_R - x^2_R[/tex] where "E" stands for Earth and "R" stands for Rocket. I am attempting to solve for [tex]t_E[/tex]. [tex]t_R[/tex] is equal to 0 since this is the reference event for the problem.

    But before I could substitute numbers to this equation, I had to find the value of [tex]x_R[/tex] at event 2 and [tex]x_E[/tex] at event 3.

    For [tex]x_R[/tex], I used the value found for [tex]t_R[/tex] at event 2.

    [tex]t^2_R = t^2_E - x^2_E [/tex]

    [tex]t^2_R = 101^2 - 99^2 [/tex]

    [tex]t^2_R = 400[/tex]

    [tex] t_R = 20[/tex]

    Multiply this distance times the rocket speed (which is 99/101 = .98) to give us 19.6 years in distance between the rocket at canopus and the earth. It is also the distance between the S1 clock in event 3 as it passes

    the earth. Back to our main equation, [tex]t^2_E - x^2_E = t^2_R - x^2_R[/tex], we now have the value of [tex]x_R[/tex] (19.6)

    For [tex]x_E[/tex], I used the value found for [tex]t_E[/tex] at event 3.

    [tex]t^2_E = t^2_R - x^2_R [/tex]

    [tex]t^2_E = 20^2 - 19.6^2 [/tex]

    [tex]t^2_E = 15.84[/tex]

    [tex] t_R = 3.98[/tex]

    Finally substituting all of our values into [tex]t^2_E - x^2_E = t^2_R - x^2_R[/tex], I have [tex]t^2_E - 3.98^2 = 0 - 19.6^2[/tex]. However, I end up getting the squareroot of a negative number. Can someone please show me where things went wrong?
  2. jcsd
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