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Opposite forces $\pm F\mathbf{e}$ applied to the fluid at $\pm (a/2)\mathbf{e}$ have no net force or torque. Their leading Stokeslet difference is $-Fa(\mathbf{e}\cdot \nabla )[J(\mathbf{r})\mathbf{e}]$, equal to the axial stresslet with strength $S=Fa$. Positive $S$ drives outward flow on the axis and inward flow in the equatorial plane, corresponding to a pusher microswimmer; negative $S$ gives a puller microswimmer.

A stresslet whose centre follows a circle and whose director stays tangent has a leading far-field mean stresslet centred at the circle centre. Orientation averaging of an axisymmetric stresslet gives normal-axis strength $-S/2$. Thus a pusher microswimmer becomes a leading mean puller microswimmer, and conversely. The radius and traversal frequency enter higher-order spatial corrections or averaging time, but not the leading dipole strength.

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.

Let the unit normal $\mathbf{n}$ point from the fluid toward the free surface, let $\mathbf{t}$ lie along the surface, and write the swimmer direction asReflecting the swimmer position across the plane and reflecting its orientation toconstructs the free-surface image of a force dipole. At the surface the two dipoles have equal tangential velocity and opposite normal velocity. Their sum therefore satisfies the no-penetration boundary condition; tangential velocity is even across the plane and normal velocity is odd, so the tangential traction vanishes and the stress-free boundary condition is also satisfied.

The vector from the image to the swimmer is $\mathbf{r}=-2h\mathbf{n}$, for which $({\mathbf{p}}^{\ast }\cdot \mathbf{r}{)}^2=4h^2{\sin }^2\theta$. Evaluating the image force-dipole flow at the swimmer givesThere is no tangential image velocity at the swimmer in this point-dipole approximation. With the convention that $\mathcal{P}>0$ is an extensile pusher microswimmer, a nearly parallel pusher is attracted toward the surface, whereas a nearly parallel contractile puller microswimmer is repelled. For $|\sin \theta |>1/\sqrt{3}$ the normal drift reverses because the image samples the axial rather than equatorial part of the dipolar flow.