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For a sphere of radius $R$ moving through a membrane tube with a uniform narrow gap $\delta \ll R$, lubrication theory givesIt exceeds the unbounded-fluid Stokes drag law coefficient by $2R^2/{\delta }^2$.

The dimensional linear friction coefficient relates drag force and velocity, $F_d=-\zeta v$, and has units of force divided by speed. It differs from the dimensionless aerodynamic drag coefficient. For an isolated sphere in Stokes flow, the Stokes drag law gives $\zeta =6\pi \eta a$.

A Reynolds number based on particle diameter and relative speed in the carrier fluid. Small values justify Stokes drag law; a high-Reynolds-number regime with nearly constant drag coefficient instead gives quadratic drag. The drag law must be evaluated in the appropriate regime for every species.

Let the particle and carrier-fluid mass densities be ${\rho }_p$ and ${\rho }_f$, with identical density contrast for both species. At small particle Reynolds number, balance Stokes drag law against submerged weight:ConsequentlyAt large particle Reynolds number, use approximately constant drag coefficient $C_D$ and quadratic drag. Balancing $\frac{1}{2}{C}_D{\rho }_f\pi r^2{W}_s^2$ against the same submerged weight gives $W_s^2=8({\rho }_p-{\rho }_f)gr/(3C_D{\rho }_f)$, soBoth particles must lie in the constant-$C_D$ regime for this simple square-root ratio. More generally it contains an additional factor $\sqrt{C_D(r_2)/C_D(r_1)}$. The speeds are positive downward magnitudes; the signs of the shock velocities below depend on the chosen spatial coordinate.

Continue with downward coordinate $y=H-z$ and $W=W_s(a)>0$. By Stokes drag law, the small-particle isolated settling velocity is $W/4$. Normalize concentrations by ${\varphi }_{\max }$: write $q_i={\varphi }_i/{\varphi }_{\max }$. In suspension the specified independent hindered settling laws give fluxes $j_1/{\varphi }_{\max }=W{q}_1(1-q_1)$ and $j_2/{\varphi }_{\max }=(W/4)q_2(1-q_2)$. The initial state is $(q_1,q_2)=(1/8,1/8)$.

Two upper sedimentation shocks clear the species separately. The faster front removes large particles, leaving $(0,1/8)$ behind it; the small concentration is unchanged across this front because the species fluxes are independent. The slower front separates that small-only suspension from clear fluid. Their downward velocities areThere is also a growing mixed deposit at the bottom. Let its normalized solid concentrations be $(p_1,p_2)$, with $p_1+p_2=1$, and let its upward speed magnitude be $s=-{\hat{U}}_3>0$. Applying the Rankine-Hugoniot condition to each species givesAdding these relations provesThe deposited flux is zero even though each species occupies only part of the packing: the settling law applies to the suspension, while the deposit is a separate stationary state. These jump relations determine the deposit composition from sedimentation jump conditions.

The large-particle clearing front has height $H-7Wt/8$ and meets the mixed-deposit front at height $35Wt/192$. ThusAt this merger all large particles have entered the deposit. Only the small particles remain suspended, at normalized concentration $1/8$. The new material deposited above the mixed layer is pure small-particle solid, at total concentration ${\varphi }_{\max }$. Its sedimentation shock therefore has downward speedFor $t>t_1$, the deposit top and the remaining clearing front have heightsTheir meeting, or the total solid-volume balance, givesThis is two-stage bidisperse batch sedimentation: the fast species finishes first, followed by deposition of the remaining slow species.

The final solid composition by volume, and also by mass because the particle densities are identical, isThe upper pure-small layer has thickness $9H/116$. As checks, $(29/40)h_1=H/8$ accounts for all large-particle solid volume, and $(11/40)h_1+(h_2-h_1)=H/8$ accounts for all small-particle solid volume, in units of ${\varphi }_{\max }$. If “percentage of particles” means number rather than volume, a large sphere occupies eight times the volume of a small sphere: the mixed bottom layer contains $29/117\simeq 24.8\%$ large particles by number and $88/117\simeq 75.2\%$ small. The upper layer remains entirely small under either convention.

