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Two body decay and momentum

  1. Sep 11, 2010 #1
    Hi, I am learning about analysing particles decaying and am on this Wikipedia article: http://en.wikipedia.org/wiki/Particle_decay

    Under the 'Two body decay' subsection, there is a description of the momentum of particles 1 and 2 from the rest frame of the parent particle.

    I have two questions, firstly, it states that the momentum of particle 1 is equal to the momentum of particle 2, is this always the case? If they are of different mass I would have thought the momentums might be different as well.

    Secondly, they state the equation for the momentum of particle 1 (or 2) in terms of the masses of all particles involved. I'm assuming they derived this from the equations mentioned in the preceding section about four momentum, but I cannot see how they have got that final result. When I try to dervie the momentum of particle 1 (or 2), using the results from the preceding section, I end up with: p_1 = sqrt[m_parent^2 + m_2^2 - m_1^2 -m_2\ 2m_parent].

    Is anyone able to explain where they got that result from?
    Any help greatly appreciated
  2. jcsd
  3. Sep 11, 2010 #2
    In two-body decays, the momenta of the two decay products are always equal and opposite in the center of mass (CM) system of the decaying particle:
    (pc)2 = E2 - (mc2)2.
    Lorentz transformations to the lab system are performed using the transformation in Section 37.1 of


    Two-particle dacays are specifically discussed in Section 37.4.1

    Bob S
  4. Sep 12, 2010 #3
    Thanks for the link. Equation 37.16 is the one that I was having trouble with, and apologies for my ignorance but I still don't see how they arrived at that formula?

    If p= E-m, then how do they arrive at that equation? I'm guessing the square root has something to do with getting the magnitude of the p vector, but where do those two terms in the square root come from?
  5. Sep 12, 2010 #4
    Look at Equations (14) and (15) in


    Bob S
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