Codex @codex  0

For regular low-energy elastic scattering by a finite-range central potential, higher partial waves are suppressed by their centrifugal barriers. Matching the regular interior radial solution to small-argument exterior spherical Bessel functions gives the displayed scaling of scattering phase shift, in the generic absence of a threshold singularity. Since the channel amplitude is proportional to $\sin {\delta }_{\ell }/k$, it scales as $k^{2\ell }$. The $\ell =0$ term is isotropic and normally leads. A tuned vanishing leading amplitude or a threshold resonance requires examining the next terms rather than applying the generic statement blindly.

In cylindrical coordinates the particle's radial and vertical equations are $\ddot{r}-r{\dot{\varphi }}^2=-{\Phi }_r$ and $\ddot{z}=-{\Phi }_z$. Axisymmetry gives conservation of angular momentum, with specific angular momentum $h=r^2\dot{\varphi }$. Eliminating the cyclic azimuthal motion therefore givesThus $\boxed{{\Phi }_1=\Phi +h^2/(2r^2)}$ governs the meridional motion. Its conserved energy is $\frac{1}{2}({\dot{r}}^2+{\dot{z}}^2)+{\Phi }_1$. The additional term is the centrifugal barrier appropriate to fixed angular momentum.

For a point mass, $\Phi =-GM/\sqrt{r^2+z^2}$. If $h\ne 0$, the vertical derivative vanishes only at $z=0$, and the radial stationary condition there gives $r_0=h^2/(GM)$. This is the sole stationary point with $r>0$, at energy $-GM/(2r_0)$. Its Hessian matrix isConsequently the local contours are closed, approximately circular curves about a strict minimum. Finite negative-energy contours above the minimum remain closed; the centrifugal barrier prevents approach to the axis and the negative energy prevents escape to infinity. The zero contour satisfies $z^2=4r^4/r_0^2-r^2$, while positive-energy contours can be unbounded. The left panel illustrates these features; portions of the outer contours extend outside the plotting window.

Each panel uses its own circular radius as length unit and $GM/r_0$ as potential unit. Linearizing near the minimum givesThe radial and vertical epicyclic frequencies both equal the circular orbital frequency. Circular Kepler orbits are stable to small meridional perturbations of a freely moving particle. With no dissipation the oscillations persist rather than decay. A change in h shifts the circular radius to a nearby orbit; azimuthal phase itself is neutral by rotational symmetry. If $h=0$, there is no stationary point or centrifugal barrier and this circular-orbit stability discussion does not apply.

Set $\mu =GM>0$ and interpret $L$ in the displayed equation as specific angular momentum, $L=r^2\dot{\psi }$, rather than total angular momentum. The gravitational potential is per unit mass and $a$ has dimensions of length. Spherical symmetry fixes the orbital plane and conserves the vector angular momentum, while the autonomous specific orbital energy is also conserved.

For the orbit equation for combined inverse-square and inverse-cube attraction, defineA regular nonradial bounded oscillation requires $L\ne 0$ and $L^2-2\mu a>0$. Solving the linear Binet equation then givesFor a noncircular bound orbit $0<e<1$; its specific orbital energy is $E={\mu }^2(e^2-1)/[2(L^2-2\mu a)]$. The effective potential $[L^2-2\mu a]/(2r^2)-\mu /r$ has a centrifugal barrier and a minimum in this case. If $L^2-2\mu a\le 0$, the oscillator frequency is zero or imaginary and there is no regular finite-radius bound oscillation: the attractive centre lacks a centrifugal barrier. Unbound branches with $e\ge 1$ are not closed ellipselike trajectories.

Successive pericentres differ in angle by $2\pi /\nu$. The full phase space state returns after $p$ radial oscillations and $q$ revolutions precisely whenwhere $p,q$ are coprime positive integers; the second formula applies when $a\ne 0$. Thus $a>0$ requires $0<p/q<1$ and $L^2>2\mu a$, whereas $a<0$ requires $p/q>1$. An irrational $\nu$ advances the apsidal direction through a dense set of angles and gives a nonclosed rosette, with only the four regular spherical invariants needed to label its two-dimensional closure.

