Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 71 / 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 $\alpha /({\beta }^2-1)\ll 1$.