Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 71 (Biological Physics)

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 $\alpha /({\beta }^2-1)\ll 1$.

Take a long periodic section of length $L$, or neglect end effects and choose an integer number of wavelengths. The amplitude $u_q$ here is a single real sine amplitude, fixing the normalization of the thermal membrane undulation spectrum. For a lipid vesicle with an axisymmetric membrane deformation, conservation of the enclosed volume givesThe mean radius therefore decreases at second order. This adjustment is essential: fixing the mean radius instead of the volume would miss the unstable term in the fixed-volume capillary spectrum of a cylindrical membrane.

The surface area of an axisymmetric graph is $\mathcal{S}=2\pi \int R\sqrt{1+R_z^2}dz$. Expanding for $|u_q|\ll R_0$ and $q|u_q|\ll 1$ givesMultiplication by the membrane tension yields a quadratic potential energy $\Delta E=K_q{u}_q^2/2$, whereFor $q{R}_0>1$, the equipartition theorem gives the stable-mode fixed-volume capillary spectrum of a cylindrical membraneHere $k_B$ is the Boltzmann constant and $T$ is the temperature. The cosine amplitude has the same variance and is an independent real coordinate at quadratic order. If instead $R-{\rho }_0={\sum }_{q\ne 0}{a}_q{e}^{iqz}$ with $a_{-q}=a_q^{\ast }$, the corresponding positive-$q$ complex coefficient has $⟨|a_q{|}^2⟩=k_{B}T{R}_0/[2\pi \sigma L((q{R}_0{)}^2-1)]$. This is the same thermal membrane undulation spectrum in a different Fourier mode normalization.

For $q{R}_0<1$, the cylinder is unstable rather than having a negative fluctuation variance. The quadratic surface area change is negative: a sufficiently long-wavelength modulation reduces area while retaining volume. This is the Rayleigh–Plateau instability, expressed here as pearling of a tension-dominated lipid vesicle. A cylinder supporting such modes has no unconstrained harmonic thermal equilibrium about the uniform state. The $q=0$ uniform radius change is forbidden by fixed volume, and $q{R}_0=1$ is marginal in the tension-only approximation. The finite length and endpoint constraints determine which nonzero Fourier modes are allowed.

The large membrane tension assumption controls modes with $q{R}_0$ of order one. For large $q$ the omitted membrane bending modulus contribution grows approximately as $\kappa q^4$ and overtakes the tension term when $\kappa q^2\gtrsim \sigma$. Thus the displayed tension-only spectrum requires $q\ll \sqrt{\sigma /\kappa }$ as well as small amplitudes and a positive stiffness. Close enough to the marginal wavelength, bending corrections must also be retained; their relative scale at $q{R}_0$ of order one is $\kappa /(\sigma R_0^2)$.

For the circular phase boundary in a lipid bilayer, the energy is its perimeter times the line tension $\gamma$. Let $\phi$ be the polar angle and write $r(\phi )={\rho }_0+u_m\sin (m\phi )$, where the periodicity requires an integer $m=q{R}_0\ge 1$. Fixed enclosed area impliesThe perimeter, including the mean-radius change, isThus the fixed-area capillary spectrum of a circular boundary, again for a single real sine or cosine amplitude, isThe complex Fourier mode coefficient convention again divides this result by two. The mode $m=0$ is excluded by the fixed-area constraint, so the negative-stiffness interval $0<m<1$ is not an allowed mode of a closed circular boundary.

At $q{R}_0=1$, the zero stiffness is the translation mode of a circular boundary, not a shape instability. Displacing the centre by a small distance $h$ changes the radius to first order by $h\cos \phi$ or $h\sin \phi$ while leaving area and perimeter unchanged. For example, an exactly translated circle hasThe higher harmonics complete a true translation mode of a circular boundary. The $m=1$ denominator therefore represents free centre motion: an unconfined domain's centre has no restoring force or finite equilibrium variance in an infinite membrane. Radius fluctuations about a recentered domain omit this mode; external confinement would give it a separate restoring stiffness.

