Past exam of the particle physics course of the University of Oxford / 2022 / 6 / c / Solution

Using the two-frequency approximation in a and $L$ in kilometres, the survival probability isThe slow coefficient is ${\cos }^4(9^{\circ }){\sin }^2(60^{\circ })=0.71374$; the fast one is ${\sin }^2(18^{\circ })=0.09549$. Since ${\sin }^{2}x$ has period $\pi$, the two individual oscillation periods areThe first broad trough is near $46.5$ km and the second near $139.5$ km; the slow component returns to its crest near $93$ km. Small, rapid ripples of depth about $0.0955$ sit on the broad envelope. Averaging over unresolved fast oscillations replaces their squared sine by $1/2$, giving $⟨P_{ee}⟩\simeq 0.95225-0.71374{\sin }^2(\pi L/92.99)$.

There is no single short period for the combined curve. If the supplied ratio $0.003/0.00008=75/2$ is taken as exact, its mathematical common period is $2T_{12}=75T_{23}\simeq 186$ km, outside the plotted range; the approximate physical data do not warrant interpreting that commensurability as exact.

The figure also shows the unmerged expressionFor that comparison take $\Delta m_{23}^2=0.003$ and enforce the identity $\Delta m_{13}^2=\Delta m_{23}^2+\Delta m_{12}^2=0.00308{\mathrm{eV}}^2$, with a consistent sign convention. This illustrates the limitation identified in a: the simplified envelope is useful, but its individual fast ripples need not match the three-frequency result over 150 km. The supplied ${\theta }_{23}$ has no effect on vacuum electron-flavour disappearance.