\frac{G^2_F u^2}{8 \pi^2 s} = \frac{\sigma_0 u^2}{8 \pi s^2} \, \, \, ,$$ which, integrated over the solid angle, results in $$\sigma_{\rm tot} = \frac{\sigma_0}{6} \, \, \, .$$

\[sec:D3\]Scattering of neutrini of the same flavour
----------------------------------------------------

The peculiarity of this case is that the particles in the final state are indistinguishable. For the scattering of electron neutrini, the graphs of Fig. \[fig:NuNu\](a) and (b) are relevant.

Starting from Eq. (\[eq:EQ100\]), one may write the scattering amplitude as $$\begin{aligned}
 \label{eq:EQ140}
\mathscr{T} = \frac{G_F}{2\sqrt{2}} &\big( \bar{u}(\nu_a) \gamma^\mu (1 - \gamma_5) u(\nu_a) \bar{u}(\nu_b) \gamma_\mu (1 - \gamma_5) u(\nu_b)\nonumber\\
&+ \bar{u}(\nu_b) \gamma^\mu (1 - \gamma_5) u(\nu_a) \bar{u}(\nu_a) \gamma_\mu (1 - \gamma_5) u(\nu_b) \big) \, \, \, .\end{aligned}$$ Let me rewrite the second term within the brackets in the form $$( \bar{\chi} \gamma^\mu (1 - \gamma_5) \psi ) \, ( \bar{\psi} \gamma_\mu (1 - \gamma_5) \chi ) \, \, \, ,$$ where $\chi=u(\nu_b)$ (hence $\bar{\chi}=\bar{u}(\nu_b)$) and $\psi=u(\nu_a)$ (hence $\bar{\psi}=\bar{u}(\nu_a)$). By invoking two of the Fierz identities, namely $$\begin{aligned}
 \label{eq:EQ150}
(\bar{\chi}\gamma^\mu \psi)(\bar{\psi} \gamma_\mu \chi)&=(\bar{\chi}\chi)(\bar{\psi}\psi)-\frac{1}{2}(\bar{\chi}\gamma^\mu \chi)(\bar{\psi}\gamma_\mu \psi)\nonumber\\
&-\frac{1}{2}(\bar{\chi}\gamma^\mu \gamma_5 \chi)(\bar{\psi}\gamma_\mu \gamma_5 \psi)-(\bar{\chi}\gamma_5 \chi)(\bar{\psi}\gamma_5 \psi) \, \, \, ,\nonumber\\
(\bar{\chi}\gamma^\mu \gamma_5 \psi)(\bar{\psi} \gamma_\mu \gamma_5 \chi)&=-(\bar{\chi}\chi)(\bar{\psi}\psi)-\frac{1}{2}(\bar{\chi}\gamma^\mu \chi)(\bar{\psi}\gamma_\mu \psi)\nonumber\\
&-\frac{1}{2}(\bar{\chi}\gamma^\mu \gamma_5 \chi)(\bar{\psi}\gamma_\mu \gamma_5 \psi)+(\bar{\chi}\gamma_5 \chi)(\bar{\psi}\gamma_5 \psi) \, \, \, 