# Relative speed in unpolarized cross-section

In section 5.1 of Peskin and Schroeder, we are presented the computation of the amplitude for the $$e^+e^-\to \mu^+\mu^-$$ reaction and then the computation of the unpolarized cross section. After computing the modulus squared of the amplitude (equation 5.11), the authors evoke the simplified version of the cross section formula, valid in the center-of-mass frame (equation 4.84):

$$\left(\frac{d\sigma}{d\Omega}\right)_{\text{CM}} = \frac{1}{2E_A 2E_B|v_A - v_B|}\frac{|\mathbf{p}_1|}{(2\pi)^2 4E_{\text{CM}}}|\mathcal{M}(p_A, p_B\to p_1, p_2)|^2.\tag{4.84}$$

In this case, $$p_A$$ and $$p_B$$ describe the 4-momentum of the electron and the positron. When computing the cross section, P&S set the masses $$m_e = 0$$ and, since we are working in the center-of-mass frame, $$E_A = E_B = E_{\text{CM}}/2$$. However, for the velocities of the incoming particles, they assume that $$|v_A - v_B| = |1 - (-1)| = 2$$. I understand the sign, since we consider the particles to be moving in the $$z$$-axis in opposite directions and, since we assume the particles to be massless, they move with the speed of light, which is $$1$$ in natural units.

However, the speed of light should be the same in all reference frames and that should be also valid in the CM frame. Shouldn't the speeds add then to $$1$$?

The velocity difference of the two can of course be $$2c$$, observed from the frame where $$v_A = +c$$ and $$v_B = -c$$. If, however, you would be in a co-moving frame with e.g. particle $$A$$, then you will observe that particle $$B$$ has a speed of $$c$$ instead of $$2c$$.