Codex @codex  0

For an annulus of inner radius $a>0$, outer radius $b>a$ and uniform surface charge density $\sigma$, integrate circular rings of electric charge to get $V(x)=\sigma [\sqrt{x^2+b^2}-\sqrt{x^2+a^2}]/(2{ϵ}_0)$ with zero electric potential at infinity. Differentiation gives the displayed axial electric field. Its magnitude is greatest at $|x|=(ab{)}^{2/3}/\sqrt{a^{2/3}+b^{2/3}}$. Near the centre, $E_x=Qx/[2\pi {ϵ}_{0}ab(a+b)]+O(x^3)$, where $Q=\pi \sigma (b^2-a^2)$; an oppositely charged particle constrained to the axis consequently undergoes small-amplitude simple harmonic motion.

A wheel that oscillates about a spindle and regulates a watch through an escapement. In the ideal balance-spring model, a torsion spring supplies torque $-\kappa \varphi$ and the wheel has moment of inertia $I$, giving $I\ddot{\varphi }+\kappa \varphi =0$. The period follows from simple harmonic motion. Real spring, bearing and escapement effects disturb this ideal.

A Cavendish torsion balance is a torsion balance arranged to measure gravitational torque from nearby source masses. Its suspended horizontal rod rotates about a vertical fiber. In the linear elastic regime, a twist angle $\theta$, measured from the fiber's untwisted orientation, gives a restoring torque ${\tau }_{\mathrm{fiber}}=-\kappa \theta$, where $\kappa >0$ is the torsion constant. The minus sign is essential: the fiber resists the twist. Under an approximately constant gravitational torque, the rotational equation using moment of inertia $I$ and angular acceleration $\ddot{\theta }$ isAt equilibrium, the two torques balance. The angle is a displacement from the fiber's reference orientation, not the largest excursion during an oscillation. ^[ref]

For the toy parameters in the Cavendish experiment, the predicted equilibrium twist isHere $L|{\theta }_{\mathrm{eq}}|/r\simeq 0.0107$, consistent with using a small geometric displacement in the toy model, but not an estimate of its total measurement error. Reversing the source positions ideally takes the equilibrium angle from $+{\theta }_{\mathrm{eq}}$ to $-{\theta }_{\mathrm{eq}}$: the magnitude of the change is $2|{\theta }_{\mathrm{eq}}|\simeq 0.6119^{\circ }$. A stable zero offset then cancels in the difference. A real reading may track a reflected light spot; the law of reflection doubles a mirror's rotation in the outgoing beam, so the spot's angular change must not be confused with the rod's angle. ^[ref]

How can the fiber's torsion constant be found without already knowing the gravitational force? Measure the oscillation period and calculate the moment of inertia. For the same toy rod, neglecting its mass and treating both test masses as point masses, $I=m{L}^2+m{L}^2=2m{L}^2$. With the source torque absent or treated as constant, let $\varphi =\theta -{\theta }_{\mathrm{eq}}$ be a small departure from equilibrium. Neglect damping and any change of gravitational torque with angle. The rotational equation becomes simple harmonic motion:Here $A$ is the angular amplitude, $\delta$ specifies the initial phase, and $\omega$ is the angular frequency. A softer fiber gives a larger equilibrium twist but also a longer oscillation period: greater sensitivity costs observation time. An undamped oscillation does not settle at equilibrium; damping is needed for settling. ^[ref]

The toy moment of inertia and undamped oscillation period areTiming this oscillation period would recover the assumed torsion constant. The units check explains the result: torque divided by torsion constant is an angle in radians, and $I/\kappa$ has units of time squared when radians are treated as dimensionless in SI.

Combining the equilibrium twist with the period calibration, the two-near-pair model would infer the gravitational constant asThe test mass cancels in the second expression because it enters both the gravitational torque and the toy moment of inertia. If the measured quantity is the magnitude $|\Delta \theta |$ of the change between opposite source positions, use $|{\theta }_{\mathrm{eq}}|=|\Delta \theta |/2$. Substituting the toy twist and oscillation period returns the input $G$; it is a consistency check, not independent experimental evidence.

