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Relativistic Centre of Momentum Frame

  1. Jan 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A partical of mass m ha speed 0.546c relative to inertial frame S. The particle collides with an identical particle at rest in the inertial frame S. Relative to S and in terms of c, what is the speed of S' in which the total momentum of these particles is 0.
    2. Relevant equations


    3. The attempt at a solution
    I can see that in the S' frame, Pa = - Pb so that means that Va = - Vb since both masses are equal. But I just can't get my head around why i need to use special relativity because everything is in reference to the S frame anyway.
     
  2. jcsd
  3. Jan 4, 2016 #2

    andrewkirk

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    That's correct, but it doesn't answer the question. They want you to tell them the value of Va. THe naïve non-relativistic answer is that it's half of 0.546c, ie 0.273c but that answer is wrong because it doesn't allow for the fact that, if you choose a frame that has velocity 0.273c relative to the first one, the velocities of the two particles will NOT be 0.273c and -0.273c respectively, because of the relativistic corrections.
     
  4. Jan 4, 2016 #3

    mfb

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    Pa=-Pb and Va=-Vb are not in S.
     
  5. Jan 4, 2016 #4
    S' frame is moving at a velocity v w.r.t S frame.
    Let velocity of Particle 1 = U1 in S frame and particle 2 = U2 in S frame therefore Particle 1 = U1' and particle 2 = U2' in S' Frame
    Since particle 2 is at rest in S frame, its velocity in S' frame is equal to the velocity of the S' frame relative to S i.e. v=u2'
    So in a centre of momentum frame u2'= - u1' = v

    Using Einstein velocity addition formulas and setting c to 1

    u2' = ( u2 - v ) / (1-v/c) = v

    When solved gives v = 0.297c

    Does this seem right?
     
  6. Jan 4, 2016 #5

    mfb

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    Right.
     
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