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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.

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.

A freely swimming body has no net external force or torque in Stokes flow. Its force-free condition removes the $r^{-1}$ Stokeslet monopole, and its torque-free condition removes the antisymmetric force-dipole contribution. The leading generic far field is therefore a symmetric traceless rank-two tensor stresslet, decaying as $r^{-2}$. Its axial strength may vanish for special swimmers, in which case higher-order singularities dominate.

The axisymmetric stresslet from a Stokeslet pair provides an explicit construction and fixes the sign convention. Place forces $+F\mathbf{e}$ and $-F\mathbf{e}$ on the fluid at $+(a/2)\mathbf{e}$ and $-(a/2)\mathbf{e}$. The Stokeslet tensor isExpanding their sum for $r\gg a$ givesSincethe dipole strength is $S=Fa$ and the resulting stresslet is the given one. More general surface-traction distributions produce the same leading form through their symmetric traceless first force moment.

Writing $\cos \alpha =\mathbf{e}\cdot \hat{\mathbf{r}}$, its radial velocity isConsequentlyThis classifies pusher microswimmers and puller microswimmers in the force-on-fluid convention used by the given formula.

For the circular swimmer let $\phi =\omega t$, $\mathbf{p}=(\cos \phi ,\sin \phi ,0)$, ${\mathbf{x}}^{\prime }=R\mathbf{p}$, and choose the tangent director $\mathbf{e}=(-\sin \phi ,\cos \phi ,0)$. Its sign is immaterial to the stresslet. Write $\mathbf{r}=\mathbf{x}$ for the observation vector from the circle centre, and assume $r\gg R$. The leading far-field orbit average of a tangent stresslet is obtained by replacing $\mathbf{x}-{\mathbf{x}}^{\prime }$ by $\mathbf{r}$ while retaining the time-dependent director.

The orientation averaging of an axisymmetric stresslet over a full revolution givesSubstitution givesThuswith $\tilde{\mathbf{e}}$ equivalently $-\hat{\mathbf{z}}$. The effective axis is normal to the orbit plane. A positive original strength becomes negative, and a negative strength becomes positive: a pusher averages to a puller, and a puller averages to a pusher, at leading far-field order. The result is independent of $R$ and $\omega$ at this order.

The accuracy of the centre replacement can be made explicit. Half a revolution changes $\mathbf{p}$ and $\mathbf{e}$ to their negatives; the stresslet is even in $\mathbf{e}$. Pairing those times replaces the displaced field byHenceThe finite-radius average is not exactly a point stresslet at every location. For example, on the positive $z$ axis the exact average of the supplied model iswhich has the derived point-stresslet limit for $z\gg R$. This even displacement correction in an orbit-averaged stresslet is consistent with the far-field restriction.

Let $e=(\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)/2$ be the rate-of-strain tensor and $\mathbf{\sigma }=-pI+2\mu e$ the Newtonian fluid stress tensor. The viscous dissipation is $D[\mathbf{u}]=2\mu {\int }_{V}e:edV$. The Minimum-dissipation theorem for Stokes flow compares the Stokes flow with every sufficiently regular incompressible flow in the same domain having the same prescribed boundary velocity, with no body force or with a conservative body force absorbed into the fluid pressure. A trial field need not satisfy the Stokes flow equations. Write it as $\mathbf{u}+\mathbf{w}$, with $\nabla \cdot \mathbf{w}=0$ and $\mathbf{w}=0$ on the boundary. Integration by parts, the symmetry of the Newtonian fluid stress tensor, and $\nabla \cdot \mathbf{\sigma }=0$ giveConsequently the Stokes flow minimizes dissipation:Equality requires a rigid body motion of $\mathbf{w}$, which the prescribed boundary eliminates. Extending the flow inside the inserted particle as its rigid body velocity produces an admissible comparison field with zero internal rate-of-strain tensor; hence the extra dissipation due to a rigid inclusion is nonnegative.

