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For identical no-slip spheres of radius $a$ linked by a zero-rest-length spring, with $R\gg a$ and $\zeta =6\pi \mu a$, the leading Oseen tensor interaction reduces their relative spring relaxation rate by $3a/(2R)$. The midpoint follows an imposed incompressible linear velocity field. Add the two bead force balance equations to prove midpoint advection; subtract them and use $\mathsf{G}(\mathbf{R})\mathbf{R}=\mathbf{R}/(4\pi \mu R)$ to obtain the displayed equation. This deterministic far-field model omits bead inertia, thermal forcing, connector drag and near-contact lubrication.

The Oseen tensor is the unbounded three-dimensional incompressible Stokes flow Green tensor relating a point force to the velocity it produces: $u_i={\mathsf{G}}_{ij}{F}_j$. At separation $r$, its longitudinal mobility is twice its transverse mobility. It generates the leading far-field hydrodynamic interaction between externally forced particles. The singular value at $r=0$ is not a particle's self mobility; a no-slip sphere instead has self mobility $1/(6\pi \eta a)$. Walls and near-contact lubrication require different or corrected tensors.

Neglect the inertia of the sphere and balance the applied force against Stokes drag. The sphere translates with $\mathbf{U}={\mathbf{F}}_1/\zeta$, where $\zeta =6\pi \mu a$. The applied force is also the net force transmitted to the fluid; the fluid's drag on the sphere has the opposite sign.

Translate the origin to ${\mathbf{r}}_1$. In the solution of part a, retain the terms of order $a/r$ and discard those of order $a^3/r^3$. With $\mathbf{n}=(\mathbf{x}-{\mathbf{r}}_1)/r$,This is the Stokeslet, or equivalently the Oseen tensor $\mathsf{G}(\mathbf{r})=(I+\mathbf{n}{\mathbf{n}}^T)/(8\pi \mu r)$ acting on ${\mathbf{F}}_1$. The omitted finite-size velocity is of order $a^2|{\mathbf{F}}_1|/(\mu r^3)$, so the displayed term is the leading far field for $r\gg a$. The singular Stokeslet must not be evaluated at its own centre to obtain a particle's self mobility.

Use the deterministic Hookean dumbbell model, with a zero-rest-length spring and negligible connector drag. For $\mathbf{R}={\mathbf{r}}_1-{\mathbf{r}}_2$, its elastic forces on the two beads areHere $H>0$ is the spring constant. No independent thermal forcing from Brownian motion or bead inertia is included. The unperturbed linear velocity field is $\mathbf{Lr}$, with $\mathrm{tr}\mathbf{L}=0$ by incompressibility.

The fluid velocity used in each bead's drag law is the imposed flow plus the disturbance from the other bead, with its own singular field omitted. At leading order in $a/R$, the evenness of the Oseen tensor, $\mathsf{G}(-\mathbf{R})=\mathsf{G}(\mathbf{R})$, givesThese are the pair hydrodynamic mobility matrix relations. Their signs follow from balancing each spring force with $-\zeta ({\dot{\mathbf{r}}}_i-{\mathbf{u}}_{\mathrm{ext},i})$, the fluid drag on that bead.

Adding the equations cancels both the direct spring slips and the cross-induced velocities, since ${\mathbf{F}}_1+{\mathbf{F}}_2=0$. For the midpoint ${\mathbf{r}}_c=({\mathbf{r}}_1+{\mathbf{r}}_2)/2$,The midpoint therefore follows the undisturbed linear velocity field. Subtracting the equations instead gives the hydrodynamic interaction of a Hookean dumbbell:The dimensionless interaction matrix isIts transverse and longitudinal eigenvalues are $3a/(4R)$ and $3a/(2R)$, respectively. Since the spring force is longitudinal, the separation equation can also be writtenThe relative correction to the free-draining spring relaxation rate is therefore $3a/(2R)$, explicitly proportional to $a/R$. Entrainment by the other bead reduces their relative approach for given opposing spring forces.

This is a far-field hydrodynamic interaction, valid for $R\gg a$; it is not a near-contact law and omits higher reflections and lubrication. The printed target also fixes the spring's rest length implicitly. For an ordinary spring obeying Hooke's law with nonzero rest length $R_0$, the elastic force would be $-H(R-R_0)\hat{\mathbf{R}}$, and every spring term above would acquire the factor $1-R_0/R$. The displayed result uses the zero-rest-length spring convention appropriate to the ideal Hookean dumbbell model.

A moving or externally forced colloid disturbs the surrounding fluid, changing the velocity of other colloids even without direct contact. These hydrodynamic interactions are long-ranged at small Reynolds number. In an unbounded incompressible Newtonian liquid, the Stokes flow response to a point force is the Stokeslet:Its Oseen tensor isThis is a mobility tensor, with units velocity per force, rather than a stress tensor. At separation $r$ the response parallel to the line of centres is $1/(4\pi \eta r)$; the perpendicular response is $1/(8\pi \eta r)$. Flow is not isotropic even though the solvent is. The disturbance decays as $1/r$, so treating all drag forces as independent can miss correlated motion.

For well-separated equal spheres of radius $a$, the leading hydrodynamic mobility matrix givesThe first term is the isolated-sphere self mobility; the remaining terms couple distinct particles. At thermal equilibrium the fluctuation-dissipation theorem also predicts correlated short-time increments, $⟨\Delta r_{i\alpha }\Delta r_{j\beta }⟩\simeq 2k_{B}T{\mu }_{i\alpha ,j\beta }\Delta t$ with the configuration locally fixed. Thus correlated Brownian displacements can reveal hydrodynamic coupling without an applied force.

The point-force approximation requires $r\gg a$ and negligible inertia. It diverges at the origin and cannot replace the finite self mobility. Near-contact lubrication, walls, many-body corrections, and a viscoelastic solvent require more complete mobility laws. A force-free particle can instead have a faster-decaying leading disturbance, so the $1/r$ claim specifically concerns a net-force-driven object.