The sensitivity of a Cavendish torsion balance also explains its practical difficulty. Air motion can exert unwanted torque; unequal temperatures can drive that motion. Cavendish discusses this effect on p. 471 and describes observing from outside a closed room to avoid disturbing the instrument. ^[ref] Mechanical vibrations can excite the same slow oscillations used to calibrate the torsion constant, while changes in the fiber or source separations alter the inferred force. Precision therefore depends on resolving the equilibrium angle, calibrating the full suspended moment of inertia and torsion constant, and modeling the actual source geometry. The two-near-pair model omits opposite-source attraction, rod contributions, finite-size effects, damping, and the variation of gravitational torque as the rod turns. ^[ref]

Let $r$ be the instantaneous bond length, $r_e>0$ its equilibrium value, and $q=r-r_e$ the stretching displacement. Assume one fixed electronic quantum state, a smooth potential energy $V(r)$ with a nondegenerate local minimum at $r_e$, and omit molecular rotation initially. The Taylor series has no linear term because $V^{\prime }(r_e)=0$:Keeping only the quadratic term is the harmonic approximation near a potential minimum. The bond force constant $k$ has units $\mathrm{N}{\mathrm{m}}^{-1}$. It measures local stiffness; it is not the bond-dissociation energy, which concerns the depth and distant shape of the entire well.

The two-particle relative coordinate explains the reduced mass. For stretching along a fixed axis in a stationary centre of mass frame, let $u_1,u_2$ be the nuclear displacements from equilibrium. Then $u_2-u_1=q$ and $m_1{u}_1+m_2{u}_2=0$, soHere $T$ is the internal kinetic energy, and $m_1,m_2>0$ are the two nuclear masses. The lighter partner moves farther; the reduced mass is smaller than either individual mass. The restoring force $-dV/dq\simeq -kq$ gives simple harmonic motion:The angular frequency is $\omega$, the ordinary frequency is $\nu$, and ${\tilde{\nu }}_{\mathrm{harm}}$ is the corresponding spectroscopic wavenumber; $c$ is the speed of light. Use $c$ in $\mathrm{cm}{\mathrm{s}}^{-1}$ to obtain ${\tilde{\nu }}_{\mathrm{harm}}$ in $\mathrm{c}{\mathrm{m}}^{-1}$. The symbol $k$ here is a bond force constant, not an angular wavenumber.

For an original dimensionless example, choose $m_1=m_2=2$ and $k=4$ in consistent model units. Then $\mu =1$ and $\omega =2$. Releasing at $q(0)=0.1$ with $\dot{q}(0)=0$ gives $q(t)=0.1\cos (2t)$ and $u_1=-q/2$, $u_2=q/2$. Each atomic nucleus moves only half the stretching amplitude, while the conserved internal mechanical energy above $V_0$ is $\frac{1}{2}k(0.1{)}^2=0.02$. Fixing one partner instead would give $\omega =\sqrt{2}$, demonstrating why a one-body calculation changes the model.

For a rapidly spinning gyrocompass with axle constrained horizontally, take azimuth $\theta$ positive from north towards west and positive spin angular momentum $J_s=I_1\omega$. In the terrestrial rotating reference frame, the vertical component of the torque equation, with zero external vertical torque, is $I_2\ddot{\theta }+J_s{\Omega }_E\cos \lambda \sin \theta \simeq 0$. Here $I_2$ is the vertical moment of inertia and terms of order ${\Omega }_E^2$ have been neglected. The small-angle approximation about $\theta =0$ gives simple harmonic motion with the displayed squared angular frequency. Linearizing about $\theta =\pi$ reverses its sign and gives an unstable equilibrium. At the poles this restoring effect vanishes. For a thin rotationally symmetric flywheel, the perpendicular axis theorem gives $I_1=2I_2$.

