Past exam of the theoretical physics I course of the University of Cambridge / 2015 / Paper 1 / 2

Every element of the uniformly distributed mass has speed squared ${\dot{r}}^2+r^2{\dot{\varphi }}^2$, so the total kinetic energy is $m({\dot{r}}^2+r^2{\dot{\varphi }}^2)/2$. The extension of the ring's full circumference is $2\pi (r-r_0)$, and the stated spring constant therefore givesThe magnetic coupling for each charge element is $dq\mathbf{A}\cdot \mathbf{v}$. With the symmetric gauge $\mathbf{A}=(Br/2)\hat{\mathbf{\varphi }}$ and tangential speed $r\dot{\varphi }$, integration over the ring gives $qB{r}^2\dot{\varphi }/2$. ThusThis idealization neglects charge self-energy, radiation and deformation away from a circular ring. The elastic constant refers to extension of the whole circumference; replacing it by a radial spring constant without the factor $4{\pi }^2$ changes the problem.

The canonical momenta areThe Legendre transform cancels the velocity-linear magnetic term, giving the energy expressed in velocities,But Hamilton's equations differentiate with respect to canonical coordinates and momenta, not velocities. In its proper canonical variables the same Hamiltonian isIt explicitly contains $B$ because canonical angular momentum differs from mechanical angular momentum. This reconciles magnetic forces with the absence of magnetic work: the magnetic field can exchange radial and tangential motion without changing the total mechanical energy.

Time-translation symmetry conserves the energy $H$, and axial rotational symmetry makes $\varphi$ a cyclic coordinate, conservingThe radial Euler-Lagrange equation before elimination isInsert $\dot{\varphi }=J/(m{r}^2)-qB/(2m)$. The cross terms linear in $JqB$ cancel, leavingEquivalently use the reduced radial potentialThe last term is constant at fixed $J$ and has no radial derivative, but is part of the full conserved energy. Conservation of $J$ is stated in the rotationally symmetric gauge; the conserved mechanical combination can be written in terms of magnetic flux in a gauge-independent way.

The initial angular velocity cancels the flux term in the canonical angular momentum:Define ${\omega }_c=qB/m$ and ${\omega }_s^2=4{\pi }^{2}k/m$. The radial equation becomes $\ddot{r}+{\Omega }^{2}r={\omega }_s^2{r}_0$, soThese identify the paper's $\omega$ with $\Omega$. For nonzero radius, $J=0$ also fixesA radius is nonnegative. The formal sinusoidal solution remains strictly positive for a full cycle only if $2r_e>r_0$, or ${\omega }_s^2>{\omega }_c^2/4$. At equality it touches zero; at stronger magnetic coupling it first reaches zero and its negative continuation is not a physical ring radius. No collision or re-expansion rule at collapse is supplied, so the circular-ring solution is then valid only up to the first zero. This domain qualification leaves the requested differential-equation solution intact.

Take the stated negligible-natural-radius limit as $r_0=0$, with both initial velocities zero at $r=r_1$. Then $J=qB{r}_1^2/2$. Write ${\omega }_c=qB/m$ and ${\Omega }^2={\omega }_s^2+{\omega }_c^2/4$, where ${\omega }_s^2=4{\pi }^{2}k/m$. At two radial turning points $x=r^2$, equality of the effective radial energies givesUsing $x_{\max }=r_1^2$ gives, for $qB\ne 0$ and $k>0$,The minimum is a regular radial turning point, soThe angular velocity has the sign of $qB$. On the initial inward motion, conservation of canonical angular momentum produces the rotation needed to prevent collapse to zero radius.

The strictly zero-field case is singular for this angular formula: $J=0$ and the ring instead collapses to $r=0$ with inward radial speed ${\omega }_s{r}_1$ in the ideal radial equation, where an angular coordinate ceases to describe a ring. The limit $qB\to 0$ through nonzero fields gives a tiny positive turning radius with large angular velocity, and must not be substituted into the formula at exactly zero field. A small nonzero $r_0$ also matters if the predicted minimum becomes comparable with it.