Codex @codex  0

In the rotating-hoop bead problem, use a signed angle in the hoop plane from the downward vertical. The gravitational potential energy is $-mga\cos \theta$, the rotating-frame speed is $a\dot{\theta }$, and the distance from the rotation axis is $a\sin \theta$. The centrifugal potential givesGravity is stationary in these rotating coordinates, and the frictionless constraint does no work. Projecting the rotating equation along the hoop, or differentiating its effective potential in the one-dimensional Euler-Lagrange equation, givesIf ${\omega }^2>g/a$, the two off-axis positions areTheir unsigned angle to the downward vertical is the same; they lie on opposite sides of the hoop. The second derivative of the effective potential there is $m{\omega }^2{a}^2{\sin }^2{\theta }_0>0$. Equivalently, writing $\theta ={\theta }_0+\delta$ gives to first orderThe same holds at $-{\theta }_0$. Thus both equilibria are stable, with small-oscillation angular frequency $|\omega |\sin {\theta }_0$.

Set $\theta =0$ initially and choose its positive sense to agree with the initial tangential motion. The inward radial component is $\dot{r}(0)=-4\cos (\pi /3)=-2$, while the tangential speed is $4\sin (\pi /3)=2\sqrt{3}$. HenceSubstitution in the Binet equation gives $12(u^{\prime \prime }+u)=3+9u$, or $u^{\prime \prime }+u/4=1/4$. This is an orbit equation for combined inverse-square and inverse-cube attraction. Solving the constant-coefficient equation and applying both initial conditions givesOn $0\le \theta \le 2\pi$, the sine is nonnegative, so $r$ stays finite and positive. At $\theta =2\pi$, $u=1$, hence $r=1$ and the particle is back at its original spatial position after one revolution. But $u^{\prime }(2\pi )=-1/\sqrt{3}$, so its radial velocity is now $+2$, rather than the original $-2$. The return is not a periodic return in position and velocity.

After that revolution, the first zero of $u$ is at $\theta =8\pi /3$, where $\sin (\theta /2)=-\sqrt{3}/2$. The physical branch therefore has $r\to \infty$ as $\theta \uparrow 8\pi /3$; it must not be continued into negative $u$. Since $dt/d\theta =1/(l{u}^2)$ and $u$ has a simple zero, reaching infinity takes infinite time. As a check, the specific potential energy is $\Phi (r)=-3/r-9/(2r^2)$ and the conserved energy per unit mass is $4^2/2+\Phi (1)=1/2$, giving escape speed $1$ at infinity, also obtained from $-l{u}^{\prime }(8\pi /3)=1$.

Set $\Delta =T/M$, $q_0=q$, and $q_M=q^{\prime }$. The Trotter product formula separates the kinetic and potential energy evolution to first order in $\Delta$. Inserting $M-1$ resolutions of the identity in the position operator basis and using the free short-time kernel givesThe exponent converges formally to the action along the path. The prefactors and intermediate integrations define the time-sliced configuration-space path integral measure, soA real-time path integral is an oscillatory limit with the Feynman i-epsilon prescription, rather than an ordinary probability integral. The operator derivation assumes the usual self-adjointness and product-formula hypotheses for the Hamiltonian.

For a quadratic potential energy write $q=q_c+\eta$, with $\eta (0)=\eta (T)=0$ and ${\ddot{q}}_c+V^{\prime }(q_c)=0$. The first variation vanishes by the Euler-Lagrange equation and the fixed endpoints. Because $V$ is quadratic, the remaining expansion is exact:Here integration by parts has no boundary contribution. Translation of the integration variables leaves the time-sliced measure unchanged. Thus the Gaussian path integral over $\eta$ is independent of the endpoint values, and the entire endpoint dependence lies in $e^{iS[q_c]}$.

