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The dynamical method. In Lagrangian mechanics, choose independent generalized coordinates $q_j$ consistent with the constraints, form the Lagrangian $\mathcal{L}=T-V$, and solve the Euler-Lagrange equations,This form applies to a Newtonian system with ideal constraints, so constraint forces do no allowed virtual work, and with the remaining forces derivable from the stated potential. Smooth holonomic constraints can be built into the coordinates. Additional nonconservative generalized forces would appear on the right. Prescribed shaft rotation is a driven constraint; its constant $\Omega$ is a parameter, not a conserved angular momentum.

Energy of the two balls. The diagram labels each ball $m/2$, so their combined mass is $m$. It also shows $\theta$ between an arm and the downward vertical, rather than the full angle between the arms. Each ball has horizontal radius $l\sin \theta$, vertical coordinate $-l\cos \theta$, and squared speed $l^2{\dot{\theta }}^2+l^2{\Omega }^2{\sin }^2\theta$. ThusandThe massless connecting rods and sliding collar enforce the geometry but contribute no energy.

Effective potential. Define ${\omega }_0^2=g/l$. At fixed $\Omega$,This effective potential determines the reduced $\theta$ dynamics. Its conserved effective energy is not the full lab-frame mechanical energy: the shaft drive can exchange work and angular momentum with the balls.

The change of stability. Near $\theta =0$,For $|\Omega |<{\omega }_0$, the downward state is a minimum. At $|\Omega |={\omega }_0$, its quadratic stiffness vanishes but the quartic coefficient is $+{\omega }_0^2/8$. Above that speed, the downward state becomes a maximum and stable nonzero openings appear. This is the gravity-restored centrifugal governor bifurcation.

The physical opening is nonnegative; the two minima in the signed extension represent opposite transverse displacements. The stable opening increases continuously from zero toward a right angle as $|\Omega |$ grows. The apparatus is frictionless in the model, so a displaced system oscillates rather than automatically settling at the equilibrium shown by the minima.

Equilibria. Differentiation givesHence the downward equilibrium is $\theta =0$, and the open equilibria satisfyThe signed extension includes the negative of this angle. Formally $\theta =\pi$ is also stationary, but it is an unstable inverted configuration and lies outside the usual opening range of the mechanism.

Small oscillations. Write $\theta ={\theta }_{\mathrm{eq}}+\eta$. The linear equation is $\ddot{\eta }+{\nu }^2\eta =0$, whereAt the downward state and at a nonzero open state, respectively,These are angular frequencies in ${\mathrm{s}}^{-1}$; ordinary frequencies are $\nu /(2\pi )$. Negative ${\nu }^2$ indicates exponential instability rather than an oscillation. At $\theta =\pi$, ${\nu }^2=-{\omega }_0^2-{\Omega }^2<0$. At the critical speed, the linear frequency is zero and the positive quartic potential gives amplitude-dependent nonlinear restoring motion.

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.

In a fixed point-mass Newtonian gravitational field, a test mass $m$ experiencesBecause this is a central force, its torque about the attracting centre vanishes and angular momentum is conserved. The orbit lies in a fixed plane and sweeps out equal areas in equal times. Since the force is also conservative, the specific orbital energy is constant:The last two terms form an effective potential. They show how the angular-momentum barrier competes with attraction and determines the turning radii. For nonradial motion the orbit is a conic section:Negative energy gives an ellipse, with $e<1$; zero energy gives a parabola; positive energy gives a hyperbola. The circular limit has $e=0$, $v_c=\sqrt{GM/r}$ and $\epsilon =-GM/(2r)$. For a bound ellipse of semimajor axis $a$, $\epsilon =-GM/(2a)$ and the orbital period is $T=2\pi \sqrt{a^3/(GM)}$. The local escape speed is $\sqrt{2GM/r}$. These are the point-mass, test-particle limits; extended mass distributions or noncentral perturbations change the orbit laws.