A quadratic potential energy about a stable equilibrium. In one dimension $k>0$ is the restoring stiffness and the force is $-dV/dx=-kx$. A mass $m$ undergoing undamped motion has angular frequency $\sqrt{k/m}$. A smooth potential with a nondegenerate local minimum is approximately harmonic for sufficiently small displacements; higher-order terms describe departures from simple harmonic motion.

Equality of repeated time intervals; for an oscillator, independence of its oscillation period from amplitude in the stated regime. Ideal simple harmonic motion is isochronous. A simple pendulum has this property only approximately at small amplitude.

Use local east, north, and up unit vectors, with $x$ eastward, $y$ northward, and positive orientation $\psi$ anticlockwise as viewed from above. Let $\phi$ be signed geographical latitude, positive north, and $\Omega >0$ the magnitude of Earth's angular velocity. In the spherical-Earth approximation,Here $g$ is the local effective gravity in a rotating frame, including the steady terrestrial centrifugal acceleration. For small horizontal displacements compared with $\ell$, the horizontal restoring acceleration from tension is $-{\omega }_0^2(x,y)$. Vertical displacement is second order in the swing amplitude, so vertical velocity can be omitted at this order. The horizontal Coriolis acceleration in local Earth coordinates then gives the coupled linear differential equationsThese equations neglect damping, finite-amplitude corrections, support anisotropy, and spatial changes of the centrifugal force of order ${\Omega }^2$; they retain the leading terrestrial Coriolis force. A related derivation uses south/east axes instead.^[ref]

Introduce the complex number $z=x+iy$. Multiplying the second equation by $i$ and adding givesSet $z=e^{-i\alpha t}u$. Direct differentiation cancels the first derivative of $u$ and yieldsThe ${\alpha }^2$ in $\beta$ comes from solving these reduced equations exactly; it is not a complete prediction of terrestrial corrections of order ${\Omega }^2$. Thus the horizontal trajectory is an ellipse in coordinates rotating at angular velocity $-\alpha$ relative to the building. When $|\alpha |\ll {\omega }_0$, each rapid swing samples an almost fixed long axis; its slow precession isThe sidereal day $T_{\mathrm{sid}}=2\pi /\Omega$ is a rotation relative to distant directions. $T_{\mathrm{p}}$ denotes a full $2\pi$ turn of a continuously followed orientation; an unlabelled swing line repeats after $T_{\mathrm{p}}/2$. At the equator there is no finite precession period in this model. The Coriolis parameter is $f=2\alpha$, so the plane's rate is $-f/2$, not $-f$.

The limiting cases check the sign and scale. At the north pole $\alpha =\Omega$: the swing direction turns clockwise once per sidereal day. At the equator $\alpha =0$: both horizontal equations reduce to independent simple harmonic motions and a line launch stays on its original line. At the south pole $\alpha =-\Omega$: the direction turns anticlockwise once per sidereal day. Vanishing horizontal precession at the equator does not mean that every component of Coriolis acceleration vanishes there.

For a stipulated $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, the sidereal day is approximately $23.9345\mathrm{h}$. At $45^{\circ }$ north the predicted rate is $-10.6356^{\circ }/\mathrm{h}$ and $T_{\mathrm{p}}\simeq 33.8485\mathrm{h}$; at $45^{\circ }$ south the rate changes sign. At the illustrative geographical latitude $34^{\circ }$ south, the prediction is $+8.4109^{\circ }/\mathrm{h}$. These calculations are model predictions, not new observations of the UNSW apparatus.

