Past exam of the sub-atomic physics course of the University of Oxford / 2016 / 2 / c

Use $c=1$ and let $E$ denote the total center-of-momentum energy, not one beam's energy. Opposite equal momenta give $E=2E_{\mathrm{beam}}=498$ MeV, so the kaon is at its stipulated resonance mass $m_K=498$ MeV. Write the unspecified selected exit branching fractions as $b_{\rho }$ and $b_K$. The supplied line shape givesThe distinction between inclusive and exclusive resonance branching factors matters here. For inclusive resonance formation, sum over all decays, so $b_{\rho }=b_K=1$. At $E=m_K$ the small kaon width cancels from the peak value:If instead the measured cross section is specifically resonant elastic ${\pi }^{+}{\pi }^{-}\to {\pi }^{+}{\pi }^{-}$, its exit branching fractions equal the entrance ones: $b_{\rho }=1$, $b_K=0.69$. ThenThe source says production without specifying an observed decay channel, so both conventions are made explicit rather than silently squaring a branching fraction for an inclusive measurement. The quoted kaon width and two-pion branching fraction identify the short-lived propagation state $K_S$; a pure flavor state $K^0$ is a superposition of short- and long-lived states. This exercise uses the stipulated $K_S$ resonance model and equal omitted proportionality factors.

Raising each beam energy to 250 MeV makes $E=500$ MeV. The kaon denominator is now dominated byrather than by $E^2{\Gamma }_K^2$. Substituting into a givesThe enormous change is a tiny-width effect, not a comparably abrupt change in the broad rho contribution. The center-of-momentum energy has moved about $2/(7.4\times 10^{-12})\simeq 2.7\times 10^{11}$ kaon widths from the peak. A width of $7.4\times 10^{-12}$ MeV is only $7.4\times 10^{-6}$ eV, requiring an energy-selection scale of order $10^{-14}$ relative to the several-hundred-MeV beam energy.

Real beams have vastly broader energy distributions. Averaging over such a distribution suppresses the useful narrow-resonance contribution: the finite peak value does not imply a large energy-integrated formation rate, whose resonant area shrinks with the width at fixed branching fractions. The weak inverse-decay entrance coupling and the broad strong-interaction rho background therefore make this a poor way to study kaon resonant formation. The supplied broad-resonance formula is used algebraically as requested; a practical calculation would convolve an appropriate narrow line shape with the actual beam distribution.

The beam energy is a total pion energy, so the momentum follows the relativistic relation $p^2{c}^2=E_{\pi }^2-m_{\pi }^2{c}^4$. Taking the standard charged-pion rest energy $m_{\pi }{c}^2=139.57$ MeV,^[ref]For motion perpendicular to a uniform magnetic field, the Lorentz force supplies the transverse momentum change and gives $r=p/(qB)$. With the stipulated $qB=1$ eV s m${}^{-2}$,The unit cancellation leaves length after dividing by $c$ in metres per second. Equivalently, the usual magnetic-rigidity conversion is $p[\mathrm{GeV}/c]\simeq 0.2998B[\mathrm{T}]r[\mathrm{m}]$ for unit charge. The sign of the pion charge chooses the bending direction, while the radius uses its magnitude. Treating the pion as massless would give about $0.831$ m and is not a good approximation at this modest total energy. The production target does not determine the downstream orbit once the pion momentum and magnetic field are specified.