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For the given harmonic potentials, ${\nabla }^2(\mathbf{\Phi }\cdot \mathbf{x})=2\nabla \cdot \mathbf{\Phi }$. Taking the divergence of the velocity representation therefore gives $\nabla \cdot \mathbf{u}=0$. Taking its Laplacian gives ${\nabla }^2\mathbf{u}=-2\nabla (\nabla \cdot \mathbf{\Phi })$, so the Stokes equations hold withThis is the Papkovich–Neuber representation in the sign and normalization of this problem. The potentials used in other conventions can differ by an overall factor and sign; the velocity and pressure must be scaled together.

(a) Point force. Put the singularity at the origin, $r=|\mathbf{x}|$, and choose $\mathbf{\Phi }=\mathbf{F}/(8\pi \mu r)$, $\chi =0$. The resulting Stokeslet isThe harmonic calculation applies away from the force. Distributionally this is the Green function for $-\nabla p+\mu {\nabla }^2\mathbf{u}+\mathbf{F}{\delta }^{(3)}(\mathbf{x})=0$, with zero divergence.

(b) Point source. Set $\mathbf{\Phi }=0$, $\chi =-Q/(4\pi r)$. The three-dimensional point source field isIts outward volume flux across every enclosing sphere is $Q$. The divergence is zero away from the source and equals $Q{\delta }^{(3)}(\mathbf{x})$ distributionally, so the singular point represents volume injection rather than an incompressible material point.

(c) Stresslet. Let $G_{ij}(\mathbf{x})=({\delta }_{ij}/r+x_i{x}_j/r^3)/(8\pi \mu )$ and use a symmetric trace-free stresslet tensor $S$ with the convention $u_i=-S_{jk}{\partial }_k{G}_{ij}$. Differentiating the Stokeslet and using symmetry and zero trace givesFor an axial force-dipole flow with axis $\mathbf{n}$, write $S=s(\mathbf{nn}-I/3)$. Then the radial factor is proportional to $s[3(\mathbf{n}\cdot \hat{\mathbf{x}}{)}^2-1]$. Specifying the derivative convention is essential for interpreting the signs of image strengths.

(d) Source dipole. A positive source and negative source separated by a vector $\mathbf{\ell }$ have moment $\mathbf{M}=Q\mathbf{\ell }$. Their dipole limit is minus the directional derivative of the unit source field. Hence the source dipole hasIt carries no net volume flux, and unlike a stresslet it is an irrotational potential flow.

Now set $\mathbf{r}=\mathbf{x}-d\mathbf{n}$ and $\mathbf{R}=\mathbf{x}+d\mathbf{n}$, with $d>0$. Define the unscaled kernels $K(\mathbf{R})=\mathbf{n}/R+(\mathbf{n}\cdot \mathbf{R})\mathbf{R}/R^3$, $H(\mathbf{R})=\mathbf{R}/R^3$, and let ${\partial }_n=\mathbf{n}\cdot {\nabla }_{\mathbf{R}}$. The no-slip image system of a normal Stokeslet isThe reflected Stokeslet has force $-F\mathbf{n}$. With the conventions above, the remaining image strengths areThe isotropic part of $S$ produces zero velocity, because ${\partial }_j{G}_{ij}=0$. Its inclusion simply puts the tensor in standard trace-free form. The pressure follows by superposition:

To verify the no-slip boundary condition explicitly, write $\mathbf{x}=\mathbf{\rho }$ on the plane, with $\mathbf{\rho }\cdot \mathbf{n}=0$, and let $b^2={\rho }^2+d^2$. ThenThe $b^{-5}$ terms cancel, while the remaining image contribution is $2d(\mathbf{R}-d\mathbf{n})/b^3=2d\mathbf{\rho }/b^3$. Thus both normal and tangential velocity vanish. All image singularities lie outside the fluid, so the physical half-space contains only the prescribed force.

For the far field, expand about the wall origin. The pair of opposite forces starts with $-2d{\partial }_{n}K$; the image stresslet cancels this entire $r^{-2}$ term. The first surviving field isWriting $z=\mathbf{x}\cdot \mathbf{n}$ and now $r=|\mathbf{x}|$, this isTherefore the generic fixed-angle far-field velocity decays as $r^{-3}$, rather than the $r^{-1}$ free-space Stokeslet or an uncancelled $r^{-2}$ stresslet. Along the wall it is exactly zero; at fixed height and large horizontal distance the leading tangential field instead scales as ${\rho }^{-4}$, consistent with the angular zeros of the displayed far field.

