Past exam of the advanced fluid dynamics course of the University of Oxford / 2025 / 2

The stress tensor of the incompressible Newtonian fluid is $\mathbf{\sigma }=-pI+2\mu e$, where $e=(\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)/2$ is the rate-of-strain tensor. The homogeneous Stokes equations give $\nabla \cdot \mathbf{\sigma }=0$. Apply the divergence theorem to the fluid between the body and a surrounding sphere $S_R$. If ${\mathbf{n}}_B$ points from the body into the fluid, the fluid-domain normal on the inner boundary is $-{\mathbf{n}}_B$. HenceThis is the force flux at infinity in Stokes flow. A nonzero resultant force must be supported by a stress of order $R^{-2}$ on a surface of area of order $R^2$. More precisely, the leading exterior solution is a Stokeslet with force on the fluid $\mathbf{f}=-\mathbf{F}$:Thus $\mathbf{u}=O(R^{-1})$, $p-p_{\infty }=O(R^{-2})$ and $e=O(R^{-2})$. These are leading generic orders for a translating body with nonzero drag; the integral alone does not determine every angular component, and force-free disturbances may decay faster.

For two Stokes flows in the same fluid domain with the same dynamic viscosity, symmetry of the Newtonian fluid stress tensor and incompressibility give the local reciprocal identityIn index notation, its remaining cross-gradient terms are both $2\mu e:e^{\prime }$. On the remote sphere the integrated reciprocal flux tends to zero: the velocity is $O(R^{-1})$, the stress tensor is $O(R^{-2})$ and the area is $O(R^2)$. On the body, the no-slip boundary condition makes the two velocities the constants $\mathbf{U},{\mathbf{U}}^{\prime }$. The Lorentz reciprocal theorem therefore givesBy the Linearity of Stokes flow, the sum of the two velocity-pressure fields satisfies the boundary conditions for body velocity $\mathbf{U}+{\mathbf{U}}^{\prime }$. Uniqueness of the exterior Stokes flow identifies it with the third solution. Since the stress tensor and its surface integral are linear too, $\mathbf{F}(\mathbf{U}+{\mathbf{U}}^{\prime })=\mathbf{F}(\mathbf{U})+\mathbf{F}({\mathbf{U}}^{\prime })$; scalar multiplication follows in the same way. Thus $\mathbf{F}=-A\mathbf{U}$ for a fixed-orientation hydrodynamic resistance matrix $A$. Substituting arbitrary $\mathbf{U},{\mathbf{U}}^{\prime }$ in the reciprocal identity yieldsThe minus sign is natural because viscous dissipation gives $-\mathbf{F}\cdot \mathbf{U}=2\mu \int e:edV>0$ for nonzero rigid translation in an otherwise stationary fluid; $A$ is a positive-definite matrix.

Use the free-draining approximation: neglect the hydrodynamic interaction between the widely separated spheres and neglect the connector's viscous drag. Stokes law gives ${\mathbf{f}}_i=-6\pi \mu a_i\mathbf{U}$ for each sphere. Their resultant isThis is the leading two-sphere approximation, not an exact finite-separation hydrodynamic resistance matrix. Writing $d=|{\mathbf{r}}_2-{\mathbf{r}}_1|$ and $\mathbf{n}=({\mathbf{r}}_2-{\mathbf{r}}_1)/d$, the first Stokeslet reflection gives, if the connector remains negligible,The negative correction expresses reduced drag from fluid entrainment by the other sphere. A very thin but arbitrarily long rod need not have negligible total drag: its typical transverse drag scales as $\mu d/\log (d/b)$ for rod radius $b$. Thus the simple boxed result uses the intended negligibly dragging connector limit as well as $a_i/d\ll 1$; the two printed size inequalities alone do not specify a uniform limit in which rod drag vanishes.

Hold the two-sphere object fixed in a uniform stream $\mathbf{V}$. In the free-draining approximation, its drag forces are ${\mathbf{f}}_i={\zeta }_i\mathbf{V}$, with ${\zeta }_i=6\pi \mu a_i$. The hydrodynamic torque about ${\mathbf{r}}_c$ isTo make it zero for every incoming-stream direction, choose the hydrodynamic centre of a two-sphere dumbbell:It lies closer to the larger sphere, whose larger drag requires a shorter lever arm. Uniform translation does not spin an isolated sphere, so no additional isolated-sphere spin torque enters this leading approximation.

Finite-separation hydrodynamic interactions and connector drag can change the displayed location; the calculation has the same approximation as part c.

For rotation about the hydrodynamic centre of a two-sphere dumbbell, the sphere centres have velocities ${\mathbf{v}}_i=\mathbf{\Omega }\times ({\mathbf{r}}_i-{\mathbf{r}}_c)$. Their resultant drag force in the free-draining approximation isEach sphere also experiences a spin drag torque, but its isolated rotation causes no net force. Those spin contributions therefore do not alter the cancellation.

This illustrates translation-rotation reciprocal coupling. About any fixed origin, write the fluid's rigid-body force and torque asThe transposed cross-blocks follow from the Lorentz reciprocal theorem. If pure translation causes zero torque about an origin for every $\mathbf{U}$, then $B^T=0$, hence pure rotation about that origin causes zero force. This implication is general for linear Stokes flow; the existence and leading location of such an origin here use the two-sphere geometry and the approximation in part c.