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For equal spheres of radius $a$ separated by $R\gg a$, one held fixed and the other forced with $\mathbf{F}=6\pi \mu a\mathbf{V}$, put $\zeta =6\pi \mu a$, $\mathbf{n}=\mathbf{X}/R$ and $\mathsf{A}=3a(I+\mathbf{nn})/(4R)$. The method of reflections for Stokes flow gives $\mathbf{U}=\mathbf{V}-{\mathsf{A}}^2\mathbf{V}+O(V{a}^4/R^4)$, since the fixed sphere creates the reflected Stokeslet with force $-6\pi \mu a\mathsf{A}\mathbf{V}$. The leading hydrodynamic mobility matrix correction therefore reduces longitudinal motion more strongly than transverse motion.

The shear has $E_{xy}=E_{yx}=C/2$. The separation between centers is $2(X,Y,0)$, so in the leading stresslet from part b, $q=4CXY$, $r=2R$ and the radial vector's $y$ component is $2Y$. Evaluating the other sphere's disturbance at the positive center givesThis is the first method of reflections for Stokes flow approximation. The next multipole in part b and the receiver's Faxén Laplacian contribute at $C{a}^5/R^4$; changes to the incident strain from further reflections are smaller, $C{a}^6/R^5$. The hypothesis $Y_{\infty }\gg a$ ensures well-separated spheres throughout the encounter.

The leading horizontal velocity is $\dot{X}=CY+O(C{a}^3/R^2)$. Dividing and using $Y_{\infty }$ on the right at first order givesIntegrating from negative infinity,The correction increases until $X=0$ and decreases afterward. ThusThe expansion gives $Y\to Y_{\infty }$ as $X\to +\infty$. More generally, reversible scattering of two spheres in simple shear gives no net transverse shift for this smooth, noncontact, deterministic two-sphere encounter: reflection $X\mapsto -X$ reverses the imposed shear, and time reversal restores it. Uniqueness makes the incoming and outgoing branches mirror images. Brownian motion, contact forces or finite inertia would remove the assumptions behind this conclusion.

Let $\zeta =6\pi \mu a$, $ϵ=a/R$ and define the dimensionless leading cross-mobility tensorFor a given applied force, the outgoing Stokeslet is fixed by that force, regardless of the sphere's entrainment velocity. At the other center the Faxén translation law gives its force-only velocity plus the incident Stokeslet velocity. The finite-size dipole and the Laplacian Faxén correction first contribute $O({ϵ}^3|\mathbf{V}|)$; an emitted stresslet reflection is still smaller at this order. ThereforeThere is no second-order translational term for prescribed equal forces.

The vorticity of translating-sphere Stokes flow is $\mathbf{\omega }=3a\mathbf{V}\times \mathbf{r}/(2r^3)$; its potential dipole has zero curl. Apply the torque-free Faxén rotation law to the other sphere's field, using $\mathbf{r}=-\mathbf{R}$ at sphere 1 and $\mathbf{r}=\mathbf{R}$ at sphere 2. This yieldsthrough second order in the dimensionless spin $a\mathbf{\Omega }/|\mathbf{V}|$. Neglected spin terms are higher order. This is the method of reflections for Stokes flow, with the ambient field at each application excluding that sphere's own disturbance.

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.

Let $\mathbf{e}=\mathbf{X}/R$, $s=\mathbf{e}\cdot \mathbf{Be}$, and $A_{\parallel }=\mathbf{A}\cdot \mathbf{e}$. Applied forces and couples below are forces and torques on the solid; by force balance their values are also the strengths exerted by that solid on the fluid. This fixes the signs in Faxén's first law and Faxén's rotational law. To hold the passive sphere fixed, set its translation and angular velocity to zero in those laws. The incident two-mode tensorial squirmer flow givesOnly the stresslet part has vorticity:Consequently, for fixed $\mathbf{A},\mathbf{B}$ as $a/R\to 0$,The leading force is the radial stresslet force. Its next correction, $O(\mu a^4|\mathbf{A}|/R^3)$, comes from the incident potential dipole. The next $\mathbf{B}$ correction, $O(\mu a^5\Vert \mathbf{B}\Vert /R^4)$, contains both the swimmer's $r^{-4}$ field and the Laplacian term in Faxén's first law. If the relevant coefficient vanishes, these next terms must be retained instead of calling the vanished term a nonzero leading approximation. Further exchanges in the method of reflections for Stokes flow are smaller still.