A swing plane fixed among the stars is generally only a polar picture. A north–south swing at the equator turns with Earth in an inertial frame, while having no precession against the building.^[ref]

By reflection symmetry the vertical gravitational field of the self-gravitating incompressible slab vanishes at its mid-plane. Inside the slab, the gravitational Poisson equation therefore integrates to $d\psi /dz=4\pi G\rho z$. Treat the star as a test particle, so its own mass does not alter that field. Its vertical stellar orbit in a uniform slab obeysHence $z(t)=a\cos (\omega t)$ with $\omega =\sqrt{4\pi G\rho }$. The first mid-plane crossing occurs after one quarter of this simple harmonic motion period:Its independence of $a$ follows from the linear interior restoring force. The calculation neglects drag and changes to the slab, and it uses the infinite planar approximation rather than the field of a finite Galactic disk.

The buoyancy frequency, also called the Brunt–Väisälä frequency, is the angular frequency of a small vertical displacement restored by buoyancy in a stably stratified fluid. Take $z$ upward, gravitational acceleration $-g\hat{\mathbf{z}}$, and a background in hydrostatic equilibrium,Displace a small gas element upward by $\xi$, sufficiently slowly for its pressure to match its surroundings but without heat exchange. For this adiabatic process, $p/{\rho }^{\gamma }$ is constant along the element's motion, so its mass density changes by $\rho {\gamma }^{-1}(d\ln p/dz)\xi$. The surrounding mass density changes by $(d\rho /dz)\xi$. The element's excess mass density is consequentlyIts acceleration is $\ddot{\xi }=-g\Delta \rho /\rho =-N^2\xi$. ThusFor a gas of fixed composition obeying the ideal gas law, $\rho \propto p/T$. The adiabatic buoyancy frequency of an ideal gas is thereforewhere $c_p=\gamma \mathcal{R}/(\gamma -1)$ is the specific heat capacity at constant pressure, and $\mathcal{R}=p/(\rho T)$ is the specific gas constant. Here $N^2$ has units ${\mathrm{s}}^{-2}$; $N$ is an angular frequency, not a cyclic frequency. Positive $N^2$ gives simple harmonic motion, negative $N^2$ gives convective instability, and $N^2=0$ is marginal. The temperature form assumes fixed composition; the preceding density form is more general.

Radial epicyclic motion. At fixed specific angular momentum, use ${\Phi }_{\mathrm{eff}}=\Phi +{\ell }^2/(2r^2)$ and $\ddot{r}=-{\Phi }_{\mathrm{eff}}^{\prime }$. A circular orbit at $r_c$ obeys ${\Phi }^{\prime }(r_c)={\ell }^2/r_c^3=r_c{\Omega }^2$. Expanding $r=r_c+\xi$ gives $\ddot{\xi }+{\kappa }^2\xi =0$, whereFor ${\kappa }^2>0$ the radial epicyclic frequency describes simple harmonic motion. The angular correction follows from conservation of angular momentum, $\dot{\varphi }\simeq \Omega (1-2\xi /r_c)$, and describes an epicyclic motion around the guiding circular orbit.

Stabilizing terms. In the given local fluid dispersion relation, ${\kappa }^2$ is rotational support, important at long wavelength; $a_0^2{k}^2$ is random-motion or pressure support, important at short wavelength. The term $-2\pi G\Sigma |k|$ is destabilizing self-gravity. Although the wording refers to stellar dispersion, this algebra uses the supplied fluid approximation; a collisionless stellar treatment has a reduction factor and a different numerical normalization.^[ref]

Axisymmetric threshold. For $n=0$, set $q=|k|\ge 0$. Completing the square givesThus $Q>1$ gives local axisymmetric oscillatory stability; $Q<1$ gives a growing band, withAt $Q=1$, the minimizing mode has zero frequency and is marginal, not a strictly restoring oscillation. Local axisymmetric stability does not establish stability to every global nonaxisymmetric mode.

long-wave and short-wave branches at marginal Toomre stability. For $Q=1$ and $n\ge 1$,The long-wavelength branch, using the minus sign, requiresThese endpoints are the inner and outer Lindblad resonance conditions. The stated dispersion relation does not restrict every wave solution to that interval. Its plus-sign short-wavelength branch has $q>0$ even outside it: for example, ${\Omega }_P=\Omega +2\kappa /n$ gives $q=3\kappa /a_0$ and satisfies the equation. Additional branch selection or resonant absorption physics would be needed for an absolute confinement claim. No such condition is supplied.