Let $\mathbf{n}$ point from fluid $-$ to fluid $+$, $P_s=I-\mathbf{nn}$, ${\nabla }_s=P_s\nabla$, and $\mathcal{K}={\nabla }_s\cdot \mathbf{n}$. A clean material interface without phase change has continuous velocity, with normal interface speed $V_n={\mathbf{u}}_{\pm }\cdot \mathbf{n}$. Its surface stress is $\gamma P_s$. The surface divergence of that tensor is ${\nabla }_s\gamma -\gamma \mathcal{K}\mathbf{n}$. Balancing forces on a small interfacial patch therefore gives the interfacial stress balance with variable surface tensionThe normal condition is $\mathbf{n}\cdot ({\mathbf{\sigma }}_{+}-{\mathbf{\sigma }}_{-})\mathbf{n}=\gamma \mathcal{K}$; the tangential condition is $P_s({\mathbf{\sigma }}_{+}-{\mathbf{\sigma }}_{-})\mathbf{n}=-{\nabla }_s\gamma$. On a stationary sphere with outward normal these reduce to $p_{-}-p_{+}=2\gamma /a$ when viscous normal stresses vanish. This fixes the sign convention for the Marangoni stress rather than leaving the direction ambiguous.

For the bubble take negligible internal dynamic viscosity, no contaminants, and initially the standard effectively insulating gas-bubble approximation. With heat advection neglected, the exterior temperature satisfies Laplace's equation, approaches $T_0+\mathbf{H}\cdot \mathbf{r}$ remotely, and has zero normal heat flux at $r=a$. The temperature dipole around an insulating sphere givesThe zero-flux condition follows by differentiating the radial factor. Let ${\gamma }_T=(d\gamma /dT{)}_{T_0}$. To first order in the weak gradient,

In the laboratory frame the force-free bubble can be described by the harmonic Papkovich–Neuber representation potentialsThey give constant exterior pressure and the decaying source dipole fieldAt the surface $\mathbf{u}\cdot \mathbf{n}=\mathbf{U}\cdot \mathbf{n}$, so the normal kinematic boundary condition holds for translation. In coordinates aligned with $\mathbf{U}$, $u_r=a^{3}U\cos \theta /r^3$ and $u_{\theta }=a^{3}U\sin \theta /(2r^3)$. Direct differentiation givesEquivalently, the surface tangential traction is $(3\mu /a)P_s\mathbf{U}$. The tangential stress condition therefore requiresFor the usual negative ${\gamma }_T$, the bubble moves toward hotter liquid. The normal condition is consistent as well: its viscous dipole is $-6\mu \mathbf{U}\cdot \mathbf{n}/a=3{\gamma }_T\mathbf{H}\cdot \mathbf{n}$, exactly the dipolar part of $2{\gamma }_s/a$. The constant part fixes $p_{\mathrm{gas}}-p_{\infty }=2{\gamma }_0/a$. There is no Stokeslet term, hence no net applied force or drag imbalance.

The thermal assumption matters. The problem gives the exterior thermal diffusivity but no gas thermal conductivity. For an inviscid bubble with conductivity ratio $K=k_{\mathrm{gas}}/k_{\mathrm{liquid}}$, temperature and normal heat-flux continuity instead giveso the same stress calculation givesThe insulating-bubble result uses $K\ll 1$. If instead an undisturbed surface temperature gradient is imposed, corresponding to $K=1$, the coefficient is $1/3$ rather than $1/2$. Exterior thermal diffusivity alone does not select between those thermal boundary conditions. This dependence is explicit rather than silently treating a missing gas property as a prescribed value.

