Differential cross section in the case of identical masses In the book "An Introduction to Quantum Field Theory" by M.E. Peskin and D.V. Schroeder in page $107$, they calculate the differential cross section for two particles $A$ and $B$ with initial energy $E_A$, $E_B$ and momentum $p_A$, $p_B$ to scatter and become particles $1$ and $2$ with final momentum $p_1$, $p_2$, in the center of mass. They come to this relation
$$\left(\frac{dσ}{dΩ}\right)_{CM}=\frac{1}{2E_A2E_B\vert υ_A-υ_B\vert}\frac{\vert\textbf{p}_1\vert}{(2π)^24E_{cm}}\vert M(p_A,p_B\rightarrow p_1,p_2)\vert^2\qquad\quad(4.84)$$ 
where $υ_A-υ_B$ is the relative velocity of the beams as viewed from the laboratory frame, $E_cm$ is the energy of the system in the center of mass and $M(p_A,p_B\rightarrow p_1,p_2)$ is the invariant matrix element of the process.
Then the authors make the hypothesis that the four particles have identical mass and this formula reduces to 
$$\left(\frac{dσ}{dΩ}\right)_{CM}=\frac{\vert M\vert^2}{64π^2E^2_{cm}}\qquad\quad(4.85)$$ 
My question is how did they came to eq. $(4.85)$ with this assumptions. I can not follow the maths.
Any helps?
 A: Assume the 3-velocities $v$ to be along the x-axis and express them through momenta and energies
$$v_{Ax}=\frac{{\rm d} x}{{\rm d} t}=\frac{{\rm d} x}{{\rm d} \tau}\frac{{\rm d} \tau}{{\rm d} t}=\frac{p_{Ax}}{E_A}.$$
So, (4.84) becomes $$\left(\frac{dσ}{dΩ}\right)_{CM}=\frac{1}{4\vert p_{Ax}E_B-p_{Bx}E_A\vert}\frac{\vert\textbf{p}_1\vert}{(2π)^24E_{cm}}\vert M(p_A,p_B\rightarrow p_1,p_2)\vert^2\qquad\quad(4.84)$$ 
The expression in the denominator can be simplified by choosing a frame where one of the initial particles is at rest, i.e. $E_A=m$, such that 
$$\frac{1}{4\vert p_{Ax}E_B-p_{Bx}E_A\vert}= \frac{1}{4 p_{Bx}m}.$$
Using the energy-momentum relation for $p_B$ and going to the limit $m\rightarrow 0$ this becomes $\frac{1}{2s}$.
For the $\mathbf{p_1}$ in the numerator look at the direction of the momenta of $p_1$ and $p_2$ after the collision. In CMS the two particles have the same energy but opposite momenta (e.g. along the x-axis). Hence, $s=(p_1+p_2)^2=4E_1^2$ and ${\mathbf p_1}\approx E_1=\frac{\sqrt s}{2}$.
Using $E_{CM}=\sqrt s$ one should arrive at the desired result.
