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At the initially resting instant $K_{ij}=0$, and the cold-fluid assumption removes the pressure stress. Sum the $xx$ and $yy$ components of the magnetized-fluid tensor virial theorem. The magnetic and gravitational traces areSince $I_{\perp }=I_{xx}+I_{yy}$ and $g_z=-{\partial }_z\Phi$, this yields the horizontal virial balance of a cold magnetized fluid:Being at rest sets the instantaneous velocity to zero; it does not assert equilibrium or zero acceleration. The fields throughout space contribute to this stress integral, including their vacuum exterior.

The pressure-square term in the Cowling energy principle for a rotating barotropic star is nonnegative, and ${\mathcal{N}}^2>0$ makes its stratification term nonnegative. Write $j=R^2\Omega$ for specific angular momentum. ThenFor a regular star reaching the rotation axis, $j(0)=0$. The given $dj/dR>0$ therefore makes $j>0$ for $R>0$, and consequently ${\kappa }^2>0$. Equivalently, with the usual nonnegative angular-velocity convention the sign follows directly. All three energy terms are nonnegative, soThis is stability within the Cowling approximation used throughout. The orientation-independent rotational condition is $d(j^2)/dR\ge 0$, the Rayleigh discriminant criterion. If a fluid region excludes the axis and negative angular velocity is allowed, $dj/dR>0$ alone needs the additional sign of $j$; the squared-angular-momentum criterion avoids that ambiguity.

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

Use a local, homogeneous razor-thin disk approximation in a frame rotating with constant $\Omega$. The unperturbed planar velocity is zero in this frame, and the large-scale gravitational and centrifugal forces balance. Neglect viscosity, magnetic fields, thickness and background gradients across a wavelength. Take small planar disturbances, with wavelength short compared with the galaxy's background scale but long enough for a fluid description. The barotropic closure of a razor-thin disk is $P=K{\Sigma }^{\gamma }$, with $K$ fixed and a positive derivativeSolid-body rotation has no shear and has radial epicyclic frequency $\kappa =2|\Omega |$. By rotational symmetry of the local model choose a wavevector along $x$, and write perturbations proportional to $e^{i(kx-\omega t)}$.

Let ${\Sigma }_1,u,v,{\Phi }_1$ be the surface density, $x$-velocity, $y$-velocity and Newtonian gravitational potential amplitudes. The linearized continuity equation and Euler equations with the Coriolis force areThe perturbing Newtonian gravitational potential solves the Poisson equation for Newtonian gravityIts decaying solution is proportional to $e^{-|k||z|}$. The jump in its derivative is $-2|k|{\Phi }_1(0)=4\pi G{\Sigma }_1$, giving the razor-thin disk Poisson kernelEliminating $v$ from the two momentum equations, and using ${\Sigma }_1=k{\Sigma }_{0}u/\omega$, gives the density-wave branchThe complete linear system also has a zero-frequency balanced mode; it is not the growing density-wave branch, and division by $\omega$ in this elimination excludes it. The result is the uniformly rotating gas-sheet dispersion relation: pressure opposes compression at large wavenumber, self-gravity promotes compression, and rotation provides epicyclic support.

A mode is exponentially unstable when ${\omega }^2<0$. With $\Omega =0$ and $c>0$, this occurs atThe uniform $k=0$ perturbation is marginal rather than growing in this local calculation. With $c=0$, rotation fails to stabilize sufficiently short waves:These limits show why both pressure and rotation are needed for stability at all wavelengths.

For nonzero pressure and rotation, put $q=|k|$. Complete the square:Its minimum lies at $q_{\ast }=\pi G{\Sigma }_0/c^2$. Thus a growing mode exists precisely whenThe printed inequality has the opposite physical meaning: it is the condition for no exponentially growing density wave. Equality is marginal. In the usual gas Toomre stability criterion, $Q=\kappa c/(\pi G{\Sigma }_0)=2|\Omega |c/(\pi G{\Sigma }_0)$, so instability is $Q<1$ and stability is $Q\ge 1$. The unstable wavenumber band of a rotating gas sheet iswhen $Q<1$.

