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Bodies immersed in one fluid affect each other through the flow that each produces. At large separation, a force-driven sphere generates a Stokeslet that modifies the other sphere's velocity. These interactions produce off-diagonal entries in the hydrodynamic mobility matrix.

For equal spheres of radius $a$ separated by $R\gg a$, one held fixed and the other forced with $\mathbf{F}=6\pi \mu a\mathbf{V}$, put $\zeta =6\pi \mu a$, $\mathbf{n}=\mathbf{X}/R$ and $\mathsf{A}=3a(I+\mathbf{nn})/(4R)$. The method of reflections for Stokes flow gives $\mathbf{U}=\mathbf{V}-{\mathsf{A}}^2\mathbf{V}+O(V{a}^4/R^4)$, since the fixed sphere creates the reflected Stokeslet with force $-6\pi \mu a\mathsf{A}\mathbf{V}$. The leading hydrodynamic mobility matrix correction therefore reduces longitudinal motion more strongly than transverse motion.

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.

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.

Let $\mu$ be the dynamic viscosity, $u_i={\dot{x}}_i$, and $F_i$ the signed force exerted by sphere $i$ on the fluid. For negligible sphere inertia, this also equals the externally or internally applied force on that sphere; its hydrodynamic force is $-F_i$. The Stokes drag law and the axial velocity of a Stokeslet give the leading hydrodynamic mobility matrixThe factor $1/(4\pi \mu \ell )$ follows from $(I+\hat{\mathbf{r}}\hat{\mathbf{r}})\mathbf{F}/(8\pi \mu \ell )$ when $\mathbf{F}$ is parallel to the line of centres. The longitudinal two-sphere mobility retains the first interaction in $a_i/\ell$; finite-size and repeated-reflection corrections are higher order.

For the linked force-free pair, $F_2=-F_1$. Put $X=(x_1+x_2)/2$ and $S_m=m_1+m_2$. ThensoThe coefficient depends only on the current separation. For any period $T$, $\ell (T)=\ell (0)$, and thereforeThis is a closed integral of a single-valued function of one real shape coordinate. Since $x_1=X-\ell /2$ and $x_2=X+\ell /2$, both spheres also have zero net displacement. Equal radii make $\dot{X}=0$ instantaneously; unequal radii generally permit oscillatory translation, but still no mean motion. This explicit force-free two-sphere stroke is the scallop theorem: a single-parameter reciprocal deformation in a Newtonian fluid at zero inertia cannot produce net free swimming. The timing of extension and contraction cannot change that conclusion, because Stokes flow has no inertial memory.

For the externally prescribed pair, put $q=\omega t$, ${\zeta }_i=1/m_i=6\pi \mu a_i$, andInvert the hydrodynamic mobility matrix:In a large-${\ell }_0$ expansion with $a_i,\delta$ fixed,The isolated-drag term and the constant-$h$ interaction have zero mean. The order-${\ell }_0^{-2}$ term from $h^2{u}_1$ also has zero mean, since its coefficient is constant at that order. Meanwhile,Consequently the externally driven two-sphere pump has the following phase-dependent mean force of an externally driven sphere pair:Here $⟨\cdot {⟩}_t$ denotes averaging over one period. The analogous calculation gives $⟨d{u}_1{⟩}_t={\delta }^2\omega \sin \varphi /2$, so $⟨F_2{⟩}_t=⟨F_1{⟩}_t$ at leading order, andThus the mean fluid forcing is nonzero except at $\varphi =0$ or $\pi$ modulo $2\pi$. The sign changes when the phase lag is reversed. Those exceptional strokes are reciprocal: the position-space loop collapses to a line, and their mean force vanishes by reversibility, not just by this leading expansion.

There is no contradiction with the scallop theorem. The externally imposed motion is not force-free, and the actuators prescribe two phase-shifted motions; their combined motion is not reciprocal for a generic phase. The spheres return to their prescribed positions while transferring a mean force to the fluid. This is pumping by external forcing, rather than propulsion of the freely linked one-shape-coordinate system.

The asymptotic calculation requires $\delta /{\ell }_0\ll 1$ and persistent large separation. The minimum separation of phase-shifted sphere oscillations isThe printed condition ${\ell }_0\ge \delta$ alone does not guarantee nonoverlap or the stipulated far-separated regime. The prescribed trajectories must additionally keep this minimum well above both radii. This is a compatibility qualification on the data, not a change of the phase-dependent force calculation.