Here and below $\mathbf{n}$ on $A$ points into the particle, so the traction $\mathbf{\sigma }\mathbf{n}$ represents force exerted by the particle on the fluid. Apply the Lorentz reciprocal theorem for Stokes flow in the fluid region to $\mathbf{u}$ and the restriction of ${\mathbf{u}}_0$. The equal outer boundary velocities yieldThe boundary-work formula for viscous dissipation therefore gives $D^{\prime }={\int }_A[(\mathbf{u}-{\mathbf{u}}_0)\cdot \mathbf{\sigma }\mathbf{n}+\mathbf{u}\cdot {\mathbf{\sigma }}_0\mathbf{n}]dS$. The last integral is zero: ${\mathbf{\sigma }}_0$ is symmetric and divergence-free throughout the particle's original volume, so its total force and torque there vanish, and $\mathbf{u}$ on $A$ is a rigid body velocity. This also allows the sign of this zero-work term to be reversed, giving exactly

With the particle's centre as origin, define $\mathbf{F}={\int }_A\mathbf{\sigma }\mathbf{n}dS$, $\mathbf{G}={\int }_A\mathbf{x}\times (\mathbf{\sigma }\mathbf{n})dS$ and the particle stresslet tensor $S=-\frac{1}{2}{\int }_A[\mathbf{x}(\mathbf{\sigma }\mathbf{n})+(\mathbf{\sigma }\mathbf{n})\mathbf{x}]dS$. Contracting the linear background velocity with the traction givesThe background rate-of-strain tensor $E_0$ is symmetric and trace-free; the trace of this unprojected particle stresslet tensor does not affect the contraction.

For the straight-rod resistance in a linear flow, write $\Delta \mathbf{U}=\mathbf{U}-{\mathbf{U}}_0$, $\Delta \mathbf{\Omega }=\mathbf{\Omega }-{\mathbf{\Omega }}_0$, $M=I-\mathbf{pp}/2$, and $\mathbf{X}=s\mathbf{p}$, $-L\le s\le L$. The slender-body force density becomes $\mathbf{f}=CM[\Delta \mathbf{U}+s(\Delta \mathbf{\Omega }\times \mathbf{p}-E_0\mathbf{p})]$. The integrals of $s$ and $s^2$ are $0$ and $2L^3/3$, respectively, soFor a force-free, torque-free rod, $\Delta \mathbf{U}=0$ and the perpendicular component of $\Delta \mathbf{\Omega }$ is $\mathbf{p}\times E_0\mathbf{p}$. Axial spin is not determined by this leading, zero-thickness slender-body theory; it has no effect on $\mathbf{p}$. Put $a_p=\mathbf{p}\cdot E_0\mathbf{p}$. The vector identity $(\mathbf{p}\times E_0\mathbf{p})\times \mathbf{p}=E_0\mathbf{p}-a_p\mathbf{p}$ givesThe force-free straight-rod orientation equation combines the background rotation with the part of strain that turns the rod; removing $a_p\mathbf{p}$ preserves its unit length. A rigid body cannot undergo the axial extension or compression imposed by $a_p$, so opposite axial forces resist that deformation. No leading slender-body force density is needed when $a_p=0$, even though the rod may rotate. Its particle stresslet tensor and viscous dissipation are

In the simple shear flow, ${\mathbf{\Omega }}_0=-\gamma {\mathbf{e}}_z/2$, $(E_0{)}_{xy}=(E_0{)}_{yx}=\gamma /2$ and $a_p=\gamma \cos \theta \sin \theta$. Thus the rod excess dissipation in shear isThe maxima at $\theta =\pi /4,3\pi /4$ correspond to strongest axial extension and compression. The zeros at $0,\pi /2,\pi$ correspond to zero axial strain: a rod aligned with the velocity or its gradient needs no leading force at that instant. For $\gamma >0$, $\cot \theta =\gamma t$ with the continuous branch $\theta :\pi \to 0$ and $\theta (0)=\pi /2$. Reversing the shear flow reverses the traversal. Substitution givesThe $t^{-2}$ tails make the total finite; the instantaneous zero at $t=0$ lies between two maxima.

For a dilute rod suspension, the added bulk stress is the number density times the orientation average of the particle stresslet tensor. In this ideal infinitely slender, non-interacting model, rods approach alignment with the shear flow and their excess viscous dissipation decays. A real finite-aspect-ratio rod continues to tumble; rotational diffusion and interactions also maintain a spread of orientations. One therefore expects increased effective viscosity, with alignment offering a mechanism for reduced excess viscosity at strong shear flow. This is a qualitative expectation, not a claim that a finite rod suspension has zero steady excess stress.

