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Energy of neutral pion in proton's rest mass frame

  1. Jan 21, 2015 #1
    Hello, my problem is as follows
    I've tried finding the invariant mass of the positron and pion as follows


    [itex]M^2=(E_e+E_{\pi})^2-(\mathbf{p_e}+\mathbf{p_{\pi}})\\[/itex]
    [itex]=E_e^2+E_{\pi}^2+2E_eE_{\pi}-p_e^2-p_{\pi}^2-2\mathbf{p_ep_{\pi}}\\[/itex]
    [itex]=m_e^2+m_{\pi}^2-2(E_eE_{\pi}+\mathbf{p_ep_{\pi}})\\[/itex]
    [itex]=m_e^2+m_{\pi}^2-2(E_eE_{\pi}+p_ep_{\pi}\textrm{cos}\theta)\\[/itex]
    And this is presumably equal to the mass of the proton so

    [itex]m_e^2+m_{\pi}^2-2(E_eE_{\pi}+p_ep_{\pi}\textrm{cos}\theta)=m_p^2[/itex]

    At this point I cannot see how to get any closer to the given answer. Any help with how to proceed from here, or advice on where I may have already gone wrong, would be greatly appreciated.

    Thanks
     
  2. jcsd
  3. Jan 21, 2015 #2

    DEvens

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    Gold Member

    What is the value of theta in the lab frame? What is the total momentum in the lab frame? And also, what is E_e compared to E_pi, given the answer to the question about momentum?

    Whenever you do interactions of this type (collisions, decays, scattering, etc.) the center of momentum frame (COM) is always something to keep in mind for simplifying calculations.
     
  4. Jan 21, 2015 #3
    Hi. As pointed out by DEvens, in this kind of problem you only need energy and momentum conservation laws;
    4-vector invariant is useful when going from one frame to another but here you can solve the problem by staying in the lab frame the whole time since it's also the COM frame!
     
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