Question

If the weak current had had a $V+A$ structure, $\gamma^\mu(1+$ $\gamma^5$ ), show that $$ \frac{d \sigma}{d \Omega}=\frac{G^2 s}{4 \pi^2}(1+\cos \theta)^2 $$ for both $\nu$ e and $\bar{v}$ e elastic scattering. If this were the case, then, in contrast to (12.64), we would have $$ \sigma\left(\bar{\nu}_e \mathrm{e}\right)=\sigma\left(\nu_e \mathrm{e}\right) . $$ The most striking difference between the two angular distributions, (12.59) and (12.62), is that $\bar{\nu}_e$ e scattering vanishes for $\cos \theta=1$, whereas $\nu_e$ e scattering does not. With our definition of $\theta$, see Fig. 12.9, this corresponds to backward scattering of the beam particle. We could have anticipated these results from the helicity arguments we used to interpret previous calculations. The by now familiar pictures are shown in Fig. 12.10. Backward $\bar{\nu}_e$ e scattering is forbidden by angular momentum conservation. In fact, the process $\bar{\nu}_e \mathrm{e} \rightarrow \bar{\nu}_e \mathrm{e}$ proceeds entirely in a $J=1$ state with net helicity +1 ; that is, only one of the three helicity states is allowed. This is the origin of the factor $\frac{1}{3}$ in (12.64). With our definition of $\theta$, the allowed amplitude is proportional to $d_{-11}^1(\theta)=\frac{1}{2}(1-\cos \theta)$, see (6.39), in agreement with result (12.61). Elastic $\nu_e \mathrm{e}^{-}$and $\bar{\nu}_e \mathrm{e}^{-}$scattering can also proceed via a weak neutral current interaction (see Fig. 12.11) which interferes with the charged current interaction (Fig. 12.8a). This is discussed in Section 13.5. However, high-energy neutrino beams are predominantly $\nu_\mu$ (or $\bar{\nu}_\mu$ ), and so the most accessible (charged current) purely leptonic scattering process is $$ \nu_\mu+e^{-} \rightarrow \mu^{-}+\nu_e $$ (i.e., inverse muon decay). Here, there is no neutral current contribution, and so the cross section is given just by (12.59).

   If the weak current had had a $V+A$ structure, $\gamma^\mu(1+$ $\gamma^5$ ), show that
$$
\frac{d \sigma}{d \Omega}=\frac{G^2 s}{4 \pi^2}(1+\cos \theta)^2
$$
for both $\nu$ e and $\bar{v}$ e elastic scattering. If this were the case, then, in contrast to (12.64), we would have
$$
\sigma\left(\bar{\nu}_e \mathrm{e}\right)=\sigma\left(\nu_e \mathrm{e}\right) .
$$

The most striking difference between the two angular distributions, (12.59) and (12.62), is that $\bar{\nu}_e$ e scattering vanishes for $\cos \theta=1$, whereas $\nu_e$ e scattering does not. With our definition of $\theta$, see Fig. 12.9, this corresponds to backward scattering of the beam particle. We could have anticipated these results from the helicity arguments we used to interpret previous calculations. The by now familiar pictures are shown in Fig. 12.10. Backward $\bar{\nu}_e$ e scattering is forbidden by angular momentum conservation. In fact, the process $\bar{\nu}_e \mathrm{e} \rightarrow \bar{\nu}_e \mathrm{e}$ proceeds entirely in a $J=1$ state with net helicity +1 ; that is, only one of the three helicity states is allowed. This is the origin of the factor $\frac{1}{3}$ in (12.64). With our definition of $\theta$, the allowed amplitude is proportional to $d_{-11}^1(\theta)=\frac{1}{2}(1-\cos \theta)$, see (6.39), in agreement with result (12.61).
Elastic $\nu_e \mathrm{e}^{-}$and $\bar{\nu}_e \mathrm{e}^{-}$scattering can also proceed via a weak neutral current interaction (see Fig. 12.11) which interferes with the charged current interaction (Fig. 12.8a). This is discussed in Section 13.5. However, high-energy neutrino beams are predominantly $\nu_\mu$ (or $\bar{\nu}_\mu$ ), and so the most accessible (charged current) purely leptonic scattering process is
$$
\nu_\mu+e^{-} \rightarrow \mu^{-}+\nu_e
$$
(i.e., inverse muon decay). Here, there is no neutral current contribution, and so the cross section is given just by (12.59).
Show more…
Quarks and leptons: introductory course in modern particle physics
Quarks and leptons: introductory course in modern particle physics
Francis Halzen, Alan… 1st Edition
Chapter 12, Problem 16 ↓

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This structure implies that both vector and axial-vector currents contribute equally. In the context of neutrino-electron scattering, this modifies the interaction vertex compared to the standard $V-A$ structure used in the Standard Model.  Show more…

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If the weak current had had a $V+A$ structure, $\gamma^\mu(1+$ $\gamma^5$ ), show that $$ \frac{d \sigma}{d \Omega}=\frac{G^2 s}{4 \pi^2}(1+\cos \theta)^2 $$ for both $\nu$ e and $\bar{v}$ e elastic scattering. If this were the case, then, in contrast to (12.64), we would have $$ \sigma\left(\bar{\nu}_e \mathrm{e}\right)=\sigma\left(\nu_e \mathrm{e}\right) . $$ The most striking difference between the two angular distributions, (12.59) and (12.62), is that $\bar{\nu}_e$ e scattering vanishes for $\cos \theta=1$, whereas $\nu_e$ e scattering does not. With our definition of $\theta$, see Fig. 12.9, this corresponds to backward scattering of the beam particle. We could have anticipated these results from the helicity arguments we used to interpret previous calculations. The by now familiar pictures are shown in Fig. 12.10. Backward $\bar{\nu}_e$ e scattering is forbidden by angular momentum conservation. In fact, the process $\bar{\nu}_e \mathrm{e} \rightarrow \bar{\nu}_e \mathrm{e}$ proceeds entirely in a $J=1$ state with net helicity +1 ; that is, only one of the three helicity states is allowed. This is the origin of the factor $\frac{1}{3}$ in (12.64). With our definition of $\theta$, the allowed amplitude is proportional to $d_{-11}^1(\theta)=\frac{1}{2}(1-\cos \theta)$, see (6.39), in agreement with result (12.61). Elastic $\nu_e \mathrm{e}^{-}$and $\bar{\nu}_e \mathrm{e}^{-}$scattering can also proceed via a weak neutral current interaction (see Fig. 12.11) which interferes with the charged current interaction (Fig. 12.8a). This is discussed in Section 13.5. However, high-energy neutrino beams are predominantly $\nu_\mu$ (or $\bar{\nu}_\mu$ ), and so the most accessible (charged current) purely leptonic scattering process is $$ \nu_\mu+e^{-} \rightarrow \mu^{-}+\nu_e $$ (i.e., inverse muon decay). Here, there is no neutral current contribution, and so the cross section is given just by (12.59).
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