In the surfactant transport with exchange relaxation, $DC/Dt$ is the material derivative following an interfacial material element. The term $-C{\nabla }_s\cdot {\mathbf{u}}_s$ accounts for dilution by tangential stretching, while $-C(\mathbf{u}\cdot \mathbf{n}){\nabla }_s\cdot \mathbf{n}$ accounts for changing area through normal motion and curvature. The surface diffusion term redistributes surfactant down concentration gradients. The exchange term relaxes its concentration towards $C_0$, through adsorption or desorption from a surrounding reservoir, on time scale $k^{-1}$.

In the bubble frame the spherical interface is stationary and impermeable. A steady material derivative is ${\mathbf{u}}_s\cdot {\nabla }_s{C}^{\prime }$. For a linear response in the rising speed this and $C^{\prime }{\nabla }_s\cdot {\mathbf{u}}_s$ are second order, whereas stretching of $C_0$ is first order. ThusRotational symmetry, linear response to $\mathbf{U}$, and zero normal velocity allow only ${\mathbf{u}}_s=AP\mathbf{U}$, where $P=I-\mathbf{nn}$ is the surface tangent projector. This is the dipolar, axisymmetric sector; axisymmetry alone, without the linear-response assumption, would allow other angular dependence.

On a sphere $\mathbf{n}=\mathbf{x}/r$ and ${\partial }_j{n}_i=({\delta }_{ij}-n_i{n}_j)/r$. Applying the surface tangent projector to the derivative gives the sphere surface derivative identitiesFor example, taking a second surface divergence of the first identity yields the third. Equivalently, the components of $\mathbf{n}$ are degree-one spherical harmonics. Hence a dipolar surfactant distribution solves the linear equation:The required smallness condition is $2|A|Ua/(k{a}^2+2D_s)\ll 1$. Surface diffusion or exchange must smooth the concentration faster than it is redistributed by the actual interfacial velocity. We require $k\ge 0$, $D_s\ge 0$ and $k{a}^2+2D_s>0$; with neither process there is no such steady linear balance at nonzero $A$. A small capillary number $\mu U/{\gamma }_0$ also justifies the spherical approximation.

Take the normal from the inner bubble to the outer liquid, define $[\sigma {]}_{-}^{+}={\sigma }_{+}-{\sigma }_{-}$, and set $\kappa ={\nabla }_s\cdot \mathbf{n}$. The interfacial stress balance with variable surface tension isBecause ${\nabla }_s{C}^{\prime }=BP\mathbf{U}/a$ and $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$, its tangential part becomesThis is a Marangoni stress: surface-tension gradients oppose the clean-bubble circulation.

Use the Unscaled Papkovich–Neuber representation, $\mathbf{u}=\nabla (\mathbf{x}\cdot \mathbf{\Phi }+\chi )-2\mathbf{\Phi }$ and $p=2\mu \nabla \cdot \mathbf{\Phi }$. The constant vector potential gives $\mathbf{u}\to -\mathbf{U}$. In the decaying, linear axisymmetric sector, a vector monopole and scalar dipole provide the two constants needed at the sphere; other equivalent potential choices amount to a representation freedom. The PDF potential contains the factor $a$ omitted in the TeX aid. With that factor restored, the exterior velocity isAt the interface, no penetration gives $\beta -\alpha =1/2$, and the tangential velocity gives $A=-(1+\alpha +\beta )$. Using the traction coefficient printed in equation (3), the tangential stress condition gives $2\beta =A\lambda$. The requested algebraic answers are therefore

There is a factor-of-two error in the PDF traction. Direct differentiation of the preceding velocity with the Newtonian fluid stress tensor givesrather than its printed coefficient $12\mu /a$. In particular the tangential traction is $6\mu \beta P\mathbf{U}/a$. Keeping the same definition of $\lambda$, the physically consistent translating surfactant-coated bubble instead has $\beta =A\lambda$ andThe two sets must not be silently conflated. As a useful check, the no-slip limit has the usual total Stokes flow drag $6\pi \mu aU$ with the corrected traction; the printed traction would double it.

