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The canonical momenta areTaking the Legendre transform givesIts four Hamilton equations, in a polar chart with $\rho >0$, areCanonical angular momentum includes the vector-potential contribution and is not simply mechanical angular momentum.

Direct variation of the Lagrangian givesFor a static configuration, $k\rho =qE\cos \theta$ and $qE\rho \sin \theta =0$. Since $qE>0$ and the radius is positive,The magnetic force vanishes at rest and does not shift this equilibrium. In canonical variables it has $p_{\rho }=0$ but $p_{\theta }=qB{\rho }_0^2/2$, rather than zero canonical angular momentum.

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.

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.

An infinitesimal rotation about $x$ gives $\delta {\psi }_y=-ϵ{\psi }_z$ and $\delta {\psi }_z=ϵ{\psi }_y$. Both the squared transverse velocity and squared transverse slope are rotational invariants. Noether's theorem therefore givesDifferentiating and using the two wave equations givesConsequently $L_x=\int {\ell }_{x}dx$ is the angular momentum about the equilibrium string axis, conserved when the net endpoint torque flux vanishes. For a finite segment, $d{L}_x/dt=-[{\mathcal{J}}_x{]}_{\mathrm{left}}^{\mathrm{right}}$. It is not momentum associated with translating the coordinate $x$; its symmetry rotates the two physical transverse displacement components into one another.

The continuous transformation $\theta \mapsto \theta +ϵ$, ${\theta }_B\mapsto {\theta }_B+ϵ$, $z\mapsto z+bϵ$, $r\mapsto r$ moves both endpoints by the same screw motion and preserves the fixed helix. Both $\Delta$ and $u$ are unchanged, so $\delta L=0$. Noether theorem therefore givesThis is the Noether conserved quantity for a mechanical point symmetry applied to helical symmetry. In physical Cartesian terms it equals total $z$ angular momentum plus $b$ times total $z$ linear momentum: the bead contributes $M{a}^2{\dot{\theta }}_B+b(Mb{\dot{\theta }}_B)$. Neither pure axial translation nor pure axial rotation preserves the fixed coil separately. Their conserved combination is a consequence of the helical symmetry, not two independently conserved momenta. Substitution of the four equations of motion verifies $\dot{\mathcal{Q}}=-kar\sin \Delta +k(ar\sin \Delta +bu)-bku=0$.^[ref]

Constraint and example. A holonomic constraint can be written $f_{\alpha }(q_1,\dots ,q_n,t)=0$, with no irreducible dependence on velocities. A velocity relation is holonomic if it integrates to such an equation. For example, a particle constrained to a stationary circle satisfies $x^2+y^2={\ell }^2$, reducing its two Cartesian degrees of freedom to one angle. A nonholonomic constraint cannot be integrated in this way.

Kinematics and kinetic energy. Set $d=R-r>0$. In the plane of the original diagram, choose the orbital angle $\theta$ positive counterclockwise from vertically downward. The marked radius of the inner cylinder then has absolute orientation $\alpha =\theta -\varphi$ measured from the same downward direction. Its signed spin angular velocity is $\Omega =\dot{\theta }-\dot{\varphi }$, whileThe moment of inertia of a uniform solid cylinder is $I=m{r}^2/2$. Applying the kinetic energy decomposition about the center of mass gives

No slip. With tangential unit vector ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$, the inner contact point has velocity $(d\dot{\theta }+r\Omega ){\mathbf{e}}_{\theta }$. The fixed wall requires this to vanish. HenceBecause $r,R$ are constant, integration giveswhere the integration constant is zero with the marked-dot reference in the diagram and continuously unwrapped angles. This is a holonomic constraint, even though it was first obtained from a condition on velocities.

Oscillations. The gravitational potential energy above the lowest position is $V=mgd(1-\cos \theta )$. Eliminating $\varphi$ gives the reduced LagrangianThe Euler-Lagrange equation becomesThus the bottom is a stable equilibrium, and for $|\theta |\ll 1$ there is simple harmonic motion, withThe extra rotational kinetic energy reduces the frequency compared with a particle sliding without friction along the same circular path. The small oscillations inside a circular cylinder formula applies with inertia ratio $\gamma =1/2$.

