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Addition of masses in relativistic collision

  1. Nov 8, 2016 #1
    Really basic question:

    a particle, moving at speed u (u is fast enough for relativistic effects) with rest mass m0 collides with a stationary particle with rest mass m0. They coalesce to form a new particle of mass M (observer fame, not rest mass M) and move at speed v. find v in terms of y (y is gamma factor).

    Conservation of momentum:

    m0y(u)u=M0y(v)v

    Is M(the rest mass of the 2 particles): 2m0 (just adding the rest masses and multiplying by the new gamma factor)

    or

    m0y(u) +m0 (adding up the masses as they were before the collision)

    Thank you
     
  2. jcsd
  3. Nov 8, 2016 #2

    PeroK

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    Where are you learning SR? You don't see to have much idea about the basic concepts.
     
  4. Nov 8, 2016 #3

    Ray Vickson

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    These problems are always easier and less confusing if you analyze them in the center-of-momentum (CM) frame.

    In a frame whose origin moves with velocity ##(v,0)## you can work out the initial (pre-collision) momenta of the two particles, and equate the total momentum to 0; that tells you the value of ##v## that gives you the CM frame. So, in the CM frame the total momentum before and after the collision is zero. Conservation of energy says that the post-collision mass ##M_0## of the coalesced particle is given by ##M_0 c^2 = m_0 c^2 \gamma(u_1) + m_0 c^2 \gamma(-v)##, where ##u_1## and ##-v## are the pre-collision velocities of particles 1 and 2 in the CM frame. That is, ##M_0 = m_0 \gamma(u_1) + m_0 \gamma(-v).##

    After you have determined ##M_0## you can transform back to the lab frame to find the final momentum of the new particle.
     
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