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Lorentz/velocity transformation

  1. Jan 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Two spaceships approach the Earth from opposite directions. According to an observer on the Earth, ship A is moving at a speed of .753c and ship B at a speed of .851c. What is the speed of ship A as observed from ship B? Of ship B as observed from ship A?


    2. Relevant equations

    v'=(v-u)/(1-u*v/c^2)

    3. The attempt at a solution

    For the speed of ship A with respect to B I tried
    (.851c-.753c)/(1-.851c*.753c/c^2) = .2728c
    For the speed of ship B with respect to A i tried
    (.851c+.753c)/(1+.851c*.753c/c^2) = .9776c
    I am not sure of the correct answers, However I'm pretty sure these are wrong. Thanks.
     
  2. jcsd
  3. Jan 22, 2007 #2
    Let [tex] S [/tex] be a coordinate system fixed on the observer on earth, oriented such that the positive x-axis is in the direction of the motion of ship A. Then in that frame ship A has velocity [tex] 0.753c [/tex] and ship B has velocity [tex] -0.851c [/tex].

    Now, we set up a fram [tex] S' [/tex] in which ship A is stationary and which has x-axis oriented in the same way as fram [tex] S [/tex]. Then in your formula [tex] v' [/tex] is the velocity of ship B as seen from ship A, [tex] u [/tex] is the velocity of ship A w.r.t. [tex] S [/tex] (earth) and [tex] v [/tex] is the velocity of ship B w.r.t. earth.

    Then the velocity of ship B as seen from ship A is

    [tex] v' = \frac{-0.851-0.753}{1+0.753\times 0.851} c [/tex]

    If you instead consider an observer in ship B then you will get ([tex] v' [/tex] is the velocity of ship A as seen from ship B, [tex] v [/tex] is the velocity of ship A as seen from earth and [tex] u [/tex] is the velocity of ship B as seen from earth

    [tex] v' = \frac{0.753+0.851}{1+0.753\times 0.851} c [/tex]

    and both these are equal to (except for signs) [tex] 0.9776 c [/tex] .
     
    Last edited: Jan 22, 2007
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