photon orbit and the Newtonian limit. Write $r_g=GM/c^2$ and $e=E/(m{c}^2)$. For a timelike circular Schwarzschild geodesic,The denominator tends to zero at $r=3r_g$, so the energy per unit rest mass diverges as a massive circular orbit approaches that radius. The corresponding local orbital speed measured by a static observer satisfies $v_{\mathrm{loc}}^2/c^2=r_g/(r-2r_g)$ and tends to one. This suggests a null circular orbit, but one must not simply insert $m=0$ into a massive-particle formula. For a photon, the radial effective potential is proportional to $(1-2r_g/r)/r^2$; its derivative vanishes at $r=3r_g$ and its second derivative there is negative. Thus the photon sphere is at $3GM/c^2$, and its circular null orbits are unstable.

At $r\gg r_g$, expansion givesThe second term is the total Newtonian circular-orbit energy, kinetic plus potential; the relativistic energy also includes rest energy. Here $E$ is the conserved energy referred to a stationary observer at infinity, not the local kinetic energy.

The stable inner edge. At fixed angular momentum, write $\ell =L/(mc)$. The timelike radial geodesic equation can be writtenThe circular condition $V_{\mathrm{eff}}^{\prime }=0$ gives ${\ell }^2=r_g{r}^2/(r-3r_g)$. Substituting into the second derivative taken at fixed $\ell$ givesTimelike circular orbits exist for $r>3r_g$; they are stable against radial perturbations for $r>6r_g$ and unstable for $3r_g<r<6r_g$. ThereforeThe boundary orbit is marginally stable. Differentiating $e(r)$ also gives a minimum there, consistently with this fixed-angular-momentum stability calculation.

accretion efficiency and black-hole spin. Matter supplied slowly from far away initially brings rest energy $d{m}_0{c}^2$. A radiatively efficient, zero-inner-torque thin disk releases the binding energy down to the innermost stable circular orbit; the remaining energy $e_{\mathrm{ISCO}}d{m}_0{c}^2$ is carried into the hole during the plunge. ThusHere $\eta$ denotes the efficiency called $ϵ$ in this question. The ideal relation assumes negligible additional dissipation during plunge and neglects photon capture and advection. A hot radiatively inefficient flow need not have this efficiency.

For a Kerr black hole, an equatorial prograde disk has a smaller innermost stable circular orbit and greater binding-energy efficiency than the Schwarzschild value; a retrograde disk has the opposite trend. The formal extremal prograde limits are $r_{\mathrm{ISCO}}\to r_g$ and $\eta \to 1-1/\sqrt{3}\simeq 0.423$. These are ideal limiting values, not a universal realized efficiency. photon capture and the accretion geometry affect the achievable black-hole spin and escaping radiation.

radiation force and the limiting luminosity. For isotropic luminosity $L$, the outward radiation momentum flux at radius $r$ is $L/(4\pi r^{2}c)$. In fully ionized hydrogen, Thomson scattering acts mainly on the electron, while electrostatic coupling makes the electron and proton move together. Neglecting the electron mass, one electron–proton pair experiencesEquating them yieldsThis Eddington luminosity uses electron-scattering opacity, spherical or isotropic force balance and negligible relativistic corrections at the force-comparison radius. Composition, other opacity, anisotropy and photon trapping can change the relevant limit.

Exponential growth and doubling. Distinguish rest-mass supply ${\dot{m}}_0$ from black-hole mass growth $\dot{M}$:At the Eddington limit,Using the printed scattering cross section gives $t_E\simeq 4.50\times 10^8\mathrm{yr}$. The Salpeter time $t_S$ is an e-folding time; the Eddington-limited doubling time isfor the Schwarzschild thin-disk efficiency. If one approximates $\dot{M}\simeq {\dot{m}}_0$, the factor $1/(1-\eta )$ is dropped and the estimate is about $1.79\times 10^7\mathrm{yr}$. The retained-energy version is the mass-conserving expression.

