Past exam of the particle physics course of the University of Oxford / 2019 / 2 / b / Solution

Write $z=\sqrt{ϵ}\omega \cos \psi$, with $\omega =\sqrt{\beta }>0$ and $\psi =\varphi -{\varphi }_0$. Differentiating twice givesThe coefficients must vanish for arbitrary initial phase ${\varphi }_0$. The sine coefficient givesThe conventional Courant-Snyder parameters normalize the phase so that $C=1$; any positive constant $C$ can be absorbed into the scale assigned to $\beta$ and $ϵ$. Explicitly, $\tilde{\beta }=\beta /C$ and $\widetilde{ϵ}=Cϵ$ preserve the orbit while making $\tilde{\beta }{\varphi }^{\prime }=1$. With this convention,The second expression is the Ermakov-Pinney equation for the beam envelope. It follows by substituting ${\varphi }^{\prime }=1/{\omega }^2$ into the cosine coefficient.

The printed equation omits the factor $\omega$ multiplying $k(s)$. For a concrete counterexample, take the constant periodic focusing strength $k=1/16$ and the constant envelope $\omega =2$. Then ${\varphi }^{\prime }=1/4$ and $z=2\sqrt{ϵ}\cos (s/4-{\varphi }_0)$ solve $z^{\prime \prime }+z/16=0$. The correct envelope equation is $0-1/8+(1/16)2=0$, whereas the printed equation gives $-1/8+1/16=-1/16$.

The associated invariant isThus $ϵ$ fixes the phase-space ellipse of a trajectory; a beam's rms geometric emittance uses the corresponding second-moment convention. Factors of $\pi$ depend on whether emittance means ellipse area or area divided by $\pi$ and do not alter the envelope equation.