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Help with derivation of two-body relativistic cross section

  1. Dec 14, 2016 #1
    1. The problem statement, all variables and given/known data
    I am trying to derive the 2 body cross section, as given in
    https://web2.ph.utexas.edu/~schwitte/PHY362L/QMnote.pdf

    dσ/dΩ

    2. Relevant equations

    I am stuck on

    d√s/dp=v

    where

    √s=(Ea+Eb)=E

    3. The attempt at a solution

    p=Ev (relativisitic energy-momentum relationship)
    √s=E

    E=Ec+Ed=p/vc + p/vd

    pc = pd = pf

    d√s/dp = d(Ec+Ed)/dp = d(p/vc + p/vc)/dp=(1/vc + 1/vd)=(vc+vd)/(vc*vd)=vf/vc*vd




    .
     
    Last edited: Dec 14, 2016
  2. jcsd
  3. Dec 14, 2016 #2

    vela

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    Did you have a specific question?
     
  4. Dec 14, 2016 #3
    I am trying to derive the 2+2 body relativisitic cross section, as given in
    https://web2.ph.utexas.edu/~schwitte/PHY362L/QMnote.pdf

    a+b -> c+d

    I want the form of the cross section

    dσ/dΩ ~ 1/((v_i)(v_f))

    I am stuck on step (13)-(14)
    and I don't understand why d√s/dp_f=v_f

    Here, s is the mandelstam variable √s=(Ec+Ed)=(Ea+Eb)
    and p_f is the (amplitude) of the final (I assume 3 vector ) momentum
     
  5. Dec 14, 2016 #4

    vela

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    ##\vec{p}_\text{f}## is the three-momentum of particle C.

    Try differentiating ##-m_c^2 = p_c^2-E_c^2## to calculate ##dE_c/dp_c##. I think the problem with your attempt is that you're assuming ##dv_c/dp_c = 0## and ##dv_d/dp_c = 0##.
     
  6. Dec 14, 2016 #5
     
  7. Dec 14, 2016 #6
    Thanks ! That makes perfect sense. this is a great forum
     
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