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Use a rotlet as the auxiliary Stokes flow, with a rigid sphere at its center. Its exterior-domain boundary torque is the imposed auxiliary couple. In the Lorentz reciprocal theorem, the actual sphere's zero torque removes the other surface term, while the volume term is the rotlet dotted with the body force. Varying the auxiliary couple gives the displayed angular velocity, assuming the integrals and the far boundary limit converge.

Apply the reciprocal identity with flow1 the actual body-force-driven flow and flow2 the rotlet auxiliary field. Its body force is zero. The large-sphere boundary terms vanish under the usual decay and convergence assumptions for this unbounded body-force problem.

On the sphere, the actual rigid velocity is $\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}$. The auxiliary force is zero and its torque with the outward fluid normal is $\mathbf{G}$. The actual torque with that normal is the negative of the hydrodynamic torque on the sphere, hence zero for a torque-free sphere. Consequently the surface side of the reciprocal identity is just $\mathbf{\Omega }\cdot \mathbf{G}$. The volume side isSince the auxiliary couple is arbitrary,An unknown translation contributes nothing because the auxiliary rotlet has zero net force. This is body-force-driven rotation of a torque-free sphere.

Subtract the ambient velocity and stress, writing ${\mathbf{u}}^{\prime }=\mathbf{u}-{\mathbf{u}}_{\infty }$. This perturbation decays at infinity and on the sphere has velocity $\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}-{\mathbf{u}}_{\infty }$. The ambient flow is regular through the ball and has no body force. Its net torque on the imaginary spherical surface is zero: the divergence theorem converts the torque to the volume integral of the stress divergence plus its antisymmetric part, both zero. Thus the perturbation has zero torque too.

Reciprocity with the auxiliary rotlet givesUsing ${\mathbf{\sigma }}_G{\mathbf{n}}_D=3(\mathbf{G}\times \mathbf{x})/(8\pi a^4)$ yieldsOn this sphere $\mathbf{x}=a\mathbf{n}$, so the vector divergence theorem gives $\int \mathbf{x}\times {\mathbf{u}}_{\infty }dS=a{\int }_{r<a}\nabla \times {\mathbf{u}}_{\infty }dV$. Taking the curl of the homogeneous Stokes equation makes ${\mathbf{\omega }}_{\infty }$ harmonic. Its ball average is its central value: the derivative of each spherical mean is proportional to the volume integral of its Laplacian, hence zero, and its small-radius limit is its value at the center. ThereforeThis proves Faxén rotation law directly from the reciprocal theorem, including why no finite-radius Laplacian correction is needed for a body-force-free ambient flow. Regularity of the ambient field throughout the sphere's prospective interior is part of this argument.

Let $\mathbf{n}=\mathbf{R}/R$. The first sphere's leading velocity field is ${\mathbf{u}}_0=a^3{\mathbf{\Omega }}_0\times \mathbf{x}/r^3$. Taking its curl givesThe second sphere is torque-free, soIts own force-free and torque-free motion emits no Stokeslet or rotlet; further reflections are smaller at large separation. In particular it spins in the opposite direction for a primary rotation perpendicular to the line of centers, and in the same direction with twice that magnitude for a parallel rotation.

Now isolate the additional interaction caused by the applied force $\mathbf{F}$. Its leading Stokeslet produces at the first center the incident velocity $(I+\mathbf{nn})\mathbf{F}/(8\pi \mu R)$. The Stokes drag law makes the increment of the force required to hold that sphere fixedThe sign opposes the incident velocity. This holding force acts on the fluid and generates a reflected Stokeslet. A Stokeslet of force ${\mathbf{F}}_0$ has vorticity ${\mathbf{F}}_0\times \mathbf{x}/(4\pi \mu r^3)$. Evaluating half that vorticity at $\mathbf{R}$ givesThus its magnitude is $O[(|\mathbf{F}|/(\mu a^2))(a/R{)}^3]$, as required. This is forced-sphere rotation reflected from a held sphere. Longitudinal forcing has zero leading spin by symmetry. Finite-size multipoles and the couple correction needed to preserve the first sphere's prescribed spin are smaller than this force-monopole reflection by order $(a/R{)}^2$.

Use the Unscaled Papkovich–Neuber representationwith harmonic vector and scalar potentials. For a torque-driven sphere chooseIts components are harmonic for $r>a$, and both $\mathbf{x}\cdot \mathbf{\Phi }$ and $\nabla \cdot \mathbf{\Phi }$ vanish. HenceThe last equality follows from the no-slip boundary condition at $r=a$. To check that $\mathbf{G}$ is indeed the applied couple, the traction on the sphere, with $\mathbf{n}=\mathbf{x}/a$ directed into the fluid, is ${\mathbf{\sigma }}_G\mathbf{n}=-3\mu {\mathbf{\Omega }}_G\times \mathbf{n}$. Its hydrodynamic torque isHere $\int \mathbf{nn}dS=(4\pi a^2/3)I$. The applied couple balances this resisting torque. The resulting field is the rotlet of a rotating sphere in Stokes flow.