Let $\mu$ be the dynamic viscosity, $u_i={\dot{x}}_i$, and $F_i$ the signed force exerted by sphere $i$ on the fluid. For negligible sphere inertia, this also equals the externally or internally applied force on that sphere; its hydrodynamic force is $-F_i$. The Stokes drag law and the axial velocity of a Stokeslet give the leading hydrodynamic mobility matrixThe factor $1/(4\pi \mu \ell )$ follows from $(I+\hat{\mathbf{r}}\hat{\mathbf{r}})\mathbf{F}/(8\pi \mu \ell )$ when $\mathbf{F}$ is parallel to the line of centres. The longitudinal two-sphere mobility retains the first interaction in $a_i/\ell$; finite-size and repeated-reflection corrections are higher order.

For the linked force-free pair, $F_2=-F_1$. Put $X=(x_1+x_2)/2$ and $S_m=m_1+m_2$. ThensoThe coefficient depends only on the current separation. For any period $T$, $\ell (T)=\ell (0)$, and thereforeThis is a closed integral of a single-valued function of one real shape coordinate. Since $x_1=X-\ell /2$ and $x_2=X+\ell /2$, both spheres also have zero net displacement. Equal radii make $\dot{X}=0$ instantaneously; unequal radii generally permit oscillatory translation, but still no mean motion. This explicit force-free two-sphere stroke is the scallop theorem: a single-parameter reciprocal deformation in a Newtonian fluid at zero inertia cannot produce net free swimming. The timing of extension and contraction cannot change that conclusion, because Stokes flow has no inertial memory.

For the externally prescribed pair, put $q=\omega t$, ${\zeta }_i=1/m_i=6\pi \mu a_i$, andInvert the hydrodynamic mobility matrix:In a large-${\ell }_0$ expansion with $a_i,\delta$ fixed,The isolated-drag term and the constant-$h$ interaction have zero mean. The order-${\ell }_0^{-2}$ term from $h^2{u}_1$ also has zero mean, since its coefficient is constant at that order. Meanwhile,Consequently the externally driven two-sphere pump has the following phase-dependent mean force of an externally driven sphere pair:Here $⟨\cdot {⟩}_t$ denotes averaging over one period. The analogous calculation gives $⟨d{u}_1{⟩}_t={\delta }^2\omega \sin \varphi /2$, so $⟨F_2{⟩}_t=⟨F_1{⟩}_t$ at leading order, andThus the mean fluid forcing is nonzero except at $\varphi =0$ or $\pi$ modulo $2\pi$. The sign changes when the phase lag is reversed. Those exceptional strokes are reciprocal: the position-space loop collapses to a line, and their mean force vanishes by reversibility, not just by this leading expansion.

There is no contradiction with the scallop theorem. The externally imposed motion is not force-free, and the actuators prescribe two phase-shifted motions; their combined motion is not reciprocal for a generic phase. The spheres return to their prescribed positions while transferring a mean force to the fluid. This is pumping by external forcing, rather than propulsion of the freely linked one-shape-coordinate system.

The asymptotic calculation requires $\delta /{\ell }_0\ll 1$ and persistent large separation. The minimum separation of phase-shifted sphere oscillations isThe printed condition ${\ell }_0\ge \delta$ alone does not guarantee nonoverlap or the stipulated far-separated regime. The prescribed trajectories must additionally keep this minimum well above both radii. This is a compatibility qualification on the data, not a change of the phase-dependent force calculation.

Write $Q=U_j{x}_j$ and use repeated-index summation. Direct differentiation givesTracing yields ${\partial }_i{u}_i=0$. Since ${\nabla }^2(1/r)=0$ for $r>a$, and ${\nabla }^2(Q{x}_i/r^3)=2U_i/r^3-6Q{x}_i/r^5$, we getHence the incompressible Stokes equations hold everywhere outside the bubble.

