Show that
$$
\sigma\left(\nu_e \mathrm{e}^{-}\right) \approx\left(E_\nu \text { in } \mathrm{GeV}\right) \times 10^{-41} \mathrm{~cm}^2,
$$
where $E_v$ is the laboratory energy of the neutrino.
The Feynman diagram for $\tilde{\nu} \mathrm{e}^{-} \rightarrow \mathrm{e}^{-} \tilde{\nu}_e$ is shown in Fig. 12.8b. We see that it can be obtained by crossing the neutrinos in $\nu_e \mathrm{e}^{-} \rightarrow \mathrm{e}^{-} \nu_e$ of diagram (a); see Sections 4.6 and 4.7. We therefore simply replace $s$ by $t$ in (12.57):
$$
\begin{aligned}
\frac{1}{2} \sum_{\text {spins }}|\mathscr{T}|^2 & =16 G^2 t^2 \\
& =4 G^2 s^2(1-\cos \theta)^2,
\end{aligned}
$$
where $\theta$ is the angle between the incoming $\bar{\nu}_e$ and the outgoing $\mathrm{e}^{-}$(see Fig. 12.9), and
$$
t=-\frac{s}{2}(1-\cos \theta),
$$
see (4.45). From (12.61), we obtain
$$
\frac{d \sigma}{d \Omega}\left(\bar{\nu}_e \mathrm{e}^{-}\right)=\frac{G^2 s}{16 \pi^2}(1-\cos \theta)^2,
$$
and integrating over angles yields
$$
\sigma\left(\bar{v}_e \mathrm{e}^{-}\right)=\frac{G^2 s}{3 \pi} .
$$
Comparing with (12.60) gives
$$
\sigma\left(\bar{v}_e \mathrm{e}^{-}\right)=\frac{1}{3} \sigma\left(\nu_e \mathrm{e}^{-}\right) .
$$
Results (12.59), (12.62), and (12.64) expose the $\gamma^\mu\left(1-\gamma^5\right)$ structure of the weak current in a way that can be experimentally checked. We can convince ourselves of this important statement by comparing the results with those obtained for the $\gamma^\mu$ vertex in the electromagnetic process $\mathrm{e} \mu \rightarrow \mathrm{e} \mu$ or by performing the following exercise.