The passive sphere reflects primarily a Stokeslet with strength ${\mathbf{F}}_h$. At the swimmer, its velocity is $(\mathbf{I}+\mathbf{ee}){\mathbf{F}}_h/(8\pi \mu R)$. The unchanged surface slip velocity adds the free swimming velocity to the incident-flow contribution in Faxén's first law. ThusThe leading change is $O(a^3\Vert \mathbf{B}\Vert /R^3)$; its next correction is the displayed $O(a^4|\mathbf{A}|/R^4)$ term. Finite-radius Faxén's first law corrections, the passive sphere's reflected stresslet, and its rotlet contribute at order $a^5\Vert \mathbf{B}\Vert /R^5$ to translation.

For a Stokeslet at the origin,The curl of the first term is $\mathbf{F}\times \mathbf{x}/r^3$; that of the second is also $\mathbf{F}\times \mathbf{x}/r^3$, before multiplying by $1/(8\pi \mu )$. Hence ${\mathbf{\omega }}_F=\mathbf{F}\times \mathbf{x}/(4\pi \mu r^3)$. At the swimmer use displacement $-\mathbf{X}$. The leading $\mathbf{B}$ force is parallel to $\mathbf{X}$, so its Stokeslet has zero vorticity there. This eliminates the putative $O(a^3\Vert \mathbf{B}\Vert /R^4)$ rotation. The potential dipole itself has no vorticity, but the holding force it induces is not radial. Applying Faxén's rotational law to its reflected Stokeslet yieldsThe passive sphere's rotlet and reflected stresslet can affect that last order, so they cannot restore the larger missing rotation. When $\mathbf{A}\times \mathbf{X}=0$, the boxed coefficient is zero and the higher-order terms determine any rotation.

Write $r=|\mathbf{x}|$ and take $\mu$ to be the dynamic viscosity. The Papkovich–Neuber representation of a homogeneous Stokes flow uses a harmonic vector potential $\mathbf{\Phi }$ and harmonic scalar potential $\chi$:These formulas give incompressibility because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and give $\mu {\nabla }^2\mathbf{u}=\nabla p$, the Stokes equation away from forcing.

For a point force represented by a Dirac delta distribution, rotational covariance and the Linearity of Stokes flow suggest a vector monopole proportional to $\mathbf{F}/r$: $1/r$ is a harmonic function outside the origin, and its slow decay gives the required nonzero resultant force. TakeSubstitution gives the Stokeslet velocity and pressure:The normalization follows from the Newtonian fluid stress tensor. Differentiating the velocity gives the rate of strain and vorticity of a Stokeslet:Here $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$. Thus ${\sigma }_{ij}=-p{\delta }_{ij}+2\mu e_{ij}=-3(\mathbf{F}\cdot \mathbf{x})x_i{x}_j/(4\pi r^5)$. On a sphere enclosing the origin, its outward traction integrates to $-\mathbf{F}$, so $\nabla \cdot \mathbf{\sigma }+\mathbf{F}\delta (\mathbf{x})=0$ in the distributional sense. All displayed fields away from the point force have $r>0$.

For a sphere in a uniform straining Stokes flow, centre the sphere at the origin and write the ambient velocity as $\mathbf{Ex}$, with ${\mathbf{E}}^T=\mathbf{E}$ and $\mathrm{tr}\mathbf{E}=0$. Faxén's first law and Faxén's rotational law give zero translation and rotation. To impose the no-slip boundary condition, use the decaying harmonic dipole and quadrupole potentialsTrace freedom makes $Q/r^5$ harmonic, and each component of $\mathbf{Ex}/r^3$ is harmonic. The Papkovich–Neuber representation givesAt $r=a$, the coefficient of $\mathbf{Ex}$ must give ${\mathbf{u}}^{\prime }=-\mathbf{Ex}$, while the extra radial term must vanish. Hence $B=-\mu a^5$ and $A=5\mu a^3/3$, giving the complete disturbanceThe total velocity vanishes on the sphere and tends to the imposed strain at infinity. This field has no Stokeslet or rotlet contribution, consistent with a force-free, torque-free sphere. Its leading far field is the radial stresslet $-5a^{3}Q\mathbf{x}/(2r^5)$.

