In a planar force-driven encounter with a fixed equal sphere and positive impact parameter , the mobility correction from a fixed distant sphere gives . The moving sphere reaches maximum deflection and rotates clockwise through to leading order. The spin is obtained by integrating half the reflected Stokeslet vorticity, as prescribed by Faxén rotation law. These numerical formulas require a distant encounter.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 73 1 b Solution Created 2026-10-03 Updated 2026-10-06
Let , , and letUse the method of reflections for Stokes flow. At the fixed sphere, the first sphere produces the incident Stokeslet velocity to leading order. In Faxén translation law, set the second sphere's translational velocity to zero. The applied holding force is thereforeThe first sphere's finite-radius potential dipole adds to the incident velocity, as does the Laplacian term in Faxén translation law at the fixed sphere. Multiplying by gives the stated next correction. There is no intermediate term. This is the holding force and torque for a sphere in a distant Stokeslet.
The vorticity of a Stokeslet at displacement is . At the fixed sphere, , soSet its angular velocity to zero in Faxén rotation law. The applied holding couple isThe displayed sign is the external couple needed to oppose the ambient rotation, rather than the hydrodynamic couple on the sphere.
The leading reflected flow at the first sphere is . Apply Faxén translation law there with its prescribed force . Since , the mobility correction from a fixed distant sphere isThe next correction comes from finite-radius terms in the incident/reflected flow and in Faxén translation law, together with the fixed sphere's induced stresslet and holding-couple rotlet. Each gives at the first sphere. The symbol is its isolated-sphere velocity scale, not its actual velocity in the two-sphere problem.
The first sphere has no applied couple. Its angular velocity follows from Faxén rotation law and the reflected Stokeslet:The component of parallel to contributes no vorticity. The next reflected rotlet and stresslet give the stated error order.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 73 1 c Solution Created 2026-10-03 Updated 2026-10-06
The mobility correction from a fixed distant sphere givesTaking their ratio and keeping the first nonzero transverse correction yieldsto the stated leading order. Set . The deflection is , so replacing by on the right introduces only higher-order errors. Integrating from givesThe maximum occurs at , and the deflection and spin in a distant sphere encounter areFor the rotation, use and to leading order in the angular velocity from part (b). Its signed angle about the positive axis isThus the rotation is clockwise when viewed from positive , with the magnitude of the displayed leading term.
The deflection tends back to zero downstream: . More generally, kinematic reversibility of Stokes flow combined with reflection in the plane 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 . Hence is even in and is odd in . Uniqueness of the trajectory through then gives .
For , 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.