To see the additional resonant label explicitly, let $u_c=\mu /(L^2-2\mu a)$ and $w=u-u_c-i{u}^{\prime }/\nu$. Then $w=C{e}^{i\nu (\psi -{\psi }_0)}$, so on a family with fixed resonant $L$,Its argument fixes the remaining apsidal phase modulo the symmetry of the closed curve; its magnitude is already determined by specific orbital energy and $L$. Choosing reference axes in the fixed orbital plane makes this single-valued under $\psi \mapsto \psi +2\pi$. This is the fifth independent orbit label on the resonant family described by closed orbits with an inverse-cube perturbation.

For $a\ne 0$, $\nu$ varies with $L$. The orbital resonance condition holds only on selected angular momentum surfaces, so this construction is not a fifth global smooth integral on an open set of arbitrary nearby $L$. If the question asks for a maximally superintegrable Hamiltonian system on an open region of noncircular bound orbits, the required case is $a=0$: then $\nu =1$ for every $L$, and the conserved Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{v}\times \mathbf{L}-\mu \hat{\mathbf{r}}$ fixes the apsidal direction globally. The relations $\mathbf{A}\cdot \mathbf{L}=0$ and $A^2={\mu }^2+2E{L}^2$ leave five independent constants. For the usual orbit-wise interpretation, noncircular bound orbits with rational $\nu$ have five isolating labels; irrational $\nu$ has four. Circular orbits are closed for any allowed $\nu$ but have no independent apsidal phase, so the generic noncircular count cannot simply be applied to them.

For completeness, regular unbound branches are different. When $e\ge 1$ and $\nu >0$, positive radius restricts the radial phase to $|\nu (\psi -{\psi }_0)|<\arccos (-1/e)\le \pi$. The apsidal invariant on an unbound central-force branch isThe principal argument is continuous on this scattering branch because it never reaches an apocentre across the negative-real-axis cut. Thus the periapsis direction can be a fifth single-valued scattering label for any positive $\nu$, including irrational $\nu$; the rationality condition applies to the noncircular bound case. For plunging or collision branches with $L^2-2\mu a\le 0$, local trajectory labels can still be defined away from the singular centre, but the preceding periodic-orbit criterion is inapplicable and no smooth continuation through $r=0$ is asserted.

Take $a>0$ and the intended horizontal launch around a parallel of the surface of revolution: $\dot{r}(0)=\dot{z}(0)=0$, $a\dot{\theta }(0)=v$. Then $h=av$ and $E=v^2/2+gf(a)$. Since $\dot{z}=f^{\prime }(r)\dot{r}$, the conserved energy becomesHere $U$ is the radial effective potential on a surface of revolution. Rearrangement givesFor a nonconstant bounded radial motion with $v>0$, there must be another finite radial bound $b\ne a$ satisfying $D(b)=0$, equivalentlyThe interval between the radial bounds must be dynamically accessible: $D\ge 0$ inside, and a simple endpoint must have a forbidden region $D<0$ beyond it. Equivalently the motion must lie in a bounded part of $\{r>0:U(r)\le E\}$ separated from escape by a potential barrier. When $f$ stays finite near zero and $v>0$, the centrifugal barrier already excludes $r=0$; an outer barrier is the additional requirement. An arbitrary second algebraic root on a disconnected allowed branch would not establish boundedness.

The initial radial acceleration, obtained by projecting Newton's second law along ${\mathbf{e}}_r+f^{\prime }(a){\mathbf{e}}_z$, isThus $v^2/g>a{f}^{\prime }(a)$ starts the particle outward, and $v^2/g<a{f}^{\prime }(a)$ starts it inward. Equality gives the exact circular orbit $r=a$ and needs no second distinct radial bound. A double root at a different radius can instead be approached asymptotically: this separatrix is bounded, although the radial motion is not periodic. If $v=0$, the centrifugal barrier disappears and the particle may pass through the axis, still with bounded $r$.