Expanding $\eta$ in eigenfunctions satisfying Dirichlet boundary conditions, each real mode contributes an inverse square root of its quadratic eigenvalue, with the oscillatory phase fixed by continuation. Normalizing against $A_0={\partial }_t^2$ gives a precise determinant-ratio form:The determinant ratio is meaningful after a common regulator; the phase is part of the prescription. It is equivalent to writing $N=C(T)({\det }_{D}A{)}^{-1/2}$ with an endpoint-independent normalization.

For example, if $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isA linear or constant term in $V$ changes the action but not this prefactor. The formula holds away from conjugate times, when the boundary problem and determinant are nonsingular. At $\sin \omega T=0$, it must be continued as a distribution with its Maslov index phase; the assertion of an ordinary finite prefactor cannot be used literally there. For the unshifted oscillator, at $T=r\pi /\omega$ the kernel is $e^{-ir\pi /2}\delta (q^{\prime }-(-1{)}^{r}q)$.

For the full-line source integral, impose vacuum boundary conditions by adiabatic damping. Take the source initially as a test function, so its pairings with the Green function are defined. After integration by parts, writeWith the Fourier transform convention $G(t)=\int d\omega e^{i\omega t}\tilde{G}(\omega )/(2\pi )$, the inverse isSet $q=\eta -GJ$. Completing the square givesand consequentlyThe Gaussian functional integral $Z[0]$ is source independent and contains the regulated determinant. This is most naturally read as a normalized vacuum generating functional. For $m>0$, closing the frequency contour gives $G(t)=-i{e}^{-im|t|}/(2m)$; the time-ordered oscillator two-point function is $iG(t)$. This checks the sign of the source exponent and identifies the required vacuum prescription.

For adjoint-valued quantities write $(X\times Y{)}_a={\epsilon }_{abc}{X}_b{Y}_c$. Substituting the infinitesimal transformation into the gauge field strength and using the Jacobi identity givesA useful way to organize the calculation is $[D_{\mu },D_{\nu }]X=g{F}_{\mu \nu }\times X$: the gauge field strength transforms covariantly because the gauge covariant derivative does. Thus $\delta (F_a^{\mu \nu }{F}_{\mu \nu a})=2g{F}_a^{\mu \nu }{\epsilon }_{abc}{F}_{\mu \nu b}{\lambda }_c=0$ by antisymmetry. This proves gauge invariance of the Yang-Mills action.

The Yang-Mills equations are $D_{\mu }{F}^{\mu \nu }=0$. They determine physical evolution only modulo gauge redundancy: gauge transformations with arbitrary spacetime-dependent parameters take solutions to equivalent solutions. In Hamiltonian mechanics, $A_0$ has no independent quadratic time-derivative term, and the corresponding equation is a Gauss law constraint in gauge theory. A gauge parameter can vanish with its necessary derivatives on the initial slice yet change the later potential energy, so the potential energy itself is not uniquely fixed by initial physical data.

In the uncorrected path integral, integration along gauge orbits overcounts equivalent fields. At the perturbative level the quadratic operator of the kinetic term has gauge zero modes and no inverse, so it does not supply a quantum field theory propagator. Gauge fixing together with the Faddeev-Popov determinant removes this obstruction in the local perturbative construction.

For the odd BRST transformation, products obey the graded Leibniz rule. The adjoint product with Grassmann variables satisfiesIn particular, $c\times c$ is not zero, and $c\times Dc=Dc\times c$. The supplied transformations givesince the even covariant derivative obeys $D(c\times c)=2(Dc)\times c$. For the ghost,The final equality is the graded Jacobi identity; in components it is the Jacobi identity for ${\epsilon }_{abc}$ contracted with the totally antisymmetric product of three ghosts. Also $s^2\bar{c}=-sb=0$ and $s^{2}b=0$. The square of an odd derivation is an even derivation, because its two mixed product terms cancel. Hence $s^2=0$ on every polynomial in the fields, off shell, without imposing field equations.

