Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 355 / 2

For a cylinder of radius $r$ and length $L$, the mean curvature is $H=1/(2r)$ and the area is $2\pi rL$. Its Helfrich membrane energy per unit length is thereforeThe first term favors a narrow tube and the second penalizes its curvature. Setting the derivative with respect to $r$ to zero givesThe positive second derivative confirms a minimum.

In the high-tension, narrow-gap regime, the almost spherical membrane has area $4\pi (R+\delta {)}^2$. Its spherical bending energy $8\pi k_c$ is independent of radius, while Helfrich repulsion over area $4\pi R^2$ costsUp to terms independent of $\delta$, a suitable free energy is consequentlyIts stationary point satisfiesUsing $\delta /R\ll 1$ gives the fluctuation-supported membrane--particle gapThe inverse-square entropic repulsion prevents contact, while membrane tension limits the area gained by opening the gap.

Let $z$ measure distance across the narrow gap and let $\theta$ be polar angle about the tube axis. In lubrication theory, radial velocity and radial pressure variation are negligible. Axisymmetric incompressible flow is obtained frombecause ${\partial }_{\theta }(u_{\theta }\sin \theta )=0$. The tangential Stokes equation then separates:and hence, after choosing an irrelevant pressure constant,

In the sphere frame, $u_z=-u_{\theta }\sin \theta =-\varphi$. The no-slip boundary condition gives $\varphi (0)=0$ on the sphere and $\varphi (\delta )=U$ on the membrane translating backward relative to it. The sphere-frame volume flux inherited from the narrow remote tube is $-\pi r_0^{2}U$, soUnder the asymptotic condition $r_0^2\ll R\delta$, the right-hand side is negligible at leading order. Solving the quadratic profile subject to the two wall values and zero leading-order integral givesChanging the chosen positive tube direction reverses both signs but leaves the drag magnitude unchanged.

The pressure scale is $p\sim \mu UR/{\delta }^2$, whereas the viscous shear scale is $\tau \sim \mu U/\delta$. After multiplication by comparable areas, pressure drag exceeds shear drag by $R/\delta \gg 1$, an instance of lubrication pressure dominates shear stress. Put $c=\cos \Delta \theta$. The axial pressure force isAs $\Delta \theta \to 0$, the bracket tends to $-2$, and thereforeThe resulting confined-sphere drag coefficient isThus it exceeds the free Stokes drag law coefficient by $2R^2/{\delta }^2$. The Stokes–Einstein relation then gives