Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 73 / 1

One convention for the Papkovich–Neuber representation uses a harmonic vector potential $\mathbf{\Phi }$ and a harmonic scalar potential $\chi$:These are harmonic functions away from any singular force point. Since ${\nabla }^2(\mathbf{r}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, the representation gives $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=\nabla p$, the equations of homogeneous Stokes flow.

Place the point force at the origin. A velocity linear in $\mathbf{F}$, decaying as $r^{-1}$ and having the rotational symmetry of a point force is obtained from $\mathbf{\Phi }=c\mathbf{F}/r$, $\chi =0$. The vector components are harmonic functions for $r>0$; scalar dipole potentials would instead generate higher-order decaying singularities. The force normalization fixes $c=-1/(4\pi )$. Indeed, substitution gives the Stokeslet:To verify its strength, the Newtonian fluid stress tensor is ${\sigma }_{ij}=-3(\mathbf{F}\cdot \mathbf{r})r_i{r}_j/(4\pi r^5)$. Its outward traction integrated over any sphere surrounding the origin is $-\mathbf{F}$, because $\int \mathbf{nn}d\Omega =(4\pi /3)I$. Thus the localized force applied to the fluid is $\mathbf{F}$, as required. Translation of the origin gives the same Stokeslet centered at any prescribed force point.

Let $\mathbf{n}=\mathbf{X}/R$, $\zeta =6\pi \mu a$, and letUse the method of reflections for Stokes flow. At the fixed sphere, the first sphere produces the incident Stokeslet velocity ${\mathbf{u}}_{\infty }(0)=\mathsf{A}\mathbf{V}$ to leading order. In Faxén translation law, set the second sphere's translational velocity to zero. The applied holding force is thereforeThe first sphere's finite-radius potential dipole adds $O(V{a}^3/R^3)$ to the incident velocity, as does the Laplacian term in Faxén translation law at the fixed sphere. Multiplying by $\zeta$ gives the stated next correction. There is no intermediate $O(\mu V{a}^3/R^2)$ term. This is the holding force and torque for a sphere in a distant Stokeslet.

The vorticity of a Stokeslet at displacement $\mathbf{r}$ is $\mathbf{F}\times \mathbf{r}/(4\pi \mu r^3)$. At the fixed sphere, $\mathbf{r}=-\mathbf{X}$, soSet its angular velocity to zero in Faxén rotation law. The applied holding couple isThe displayed sign is the external couple needed to oppose the ambient rotation, rather than the hydrodynamic couple on the sphere.

The leading reflected flow at the first sphere is $\mathsf{G}(\mathbf{X}){\mathbf{F}}_2=-{\mathsf{A}}^2\mathbf{V}$. Apply Faxén translation law there with its prescribed force $\mathbf{F}=\zeta \mathbf{V}$. Since $(I+\mathbf{nn}{)}^2=I+3\mathbf{nn}$, the mobility correction from a fixed distant sphere isThe next correction comes from finite-radius terms in the incident/reflected flow and in Faxén translation law, together with the fixed sphere's induced stresslet and holding-couple rotlet. Each gives $O(V{a}^4/R^4)$ at the first sphere. The symbol $\mathbf{V}$ is its isolated-sphere velocity scale, not its actual velocity in the two-sphere problem.

The first sphere has no applied couple. Its angular velocity follows from Faxén rotation law and the reflected Stokeslet:The component of ${\mathbf{F}}_2$ parallel to $\mathbf{X}$ contributes no vorticity. The next reflected rotlet and stresslet give the stated error order.

The mobility correction from a fixed distant sphere givesTaking their ratio and keeping the first nonzero transverse correction yieldsto the stated leading order. Set $b=Y_{\infty }\gg a$. The deflection is $O(a^2/b)$, so replacing $Y$ by $b$ on the right introduces only higher-order errors. Integrating from $X=-\infty$ givesThe maximum occurs at $X=0$, and the deflection and spin in a distant sphere encounter areFor the rotation, use $dt=dX/V$ and $Y=b$ to leading order in the angular velocity from part (b). Its signed angle about the positive $z$ axis isThus the rotation is clockwise when viewed from positive $z$, with the magnitude of the displayed leading term.

The deflection tends back to zero downstream: $Y(+\infty )=Y_{\infty }$. More generally, kinematic reversibility of Stokes flow combined with reflection in the plane $X=0$ makes a passing trajectory fore-aft symmetric. One can see this without using the distant-sphere approximation: the relevant translational hydrodynamic mobility matrix has the form $m_{\perp }(R)I+[m_{\parallel }(R)-m_{\perp }(R)]\mathbf{nn}$. Hence $\dot{X}$ is even in $X$ and $\dot{Y}$ is odd in $X$. Uniqueness of the trajectory through $X=0$ then gives $Y(X)=Y(-X)$.

For $Y_{\infty }=a/100$, the numerical deflection and spin approximations above are invalid: the encounter enters a narrow gap and requires lubrication theory. However, the same return to the incoming offset holds for an ideal passing encounter of perfectly smooth spheres in Stokes flow. Lubrication resistance prevents finite-time contact under a bounded force, and does not itself destroy kinematic reversibility of Stokes flow. Contact, surface roughness or nonhydrodynamic forces could change that conclusion; they are additional physics, not part of the ideal model. This distinction is the fore-aft symmetry of a sedimenting-sphere encounter.