If ${\Sigma }_0$ and $\Omega \ne 0$ remain fixed while $c^2$ decreases slowly, first reach marginality atThe first wavelength to become unstable just below this threshold is the marginal fragmentation wavelength of a rotating sheet:Density maxima are separated by approximately this wavelength. The expected fragment size is of this order; an overdense half-wave has width about ${\ell }_{\mathrm{crit}}/2$. Linear theory fixes a preferred wavelength, not an exact nonlinear clump radius or shape. Further cooling shifts the fastest-growing wavelength to $2c^2/(G{\Sigma }_0)$. A rough fragment mass is consequently of order ${\Sigma }_0{\ell }_{\mathrm{crit}}^2$, with a geometrical factor depending on the nonlinear fragmentation pattern.

Put $\mu =GM$ and use energy and angular momentum per unit particle mass. The Kepler orbit equation is $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$. HenceThus $E$ and $\mathbf{L}=\mathbf{r}\times \dot{\mathbf{r}}$ are integrals of motion. Splitting the velocity into radial and transverse parts, $|\dot{\mathbf{r}}{|}^2={\dot{r}}^2+L^2/r^2$, gives the Kepler radial energy equationHere $\dot{r}$ denotes the derivative of the scalar radius, unlike $\dot{\mathbf{r}}$.

For the Laplace-Runge-Lenz vector, differentiate directly:The vector triple-product identity yieldsThe terms cancel, proving every component of $\mathbf{A}$ is conserved.

Write $q=\bar{r}(r)$ and $\tau =\bar{t}$ for a monotone radial orbit time transformation. The chain rule givesImpose $q^{\prime }=(r/q)\dot{r}$, equivalentlyMultiplying the radial energy equation by $r^2/q^2$ then givesThis is a transformation along each trajectory; the new clock is obtained by integrating its radius-dependent rate.

For the Kepler–harmonic radial duality, take the old inverse-radius term as the new constant energy, $\bar{E}=\mu r/q^2$. On a positive-radius branch $\bar{E}>0$, soThe remaining term is minus the new Newtonian gravitational potential:For $E<0$ this is an attractive isotropic harmonic oscillator with ${\bar{\omega }}^2=-2E(\bar{E}/\mu {)}^2$. For $E>0$ it is an inverted oscillator, and for $E=0$ the Newtonian gravitational potential vanishes. Calling it a confining oscillator therefore requires a bound original Kepler orbit.

For the angular reconstruction of a radial orbit transformation, choose the new angle by $d\bar{\theta }/d\tau =L/q^2$. The same $L$ then supplies its centrifugal term. From $d\theta /dt=L/r^2$, the transformation is $d\bar{\theta }/d\theta =(r/q)(dq/dr)$; in the harmonic case this gives $\bar{\theta }=\theta /2$ up to a constant. Keeping the old angle unchanged would not in general preserve the angular momentum used in the displayed transformed radial equation.

For the second transformation use the explicit displayed split of the energy terms. It assigns weight $\lambda$ to $E$ and $1-\lambda$ to $\mu /r$; the surrounding printed prose interchanges these weights. Define the transformed quantities consistently byTheir difference is exactly $(r^2/q^2)(E+\mu /r)$, so they give the same transformed radial energy equation.

Let $a=(1-\lambda )\mu$, $h=\lambda E$, and $\epsilon =\bar{E}$. Constancy of the new energy imposesFor the standard regular isochrone branch take $h>0$, $\epsilon >0$, and $a\ge 0$. The positive-radius root isFor $q>0$ it is monotone. The corresponding clock is also regular there:With $\alpha =\sqrt{\epsilon /h}$ and $s=\sqrt{q^2+b^2}$, rationalization gives $r/q^2=\alpha /(s+b)$. Eliminate $r^2/q^2$ using $h{r}^2/q^2=\epsilon -ar/q^2$:where one possible identification isThis is the Kepler–isochrone radial transformation to the spherical isochrone model, up to the additive energy constant $k$.