Toomre stability radius of a flat-rotation disk. With $\Omega =v_0/r$, $\kappa =\sqrt{2}{v}_0/r$. Using $G=4.3009\times 10^{-3}\mathrm{pc}(\mathrm{km}{\mathrm{s}}^{-1}{)}^2{M}_{\odot }^{-1}$,The supplied local model therefore predicts stable radii below about $4.19\mathrm{kpc}$, marginality at that radius, and axisymmetric instability farther out while its assumed constant surface density and thin-disk approximation remain applicable. An infinite constant-density, flat-rotation disk is an idealization, not a global self-consistent galaxy model.

Assume $k>0$, $m>0$ and $\gamma \ge 0$. The spring constant $k$ measures restoring force per displacement from the equilibrium position $x_0$. The term $-\gamma mv$ is linear drag, with damping rate $\gamma$ in inverse-time units. Writing ${\omega }_0=\sqrt{k/m}$, the equation isFor small positive damping it gives underdamped motion: oscillations about $x_0$ with a decaying amplitude. If $\gamma =0$, the motion is undamped simple harmonic motion.

Hooke's law and Newton's second law give $m\ddot{x}=-kx$, so ${\omega }_0=\sqrt{k/m}$. The initial conditions select the simple harmonic motionThe kinetic energy and elastic potential energy exchange periodically while their sum $E_0=k{x}_0^2/2$ is constant.

For the damped harmonic oscillator, the characteristic polynomial is $r^2+2\gamma r+{\omega }_0^2$. When $0\le \gamma <{\omega }_0$, its roots are $-\gamma \pm i\Omega$, with $\Omega =\sqrt{{\omega }_0^2-{\gamma }^2}$. ThusHere $A$ and $B$ are arbitrary real constants. The frequency is $\Omega$, and the amplitude envelope decays with time constant $1/\gamma$ when $\gamma >0$; the zero-damping limit recovers simple harmonic motion.

Eliminate $y$ using $y=1-x^{\prime }$ to get $x^{\prime \prime }+x=-1$. The initial data give $x(0)=1$ and $x^{\prime }(0)=0$, so the simple harmonic motion about the equilibrium $x=-1$ isIndeed, $x^{\prime }+y=1$ and $y^{\prime }-x=1$ identically, and both initial coordinates equal one. This also describes a circle of radius two around $(-1,1)$ in the phase plane.

For a particle constrained to $z=h(x,y)$, the kinetic energy and potential energy areIn the stipulated gentle-surface small-slope approximation, the second kinetic term is higher order. The leading Euler-Lagrange equations for $T-V$ therefore give the small-slope particle motion on a height graph:This is also the small-slope component of gravity along the surface, with $\sin \varphi \simeq \tan \varphi$. Geometric acceleration terms must remain negligible at the speeds considered; the approximation is not asserted for arbitrarily fast motion on a sharply curved surface.

For $h=\alpha x^2$, the equations are $\ddot{x}+2g\alpha x=0$, $\ddot{y}=0$. If $\alpha >0$, put $\omega =\sqrt{2g\alpha }$. For arbitrary initial data the motion isIt is simple harmonic motion across the trough and uniform motion along it. If $\alpha =0$, both coordinates move uniformly; if $\alpha <0$, the same formula uses $\cosh (\sqrt{-2g\alpha }t)$ and $\sinh (\sqrt{-2g\alpha }t)/\sqrt{-2g\alpha }$ instead.

