Chapter 12 - weak interaction of leptons

At high energy, $s \approx 2m_N E_\nu$

Lepton universality

Muon decay rate: $\Gamma(\mu^- \rightarrow e^- \bar{\nu}_e \nu_\mu) = \frac{1}{\tau_\mu} = \frac{G_F^{(e)}G_F^{(\mu)}m_\mu^5}{192 \pi^3}$

$G_F$ for all three leptons are the same!

(muon) Neutrino scattering

low $Q^2$, quasi elastic: $\nu_\mu n \rightarrow p \mu^-$.

At several GeV, resonance inelastic: $\nu_{\mu} n \rightarrow \mu^- \Delta^+ \rightarrow \mu^- p \pi^0$

At higher energies (> 30 GeV): deep inelastic scattering (DIS)

Neutrino DIS

available $Q^2$

For fixed target N, $s = (p_1 + p_2)^2 = 2m_N E_\nu + m_N^2$

Maximum $Q^2$ is restricted by $Q^2 = (s - m_N^2) xy = 2 m_N E_\nu xy \leq 2m_N E_\nu$

For accelerator neutrino with E < 400 GeV, W propagator can be approximated as $\frac{ig_{\mu\nu}}{m_W^2}$

At high energy, $s \approx 2m_N E_\nu$

neutrino-quark scattering

For $\nu_\mu d \rightarrow \mu^- u$, in CM frame, $M_{fi} = \frac{g_W^2}{m_W^2} \hat{s}$

Both initial neutrino and quark are LH helicity states -> S=0 state -> no angular dependence.

$\hat{s} = (2E)^2$ in CM frame.

Total cross section: $\sigma_{\nu q} = \frac{G_F^2}{\pi}\hat{s}$

Similarly, for $\bar{\nu}q$, $\sigma_{\bar{\nu}q} = \frac{G_F^2}{3\pi} \hat{s}$

Therefore, $\frac{\sigma_{\bar{\nu}q}}{\sigma_{\nu_q}} = \frac{1}{3}$

neutrino-nucleon scattering

Use kinematic variable $y$ since it can be measured by outgoing muon energy and neutrino energy.

$\frac{d\sigma_{\nu q}}{dy} =\frac{d\sigma_{\bar{\nu}\bar{q}}}{dy} = \frac{G_F^2}{\pi}\hat{s}$

$\frac{d\sigma_{\nu \bar{q}}}{dy} =\frac{d\sigma_{\bar{\nu}q}}{dy} = \frac{G_F^2}{\pi}(1-y)^2\hat{s}$

has angular distribution due to helicity structure

Note: neutrino only scatters of $d, \bar{u}$; anti-neutrino only scatters off $u, \bar{d}$!!!

for differential cross section of $\nu/\bar{\nu} - p/n$, use PDF $u(x), d(x), \bar{u}(x), \bar{d}(x)$

Finally, $\nu N$ cross section is average of $\nu p, \nu n$

$\frac{d\sigma^{\nu N}}{dx dy} = \frac{G_F^2 m_N }{\pi} E_\nu x [u(x) + d(x) + (1-y)^2(\bar{u}(x) + \bar{d}(x))]$, $\frac{d\sigma^{\bar{\nu} N}}{dx dy} = \frac{G_F^2 m_N }{\pi} E_\nu x [\bar{u}(x) + \bar{d}(x) + (1-y)^2(u(x) + d(x))]$,

$\frac{d\sigma^{\nu N}}{dy} = \frac{G_F^2 m_N}{\pi} E_\nu (f_q + (1-y)^2 f_{\bar{q}})$, $\frac{d\sigma^{\bar{\nu} N}}{dy} = \frac{G_F^2 m_N}{\pi} E_\bar{\nu} (f_\bar{q} + (1-y)^2 f_{q})$

$\sigma^{\nu N} = \frac{G_F^2 m_N}{\pi} E_\nu (f_q + \frac{1}{3}f_{\bar{q}})$, $\sigma^{\bar{\nu} N} = \frac{G_F^2 m_N}{\pi} E_\bar{\nu} (f_\bar{q} + \frac{1}{3} f_{q})$

integrated over x

integrated over y

Ratio: $\frac{\sigma^{\nu N}}{\sigma^{\bar{\nu}N}} = \frac{3f_q + f_\bar{q}}{3 f_\bar{q} + f_q}$

CDHS experiment (CERN, $\nu_\mu$ beam, 30 - 200 GeV; $E_\nu = E_\mu + E_{cascade}$; calculate y from it.

Result: ratio ~ 2 due to anti-quarks in the nucleon! If no anti-quarks, predict ratio to be 3.

Measured $f_q = 0.41, f_\bar{q} = 0.08$. Only half of momentum is carried by quarks/antiquarks! Others are carried by gluons (don't interact with W boson)

Structure function

3 structure functions to allow parity violation: $F_1^{\nu N} = 2x F_2^{\nu N} = x[u + d + \bar{u} + \bar{d}]$, $x F_3 ^{\nu N} = x[u + d - \bar{u} - \bar{d}]$

Therefore, $F_2 - xF_3 = 2x[\bar{u} + \bar{d}]$. Probes anti-quarks in nucleon!

Experimental results: anti-quark fraction rises at low x (due to sea quark)

In addition, $F_3 = u_V + d_V$, probing valence quarks alone. Integral = 3. Agrees with experiment.