The small parameters can now be stated precisely. With $U_{\ast }=a|{\gamma }_T||\mathbf{H}|/(2\mu )$ and $\nu =\mu /\rho$, requireSmall Reynolds number justifies Stokes flow; small Péclet number allows quasi-static heat conduction without advective distortion; small capillary number and fractional surface-tension variation support a nearly spherical interface and a weak perturbation of its base surface tension. These last two conditions are proportional for this velocity scale but describe different approximations. Linearizing $\gamma (T)$ also needs $a|\mathbf{H}||{\gamma }_{TT}/{\gamma }_T|\ll 1$ if its derivative varies appreciably. Negligible internal viscosity and the insulating specialization require the corresponding gas-to-liquid viscosity and conductivity ratios to be small. Gravity has been excluded; if restored, its buoyancy-induced migration must be small compared with $U_{\ast }$, not merely small enough to avoid deformation.

Finally, surfactants change both the constitutive surface tension and the surface stress dynamics. The surface flow driven by the thermal Marangoni effect carries insoluble surfactant toward the high-tension side, lowering its tension and producing an opposing concentration-driven gradient. When surface diffusion or exchange is slow, this tends to immobilize the interface, suppressing the clean-bubble thermocapillary migration of an insulating bubble, sometimes nearly stopping it. If the surfactant stays uniformly equilibrated, it can instead change ${\gamma }_T$ without a large opposing gradient. Its concentration, surface diffusivity and adsorption kinetics must be supplied to compute a replacement speed; no universal numerical reduction factor follows merely from its presence.

Put $r=|\mathbf{x}|$ and $s=\mathbf{F}\cdot \mathbf{x}$. In the Unscaled Papkovich–Neuber representation, take the harmonic potentials $\mathbf{\Phi }=-\mathbf{F}/(8\pi \mu r)$ and $\chi =0$. Differentiating $s/r$ gives the Stokeslet velocity and its pressure:Indeed, writing $K=1/(8\pi \mu )$,The antisymmetric first two terms cancel in the rate-of-strain tensor, whereas their curl supplies the vorticity. ThusThe rate-of-strain tensor has zero trace, as required by incompressibility. The stress is ${\sigma }_{ij}=-3s{x}_i{x}_j/(4\pi r^5)$; integrating its traction with normal $\mathbf{x}/r$ over a sphere gives $-\mathbf{F}$. This fixes the force normalization of the Stokeslet.

For a sphere in a uniform straining Stokes flow, write the ambient velocity as $\mathbf{Ex}$, where $\mathbf{E}$ is symmetric and trace-free, and set $q=\mathbf{x}\cdot \mathbf{Ex}$. Symmetry gives zero translation and rotation of the centred force-free, torque-free sphere. Seek a decaying disturbance using the Unscaled Papkovich–Neuber representation withThe components of $\mathbf{Ex}/r^3$ are derivatives of $1/r$, and $q/r^5$ is a trace-free harmonic quadrupole, so both potentials satisfy Laplace's equation outside the sphere. Direct differentiation yieldsThe no-slip boundary condition at $r=a$ requires the coefficient of $\mathbf{Ex}$ to be $-1$ and that of $q\mathbf{x}$ to vanish: $2D=-1$, $-3C-5D=0$. Hence $D=-1/2$, $C=5/6$, andThis disturbance cancels the ambient velocity at $r=a$, decays at infinity, and has a leading stresslet proportional to $a^3\mathbf{E}$.

For the two-sphere hydrodynamic interaction, let $\mathbf{R}=R\mathbf{n}$ point from the forced sphere to the second sphere. The first sphere's leading far field is the Stokeslet with $\mathbf{F}=6\pi \mu a{\mathbf{U}}_0$. Its incident rate-of-strain tensor at the second sphere isThe force-free sphere translates with the local uniform incident flow, and the torque-free sphere rotates with half the incident vorticity. Those parts cause no leading Stokeslet or rotlet disturbance. The remaining incident strain produces the stresslet just calculated. At displacement $-\mathbf{R}$ from the second sphere its velocity isThe first sphere translates in this returned incident flow. Finite-size sampling and higher incident multipoles enter only beyond the retained reflection order. Thus the stresslet reflection between two force-free and forced spheres givesThe factor $|{\mathbf{U}}_0|$ makes explicit the velocity scale implicit in the printed remainder.

