Past exam of the biological physics course of the University of Oxford / 2018 / 1

For the driven two-state rotary motor, use positive angular velocity $\omega >0$ and extend all angular functions with period $2\pi$. In the first half-turn the source figure has $k_{AB}=k_0$, $k_{BA}=0$, so the stationary equation is $\omega P_A^{\prime }=-k_0{P}_A$. In the second half-turn the rates are interchanged, giving $\omega P_A^{\prime }=k_0(1-P_A)$. HenceContinuity at $\pi$ gives $1-C=\alpha e^{-\beta \pi }$, while periodicity gives $\alpha =1-C{e}^{-\beta \pi }$. Solving the two linear conditions gives $C=\alpha$ andIn particular $P_A(0)=P_A(2\pi )=\alpha$ and $P_A(\pi )=1-\alpha$, so the two pieces fill in the apparent endpoint gap in the printed piecewise formula. The probability decreases exponentially during the detachment half-cycle and rises exponentially during the attachment half-cycle; $P_B$ is its complement. Both remain between zero and one. They are state probabilities conditional on rotor angle, not separate angular densities normalized individually to one.

For $\omega \ll k_0$, attachment and detachment almost finish within their allotted half-turns: $P_A$ is nearly zero through most of $0<\theta <\pi$ and nearly one through most of $\pi <\theta <2\pi$, with angular boundary layers of width $\omega /k_0$. For $\omega \gg k_0$, it stays near one half with a small nearly triangular modulation. The upper panels of the original figure in part (b) sketch both limits. At exactly zero speed the advective stationary problem is singular; these statements concern the positive-speed limit.

The source potential rises from zero to $U_0$ over the half-turn from $s$ to $s+\pi$, then falls over the next half-turn. Its slopes are $\pm U_0/\pi$. Torque is minus the angular derivative of potential; in the detached state $U_B$ is angle-independent and exerts no torque. A uniform set of motor phases therefore givesHere $U_0$ is an energy and $T_0$ a torque because the angle is in radians. The average over a full turn divides by $2\pi$, rather than by the bound-state occupation alone.

Differentiating the moving integration limits and using periodicity givesThe optimal phase of a two-state rotary motor follows on $0<s<\pi$: $P_A$ decreases, so the stationary point is a maximum and satisfiesThe crossing displaced by $\pi$ is the corresponding minimum. For slow positive rotation, $s_{\max }\simeq (\omega /k_0)\ln 2$ lies just after the detachment switch. The potential then rises mainly while the motor is detached and falls mainly while it is bound. For fast rotation, $s_{\max }\simeq \pi /2-{\pi }^2{k}_0/(8\omega )$, with only a small phase-dependent occupation available to produce torque. The potential should still be shifted by this optimum, rather than aligned arbitrarily with a rate switch.

The half-turn symmetry gives ${\int }_s^{s+2\pi }{P}_{A}d\theta =\pi$. At the optimum, direct integration of the two exponential pieces gives ${\int }_s^{s+\pi }{P}_{A}d\theta =s_{\max }$. Thus a useful exact check isFor fast rotation, ${\overline{T}}_{\max }\simeq T_0\pi k_0/(8\omega )$ instead tends to zero. The original plots below show the calculated probabilities and optimized potential shapes. The imposed switching rates constitute driven kinetics: their strictly one-way windows are not equilibrium detailed balance at a fixed angle. Sustained work requires the motor's chemical/ion-gradient energy supply, not extraction of work from a passive single thermal bath.