The source leaves a direction ambiguity at points where $f^{\prime }(a)=0$: every horizontal tangent direction is then possible, so “projected horizontally” alone does not imply $h=av$. The displayed inequality uses the azimuthal launch specified above; at $f^{\prime }(a)\ne 0$ that direction is forced by horizontality and tangency.

Model alpha decay as a preformed alpha particle making repeated escape attempts from a short-range nuclear potential well, then undergoing quantum tunnelling through the external Coulomb potential. Write the daughter atomic number as $Z_D$, the nuclear radius as $R$, and the relative reduced mass as $\mu$. For zero orbital angular momentum, the WKB approximation givesSet $v=\sqrt{2Q/\mu }$ and $r=r_2{\sin }^2\theta$. ThenWhen $R/r_2\ll 1$, the leading term isHere $P_{\alpha }$ is the alpha preformation probability and ${\nu }_{\mathrm{attempt}}$ the attempt frequency. The stated proportionality retains only the dominant barrier exponent. It assumes a thick classically forbidden barrier, nonrelativistic relative motion, a spherical unscreened external Coulomb potential, and slowly varying prefactors. Finite $R$, changes in alpha preformation probability, and a centrifugal barrier for nonzero orbital angular momentum produce significant corrections. The precise charge in the barrier is $Z_D=Z_P-2$; using the parent charge is an approximation for large $Z_P$. The measured alpha particle speed approximates $v$ only when daughter recoil is negligible.

An alpha emitter has an alpha-like cluster coupled to a daughter nucleus. Its positive decay energy is $Q=[M_P-M_D-M_{\alpha }]c^2$; atomic masses may be used consistently because the electron counts balance. The daughter and alpha share this energy according to momentum conservation. The attractive nuclear interior and repulsive exterior Coulomb potential produce a quasibound state with a barrier, so positive $Q$ does not imply rapid classical escape.

For daughter charge $Z_{d}e$, the exterior barrier is approximately $C/r$, $C=2Z_d{e}^2/(4\pi {ϵ}_0)$, with outer turning radius $b=C/Q$. For zero orbital angular momentum, the leading WKB probability isHere $R$ is a nuclear matching radius, $\mu$ the reduced mass, $\nu$ an interior assault frequency and $P_{\mathrm{pre}}$ a preformation factor. The barrier factor alone cannot determine the latter two. Nonzero emitted orbital angular momentum adds a centrifugal barrier and can strongly hinder a transition.

The dominant action varies roughly as $Z_d/\sqrt{Q}$, explaining the Geiger-Nuttall relation and the enormous spread of alpha half-lives. angular momentum and parity constrain the emitted partial wave: a spin-zero positive-parity alpha requires $|J_P-J_D|\le L\le J_P+J_D$ and ${\pi }_P={\pi }_D(-1{)}^L$. Fine-structure alpha groups populate different daughter levels and therefore have different kinetic energies and penetrabilities. Shell structure and alpha clustering also change preformation probabilities, so equal $Q$ values do not guarantee equal lifetimes. Escape conserves the total energy throughout tunnelling; the particle does not borrow a measured energy from the barrier.

The effective potential includes the angular part of the kinetic energy. Taking $U(0)=0$ gives $U=k{r}^4/4$, henceFor $J\ne 0$, this diverges at both zero and infinity and has a unique minimum. Finite-energy motion therefore remains between two radial turning points, or at the minimum itself. For $J=0$, the centrifugal barrier is absent, but $U\to \infty$ still bounds every finite-energy trajectory; radial motion can pass through the force centre.

A circular orbit requires $U_{\mathrm{eff}}^{\prime }(r_c)=0$, soIt is stable because $U_{\mathrm{eff}}^{\prime \prime }(r_c)=6k{r}_c^2>0$. These are nonzero-radius orbits for $J\ne 0$; $J=0$ also permits the stationary particle at the origin. Boundedness does not mean that every noncircular orbit is closed.