The efficiency bound. A growth factor $10^8$ requires $N=\log (10^8)=18.4207$ e-folds. Eddington-limited accretion is the fastest growth allowed by the adopted luminosity bound at fixed efficiency, so growth within a time $\Delta t$ requiresThis efficiency upper bound from Eddington-limited growth is an upper bound: higher efficiency radiates more of each supplied rest mass, and the luminosity ceiling therefore allows less mass to enter per unit time. A duty cycle below unity or sub-Eddington luminosity tightens it. The approximate convention $\dot{M}={\dot{m}}_0$ would instead give $\eta \lesssim 0.121$; both figures refer to the same effect, with different treatment of the radiated mass deficit. Mergers, super-Eddington growth or a different seed history would change the premises.

Coherent prograde accretion spins up the hole, moves the inner edge inward and increases $\eta$. For fixed $L/L_{\mathrm{Edd}}$ the mass-growth e-folding time then grows, so the logarithmic growth curve becomes less steep; one cannot simply retain the initial Schwarzschild efficiency throughout. Episodic misaligned accretion can have a different black-hole spin history. In general,

energy available to affect the galaxy. Let $M_g$ denote galaxy mass, keeping it distinct from black-hole mass. At constant efficiency, negligible seed mass and no merger contribution, the emitted energy in building $M_{\mathrm{BH}}=0.002M_g$ isFor $v=300\mathrm{km}{\mathrm{s}}^{-1}$ and the nonspinning efficiency this is approximately $121$. The common small-efficiency approximation gives $2000\eta \simeq 114$, and a fiducial $\eta =0.1$ would give about $222$ with retained mass included. These differences do not alter the conclusion: the available accretion energy substantially exceeds the stated galactic binding-energy estimate.

Only a coupling fraction of order one percent would then be energetically sufficient to affect the gas strongly. radiation, winds or jets can heat gas, drive outflows and suppress further cooling and star formation, providing active-galactic-nucleus feedback and possible self-regulation of black-hole growth. energy availability alone does not prove that the whole galaxy is expelled: most radiation can escape, jets can be collimated, heated gas can cool, and collisionless stars couple differently from gas.

Put $f=1-2\mu /r$ and let dots now denote an affine parameter derivative. The null geodesic lagrangian in the orbital plane is $L=(f{c}^2{\dot{t}}^2-{\dot{r}}^2/f-r^2{\dot{\varphi }}^2)/2$. The time and azimuthal Euler-Lagrange equations integrate once; the null normalization givesFor completeness, the differential radial geodesic equation isthe remaining differential equations are $\ddot{t}+(f^{\prime }/f)\dot{t}\dot{r}=0$ and $\ddot{\varphi }+2\dot{r}\dot{\varphi }/r=0$. Spherical symmetry permits the equatorial choice.

Eliminating $\dot{t},\dot{\varphi }$ yields ${\dot{r}}^2+h^{2}f/r^2=c^2{k}^2$. For $h\ne 0$, divide by $h^2$:The impact parameter is unchanged by a common rescaling of the affine parameter. At infinity it is $c$ times angular momentum divided by energy. Radial photons have $h=0$ and obey the undivided equation ${\dot{r}}^2=c^2{k}^2$.

The effective potential is zero at $r=2\mu$, positive outside, and tends to zero as $r\to \infty$. Its derivative is $V_{\mathrm{eff}}^{\prime }=2(3\mu -r)/r^4$. ThusThe circular null geodesic lies at a maximum, since $V_{\mathrm{eff}}^{\prime \prime }(3\mu )=-2/(81{\mu }^4)<0$. It is unstable: a small radial displacement away from the maximum permits motion away from the orbit. For incoming photons from infinity in the exterior of a black hole, an outer turning point exists when $b>b_{\mathrm{crit}}$; for $b<b_{\mathrm{crit}}$ the photon can cross the event horizon. For a material central body, the vacuum circular orbit is relevant only if $3\mu$ lies outside its surface.

