Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Momentum conservation in collisions.

  1. Dec 1, 2009 #1

    I have a problem that is making me crazy. Consider the following collision

    [tex]A + B \rightarrow C[/tex]

    which results in both particles (A and B) being destroyed and C being created.
    I know the rest mass of all particles. Also, in the lab system, B is stationary and A is moving toward B. What is asked for is the momentum of particle A needed to allow for particle C to be created.

    So I used the following squared four vectors:

    [tex](E_A + m_B, p_A)^2 = (m_C, 0)^2[/tex]


    [tex]E_A = \sqrt{(p_A^2 + m_A^2)}[/tex]

    and solved for [tex]p_A[/tex]. What I got was different from what I got when I used energy conservation with

    [tex]E_A + E_B = E_C \Leftrightarrow \sqrt{(p_A^2 + m_A^2)} + m_B = m_C[/tex]

    Then I realized that I completely neglected the momentum on the right side in my first approach, so instead I did

    [tex](E_A + m_B, p_A)^2 = (m_C, p_A)^2[/tex]

    ([tex]p_A[/tex] on the right side due to momentum conversion)

    and got to the same result as with the second approach. Which is correct?

    To make things even more confusing, I remembered an example our professor gave us:

    [tex]\gamma + p \rightarrow p + \pi_- + \pi_+[/tex]

    and he used the following squared four vectors:

    [tex](E_\gamma + m_p, p_\gamma)^2 = (m_p + 2*m_\pi, 0)^2[/tex]

    Note that he completely neglected the momentum on the right side. If neglecting the momentum was wrong in my first example, why is it correct here? I don't get it, please, someone enlighten me :)

    Thanks in advance
  2. jcsd
  3. Dec 1, 2009 #2


    User Avatar
    Science Advisor

    The threshold energy to produce a final state is the square of the sum of the produced masses, as in the RHS of your profs equation.
    [tex](E_\gamma+m_p)^2-p_\gamma^2[/tex], the invariant he has on the left hand side equals the right hand side. Since the LHS is an invariant, its rest frame value equals its cm value where P=0. You can apply that to any case.
  4. Dec 1, 2009 #3
    Okay, but what about my first example then? Can I use
    (E_A + m_B, p_A)^2 = (m_C, 0)^2
    then with the same reasoning to calculate the needed momentum of particle A, and if so, why don't I get the same result if I use pure energy conservation laws?
  5. Dec 1, 2009 #4

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That equation gives the correct threshold lab energy E_A.

    This equation is wrong for the reason you mentioned. In the lab system, conservation of energy requires [tex]\sqrt{p_A^2+m_c^2}[/tex] for E_C.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook