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 $E_{\mu}=E_{e}$??


2 Answers 2


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.


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: $$\pmb{p_{\mu}} = - \pmb{p_{e}} \Rightarrow \lvert p_{\mu}\rvert = \lvert p_{e}\rvert = p_i$$

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 $$E_{e} = E_{\mu}$$

  • 1
    $\begingroup$ Excellent work. Nice job taking what was stated in the answer and putting it through the equations you have until it makes sense. That helped you, and it's now helping me! $\endgroup$ May 19, 2023 at 21:42

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