Past exam of the theoretical physics I course of the University of Cambridge / 2022 / 1

Use the angular coordinate $\theta$ for the position on the ring. The kinetic energy is $T=m{a}^2{\dot{\theta }}^2/2$. The gravitational potential energy is constant because the ring is horizontal, and the ideal constraint force has no tangential component. Thus the Lagrangian has no dependence on $\theta$ itself: it is a cyclic coordinate.

The Euler-Lagrange equation givesEquivalently, the conserved conjugate momentum $p_{\theta }=m{a}^2\dot{\theta }$ is the vertical angular momentum, as required by rotational symmetry.

Let $\theta$ and $\phi$ be the lower and upper azimuthal angles, measured from a common fixed direction. The squared spring length isBy Hooke's law, a spring with zero natural length has potential energy $k{d}^2/2$. ThereforeConstant gravitational energies can be omitted. It is useful to define the moments of inertia $I_1=m{a}^2$, $I_2=M{A}^2$ and the angular coupling $\kappa =kaA$. The part of the potential that affects motion is simply $-\kappa \cos (\theta -\phi )$. In particular, $h$ changes the constant spring energy but not the angular equations, because the spring has zero natural length.

The two continuous, time-independent symmetries are time translation and a common rotation of both angles. Their Noether charges are respectively the total energy and vertical angular momentum:For the second symmetry, $\theta \mapsto \theta +ϵ$ and $\phi \mapsto \phi +ϵ$, so the Noether theorem gives the sum of their conjugate momenta. Independent rotations are not symmetries of the spring coupling.

There are also discrete symmetries: simultaneous reflection $(\theta ,\phi )\mapsto (-\theta ,-\phi )$, and time reversal in classical mechanics, which reverses both angular velocities. These leave the dynamics invariant but provide no additional infinitesimal Noether charges. Combined with common rotations, the reflections give the full planar orthogonal symmetry of the angular geometry. If $I_1=I_2$, interchange of the two angular coordinates is an additional discrete symmetry of the reduced dynamics.

There is a further useful time-dependent symmetry, up to a total derivative, on a continuous lift of the angles: adding the same constant angular velocity, $(\theta ,\phi )\mapsto (\theta +vt,\phi +vt)$. It changes $L$ byIts conserved quantity is $\boxed{I_1\theta +I_2\phi -tJ}$. It is the integration constant for free common rotation, rather than a new globally single-valued angle charge on the configuration torus. The globally defined continuous charges are $E$ and $J$ above.

The angular Euler-Lagrange equations areA static equilibrium point has both angular velocities zero and $\sin (\theta -\phi )=0$. Hence the complete equilibrium families, with angles taken modulo $2\pi$, areThe first family is aligned in azimuth and minimizes the spring length; the second is antipodal and maximizes it. Their potentials are $k[(a-A{)}^2+h^2]/2$ and $k[(a+A{)}^2+h^2]/2$, respectively. Common rotations generate a whole family rather than isolated equilibria.

Let ${\delta }_0={\theta }_0-{\phi }_0$ be $0$ or $\pi$, and write the angular displacements as ${\eta }_1,{\eta }_2$. To quadratic order the restoring potential energy is $\kappa \cos {\delta }_0({\eta }_1-{\eta }_2{)}^2/2$. The linearized normal mode equation isThe determinant gives ${\omega }^2[{\omega }^2{I}_1{I}_2-\kappa \cos {\delta }_0(I_1+I_2)]=0$. Thus the normal modes of two spring-coupled coaxial rotors areFor the aligned family, the relative mode oscillates with positive frequencyFor the antipodal family, ${\omega }_{\mathrm{rel}}^2=-{\Omega }^2$, so that displacement has solutions $e^{\pm \Omega t}$ and the equilibrium is unstable.

The common zero mode has ${\eta }_1={\eta }_2=C+Dt$. It describes a neutral displacement or uniform common rotation, not a restoring oscillation. Thus the aligned family is stable with respect to relative motion and has orbital stability modulo common rotations. A particular static point is not strictly Lyapunov stable against unrestricted small velocity perturbations: arbitrarily small nonzero $J$ eventually moves the common angle far from that point. With $J=0$, the common angle is fixed and the aligned relative equilibrium point is stable. This distinguishes the symmetry-induced drift from the exponential instability of the antipodal family.

A common rotation changes neither the separation nor the spring energy. Its infinitesimal direction is $(1,1)$, exactly the zero-frequency normal mode. Thus the continuum of rotated equilibrium points requires that zero mode; a nonzero restoring frequency in that direction would contradict rotational symmetry.

The second mode has $I_1{\eta }_1+I_2{\eta }_2=0$, so it changes the relative angle while leaving the inertia-weighted common angle unchanged. It therefore carries zero perturbation in the conserved angular momentum associated with common motion. Indeed the coordinatesseparate the Lagrangian exactly:The cyclic coordinate $q$ is free, while $\delta$ has restoring or inverted curvature. This also explains why all members of either equilibrium family have the same frequencies. Reflection symmetry makes the potential even in $\delta$, and the stable oscillatory mode is consistent with time reversal in classical mechanics through its two opposite frequency signs.