= Solution
The dimensional <linear friction coefficient> is $\zeta$. Put $y=x-v_Tt$ 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. \b[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 condition
$$
\boxed{v_T>\max_y u(y)=\max_y F(y)/\zeta}
$$
ensures 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 gives
$$
\boxed{\Delta t=\int_{-X_L}^{X_R}\frac{dy}{v_T-u(y)},\qquad
\Delta x=\int_{-X_L}^{X_R}\frac{u(y)}{v_T-u(y)}\,dy
=v_T\Delta t-w.}
$$
These 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>: $F=-U'$ and $U$ has the same value outside both ends, hence $\int F\,dy=0$. <Compact support> of the <force> alone does not ensure that condition. Under the equal-endpoint condition,
$$
\frac{u}{v_T-u}=\frac{u}{v_T}+\frac{u^2}{v_T(v_T-u)},\qquad
\boxed{\Delta x=\frac1{v_T}\int_{-X_L}^{X_R}\frac{u(y)^2}{v_T-u(y)}\,dy>0}
$$
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 trajectory
$$
\frac{dU(y)}{dt}=-\zeta\dot x^2+\zeta v_T\dot x,
\qquad
\boxed{\zeta v_T\Delta x=\zeta\int_{\mathrm{interaction}}\dot x^2\,dt.}
$$
The moving <optical trap> supplies the positive work lost to <linear drag>, although the initial and final <potential energy> are equal.
For $v_T\gg\|u\|_\infty$, a uniformly convergent expansion of the passage integrals gives
$$
\Delta x=\frac1{v_T}\int u\,dy+\frac1{v_T^2}\int u^2\,dy+O(v_T^{-3}).
$$
For a localized <potential energy> well the first integral vanishes, so
$$
\boxed{\Delta x\sim\frac1{\zeta^2v_T^2}\int_{-X_L}^{X_R}F(y)^2\,dy,\qquad
\Delta t=\frac w{v_T}+\frac1{\zeta^2v_T^3}\int F(y)^2\,dy+O(v_T^{-4}).}
$$
The 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 is
$$
T_{\mathrm{kick}}=\Delta t+\frac{L-w}{v_T},\qquad
\boxed{f_p=\frac{\Delta x/L}{\Delta t+(L-w)/v_T}
=\frac{(\Delta x/L)f_T}{1+f_T\Delta t-w/L}
=\frac{\Delta x}{L+\Delta x}f_T.}
$$
Here $f_T$ 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 $Lf_p=\Delta x(f_T-f_p)$.
The precise condition for the stated approximation is $|\Delta x|/L\ll1$, in addition to finite passage and separated encounters. A sufficient high-speed regime is $v_T\gg\|F\|_\infty/\zeta$ with fixed $w<L$, for which the asymptotic displacement divided by $L$ tends to zero. Then
$$
\boxed{f_p\simeq\frac{\Delta x}{2\pi R}f_T.}
$$
This 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 $v_c=F/\zeta=Lf_c$. In the passing regime $v_T>v_c$, the approach side has <velocity> $-v_c$ and the trailing side has <velocity> $+v_c$. Consequently
$$
\Delta t=\frac{X_0}{v_T+v_c}+\frac{X_0}{v_T-v_c}
=\frac{2X_0v_T}{v_T^2-v_c^2},\qquad
\Delta x=\frac{2X_0v_c^2}{v_T^2-v_c^2}.
$$
With $\alpha=2X_0/L=X_0/(\pi R)$ and $\beta=v_T/v_c=f_T/f_c$, the <repeated kicks from a circular optical trap> formula becomes
$$
\boxed{\frac{f_p}{f_c}=\frac{\alpha\beta}{\beta^2-1+\alpha}\quad(\beta>1).}
$$
For $0<\beta<1$ 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 $\beta\gg1$,
$$
\boxed{f_p/f_c\sim\alpha/\beta,}
$$
and the more general kick approximation requires $\alpha/(\beta^2-1)\ll1$.
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