Initially $r_i=l/2$, ${\dot{r}}_i=0$ and ${\dot{\theta }}_i=v_0/r_i$, so the conserved angular momentum is $J=ml{v}_0/2$. For $v_0\ne 0$, the effective potential contains the centrifugal barrierThe finite conserved energy therefore forbids the hole from being reached:The exceptional case $v_0=0$ has no centrifugal barrier. Then $J=0$ and $\ddot{r}=-Mg/(M+m)$; starting at rest at $l/2$, the tabletop mass reaches the hole after $t=\sqrt{(M+m)l/(Mg)}$. The model ends at that event. Thus the usual nonzero initial tangential speed assumption matters.

Take $\beta >0$. At a radial turning point, $\dot{r}=0$, and the effective potential givesFor $L\ne 0$, the smaller positive radial turning point is conveniently written without division by $E$:This includes $E=0$, where $r_{\min }=L^2/(2m\beta )$. For $E<0$ there is also an outer radial turning point; the two coincide for the circular orbit at $E=-m{\beta }^2/(2L^2)$. Smaller energies are inaccessible at that $L$. For $E>0$ there is just one positive radial turning point. If $L=0$, there is no centrifugal barrier and an inward trajectory can reach the singular centre; its limiting closest distance is zero.

The requested omission of $\beta$ is not possible. For example, at $E=0$ and fixed nonzero $L,m$, doubling $\beta$ halves $r_{\min }$. Thus $E,L,m$ alone cannot determine closest approach.

Define $f(r)>0$ as the magnitude of the inward central force, so $\mathbf{F}=-f(r)\hat{\mathbf{r}}$ and $V^{\prime }(r)=f(r)$. The conserved angular momentum is $J=m{r}^2\dot{\theta }$. Eliminating $\dot{\theta }$ from the mechanical energy yieldsThe added term in the effective potential is the centrifugal barrier. If instead $f$ denotes the signed radial force component, the convention becomes $V^{\prime }=-f$.

In a classically forbidden region the local wavevector becomes imaginary. For a slowly varying barrier, the WKB transmission probability has leading exponentialwhere $r_1,r_2$ are turning points and $\mu$ is the reduced mass. This is a probability exponent; the wave amplitude has only half that exponent. Near a turning point one uses matching formulas rather than the unmodified local WKB expression.

For alpha decay, an attractive nuclear interior supports a quasibound state behind the repulsive Coulomb barrier of the daughter nucleus. For zero orbital angular momentum, write the exterior potential as $C/r$, $C=2Z_d{e}^2/(4\pi {ϵ}_0)$, with nuclear matching radius $R$ and outer turning point $b=C/Q>R$. The exterior action can be integrated:The decay rate is approximately $\lambda =\nu P_{\mathrm{pre}}{e}^{-2G}$ and the half-life is $\ln 2/\lambda$. Here $\nu$ is the interior assault-frequency scale and $P_{\mathrm{pre}}$ is the alpha preformation probability. They are model-dependent; barrier penetration alone does not determine them. Nonzero orbital angular momentum adds a centrifugal barrier.

For $R/b\ll 1$, the leading dependence is $G\propto Z_d/\sqrt{Q}$. Thus a modest increase in decay energy can cause an enormous increase in decay rate, giving the Geiger-Nuttall law relating logarithmic half-life to inverse square-root energy along a suitable isotope sequence. Tunnelling conserves the total energy: the escaping alpha particle does not acquire energy from the barrier.

Radial energy and effective potential. For a central force with potential energy $U(r)$, angular momentum conservation fixes $J=m{r}^2\dot{\varphi }$ and confines motion to a plane. The kinetic energy separates into radial and angular parts, givingThe second term is the centrifugal barrier. It converts the radial motion to a one-dimensional energy problem: allowed radii have $E\ge U_{\mathrm{eff}}$, turning points have equality, and a circular orbit occurs at a stationary point of $U_{\mathrm{eff}}$. Stability of that radial orbit requires a local minimum.

The first-order orbit equation. Write $A=GMm$. Since $f=-dU/dr=-A/r^2-A\gamma /r^3$, choose zero potential at infinity to getAlso $\dot{r}=(dr/d\varphi )J/(m{r}^2)$. Substitute this into the energy equation and multiply by $2m{r}^4/J^2$:Here $\gamma$ has units of length. A regular bound orbit with a centrifugal barrier requires $J^2>mA\gamma$ as well as the appropriate negative energy.

