Hookean dumbbell model ($U(\mathbf{R})=H|\mathbf{R}{|}^2/2$)

Represent a dilute polymer by two beads connected by a spring obeying Hooke's law, with stiffness $H>0$ and bead drag coefficient $\zeta >0$. Neglect bead inertia and hydrodynamic interactions. For component-first velocity gradient $K_{ij}={\partial }_j{u}_i$, the configuration Fokker-Planck probability current isThe factor two comes from relative motion of both beads. At rest its normalized probability density is ${\psi }_{\mathrm{eq}}=(H/(2\pi k_{B}T){)}^{3/2}{e}^{-H{R}^2/(2k_{B}T)}$. The conformation tensor $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$ has dimensions of length squared; $(H/(k_{B}T))C$ is its dimensionless version. The Hookean dumbbell in planar extension is a particular imposed flow in this model.

For a locally uniform distribution of labelled connectors, a connector crosses a small surface when its midpoint lies in a slab of thickness $|\mathbf{R}\cdot \mathbf{n}|$. This gives the displayed unsigned count per unit area. Omitting the absolute value gives a signed count, or requires a restriction to one orientation. The signed spring force transmitted from the positive side to the negative side is $H\mathbf{R}\mathrm{sgn}(\mathbf{R}\cdot \mathbf{n})$, so multiplication by the unsigned count yields traction $nH⟨\mathbf{R}{\mathbf{R}}^T⟩\mathbf{n}$. Each labelled dumbbell is counted once.

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.

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.

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.

If the relative bead velocity $\mathbf{v}$ is independent of its separation and has a centered Maxwell-Boltzmann distribution with $⟨v_i{v}_j⟩=(2k_{B}T/m){\delta }_{ij}$, averaging the drag dissipation of a bead-spring pair gives the displayed gross loss. The cross term vanishes by the zero conditional mean. At zero velocity gradient, this remains $3n\zeta k_{B}T/m>0$, although mechanical stress power is zero. A thermal bath replenishes microscopic kinetic energy at equilibrium. Equating this gross loss to polymer stress power omits heat-bath input and cannot follow from the independently specified Maxwellian velocity distribution. The overdamped free-energy balance of a Hookean dumbbell uses a probability current velocity instead.

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.

Let $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$ be the unnormalized conformation tensor, $G_{ij}={\partial }_j{u}_i$ the velocity gradient, and $M/\zeta$ a symmetric positive hydrodynamic mobility matrix independent of the individual connector. The connector Fokker-Planck equation has drift $G\mathbf{R}-(2H/\zeta )M\mathbf{R}$ and diffusion matrix $(2k_{B}T/\zeta )M$. Integrating its second moment twice by parts gives the displayed upper-convected derivative, $C^{\nabla }=D_{t}C-GC-C{G}^T$. If $M$ is a polynomial in $C$, the two matrices commute and the relaxation term simplifies to $-(4H/\zeta )M(C-k_{B}TI/H)$. No Gaussian distribution assumption is required for this moment closure.