= Solution
For fixed quadrupole fields and fractional <momentum> offset $\delta$, $q(\delta)=q_0/(1+\delta)$. Differentiate $\cos\mu=1-L_1L_2q^2/2$ at zero to obtain $d\mu/d\delta=-L_1L_2q_0^2/\sin\mu=-2\tan(\mu/2)$. Thus
$$
\boxed{\xi_{\mathrm{cell}}=\frac1{2\pi}\frac{d\mu}{d\delta}=-\frac1\pi\tan\frac\mu2.}
$$
Both 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.
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