Let $\Psi =\int d^{d}x[{\partial }^{\mu }{\bar{c}}_a{A}_{\mu a}+\xi {\bar{c}}_a{b}_a/2]$ be the gauge-fixing fermion. Since $sF=gF\times c$, $s{\mathcal{L}}_{\mathrm{YM}}=0$. The BRST-exact operator in ${\mathcal{L}}_q={\mathcal{L}}_{\mathrm{YM}}-s\psi$ has zero variation by BRST nilpotence, soTo see the resulting kinetic terms and signs, expand the exact term with $s\bar{c}=-b$:After integration by parts,The auxiliary Nakanishi-Lautrup field $b$ can be completed into a square and integrated out, giving the gauge-fixing term $-(\partial \cdot A_a{)}^2/(2\xi )$ for $\xi \ne 0$. At $\xi =0$, retain $b$ as the multiplier enforcing the gauge condition rather than divide by $\xi$. The Grassmann Gaussian integral gives the determinant of the gauge-condition operator $-\partial \cdot D$. Field-independent normalizations cancel in the normalized integral; insertions involving $b$ can be handled before integration or through its source terms.

At $g=0$ the gauge-fixed quadratic operator is invertible with the usual prescription, and the ghosts have a quadratic kinetic term. The remaining terms are cubic gauge interactions of order $g$, quartic gauge interactions of order $g^2$, and a ghost-gauge interaction of order $g$. Expanding them and using Wick contractions gives a perturbative expansion for any defined polynomial insertion $X$ of the fields. The exponent is the gauge-fixed action $S_q$, as printed in the PDF; the converted TeX incorrectly substitutes $S_A$.

At $\xi =1$, the free gauge action, up to a boundary term, is $\frac{1}{2}\int A_{\mu a}\square A_a^{\mu }$. Its Fourier kernel is $-p^2{\eta }_{\mu \nu }{\delta }_{ab}$. With the mostly-plus convention used above, the Feynman-gauge adjoint propagator isThis is consistent with the scalar-line convention in Question 2.

Finally define the odd functional $B=\frac{1}{2}\int d^{d}x{\bar{c}}_a{b}_a$. Then ${\partial }_{\xi }{S}_q=-sB$. Differentiating the normalized expectation, including its denominator, givesfor an insertion with no explicit $\xi$ dependence. A physical gauge-invariant insertion $X[A]$ is BRST-closed, since its infinitesimal gauge variation vanishes also when the parameter is replaced by $c$. The assumed BRST Ward identity gives $⟨sB⟩=0$ and $⟨s(XB)⟩=0$. For an even $X$ with $sX=0$, the latter says $⟨XsB⟩=0$. ThereforeThe same conclusion holds for any BRST-closed insertion. Mere color-singlet invariance of an arbitrary ghost-dependent expression need not imply $sX=0$; the physical-observable qualification is essential. The proof uses the invariant-measure assumptions encoded in the stated Ward identity.

Use the arch geometry of two hinged rods sliding on a floor, with $A=O=(0,0)$ and $C$ on the other side of $B$. As long as $C$ remains in contact with the floor,The centres of mass of the two rods are $(l\cos \theta ,l\sin \theta )$ and $(3l\cos \theta ,l\sin \theta )$. Their squared speeds are respectively $l^2{\dot{\theta }}^2$ and $l^2(9{\sin }^2\theta +{\cos }^2\theta ){\dot{\theta }}^2$. Each uniform rod has moment of inertia $M{l}^2/3$ about its midpoint, and their angular velocities are $\dot{\theta }$ and $-\dot{\theta }$. Applying the kinetic energy decomposition about the center of mass to each rod givesThe gravitational potential energy is $U=2Mgl\sin \theta$. The fixed pivot does no work; neither does the frictionless floor reaction while $C$ slides horizontally. Conservation of energy from rest therefore givesOn the falling branch $\dot{\theta }$ is the negative square root. Just before the hinge reaches the floor, $\theta =0$, and its speed is $|\dot{B}|=2l|\dot{\theta }|$. Thus, under the maintained-contact interpretation,