Friction and energy. Let $F_t$ be the signed tangential contact force in the ${\mathbf{e}}_{\theta }$ direction. Applying Newton's second law to translation and the torque equation to rotation givesUsing the rolling without slipping condition yieldsExcept at special instants, static friction is needed to produce the spin acceleration. There is no sliding at the contact point, whose instantaneous velocity is zero; therefore this force does no total work. Its translational and rotational powers cancel: $F_{t}d\dot{\theta }+F_{t}r\Omega =0$. The stationary normal reaction likewise does no work. With a rigid wall, adequate static friction, and no rolling resistance, conservation of energy gives

Angular momentum. The cylinder's spin angular momentum $I\Omega$ is generally not conserved because the contact friction exerts a torque about its center of mass. About the fixed outer-cylinder center $O$, the angular momentum decomposition about the centre of mass givesGravity and friction exert the external torqueThus the rolling cylinder does not generally conserve angular momentum, although it conserves mechanical energy. There is a special geometric cancellation at $R=3r/2$: then $L_O$ and its torque vanish identically for this planar motion. Angular momentum of a larger isolated system including the wall and Earth is a separate conservation statement.

Scope of the holonomic reduction. The integrated constraint is exact at every angle for planar rolling without slipping with sustained contact; it is not restricted to small oscillations. Completing a full turn requires tracking angle winding rather than reducing each angle separately modulo $2\pi$. This global bookkeeping does not make the local velocity relation nonintegrable.

However, arbitrary initial conditions need not sustain this ideal rolling motion. The inward normal reaction and the required friction areContact requires $N\ge 0$, and a finite coefficient ${\mu }_s$ of static friction requires $|F_t|\le {\mu }_{s}N$. If either fails, the cylinder detaches or slips and the one-coordinate rolling Lagrangian no longer applies. Slipping contact allows $\theta$ and $\varphi$ to be independent; detachment removes the fixed center-distance constraint. These changes alone do not establish a nonholonomic constraint. Allowing general three-dimensional reorientation can introduce nonintegrable rolling constraints, but that motion is outside the diagram's two-angle planar model. Thus the stated planar rolling constraint is holonomic throughout its regime of validity; it cannot be imposed on every physically possible motion.

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.

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.

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.

Let $\Delta ={\Omega }_0-\omega$ and $R(t)=e^{-i\omega t{S}_z/\hslash }$. Differentiating $|\Psi ⟩=R|{\Psi }_R⟩$ gives the rotating-frame Hamiltonian operator $R^{†}HR-i\hslash R^{†}\dot{R}$. The angular momentum commutator yields $R^{†}{S}_{\pm }R=e^{\pm i\omega t}{S}_{\pm }$, while $-i\hslash R^{†}\dot{R}=-\omega S_z$. Thus the rotating-frame Rabi Hamiltonian isThis transformation is exact for the circular drive; no rotating-wave approximation has been used.

The angular momentum commutators used in the frame transformation are representation-independent. Therefore $H_R=\Delta S_z+{\Omega }_R{S}_x$ still holds. Rotating its axis onto $z$ gives $2s+1$ eigenvalues and exact laboratory solutionsNow $R(T)=(-1{)}^{2s}I$, so $e^{i{\Phi }_{\mu }}=(-1{)}^{2s}{e}^{-i\mu \Omega T}$. In the adiabatic limit the dynamical phase is $-\mu BT$ andIndeed $2\pi s+2\pi \mu \cos \theta$ and $-2\pi \mu (1-\cos \theta )$ differ by $2\pi (s+\mu )$, an integer multiple of $2\pi$. Integer spins return with a positive rotation sign and half-integer spins with a negative one; the geometric phase need not vanish for either.

Use $[a_i,a_j^{†}]={\delta }_{ij}$ and define $E_{ij}=a_i^{†}{a}_j$. A direct multiplication givesWith $S_{+}=\hslash a_{\uparrow }^{†}{a}_{\downarrow }$, $S_{-}=\hslash a_{\downarrow }^{†}{a}_{\uparrow }$, this implies $[S_{+},S_{-}]=2\hslash S_z$ and $[S_z,S_{\pm }]=\pm \hslash S_{\pm }$. Resolving the raising and lowering combinations givesThese operators realize the Schwinger boson representation of angular momentum. Each commutes with $\hat{N}=n_{\uparrow }+n_{\downarrow }$, so fixed total occupation is an invariant subspace.

