Differential cross section formula of electron-positron pair production.

[salparadise]

Homework Statement

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.

Homework 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.

The Attempt at a Solution

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

Thanks in advance

 

[salparadise]

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.

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