The attractive branch of the Kepler–isochrone transformation requires a suitable parameter domain. The displayed square root needs $\lambda E\bar{E}>0$, and the positive-root branch used above needs $\lambda E>0$, $\bar{E}>0$. An attractive regular isochrone also needs $b\ge 0$ and $\bar{M}>0$. For bound original orbits a convenient domain is $E<0$, $\lambda <0$, $\bar{E}>0$: it satisfies all these requirements and gives $\bar{E}-k=\bar{E}/\lambda <0$. For $E>0$ one may instead take $1/2<\lambda \le 1$, with $\bar{E}>0$; these map to unbound attractive isochrone orbits. Other parameter choices can produce a repulsive Newtonian gravitational potential or a different branch, so no unrestricted real-parameter statement is justified. The singular algebraic case $\lambda =0$ is the harmonic transformation already handled separately. The angular reconstruction $d\bar{\theta }/d\tau =L/q^2$ likewise completes the isochrone radial solution into a central-force orbit.

Synchronous radius and its limits. Kepler's third law for a circumplanetary orbit givesHere $G$ is the gravitational constant. A planet-synchronous orbit repeats after one spin period. A planet-stationary orbit additionally requires a circular orbit, zero orbital inclination to the equator and prograde motion. Then an antenna fixed on the ground points continuously at the same satellite; merely matching the orbital period does not give this property.

Write the planetary radius as $R_p=(3M/(4\pi \rho ){)}^{1/3}$. Requiring the semi-major axis to exceed $R_p$ gives $P^2>3\pi /(G\rho )$. The stellar tidal force restricts the circumplanetary orbit to the Hill sphere, of radius $r_H=a(M/(3M_{\star }){)}^{1/3}$. ThusThese are the surface and Hill-radius constraints in the idealized spherical, small-$M/M_{\star }$ model. Long-term prograde stability generally requires a radius appreciably inside the Hill sphere; for nonzero orbital eccentricity, the surface constraint applies to the periapsis, not just to the semi-major axis.

Collision time and the launch population. A phase-mixed isotropic swarm occupies a shell of radial scale $e{a}_s$. Its number density scales as $N_s/(e{a}_s^3)$, its relative speed as $a_s/P$, and its geometric collision cross-section as $r^2$. Consequently the total collision rate scales as $N_s^2{r}^2/(Pe{a}_s^2)$. To give the numerical normalization used here, adopt an effective shell volume $V=4\pi e{a}_s^3$. At a fixed position the velocity directions are uniformly distributed in the tangent plane, so their mean relative speed is $4v_K/\pi =8a_s/P$. For an unordered pair, the collision cross-section is $4\pi r^2$, givingThus the mean interval between collisions is $A/N_s^2$ in this large-population kinetic model. The effective shell width fixes an order-one coefficient: small orbital eccentricity and random planes alone do not specify a unique radial probability density. Phase mixing, negligible gravitational focusing, $r\ll e{a}_s$, and uncorrelated encounters are implicit in this estimate. Exactly identical orbital periods with perfectly fixed phases do not themselves produce a memoryless collision process.

For nearly planet-stationary orbits with $e\ll I\ll 1$, the swarm volume is smaller by a factor of order $I$, while the relative speed is smaller by the same factor, since vertical motion of scale $I{v}_K$ dominates the eccentric motion. The two changes cancel in the rate $n\sigma v$: there is no parametric factor $I$ in the collision time within the same phase-mixed kinetic approximation. Numerical factors and phase correlations can differ. This is not an argument that bringing all satellites into one nearly circular plane makes their phases random.

