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

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 $F_z=-(9\mu aU/2)\int {\cos }^2\theta d\Omega$.