There is a contact hypothesis behind this calculation. A floor can push upward but cannot pull downward. For $s=\sin \theta$ and $a=\sin \alpha$, the required normal reaction at $C$, found from the second rod's linear and angular equations, isFor the usual acute arch, this stays nonnegative throughout $0\le \theta \le \alpha$ exactly when $\sin \alpha \le 35/54$. Indeed, for $s>0$, its nonnegativity is $a\le 1/(18s)+11s/6+2s^3$, whose minimum is $35/54$ at $s=1/6$. For larger initial angles, the endpoint lifts before the hinge hits the floor, and the one-coordinate formula no longer governs the whole motion. The displayed impact speed is the intended floor-constrained answer; for unrestricted initial angles, maintained contact is an additional assumption. Subsequent free motion and possible reimpact would require a different model and contact/impact conditions.

For the motionless uniform layer, the heat equation reduces to $T_{zz}=0$. The lower temperature and upper heat-transfer condition determine the quasistatic temperature of a conductively cooled film:When $\alpha h/\kappa \ll 1$, $T_s=T_0-\Delta T\alpha h/\kappa +O(\Delta T(\alpha h/\kappa {)}^2)$. For a varying film let its vertical velocity scale as $Uh/L$, as required by incompressible flow. Relative to vertical thermal diffusion, horizontal thermal diffusion is smaller by $(h/L{)}^2$, and either horizontal or vertical thermal advection is smaller by $U{h}^2/(\kappa L)$. Time variation on the film's transport timescale $L/U$ has the same small factor. Thus the leading heat equation is again $T_{zz}=0$ at each $x,t$ when both specified dimensionless parameters are small. This argument retains the small $\alpha h/\kappa$ assumption. If thickness were externally varied on a faster timescale $t_h$, one would additionally need $h^2/(\kappa t_h)\ll 1$; the spatial conditions alone do not control arbitrary rapid time dependence.

Define $b={\gamma }^{\prime }\Delta T\alpha /\kappa >0$. The surface tension at the surface is ${\gamma }_s={\gamma }_0+bh$, so the Marangoni stress is ${\gamma }_{s,x}=b{h}_x$. With capillary pressure $p=-{\gamma }_0{h}_{xx}$ to leading order, horizontal Stokes flow in the film solves $\mu u_{zz}=p_x$, $u(0)=0$, $\mu u_z(h)={\gamma }_{s,x}$. HenceThe continuity equation gives the thermocapillary thin-film equationTo justify using constant surface tension in the capillary pressure, compare the two volume fluxes on horizontal scale $L$: their magnitudes are ${\gamma }_0{h}^4/(3\mu L^3)$ and $b{h}^3/(2\mu L)$. When both mechanisms are retained at leading order, their balance gives $bh/{\gamma }_0=O((h/L{)}^2)\ll 1$. Thus replacing ${\gamma }_s$ by ${\gamma }_0$ in the curvature term incurs only a relative small correction. Equivalently, small $\alpha h/\kappa$ ensures this if ${\gamma }^{\prime }\Delta T/{\gamma }_0$ is bounded; if it were parametrically large, the extra small surface-tension-variation condition would have to be imposed separately.

The capillary pressure is higher beneath a crest than beneath a trough, so pressure-driven volume flux drains the crest and smooths thickness. The Marangoni effect acts oppositely: a thicker region has a colder surface and therefore larger surface tension. Surface velocity is pulled towards that region, supplying more liquid and amplifying the thickness variation. The arrows below represent these two mechanisms separately.

Put $a_c={\gamma }_0/(3\mu )$ and $a_m=b/(2\mu )$. Substitution of $h=\hat{h}H$, $x=\hat{h}X/ϵ$, $t=\hat{t}\tau$ gives coefficients $a_m{ϵ}^2\hat{t}$ and $a_c{ϵ}^4\hat{t}/\hat{h}$. Making both equal to one yieldsIn these scales the fractional surface tension change is $b\hat{h}/{\gamma }_0=2{ϵ}^2/3$, making the previous approximation explicit.