Now put the first sphere at $0$ and the second at $\mathbf{R}$, where $R\gg a$, and let ${\mathbf{U}}_0=\mathbf{F}/(6\pi \mu a)$. The first sphere's leading disturbance is its Stokeslet. The rate-of-strain tensor and vorticity incident on the second sphere areThe second sphere translates with the local flow and rotates with half its local vorticity. Its force-free and torque-free conditions therefore remove any reflected force or torque monopole. The incident strain produces the stresslet just obtained. Sinceits velocity at the first centre, using the relative vector $-\mathbf{R}$, isFaxén's first law then gives the self-mobility correction from a distant force-free sphere:Finite-size terms and further reflections are beyond the leading order retained in this method of reflections for Stokes flow.

Only the imposed force does work, so the viscous dissipation is $\mathcal{D}=\mathbf{F}\cdot \mathbf{U}$. With $\zeta =6\pi \mu a$, define ${\epsilon }_R=a/R$ and $\gamma =15a^4/(4R^6)$. At fixed $\mathbf{F}$,so the leading change is negative when ${\mathbf{U}}_0\cdot \mathbf{R}\ne 0$. The direction is held fixed in taking ${\epsilon }_R\to 0$.

The fixed-force comparison of minimum viscous dissipation distinguishes the prescribed force from prescribed boundary velocity. Indeed,and henceAt the actual reduced velocity, adding the second sphere increases the dissipation relative to an isolated translating sphere at that same velocity, exactly as required by the Minimum-dissipation theorem for Stokes flow. One can extend the two-sphere velocity rigidly into the second sphere, adding zero strain and producing an admissible trial field in the domain without it. The reflected angular velocity of the first sphere contributes only a higher-order squared rotational power to this comparison.

Finally hold the second sphere's angular velocity at zero. Faxén's rotational law requires the applied coupleThe resulting rotlet adds at the first centreThis is the additional rotational constraint correction to sphere mobility. Including the strain reflection, the total leading change relative to the isolated velocity isFor motion parallel to the line of centres the constraint adds nothing at this order; for transverse motion it supplies an extra resisting correction.

To hold sphere 2 fixed against the incident angular velocity from part ii, its applied couple must generate the bare rotationThe resulting rotlet advects the force-free sphere 1. Since the field is harmonic, Faxén's first law gives

The returned rotlet has velocity $O(\Omega a^6/R^5)$ and rate of strain $O(\Omega a^6/R^6)$ at sphere 1. That strain induces a stresslet of strength $O(\mu \Omega a^9/R^6)$, whose velocity at sphere 2 is $O(\Omega a^9/R^8)$. This method of reflections for Stokes flow gives the stated order of the next correction to ${\mathbf{U}}_2$.

To the requested order, sphere 1 has velocity $\mathbf{V}$ and creates the translating-sphere field from part a. At the centre of sphere 2, a vector distance $\mathbf{R}$ away, this incident field isSphere 2 is force free, so Faxén's first law givesThe source-dipole term is harmonic, whileThereforeThis is the two-sphere Rotne--Prager mobility through order $a^3/R^3$.

The incident strain at sphere 2 is $O(Va/R^2)$. A force-free sphere in this strain creates a stresslet of size $O(\mu a^{3}Va/R^2)=O(\mu V{a}^4/R^2)$, whose velocity back at sphere 1 is $O(V{a}^4/R^4)$. HenceThe nearly uniform returned flow merely advects sphere 1 and does not change its fixed Stokeslet strength, because its applied force remains fixed. The next scattered disturbance is therefore generated by the returned velocity gradient, of order $V{a}^4/R^5$. It induces a stresslet of size $O(\mu V{a}^7/R^5)$ at sphere 1 and hence velocity $O(V{a}^7/R^7)$ at sphere 2. This method of reflections for Stokes flow explains both the absence of an $O(V{a}^5/R^5)$ term and the next order $O(V{a}^7/R^7)$.

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.