Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 80 / 3 / Solution

A freely swimming body has no net external force or torque in Stokes flow. Its force-free condition removes the $r^{-1}$ Stokeslet monopole, and its torque-free condition removes the antisymmetric force-dipole contribution. The leading generic far field is therefore a symmetric traceless rank-two tensor stresslet, decaying as $r^{-2}$. Its axial strength may vanish for special swimmers, in which case higher-order singularities dominate.

The axisymmetric stresslet from a Stokeslet pair provides an explicit construction and fixes the sign convention. Place forces $+F\mathbf{e}$ and $-F\mathbf{e}$ on the fluid at $+(a/2)\mathbf{e}$ and $-(a/2)\mathbf{e}$. The Stokeslet tensor isExpanding their sum for $r\gg a$ givesSincethe dipole strength is $S=Fa$ and the resulting stresslet is the given one. More general surface-traction distributions produce the same leading form through their symmetric traceless first force moment.

Writing $\cos \alpha =\mathbf{e}\cdot \hat{\mathbf{r}}$, its radial velocity isConsequentlyThis classifies pusher microswimmers and puller microswimmers in the force-on-fluid convention used by the given formula.

For the circular swimmer let $\phi =\omega t$, $\mathbf{p}=(\cos \phi ,\sin \phi ,0)$, ${\mathbf{x}}^{\prime }=R\mathbf{p}$, and choose the tangent director $\mathbf{e}=(-\sin \phi ,\cos \phi ,0)$. Its sign is immaterial to the stresslet. Write $\mathbf{r}=\mathbf{x}$ for the observation vector from the circle centre, and assume $r\gg R$. The leading far-field orbit average of a tangent stresslet is obtained by replacing $\mathbf{x}-{\mathbf{x}}^{\prime }$ by $\mathbf{r}$ while retaining the time-dependent director.

The orientation averaging of an axisymmetric stresslet over a full revolution givesSubstitution givesThuswith $\tilde{\mathbf{e}}$ equivalently $-\hat{\mathbf{z}}$. The effective axis is normal to the orbit plane. A positive original strength becomes negative, and a negative strength becomes positive: a pusher averages to a puller, and a puller averages to a pusher, at leading far-field order. The result is independent of $R$ and $\omega$ at this order.

The accuracy of the centre replacement can be made explicit. Half a revolution changes $\mathbf{p}$ and $\mathbf{e}$ to their negatives; the stresslet is even in $\mathbf{e}$. Pairing those times replaces the displaced field byHenceThe finite-radius average is not exactly a point stresslet at every location. For example, on the positive $z$ axis the exact average of the supplied model iswhich has the derived point-stresslet limit for $z\gg R$. This even displacement correction in an orbit-averaged stresslet is consistent with the far-field restriction.