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For a smooth passing encounter of a force-driven sphere with a fixed sphere, isotropy and reflection symmetry imply $\mathbf{U}=[m_{\perp }(R)I+(m_{\parallel }(R)-m_{\perp }(R))\mathbf{nn}]\mathbf{F}$. Along a force axis $X$, $U_X$ is even and the transverse velocity is odd in $X$. Uniqueness then makes the trajectory fore-aft symmetric and restores the incoming impact parameter downstream. This consequence of kinematic reversibility of Stokes flow remains valid with lubrication resistance, provided the particles remain smooth, noncontacting and subject only to the stated forces. Surface roughness and contact can invalidate these assumptions.

For $x=ϵa\cos \omega t$ and a tangent $(\sin \theta ,\cos \theta )$ with $\theta =ϵ\cos (\omega t+\varphi )$, the leading force on the fluid is $⟨F_y⟩=-(c_{\perp }-c_{\parallel })L{ϵ}^{2}a\omega \sin \varphi /2$. The sign changes under reversal of the tilt convention. Quadrature strokes maximize its magnitude; reciprocal strokes give zero, consistent with kinematic reversibility of Stokes flow.

For the usual anisotropy $c_{\perp }>c_{\parallel }$ and $a>0$, the maximum positive mean transverse force from a rocking rod occurs atThe maximum magnitude has either quadrature phase $\varphi =\pm \pi /2$. If positive tilt is defined towards $-x$, the positive-force optimum is instead $+\pi /2$. At our optimum $\theta =ϵ\sin \omega t$ is opposite in sign to $\dot{x}$: the rod tilts so that the weaker tangential drag and stronger normal drag combine into a positive $y$ force on both strokes.

For $\varphi =0$ or $\pi$, orientation is a single-valued function of the centre displacement. The configuration retraces its path under time reversal, and kinematic reversibility of Stokes flow makes the force impulse cancel. Equivalently, $⟨\theta \dot{x}⟩=0$. The enclosed loop in the displacement-orientation plane is largest in quadrature and vanishes for a reciprocal stroke, illustrating the principle behind the scallop theorem.

The mobility correction from a fixed distant sphere givesTaking their ratio and keeping the first nonzero transverse correction yieldsto the stated leading order. Set $b=Y_{\infty }\gg a$. The deflection is $O(a^2/b)$, so replacing $Y$ by $b$ on the right introduces only higher-order errors. Integrating from $X=-\infty$ givesThe maximum occurs at $X=0$, and the deflection and spin in a distant sphere encounter areFor the rotation, use $dt=dX/V$ and $Y=b$ to leading order in the angular velocity from part (b). Its signed angle about the positive $z$ axis isThus the rotation is clockwise when viewed from positive $z$, with the magnitude of the displayed leading term.

The deflection tends back to zero downstream: $Y(+\infty )=Y_{\infty }$. More generally, kinematic reversibility of Stokes flow combined with reflection in the plane $X=0$ makes a passing trajectory fore-aft symmetric. One can see this without using the distant-sphere approximation: the relevant translational hydrodynamic mobility matrix has the form $m_{\perp }(R)I+[m_{\parallel }(R)-m_{\perp }(R)]\mathbf{nn}$. Hence $\dot{X}$ is even in $X$ and $\dot{Y}$ is odd in $X$. Uniqueness of the trajectory through $X=0$ then gives $Y(X)=Y(-X)$.

For $Y_{\infty }=a/100$, the numerical deflection and spin approximations above are invalid: the encounter enters a narrow gap and requires lubrication theory. However, the same return to the incoming offset holds for an ideal passing encounter of perfectly smooth spheres in Stokes flow. Lubrication resistance prevents finite-time contact under a bounded force, and does not itself destroy kinematic reversibility of Stokes flow. Contact, surface roughness or nonhydrodynamic forces could change that conclusion; they are additional physics, not part of the ideal model. This distinction is the fore-aft symmetry of a sedimenting-sphere encounter.

For each incompressible Stokes flow without body force, ${\partial }_j{\sigma }_{ij}=0$ andThe divergence theorem givesIncompressibility removes the pressure term, antisymmetric velocity-gradient parts do not contract with the symmetric stress, and the remaining integrand is $2\mu e_{ij}^{(1)}{e}_{ij}^{(2)}$, symmetric under $1\leftrightarrow 2$. This proves the Lorentz reciprocal theorem for Stokes flow:

By Linearity of Stokes flow, the drag on a translating body is linear in its velocity, so $F_i=A_{ij}{U}_j$. Apply the reciprocal theorem to two translation solutions with velocities ${\mathbf{U}}^{(1)}$ and ${\mathbf{U}}^{(2)}$. Their surface velocities are constant and the far-field contribution vanishes, yieldingSince the two vectors are arbitrary,Thus the hydrodynamic resistance matrix is symmetric and, for fixed fluid viscosity, depends only on body geometry.

A straight rod has fore-aft symmetry about its centre. Gravity acts through that centre, so the translational-rotational coupling and applied torque vanish. Kinematic reversibility of Stokes flow and Uniqueness of Stokes flow then rule out spontaneous rotation: the inclination $\theta$ remains constant.

Let the speed for perpendicular translation be $U_0$. The given factor of two makes the parallel speed $2U_0$ for the same force component. Resolving gravity parallel and perpendicular to the rod givesResolving these velocities horizontally and vertically givesTherefore

For an incompressible Newtonian fluid with negligible inertia and no body force, the Stokes equations areThey are linear because any linear combination of solutions corresponds to the same linear combination of boundary data and applied forces. They are reversible because replacing all imposed velocities and forces by their negatives sendsFluid particles therefore retrace their paths under reversal. This is kinematic reversibility of Stokes flow.

Reflect the configuration in a plane perpendicular to the tube axis. The geometry and radial displacement are unchanged, whereas the imposed pressure gradient and every axial velocity reverse. By kinematic reversibility of Stokes flow, a radial migration velocity would also have to reverse; the spatial reflection leaves that component unchanged. Uniqueness of Stokes flow therefore forces it to vanish. Reflection in the meridional plane through the two axes similarly excludes azimuthal drift, so $c$ remains constant.