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The transverse transport of a charged-particle beam relative to a reference trajectory. Linear focusing gives betatron oscillations, described by Hill's equation and Courant-Snyder parameters; the beam envelope depends on its geometric emittance.

For a centered beam, the displayed rms area in $(z,z^{\prime })$ coordinates. A matched beam has $⟨z^2⟩=ϵ\beta$, where $\beta$ is one of the Courant-Snyder parameters. Under ideal acceleration, adiabatic damping of geometric emittance decreases this area even when the canonical phase-space area is preserved.

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.

For a real determinant-one map, bounded generic repeated transport requires eigenvalues $e^{\pm i\mu }$, equivalently $|\mathrm{tr}M|<2$. Here both planes have $\mathrm{tr}M=2-L_1{L}_2/f^2$, soFor positive drift lengths and $f>0$, this means $f>\sqrt{L_1{L}_2}/2$. The equalities are marginal and generally produce a Jordan-block growth rather than stable motion. Moving the reference point changes the map by a similarity transformation, $M\to PM{P}^{-1}$, so the eigenvalues and full-cell phase advance are unchanged.

Write the stable map using the Courant-Snyder parameters, also called Twiss parameters:At the chosen reference point $\beta =M_{12}/\sin \mu$, $\alpha =(M_{11}-M_{22})/(2\sin \mu )$, and ${\gamma }_T=-M_{21}/\sin \mu$. Repeated traversal givesIf the full ring contains $N_c$ identical cells, its one-turn phase is $N_c\mu$ modulo $2\pi$; $n$ ring turns use $n{N}_c\mu$. The displayed $M^n$ is for $n$ repetitions of the specified cell.

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.

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