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More surfactant is encountered on the side toward which the far-field concentration $\Gamma$ increases. Adsorption lowers the surface tension there, so the resulting Marangoni stress drives interfacial flow toward the cleaner, higher-tension side. The reaction force propels the bubble along the concentration gradient; this is chemophoresis.

The bulk concentration obeys the advection-diffusion equationAt $r=a$,equates the outward bulk diffusive flux to minus the net adsorption rate: $C>b\Gamma$ favors desorption into the bulk, while $C<b\Gamma$ favors adsorption onto the interface.

With advection neglected, $\Gamma$ is harmonic. Rotational symmetry about $\mathbf{G}$ and the imposed far-field gradient select the dipolar formWhen $kab\ll D$, the boundary flux is smaller than the characteristic diffusive flux, so the leading boundary condition is ${\partial }_r\Gamma =0$ at $r=a$. This gives $B=a^3/2$, and hence

For an interface with unit normal directed from the bubble into the exterior liquid, the interfacial stress balance with variable surface tension may be writtenwith signs tied to the stated curvature convention. Put $C^{\prime }=A\mathbf{G}\cdot \mathbf{n}$ and $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$. The constant part of the normal traction is balanced by the uniform bubble pressure. Since $\kappa =2/a$ andthe remaining exterior traction isIts resultant vanishes becauseand therefore the integrals of the two terms cancel. This is required because the bubble and its interfacial stresses exert no external body force on the combined bubble–fluid system.

The traction is a first spherical harmonic, so the decaying, force-free Stokes flow has no Stokeslet and is generated by the indicated Papkovich–Neuber representation. Comparing the supplied tractionwith the capillary traction givesIn the convention for these potentials, the normal velocity at $r=a$ is $(\beta /\mu )\mathbf{G}\cdot \mathbf{n}$. The kinematic boundary condition for a translating sphere is $\mathbf{u}\cdot \mathbf{n}=\mathbf{U}\cdot \mathbf{n}$, so

Linearize the surface transport equation about $C=C_0$ by writing $C=C_0+C^{\prime }$ and neglecting products of small perturbations. The tangential velocity relative to the translating bubble obtained from the same potential isUsing the identities supplied in the question,The bulk result gives ${\Gamma }^{\prime }=\frac{3}{2}a\mathbf{G}\cdot \mathbf{n}$ on the surface. The linearized equationthen yieldsThus

Increasing $k$ strengthens exchange with the imposed bulk gradient, so $A$ and $U$ increase toward a saturation value. Increasing $D_s$ smooths surface-concentration differences and decreases $U$. Increasing $C_0$ strengthens advective redistribution of the background surfactant; the resulting feedback opposes the imposed dipole, so $U$ decreases.

For each Volvox colony, the excess gravitational force isFor a normal Stokeslet a distance $R$ below a flat stress-free interface, the image is an oppositely directed Stokeslet a distance $R$ above it. The real force creates no lateral velocity at the other colony because both centres have the same height. Their separation from the image is $(x,-2R)$, however, so each colony moves toward the other with speedBoth colonies move, and therefore

DefineThe dimensionless dynamics areThis gradient flow descends an even attractive potential with minimum $V(0)=-1$ and $V\to 0^{-}$ as $|\xi |\to \infty$.

For $\xi \gg 1$, $d\xi /d\tau \simeq -1/{\xi }^2$, so the approach from ${\xi }_0=x_0/(2R)$ takes ${\tau }_c\simeq {\xi }_0^3/3$. Restoring dimensions,With water viscosity $\mu \simeq 10^{-3}\mathrm{Pa}\mathrm{s}$, ${\rho }_w\simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, $ϵ=0.03$, $R=200\mu \mathrm{m}$, and $x_0=5R$, this givesBecause $x_0/(2R)=2.5$ is only moderately large, this is a far-field estimate rather than the exact collision time.

For equal oscillation amplitudes $\delta$, frequency $\omega$, mean separation ${\ell }_0$ and phase lag $\varphi$, the leading mean total force on the fluid is $9\pi \mu a_1{a}_2{\delta }^2\omega \sin \varphi /{\ell }_0^2$. Each sphere supplies half. The large-separation calculation expands the Stokeslet interaction; reciprocal in-phase and antiphase motions have zero mean force.

Differentiating the Stokeslet gives its rate-of-strain tensor and vorticity, displayed above for $r=|\mathbf{x}|>0$. The pressure is $p=\mathbf{F}\cdot \mathbf{x}/(4\pi r^3)$. Integrating the Newtonian fluid stress tensor over a sphere enclosing the origin gives traction resultant $-\mathbf{F}$, fixing the point-force normalization.

