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When the midpoint follows a linear velocity field, the two beads' slip velocities are $\pm (\dot{\mathbf{R}}-K\mathbf{R})/2$. Summing their individual drag powers $\zeta |{\mathbf{v}}_{\mathrm{slip}}{|}^2$ proves the displayed expression. Using a probability current velocity in place of the instantaneous relative velocity defines the corresponding overdamped dissipative functional; adding independent thermal velocities is a different stochastic description.

For constant number density $n$, constant temperature, a homogeneous incompressible linear velocity field and decaying configuration tails, defineThis free energy is a relative entropy times $n{k}_{B}T$. Differentiate it using the configuration Fokker-Planck probability current. One integration by parts gives $\dot{\mathcal{F}}=n\int \psi (H\mathbf{R}+k_{B}T\nabla \log \psi )\cdot \mathbf{V}d\mathbf{R}$. Substituting the probability current velocity gives the displayed balance: the diffusive affine contribution is $-n{k}_{B}T\mathrm{tr}K=0$. The polymer stress tensor ${\sigma }_p=n(HC-k_{B}TI)$ has mechanical power ${\sigma }_p:K=nHC:K$. Its instantaneous power need not be nonnegative, because it includes changes of stored free energy. The nonnegative quantity is $\mathcal{D}$, and equality of stress power and dissipation requires a stationary free energy.

Multiply the Hookean dumbbell model configuration equation by $R_i{R}_j$ and integrate twice by parts, with sufficient decay at large separation. The linear velocity field contributes $KC+C{K}^T$, the spring drift contributes $-4HC/\zeta$, and diffusion contributes $4k_{B}TI/\zeta$. Thus the dimensionless conformation tensor $A=HC/(k_{B}T)$ obeys $\dot{A}-KA-A{K}^T=-(A-I)/\lambda$, with polymer relaxation time $\lambda =\zeta /(4H)$. Its left side is an upper-convected derivative for this gradient convention.

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.

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.