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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, $0<\alpha <1$.