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.

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.

A triangular localized potential energy has forces $+F$ behind and $-F$ ahead of the trap centre. With $v_c=F/\zeta =L{f}_c$, $\alpha =2X_0/L$ and $\beta =f_T/f_c$, its circular response is $f_p/f_c=\alpha \beta /({\beta }^2-1+\alpha )$ for $\beta >1$. A trapped particle has $f_p/f_c=\beta$ for $0\le \beta \le 1$, interpreting the discontinuous force at the cusp by a sticking or rounded-well limit.

For a translating potential energy profile with equal values at the two ends and linear drag force $-\zeta \dot{x}$, a finite passing trajectory satisfies $v\Delta x=\int {\dot{x}}^{2}dt\ge 0$. Equivalently, since $\int Fdy=0$, $\Delta x=\int u^2/[v(v-u)]dy$. Compact support of the force alone is insufficient. At high speed the displacement is $v^{-2}\int (F/\zeta {)}^{2}dy+O(v^{-3})$.