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Momentum conservation in collisions.

  1. Dec 1, 2009 #1
    Hello.

    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]

    with

    [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

    clem

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

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    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.
     
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