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3 body final state kinematics

  1. Mar 6, 2009 #1
    Hello,

    It may not be the best place to ask but that is the best place I found :)

    So I have been trying to calculate kinematic variables for the following process (incoming electron with energy ~360MeV and neutron at rest) :

    [tex]e^- + n \rightarrow e^- + \Delta^0 \rightarrow e^- + \pi^- + p [/tex]

    So I tried to find the final pion energy, [tex]E_\pi[/tex] as a function of it's solid angle ([tex]\theta_\pi[/tex],[tex]\phi_\pi[/tex]),the electron initial and final 4-momenta as well as [tex]Q^2[/tex] and the various masses...

    The problem is that I found 2 solutions for [tex]E_\pi[/tex] as I end up with a second order polynomial and both solutions seems physical to me... is it normal or I have been doing something wrong? I can post some detailed calculations if needed.
     
  2. jcsd
  3. Mar 6, 2009 #2

    clem

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    Science Advisor

    A three body final state is described by a Dalitz plot. If you plot E_e on one axis and E_pi on the other, any final state will fall within a circle in the non-relativistic case, and a more complicated shape for the relativistic case. Your two solutions for E_pi represent the two limits on the edge of the Dalitz plot. You could find more about th Dalitz plot in an older particle physics book.
     
  4. Mar 6, 2009 #3
    Thanks, that seems very helpful :)

    So in practice, if I want to run a Monte Carlo simulation of the reaction given a incident beam energy etc., what solution should I use at the end? a random value between those 2 ?
     
  5. Mar 6, 2009 #4
    How off shell is the delta ? Looks to me like it should be quite on shell. That would render the problem most easy.

    In any case, you can go to the pi proton center of mass and enforcing energy conservation should straightforwardly give you the (common) momentum shared. Then boost back to the lab.
     
  6. Mar 6, 2009 #5
    In CM frame, we have this

    [tex]\sqrt{p_e^2+m_e^2} + \sqrt{p_{\pi}^2+m_{\pi}^2} + \sqrt{p_e^2 + p_{\pi}^2 + 2 p_e p_{\pi} cos \theta + m_p^2} = E_{CM}[/tex]

    where [tex]\theta[/tex] is the angle between the pion and the electron. If we hold [tex]p_e[/tex] and [tex]\theta[/tex] constant, after some tedius algebra this indeed results in a quadratic equation for [tex]p_{\pi}[/tex] which may have one or two positive solutions. If the angle is less than 90 degrees, there should be only one positive solution. If it's greater than 90, for some [tex]p_e[/tex] and [tex]E_{CM}[/tex], there may be two. If you're doing monte carlo, you should use both.
     
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