Electron Muon scattering in the CM frame, question about the energy I calculated the spin-averaged amplitude for the $e^{-}\mu^{-} \rightarrow e^{-}\mu^{-}$ scattering in the CM frame in the high energy regime($m_{e}$,$m_{\mu} \rightarrow 0$) following the hints provided in Griffiths, and found the correct expression:
$$\left<|M|^2\right> = 2{g_e}^2 \left(\frac{1+cos^4(\theta/2)}{sin^4(\theta/2)}\right)$$
Plugging this in the expression for the differential cross-section of two bodys in the CM frame($1+2\rightarrow 1+2$), ie
$$\frac{d \sigma}{d\Omega} = \left(\frac{\hbar c}{8 \pi}\right)^2 \frac{S\left<|M|^2\right>}{(E_1+E_1)^2} \frac{|\pmb{p_i}|}{|\pmb{p_f}|}$$
I found the expression Griffths gives as the answer:
$$\frac{d \sigma}{d\Omega} = \left(\frac{\hbar c}{8 \pi}\right)^2 \frac{{g_e}^4}{2E^2} \left(\frac{1+cos^4(\theta/2)}{sin^4(\theta/2)}\right)$$
where $E$ is the energy of the electron.
My question is: in order to obtain this last relation I used that $E_1=E_2$, or in other notation, $E_{e}=E_{\mu}$. This gives me the correct expression but I'm not sure why I can do that. I found a lecture about the
$e^{-}e^{+} \rightarrow \mu^{-}\mu^{+}$ scattering who says:

because of working in the center of mass frame at the massless limit:
$$E_{e^+}=E_{e^-}=E_{\mu^-}=E_{\mu^+}$$

which is similar to what I have in the scattering I calculated.
Does anyone understand why we can set $E_{\mu}=E_{e}$??
 A: The total momentum of the collision is
$$
\vec p_\text{total} = \gamma_e m_e \vec v_e + \gamma_\mu m_\mu \vec v_\mu
$$
You are working in the limit $\gamma \gg 1$, in which $v\approx c$, and you have chosen a frame where this total momentum vanishes:
$$
0 = \gamma_e m_e c - \gamma_\mu m_\mu c
$$
I don’t remember whether this calculation involves the total energy $\gamma m c^2$ or just the kinetic energy $(\gamma - 1) mc^2 = \gamma mc^2 \cdot \left( 1-\frac1\gamma \right)$, but that difference also vanishes at high energy.
A simpler argument is that the zero-mass limit corresponds to $E^2=p^2$, and so the energies are equal in the center-of-mass frame by construction.
A: Reading Rob's answer I think I've come to an understanding.
The relativistic dispersion relation states that:
$$E^2 = m^2c^4 + p^2c^2$$
So, outside the high energy regime, we have
$$E^2_{e} = m_{e}^2c^4 + p_{e}^2c^2$$
$$E^2_{\mu} = m_{\mu}^2c^4 + p_{\mu}^2c^2$$
As Rob have said, in CM the total momentum vanishes so:
$$\pmb{p_{\mu}} = - \pmb{p_{e}} \Rightarrow p_{\mu} = p_{e} = p_i$$
and from the dispersion relation, we get
$$E^2_{e} = m_{e}^2c^4 + p_i^2c^2$$
$$E^2_{\mu} = m_{\mu}^2c^4 + p_i^2c^2$$
But, in the high energy regime, we have
$$m_e, m_{\mu} \rightarrow 0$$
So
$$E^2_{e} =   p_i^2c^2$$
$$E^2_{\mu} =  p_i^2c^2$$
and because of this, we can see that
$$E_{e} =  E_{\mu}$$
