Passage through a translating optical trap ($\Delta t,\Delta x$)

For a force profile supported on an interval of width $w$, set $u=F/\zeta$. If a trap of speed $v$ overtakes an overdamped particle with $v-u>0$, then $\Delta t=\int dy/(v-u)$ and $\Delta x=v\Delta t-w$. These expressions follow by changing to the trap-frame coordinate, whose velocity is $u-v$. Smooth zeroes of $v-u$ can prevent finite passage.

If the maximum force-induced speed is smaller in magnitude than $v_T$, expand the passage denominator as a uniformly convergent geometric series. For a localized conservative well with equal outside potential levels, the first force moment vanishes and displacement starts at order $v_T^{-2}$. An odd force has no cubic moment, giving the next correction at order $v_T^{-4}$. Compact force support alone does not guarantee a vanishing first moment or positive displacement.

In the trap frame, a particle under linear drag has relative speed $F(y)/\zeta -v_T$. A uniform strictly negative relative speed guarantees finite passage through a compact support. For a bounded continuous force, $v_T>\sup F/\zeta$ supplies that condition. For ordinary smooth forces, an encountered zero of the relative speed is an equilibrium or a passage bottleneck, not a finite crossing. Negative forces do not impede forward trap passage.

For a passing optical trap on a circle of circumference $L$ with nonoverlapping support, a particle that is stationary between encounters receives displacement $\Delta x$ in each encounter. The trap revolution frequency $f_T$ and particle revolution frequency obey $L{f}_p=\Delta x(f_T-f_p)$. This corrects the naive independent-kick estimate when the particle motion changes the encounter rate.

The exact separated-kick response is $f_p=f_T\Delta x/(C+\Delta x)$ for circumference $C=2\pi R$. Its relative correction to the independent-kick estimate is controlled by $\Delta x/C$. For a localized well with force-induced speed bounded by $ϵv_T$, $\Delta x/C\le (w/C){ϵ}^2/(1-ϵ)$. Thus high speed is sufficient when the support width is below one circumference. The local straight-force approximation also requires a narrow trap on the curved track.

A triangular localized potential energy has a force of magnitude $F>0$ on each side of its cusp, directed toward the center. Let its total width be $2X_0<L$, where $L$ is the circumference of the track, and let linear drag have coefficient $\zeta >0$. Define $v_c=F/\zeta$, $f_c=v_c/L$, $\alpha =2X_0/L$ and $\beta =f_T/f_c$. Here $f_c$ is a critical drive frequency, distinct from the thermal calibration corner frequency used elsewhere.

For a passing trap with speed $v_T=L{f}_T>v_c$, the two equal-width sides giveSubstituting this displacement into the exact repeated kicks from a circular optical trap relation yieldsFor $0\le \beta \le 1$, a captured particle can follow the trap with $f_p/f_c=\beta$. At the discontinuous cusp, this statement uses a sticking or smoothly rounded-well interpretation. Finite passage and the captured branch meet continuously at $\beta =1$.

Consider an overdamped particle obeying linear drag in a translating potential energy profile: $\zeta \dot{x}=F(x-vt)$, where $\zeta >0$, $v>0$ and $F(y)=-U^{\prime }(y)$. Assume a finite passing trajectory starts and ends at equal values of $U$.

With $y=x-vt$, the chain rule givesIntegrating between the equal-potential endpoints yieldsThe net displacement is therefore forward, even though the instantaneous force changes sign across a well. The result follows from dissipation and the endpoint condition; compact support of the force alone does not imply equal endpoint potentials.

Writing $u=F/\zeta$ and using $\int udy=0$ gives the equivalent passing-regime identityFor $\Vert u{\Vert }_{\infty }<v$, expanding the denominator gives $\Delta x=v^{-2}\int u^{2}dy+O(v^{-3})$.