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Measure polar angle $\theta$ from the downward vertical on the lower hemisphere of radius $R$. A particle released from rest at the rim and sliding without energy loss has speed $\sqrt{2gR\cos \theta }$. The spherical arc length along $\varphi =\varphi (\theta )$ is $R\sqrt{1+{\sin }^2\theta {\varphi }^{\prime 2}}|d\theta |$. Thus its travel time has integrand proportional to $\sqrt{(1+{\sin }^2\theta {\varphi }^{\prime 2})/\cos \theta }$. Since the azimuth is a cyclic coordinate, the Euler-Lagrange equation givesSolving for ${\varphi }^{\prime 2}$ gives the displayed first integral on a segment where $\theta$ is a valid path parameter. For finite $C$, approaching the rim $\theta \uparrow \pi /2$ gives ${\varphi }^{\prime }\to 0$. Thus a regular stationary path begins along the meridian toward the bottom. This is an initial direction condition; a path joining other rim points can turn before reaching the bottom. Monotone segments use positive $|d\theta |$ in the time integral, and a turning point requires another path parameter or a split into segments.

Measure $\theta \in [-\pi /2,\pi /2]$ from the vertically downward direction on the lower half of a cylinder of radius $R$, and let $z$ be axial distance. From rest at an edge, conservation of energy gives speed $\sqrt{2gR\cos \theta }$. The travel time is $T=(2gR{)}^{-1/2}\int \sqrt{R^2+z^{\prime 2}}/\sqrt{\cos \theta }d\theta$. Since $z$ is a cyclic coordinate, its conjugate quantity $z^{\prime }/[\sqrt{\cos \theta }\sqrt{R^2+z^{\prime 2}}]=K$ is constant, giving the displayed formula. For finite positive axial separation, $0<K<1$ is fixed uniquely by integrating $z^{\prime }$ between the edges: the separation grows strictly from zero to infinity as $K$ increases from zero to one. The time integrand is strictly convex in $z^{\prime }$, so its supporting-line inequality proves global minimality for the fixed endpoints. As the starting edge is approached, $z^{\prime }\to 0$: the initial direction is directly down the curved surface, with no axial component. This differs from the constant-pitch shortest geodesic on the unrolled cylinder.

For a massive test particle with electric charge $e$, choose the electromagnetic four-potential $A_t=-Q/r$ and proper time $\tau$. The cyclic coordinates $t,\varphi$ in the equatorial Lagrangian give the conserved quantities $E=mF\dot{t}+eQ/r$ and $l=r^2\dot{\varphi }$. Here $l$ is angular momentum per unit mass. Substituting these quantities into the four-velocity normalization $g(U,U)=-1$ givesThe energy includes the gauge-dependent electrostatic term; the displayed convention fixes that gauge by taking the potential to vanish at infinity.

Let a planar Lagrangian have the form $L=a(r){\dot{r}}^2+b(r){\dot{\theta }}^2+c(r)$, with smooth coefficients and $b(r)>0$. Its cyclic coordinate $\theta$ has constant conjugate momentum $p_{\theta }=2b(r)\dot{\theta }$. For a circular orbit $r=R$, the radial Euler-Lagrange equation reduces to $b^{\prime }(R){\dot{\theta }}^2+c^{\prime }(R)=0$. If $b^{\prime }(R)\ne 0$, eliminating $\dot{\theta }$ gives the displayed condition. A real orbit requires $-c^{\prime }(R)/b^{\prime }(R)\ge 0$; the radial kinetic coefficient $a$ does not enter this constant-radius condition, although it affects radial perturbations. If $b^{\prime }(R)=0$, the original condition must be used rather than divided by zero.

For a positive refractive index depending only on $x$, Fermat principle applied to $\int \mu (x)\sqrt{1+y^{\prime 2}}dx$ has $y$ as a cyclic coordinate. The Euler-Lagrange equation therefore makes the displayed quantity constant along a stationary ray. When ${\mu }^2>p^2$, its slope is $y^{\prime }=p/\sqrt{{\mu }^2-p^2}$. If $\theta$ is the angle to the $x$ axis, this is conservation of $\mu \sin \theta$, the stratified-medium form of Snell's law. For fixed endpoint values, the integrand has second derivative $\mu (1+y^{\prime 2}{)}^{-3/2}>0$ with respect to $y^{\prime }$. Its strict convexity and the constant momentum show, by the tangent-line inequality and integration, that any such ray is the unique minimum among competing graph paths with the same endpoints.