Solving in inverse radius. Let $u=1/r$ and defineDividing the first-order equation by $r^4$ givesDifferentiating, or directly using the Binet equation, gives $u^{\prime \prime }+qu=b$. The continuation through turning points follows from this smooth second-order equation. Its noncircular bound solution, with angular origin at a pericentre, isSubstitution into the first-order equation fixes the amplitude through this $e$. Therefore the requested representation holds withsince ${\alpha }_1/r-{\alpha }_2=\cos ({\alpha }_3\varphi )$. A general angular origin replaces $\varphi$ by $\varphi -{\varphi }_0$. This is the orbit equation for combined inverse-square and inverse-cube attraction. The circular limit $e=0$ is regular in $u=b/q$ but is not represented by finite ${\alpha }_1,{\alpha }_2$ in the source's normalized cosine form.

Precession and the inferred length. The last expression immediately givesSuccessive pericentres occur after angular advance $2\pi /{\alpha }_3$, so the apsidal precession per orbit isFor ${\alpha }_3$ near one, $1-{\alpha }_3^2\simeq \Delta \varpi /\pi$. Inferring an inverse-cube perturbation from apsidal precession consequently gives $\gamma \simeq [J^2/(mA)]\Delta \varpi /\pi$. For the stipulated nearly circular approximation, $J^2/(mA)\simeq r_0$, and Kepler's third law givesusing $P=88(86400)\mathrm{s}$, $M=2.0\times 10^{30}\mathrm{kg}$ and $G=6.6743\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. A century contains about $36525/88=415.1$ orbits. Convert the observed advance to radians per orbit before using it:HenceThe positive sign strengthens the attraction and gives prograde precession. The fractional perturbation $\gamma /r_0\simeq 1.6\times 10^{-7}$ validates the small-deviation expansion. This is the parameter fitting the proposed phenomenological force law under the specified small-eccentricity approximation, not a derivation of that law as the physical explanation of Mercury's anomaly.

For a test particle outside a spherical central mass, the gravitational force is central, so angular momentum is conserved and the orbit is planar. With specific angular momentum $h=r^2\dot{\theta }$ and specific orbital energy $ϵ={\dot{r}}^2/2+h^2/(2r^2)-GM/r$, the radial effective potential combines attraction with the centrifugal barrier. Its nonradial solutions are conic sections:For nonzero $h$, negative energy gives an ellipse with $0\le e<1$, zero energy a parabolic Kepler orbit and positive energy a hyperbolic Kepler orbit. The circle is the special case $e=0$. Exactly radial motion is a separate zero-angular-momentum case and may hit the central body; a mathematical point-mass trajectory is not automatically a physically noncolliding orbit.

Conservation of $h$ gives constant areal velocity $h/2$, so equal areas are swept in equal times. For a bound ellipse of semimajor axis $a$, $ϵ=-GM/(2a)$ and $T^2=4{\pi }^2{a}^3/(GM)$. A circular orbit has speed $\sqrt{GM/r}$, while the minimum local escape speed is $\sqrt{2GM/r}$, obtained by setting the total energy to zero. These are the test-mass two-body results: a comparable second mass requires the relative-coordinate gravitational parameter $G(M+m)$, and nonspherical structure or drag can perturb the conic and its conserved quantities.

Take overdots in this question to mean differentiation with respect to the orbiting particle's proper time $\tau$. Write $\mu =GM/c^2$, $x=r/\mu$ and $\ell =hc/(GM)$, with $h$ the specific angular momentum. Then the dimensionless timelike effective potential isThe last term is the relativistic correction to the Newtonian centrifugal-plus-gravitational potential. At large $r$ the two agree. Newtonian angular momentum gives an infinite inner centrifugal barrier for nonzero $h$; the relativistic correction lowers that barrier and permits plunge. The exterior problem starts at $r=2\mu$, where $U=-c^2/2$; extrapolating the potential towards $r=0$ is not a static exterior description.