For the new coordinates, set $q=2\mu /r$, so $f=1-q$. The geodesic Lagrangian is $L^{\prime }=[f{c}^2{\dot{t}}^{\prime 2}-2qc{\dot{t}}^{\prime }\dot{r}-(1+q){\dot{r}}^2-r^2{\dot{\varphi }}^2]/2$. Cyclic-coordinate conservation and the null condition give the required analogous equations:Here $k$ denotes the same conserved time-translation quantity as before. The transformation is $t^{\prime }=t+(2\mu /c)\log |r/(2\mu )-1|$. In particular, the displayed time coordinate belongs to Ingoing Kerr-Schild coordinates for Schwarzschild spacetime; the usual advanced null coordinate of ingoing Eddington-Finkelstein coordinates is $v=t^{\prime }+r/c$. Using a metric tensor with a vanishing $d{r}^2$ coefficient while retaining the name $t^{\prime }$ would therefore solve a different coordinate representation.

An elimination that stays finite at the horizon is especially simple:The middle equality uses $q^2+f(1+q)=1$. Hence the radial energy equation and $V_{\mathrm{eff}}(r)$ are unchanged, including the horizon limit. The radial metric tensor block has determinant $-c^2$ and is regular at $r=2\mu$. Its radial null directions are $dr/(cd{t}^{\prime })=-1$ and $dr/(cd{t}^{\prime })=f/(1+q)$, as shown in the original schematic.

This coordinate invariance of the Schwarzschild photon radial potential follows because the transformation changes the time slicing but preserves the areal radius and the stationary Killing vector field ${\partial }_{t^{\prime }}={\partial }_t$. Consequently the conserved Killing energy, angular momentum and impact parameter are the same. photon capture, the circular orbit and the radial turning points cannot depend on coordinate labels; coordinate velocities can.

Put $f(r)=1-2\mu /r$, where $\mu =GM/c^2$, and work in the orbital plane. A dot denotes an affine-parameter derivative; for a material particle choose that parameter to be proper time.

First integrals and normalization. The geodesic Lagrangian isTime translation and axial rotation make $t$ and $\varphi$ cyclic, givingThe tangent norm is constant along an affine geodesic because the Levi-Civita connection preserves the metric. ThereforeFor proper-time parametrization of a timelike worldline this norm is $c^2$. For a null geodesic it is zero, for every affine normalization. Thus $\mathcal{E}=c^2$ for the chosen material-particle normalization and $\mathcal{E}=0$ for light. Proper time cannot itself parametrize a null ray, despite the same printed symbol $\tau$ being used for its affine parameter.

Null normalization and impact parameter. Under $\tau \mapsto A\tau +B$ with $A>0$, both $k$ and $h$ scale as $1/A$, while the null path is unchanged. Their ratio is invariant. At large radius the spatial speed with respect to asymptotic time is $c$, and orbital angular momentum per affine tangent givesFor an orientation chosen with $h>0$, this is $h=bc$. Setting $k=1$ is a choice of affine scale, not a claim that all photons have equal measured energy. Once a tangent is identified with a physically normalized four-momentum, its energy fixes that scale.

Orbit shape. Substituting the constants in the norm and multiplying by $f$ givesFor $u=1/r$ and $h\ne 0$, $du/d\varphi =-\dot{r}/h$. HenceThe $h=0$ radial orbit must be treated with the radial equation, because azimuth is then constant.

Circular photons and stability. For a null ray define $W(r)=f(r)/r^2$. Its radial equation becomesThe circular orbit requires $\dot{r}=0$ and $W^{\prime }=0$, soIt is at a maximum of the null effective potential and is unstable. Equivalently the Binet equation $u^{\prime \prime }+u=3\mu u^2$ linearizes about $u_0=1/(3\mu )$ to $\delta u^{\prime \prime }=\delta u$, admitting exponentially growing radial deviations in azimuth.

Capture from infinity. Divide the null radial equation by $k^2{c}^2$. The quantity opposing inward motion is $b^{2}W(r)$, whose exterior maximum is $b^2/(27{\mu }^2)$. There is no outer radial turning point precisely whenAt equality an exactly critical ray from infinity asymptotes to the photon sphere; it does not cross the horizon at a finite affine parameter. Above the threshold an initially inward ray from infinity reaches an outer turning point and scatters. The radial case $b=0$ is also captured. These statements refer to inward rays launched from the asymptotically flat exterior, not arbitrary initial data inside the angular-momentum barrier.