Both versions give $A=-1/2$, $\alpha =-1/2$, $\beta =0$ when $\lambda \to 0$: the interface is tangentially stress-free and circulates as a clean inviscid bubble. For $\lambda \to \infty$, $A\to 0$, $\alpha \to -3/4$, $\beta \to -1/4$: the interface is immobile in the bubble frame and behaves as a no-slip sphere. The Marangoni stress remains finite because $\lambda A$ has a nonzero limit. The concentration smallness condition still has to hold; large $\lambda$ is not by itself a complete linearization criterion.

The normal component is balanced by the inner gas pressure and the normal capillary stress $\gamma \kappa$, including the variation of $\gamma$, together with the hydrostatic pressure difference responsible for buoyancy. A spatially uniform inner pressure alone cannot balance the dipolar traction. To make the normal stress balance on a translating bubble explicit, let $\Delta {\rho }_b$ be the outer-minus-inner mass density. After removing the equilibrium Laplace pressure, the corrected outer normal traction has dynamic part $-3\mu (1+\lambda )(\mathbf{U}\cdot \mathbf{n})/[a(1+2\lambda )]$. Its degree-one balance isThus, if a terminal speed is desired, $U=\Delta {\rho }_{b}g{a}^2(1+2\lambda )/[3\mu (1+3\lambda )]$, interpolating between the clean-bubble and rigid-sphere speeds. A degree-one shape displacement is merely a translation of the sphere, so the balance determines its rise speed, rather than a dipolar shape distortion.

The hydrostatic approximation gives $p=\rho g(h-z)$ above the horizontal floor, up to atmospheric pressure. The horizontal Stokes flow equation is $\mu {\mathbf{u}}_{zz}=\rho g\nabla h$, with the no-slip boundary condition at $z=0$ and zero tangential stress on the free surface. Integrating twice gives $\mathbf{u}=(\rho g/\mu )\nabla h(z^2/2-hz)$ and the lubrication gravity-current fluxFor the squeegee lubrication model, use $h_0$ vertically, $\ell =\rho g{h}_0^3/(\mu U)$ horizontally, and $\ell /U$ in time. If $X$ is the laboratory coordinate in the direction of travel, set $x=(X-Ut)/\ell$, $y=Y/\ell$, $t=U{t}_{\mathrm{lab}}/\ell$, and $h=h_{\mathrm{lab}}/h_0$. The blade half-length is $\alpha =L/\ell$. The moving-frame flux is $\mathbf{q}=-h{\mathbf{e}}_x-h^3\nabla h/3$. Conservation of volume gives the dimensionless lubrication equationUsing these scales as representative of the actual flow, lubrication theory requires a small aspect ratio and negligible inertia relative to vertical viscous resistance:The second is the Reynolds number based on $h_0$, multiplied by $h_0/\ell$. Surface tension is neglected as assumed. Large piles, rounded gaps, and narrow end regions need their own local slope and inertia checks if these representative scales cease to describe them.

Write $d_1={\rho }_0-{\rho }_1>0$, $d_2={\rho }_2-{\rho }_0>0$, and choose the common undisturbed interface pressure $P$. The hydrostatic pressure is $p_1=P-{\rho }_{1}gz$ in the upper fluid and $p_2=P-{\rho }_{2}gz$ in the lower fluid. Continuity of hydrostatic pressure across both current interfaces givesThus $h_1=d_{2}h/(d_1+d_2)$ and $h_2=-d_{1}h/(d_1+d_2)$. This isostatic depth partition of an interfacial current expresses local vertical buoyancy balance: a displaced column with an imbalance would move vertically until the partition was restored. With the hydrostatic pressure $P-{\rho }_{0}gz$ subtracted, the pressure within the current is $\Delta \rho gh$, whereThe equivalent body force is outward for a profile decreasing with $r$. Its magnitude is $\Delta \rho g|h_r|$; the derivative without an absolute value is a signed quantity.

