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Past exam of the particle physics course of the University of Oxford

2018

For the longitudinal state,
$$
\int |\mathcal M_0|^2d\Omega=m_W^2g^2\,2\pi\int_{-1}^{1}\frac{1-z^2}{2}\,dz=\frac{4\pi}{3}m_W^2g^2.
$$
Insert this and $p^*=m_W/2$ into the <Lorentz-invariant phase space>[phase-space] formula:
$$
\boxed{\Gamma_\ell=\frac{g^2m_W}{48\pi}.}
$$
Rotational invariance makes the integrated <Decay rate>[decay probability] independent of the orientation of the initial <spin>. Both transverse <Particle polarization>[polarizations] therefore have the same rate. An unpolarized sample averages these three equal rates; it does not add three copies. Hence its <decay width>[width] into one specified charged-lepton <Flavor eigenstate>[flavour] is also $g^2m_W/(48\pi)$.

Solution

Past exam of the particle physics course of the University of Oxford / 2018 / 1 / c / Solution

Past exam of the particle physics course of the University of Oxford / 2018 / 1 / c / Solution

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