For a uniform solid cylinder of mass $M_c$ and a frictionlessly sliding wedge of mass $M_b$, take the horizontal wedge position $X$ and uphill contact coordinate $s$. Contact and rolling without slipping leave two degrees of freedom. Up to a constant, the Lagrangian is $L=\frac{1}{2}(M_b+M_c){\dot{X}}^2+M_c\cos \alpha \dot{X}\dot{s}+\frac{3}{4}{M}_c{\dot{s}}^2-M_{c}gs\sin \alpha$.

The cyclic coordinate $X$ gives $(M_b+M_c)\ddot{X}+M_c\cos \alpha \ddot{s}=0$. The other Euler-Lagrange equation is $\cos \alpha \ddot{X}+(3/2)\ddot{s}=-g\sin \alpha$. Elimination proves the displayed acceleration, with $\ddot{X}=-M_c\cos \alpha \ddot{s}/(M_b+M_c)$. For $M_b\gg M_c$, the cylinder approaches the fixed-incline acceleration $-(2/3)g\sin \alpha$.

Set the inertial cylindrical angle to $\Phi =\varphi +\omega t$. The Minkowski metric becomes $d{s}^2=(c^2-{\omega }^2{r}^2)d{t}^2-2\omega r^{2}dtd\varphi -d{r}^2-r^{2}d{\varphi }^2-d{z}^2$. For a massive geodesic parametrized by proper time, its cyclic coordinates give conserved covariant four-momentum components $p_t=m[(c^2-{\omega }^2{r}^2)\dot{t}-\omega r^2\dot{\varphi }]$, $p_{\varphi }=-m{r}^2(\dot{\varphi }+\omega \dot{t})$ and $p_z=-m\dot{z}$. Their combination gives $\dot{t}=(p_t-\omega p_{\varphi })/(m{c}^2)$. The radial Euler-Lagrange equation gives the displayed coordinate acceleration, although the free worldline is straight in inertial coordinates. The conserved inertial energy is $p_t-\omega p_{\varphi }$, and the inertial axial angular momentum is $-p_{\varphi }$.

By spherical symmetry, choose the orbital plane to be $\theta =\pi /2$. The geodesic Lagrangian of the Schwarzschild metric isFor a massive test particle, dots denote proper time derivatives and $2\mathcal{L}=-c^2$. The cyclic coordinates $t$ and $\varphi$ give, through the Euler-Lagrange equations, two constants:At infinity, $\mathcal{E}={\gamma }_{\infty }=1/\sqrt{1-v_{\infty }^2/c^2}$ is the Lorentz factor. The constant $\ell$ is the specific angular momentum, and the energy per unit rest mass is $E_{\infty }/m=c^2{\gamma }_{\infty }$. The normalization of the four-velocity givesUse $dr/d\varphi =\dot{r}{r}^2/\ell$ and define the energy-normalized angular momentum for a Schwarzschild geodesicThenFor an incoming massive test particle with impact parameter $b$, $L_{\infty }=m{\gamma }_{\infty }{v}_{\infty }b$, so $J=b{v}_{\infty }$. In particular, $J$ is not the same as the specific angular momentum $\ell$ unless ${\gamma }_{\infty }=1$.

A photon has no proper time parametrization and no finite Lorentz factor. For it, use an affine parameter in the same geodesic Lagrangian and impose $2\mathcal{L}=0$. The ratio $J=c^2{L}_{\infty }/E_{\infty }=bc$ remains well defined, and the null result isEquivalently this is the ultrarelativistic limit of the massive formula at fixed $J$, with $1/{\gamma }_{\infty }^2\to 0$. The printed massive formula needs this interpretation before it is applied to a photon.

On the equatorial path, the Lagrangian per unit mass isBoth $t$ and $\varphi$ are cyclic coordinates. The Euler-Lagrange equations therefore conserve their canonical momenta:ConsequentlyHere $E$ is the total conserved energy in the specified gauge of the electromagnetic four-potential. The paper's $l$ is the angular momentum per unit mass; the physical angular momentum is $ml$. The equatorial choice is consistent with the $\theta$ Euler-Lagrange equation, since its angular force term contains $\sin \theta \cos \theta =0$ at $\theta =\pi /2$ and the electromagnetic four-potential has no angular component.