Let the electrolyte permittivity be $\epsilon$ and let ${\kappa }_D=1/{\ell }_D$ be the inverse Debye–Hückel screening length. We reserve $k$ for the imposed lateral wavevector. In the Debye–Hückel approximation, linearization about an electrically neutral bulk electrolyte givesaway from fixed charges. For ionic species of valence $z_i$ and bulk number density $n_i$, ${\kappa }_D^2={\sum }_i{n}_i{z}_i^2{e}^2/(\epsilon k_{B}T)$; the approximation requires $|z_{i}e\varphi |\ll k_{B}T$. Substitution of a lateral cosine into this screened Poisson equation leaves a vertical decay constantThe screened sinusoidal surface-charge mode thus decays faster than a laterally uniform charge mode. A nonzero $k$ retains a finite decay length even in the zero-salt limit.

The exterior media and their boundary conditions are not specified in the paper. Specifying the two surface charge densities alone does not fix the normal derivative on the inside unless the outside response is also fixed. We first take the confined-gap idealization, with the prescribed charge supplying all the displacement flux into the electrolyte between the sheets. The electrostatic interface boundary conditions are thenThis is, for example, the limit of negligible exterior displacement admittance. The electrostatic potential solving these conditions and the Debye–Hückel approximation equation isDifferentiation checks the two surface signs directly. This solution allows a sine as well as a cosine lateral component when the phase is nonzero.

In linear screened electrostatics, the quadratic charging free energy is $\frac{1}{2}\int \sigma \varphi dA$. It is equivalently $\frac{\epsilon }{2}\int [(\nabla \varphi {)}^2+{\kappa }_D^2{\varphi }^2]dV$ with the stated boundary conditions. The second term represents the linearized ionic response; integrating only the bare electric field energy would omit it. If $\mathcal{A}$ denotes membrane area, the wavelength-averaged energy density isUsing $⟨{\cos }^2(kx){⟩}_x=1/2$ and $⟨\cos (kx)\cos (kx+\theta ){⟩}_x=\cos \theta /2$ gives the phase registration of screened charged sheets energyThe positive cosine coefficient shows that, for nonzero charge amplitude and finite separation,At the minimum the electrostatic potential simplifies to $\varphi =\alpha \cos (kx)\sinh (pz)/[\epsilon p\cosh (pd/2)]$. Positive and negative charge patches face opposite signs across the gap. The two charge patterns therefore favor a half-wavelength lateral offset rather than like-charge alignment.

For comparison, if identical electrolyte fills all of space on both sides of infinitesimally thin charge sheets, the electrostatic interface boundary conditions are continuity of $\varphi$, decay at infinity, and the charge-induced derivative jumpsSuperposing the two screened sinusoidal surface-charge modes gives, in particular between the sheets,Its full-space continuation replaces each vertical distance by the corresponding absolute distance. Evaluating the electrostatic potential on both sheets now yieldsThe constant term is the two isolated-sheet self energies, and the cosine term is their screened interaction. The energy depends on the exterior convention, but both stated physical idealizations give the same phase registration of screened charged sheets: unlike charge patches oppose each other. In the full-space convention the optimum interaction energy is $-{\alpha }^2{e}^{-pd}/(4\epsilon p)$, giving an attractive normal force density $-{\alpha }^2{e}^{-pd}/(4\epsilon )$. Phase sensitivity becomes exponentially weak for $pd\gg 1$; if $\alpha =0$ or the separation tends to infinity there is no selected phase.

One can also display how an exterior dielectric response interpolates between these cases. Let $Y_o\ge 0$ be its normal displacement admittance for this lateral Fourier mode: $Y_o=0$ for the confined-gap idealization, $Y_o=\epsilon p$ for identical exterior electrolyte, and $Y_o={\epsilon }_o|k|$ for an ion-free exterior dielectric. DefineThe symmetric and antisymmetric surface-charge combinations giveSince $D_s<D_a$, its cosine coefficient is positive. This makes the phase-minimizing conclusion robust while exposing the boundary information needed for an absolute electrostatic energy.