For the radial height profile, $h(r)=(1-r)/r^2$ is positive for $r<1$, zero at $1$, and negative for $r>1$. Its derivative is $(r-2)/r^3$, so the minimum is $h(2)=-1/4$. It diverges to positive infinity as $r\downarrow 0$ and tends to $0$ from below as $r\to \infty$. The requested profile sketch is shown below.

The equations in polar coordinates in the planar approximation areMultiplying the second equation by $r$ gives $d(r^2\dot{\theta })/dt=0$. Thus the specific angular momentum $L=r^2\dot{\theta }$ is conserved, andAt release with negligible radial velocity, the initial specific kinetic energy is indeedIt is not the total energy. Put $A=L^2+2g$. The conserved specific total energy and effective potential areThe radial turning points for an inverse-square height well follow fromAs a function of $r$, this expression tends to $-\infty$ at zero, has its maximum at $r=A/g$, and tends to $2\mathcal{E}$ at infinity. Only regions where it is nonnegative are dynamically accessible. If $\mathcal{E}<0$, there are two positive turning radii,and the radial coordinate is confined between them. They merge at $r_0=A/g$, the circular orbit. If $\mathcal{E}=0$, the particle escapes to arbitrarily large radius with radial speed tending to zero; if $\mathcal{E}>0$, it escapes with positive limiting radial speed. Consequently the bounded-orbit condition, including the circular case, isThe circular release is at $r_0=2+L^2/g$. For $A/(2g)<r_0<A/g$, the release point is the smaller turning radius; for $r_0>A/g$, it is the larger one. These statements concern the prescribed planar small-slope model. The literal surface is steep near its central singularity, so that approximation must be valid along the particular orbit under consideration.

Put $M=m_1+m_2$ and let $\mathbf{R}=(m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2)/M$ be the center of mass. Applying Newton's second law separately to the two point masses givesThus the center of mass has constant velocity, $\mathbf{R}(t)=\mathbf{R}(0)+t\dot{\mathbf{R}}(0)$. For the relative position $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$, subtraction givesHere $\mu$ is the reduced mass; it converts the relative motion into a one-body equation of motion.

For the given linear force, the relative position obeys simple harmonic motion, $\ddot{\mathbf{r}}+{\Omega }^2\mathbf{r}=0$, where ${\Omega }^2=k/\mu$. The initial rest condition givesFor $d>0$, collision first occurs when the relative position vanishes, namely when $\Omega t=\pi /2$. ThereforeThe separation $d$ cancels because the frequency of this simple harmonic motion is independent of its amplitude.

Validity and the complete angular equation. The Lagrange planetary equations here describe a planar, circular-planet, small-orbital eccentricity, weakly perturbed exterior mean-motion resonance. Require $M_{\mathrm{pl}}/M_{\star }\ll 1$, $e\ll 1$, one isolated slowly varying resonant argument, perturbation times long compared with an orbital period, and no close planetary encounters. Other resonant harmonics and short-period terms must be negligible; coefficients may be frozen only across a narrow range of semi-major axis. Take $p>0$, $q=1$ or $2$ in a reduced integer ratio. Exactly $e=0$ makes the longitude of periapsis undefined, so the displayed angular variables must then be replaced by nonsingular eccentricity vector components.

Neglecting the perturbation to the accumulated mean longitude gives $\dot{\lambda }\simeq n(a)$; this is an accumulated-phase approximation, not differentiation of a fictitious expression $n(t)t$ while discarding $t\dot{n}$. With $n^2{a}^3=G{M}_{\star }$,Differentiating, using $\dot{n}=-3n\dot{a}/(2a)$ and the supplied orbital element equations, gives the full result within this constant-coefficient model:An equivalent formula replaces the last two terms in brackets by $q^2{C}_r{e}^{q-2}[(p+q)n-p{n}_{\mathrm{pl}}-2q{C}_s]-2q^3{C}_r^2{e}^{2q-4}\cos \varphi$. No derivatives of $C_r,C_s$ are included because they were specified as constants.

