 Solution

For ultrarelativistic particles with fixed magnet settings, $A(\delta )=A_0(1-\delta )$ and $B(\delta )=B_0(1-\delta )$. Hence the half-trace isOff-energy motion is stable when $|C(\delta )|<1$, separately in each transverse plane. Where $\sin {\mu }_0\ne 0$, its first-order phase shift isFor equal opposite strengths,Thus positive momentum error reduces the focusing and phase advance; negative error moves toward the overfocused unstable boundary. The unexpanded symmetric stability condition is $|f_0|(1+\delta )>d/2$, with $1+\delta >0$, within the ultrarelativistic energy scaling. Its natural chromaticity per cell is $d(\mu /2\pi )/d\delta =-\tan ({\mu }_0/2)/\pi$. Away from the ultrarelativistic limit replace the displayed first-order energy error by $\delta /{\beta }_0^2$. Linearizing at a stability boundary is singular and needs separate treatment.