With $N_s=Rt$, an Inhomogeneous Poisson process has cumulative hazard functionThe survival function of the first collision is $\exp [-H(t)]$. Setting the expected number of collisions to one givesThis is a characteristic first-event population, with probability $1-e^{-1}$ of an earlier event. It is not a median: the median has an additional factor $(\log 2{)}^{1/3}$.

Which population collides next? Immediately after the first disruption let $S=N_{s0}-2$ and $F=2x^{-3}$. The latter follows from mass conservation for equal-density spherical pieces. Using the same unordered-pair counting and geometric collision cross-sections as above, define $b=(1+x{)}^2/2$; thenHere $S^2,F^2$ stand for the usual large-population approximations to $S(S-1),F(F-1)$. The factor two distinguishing identical and different species is essential. The probability that the next collision is fragment–fragment isIt is the most likely type if $S<4/[x(1+x{)}^2]$ and $S<2/x^2$. It has probability greater than one half ifFor $x\ll 1$, a strongly fragment-dominated next event therefore requires $Sx\ll 4$; the fragment–satellite comparison is more restrictive than the satellite–satellite comparison. If $S=0$, the next event must be fragment–fragment, provided fragments remain.

Population equations and the normalization discrepancy. Every satellite–satellite event produces $2x^{-3}$ fragments and destroys two satellites. Every satellite–fragment event produces a net $x^{-3}-1$ fragments and destroys one satellite; every fragment–fragment event destroys two fragments. Consistent collision counting therefore givesFor $x\ll 1$, $b\simeq 1/2$ and $x^{-3}-1\simeq x^{-3}$. The last two terms in $A{\dot{N}}_f$ are then twice those in the printed equation. This is a genuine factor-of-two inconsistency: equal-size fragment pairs have a collision cross-section smaller by $x^2$, so their event rate must be $x^2{N}_f^2/A$ if the satellite event rate is $N_s^2/A$; destroying both fragments necessarily gives the sink $-2x^2{N}_f^2/A$.

If the printed equation is taken as a prescribed approximate rate model instead, its implicit event rates are $N_s^2/A$, $N_f{N}_s/(4A)$ and $x^2{N}_f^2/(2A)$. Under precisely that mixed normalization, its corresponding satellite equation isThe printed source term also neglects the consumed fragment in a satellite–fragment event, a legitimate relative $O(x^3)$ approximation. That approximation does not repair the pair-counting discrepancy.

The ensuing cascade. This is a two-size fragmentation cascade. First use the printed approximate model, dropping satellite–satellite events as requested. Put $S=N_s$, $F=N_f$, $k=4x^2$ and take $x<1/2$. ThenIntegrating this linear differential equation, with $F_0=2x^{-3}$ and $S_0=N_{s0}-2>0$, givesInitially the collisional cascade grows if $S_0>8x^2$, with approximate early exponential growth time $4A{x}^3/S_0$ when the satellite population is nearly fixed. The fragment population reaches its maximum atAfterwards satellites are depleted and fragment–fragment losses dominate. For $S_0>0$ the continuum solution has $S\to 0$, $F\to 0$, withWhen no satellites remain initially, $F(t)=F_0/[1+x^2{F}_{0}t/A]$ directly. Small integer populations eventually invalidate these deterministic differential equations.

The consistently counted model has, to leading order in $x$, exactly the same curve $F(S)$ and peak, but both retained time derivatives are twice as large. Its growth time is $2A{x}^3/S_0$, and $F(t)\sim A/(2x^{2}t)$. Keeping $(1+x{)}^2$ and the consumed fragment replaces $k$ by $4x^2/(1+x{)}^2$ and $x^{-3}$ in the curve by $x^{-3}-1$. This explicitly separates the physical collision bookkeeping from the printed normalization while giving the evolution under both conventions.

