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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.

Let a nonzero constant uniform magnetic field point along $z$, and choose the symmetric gaugeFor the charged-particle electromagnetic Lagrangian, the canonical momentum parallel to the field is $p_z=m\dot{z}+q{A}_z=m\dot{z}$, equal to the mechanical momentum in this gauge. Its equation isThus $z$ is a cyclic coordinate and this canonical momentum is conserved when the scalar electric potential is independent of $z$, equivalently when $E_z=0$ here. A uniform magnetic field alone is insufficient: choosing $\varphi =-E_{0}z$ adds a parallel electric field and gives $m\ddot{z}=q{E}_0$.

The Lorentz force from the magnetic field has no component along that field. This physical statement is gauge independent, whereas equality of canonical momentum and mechanical momentum uses the stated gauge; a gauge transformation can change a canonical momentum component by $q{\partial }_z\chi$.

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 $-p_{\varphi }$.

Noether theorem associates conservation laws with continuous action symmetries of an action. The symmetry must preserve the action up to an allowed boundary term; a symmetry of an equation alone is not the full hypothesis. Emmy Noether proved the general relation in her 1918 paper, whose first theorem concerns finite-parameter groups and whose second concerns transformations depending on arbitrary functions.^[ref]

A mechanical version is especially transparent. Let $L(q,\dot{q},t)$ change under an infinitesimal point transformation at fixed time by $\delta L=ϵdB/dt$, and let $\delta q_i=ϵ{\xi }_i$. Along an Euler-Lagrange equation solution, with $p_i=\partial L/\partial {\dot{q}}_i$,Consequently $J={\sum }_i{p}_i{\xi }_i-B$ is constant. A cyclic coordinate gives its conjugate momentum; rotational symmetry gives angular momentum. Allowing the time coordinate to transform gives the energy conservation associated with time-translation symmetry.

The field-theoretic version gives a conserved Noether current on solutions. Noether second theorem instead produces off-shell identities among field equations for arbitrary local gauge parameters. These identities and the associated constraints should not be confused with independent physical conserved charges for every local parameter.

In an isotropic medium with smooth positive refractive index $n$, the Fermat principle makes an optical path stationary for the optical path length $\int nds$. In cylindrical coordinates, on a segment parametrized by increasing axial coordinate $z$,This gives a two-function Euler-Lagrange equation problem with fixed spatial endpoints. The ray approximation assumes spatial variation on scales large compared with the optical wavelength; stationarity does not guarantee a globally shortest path.^[ref]

If $n$ is independent of $\theta$, the angle is a cyclic coordinate andis constant. If $n$ is also independent of $z$, a Beltrami identity supplies a second first integral. These symmetries reduce the ray equations and permit a helical extremal when the radial Euler-Lagrange equation also holds. A geometric helix by itself is not enough to be an optical extremal.

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.

Parametrize a nonconstant hyperbolic geodesic by unit speed. Its energy is $({\dot{x}}^2+{\dot{y}}^2)/(2y^2)$, and the Euler-Lagrange equation in the cyclic coordinate $x$ gives $\dot{x}/y^2=c$. Unit speed gives ${\dot{x}}^2+{\dot{y}}^2=y^2$. If $c=0$ the curve is vertical. Otherwise, on either side of a possible highest point,The two branches join smoothly at the top. Conversely these curves with their unit-speed parametrizations satisfy the energy equations. Thus the unparametrized complete geodesics are vertical half-lines and upper semicircles centered on the real axis.

Let the tractrix point have positive height $r(\theta )$. Its tangent segment to the axis has length one and makes angle $\theta$, so $r=\sin \theta$. Since its slope is $\tan \theta$, $h^{\prime }(\theta )={\cos }^2\theta /\sin \theta$ for its axial coordinate. Rotating $X(\theta ,v)=(h(\theta ),\sin \theta \cos v,\sin \theta \sin v)$ gives orthogonal coordinate vectors with squared lengths ${\cot }^2\theta$ and ${\sin }^2\theta$. Hence the stated first fundamental form follows.

Now put $x=v$, $y=\csc \theta$. Direct substitution givesOn the specified intervals this coordinate map is a diffeomorphism onto the stated rectangle, so is an isometry there and a local isometry on the tractoid.

Set $b=\csc \beta$, $a=\csc \alpha$. In the unwrapped coordinate, connect $(x_0,b)$ to $(x_0+2\pi ,b)$ along the semicircle centered at $(x_0+\pi ,0)$ of radius $\sqrt{b^2+{\pi }^2}$. Its height lies between $b$ and that radius. The hypothesis $a^2-b^2>{\pi }^2$ puts the whole arc in the closed height strip $b\le y\le a$. Projecting the angular coordinate modulo $2\pi$ therefore gives a geodesic segment in the closed tractoid annulus which winds once and returns to its starting point. Taking $x_0$ to represent any chosen angular position proves the assertion for every point of its $\theta =\beta$ boundary. The returned tangent need not equal the starting tangent, so this proves the claimed returning segment without asserting a periodically smooth closed geodesic.

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.