For an ambient-controlled interfacial plug current, integrating the body force through depth gives radial force per area $O(\Delta \rho g{H}^2/R)$. The ambient Stokes flow varies over distance $R$, exerting traction $O(\mu u/R)$, so $u\sim \Delta \rho g{H}^2/\mu$. The current's internal shear supports the ambient traction with a velocity difference $\Delta u\sim (\mu u/R)H/(\lambda \mu )$; $\Delta u/u\sim H/(\lambda R)\ll 1$ requires $\lambda \gg H/R$. The force per area from its in-plane viscous extension is $O(\lambda \mu uH/R^2)$, negligible against ambient traction when $\lambda \ll R/H$. Together these give the viscosity window for an interfacial plug currentFixed volume gives $V\sim H{R}^2$, and the spreading velocity has the scale $\dot{R}$. Thus $\dot{R}\sim \Delta \rho g{V}^2/(\mu R^4)$ andChoosing $A$ so that $R=A{t}^{1/5}$ exactly, the similarity solution has $h=(V/A^2)t^{-2/5}F(\eta )$, $\eta =r/(A{t}^{1/5})$, and $u=A{t}^{-4/5}U(\eta )$. In radial conservation $h_t+r^{-1}(rhu{)}_r=0$, substitution gives $(\eta FU{)}^{\prime }=({\eta }^{2}F{)}^{\prime }/5$. Regularity and no source at the centre remove the integration constant. Therefore the radial plug velocity is linear:

For the disc-traction analogy for an interfacial current, subtract the imposed straining flow from the stationary-disc solution. The perturbation has velocity $-Er{\mathbf{e}}_r$ on the disc and zero far-field velocity. Reverse its sign: the resulting Stokes flow is driven by a radially expanding disc with velocity $Er{\mathbf{e}}_r$ in otherwise quiescent fluid. The background straining flow contributes no radial shear traction on $z=0$, so the given disc traction changes sign. This step uses Linearity of Stokes flow, Uniqueness of Stokes flow, and the no-slip boundary condition. The thin plug current has the same leading boundary velocity and hence the same ambient resistance. Each face contributes $-8\mu Er/[\pi \sqrt{R^2-r^2}]$; both faces must be counted. With $E=1/(5t)$, radial force balance isIntegrate with $h(R,t)=0$. The interfacial plug-current similarity profile isVolume integration gives $V=2\pi {\int }_0^{R}rhdr=4\pi H{R}^2/5$. Eliminating $H$ gives the full prefactor:The formal edge slope diverges, so a narrow edge region lies outside the thin-current approximation. Its traction and volume contributions are integrable, and this is a leading thin-current solution rather than a uniformly valid edge description.

For a diffusion-controlled solute gravity current, excess solute mass, not enriched-fluid volume, is conserved. At late times the effective mass density excess is proportional to solute concentration, so $\Delta \rho (t)R^{2}H\sim \Delta {\rho }_0{V}_0$, and $H\sim (Dt{)}^{1/2}$. The same ambient resistance scaling givesThis scaling does not determine a universal numerical prefactor. The mass diffusivity $D$ controls the growing depth, while dilution weakens buoyancy; assuming a fixed current volume would incorrectly retain the $t^{1/5}$ law. The thin-layer requirement becomes $(\mu D/(\Delta {\rho }_{0}g{V}_0){)}^{1/3}\ll 1$; when only dissolved solute changes the upper fluid, its viscosity is of order the ambient viscosity, consistent with the plug-current window.

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.

Linearity of Stokes flow and rotational symmetry, including reflection in mathematics invariance of the spherical geometry, determine the possible rigid motions before any calculation. Translation is a polar vector; its linear dependence on a single vector $\mathbf{A}$ must be $c\mathbf{A}$. A traceless second-rank tensor $\mathbf{B}$ cannot produce a polar vector by an isotropic linear map: tensor contraction with the identity matrix vanishes and tensor contraction with the totally antisymmetric tensor vanishes because $\mathbf{B}$ is a symmetric second-rank tensor. Angular velocity is an axial vector. Neither $\mathbf{A}$ nor a symmetric $\mathbf{B}$ can produce one by an isotropic linear map. Products such as $\mathbf{A}\times \mathbf{BA}$ are excluded by Linearity of Stokes flow. Thus $\mathbf{\Omega }=0$.