For an arbitrary smooth variation $q_i\mapsto q_i+\epsilon {\eta }_i$ vanishing at both endpoints,The endpoint term in integration by parts is zero. stationarity for every variation and the fundamental lemma of the calculus of variations imply the Euler-Lagrange equations $d(L_{{\dot{q}}_i})/dt=L_{q_i}$. Least action here means stationary action, not necessarily a global minimum. The conjugate momentum is $p_i=L_{{\dot{q}}_i}$; a cyclic coordinate, $L_{q_i}=0$, makes $p_i$ constant.

For the symmetric top, $\psi$ and $\varphi$ are cyclic. Thusare constant, and consequently so is ${\omega }_3$. Conservation of energy from time-translation invariance also supplies the conserved energyThese give the requested additional conserved quantities $p_{\varphi }$ and $E$. The $\theta$ equation isFor constant $\theta$ and constant precession rate $\dot{\varphi }$, it reduces toConversely, choose constants satisfying this balance and set $\dot{\psi }={\omega }_3-\dot{\varphi }\cos \theta$. The cyclic equations and $\theta$ equation are then all satisfied, so this constructs uniform precession wherever the Euler-angle chart is regular.

Take $\theta$ as the inclination of the symmetry axis to the upward vertical, $\varphi$ as its azimuth, and $\psi$ as rotation of the body about that axis relative to the line of nodes. The kinetic energy and potential energy areThe cyclic coordinates $\psi ,\varphi$ and time independence give three conserved quantities:Their meanings are the component of angular momentum along the body axis, the component along the vertical, and the total energy.

For a vanishingly small perturbation of the upright spinning state, $L\simeq Cn$, $p_{\varphi }\simeq L$, and $E\simeq L^2/(2C)+mg\ell$. Eliminating $\dot{\varphi }$ gives the limiting effective potentialIts quadratic change near zero is $\{L^2/(8A)-mg\ell /2\}{\theta }^2$. Thus if $C^2{n}^2>4mg\ell A$, zero is a strict local minimum and a sufficiently small perturbation remains near it.

In the unstable case, the limiting energy equation becomesAt the nonzero turning point the bracket vanishes, yieldingThe approximation is the small-perturbation limit, not an assertion that arbitrary disturbances preserve the upright state's exact conserved quantities.

A Hamiltonian system on a $2d$-dimensional symplectic manifold is Liouville integrable when it has $d$ first integrals, including the Hamiltonian, whose differentials are independent on a dense regular region and which pairwise have zero Poisson bracket. The Arnold-Liouville theorem states that a compact connected regular common level set is a $d$-torus; near it there are action-angle variables $(I,\theta )$ in which the Hamiltonian depends only on $I$ and the motion is $\dot{I}=0$, $\dot{\theta }={\nabla }_{I}H$. Without compactness, the torus conclusion needs modification.

Here $q_1$ is a cyclic coordinate, so $p_1=\ell$ is conserved and $\{H,p_1\}=0$. The Hamilton equations areThe differentials $dH,d{p}_1$ are independent away from the circular-orbit critical set, establishing integrability. Put $r=q_2$; on a fixed angular-momentum level,For a bounded noncollision orbit require $\ell \ne 0$ and $-k^2/(2{\ell }^2)<E<0$. Its two turning radii areThey are positive and $\alpha <\beta$, and $p_2^2=(-2E)(r-\alpha )(\beta -r)/r^2$. In terms of $u=1/r$ and the angular coordinate, eliminating time gives $u^{\prime \prime }+u=k/{\ell }^2$. ThusAfter an angular change of $2\pi$ the phase-space state returns to itself, and the elapsed time is finite because $r$ is bounded away from zero and infinity. This proves periodicity; a torus alone would only guarantee quasiperiodic motion. At $E=-k^2/(2{\ell }^2)$ the orbit is circular and remains periodic. For $E\ge 0$ the allowed radial motion is unbounded. The collision case $\ell =0$ reaches the excluded boundary $r=0$ and is not a periodic orbit in the given phase space.

With the usual action convention $(2\pi {)}^{-1}\oint pdq$, the angular cycle gives $I_1=\ell$, and the two radial legs giveFor completeness, substitute $r=(\alpha +\beta )/2+(\beta -\alpha )\cos \theta /2$ to evaluate the integral as $\pi [(\alpha +\beta )/2-\sqrt{\alpha \beta }]$. Therefore $I_2=k/\sqrt{-2E}-|\ell |$ and $H=-k^2/[2(I_2+|I_1|{)}^2]$. The two frequencies have equal magnitude, consistent with the closed Kepler orbits. The circular limit has $I_2=0$ and $\alpha =\beta$.

