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Homework Help: Differential cross section formula of electron-positron pair production.

  1. Jul 23, 2010 #1
    1. The problem statement, all variables and given/known data
    I need to calculate the differential cross section in order of Mandelstam variable [tex]t[/tex], instead of the angle [tex]\theta[/tex]. My problem is with the change of variable not the amplitude of the process. I'm getting a global minus sign which can only be wrong.

    It seems I'm making a very basic error but I cannot find it.


    2. Relevant equations
    Starting from (p1+p2->p3+p4):

    [tex]\frac{d \sigma}{d\Omega}=\frac{1}{64\pi^2s}\frac{\left|\vec{p}_3^{CM}\right|}{\left|\vec{p}_1^{CM}\right|}\left|M\right|^2[/tex]

    And knowing that for this particular process we have ([tex]t=(p_1-p3)^2[/tex]):

    [tex]t=m^2-2\left(E_{1}^0 E_{3}^0-\left|\vec{p}_3^{CM}\right| \left|\vec{p}_1^{CM}\right| cos(\theta)\right)=m^2-\frac{s}{2}+\frac{1}{2}\sqrt{s(s-4m^2)}cos(\theta)[/tex]

    I then calculate:

    [tex]d\theta=-\frac{2}{\sqrt{s(s-4m^2)}sin(\theta)}[/tex]

    And use this in:

    [tex]d\Omega=sin(\theta)d\theta d\phi[/tex]

    This global minus sign propagates then into the differential cross section [tex]\frac{d\sigma}{dt}[/tex] and into the total cross section.

    3. The attempt at a solution

    Can someone please help me find where are my calculations failing?

    Thanks in advance
     
    Last edited: Jul 23, 2010
  2. jcsd
  3. Jul 23, 2010 #2
    I've thought about it again, and I see that with the change of variable [tex]\theta \rightarrow t[/tex], the new limits of integration are [tex][t_{max},t_{min}][/tex], since [tex]t[/tex] decreases with [tex]\theta[/tex] in the interval [tex][0,\pi][/tex], which accounts for an extra minus sign.

    I would delete the first post, but I don't think it's possible.
     
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