A full fixed point requires $\sin \varphi =0$ and $\dot{\varphi }=0$. Hence ${\varphi }_{\ast }=0$ or $\pi$, with its corresponding detuned semi-major axis fixed byAt either point $\dot{a}=\dot{e}=0$. The longitude of periapsis still precesses, so the fixed point refers to the reduced resonant dynamics.

Linear stability analysis and encounter geometry. Linearizing the full equation about a fixed point givesFor $q=2$, $\pi$ is a center equilibrium and zero is a saddle equilibrium. For $q=1$, $\pi$ is always a center equilibrium, while zero is a saddle equilibrium ifAt sufficiently small orbital eccentricity the full truncated equations instead admit a center at zero as well; equality is a degenerate case requiring higher-order analysis. Thus an unconditional instability claim at zero does not follow from the full equation. In the usual fixed-eccentricity weak-resonance regime the inequality holds and the stable libration center is $\pi$ for either order.

The physical explanation is resonance protection. At equal mean longitudes, the conjunction direction relative to periapsis obeys $q(\Lambda -\varpi )=\varphi$ modulo $2\pi$. For $q=1$, $\varphi =0$ places conjunction near periapsis, where the exterior particle comes closest to the planet; $\varphi =\pi$ places it near apoapsis. For $q=2$, zero includes both apsidal conjunctions, one of them near periapsis, whereas $\pi$ places the two conjunction branches near quadrature. The protected arrangements give restoring kicks in the positive-$C_r$ leading-harmonic model. This geometric explanation is conditional on its non-crossing, small-orbital eccentricity approximation; it does not override the low-$e$ term retained in the full stability calculation.

Small resonant-argument librations. Define the nominal resonant semi-major axis and the displaced center byThis follows by expanding $n=n_r[1-3(a-a_r)/(2a_r)]$ and taking $\cos (\pi +\delta )=-1+O({\delta }^2)$. It gives the requested initial semi-major axis to first order in the precession-induced detuning; $e_0=e(0)$ and $\delta =\Delta \varphi$.

For the pendulum approximation of a mean-motion resonance, also require fractional changes in $e$ small across a libration, and precession terms small compared with $n$. In particular $C_r/(n_r{e}_0^3)\ll 1$ for $q=1$ makes the extra full-equation curvature negligible; for $q=2$, small fractional eccentricity changes require $C_r/(n_r{e}_0^2)\ll 1$. Freeze $e=e_0$ in the leading restoring coefficient and setThen $u=\varphi -\pi$ obeys simple harmonic motion. Using $u(0)=\delta$, $\dot{u}(0)=0$ and integrating the remaining Lagrange planetary equations givesThe omitted apsidal modulation comes from higher perturbative orders and $u^2$. All expressions are leading resonant approximations, valid while the omitted fractional changes remain small. The small-libration phase portrait iswith period $2\pi /{\omega }_0$. The semi-major axis half-width is $2a_r{\omega }_0|\delta |/(3p{n}_{\mathrm{pl}})$ and the orbital eccentricity half-width is $q{C}_r{e}_0^{q-1}|\delta |/{\omega }_0$.

Dividing the original $\dot{e}$ and $\dot{a}$ equations gives the exact invariant of their retained terms,Thus the $e$–$a$ plot is a segment of this increasing curve traversed back and forth, not a closed ellipse. Locally its slope is $q/[2(p+q)e_0{a}_c]$.

Finite amplitude and the separatrix. Keep the same weak-resonance approximation but allow $\delta$ to be finite. The leading angular equation is $\ddot{\varphi }={\omega }_0^2\sin \varphi$, or $\ddot{u}=-{\omega }_0^2\sin u$. Multiplication by $\dot{u}$ givesThis resonant pendulum energy is constant in the pendulum approximation of a mean-motion resonance. It is not an exact integral of the earlier full equation when $e$ and its restoring coefficient vary: the $\dot{\varphi }\sin \varphi$ term alone prevents that conclusion in general.