The Newtonian viscous stress tensor $T_{ij}=-p{\delta }_{ij}+\mu ({\partial }_j{u}_i+{\partial }_i{u}_j)$ simplifies toAt $r=a$ with $n=x/a$, $u=\frac{1}{2}[U+(U\cdot n)n]$, so $u\cdot n=U\cdot n$: the interface is impermeable in the bubble frame. Also $Tn=-(3\mu /a)(U\cdot n)n$ is purely normal, so the tangential traction vanishes, appropriate to a clean bubble with negligible internal viscosity. The tangential fluid velocity need not equal the bubble's velocity. At infinity $u,p\to 0$. The calculation assumes the usual spherical, negligible-deformation limit; exact normal-stress balance includes interior pressure, surface tension and any buoyancy sustaining the motion.

Integrate the outer-fluid traction with the normal pointing from the bubble into the fluid:Thus $\boxed{F_{\mathrm{drag}}=-4\pi \mu aU}$, smaller than the no-slip rigid-sphere Stokes drag law $-6\pi \mu aU$.

The Papkovich–Neuber representation writes a homogeneous incompressible Stokes flow in terms of a harmonic vector field $\mathbf{\Phi }$ and harmonic scalar $\chi$:with ${\nabla }^2\mathbf{\Phi }=0$ and ${\nabla }^2\chi =0$. The representation satisfies incompressibility because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and substitution then verifies the Stokes equation.

For a sphere translating with constant vector velocity $\mathbf{U}$, rotational covariance and decay at infinity suggest a harmonic vector monopole and scalar dipole:Substitution gives the translating sphere in Stokes flowAt $r=a$ the radial tensor terms cancel and $\mathbf{u}=\mathbf{U}$, while $\mathbf{u}\to 0$ as $r\to \infty$, so the no-slip boundary condition and far-field condition hold. The resulting traction integrates to the Stokes drag law $6\pi \mu a\mathbf{U}$ in magnitude.

Let $z$ measure distance across the narrow gap and let $\theta$ be polar angle about the tube axis. In lubrication theory, radial velocity and radial pressure variation are negligible. Axisymmetric incompressible flow is obtained frombecause ${\partial }_{\theta }(u_{\theta }\sin \theta )=0$. The tangential Stokes equation then separates:and hence, after choosing an irrelevant pressure constant,

In the sphere frame, $u_z=-u_{\theta }\sin \theta =-\varphi$. The no-slip boundary condition gives $\varphi (0)=0$ on the sphere and $\varphi (\delta )=U$ on the membrane translating backward relative to it. The sphere-frame volume flux inherited from the narrow remote tube is $-\pi r_0^{2}U$, soUnder the asymptotic condition $r_0^2\ll R\delta$, the right-hand side is negligible at leading order. Solving the quadratic profile subject to the two wall values and zero leading-order integral givesChanging the chosen positive tube direction reverses both signs but leaves the drag magnitude unchanged.

The pressure scale is $p\sim \mu UR/{\delta }^2$, whereas the viscous shear scale is $\tau \sim \mu U/\delta$. After multiplication by comparable areas, pressure drag exceeds shear drag by $R/\delta \gg 1$, an instance of lubrication pressure dominates shear stress. Put $c=\cos \Delta \theta$. The axial pressure force isAs $\Delta \theta \to 0$, the bracket tends to $-2$, and thereforeThe resulting confined-sphere drag coefficient isThus it exceeds the free Stokes drag law coefficient by $2R^2/{\delta }^2$. The Stokes–Einstein relation then gives

For spheres with the same density contrast, balance submerged weight, proportional to $r^3$, against Stokes drag law, proportional to $r{W}_s$, to obtain $W_s\propto r^2$. With constant drag coefficient, quadratic drag is proportional to $r^2{W}_s^2$, giving $W_s\propto r^{1/2}$. The two radius laws require their corresponding particle Reynolds number limits.

The Stokes settling velocity is the terminal relative velocity obtained by balancing gravity, buoyancy, and the Stokes drag law for a dilute isolated sphere.

A sphere of radius $a$ translating with velocity $\mathbf{U}$ through otherwise stationary fluid hasIts surface traction integrates to the Stokes drag law.