 Solution

In a uniform-field region, $z$ is cyclic and $\gamma$ is constant. Its equation is $d(\gamma m\dot{z})/dt=0$. The radial Euler-Lagrange equation isso with $\dot{\theta }=-\Omega$ and $qB/(\gamma m)=2\Omega$,The radial equation applies on intervals with a regular signed cylindrical-coordinate description; the Cartesian solution handles axis crossings.

In a field-free region, the equations areFor this particle $p_{\theta }=0$, so $\dot{\theta }=0$ and $\ddot{\rho }=0$ outside, until a coordinate change is needed at an axis crossing.

Axial velocity is constant within each uniform region, but not across the ends. In a smooth fringe,The entrance develops an azimuthal speed $|\Omega |{\rho }_0$ while radial velocity remains zero. Conservation of total speed givesForward passage presumes the radicand is positive. The exit fringe changes axial velocity again. Neglecting these small changes is a paraxial approximation, requiring $|\Omega |{\rho }_0\ll v$, rather than an exact consequence of cylindrical symmetry.