On periodic ring states the angular-momentum index $m$ is an integer. At the stated resonance,so $U_T$ is the identity on every orbital state. After $N$ steps from the normalized ground state,Its phase varies in space but its position probability density remains uniform. Periodicity permits integration by parts, givingConsequentlyEquivalently, the angular-momentum probabilities are $P_m=J_m(N\eta {)}^2$, with ${\sum }_m{m}^2{J}_m(z{)}^2=z^2/2$. The mean angular momentum remains zero while its variance grows as $N^2$. This quantum resonance of the kicked rotor gives coherent ballistic energy growth. The Bessel second-moment identity connects the momentum spreading to that quadratic energy law; an incoherent random walk would instead give linear energy growth.

The continuous internal symmetry is local $U(1)$ gauge invariance. With the source's gauge covariant derivative convention it acts asso $D_{\mu }\phi$ transforms like $\phi$ and $F_{\mu \nu }$ is unchanged. Constant phases are its global subgroup, not an additional independent internal rotation group.

In flat four-dimensional space-time the continuous external symmetry is the Poincaré group: four translations and six Lorentz transformations. Noether's theorem associates translations with conserved energy and momentum, spatial rotations with angular momentum, and boosts with the relativistic center-of-energy/boost charges. In tensor notation the latter two are the six components of the conserved total angular-momentum tensor $M^{\mu \nu }$, including field spin contributions where appropriate. The nonzero dimensionful parameter ${\mu }^2$ prevents exact scale invariance. Discrete transformations, such as parity, are outside the requested continuous-symmetry list.

The action is dimensionless and $[d^{4}x]=-4$, so $[\mathcal{L}]=4$. A canonical scalar kinetic term then requires $2+2[\varphi ]=4$, and the electromagnetic kinetic term requires $2[F_{\mu \nu }]=4$. ThusThe brackets here denote mass dimension. The Lagrangian has no explicit coordinate dependence, and all indices are contracted as Lorentz scalars. Its continuous spacetime symmetry is the Poincaré group. Noether's theorem gives conserved four-momentum from translations and conserved angular momentum, including the boost generators, from Lorentz transformations. The nonzero mass $M$ and dimensionful couplings prevent a dilation symmetry.

Sign qualification: the printed electromagnetic kinetic term has the opposite healthy relative sign to the scalar kinetic terms. With the stated mostly-minus Minkowski metric, a physical Maxwell Lagrangian is $-F_{\mu \nu }{F}^{\mu \nu }/4$. The printed plus sign instead gives photon Hamiltonian density $-({\mathbf{E}}^2+{\mathbf{B}}^2)/2$; a nonzero purely electric configuration already has negative energy, unbounded below as its amplitude grows. The physical decay widths below use the healthy Maxwell sign with the given interaction normalization. The dimensional and spacetime-symmetry conclusions themselves do not depend on this sign correction.

The rovibrational spectroscopy branch in which absorption lowers the total rotational angular momentum quantum number by one. A label $P(J)$ conventionally specifies the lower-state value $J$.

Let $p$ be the perpendicular distance from the force centre to the tangent of a regular planar orbit. Conserved physical angular momentum has magnitude $|h|=mpv$, while conserved energy is $E=m{v}^2/2+V(r)$. Eliminating speed gives the displayed equation when $h\ne 0$. For a differentiable polar curve, $p=r^2/\sqrt{r^2+(dr/d\theta {)}^2}$, connecting the geometric tangent distance to the radial energy equation.

A rigid rotor restricted to one angular coordinate $\varphi$. Single-valued spatial states $e^{im\varphi }$ have $m\in \mathbb{Z}$ and energy $E_m={\hslash }^2{m}^2/(2I)$, with moment of inertia $I$. The zero-angular-momentum level has state degeneracy one; every nonzero $|m|$ has two directions. For labelled molecular ends its high-temperature rotational partition function is $\sqrt{\pi T/{\theta }_{\mathrm{rot}}}$, where ${\theta }_{\mathrm{rot}}={\hslash }^2/(2I{k}_{\mathrm{B}})$. Equivalent orientations and nuclear spin statistics require a further molecular counting convention. One quadratic angular momentum gives rotational heat capacity $k_{\mathrm{B}}/2$ per molecule, rather than $k_{\mathrm{B}}$ for an unrestricted linear rigid rotor.