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

1. Jul 23, 2010

1. The problem statement, all variables and given/known data
I need to calculate the differential cross section in order of Mandelstam variable $$t$$, instead of the angle $$\theta$$. 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):

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

And knowing that for this particular process we have ($$t=(p_1-p3)^2$$):

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

I then calculate:

$$d\theta=-\frac{2}{\sqrt{s(s-4m^2)}sin(\theta)}$$

And use this in:

$$d\Omega=sin(\theta)d\theta d\phi$$

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

3. The attempt at a solution

I've thought about it again, and I see that with the change of variable $$\theta \rightarrow t$$, the new limits of integration are $$[t_{max},t_{min}]$$, since $$t$$ decreases with $$\theta$$ in the interval $$[0,\pi]$$, which accounts for an extra minus sign.