A buoyantly rising inviscid bubble in otherwise quiescent unbounded liquid has an exterior Stokeslet with $F=4\pi \mu aU=4\pi \rho g{a}^3/3$. Its tangential traction vanishes on the sphere. The dynamic normal traction is $-3F\cos \theta /(4\pi a^2)$, while the hydrostatic normal-traction variation is $\rho ga\cos \theta$. Their dipoles cancel at the terminal velocity, leaving a constant total normal stress compatible with uniform gas pressure even when surface tension is zero. This is an exact spherical solution in the stated Stokes model, not a proof of shape stability under arbitrary perturbations.

For a two-mode tensorial squirmer flow of radius $a$ and a passive equal sphere held at $X=Re$, with $a/R\ll 1$, let $s=e\cdot Be$. Faxén's first law gives holding force $F_h=-9\pi \mu a^{3}se/R^2+O(\mu a^4|A|/R^3+\mu a^5\Vert B\Vert /R^4)$. Its reflected Stokeslet changes swimming velocity by $-9a^{3}se/(4R^3)$. Since this leading force is radial, its vorticity at the swimmer vanishes. The next holding force, generated by the potential dipole, yields rotation $a^{4}A\times X/(4R^6)$. Terms from the passive sphere's finite radius, reflected stresslet and rotlet are needed at higher order. These formulas assume fixed slip and no externally applied swimmer force or torque.

With body-outward normal $n$, put $F=-\int \sigma ndS$, ${\Sigma }_{ij}=\int x_i(\sigma n{)}_{j}dS$, $U_{ij}=\int u_i{n}_{j}dS$, and $M=\Sigma -2\mu U$. Taylor expansion of the exterior boundary integral representation of Stokes flow gives a Stokeslet from $F$, a rotlet from $G_j=-{ϵ}_{jkl}{\Sigma }_{kl}$, a source from $Q=\mathrm{Tr}U$, and a stresslet from $S=(M+M^T)/2-I\mathrm{Tr}M/3$. Thus $F$ and $G$ are the force and torque exerted on the fluid, while $Q$ is the instantaneous displaced-volume rate. A volume-preserving deforming body has $Q=0$.

Place a normal Stokeslet below a plane and an equal, oppositely directed image at the reflected point above it. The normal velocity of their sum is odd across the plane and the tangential velocity is even. The plane therefore has zero normal velocity and zero normal derivative of tangential velocity, giving zero tangential stress. For a sphere of radius $a$ translating towards that plane at speed $U$ from distance $d\gg a$, the leading radial surface flow is $3aUrd/[2(r^2+d^2{)}^{3/2}]$, directed outward. The image's incident flow at the sphere changes its required force by relative order $a/d$, so the first correction to surface velocity at fixed $r/d$ is $O(U(a/d{)}^2)$; the isolated sphere's potential-dipole correction occurs only at third order.

Consider two widely separated rigid spheres of equal radius $a$, with separation $R\mathbf{n}$, in Stokes flow. Apply a force $6\pi \mu a{\mathbf{U}}_0$ to the first sphere and let the second be force-free and torque-free. The second sphere samples the rate-of-strain tensor of the first sphere's Stokeslet, $\mathbf{E}=3a({\mathbf{U}}_0\cdot \mathbf{n})(\mathbf{I}-3\mathbf{nn})/(4R^2)$. Its leading disturbance is the stresslet of a sphere in a uniform straining Stokes flow. Evaluating that disturbance back at the first sphere givesThe hydrodynamic mobility matrix decreases at fixed force, consistently with the increase of resistance caused by an additional rigid inclusion.

For a forced sphere and a distant equal force-free, torque-free sphere, the leading incident Stokeslet strain is proportional to $(\mathbf{U}\cdot \mathbf{n})(I-3\mathbf{nn})$. Its reflected stresslet is axisymmetric about the line of centers and has zero vorticity on that line. The nominal fifth-order self-spin coefficient therefore vanishes. Finite-size incident-strain corrections and higher force-free scattering multipoles first supply a generic seventh-order angular velocity. Purely longitudinal forcing has zero spin exactly by axial symmetry. The leading translation correction is fourth order for a generic orientation, but sixth order for purely transverse forcing. These cancellations matter when using the method of reflections for Stokes flow to infer orders from far-field decay alone.

For $\Phi =U+\alpha aU/r$ and $\chi =\beta a^{3}U\cdot \nabla (1/r)$ in the Unscaled Papkovich–Neuber representation, the Newtonian fluid stress tensor gives the displayed traction at $r=a$. The $\alpha$ contribution is the Stokeslet stress; the $\beta$ contribution is twice the viscosity times the Hessian of its harmonic scalar potential. The prefactor is $6\mu /a$, not $12\mu /a$. The rigid translating-sphere limit is a direct normalization check.