Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 73 / 1 / b

Write $\mathbf{n}=\mathbf{R}/R$. At fixed geometry, Linearity of Stokes flow makes the translational velocities linear in the couples. With zero forces and couples perpendicular to $\mathbf{R}$, isotropy and reflection symmetry allow only terms proportional to ${\mathbf{G}}_j\times \mathbf{n}$ in each ${\mathbf{U}}_i$. There is no longitudinal term: reflect in the plane containing $\mathbf{R}$ and the common torque axis, taking account of the axial-vector transformation of torque, and use linearity under torque reversal. HenceThe direction of $\mathbf{R}$ can still change; constant separation does not mean stationary centers.

When the couples are equal, a rotation through $\pi$ about their common axis exchanges the identical spheres and reverses the in-plane translational velocities while leaving the axial angular velocities unchanged. Uniqueness of Stokes flow then givesThe midpoint is stationary and the pair can orbit it. The body spin need not equal the angular rate of that orbit. These symmetry statements hold at arbitrary noncontact separation, not just in a distant rotlet approximation.

The minimum dissipation theorem fixes the boundary velocities and the far-field velocity: among admissible incompressible velocity fields, the Stokes flow minimizesFor a competitor $\mathbf{u}+\mathbf{v}$ with zero boundary data for $\mathbf{v}$, integration by parts eliminates the cross term and leaves $\mathcal{D}[\mathbf{u}+\mathbf{v}]-\mathcal{D}[\mathbf{u}]=2\mu \int \mathbf{e}(\mathbf{v}):\mathbf{e}(\mathbf{v})dV\ge 0$.

Extend the actual two-sphere velocity rigidly through sphere 2. It is an admissible field in the one-sphere exterior, with the same translation and rotation of sphere 1. It is continuous across the filled boundary, incompressible, and adds zero strain dissipation inside. Applying the theorem to the exact isolated-sphere flow therefore givesWith all applied couples and the second force zero, the boundary-work identity is ${\mathcal{D}}_{\mathrm{pair}}={\mathbf{F}}_1\cdot {\mathbf{U}}_1$. Thus $6\pi \mu a|{\mathbf{U}}_1{|}^2\le {\mathbf{F}}_1\cdot {\mathbf{U}}_1\le |{\mathbf{F}}_1||{\mathbf{U}}_1|$, which provesThis fixed-force comparison of minimum viscous dissipation uses a comparison at fixed actual velocity first; directly comparing different-force or different-velocity solutions would not justify the result.

If ${\mathbf{G}}_1\ne 0$, the power is instead ${\mathbf{F}}_1\cdot {\mathbf{U}}_1+{\mathbf{G}}_1\cdot {\mathbf{\Omega }}_1$, so the preceding estimate no longer bounds the force contribution alone. The two-sphere hydrodynamic mobility matrix generally has a nonzero self translation-rotation coupling: the freely moving second sphere reflects the first sphere's torque field. In the planar geometry this coupling produces a velocity perpendicular to $\mathbf{R}$; by choosing the sign and magnitude of the couple when the force has a component in that direction, its contribution to ${\mathbf{F}}_1\cdot {\mathbf{U}}_1$ can exceed the isolated force-only value. There is no universal force-only inequality with an additional applied couple. Special geometries can eliminate the coupling, but do not restore a general theorem.

Let $\zeta =6\pi \mu a$, $ϵ=a/R$ and define the dimensionless leading cross-mobility tensorFor a given applied force, the outgoing Stokeslet is fixed by that force, regardless of the sphere's entrainment velocity. At the other center the Faxén translation law gives its force-only velocity plus the incident Stokeslet velocity. The finite-size dipole and the Laplacian Faxén correction first contribute $O({ϵ}^3|\mathbf{V}|)$; an emitted stresslet reflection is still smaller at this order. ThereforeThere is no second-order translational term for prescribed equal forces.

The vorticity of translating-sphere Stokes flow is $\mathbf{\omega }=3a\mathbf{V}\times \mathbf{r}/(2r^3)$; its potential dipole has zero curl. Apply the torque-free Faxén rotation law to the other sphere's field, using $\mathbf{r}=-\mathbf{R}$ at sphere 1 and $\mathbf{r}=\mathbf{R}$ at sphere 2. This yieldsthrough second order in the dimensionless spin $a\mathbf{\Omega }/|\mathbf{V}|$. Neglected spin terms are higher order. This is the method of reflections for Stokes flow, with the ambient field at each application excluding that sphere's own disturbance.

Put ${\mathbf{W}}_2={\mathbf{F}}_2/\zeta$. To the needed accuracy the Faxén translation law gives ${\mathbf{U}}_2={\mathbf{W}}_2+\mathsf{M}\mathbf{V}+O({ϵ}^3|\mathbf{V}|)$. Enforcing its prescribed velocity therefore givesHere the force on sphere 1 remains prescribed; no extra second-order inversion term is needed in ${\mathbf{F}}_2$. Substitute this force into the incident field at sphere 1:The negative second-order term reflects the reduced force required on sphere 2 because sphere 1 already entrains it.

Sphere 2 is now force-free and torque-free. Its translation and spin follow the incident rigid-motion components, so it emits neither a Stokeslet nor a rotlet. Its first disturbance is a stresslet from the incident rate of strain, of magnitude $\mu a^3(a|\mathbf{V}|/R^2)$. Its return velocity at sphere 1 is consequently $O(a^4|\mathbf{V}|/R^4)$:More precisely, that leading correction is $-(15/4){ϵ}^4(\mathbf{V}\cdot \mathbf{n})\mathbf{n}$. It vanishes for purely transverse forcing, for which the first translational correction is of order ${ϵ}^6$.

The first-return spin requires an additional cancellation. The leading incident strain at sphere 2 isIt is axisymmetric about the line of centers, so the reflected stresslet has zero vorticity at sphere 1 on that axis. A naive vorticity estimate of order $(|\mathbf{V}|/a){ϵ}^5$ therefore has zero coefficient. The finite-size correction to the incident strain and the next force-free scattering multipoles are two powers of $ϵ$ smaller; these generically giveThis suppressed spin from a force-free distant sphere is also consistent with the reciprocal torque-to-self-translation coupling. Purely longitudinal forcing gives zero spin exactly by axial symmetry. Distinguishing the stresslet's on-axis curl from its generic off-axis scale is essential here.