Specific orbital energy, orbital eccentricity and the distribution. Let $\mu =G{M}_{\star }$, $v_K=\sqrt{\mu /a}$ and $c=\cos \theta$. Choose positive $\sin \theta$ to mean an outward radial kick. At release the tangential and radial velocities are $v_t=v_K(1+\gamma c)$ and $v_r=\gamma v_K\sin \theta$. The specific orbital energy and specific angular momentum areUsing ${\epsilon }^{\prime }=-\mu /(2a^{\prime })$ and $e^{\prime 2}=1+2{\epsilon }^{\prime }{h}^{\prime 2}/{\mu }^2$ therefore givesThe denominator must be positive for an elliptic Kepler orbit. A vanishing denominator describes a parabolic Kepler orbit; negative $a^{\prime }$ is the signed semi-major axis of a hyperbolic Kepler orbit.

Differentiating the squared orbital eccentricity givesIn the small-kick regime $0<\gamma <1$, $1+\gamma c>0$. Thus the minimum is at $c=-\gamma /3$, withThe entire expression, including the ${\gamma }^4/27$ term, lies under the square root in the PDF. This minimum is not a universal statement for arbitrary kick size. For example $\gamma =2$, $c=-1$ reverses the circular velocity and gives a retrograde circular orbit with $e^{\prime }=0$; the displayed stationary value is then not the global minimum.

For $0<\gamma <\sqrt{2}-1$, every kick direction gives a bound prograde Kepler orbit. The two endpoint maxima, one at each end of the accessible curve, areThe positive endpoint is the global maximum. There is one interior minimum, at the location just found. Uniform $\theta$ gives the kick-orbit distributionBoth signs of the radial kick occupy the same $e^{\prime }$–$a^{\prime }/a$ curve; they have opposite apsidal orientations. The distribution is concentrated near its endpoints, with integrable square-root singularities. For larger kicks retain the parametric curve and its bound portion $c<(1-{\gamma }^2)/(2\gamma )$; an unbound branch begins at $e^{\prime }=1$, $a^{\prime }/a\to \pm \infty$. In particular, a sketch of a wholly elliptic population presupposes the bound-kick restriction above.

The eccentricity vector at release has radial component $v_t^2/v_K^2-1$ and tangential component $-v_r{v}_t/v_K^2$. With the release radius chosen as zero longitude,This also directly verifies the squared orbital eccentricity above and fixes the sign convention for the longitude of periapsis.

The 1:1 encounter window. Exact equality of orbital periods requires $a^{\prime }=a$, hence $c_0=-\gamma /2$. There are two kick directions with this cosine when $0<\gamma <2$. To quantify a finite window one must specify a return time: near exact commensurability, consider the particle's first complete return to its release point. During that time the source advances through $2\pi (a^{\prime }/a{)}^{3/2}$. Its longitudinal miss distance is, to first order,Thus $d<\Delta$ gives the two-sided cosine windowIntegrating $f_c$ over that window gives, to leading order for a narrow window away from $\gamma =2$,The printed fraction is half this value. It is obtained by retaining only one side of the semi-major axis window, or only one of the two kick-angle branches. Neither restriction is in the question. Thus the printed coefficient cannot be shown for the stated uniform $0\le \theta <2\pi$ population and a symmetric first-return distance criterion. This already exhibits the discrepancy under the usual longitudinal approximation; minimizing the distance over the encounter interval introduces a further velocity-direction correction, not a missing branch. With arbitrarily long observation times, noncommensurate bound trajectories can also return arbitrarily close to the common release point, so an eventual-encounter fraction is a different, time-dependent question.