Approach to the singularity. A captured nonradial ray bends inward, crosses $r=3\mu$ and then the event horizon $r=2\mu$, and continues toward the curvature singularity. A plot in $(r\cos \varphi ,r\sin \varphi )$ is only a coordinate projection, not a Euclidean spatial slice inside the horizon. For large $u$ the cubic term in the shape equation dominates:for the inward branch with $h>0$. Its integral isThus the azimuth tends to a finite limit at the singularity, rather than winding infinitely often there. Near-critical rays can accumulate many turns near $3\mu$ before capture, which is a separate phenomenon. A strictly radial ray has constant azimuth and does not admit this azimuthal asymptotic form. The radial equation also gives $r\propto ({\tau }_{\ast }-\tau {)}^{2/5}$ for a nonradial null ray, confirming finite remaining affine parameter.

Material capture and the condition at infinity. To recover the quoted threshold, take the usual marginally bound orbit, initially at rest at infinity: $k=1$. With $\mathcal{E}=c^2$,The turning-point quadratic has discriminant $h^2(h^2-16{\mu }^2{c}^2)$. It is positive for large $r$ and has no real positive root when $0<|h|<4\mu c$, while the radial case also falls in. ThereforeAt equality the quadratic is proportional to $(r-4\mu {)}^2$ and the orbit asymptotes to the unstable marginally bound circular orbit at $r=4\mu$. For larger $|h|$, an incoming marginally bound particle encounters the outer turning point and scatters.

The source says only “from infinity”, without explicitly fixing the asymptotic speed. For a general incoming speed $v_{\infty }$, $k=(1-v_{\infty }^2/c^2{)}^{-1/2}\ge 1$ andThe same bound $|h|<4\mu c$ remains sufficient for capture, since the added constant is nonnegative. It is the exact threshold only for $k=1$. At fixed $k>1$, the critical unstable orbit satisfiesThen $h_c>4\mu c$. This states both the printed sufficient condition and the additional assumption needed to call it a sharp capture boundary.

Write $A(r)=1-2\mu /r$, with $\mu =GM/c^2$. On the equatorial plane, the geodesic Lagrangian isThe coordinates $t$ and $\varphi$ are cyclic, giving $\boxed{A\dot{t}=k,r^2\dot{\varphi }=h}$. The conserved physical energy per unit rest mass is $c^{2}k$, while $h$ is the signed angular momentum per unit rest mass. The radial Euler-Lagrange equation, with $A^{\prime }=2\mu /r^2$, isThese are the requested geodesic equations. The equatorial condition is consistent with the polar equation by the spherical symmetry of the Schwarzschild metric.

The normalization of the four-velocity is $c^2=c^{2}A{\dot{t}}^2-A^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2$. Insert the two conserved quantities and multiply by $A$ to obtainThis is a one-dimensional radial energy equation. Half of its left side is radial kinetic energy per unit mass plus an effective potential: the Newtonian centrifugal and gravitational terms acquire the extra attractive contribution $-\mu h^2/r^3$.

A timelike circular Schwarzschild geodesic requires $\dot{r}=\ddot{r}=0$. Equivalently, set the derivative of $V(r)=h^2/r^2-2\mu h^2/r^3-2\mu c^2/r$ to zero. At $r=R$ this gives $h^2(R-3\mu )=\mu c^2{R}^2$, and substitution in the radial energy equation yieldsSuch timelike circular orbits require $R>3\mu$. Their coordinate angular velocity satisfies ${\Omega }^2=\mu c^2/R^3$, and their proper clock rate is $d{\tau }_s/dt=A/k=\sqrt{1-3\mu /R_s}$.

For the static North-Pole clock, $d{\tau }_E/dt=\sqrt{1-2\mu /R_E}$. Consecutive overhead passages occur one coordinate orbital period apart, and the repeated Earth-satellite geometry gives equal signal travel delays. The comparison of circular-orbit and static Schwarzschild clocks therefore givesThis includes both gravitational and orbital-motion effects; a separate Newtonian time-dilation correction must not be added again.

