Let's tackle the problems step by step: --- ### (a) **Matrix Element and Cross-section for \(\nu_e e^- \to e^- \nu_e\)** #### **Step 1: Feynman Diagram and Matrix Element** - This is **charged-current scattering** mediated by the **W boson** (since flavor is changing). - The Feynman diagram is a **t-channel** exchange: \(\nu_e\) and \(e^-\) in, \(e^-\) and \(\nu_e\) out. #### **Step 2: Write the Matrix Element** The charged-current Lagrangian (neglecting masses) is: \[ \mathcal{L}_{CC} = -\frac{g}{\sqrt{2}} \left( \bar{\nu}_e \gamma^\mu P_L e W^+_\mu + \bar{e} \gamma^\mu P_L \nu_e W^-_\mu \right) \] The matrix element (\(\mathcal{M}\)) is: \[ \mathcal{M} = \left( \frac{-ig}{\sqrt{2}} \right)^2 \left[ \bar{u}(k') \gamma^\mu P_L u(p) \right] \frac{-i}{q^2 - M_W^2} \left[ \bar{u}(p') \gamma_\mu P_L u(k) \right] \] where: - \(p\): initial \(\nu_e\) - \(k\): initial \(e^-\) - \(p'\): final \(\nu_e\) - \(k'\): final \(e^-\) - \(q = p - p'\) is the momentum transfer (\(t\)-channel: \(t = q^2\)) - \(P_L = \frac{1-\gamma^5}{2}\) is the left-handed projector. #### **Step 3: Large \(M_W\) Limit (Contact Interaction)** When \(M_W^2 \gg |t|\), the propagator simplifies: \[ \frac{1}{q^2 - M_W^2} \approx -\frac{1}{M_W^2} \] So, \[ \mathcal{M} = \frac{g^2}{2 M_W^2} \left[ \bar{u}(k') \gamma^\mu P_L u(p) \right] \left[ \bar{u}(p') \gamma_\mu P_L u(k) \right] \] Recall the Fermi constant: \[ \frac{G_F}{\sqrt{2}} = \frac{g^2}{8 M_W^2} \implies \frac{g^2}{2 M_W^2} = 4 \frac{G_F}{\sqrt{2}} \] Thus, \[ \boxed{ \mathcal{M} = 4 \frac{G_F}{\sqrt{2}} \left[ \bar{u}(k') \gamma^\mu P_L u(p) \right] \left[ \bar{u}(p') \gamma_\mu P_L u(k) \right] } \] #### **Step 4: Matrix Element Squared** Averaging/summing over spins, neglecting fermion masses: \[ |\mathcal{M}|^2 = 32 G_F^2 \left[ (k' \cdot p)(p' \cdot k) \right] \] Expressing in Mandelstam variables: - \(s = (p + k)^2\) - \(t = (p - p')^2 = (k' - k)^2\) - \(u = (p - k')^2\) Here, - \(k' \cdot p = \frac{(u - m_e^2)}{2} \approx \frac{u}{2}\) - \(p' \cdot k = \frac{(s - m_e^2)}{2} \approx \frac{s}{2}\) So, \[ |\mathcal{M}|^2 = 32 G_F^2 \left( \frac{u}{2} \right) \left( \frac{s}{2} \right) = 8 G_F^2 u s \] #### **Step 5: Differential Cross-section in CM Frame** The general formula: \[ \frac{d\sigma}{dt} = \frac{1}{16\pi s^2} |\mathcal{M}|^2 \] Thus, \[ \boxed{ \frac{d\sigma}{dt} = \frac{1}{16\pi s^2} \cdot 8 G_F^2 u s = \frac{G_F^2}{2\pi s} u } \] This is your answer in terms of Mandelstam variables. --- ### (b) **\(\bar{\nu}_e e^- \rightarrow e^- \bar{\nu}_e\) and Backward Scattering** #### **Symmetry Argument** For \(\bar{\nu}_e e^- \rightarrow e^- \bar{\nu}_e\), the process is similar, but the initial neutrino is an **antineutrino**. The matrix element is nearly the same, but the spinor structure is different due to the different chiralities. #### **Matrix Element Squared** You can show (by crossing symmetry) that: \[ |\mathcal{M}|^2 = 8 G_F^2 t s \] where now the relevant invariants are \(t\) and \(s\). So the differential cross-section: \[ \boxed{ \frac{d\sigma}{dt} = \frac{G_F^2}{2\pi s} t } \] #### **Backward Scattering** - **Backward scattering** corresponds to \(t \to 0\) (\(\theta = \pi\)). - For this process, \(d\sigma/dt \propto t\), so for \(t = 0\) (exact backward), the cross-section vanishes. **Physical Reason:** Weak charged-current only couples to left-handed particles and right-handed antiparticles. In the backward direction, helicity conservation is violated, so the amplitude vanishes. --- ### (c) **Physical Reason for Different Cross-sections** - **Neutrino-electron (\(\nu_e e^-\)) scattering**: Both particles are left-handed; their spins can be aligned so that total angular momentum is 0 in the CM frame, allowing for scattering at all angles. - **Antineutrino-electron (\(\bar{\nu}_e e^-\)) scattering**: The antineutrino is right-handed; the spins of the two particles cannot be aligned to give 0 total angular momentum in the CM frame, so **backward scattering is forbidden**. **Summary Table:** | Process | \(d\sigma/dt\) | Backward Scattering? | |------------------------------|-----------------------|----------------------| | \(\nu_e e^- \to e^- \nu_e\) | \(\frac{G_F^2}{2\pi s} u\) | Allowed | | \(\bar{\nu}_e e^- \to e^- \bar{\nu}_e\) | \(\frac{G_F^2}{2\pi s} t\) | Forbidden | --- ## **Final Answers (Summary)** ### (a) \[ \boxed{ \frac{d\sigma}{dt} = \frac{G_F^2}{2\pi s} u } \] for \(\nu_e e^- \to e^- \nu_e\). --- ### (b) \[ \boxed{ \frac{d\sigma}{dt} = \frac{G_F^2}{2\pi s} t } \] for \(\bar{\nu}_e e^- \to e^- \bar{\nu}_e\). Backward scattering (\(t=0\)) is forbidden. --- ### (c) - The difference arises from the chiral nature of the weak interaction and helicity conservation. Neutrinos are left-handed, antineutrinos are right-handed, leading to different angular distributions and forbidding backward scattering for antineutrinos. --- If you need explicit integration over angles or more details, let me know!