The resulting input power, equal to the viscous dissipation, isIt decreases at fixed force. The Minimum-dissipation theorem for Stokes flow instead compares admissible fields at fixed first-sphere velocity: extending a two-sphere field rigidly into the second sphere gives an admissible one-sphere field, so the extra rigid inclusion can only increase the minimum resistance. Allowing the second sphere to choose its translation and rotation minimizes over its rigid motions, producing zero force and couple there. An increase of this effective hydrodynamic resistance matrix decreases its inverse hydrodynamic mobility matrix, and hence decreases fixed-force power. The negative mobility correction is therefore consistent with minimum dissipation.

If the second sphere is prevented from rotating, its angular velocity relative to the incident fluid becomesUse the rotating sphere in Stokes flow at $-\mathbf{R}$. The additional reflected rotlet givesAdding this rotation-clamped reflection between two spheres to the strain reflection,The holding couple supplies no power because the held sphere's angular velocity is zero.

The fluid is outside the body, so its normal on the inner boundary is $-n^{\prime }$. Let $t=\sigma n^{\prime }$ be traction on the body's surface from the fluid. For an exterior point and a flow vanishing at infinity, the boundary representation readsFor $r=|y|\gg a$, expand the single layer through its first moment and the double layer through its zeroth moment:The next single-layer and double-layer terms are both $O(r^{-3})$ at fixed body size and boundary data. Differentiating $J$ givesHence the dipole terms combine intoThe antisymmetric part obeys $(\Sigma -{\Sigma }^T)y=G\times y$ for $G_j=-{ϵ}_{j\alpha \beta }{\Sigma }_{\alpha \beta }$. For the symmetric trace-free tensor $S$, $y\cdot My=y\cdot Sy+(\mathrm{Tr}M)r^2/3$. Also $\mathrm{Tr}\Sigma -\mathrm{Tr}M=2\mu \mathrm{Tr}U=2\mu Q$. Substitution givesThis derives the Stokes far-field multipoles of a deforming body. $F$ is the force exerted by the body on the fluid, the negative of the integrated fluid traction. $G$ is its torque on the fluid: $G=-\int x\times tdS$. Their kernels are the Stokeslet and rotlet. $Q=\int u\cdot n^{\prime }dS$ is the rate of increase of the body's volume, by the Reynolds transport theorem. It gives a radial source flow with volume flux $Q$. A volume-preserving body has $Q=0$. The trace-free tensor $S$ gives the stresslet, which can remain nonzero for a force-free, torque-free swimmer.

Use the Newtonian fluid stress tensor $\sigma =-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$. The Stokes equations for a point force on the fluid are $\nabla \cdot u=0$ and $\nabla \cdot \sigma +F\delta (x)=0$. Put $r=|x|$. Away from the origin, a harmonic vector potential proportional to $F/r$ in the Unscaled Papkovich–Neuber representation gives the rotationally covariant decaying solution. Specifically, take $\Phi =-F/(8\pi \mu r)$ and $\chi =0$. Differentiation yieldsThis satisfies the homogeneous Stokes equations for $r>0$. Its stress isTo verify the point-force normalization, on a sphere $r=\epsilon$ the traction is $\sigma n=-3(F\cdot n)n/(4\pi {\epsilon }^2)$. Since ${\int }_{S^2}{n}_i{n}_{j}d\Omega =4\pi {\delta }_{ij}/3$, its surface integral is $-F$. Thus integrating momentum balance over the small ball gives exactly the applied force $F$, fixing the normalization of the Stokeslet. Its radial velocity is $(F\cdot n)/(4\pi \mu r)$, soThere is no net volume or mass source at the singularity. The kernels have the parity $J(-x)=J(x)$ and $K(-x)=-K(x)$.

For two Stokes flows with the same viscosity and body forces $f^{(1)},f^{(2)}$, the Lorentz reciprocal theorem including body forces isIndeed, apply the divergence theorem to $u_i^{(1)}{\sigma }_{ij}^{(2)}-u_i^{(2)}{\sigma }_{ij}^{(1)}$. The cross-gradient terms cancel: incompressibility removes each pressure contraction and the remaining terms are both $2\mu e^{(1)}:e^{(2)}$. The two stress divergences give the body-force terms shown.