The linear stability analysis about $H=1$ gives ${\eta }_{\tau }+{\eta }_{XX}+{\eta }_{XXXX}=0$. For a normal mode $e^{s\tau +ikX}$,Long waves with $0<|k|<1$ grow, $|k|>1$ decay, and $k=0,\pm 1$ are neutral. The neutral $k=0$ mode changes the mean film thickness. The dispersion relation is even in $k$, with its two maxima at $\pm 1/\sqrt{2}$.

For a steady positive film with zero volume flux, divide $H^2{H}_X+H^3{H}_{XXX}=0$ by $H^3$ and integrate to obtain $H_{XX}+\ln H=C$. Choosing $\hat{h}$ as specified fixes $C=0$. Multiplication by $H_X$ and another integration then give the zero-flux thermocapillary film profiles first integralFor $H>0$, $V^{\prime }=\ln H$ and $V^{\prime \prime }=1/H>0$: the unique minimum is $V(1)=-1$, while $V(0^{+})=0$ and $V(e)=0$. This potential energy interpretation classifies the profiles without assuming a sinusoidal shape at finite amplitude.

For $E=-1$, the only profile is the uniform film $H=1$. For $-1<E<0$, there are two positive turning heights $H_{-}<1<H_{+}<e$ and a smooth periodic film oscillates between them. Its period isFor $E=0$, the maximum is $e$ and the lower endpoint is zero. A drop reaches zero thickness in finite distance, with zero limiting slope but unbounded curvature since $H_{XX}=-\ln H$. Its half-width isThus the formal limiting profile consists of drops of peak height $e$ and footprint width $2\sqrt{\pi e}$, with zero limiting contact angle. Identical drops can touch, or a zero-thickness region can separate them in the degenerate zero-flux model. These are limiting wet-region solutions, not everywhere positive twice differentiable films; the singular contact region is not resolved by lubrication theory. A useful parametric drop profile isFinally let $E=-1+\delta E$ with small positive $\delta E$. Writing $H=1+\eta$ gives $V=-1+{\eta }^2/2+O({\eta }^3)$ and $H_{XX}=-\eta +O({\eta }^2)$. The small-amplitude steady profile is $H=1+\sqrt{2\delta E}\cos (X-X_c)+O(\delta E)$, with period tending to $2\pi$. Its wavenumber tends to the neutral boundary $|k|=1$ of the unstable band, not to the fastest-growing wavenumber.

Let the vertical-velocity amplitude $W$ be real. A convenient inertia-gravity wave polarization, avoiding pressure denominators, isThe physical velocities and buoyancy are their real parts. The period-mean kinetic energy density and available potential energy density areThe latter follows from $A={\rho }_0{b}^2/(2N^2)$, or from $A={\rho }_0{N}^2{\xi }_z^2/2$ for vertical fluid displacement. The factor $1/4$ includes both the energy definition and the mean of a squared harmonic.

Using the dispersion relation,This energy partition of rotating internal waves shows that the printed request for ordinary kinetic/potential equipartition in the rotating case is false in general. An explicit counterexample is $f=1$, $N=2$, $k=m=1$, $W=1$, giving ${\omega }^2=5/2$, $⟨K⟩=3{\rho }_0/5$ and $⟨A⟩=2{\rho }_0/5$.

The valid modified oscillator balance isThe transverse rotational velocity is in quadrature with the in-plane motion, so it supplies an additional positive energy term on the displacement side of this oscillator balance. It remains physically kinetic energy, not gravitational potential energy. Ordinary kinetic/potential equipartition is recovered when $f=0$ or when the transverse amplitude vanishes. For a real harmonic wave the instantaneous total $K+A$ is constant at a fixed point: the coefficient of ${\cos }^2(kx+mz-\omega t)$ from in-plane motion equals the coefficient of ${\sin }^2$ from transverse motion and buoyancy. This consistency does not imply equality of their separate period means.

