Past exam of the advanced fluid dynamics course of the University of Oxford / 2019 / 2

Work in the laboratory frame, with the fluid at rest at infinity. SetAt $r=a$, $f=1$ and $g=0$, so the no-slip boundary condition is $\mathbf{u}=\mathbf{U}$ on the translating sphere. As $r\to \infty$, $\mathbf{u}=O(r^{-1})\to 0$ and the gauge-fixed pressure tends to zero.

It is also possible to check the exterior Stokes equations directly. Differentiating the radial functions givesandThus the Newtonian fluid stress tensor $\sigma =-pI+\mu (\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)$ satisfies $\nabla \cdot \sigma =0$ throughout $r>a$.

Let $\mathbf{n}$ point away from the common centre on each concentric sphere. Applying the divergence theorem to the intervening fluid shell, whose inner normal is $-\mathbf{n}$, yieldsThis is the force flux at infinity in Stokes flow. It is legitimate even though the stress tensor tends to zero pointwise, because the area of the large sphere grows as $R^2$.

For an explicit evaluation, put $A=3a/4$ and $\mathbf{n}=\mathbf{x}/R$. The leading terms are $\mathbf{u}=A[\mathbf{U}+(\mathbf{U}\cdot \mathbf{n})\mathbf{n}]/R$ and $p=2\mu A(\mathbf{U}\cdot \mathbf{n})/R^2$. Their velocity gradient isAfter symmetrizing and subtracting the pressure contribution,By rotational symmetry, $\int n_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$. HenceThis is Stokes's law, with the force on the sphere opposite its velocity. For a component calculation with $\mathbf{U}=U{\mathbf{e}}_z$, the same integral is $F_z=-(9\mu aU/2)\int {\cos }^2\theta d\Omega$.

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.