Choose flow 1 to be the desired body-force-free field and flow 2 to be a Stokeslet with arbitrary force vector $q$ at $y$. For $y\in V$, the body-force term is $-q\cdot u(y)$. Rearrange the reciprocal identity and use the parity of $J,K$; equality for every $q$ givesFor $y$ outside the volume there is no point force in $V$, so the same two integrals sum to zero. To obtain the smooth boundary value, exclude a small hemisphere about $y$, use continuity of $u$ to replace it by $u(y)$ on that hemisphere, and take the radius to zero. A full small sphere contributes $u(y)$; a hemisphere contributes half of it. Equivalently, the local solid-angle factor is $2\pi /(4\pi )$. The result is the boundary integral representation of Stokes flow with right side $\frac{1}{2}u(y)$ on a smooth boundary, and with the double layer interpreted by its boundary principal value. A nonsmooth corner would have a different solid-angle coefficient.

The material derivative follows surfactant carried by the moving interface. The term $-C{\nabla }_s\cdot u_s$ is dilution or concentration by tangential area expansion or contraction. The term $-C(u\cdot n){\nabla }_s\cdot n$ is the additional area change produced by normal motion of a curved interface. The term $D_s{\nabla }_s^{2}C$ represents surface diffusion, while $-k(C-C_0)$ relaxes the concentration towards a reservoir value through exchange. All these terms refer to concentration per surface area, as in surfactant transport with exchange relaxation.

In the steady bubble frame, the spherical interface has $u\cdot n=0$. Write $C=C_0+C^{\prime }$ with uniform $C_0$ and regard both $u$ and $C^{\prime }$ as first order in the rising speed. Advection of $C^{\prime }$ and its product with surface dilation are second order. Hence the linear steady equation isSpherical symmetry and linearity leave only the tangential projection of the imposed translation vector: $u_s=A{I}_{s}U$. An azimuthal term $n\times U$ would change sign under reflection and is excluded for this non-chiral rising bubble. The degree-one forcing does not excite other spherical-harmonic degrees in the linear problem.

On $r=a$, differentiating $n=x/r$ givesEach component of $n$ is a degree-one spherical harmonic, soOne may also obtain the coefficient from $n\cdot n=1$: its surface Laplacian gives $n\cdot {\nabla }_s^{2}n=-|{\nabla }_{s}n{|}^2=-2/a^2$, while rotational symmetry makes the vector Laplacian parallel to $n$. Expanding $I_{s}U=U-n(U\cdot n)$ then yieldsThe candidate $C^{\prime }=BU\cdot n$ therefore solves the linear transport equation ifIts spherical mean is zero. This is the dipolar surfactant distribution. Assuming $k{a}^2+2D_s>0$, its consistency condition isDiffusion or exchange must restore uniform concentration faster than the interfacial flow redistributes it: this holds with a small surface Péclet number, fast exchange, or sufficiently immobilized surface velocity. These same conditions justify neglecting the perturbation's advection. Small capillary number and sufficiently small buoyancy-induced shape distortion are separate requirements for a spherical interface.

Take $+$ to mean the outer liquid, $-$ the bubble interior and $n$ outward from the bubble. The general interfacial stress balance with variable surface tension isFor $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$, its tangential part is $-{\nabla }_s\gamma ={\gamma }_{1}B{I}_{s}U/a$. Thus the requested form isThis positive parameter measures the strength of the restoring Marangoni stress relative to viscous stress, including the relaxation of concentration by diffusion and exchange.

The far velocity $-U$ is produced by the constant harmonic vector potential $\Phi =U$. The two independent decaying degree-one modes are the force-carrying vector potential $\alpha aU/r$ and the scalar dipole $\chi =\beta a^{3}U\cdot \nabla (1/r)$. They are harmonic outside the bubble and have precisely the symmetry of uniform translation. The Unscaled Papkovich–Neuber representation gives, at arbitrary radius,The velocity at $r=a$ agrees with the printed expression. However, the original PDF's traction formula has a factor-of-two error. Direct differentiation with $\sigma =-pI+\mu (\nabla u+\nabla u^T)$ givesnot the printed prefactor $12\mu /a$. This is traction of translating-sphere Papkovich–Neuber potentials. The $\alpha$ term follows immediately from the Stokeslet stress, and the scalar-potential part gives $6\mu \beta [U-3(U\cdot n)n]/a$; neither contribution contains the extra factor two.

