Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 77 / 3

A Stokeslet is the flow produced by a point force. With $C=1/(8\pi \mu )$, its tensor is $G_{ij}=C({\delta }_{ij}/r+r_i{r}_j/r^3)$. A separated pair of equal and opposite forces has no force monopole. Expanding their difference in the separation gives a directional derivative of this tensor. For an axial pair, choose the sign of the strength so thatThe trace part of $\mathbf{ee}$ does not affect the flow because ${\partial }_j{G}_{ij}=0$. The remaining symmetric force-dipole contribution is the stresslet, decaying as $r^{-2}$ instead of the Stokeslet's $r^{-1}$. A force-free swimming cell has thrust balanced by drag, so this is typically its leading far field. With $S>0$ it is a pusher microswimmer: axial outflow and transverse inflow. It is distinct from the antisymmetric torque singularity, the rotlet.

On the axis of an identical stresslet, $\mathbf{r}=\ell \mathbf{e}$ gives velocity $S\mathbf{e}/(4\pi \mu {\ell }^2)$, while the point on the opposite axis has its negative. Their identical intrinsic swimming velocities cancel from the relative motion. The axial repulsion of pusher stresslets obeysThis point-singularity approximation assumes the bodies are separated by distances large compared with their own sizes.

For a flat impermeable free surface with zero tangential stress, the image of a parallel stresslet is a stresslet of the same strength and parallel orientation at the reflected point. Reflection makes the normal velocity odd and tangential velocity even across the plane, enforcing zero normal flow and zero tangential shear. This is the free-surface image of a force dipole; it is not the no-slip-wall image system.

The cell at $z=-h$ lies a distance $2h$ vertically below its image. Since $\mathbf{e}\cdot \mathbf{r}=0$ there, its induced vertical velocity is $S/(32\pi \mu h^2)$ towards the surface. There is no image-induced tangential velocity, vorticity or tangent-normal strain at this point, so a parallel orientation remains parallel in this idealization. Therefore the finite-time free-surface approach of a point stresslet isThe trajectory combines its parallel self-propulsion with this normal drift. Finite-time contact is the point-model prediction for $S>0$, carried over from part (b). A puller with $S<0$ moves away instead. For a finite cell, the far-field approximation fails before $h$ reaches zero; surface deformation and near-contact physics can change the final encounter.

Let $C=S/(8\pi \mu )$. The image of the other stresslet is at relative vector $(\ell ,0,-2h)$, for which $r^2={\ell }^2+4h^2$. Summing direct and image interactions gives the free-surface interaction of two parallel stressletsThe direct neighbour gives no vertical velocity. Its image gives a vertical correction of order $Ch/{\ell }^3$, compared with the own-image attraction of order $C/h^2$. Thus for $h/\ell \ll 1$,The height evolution is unchanged to leading order, while the repulsion is twice the unbounded-fluid value: the neighbouring image lies almost at the same horizontal separation as the real neighbour. At this order ${\ell }^3={\ell }_0^3+3St/(\pi \mu )$ and the height follows part (c). The equality for height is asymptotic, not exact; weak neighbour-induced orientation changes are also neglected in the stated parallel configuration.

The rotlet is the point-torque fundamental solution of Stokes flow. Its torque strength $\mathbf{R}$ is an axial vector, and its circulating velocity decays as $r^{-2}$. A freely swimming cell with no external torque is torque-free. The motor applies opposite internal torques to its body and flagellar apparatus, so the net torque monopole satisfiesIndividual body and flagellar torques need not vanish. A spatially separated pair can therefore leave a rotlet dipole even though there is no far-field rotlet monopole.

A flagellar bundle and the counterrotating body supply oppositely signed axial torques separated along the swimming direction. Their leading torque moments cancel, and the next term is a rotlet dipole, decaying as $r^{-3}$. For the ideal axisymmetric bacterium the torque vector is parallel or antiparallel to $\mathbf{e}$.

To fix the sign convention, define a signed dipole strength $D$ with units torque times length and setDifferentiating givesMore generally, differentiating $\mathbf{R}\times \mathbf{r}/r^3$ gives $\mathbf{R}\times \mathbf{e}/r^3-3(\mathbf{e}\cdot \mathbf{r})(\mathbf{R}\times \mathbf{r})/r^5$; the first term vanishes when $\mathbf{R}\parallel \mathbf{e}$. A derivative with respect to source position instead of observation position changes the definition of the signed moment. The physical handedness depends on which torque is ahead, so it cannot be fixed without that convention.

Take $\mathbf{e}={\mathbf{e}}_x$ and the surface normal ${\mathbf{e}}_z$. Under reflection in the free surface, polar vectors transform by $M=\mathrm{diag}(1,1,-1)$ and an axial torque transforms by $\det (M)M$. A parallel torque therefore reverses sign, while the separation direction ${\mathbf{e}}_x$ does not. The free-surface image of a rotlet dipole is a dipole of strength $-D$ at $z=+h$.

The observation point at the cell is directly below the image, so $\mathbf{e}\cdot \mathbf{r}=0$ and the image velocity vanishes. A symmetry proof of the vanishing normal drift is also useful: reflection in the vertical plane containing $\mathbf{e}$ reverses the axial dipole moment, preserves the normal component of a polar velocity, and leaves the geometry unchanged. Linearity would make that same component change sign with $D$, forcing it to be zero. Thus this singularity alone produces no attraction or repulsion.

Write $X,Y,Z$ relative to the image, so the cell is at $(0,0,-2h)$. The image field isAlthough its value at the cell is zero, its velocity gradient is not. In particular, the surface-induced yaw of a rotlet dipole hasA spherical torque-free body rotates with the local fluid angular velocity ${\Omega }_z={\omega }_z/2$. Its swimming direction therefore turns in the surface plane and, at fixed height and speed, it traces a circle with radius $V_{\mathrm{swim}}/|{\Omega }_z|$. The sign of $D$ sets the sense of turning in this convention. An elongated swimmer also responds to strain; its rotation coefficient depends on its shape and need not equal one half of the vorticity. When the stresslet is included, its changing height produces a changing turning rate rather than an exactly fixed-radius circle.