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4-momenta problem

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data
    For a scattering problem A + A - > B + B in the Toy Theory,

    Need to show

    [tex](p_4-p_2)^2-m_C^2c^2=p_4^2+p_2^2-2p_4\cdot p_2 = -\vec{\textbf{p}}^2(1-\cos\theta)[/tex]


    2. Relevant equations

    this is assuming [itex]m_A=m_B=m[/itex] and [itex]m_C=0[/itex]


    3. The attempt at a solution
    [tex]p_4^2+p_2^2=2m^2c^2[/tex]

    but i'm not sure how to do

    [tex]p_4\cdot p_2[/tex]

    do i work with

    [tex]\sqrt{\left( \frac{E_4^2}{c^2} - \vec{\textbf{p}}_4^2 \right) \left( \frac{E_2^2}{c^2} - \vec{\textbf{p}}_2^2 \right)}[/tex]

    ?
     
  2. jcsd
  3. Apr 6, 2013 #2
    please mention what is mc.Also the expression so far is correct only in CM frame.also it seems that a factor of 2 is missing in p2(1-cosθ).
     
  4. Apr 6, 2013 #3
    It's been awhile since I have done this problem! Have you tried,

    $$
    p_2 p_4 = \frac{E_2 E_4}{c^2} - \vec{p_2} \cdot \vec{p_4}
    $$
     
    Last edited: Apr 6, 2013
  5. Apr 6, 2013 #4
    Terribly sorry, I should've made my question clearer. All of the [itex]p[/itex]s in my post refers 4-momenta while [itex]\vec{\textbf{p}}[/itex] refers to 3 momenta.

    I'm working in the CM frame.

    In this scattering reaction the subscripts refers to the particle in the reaction as follows: 1 + 2 -> 3 + 4

    @andrien, you are right, I missed the factor 2. [itex]m_c[/itex] is mass of the mediating particle in the lowest order Feynman diagram of this particular scattering reaction. In this problem, I guess it could just be ignored.

    @Shinobii, does your [itex]\vec{p_2}[/itex] and [itex]\vec{p_4}[/itex] denote 3-momenta or 4-momenta? Both particles 2 and 4 are not massless...
     
  6. Apr 6, 2013 #5

    BruceW

    User Avatar
    Homework Helper

    well, they are both 4-vectors in minkowski spacetime. So how would you normally take the inner product? (Maybe think of a general 4-vector first, then apply it to the specific vectors you have here)
     
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