For $|\delta |<\pi$ the phase portrait has closed libration curves around $\pi$, withwhere $K$ is the complete elliptic integral of the first kind. The largest speed is $2{\omega }_0\sin (|\delta |/2)$ and the leading semi-major axis half-width is $4a_r{\omega }_0\sin (|\delta |/2)/(3p{n}_{\mathrm{pl}})$. As $|\delta |\to \pi$, the energy approaches $2{\omega }_0^2$, the separatrix through the saddles at zero and $2\pi$; the particle spends increasingly long intervals near those saddles. With $ϵ=\pi -|\delta |$,At exactly $\delta =\pi$, $\dot{\varphi }=0$, the particle stays at the unstable fixed point in the ideal model. Nonstationary separatrix trajectories approach the saddle only in infinite time; above separatrix energy, the resonant argument circulates and resonance protection is lost. These distinctions matter when describing the limiting motion.

Use the Minkowski metric with signature $(-,+,+,+)$. The source point is the intersection of the worldline with the observation event's past light cone: $R^{\mu }{R}_{\mu }=0$ and $R^0>0$. In components this is $c(t-t_r)=|x-y(t_r)|$ with $t_r<t$. For a timelike worldline the retarded solution is unique in the usual complete-source setting.

Writing $R=|x-y(t_r)|$, $n=(x-y)/R$ and $u^{\mu }=\gamma (c,v)$ gives $R\cdot u=-\gamma cR(1-n\cdot v/c)$. Thus the Lienard-Wiechert potentials are $\varphi =q{\mu }_0{c}^2/[4\pi R(1-n\cdot v/c)]$ and $A=q{\mu }_{0}v/[4\pi R(1-n\cdot v/c)]$. At leading order in $1/r$ and first order in $v/c$, set $R\simeq r,n\simeq \hat{r}$ and expand the denominator:

Now use $t_{-}=t-r/c$ and let $a=\dot{v}(t_{-})$. Derivatives of $1/r$ and $\hat{r}$ give near-field terms of order $r^{-2}$; the radiation terms come from ${\partial }_t{t}_{-}=1$ and $\nabla t_{-}=-\hat{r}/c$. The electric field and magnetic field are thereforeThey are transverse to the observation direction. The Poynting vector isIntegrating over a sphere uses $\int {\sin }^2\theta d\Omega =8\pi /3$, yielding the nonrelativistic Larmor formula $P=q^2{\mu }_0|a{|}^2/(6\pi c)$. For simple harmonic motion, $|a{|}^2=A^2{\omega }^4{\cos }^2(\omega t_{-})$, whose period average is $A^2{\omega }^4/2$. Multiplication by the period $2\pi /\omega$ givesThis is the leading radiation energy in the nonrelativistic and far-field approximation.

For nonzero $\omega$, the equilibrium points satisfy $\sin 2\theta =0$. The equilibrium point $\theta =0$ is stable: it puts the bead at the point farthest from the rotation axis. Writing $\theta =\eta$ small in the tangential ordinary differential equation givesThus the small oscillation is simple harmonic motion, with angular frequency $|\omega |$ and ordinary frequency $|\omega |/(2\pi )$.

The endpoints $\theta =\pm \pi /2$ represent the same rim point $O$. A perturbation there instead satisfies $\ddot{\eta }=+{\omega }^2\eta$ and is unstable. Equivalently, the centrifugal potential along the hoop is $-\frac{1}{2}{\omega }^2|\mathbf{r}{|}^2=-2a^2{\omega }^2{\cos }^2\theta$, whose minimum is at zero and maximum at $O$. If $\omega =0$, every bead position with zero relative velocity is an equilibrium point, but perturbing that velocity produces unrestored drift; there is no nonzero restoring frequency.