For the Lorentz reciprocal theorem for Stokes flow, let ${\mathbf{n}}_D$ point out of the fluid domain $D$ and take the same viscosity in both flows. With $\nabla \cdot {\mathbf{\sigma }}_i+{\mathbf{f}}_i=0$, the theorem isIt follows by taking the divergence of the boundary integrand: incompressibility removes pressure-work terms, the two Newtonian strain contractions are equal, and the remaining stress divergences give the displayed body-force terms. This convention matters: on the inner boundary of an exterior domain, ${\mathbf{n}}_D=-\mathbf{n}$.

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.

Write $\mathbf{n}=\mathbf{R}/R$. At fixed geometry, Linearity of Stokes flow makes the translational velocities linear in the couples. With zero forces and couples perpendicular to $\mathbf{R}$, isotropy and reflection symmetry allow only terms proportional to ${\mathbf{G}}_j\times \mathbf{n}$ in each ${\mathbf{U}}_i$. There is no longitudinal term: reflect in the plane containing $\mathbf{R}$ and the common torque axis, taking account of the axial-vector transformation of torque, and use linearity under torque reversal. HenceThe direction of $\mathbf{R}$ can still change; constant separation does not mean stationary centers.

When the couples are equal, a rotation through $\pi$ about their common axis exchanges the identical spheres and reverses the in-plane translational velocities while leaving the axial angular velocities unchanged. Uniqueness of Stokes flow then givesThe midpoint is stationary and the pair can orbit it. The body spin need not equal the angular rate of that orbit. These symmetry statements hold at arbitrary noncontact separation, not just in a distant rotlet approximation.

Sphere 2 is now force-free and torque-free. Its translation and spin follow the incident rigid-motion components, so it emits neither a Stokeslet nor a rotlet. Its first disturbance is a stresslet from the incident rate of strain, of magnitude $\mu a^3(a|\mathbf{V}|/R^2)$. Its return velocity at sphere 1 is consequently $O(a^4|\mathbf{V}|/R^4)$:More precisely, that leading correction is $-(15/4){ϵ}^4(\mathbf{V}\cdot \mathbf{n})\mathbf{n}$. It vanishes for purely transverse forcing, for which the first translational correction is of order ${ϵ}^6$.

The first-return spin requires an additional cancellation. The leading incident strain at sphere 2 isIt is axisymmetric about the line of centers, so the reflected stresslet has zero vorticity at sphere 1 on that axis. A naive vorticity estimate of order $(|\mathbf{V}|/a){ϵ}^5$ therefore has zero coefficient. The finite-size correction to the incident strain and the next force-free scattering multipoles are two powers of $ϵ$ smaller; these generically giveThis suppressed spin from a force-free distant sphere is also consistent with the reciprocal torque-to-self-translation coupling. Purely longitudinal forcing gives zero spin exactly by axial symmetry. Distinguishing the stresslet's on-axis curl from its generic off-axis scale is essential here.

A Stokeslet is the flow produced by a point force. With $C=1/(8\pi \mu )$, its tensor is $G_{ij}=C({\delta }_{ij}/r+r_i{r}_j/r^3)$. A separated pair of equal and opposite forces has no force monopole. Expanding their difference in the separation gives a directional derivative of this tensor. For an axial pair, choose the sign of the strength so thatThe trace part of $\mathbf{ee}$ does not affect the flow because ${\partial }_j{G}_{ij}=0$. The remaining symmetric force-dipole contribution is the stresslet, decaying as $r^{-2}$ instead of the Stokeslet's $r^{-1}$. A force-free swimming cell has thrust balanced by drag, so this is typically its leading far field. With $S>0$ it is a pusher microswimmer: axial outflow and transverse inflow. It is distinct from the antisymmetric torque singularity, the rotlet.

The rotlet is the point-torque fundamental solution of Stokes flow. Its torque strength $\mathbf{R}$ is an axial vector, and its circulating velocity decays as $r^{-2}$. A freely swimming cell with no external torque is torque-free. The motor applies opposite internal torques to its body and flagellar apparatus, so the net torque monopole satisfiesIndividual body and flagellar torques need not vanish. A spatially separated pair can therefore leave a rotlet dipole even though there is no far-field rotlet monopole.

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.

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.

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.