For an estimate, Kepler's third law gives $R_s=42100\mathrm{km}(3/24{)}^{2/3}=10525\mathrm{km}$. The same approximate geostationary radius and period give $\mu =4{\pi }^2{R}_{\mathrm{geo}}^3/(c^2{T}_{\mathrm{geo}}^2)\simeq 4.4\mathrm{mm}$. Expanding the ratio to first order,The satellite clock consequently gains about $0.7\mu \mathrm{s}$ over a three-hour orbit in this approximation. The surface observer is taken to be static, and the spacetime model is the spherical Schwarzschild metric specified in the problem.

Set $E=0$ in part (a), substitute the two conserved quantities, and multiply the normalization by $f$:Consequently the null effective potential in Schwarzschild spacetime isOnly radii where the total radial energy is at least this effective potential are accessible.

For $h\ne 0$, the effective potential vanishes at $r=2\mu$, rises to a maximum, and then falls to zero as $r\to \infty$. Its derivative is $V_{\mathrm{eff}}^{\prime }=h^2(3\mu -r)/r^4$, so the only circular null geodesic hasThe second derivative there is $V_{\mathrm{eff}}^{\prime \prime }=-h^2/(81{\mu }^4)<0$, proving instability by the effective potential stability criterion. These circular orbits make up the photon sphere. If the formula is continued inside the horizon, the potential is negative and diverges to $-\infty$ as $r\downarrow 0$. For purely radial light, $h=0$, the potential is identically zero and there is no circular orbit.

For fixed nonzero $l=p_{\varphi }$, the radial effective potential is $V_{\mathrm{eff}}(s)=l^2/(2m{s}^2{\sin }^2\theta )+mgs\cos \theta$. Its stationary point isThe equilibrium is therefore stable. A Taylor expansion with $x=s-s_0$ and $p_x=p_s$ givesQualification: the printed “any” $p_{\varphi }$ excludes a regular interior equilibrium when $p_{\varphi }=0$. In that case $V_{\mathrm{eff}}=mgs\cos \theta$ has no stationary point for $s>0$. Its minimum is the nonsmooth apex, where these coordinates and the smooth-cone dynamics fail.

A spherical three-momentum ultraviolet cutoff givesFor $m>0$ and $\Lambda \gg m$, the asymptotic expansion isIn particular, the leading ultraviolet divergence is quartic. For $m=0$, $\mathcal{E}={\Lambda }^4/(16{\pi }^2)$ exactly.

Write $\phi =⟨\psi ⟩$. Under the proposed identification, the scalar potential from a field-dependent momentum cutoff is $V_{\mathrm{ind}}(\phi )=\mathcal{E}(|\phi |)$. It is even and nonnegative, and its derivative with respect to $|\phi |$ is $|\phi {|}^2\sqrt{{\phi }^2+m^2}/(4{\pi }^2)>0$ away from zero. Consequently its only global minimum is at $\phi =0$: this contribution alone does not cause spontaneous symmetry breaking. At large $|\phi |$ it grows quartically. At small $|\phi |$ with fixed positive $m$, however,Thus the quartic sketch is an asymptotic shape, not the exact small-field behavior for a massive original field. Identifying a regulator with a new field is an additional modeling prescription; an actual effective potential also depends on the new field's dynamics and any other allowed terms.

For central potential $V(r)=\beta r^k$ and nonzero angular momentum $J$, a circular orbit satisfies $\beta k{r}_0^{k+2}=J^2/m$, so attraction requires $\beta k>0$. Twice differentiating the radial effective potential $V+J^2/(2m{r}^2)$ gives $V_{\mathrm{eff}}^{\prime \prime }(r_0)=(k+2)J^2/(m{r}_0^4)$. Its harmonic approximation therefore has radial angular frequency ${\omega }_r=\sqrt{k+2}|J|/(m{r}_0^2)$, whereas $|{\omega }_{\theta }|=|J|/(m{r}_0^2)$. This proves the ratio and shows stability precisely when $k>-2$. At $k=-2$ the tuned effective potential is flat; at $k<-2$ the radial disturbance grows exponentially. Rationality of this local frequency ratio is only a necessary condition for all nearby orbits to close; Bertrand's theorem makes a stronger finite-amplitude assertion.

