I am reading p. 107 in Peskin and Schroeder's QFT, and I am stucked on one of the steps they took while calculation $2\rightarrow 2$ cross section. For $A+B\rightarrow 1+2$ differential cross section we have: $$\frac{d\sigma}{d\Omega}_{CM} = \frac{1}{4E_AE_B}\frac{1}{|v_A-v_B|}\frac{|p_1|}{(2\pi)^24E_{cm}}|M|^2 \tag{4.84}$$ P&S then goes on to say that if 4 masses are equal, it reduces to $$\frac{d\sigma}{d\Omega}_{CM} = \frac{|M|^2}{64\pi^2E^2_{cm}},\tag{4.85}$$ this implies $$\frac{1}{4E_AE_B}\frac{1}{|v_A-v_B|}=\frac{1}{4|p_1|E_{cm}},$$ therefore I try to calculate it myself. The result I got was $$\frac{1}{|\frac{p_A}{E_A}-\frac{p_B}{E_B}|4E_AE_B}=\frac{1}{4|E_Bp_A-(E_A(-p_A))|}=\frac{1}{4|p_1|E_{cm}}.$$The issue is that I did not get to use the fact that all 4 masses are equal. The only assumption I used was the CoM frame, so $p_A = -p_B$. What did I miss?
1 Answer
This exact calculation is done explicitly in Schwartz Quantum Field Theory and the Standard Model in section 5.1.2. The important thing is that $|\mathbf{p}_1| \neq |\mathbf{p}_A|$ unless the masses are equal. Start with the first formula, $$ \frac{d\sigma}{d\Omega}_{CM} = \frac{1}{4 E_A E_B} \frac{1}{|\mathbf{v}_A - \mathbf{v}_B|} \frac{|\mathbf{p}_1|}{(2\pi)^2 4 E_{CM}} |M|^2 $$ I am using boldface variables to clarify that the quantities involved are three vectors. In the center ofmomentum frame we have $\mathbf{p}_1 = - \mathbf{p}_2$ and $\mathbf{p}_A = - \mathbf{p}_B$, but importantly if the masses are not equal we will not necessarily have $|\mathbf{p}_1| = |\mathbf{p}_A|$. The basic idea of your calculation is correct, indeed we have $$ |\mathbf{v}_A - \mathbf{v}_B| = \left| \frac{|\mathbf{p}_A|}{E_A}+ \frac{|\mathbf{p}_A|}{E_A} \right| = |\mathbf{p}_A| \frac{E_{CM}}{E_A E_B} $$ Plugging this in we get, $$\frac{d\sigma}{d\Omega}_{CM} = \frac{1} {64 \pi^2 E_{CM}^2} \frac{|\mathbf{p}_1|}{|\mathbf{p}_A|} |M|^2$$ This is equation 5.32 in Schwartz. We see that if $|\mathbf{p}_1| = |\mathbf{p}_A|$, which only happens if all the masses are equal, we get the desired expression.
To be thorough I will justify my claim that we will only necessarily have $|\mathbf{p}_1| = |\mathbf{p}_A|$ if all the masses are equal. For simplicity let's assume that both initial states have the same mass, and both final states have the same mass i.e. $m_A =m_B$. This will not be true in general, but it suffices to show that we don't necessarily have $|\mathbf{p}_1| \neq |\mathbf{p}_A|$ in this case because the general case will be even less restrictive. Conservation of energy give, $$ E_A +E_B = 2 \sqrt{m_A^2 + |\mathbf{p}_A|^2} = 2 \sqrt{m_1^2 + |\mathbf{p}_1|^2} = E_1 + E_2$$ Thus we have, $$\sqrt{ m_A^2 - m_1^2 + |\mathbf{p}_A|^2} = |\mathbf{p}_1| $$ This only reduces to $|\mathbf{p}_1| = |\mathbf{p}_A|$ if $m_1 = m_A$.
