Past exam of the theoretical physics I course of the University of Cambridge / 2017 / 2

Write $\rho$ for cylindrical radius in this question, not the rod's mass density. A charged relativistic particle in a time-independent magnetic field has the charged-particle electromagnetic Lagrangianwith zero electrostatic potential. The cylindrical speed is ${\mathbf{v}}^2={\dot{\rho }}^2+{\rho }^2{\dot{\theta }}^2+{\dot{z}}^2$. Inside a uniform axial field, choose the symmetric magnetic vector potential $A_{\theta }=B\rho /2$, $A_{\rho }=A_z=0$. Its cylindrical curl gives $B_z={\rho }^{-1}{\partial }_{\rho }(\rho A_{\theta })=B$. HenceIn a field-free region far from either end one can use $A_{\theta }=0$, recovering the requested outside expression. The question's $V=-qB{\rho }^2\dot{\theta }/2$ is a velocity-dependent generalized potential, not electrostatic potential energy.

There is an important end-field qualification. A finite axial field that jumps to zero with no radial component would violate Gauss's law for magnetism. The consistent near-axis idealisation is the solenoid fringe field from an axisymmetric vector potential, $A_{\theta }=\rho B(z)/2$, giving $B_{\rho }=-\rho B^{\prime }(z)/2$ and $B_z=B(z)$. The piecewise Lagrangian is its hard-edge limit and retains fringe impulses through the jump in $A_{\theta }$. Those impulses are essential to the focusing calculation.

The azimuth is a cyclic coordinate, so its canonical momentum is conserved:Initially the velocity is parallel to the axis in a field-free region, so $p_{\theta }=0$. Conservation of canonical angular momentum across a solenoid fringe therefore gives, inside the uniform field,This holds away from the coordinate singularity $\rho =0$; the Cartesian orbit will be smooth there.

Time independence also conserves the energy Hamiltonian,The velocity-linear magnetic term cancels in this Legendre transform. Thus $\gamma ={\gamma }_v=(1-v^2/c^2{)}^{-1/2}$ is constant, as expected because a static magnetic Lorentz force does no work. Mechanical angular momentum can change even though $p_{\theta }$ and the total speed are conserved.

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.

Set $t=0$ just inside the entrance, $z=0$, $\rho ={\rho }_0$ and $\theta ={\theta }_0$. The stated continuity gives $\dot{\rho }(0)=0$. Until the first axis crossing,For the full orbit, including any crossings, use the transverse complex coordinate $u=x+iy$:This helical solenoid trajectory with zero incident angular momentum is a circle of radius ${\rho }_0/2$ about transverse guiding centre $u_0/2$, combined with uniform axial motion. Its cyclotron angular frequency is $|qB|/(\gamma m)=2|\Omega |$, and its axial pitch per turn is $\pi w/|\Omega |$. The speed check is $|\dot{u}{|}^2+w^2={\Omega }^2{\rho }_0^2+w^2=v^2$.

The physical radius is $|u|={\rho }_0|\cos \Omega t|$. When the cosine changes sign, a nonnegative-radius coordinate adds $\pi$ to $\theta$; this is a coordinate change, not a discontinuous turn of the particle. For $q=0$ or $B=0$, the limit is simply $u=u_0$, $z=vt$.

Define the positive focusing frequencyIn the paraxial approximation $w\simeq v$, traversal time is $T\simeq l/v$. The exit values areAcross the exit fringe, $p_{\theta }=0$ and $A_{\theta }\to 0$ remove the azimuthal velocity. The radial velocity remains continuous in the thin-fringe model. Outside, the transverse trajectory is therefore straight along a fixed azimuth, not a continuing helix. Taking the outgoing axial speed also as $v$, the distance to the axis isIt is independent of ${\rho }_0$, so all equal-speed paraxial rays focus to a common point. The condition $0<\omega l/v<\pi /2$ ensures the first focus lies beyond the exit. For a weak lens, $\boxed{f\simeq 4({\gamma }_{v}mv{)}^2/(q^2{B}^{2}l)}$.

The relativistic solenoid magnetic lens focuses either charge sign, while its rotation reverses with $qB$. Keeping the exact radius-dependent $w$, or the exit axial-speed change, produces radius-dependent corrections and spoils an exact common focus. Thus the requested focus is appropriately a paraxial result, with the end fields included even when their axial-speed changes are neglected.