At fixed specific angular momentum $j$, perturb the effective potential $\Phi (r)+j^2/(2r^2)$ about a circular orbit. Linear radial motion has $\ddot{x}+{\kappa }^{2}x=0$, where ${\kappa }^2={\Phi }^{\prime \prime }+3{\Phi }^{\prime }/r=r^{-3}d(j_c^2)/dr$. Positive ${\kappa }^2$ gives stable oscillations; axial symmetry alone does not force that sign.

For the height profile $h(r)=r^{-2}-r^{-1}$, set $A=L^2+2g$ and use the planar small-slope model. Its effective potential per unit mass is $V_{\mathrm{eff}}=A/(2r^2)-g/r$. Release with negligible radial speed at $r_0$ gives $\mathcal{E}=A/(2r_0^2)-g/r_0$ and the displayed radial-speed law. A bounded trajectory requires $\mathcal{E}<0$, equivalently $r_0>A/(2g)$. The two positive turning radii are $r_0$ and $A{r}_0/(2g{r}_0-A)$; they coincide at the circular solution $r_0=A/g$. Zero energy is marginal escape, not bounded motion.

When the radiofrequency drive is much faster than the Paul trap secular motion and Paul trap micromotion is small, averaging the oscillatory force produces the displayed effective potential. The electric field here is its peak oscillating amplitude. It is an energy, not a static electrostatic potential, so confinement does not contradict the static electrostatic no-trapping theorem. Excess Paul trap micromotion from displacement from the radiofrequency null is distinct from the intended slow Paul trap secular motion.

If a free real scalar field's three-momentum ultraviolet cutoff is identified with the absolute vacuum expectation value of another scalar field, the displayed scalar potential is even and has its unique global minimum at zero. It grows as ${\phi }^4/(16{\pi }^2)$ for $|\phi |\gg m$, but as $m|\phi {|}^3/(12{\pi }^2)$ for $|\phi |\ll m$ with $m>0$. The positive derivative with respect to $|\phi |$ proves that this contribution alone does not produce spontaneous symmetry breaking. This cutoff identification is a model assumption, distinct from deriving a complete effective potential for a dynamical field.

For $V(r)=c/r^2$ with $c>0$, nonzero physical angular momentum $h$ and positive energy $E$, put $\beta =\sqrt{1+2mc/h^2}>1$ and $r_{\min }=\sqrt{(c+h^2/(2m))/E}$. Direct integration of the effective potential radial equation gives the displayed orbit. It has one radial minimum and two straight-line asymptotes separated by polar angle $\pi /\beta$. The repulsive scattering deflection is $\pi -\pi /\beta$; no bounded or circular orbit exists at finite positive energy.

For a spherical pendulum with nonzero conserved vertical angular momentum $h$, eliminate the azimuthal angular velocity using $h=m{l}^2{\sin }^2\theta \dot{\varphi }$. The polar effective potential is $V_{\mathrm{eff}}=h^2/(2m{l}^2{\sin }^2\theta )-mgl\cos \theta$. A conical pendulum at $0<{\theta }_0<\pi /2$ is a minimum satisfying $h^2=m^{2}g{l}^3{\sin }^4{\theta }_0/\cos {\theta }_0$. Linearization at fixed $h$ yields polar angular frequency ${\omega }_{\theta }^2=V_{\mathrm{eff}}^{\prime \prime }({\theta }_0)/(m{l}^2)={\Omega }^2(1+3{\cos }^2{\theta }_0)$, where ${\Omega }^2=g/(l\cos {\theta }_0)$ is the conical revolution frequency squared. Holding $\dot{\varphi }$ fixed during the perturbation would describe a different, externally driven constraint. The polar perturbation is a nutation; the exactly planar $h=0$ limit requires separate treatment.