Past exam of the particle and nuclear physics course of the University of Cambridge / 2014 / Paper 1 / 1

In time $\Delta t$ the particle advances $\beta c\Delta t$, while a light wavelet travels $c\Delta t/n$. Coherent addition along the direction of emitted photons gives the Cherenkov cone angleThe angle is measured between photon propagation and the particle velocity. It is complementary to the angle between the conical wavefront surface and the trajectory; confusing those two angles exchanges sine and cosine. At threshold the photon cone closes, $\alpha \to 0$.

The decay is energetically open, but two spin-zero pions have total angular momentum equal to their relative orbital angular momentum $L$. Conserving the parent spin requires $L=1$. Interchanging two identical neutral pions multiplies this orbital state by $(-1{)}^L=-1$, whereas their total state must be symmetric by Bose statistics. There is no different spin or species label to supply another minus sign. Thus the two-neutral-pion final state is forbidden. Its parity for $L=1$ would actually be $(-1{)}^2(-1{)}^1=-1$, matching the parent; the decisive obstruction is identical-boson symmetry, not the energy or the shared quark-wavefunction component.

Away from thresholds and narrow resonances, photon-mediated annihilation gives one colour copy proportional to $Q_q^2$ relative to the muon cross-section. The accessible flavours are $u,d,s,c$, each with three colours, soBottom production is closed. QCD radiative corrections raise the leading parton-model result by a relative term beginning at ${\alpha }_s/\pi$; the requested estimate does not require a supplied value of ${\alpha }_s$.