Past exam of the advanced fluid dynamics course of the University of Oxford / 2024 / 2

Using the given linear momentum balance, multiply by the velocity and integrate over the fixed volume:This is integration by parts component by component, or the divergence theorem. The first term is boundary stress power. With component-first velocity gradient $K_{ij}={\partial }_j{u}_i$, the second is the double contraction $\Phi =\sigma :K$.

For a Newtonian fluid, $\sigma =-pI+2\mu e$, where $e=(K+K^T)/2$ is the rate-of-strain tensor and $\mu \ge 0$ is dynamic viscosity. Incompressibility removes the pressure contribution, and the symmetric tensor $e$ is orthogonal to the antisymmetric part of $K$. ThereforeThe displayed energy identity follows for the stated linear balance. A full advective momentum equation in a fixed control volume would also contribute an advective kinetic energy flux. The nonnegativity statement here is for a purely Newtonian viscous stress; a viscoelastic stress can exchange stored free energy with the flow.

Let $K_{ij}={\partial }_j{u}_i$, $C_{ij}=⟨R_i{R}_j⟩$, $\alpha =2H/\zeta$ and $D_R=2k_{B}T/\zeta$. The configuration equation of the Hookean dumbbell model becomesHere the paper's $k$ denotes the Boltzmann constant $k_B$. Integrating the second moment equation against $R_i{R}_j$ and discarding boundary terms at infinite separation givesFor example the spring drift differentiates each factor $R_i,R_j$, while the Laplacian gives ${\nabla }_{\mathbf{R}}^2(R_i{R}_j)=2{\delta }_{ij}$. The Hookean dumbbell conformation equation isIn the row-vector convention of $\mathbf{R}\cdot \nabla \mathbf{u}$, the displayed gradient tensor is $K^T$, giving exactly the transposes in the requested equation. The dimensionless conformation tensor $A=HC/(k_{B}T)$ consequently has upper-convected derivative $-(A-I)/\lambda$, with polymer relaxation time $\lambda =\zeta /(4H)$. This normalization matters: the unscaled $C$ has equilibrium value $(k_{B}T/H)I$, rather than $I$.

Write the centre position as $\mathbf{x}$ and the bead positions as ${\mathbf{r}}_{1,2}=\mathbf{x}\mp \mathbf{R}/2$. In the affine velocity field, $\mathbf{u}({\mathbf{r}}_{1,2})=\mathbf{u}(\mathbf{x})\mp K\mathbf{R}/2$. The midpoint follows the flow, so its two slip velocities areEach bead loses mechanical power $\zeta$ times its squared slip speed. Adding the two powers proves the drag dissipation of a bead-spring pair:The factor one half comes from two equal-and-opposite bead slips, not from assigning only one bead a drag force.

The claimed equality does not hold for the printed joint distribution. The original PDF specifies a centered Maxwell-Boltzmann distribution of $\mathbf{v}=\dot{\mathbf{R}}$ independent of $\mathbf{R}$, withSince the double average includes number density $n$, evaluating the Gaussian integrals givesThe cross term vanishes because the conditional mean of $\mathbf{v}$ is zero. In particular, set $K=0$ and take the equilibrium configuration probability density ${\psi }_{\mathrm{eq}}\propto e^{-H{R}^2/(2k_{B}T)}$. The left side is $3n\zeta k_{B}T/m>0$, while $nHC:K=0$. This is a counterexample even at thermal equilibrium, so no rearrangement of the earlier conformation tensor equation can prove the printed statement.

The discrepancy is thermal bead drag and heat-bath power: gross viscous loss of microscopic thermal motion is replenished by the thermal bath. It is not identical to mechanical polymer stress power. Moreover, $\dot{\mathbf{R}}$ in the earlier configuration continuity equation is a probability current velocity, including a distribution-dependent osmotic drift; it cannot simultaneously be identified with an independently sampled centered Maxwellian velocity.

A consistent overdamped result is the free-energy balance of a Hookean dumbbell. Define $\mathbf{q}=H\mathbf{R}+k_{B}T{\nabla }_{\mathbf{R}}\log \psi$ andDifferentiate the relative entropy and use the configuration Fokker-Planck probability current $\psi \mathbf{V}$. After integration by parts,The affine osmotic term is $n{k}_{B}T\int \nabla \psi \cdot K\mathbf{R}d\mathbf{R}=-n{k}_{B}T\mathrm{tr}K=0$. HenceThe polymer stress tensor ${\sigma }_p=n(HC-k_{B}TI)$ has ${\sigma }_p:K=nHC:K$ in an incompressible flow. It stores or releases free energy as well as dissipating it, so the total stress-work term in part a need not itself be a purely dissipative quantity for this viscoelastic material. Stress power equals $\mathcal{D}$ only for stationary free energy, and $\mathcal{D}$ uses the probability-current slip. Neither condition turns the printed independent-Maxwellian average into that expression.