With this consistent traction, no penetration requires $\beta -\alpha =1/2$, tangential velocity requires $A=-(1+\alpha +\beta )$, and tangential stress balance requires $\beta =A\lambda$. Solving,For comparison, taking the printed $12\mu /a$ literally would instead giveThose solve the supplied algebraic pair but not the Newtonian stress calculated from the supplied potentials. The factor-six result is the physically consistent translating surfactant-coated bubble solution.

As $\lambda \to 0$, $A\to -1/2$ and $\beta \to 0$: the clean bubble has a mobile interface and zero tangential traction. As $\lambda \to \infty$, $A\to 0$, $\alpha \to -3/4$ and $\beta \to -1/4$: the interface becomes immobilized and the exterior flow approaches that past a no-slip rigid sphere. The limiting tangential traction is $-3\mu I_{s}U/(2a)$, a useful check on the correct normalization.

The normal dynamic traction isapart from an arbitrary constant pressure. In the normal stress balance on a translating bubble, its degree-one part is balanced by the buoyant hydrostatic pressure difference and by the variation $-2{\gamma }_1{C}^{\prime }/a$ of normal capillary stress. A uniform internal gas pressure balances the constant Laplace-pressure term but cannot by itself balance this angular variation. If $U$ is upwards and $\Delta \rho$ is liquid minus bubble mass density, the degree-one balance can be writtenIt fixes $U=\Delta \rho g{a}^2(1+2\lambda )/[3\mu (1+3\lambda )]$, tending to the clean-bubble and rigid-sphere rise speeds. A degree-one radial displacement translates a sphere rather than changing its curvature, so a supposed dipolar shape distortion cannot replace this force balance.

Work in the laboratory frame, with fluid at rest at infinity and the instantaneous sphere centre at the origin. In the Unscaled Papkovich–Neuber representation, take the harmonic potentialsDifferentiating the two terms gives the translating sphere in Stokes flow:Here $\mathbf{n}=\mathbf{x}/r$. The potentials are harmonic outside the sphere, so they satisfy the Stokes equations. At $r=a$, the two coefficients of $\mathbf{nn}$ cancel and the coefficient of $I$ is one, proving the no-slip boundary condition $\mathbf{u}=\mathbf{V}$; the field decays at infinity. Its leading part is a Stokeslet of force $6\pi \mu a\mathbf{V}$ exerted on the fluid.

Let $\hat{\mathbf{z}}$ point upwards, $\delta =V_1-V_2$, and $D=a_2{V}_2-a_1{V}_1$. Each isolated sphere moves with velocity $-V_i\hat{\mathbf{z}}$. To leading order its correction is the other's incident Stokeslet velocity:The forces generating these fields are the fixed sedimenting loads, so their leading strengths are set by the isolated speeds. Subtraction yields the leading interaction of two sedimenting spheres:Since $\hat{\mathbf{z}}=\cos \theta \hat{\mathbf{r}}-\sin \theta \hat{\mathbf{\theta }}$, this reduces toThe stipulated higher-order errors can be appended to both equations; they do not enter this leading calculation.

For the same-sign case, interchange labels if necessary to arrange $\delta >0,D>0$, and put $c=3D/4$. The vertical separation $z=r\cos \theta$ satisfiesIf sphere one is initially above sphere two, its vertical order reverses; in every case $z\to -\infty$, implying $r\to \infty$. For a noncollinear trajectory there is no collision in the point-force equations. Indeed, divide the two polar equations and integrate:At side-by-side passage, $\theta =\pi /2$, the closest separation is the positive root of $\delta r^2+2cr=C$. Before passage $\dot{r}<0$ and after passage $\dot{r}>0$, proving overtaking and unbounded subsequent separation. This passing invariant for unequal point-force spheres describes a physical far-field passage only when the closest separation also greatly exceeds the sphere radii. Exactly collinear approach instead reaches the near-contact regime, where this leading approximation cannot describe two finite spheres passing through one another.

