 Solution

For fixed quadrupole fields and fractional momentum offset $\delta$, $q(\delta )=q_0/(1+\delta )$. Differentiate $\cos \mu =1-L_1{L}_2{q}^2/2$ at zero to obtain $d\mu /d\delta =-L_1{L}_2{q}_0^2/\sin \mu =-2\tan (\mu /2)$. ThusBoth transverse planes have the same natural result; the ring value is $N$ times this. It vanishes only in the zero-focusing limit, outside robust stable operation. Additional correcting elements, such as sextupoles where dispersion is nonzero, are required for zero corrected chromaticity.