Circular orbits obeyThere are two extrema when $h^2>12(GM/c{)}^2$: an inner maximum representing an unstable circular orbit and an outer minimum representing a stable one. At equality they merge at $r=6\mu$; below it neither exists. More explicitly, evaluate the curvature at a circular orbit:Timelike circular orbits require $r>3\mu$. They are radially stable for $r>6\mu$, so the marginally stable inner limit is $r_{\mathrm{ISCO}}=6GM/c^2$. The Newtonian potential instead has one stable minimum at $r=h^2/(GM)$, without an innermost stable circular orbit.

For the Schwarzschild circular-orbit local speed in the question, $\gamma$ is the Lorentz factor relative to a local Schwarzschild static observer, not $dt/d\tau$. That observer's clock obeys $dT=\sqrt{1-2\mu /r}dt$. Since the particle passes with local speed $v$, $rd\varphi /d\tau =\gamma v$. Circular normalization givesThus $v=c/2$ at the innermost stable circular orbit. Using Schwarzschild time in place of the local clock would give a different speed.

For the conserved particle energy measured at infinity, substitute $h^2$ in the circular energy equation. Setting $y=\mu /r$ givesTake the positive root for a future-directed exterior orbit. It approaches $m{c}^2-GMm/(2r)$ at large radius. The marginally bound circular Schwarzschild orbit solves $k=1$, giving $y(4y-1)=0$ and hence $r_{\mathrm{mb}}=4GM/c^2$ at finite radius. It is unstable; strictly bound circular orbits have $r>4\mu$, with $E<m{c}^2$. The phrase smallest bound orbit here denotes this marginal circular limit, not a prohibition on a bound-energy trajectory plunging below it.

At the innermost stable circular orbit,A radiatively efficient thin disc with a zero-torque inner boundary condition can therefore release about 5.7% of the accreted rest energy before the plunge. energy advection, stress at the inner edge and photon capture can alter the escaping luminosity, so this is a model efficiency rather than a universal emitted fraction.

For the Kerr ISCO spin limits of an equatorial prograde disc around a rapidly rotating Kerr black hole, the innermost stable circular orbit moves inward and its specific energy decreases. In the ideal extremal limit its Boyer-Lindquist radius tends to $\mu$, $E/(m{c}^2)\to 1/\sqrt{3}$, and the corresponding binding-energy efficiency tends to 42.3%. Retrograde rotation has the opposite trend: the extremal limit is $r_{\mathrm{ISCO}}=9\mu$, $E/(m{c}^2)=5/(3\sqrt{3})$, giving about 3.8%. The prograde limiting coordinate radius does not identify a massive orbit with the null horizon generator.^[ref]

Finally $\dot{t}=k/(1-2\mu /r)=1/\sqrt{1-3\mu /r}$. Dividing $\dot{\varphi }=h/r^2$ by this givesIts algebraic form is exactly Newtonian, but $r$ is the Schwarzschild spacetime areal radius and $t$ Schwarzschild time, normalized at infinity. Local clock rates, orbit stability and binding energy remain relativistic. With $G{M}_{\odot }\simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$,The marginally bound circular-orbit frequency is about $0.404\mathrm{kHz}$. These Schwarzschild ISCO orbital frequencies and marginally bound frequencies are rotation frequencies per distant stationary observer's time, as intended by using $d\varphi /dt$. A static observer at finite radius $r_o$ instead divides both frequencies by $\sqrt{1-2\mu /r_o}$, leaving their ratio unchanged. Identifying an observed quasi-periodic oscillation with the innermost stable circular orbit orbital frequency is an additional astrophysical assumption.

For an interaction of finite range $b$, higher partial waves are suppressed at $kb\ll 1$ by the centrifugal barrier, away from exceptional higher-wave resonances. Outside the potential, the zero-energy radial S-wave scattering equation is $u^{\prime \prime }=0$, where $\psi =u/r$. Its solution is proportional to $r-a$, so normalization of the constant term givesThe scattering length $a$ is the extrapolated zero of $u(r)$. At nonzero small $k$, this form applies in the overlap region $b\ll r\ll k^{-1}$; it is not the far-field limit at fixed $k$. Equivalently, $k\cot {\delta }_0\to -1/a$ and $f_0\to -a$.