Rotating-frame sketches. In the rotating reference frame of the source put $x$ radially outward, $y$ along the source's motion and $s=nt$, where $n=v_K/a$. The linear Hill equations for a stellar Kepler orbit, with initial displacement zero, give the velocity-kick epicycleThese solve $\ddot{x}-2n\dot{y}-3n^{2}x=0$, $\ddot{y}+2n\dot{x}=0$ and reproduce the initial kick. For $\theta =0$, the particle starts forward, moves outward and drifts backward; the curve has a loop each orbital cycle, with net $\Delta y=-6\pi \gamma a$ per cycle. For $\theta =\pi$, both coordinates reverse: it moves inward and drifts forward by $6\pi \gamma a$. For $\theta =\pi /2$, the leading trajectory is a closed epicyclic ellipse,It starts at the top of the ellipse moving radially outward and returns to the source after one period to this order. Its exact semi-major axis differs from $a$ at order ${\gamma }^2$, so exact closure is not implied. The local sketches require $|x|,|y|\ll a$, in particular $\gamma s\ll 1$ for the drifting cases.

The dimensional linear friction coefficient is $\zeta$. Put $y=x-v_{T}t$ and $u(y)=F(y)/\zeta$. In overdamped particle dynamics the laboratory velocity is $\dot{x}=u(y)$, whereas the trap-frame velocity is $\dot{y}=u(y)-v_T$. The particle enters the optical tweezers at $y=X_R$ and leaves at $y=-X_L$ when the trap overtakes it. Write $w=X_R+X_L$ for this support width. The printed subtraction in the later definition of the width is a sign error: the distance between these endpoints is their sum. We use $w=2X_0$ in the circular calculation.

For a bounded continuous force, the strict conditionensures finite passage through a translating optical trap: the trap-frame coordinate decreases throughout the interaction. If $u(y)=v_T$ is encountered from the right, the deterministic trajectory cannot pass this stationary trap-frame point for a locally Lipschitz force; it approaches or stays at a locked state. At a smooth maximum the critical passage time diverges. More generally the passage criterion is $v_T-u>0$ on the traversed interval together with a finite integral below. Separating the trap-frame equation givesThese formulas account for both portions of the optical tweezers, including motion against the trap direction.

The forward-displacement result uses the ordinary localized potential energy interpretation of an optical trap: $F=-U^{\prime }$ and $U$ has the same value outside both ends, hence $\int Fdy=0$. Compact support of the force alone does not ensure that condition. Under the equal-endpoint condition,for a nonzero force and a passing trajectory. This is forward displacement from a translating localized potential. A zero force gives zero displacement. If the force is negative everywhere on its support, it instead gives $\Delta x<0$; thus an arbitrary compactly supported force does not satisfy the printed assertion.

There are two complementary explanations of the forward displacement from a translating localized potential. The forward push slows the relative passage and therefore acts longer, while the backward pull speeds up the relative passage and therefore acts for less time. An exact potential energy balance makes the same point: along the trajectoryThe moving optical trap supplies the positive work lost to linear drag, although the initial and final potential energy are equal.

For $v_T\gg \Vert u{\Vert }_{\infty }$, a uniformly convergent expansion of the passage integrals givesFor a localized potential energy well the first integral vanishes, soThe displacement decreases quadratically with the trap speed, rather than linearly. Without the equal-endpoint assumption the general expansion above remains valid.

For repeated kicks from a circular optical trap, let $L=2\pi R$ and interpret the force profile locally along the arc. This description requires a tangential constraint, a nonoverlapping force support $w<L$, and, if the straight profile is used geometrically, a small support compared with $R$. The condition $R\gg a$ alone does not specify that latter width hierarchy. During an encounter the particle advances by $\Delta x$ and the trap advances by $v_T\Delta t=w+\Delta x$. Between encounters the deterministic particle is stationary and the trap travels the remaining relative distance $L-w$. Thus the time between corresponding points of successive encounters isHere $f_T$ and $f_p$ are revolution frequencies, not angular velocities; the printed relation $f_T=v_T/(2\pi R)$ fixes this convention. The encounter frequency is $f_T-f_p$, so the same result follows from $L{f}_p=\Delta x(f_T-f_p)$.