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Calculate relativistic com frame for two particles?

  1. May 13, 2012 #1
    Does anyone know of a standard way of calculating the com frame velocity for two particles moving at arbitary velocities in the lab frame?

    It's strange that this standard result isn't even in Goldstein's et al book
  2. jcsd
  3. May 13, 2012 #2

    Vanadium 50

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    E2 - p2 = m2

    βγ = p/m.
  4. May 13, 2012 #3


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    The velocity of a single particle in terms of its energy and momentum is given by

    $$\beta = \frac {pc}{E}$$

    Given this, what would you expect the velocity of the "equivalent particle" representing the motion of a system of two particles to be? Or indeed, any number of particles?

    You can get the result a bit more rigorously by using the Lorentz transformation for energy and momentum

    $$p^{\prime} c = \gamma (pc - \beta E)$$

    and requiring that the total momentum in the primed frame be zero, i.e. for a system of two particles ##p_1^{\prime} c + p_2^{\prime} c = 0##.
    Last edited: May 13, 2012
  5. May 13, 2012 #4

    Meir Achuz

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    Calulate [tex]{\bf P=p_1+p_2)}[/tex], and [tex]E=E_1+E_2[/tex].
    Then [tex]{\bf V=P}/E.[/tex].
  6. May 13, 2012 #5
    @itbell and Meir Achz, yes I see what you mean!

    The key for me was visualising the disintegration of a particle into p1 and p2, while using conserving 4-momentum.

    This really is a very elegant, beautiful result, which I can't find anywhere in my copy of Goldstein, 3rd edition, nor in any of the problems. Maybe it's mentioned in books devoted to the dynamics of particle collisions.
  7. May 13, 2012 #6
    For c=1, but in general [tex]{\bf V=P}c^2/E.[/tex]

  8. May 13, 2012 #7
    Much of the formalism, including the concept of invariant mass during particle disintegration, is included here:
  9. May 14, 2012 #8

    Meir Achuz

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    I use light years.
  10. May 24, 2012 #9
    So in Fig 39.3 it shows the constraints of the final state.

    What determines in what final state the system of equations will stabilize?
  11. May 24, 2012 #10


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    The final state is located randomly somewhere inside the shaded area, with a probability distribution that is determined by the matrix element for the process that we're dealing with.

    It's rather like asking "what determines exactly when a particular radioactive nucleus will decay?"
  12. May 24, 2012 #11
    So this is where deterministic measurement meets the road to philosophical physics.

    This is where "squeezing" occurs?
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