Noether theorem associates a conserved quantity ${\sum }_i{p}_i\delta q_i$ to a continuous symmetry of the Lagrangian (with the usual boundary-term correction if the Lagrangian changes by a total derivative). The angles $\psi$ and $\varphi$ are cyclic coordinates, so for the heavy symmetric topTime-translation symmetry gives conserved energy, whose Hamiltonian expression isIt is quadratic in the momenta; $p_{\theta }=I_1\dot{\theta }$. Eliminating the cyclic velocities and putting $\mu =\cos \theta$, so ${\dot{\mu }}^2={\sin }^2\theta {\dot{\theta }}^2$, yieldsThe first term has degree three and the second degree two, displaying the cubic governing nutation.

The internal orientation ${\alpha }_0$ is a periodic collective coordinate with period $2\pi$. Its conjugate momentum is the integrated canonical charge. The Bohr-Sommerfeld quantization for a positively oriented circuit of this angle is, in units $\hslash =1$,There is no oscillator turning point and hence no half-integer Maslov correction for this cyclic coordinate. Equivalently, a wavefunction $e^{i{Q}_{\theta }{\alpha }_0}$ must be single valued. At $\theta =0$, the allowed signed charges and leading semiclassical masses in the $T=1$ family are thereforeThe signed integers include both rotation directions and the nonrotating $\ell =0$ state. Every finite integer gives $|\omega |<m$.

For nonzero theta it is still the canonical charge that is integral, while the mechanical charge is shifted:For the stated family set $T=1$; the reversed kink gives the corresponding $T=-1$ family. This is the semiclassical charged-kink spectrum with a theta angle. One may equivalently describe the spectrum by fractional mechanical charges instead of integral canonical charges, but the two conventions must not be mixed. The theta periodicity follows by relabeling the integer when $\theta \mapsto \theta +2\pi$. These are leading semiclassical energies; the quantization argument does not assert that quantum fluctuation corrections vanish.

The canonical momentum is $p=L_{\dot{q}}$, and a regular Legendre transform in mechanics defines $H(q,p,t)=p\dot{q}-L(q,\dot{q},t)$ after solving for $\dot{q}$. Its differential isbecause the $d\dot{q}$ terms cancel. With the Euler-Lagrange equation $\dot{p}=L_q$, this gives Hamilton's equationsIn particular $H_q=-L_q$ at corresponding velocities, so a cyclic coordinate for the Lagrangian is also cyclic for the Hamiltonian.

The principle of stationary action, also called Hamilton's principle, requires $\delta {\int }_{t_1}^{t_2}Ldt=0$ for fixed-endpoint path variations. Here the original PDF has ${\dot{r}}^2$ in the radial kinetic energy; the TeX has lost that dot. With $m>0$ and $r>0$, the Lagrangian isThe Euler-Lagrange equation for each coordinate givesThe two conserved quantities are angular momentum $h=m{r}^2\dot{\varphi }$ and energy $E=\frac{m}{2}{\dot{r}}^2+\frac{m}{2}{r}^2{\dot{\varphi }}^2+km{r}^2$. By Noether's theorem, they arise respectively from rotational invariance and time-translation invariance of the action. The coordinate $\varphi$ is a cyclic coordinate, and $L$ has no explicit time dependence.

The conjugate momentum for each coordinate is $p_r=m\dot{r}$, $p_{\varphi }=m{r}^2\dot{\varphi }$. The Legendre transform $H=p_r\dot{r}+p_{\varphi }\dot{\varphi }-L$ gives the HamiltonianHamilton's equations areSetting $p_{\varphi }=h$ yields $m\ddot{r}=-V_{\mathrm{eff}}^{\prime }(r)$ with the effective potential $V_{\mathrm{eff}}=h^2/(2m{r}^2)+km{r}^2$.

