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For the optical trap, use the trap-frame coordinate $y=x-v_{T}t$ and the force-induced speed $u(y)=F(y)/\zeta$. Then $\dot{y}=u(y)-v_T$. The particle first enters the support at $y=X_R$ and exits at $y=-X_L$. Thus the actual support width is $w=X_R+X_L$. The printed difference $X_R-X_L$ is inconsistent with those endpoints and would give zero width for the symmetric example; below use the geometrically correct $w=2X_0$.

A sufficient escape condition for a translating trap isFor a bounded continuous force this makes the relative velocity strictly negative and bounded away from zero, ensuring finite passage. For an ordinary smooth well, a root of $u(y)=v_T$ met from the incident side instead prevents passage: the trajectory approaches a trap-frame equilibrium. At the limiting maximum-speed equality, the passage time can diverge. Under the strict condition, passage through a translating optical trap givesThe time is the support-crossing time; gaps where the force vanishes inside that interval are included.

There is an essential qualification to the direction claim. Compact support of $F$ alone does not imply positive displacement. For example, $F=-F_0<0$ throughout the support gives $\Delta x=-w(F_0/\zeta )/(v_T+F_0/\zeta )<0$ and satisfies the strict passing condition. A smooth negative bump is also a counterexample. The intended optical trap is a localized conservative well with $F=-U^{\prime }$ and equal outside potential values. This additional assumption gives $\int udy=0$. NowIt is strictly positive for a nonzero force on a set of positive measure. This is forward displacement from a translating localized potential. During motion in the trap direction, the relative passage is slower, so the positive-force side acts longer; opposite motion speeds passage through the negative-force side. A complementary energy argument makes the sign transparent:The moving well must supply the viscous energy loss.

For the high-speed passage expansion of a localized trap, require $\Vert u{\Vert }_{\infty }/v_T\ll 1$. Expanding the denominator uniformly givesTherefore the localized conservative well hasFor a symmetric well the force is odd, the cubic moment vanishes, and the displacement remainder improves to $O(v_T^{-4})$. For an arbitrary compact force without equal outside potential levels, its signed area instead gives the leading $v_T^{-1}$ displacement.

For circular motion, let $C=2\pi R$ and use the local arclength model, with nonoverlapping support $w<C$. In each encounter the relative position traverses $w$, while the particle advances $\Delta x$. Outside the support the particle is stationary and the trap must traverse the remaining relative arclength $C-w$. Hence the time between kicks and the repeated kicks from a circular optical trap response areThe last equality uses the single-passage identity. Equivalently, kicks arrive at the relative lap frequency $f_T-f_p$.

The precise small-correction condition for the large-speed approximation for circular-trap kicks is $\Delta x/C\ll 1$. A sufficient high-speed regime for a localized well is $v_T\gg \Vert F{\Vert }_{\infty }/\zeta$, with $w<C$ fixed: writing $ϵ=\Vert u{\Vert }_{\infty }/v_T<1$, the positive-displacement integral bounds $\Delta x/C\le (w/C){ϵ}^2/(1-ϵ)$. ThusThe locally straight force profile on a real circular track also presumes a trap width small relative to the radius; $R\gg a$ supplies the particle-size part of that approximation.

For the triangular optical-trap response, put $v_c=F/\zeta =C{f}_c$, $\beta =v_T/v_c$ and $\alpha =w/C=X_0/(\pi R)$. In the passing regime $\beta >1$, the two constant-force halves giveSubstitution into the circular formula yieldsFor $0<\beta \le 1$, the particle remains locked to the trap and $f_p/f_c=\beta$. At the ideal triangular cusp this is understood as sticking, or as the limit of a rounded well; it is not a finite passing kick. The passing branch tends continuously to one at $\beta \downarrow 1$, while at large $\beta$ it behaves as $\alpha /\beta$. The zero-speed limiting response is zero. These branches apply to separated kicks, $0<\alpha <1$.

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

The precise condition for the stated approximation is $|\Delta x|/L\ll 1$, in addition to finite passage and separated encounters. A sufficient high-speed regime is $v_T\gg \Vert F{\Vert }_{\infty }/\zeta$ with fixed $w<L$, for which the asymptotic displacement divided by $L$ tends to zero. ThenThis condition controls the encounter frequency correction; there is no need to discard the finite interaction time without accounting for the support width.

For the triangular optical-trap response, set $v_c=F/\zeta =L{f}_c$. In the passing regime $v_T>v_c$, the approach side has velocity $-v_c$ and the trailing side has velocity $+v_c$. ConsequentlyWith $\alpha =2X_0/L=X_0/(\pi R)$ and $\beta =v_T/v_c=f_T/f_c$, the repeated kicks from a circular optical trap formula becomesFor $0<\beta <1$ the usual ideal triangular-well dynamics locks at the cusp: the forces on its two sides direct the trap-frame particle towards the cusp. This is understood as the sticking limit of a rounded potential or of the overdamped differential inclusion at its discontinuous force. The particle follows the trap, giving $f_p/f_c=\beta$. At $\beta =1$ it can remain at a fixed point of the trailing segment and also has $f_p/f_c=1$. Thus the physical triangular optical-trap response is continuous at threshold, even though the isolated passing-kick displacement diverges there. For $\beta \gg 1$,and the more general kick approximation requires $\alpha /({\beta }^2-1)\ll 1$.