= Muon-neutrino electron recoil spectrum
{title2=$d\sigma_\nu/dy=K[g_L^2+g_R^2(1-y)^2]$}
For an especially simple <neutral current> example, scatter a <muon neutrino> elastically from an initially stationary unpolarized <Electron>:
$$
\nu_\mu e^-\longrightarrow\nu_\mu e^-,\qquad
y=\frac{T_e}{E_\nu},\qquad
s_W^2=\sin^2\theta_W,\qquad
g_L=-\frac12+s_W^2,\qquad g_R=s_W^2.
$$
Here $T_e$ is the final <Electron> <kinetic energy>, $E_\nu$ is the incident <neutrino> <energy>, and $\theta_W$ is the <Weinberg angle>. These are electron <chiral> couplings in the $g/\cos\theta_W$ vertex convention of <neutral-current vector and axial couplings>. The chosen <muon neutrino> elastic channel has no additional <W boson> exchange amplitude; the electron-neutrino elastic channel would have one. Other charged-current final states can become kinematically accessible at sufficiently high incident <energies> and are not part of this elastic model.
Use <natural units> $\hbar=c=1$, retain only <tree-level Feynman diagrams>, and take the contact limit $|q^2|\ll m_Z^2$ with $E_\nu\gg m_e$. Neglect <Electron>-mass terms relative to $E_\nu$, <radiative corrections> and detector response. Then
$$
K=\frac{2G_F^2m_eE_\nu}{\pi},\qquad
\frac{d\sigma_\nu}{dy}=K\bigl[g_L^2+g_R^2(1-y)^2\bigr],\qquad
\frac{d\sigma_{\bar\nu}}{dy}=K\bigl[g_R^2+g_L^2(1-y)^2\bigr],\qquad 0\leq y\leq1.
$$
$G_F$ is the <Fermi constant>. The approximate endpoint $y=1$ replaces the exact elastic endpoint $y_{\max}=2E_\nu/(m_e+2E_\nu)$. The omitted leading mass-dependent interference term in $d\sigma_\nu/dy$ is $-K g_Lg_Rm_ey/E_\nu$. Thus the approximation is not appropriate for arbitrarily soft <neutrinos>.
The shape follows from the two <chiral> contributions to the massless <scattering amplitude>. The <spin sum> gives terms proportional to $s^2$ and $u^2$ in <Mandelstam variables>; in this limit $u/s=-(1-y)$, giving the square $(1-y)^2$. The two electron <chiralities> do not interfere when the relevant mass corrections are dropped. Replacing the <neutrino> by its <antineutrino> exchanges which coupling multiplies each kinematic term. This is the same <chiral> angular weighting that appears in <neutrino scattering on an antiquark>, now with a target whose couplings are particularly simple.
Integrating the two elementary terms gives
$$
\int_0^1(1-y)^2\,dy=\frac13,\qquad
\sigma_\nu=K\left(g_L^2+\frac{g_R^2}{3}\right),\qquad
\sigma_{\bar\nu}=K\left(g_R^2+\frac{g_L^2}{3}\right).
$$
For an original teaching example choose $s_W^2=1/4$, so $g_L=-1/4$ and $g_R=1/4$. This is a stipulated parameter, not a measurement of the <Weinberg angle>. Then
$$
\frac{1}{K}\frac{d\sigma_\nu}{dy}=\frac{1+(1-y)^2}{16},\qquad
\frac{\sigma_\nu}{K}=\frac1{12},\qquad
\langle y\rangle=\frac{\int_0^1y[1+(1-y)^2]dy}{4/3}=\frac7{16}.
$$
The normalized <probability density function> is $p(y)=3[1+(1-y)^2]/4$, not the unnormalized differential <cross section>. Its endpoints have ratio two, and the fraction with $y>1/2$ is $13/32$. The <antineutrino> spectrum happens to coincide at this chosen parameter because $g_L^2=g_R^2$; this equality is not a general prediction for arbitrary <Weinberg angles>.
\Image[/muon-neutrino-electron-recoil-spectrum.png]
{title=Chosen <muon neutrino> <Electron> recoil spectra and their <chiral> contributions}
{description=Original tree-level curves divided by $K=2G_F^2m_eE_\nu/\pi$. The left panel chooses $s_W^2=1/4$; the right compares three stipulated values in $(0,1)$. All omit finite electron-mass corrections and radiative corrections; none is historical data or a fitted mixing angle.}
Unlike a visible <Electron> recoil, the scattered <neutrino> is unrecorded. A measured recoil spectrum must therefore be compared with incident <neutrino> flux, <energy> distribution, acceptance, <detection efficiency> and backgrounds. A shape or rate tests a specified theoretical model only after these experimental ingredients have been supplied.
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