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Relative Velocity Addition

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data

    A proton and an antiproton are moving toward each other in a head-on collision. If each has a speed of 0.8c with respect to the collision point, how fast are they moving with respect to each other?


    2. Relevant equations

    Ux = [U'x + V] / [1 + (v/c^2) Ux']

    3. The attempt at a solution

    I believe I have solved it correctly, but I am unsure of one assumption which I made, and want to make sure that it is alright. Instead of looking at it as the collision point being stationary, and the proton moving towards it at 0.8c, I assume that the proton is stationary, and the collision point moves towards it at 0.8c. Then the anti-proton moves towards the collision point at 0.8c as measured in the frame of the collision point.
    This gives me Ux = the speed of the antiproton in the proton frame
    Ux' = the speed of the anti-proton in the collision point frame (0.8c)
    V = Speed of collision point in the proton frame (0.8c)

    Thus, Ux = [0.8c + 0.8c] / [1 + (0.8c/c^2) 0.8c] = 1.6c / 1.64 = 0.976c

    Does this look correct?

    Thanks
     
  2. jcsd
  3. Sep 20, 2009 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Rather than thinking of the collision point, think of the centre of mass frame of reference (the frame in which the velocities are initially measured). You wish to determine the velocity of the anti-proton in the proton's frame of reference. To do this, you apply the Einstein velocity addition to translate the velocity of the anti-proton in the centre of mass frame to the proton frame, which you have done correctly.

    AM
     
  4. Sep 20, 2009 #3
    Thanks for the confirmation
     
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