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Relativity Question (Twin Paradox)

  1. Sep 16, 2006 #1
    Hi all,

    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:

    Problem
    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?
     
    Last edited: Sep 16, 2006
  2. jcsd
  3. Sep 17, 2006 #2
    Did I not include enough info? :confused:
     
  4. Sep 17, 2006 #3
    Please help! I have to turn this in tomorrow. :cry:
     
  5. Sep 17, 2006 #4
    For a moment, forget about what I posted above and let's see if I can make this simpler.

    A rocket is traveling from Earth to Canopus, which is 99 light years away. The rocket reaches canopus in 101 years (so v = 99/101 = .98). There is a clock attached to a rope that is 3.88 light years long (in the earth frame) and this rope is attaced the the back of the rocket (basically, there is a clock travling the same velocity as the rocket but is 3.88 light years behind the rocket). When the rocket reaches Canopus, what does this clock read in the earth frame?
     
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