graph of Fig. \[fig:NuNuBar\](d) is relevant. Straightforward considerations, again invoking Feynman’s interpretation of the negative-energy particle solutions, lead to a result identical to the one obtained in Section \[sec:D2\]. The DCS is of the form $$\frac{d\sigma}{d\Omega} = \frac{\sigma_0 u^2}{8 \pi s^2} \, \, \, ,$$ which, integrated over the solid angle, results in $$\sigma_{\rm tot} = \frac{\sigma_0}{6} \, \, \, .$$

\[sec:D6\]Scattering of neutrini off an electron
------------------------------------------------

For the scattering of an electron neutrino off an electron, the graphs of Fig. \[fig:NuElPos\](a) and (b) are relevant. The scattering amplitude is of the form $$\begin{aligned}
 \label{eq:EQ200}
\mathscr{T} &= \left( \frac{g_N}{2 M_Z} \right)^2 \bar{u}(\nu_e) \gamma^\mu \frac{1 - \gamma_5}{2} u(\nu_e) \bar{u}(e^-) \gamma_\mu (g_V^l - g_A^l \gamma_5) u(e^-)\nonumber\\
&+ \left( \frac{g}{\sqrt{2} M_W} \right)^2 \bar{u}(\nu_e) \gamma^\mu \frac{1 - \gamma_5}{2} u(e^-) \bar{u}(e^-) \gamma_\mu \frac{1 - \gamma_5}{2} u(\nu_e)\nonumber\\
&= \frac{G_F}{\sqrt{2}} \big( \bar{u}(\nu_e) \gamma^\mu (1 - \gamma_5) u(\nu_e) \bar{u}(e^-) \gamma_\mu (g_V^l - g_A^l \gamma_5) u(e^-)\nonumber\\
&+ \bar{u}(e^-) \gamma^\mu (1 - \gamma_5) u(\nu_e) \bar{u}(\nu_e) \gamma_\mu (1 - \gamma_5) u(e^-) \big) \, \, \, .\end{aligned}$$ After employing the two Fierz identities of Eqs. (\[eq:EQ150\]), the second term within the brackets may be rewritten as $\bar{u}(\nu_e) \gamma^\mu (1 - \gamma_5) u(\nu_e) \bar{u}(e^-) \gamma_\mu (1 - \gamma_5) u(e^-)$, resulting in $$\begin{aligned}
 