For $k>0$ and $h\ne 0$, this potential tends to infinity both as $r\downarrow 0$ and as $r\to \infty$, and has its unique stationary point atThe effective potential stability criterion proves a stable circular orbit in a quadratic central potential, with $|\dot{\varphi }|=\sqrt{2k}$. For a small radial displacement $\eta$, $\ddot{\eta }+8k\eta =0$ to first order, so radial perturbations oscillate rather than grow. The diagram shows the centrifugal and quadratic contributions and their sum in dimensionless units:

The printed paper only says that $k$ is constant. Its claimed positive-radius stable circular orbit needs the additional attractive-force assumption $k>0$ and nonzero angular momentum. For $k<0$ there is no such minimum; for $k=0,h\ne 0$ the effective potential is strictly decreasing. With $h=0,k>0$, the minimum is at the origin and is not a positive-radius circular orbit. These cases are not represented by the diagram.

The Euler-Lagrange equations areThe coordinate $\varphi$ is a cyclic coordinate, so its conjugate momentumis conserved. Substitution into the first equation givesMultiplying by $2\dot{\theta }$ shows thatThus $h$ is conserved. Since $l^2/{\sin }^2\theta ={\sin }^2\theta {\dot{\varphi }}^2$, $h$ is the squared speed on the unit sphere, equivalently twice the kinetic energy.

Define the canonical momentaWhen the velocity Hessian is nonsingular, these equations can be inverted for ${\dot{q}}_i(q,p,t)$, and the Hamiltonian is the Legendre transformIts differential isbecause the $p_{i}d{\dot{q}}_i$ terms cancel. The Euler-Lagrange equation identifies $\partial \mathcal{L}/\partial q_i={\dot{p}}_i$, so coefficient comparison gives Hamilton's equations

For the symmetric top, the momenta areThusThe Legendre transform gives

Both $\varphi$ and $\psi$ are cyclic coordinates, so $p_{\varphi }$ and $p_{\psi }$ are conserved. The Hamiltonian has no explicit time dependence and is also conserved. Since $H$ is independent of $\varphi ,\psi$,under the Poisson bracket. Hence three independent first integrals in involution are

The two Euler-Lagrange equations areDifferentiating $\kappa =f-u^{\prime }{f}_{u^{\prime }}-w^{\prime }{f}_{w^{\prime }}$ along an extremal and using these equations gives $d\kappa /dx=f_x$. Hence the Beltrami identity makes $\kappa$ constant when $f$ has no explicit dependence on $x$. If $f$ omits $u$ or $w$, that variable is a cyclic coordinate and the corresponding momentum $f_{u^{\prime }}$ or $f_{w^{\prime }}$ is an additional first integral; continuous symmetries give the analogous conserved quantities through Noether's theorem.

Put $A=1-m/u$. Since $w$ is cyclic,for a constant $\lambda$. The Beltrami integral isTherefore

If $\kappa =-{\lambda }^2$, then $u^{\prime 2}={\lambda }^{2}m/u$ andFor $z>1$, definethe hinted identity gives $F^{\prime }(z)=z^{3/2}/(z-1)$. Hence $w=C\pm mF(u/m)$. Because $F(z)\to +\infty$ as $z\to \infty$, the prescribed limit selects the minus sign:As $u\downarrow m$, the logarithm tends to $-\infty$, so every such solution satisfies $\boxed{w\to +\infty }$.

The Euler-Lagrange equations of the particle Lagrangian in a shearing sheet areThe terms coupling $x$ and $y$ are the Coriolis acceleration. The cyclic coordinate $y$ has conserved canonical momentumFor a Newtonian potential of a point mass, $q_0=3/2$ and ${\Omega }_z={\kappa }_r={\Omega }_0$. Set $p_y={\Omega }_0{x}_0/2$ and substitute $\dot{y}=p_y-2{\Omega }_{0}x$ in the radial equation. It becomes $\ddot{x}+{\Omega }_0^2(x-x_0)=0$, a harmonic oscillator equation about the epicyclic guiding center $x_0$. Integration givesThe four real constants in $A,y_0,x_0$ and the two in $B$ account for the six initial position and velocity data.

Expanding the inertial specific angular momentum gives $h=h_0+r_0{p}_y+O(2)$. Thus $p_y$ measures the angular-momentum offset from the reference circular orbit, and $x_0$ specifies the radius of its associated epicyclic guiding center.

The conserved horizontal energy in the rotating frame isIt is the horizontal part of the local Jacobi energy in a shearing sheet, rather than the inertial specific orbital energy. Its positive term is the epicyclic energy; its negative term is the energy of the background shear at guiding-center position $x_0$. Independently, the vertical harmonic oscillator has conserved energy