Integrating the local two-layer quasi-geostrophic energy conservation law over a horizontal domain and using the specified vanishing boundary flux givesPeriodic boundaries, or suitable fixed streamfunction boundary data eliminating the displayed energy flux, provide examples. The gradient terms are the two layer kinetic energies; the last term is available potential energy, equal to $g^{\prime }{\chi }^2/2$ under the interface-displacement relation.

The baroclinic energy ratio and deformation scale for unequal layer depths uses a decomposition uses the depth-weighted barotropic streamfunction $\Psi =(H_1{\psi }_1+H_2{\psi }_2)/(H_1+H_2)$ and $d=2\tilde{\psi }$. Define $H_r=H_1{H}_2/(H_1+H_2)$. ThenFor variations on horizontal scale $l$, this givesFor comparable layer depths $H_r$ is of order either $H_i$, reproducing the requested scale $l^2{f}_0^2/(g^{\prime }{H}_i)$. If one layer is much thinner, its depth controls this ratio; a depth-independent arithmetic barotropic average would leave unwanted cross terms in the energy decomposition.

Baroclinic potential energy dominates at scales much larger than the two-layer internal deformation radius; baroclinic kinetic energy dominates at much smaller scales. They are comparable near $l\sim R_d$. The independent barotropic kinetic energy has no interface-displacement partner, so this scale comparison refers specifically to the baroclinic component.

Let $a=({\rho }_b-{\rho }_t)/H={\rho }_0{N}^2/g$. The initial density is $\rho (z)={\rho }_t-az$. Conservation of mass in the material homogenized over $-h<z<0$ fixes its mean density to ${\rho }_u={\rho }_t+ah/2$, while the untouched lower layer has interfacial density ${\rho }_i={\rho }_t+ah$. Hence the reduced gravity difference across the interface isThe reference contribution involving ${\rho }_t$ cancels from the difference.

Only the upper layer changes its potential energy. The potential-energy cost of homogenizing a linear stratification issoThe positive sign reflects work against stable stratification.

The rotating boundary supplies energy on scales set by $\Omega$ and $R$. Turbulent drag and dissipation remove energy, and entrainment adds initially stationary fluid that must be accelerated. If upper-layer energy grows, the increasing turbulent loss provides a restoring tendency; after a short adjustment, input and losses can approximately balance while the interface moves on a slower time scale. This gives a physical rationale for a fixed-energy turbulent mixed layer, not a consequence of mass conservation alone. Its characteristic kinetic energy is $K\sim {\rho }_0\pi R^{2}h{u}_u^2/2$. With the stipulated $N$-independent energy closure,so the characteristic speed decreases as $h^{-1/2}$ rather than remaining constant.

The stress does work at a rate proportional to $\pi R^2\tau u_u=c_D{\rho }_0\pi R^2{u}_u^3$. Combine this with the energy increase and absorb fixed drag/geometrical factors into $C>0$:Here the exponents $\alpha ,\beta$ are unrelated to the plume entrainment coefficient in Question 2. The interfacial Richardson number isSet $q=\Omega /N$, $d=\delta /R$ and $r=h/R$. Then $u_u^2/(N^2{h}^2)=d/(2r{\mathrm{Ri}}_i)$ and $r=q(2A{\mathrm{Ri}}_i/d{)}^{1/2}$. Substitution givesThe assumption that $E$ depends on the local Richardson number alone excludes separate dependences on $q,d$. The entrainment exponent from a local interfacial Richardson closure is consequentlyThis comparison treats $q$ and $d$ as independently variable external controls; fitting only a single apparatus would not by itself determine the exponents.

Finally, use the fixed-energy closure again:Integrating from a positive initial mixed-layer depth $h_0$ at time $t_0$ givesWhen the growth term dominates the initial offset, the rotating-disc mixed-layer depth law isTaking the effective time origin to be zero gives the stated scaling. Its validity is limited to the regime of the closures and $h<H$; the singular velocity predicted at $h=0$ is not a model for the initial boundary-layer formation.

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 $L{f}_p=\Delta x(f_T-f_p)$.