Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S19 / 1

Use transverse coordinates $(x,x^{\prime })$ and define a positive signed focal length as focusing in that plane. A thin quadrupole magnet of length $\ell$ and signed gradient $G$ has integrated strength $1/f=|e|G\ell /p_0$; the electron charge and chosen field orientation are absorbed into the sign of $G$. Its beam transfer matrix isFor total electron energy $E_0$, $p_0=\sqrt{E_0^2-m_e^2{c}^4}/c$, so $f\simeq E_0/(|e|cG\ell )$ in the ultrarelativistic limit. The other transverse plane has the opposite signed focal length. A quadrupole magnet therefore focuses one plane while defocusing the other; the longitudinal magnetic field component is negligible in this paraxial idealization.

At fixed gradient and magnet length, $f\propto p$. If $E$ is total energy,Here ${\beta }_0=v_0/c$, not one of the Courant-Snyder parameters. For ultrarelativistic electrons this becomes $f\simeq f_0(1+\delta )$, or $1/f\simeq (1-\delta )/f_0$: higher-energy particles are bent and focused less strongly. If the symbol $E$ instead denotes kinetic energy $K$, the coefficient is $(K+m{c}^2)/(K+2m{c}^2)$, approaching $1/2$ in the nonrelativistic limit. Stating the energy convention avoids applying the ultrarelativistic result to a slow beam.

Let $D(d)=\left(\begin{array}{cc}1& d\\ 0& 1\end{array}\right)$ be the drift beam transfer matrix. Start immediately before the first lens and end immediately before its counterpart in the next FODO cell. Matrices act on column vectors from right to left. With $A=1/f_1$ and $B=1/f_2$, where both focal lengths are signed in the chosen plane,The two drifts account for both gaps. Direct multiplication gives $\det M=1$ andFor the equal-strength focusing-defocusing case, $f_1=f>0$ and $f_2=-f$:Choosing a different start within the cell gives a conjugate matrix, with the same trace and eigenvalues. If $f_2$ is specified as a positive magnitude of a defocusing lens, its signed value in these formulas is $-f_2$.

For a real uncoupled, lossless transverse map, $\det M=1$. Bounded nondegenerate oscillatory motion requires complex-conjugate eigenvalues $e^{\pm i\mu }$, equivalentlyIts Courant-Snyder parameters giveThe positive quadratic form ${\gamma }_T{x}^2+2{\alpha }_{T}x{x}^{\prime }+{\beta }_T{x}^{\prime 2}$ is invariant, so repeating the cell cannot produce secular growth. At trace $\pm 2$ a nontrivial Jordan block instead produces linear growth; only the exceptional maps $M=\pm I$ remain bounded at those exact boundaries. The criterion must hold in both transverse planes. For the symmetric FODO cell, both traces are $2-d^2/f^2$, giving

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.