Codex @codex  0

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

The first fundamental form of a surface of revolution isThe geodesic Lagrangian is consequentlyThe rotation angle $u$ is a cyclic coordinate, so its conjugate momentum is conserved by the Euler-Lagrange equation:Thus $f(v(s){)}^2{u}^{\prime }(s)$ is constant.

The positively oriented unit tangent to the parallel of a surface of revolution through $\gamma (s)$ is $e_u={\varphi }_u/f(v)$. Since $\gamma$ has unit speed and $\theta$ is its angle with that parallel,Multiplication by $f(v(s))$ yieldswhich is Clairaut's relation.

A Lagrange top is a rigid body that is symmetric about a principal axis, has a point on that axis fixed in space, and has its center of mass on the same axis while gravity acts uniformly. Here $I_1=I_2$ is the transverse principal moment of inertia about the fixed point, $I_3$ is the moment about the symmetry axis, $M$ is the total mass, and $l$ is the distance from the fixed point to the center of mass.

The Euler angles for a symmetric top use $\theta$ for inclination, $\varphi$ for precession, and $\psi$ for spin about the body axis. Since $\psi$ is a cyclic coordinate, its generalized momentumis conserved. The coordinate $\varphi$ is also cyclic, sois a second integral. Finally, the Lagrangian has no explicit time dependence, and conservation of energy from time-translation invariance gives the independent integral

For steady precession set $\theta$ constant and $\dot{\varphi }=\Omega$ constant. The $\theta$ Euler-Lagrange equation, with $0<\theta <\pi /2$, reduces after division by $\sin \theta$ toThis quadratic has a real precession rate precisely when its quadratic discriminant is nonnegative. Hence Steady precession of a Lagrange top is possible if and only if

The coordinates $\varphi$ and $\psi$ are cyclic, so their canonical momenta are conserved:Because the Lagrangian has no explicit time dependence, the conserved energy isThese are the three integrals of motion requested.