For opposite signs, choose $\delta >0,D<0$ and consider the invariant configuration with sphere one vertically above sphere two, $\theta =0$. The radial equation isIf $r>r_{\ast }$, separation decreases; if $r<r_{\ast }$, it increases. Thus it approaches this equilibrium along the vertical line. Linearization gives $\dot{\delta r}=-(\delta /r_{\ast })\delta r$. But a small tilt obeys $\dot{\theta }=(\delta /(2r_{\ast }))\theta$, so the vertical bound pair in the point-force sedimentation model is stable only against vertical perturbations and is a saddle in the full separation dynamics. The opposite vertical orientation has the reversed stability. Reversing gravity reverses both $\delta$ and $D$, leaves $r_{\ast }$ unchanged, and reverses every trajectory; radial attraction becomes repulsion. This is fully consistent with kinematic reversibility of Stokes flow.

There is an important restriction in the printed data. Its hierarchy $|D|\ll (a_1+a_2)|\delta |$ implies $r_{\ast }\ll a_1+a_2$, so the proposed equilibrium would overlap the physical spheres and lie outside the well-separated approximation. With that hierarchy there is no justified physical far-field bound pair. The intended far-field equilibrium instead requires the first inequality reversed, $|D|\gg (a_1+a_2)|\delta |$. The formal equilibrium and its vertical stability above show the intended mechanism, while also identifying the missing validity and transverse-stability qualifications.

For a force $\mathbf{F}\delta (\mathbf{x})$ acting on an unbounded incompressible fluid, the Stokeslet and its pressure areThey solve $-\nabla p+\mu {\nabla }^2\mathbf{u}+\mathbf{F}\delta =0$ and $\nabla \cdot \mathbf{u}=0$. Approximate a slender body by a line distribution of these point forces on its centreline. At a point labelled by arclength $s$,To extract the large logarithm, write $s^{\prime }=s+\ell$ and approximate $\mathbf{X}(s+\ell )-\mathbf{X}(s)\simeq \ell \mathbf{t}$, $\mathbf{f}(s+\ell )\simeq \mathbf{f}(s)$, where $\mathbf{t}={\mathbf{X}}^{\prime }$ is the unit tangent vector. The near-line contribution has kernel $(I+\mathbf{tt})/(8\pi \mu |\ell |)$. Cut it off at the wire radius $R$ and a macroscopic length $L$. The two sides of the line giveThe far contribution and changes in the cutoff add terms without the leading logarithm. Since $(I+\mathbf{tt}{)}^{-1}=I-\mathbf{tt}/2$, inversion proves the leading slender-body force densityThis is force on the fluid; the fluid's force on the wire has the opposite sign. It gives twice the resistance to transverse motion as to tangential motion at this order.

The stated arclength describes the circular helix $\mathbf{X}=(a\cos \theta ,a\sin \theta ,b\theta )$. The PDF omits the second $a$ in the coordinates, so the literal printed curve and its arclength are inconsistent unless $a=1$. First use the circular interpretation consistent with the given arclength. Put $q=\sqrt{a^2+b^2}$ and $c=2\pi \mu /\ln (L/R)$. ThenRotation about $z$ has velocity $\mathbf{v}=\Omega (-a\sin \theta ,a\cos \theta ,0)$, so $\mathbf{t}\cdot \mathbf{v}=\Omega a^2/q$. The axial force density is $f_z=-c\Omega a^{2}b/q^2$. Integrating over one turn givesFor translation, $\mathbf{v}=W\hat{\mathbf{z}}$ and $\mathbf{f}=cW(2\hat{\mathbf{z}}-(b/q)\mathbf{t})$. Only the tangential term contributes to axial couple. Since $(\mathbf{X}\times \mathbf{t}{)}_z=a^2/q$, its density is $-cW{a}^{2}b/q^2$, givingThese are the off-diagonal coefficients of the axial resistance matrix of a slender helix. Their equality does not depend on this particular geometry: with symmetric local resistance $D=c(2I-\mathbf{tt})$, the rotation-induced axial force is $\int \hat{\mathbf{z}}\cdot D(\hat{\mathbf{z}}\times \mathbf{X})ds$, whereas the translation-induced axial torque is $\int (\hat{\mathbf{z}}\times \mathbf{X})\cdot D\hat{\mathbf{z}}ds$. They agree pointwise by symmetry of $D$. The full Stokes-flow statement is the Lorentz reciprocal theorem.