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Chapter 12 - weak interaction of leptons - Coggle Diagram
Chapter 12 - weak interaction of leptons
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 \(Q^2 = (s - m_N^2) xy = 2 m_N E_\nu xy \leq 2m_N E_\nu\)
At high energy, \(s \approx 2m_N E_\nu\)
For accelerator neutrino with E < 400 GeV, W propagator can be approximated as \(\frac{ig_{\mu\nu}}{m_W^2}\)
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, \(\frac{\sigma_{\bar{\nu}q}}{\sigma_{\nu_q}} = \frac{1}{3}\)
Note:
neutrino
only scatters of \(d, \bar{u}\);
anti-neutrino
only scatters off \(u, \bar{d}\)!!!
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
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 y
integrated over x
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.