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?

• In the CM frame, you'll have $E_A = E_B=E_1=E_2$. As A and B are back to back you can say something about vA - vB. Lastly, you find some cancellations... Sep 5, 2016 at 19:53
• You should be able to solve it and then please answer your own question here Sep 5, 2016 at 19:54
• i think $υ_Α-υ_Β=2υ$, because $υ_Α=-υ_Β$ they are back to back as you said. $2E_1=E_1+E_1=E_1+E_2=E_{cm}$. I made this far. The rest is $p_1/(E_1*υ)$ which has to become $1$. But how? Sep 5, 2016 at 21:06
• @JonSnow Just use the standard relativistic formulas for $p$ and $E$. Sep 5, 2016 at 22:56
• Still if I write $(E_1^2-m^2)^{1/2}/(E_1*υ)$ and take the limit $m \rightarrow 0$, it will give $1/υ$. How to get rid of this? Sep 6, 2016 at 9:00

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.