Use the Unscaled Papkovich–Neuber representation, with harmonic functions $\mathbf{\Phi }$ and $\chi$:Indeed $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=\nabla p$ because each potential is a harmonic function. Put $r=|\mathbf{x}|$ and $S=\mathbf{x}\cdot \mathbf{Bx}$. The decaying harmonic functions with the necessary angular dependence are $\mathbf{A}\cdot \mathbf{x}/r^3$, $\mathbf{Bx}/r^3$ and $S/r^5$. The last is a harmonic function precisely because $\mathrm{tr}\mathbf{B}=0$. A $1/r$ Stokeslet is excluded by the force-free condition, and a rotlet by the torque-free condition. TryThe resulting Stokes flow isAt $r=a$, the coefficient of $\mathbf{n}(\mathbf{A}\cdot \mathbf{n})$ fixes $c=a^3/3$; the remaining constant vector gives $\mathbf{U}=-2\mathbf{A}/3$. In the $\mathbf{B}$ mode, matching $\mathbf{Bn}-(\mathbf{n}\cdot \mathbf{Bn})\mathbf{n}$ gives $e=a^4/2$ and $d=-a^2/2$. Therefore the complete two-mode tensorial squirmer flow isThe boundary condition is satisfied in the laboratory frame, so the Stokes flow tends to zero at infinity. The potential dipole in the $\mathbf{A}$ mode gives an irrotational flow and decays as $r^{-3}$; the leading $\mathbf{B}$ mode is a stresslet, decaying as $r^{-2}$. Neither carries a net force or torque. Direct surface integral averaging with the surface slip velocity formula gives the same translation, providing an independent check. Uniqueness of Stokes flow then identifies this decaying solution.

Put the fixed outer centre at the origin, and let the moving inner centre be $\mathbf{d}=(x,z)$. This reverses the sign of the vertical offset used in the previous parts: now $h=\Delta -\mathbf{d}\cdot \mathbf{n}$. At concentricity rotation produces no force and translation has isotropic resistance $R_0=12\pi \mu a^3/{\Delta }^3$. Balancing the external load givesThere is no inertial acceleration in this Stokes flow model; velocity is determined instantaneously by force balance.

For the subsequent motion, the full-film journal bearing assumption matters: pressure is single-valued around the whole annulus, with no specified cavitation boundary. Let $r=|\mathbf{d}|$, $e=r/\Delta$, $\varphi =\arg (x+iz)$, and ${\mathbf{e}}_r,{\mathbf{e}}_{\varphi }$ be its radial and tangential unit vectors. The squeeze resistance of an eccentric journal bearing has radial and tangential coefficientsFor completeness, rotate coordinates until $\mathbf{d}$ points along $x$, so $h=\Delta (1-e\cos \theta )$. Radial translation gives squeezing flux $q=a{v}_r\sin \theta$ and resistance proportional to $\int {\sin }^2\theta /(1-e\cos \theta {)}^3=\pi /(1-e^2{)}^{3/2}$. Tangential translation gives $q=a{v}_t[-\cos \theta +3e/(2+e^2)]$; the constant enforces periodic pressure. Its resistance usesThese expressions supply both translation directions; retaining only the vertical squeezing formula would miss the moving axle's horizontal dynamics.

The rotational fluid force from the earlier calculation points along ${\mathbf{e}}_{\varphi }$, with magnitude $R_t\Omega r/2$. Superposing rotation and translation by Linearity of Stokes flow, and resolving the load, givesFor positive $\Omega$ the axle initially moves down, then to the right because the rotation-generated force is perpendicular to the downward eccentricity. The equilibrium is displaced horizontally to the right, at the same height as the outer axis. Its eccentricity $0<e_{\ast }<1$ is the unique solutionThe rotational force then points upwards and balances the load. For negative $\Omega$ the displacement is to the left, using $|\Omega |$ in the magnitude equation. Uniqueness follows because the logarithmic derivative of $e/[\sqrt{1-e^2}(1+e^2/2)]$ has positive numerator $1-e^2/2+e^4$ after a common positive denominator. If $\Omega =0$ and $F>0$, no separated static equilibrium exists in this ideal model. If $F=0$, concentricity remains a solution.