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At fixed outer boundary velocity, extending the inclusion velocity rigidly through its interior makes the particle flow an admissible competitor for the particle-free Minimum-dissipation theorem for Stokes flow. Its interior rate-of-strain tensor vanishes. For a linear background, the increase in viscous dissipation is force times relative translation, torque times relative rotation, plus the contraction of the particle stresslet tensor with background strain.

The Minimum-dissipation theorem for Stokes flow compares fields with fixed boundary velocities. Introducing a freely moving rigid inclusion can reduce velocity and total power at fixed applied force while increasing the resistance at fixed velocity. Extending the inclusion's rigid velocity through its interior adds zero viscous dissipation and gives a trial field in the domain without that inclusion, proving the fixed-velocity comparison.

At fixed sphere rotation in an annular Stokes flow, add rigid particles with zero net force and torque. Extend their actual rigid velocities through their interiors. The extension is an admissible incompressible competitor in the particle-free annulus, with zero interior rate-of-strain tensor. The Minimum-dissipation theorem for Stokes flow gives $D\ge D_0$. The boundary-work identity gives $D=\mathbf{\Omega }\cdot \mathbf{G}$, because the stationary outer wall and the freely moving particles contribute no power. Equality would require the extension to equal the unique particle-free flow, whose strain cannot vanish on a finite particle volume when $\Omega \ne 0$. Thus the resistance component along the imposed rotation strictly increases.

Reflection of a uniform tube in a horizontal plane preserves an off-axis sphere's transverse position but reverses the axial force. By linearity, reversing the force reverses every translational velocity. By reflection symmetry, it preserves the transverse velocity. Thus the transverse velocity is zero. The Minimum-dissipation theorem for Stokes flow does not predict motion between different particle positions: it compares velocity fields in one fixed geometry with prescribed boundary velocities. Inertia, deformability or broken geometric symmetry can invalidate this no-migration conclusion.

Put $r=|\mathbf{x}|$. In the Unscaled Papkovich–Neuber representation, choose the componentwise harmonic vector fieldBoth $\mathbf{\Omega }\times \mathbf{x}$ and $(\mathbf{\Omega }\times \mathbf{x})/r^3$ are harmonic away from the origin: the latter is a linear combination of derivatives of the harmonic function $1/r$. Also $\mathbf{x}\cdot \mathbf{\Phi }=0$, so the representation gives $\mathbf{u}=f(r)\mathbf{\Omega }\times \mathbf{x}$. The no-slip boundary condition requires $f(a)=1$ and $f(b)=0$, givingThus the Rotational Stokes flow between concentric spheres isThe divergence of $\mathbf{\Phi }$ is zero: the cross-product field itself has zero divergence, and $\nabla f$ is radial and perpendicular to $\mathbf{\Omega }\times \mathbf{x}$. Hence $p=2\mu \nabla \cdot \mathbf{\Phi }=0$ in this pressure gauge.

There is also a symmetry argument. Take the rotation axis as the polar axis. Linearity of Stokes flow makes velocity and pressure perturbations change sign when the rotation reverses. Reflection in any plane containing that axis reverses the axial rotation vector. Combined with axial symmetry, this excludes meridional velocity components and forces the pressure perturbation to be both unchanged and sign-reversed by that reflection. Equivalently a purely azimuthal axisymmetric Stokes flow has no meridional component of its vector Laplacian, so its meridional pressure gradient vanishes. Thus the pressure is constant; setting that arbitrary constant to zero gives the same result without solving the radial coefficients.

For the Newtonian fluid stress tensor, let $\mathbf{q}=\mathbf{\Omega }\times \mathbf{x}$. The symmetric part of the velocity gradient isbecause the gradient of rigid rotation is antisymmetric. Consequently the full stress tensor, with $p=0$, isFor $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$, its only nonzero spherical-coordinate components are ${\sigma }_{r\varphi }={\sigma }_{\varphi r}=-3\mu B\Omega \sin \vartheta /r^3$. On the inner sphere the traction exerted by the fluid, using the sphere's outward normal $\mathbf{n}$, isIts torque isusing $\int n_i{n}_{j}d\omega =4\pi {\delta }_{ij}/3$. Therefore the applied Torque in rotational Stokes flow between concentric spheres isFor $b\gg a$, the relevant speed and length are $\Omega a$ and $a$, so negligible inertia requires the rotational Reynolds numberFor a narrow gap $d=b-a\ll a$, the viscous gradients are controlled by $d$, not $a$. A centrifugal acceleration of scale ${\Omega }^{2}a$ is then compared with viscous scale $\mu \Omega a/d^2$, rather than $\mu \Omega /a$. Their ratio is $\rho |\Omega |d^2/\mu$. This explains why the large-gap condition is not a necessary narrow-gap criterion; a thin gap greatly strengthens viscous resistance. This base-flow scaling is distinct from a criterion for stability to arbitrary short-scale disturbances.

The Minimum-dissipation theorem for Stokes flow compares a Stokes solution with all sufficiently regular incompressible trial velocities in the same fluid domain that have the same prescribed boundary velocities. With no body-force work, the dissipation functional isIf $\mathbf{v}=\mathbf{u}+\mathbf{w}$ and $\mathbf{w}=0$ on the prescribed boundary, integration by parts with $\nabla \cdot \mathbf{\sigma }(\mathbf{u})=0$ and $\nabla \cdot \mathbf{w}=0$ makes the cross term vanish. ThusThe comparison fixes velocities, rather than applied torques or forces.

Now let the particles take their actual rigid translations and rotations in the particle-containing flow. Extend the fluid velocity through each particle by its rigid velocity. No-slip boundary conditions make this extension continuous; rigid motion has zero rate-of-strain tensor and zero divergence. The extension is therefore an admissible trial field in the original particle-free annulus, and its dissipation equals that of the fluid outside the particles. Hence $D\ge D_0$.

The Stokes boundary-work identity gives $D=\mathbf{G}\cdot \mathbf{\Omega }$. The outer sphere is stationary, the maintained inner sphere has zero translational velocity, and every added particle has zero total force and torque, so none of those other boundaries contributes power. Likewise $D_0={\mathbf{G}}_0\cdot \mathbf{\Omega }$. ThereforeFor nonzero rotation and a particle of positive volume the inequality is strict: equality in the dissipation identity would force the extension to be the unique original flow, but that original flow has nonzero strain on every open particle region, whereas a rigid interior has zero strain. This proves that force-free inclusions increase rotational resistance:No dilute-suspension approximation or assumption about the particle shapes was used.

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.

For a fixed fluid domain with prescribed velocity on its whole boundary, the Minimum-dissipation theorem for Stokes flow compares the actual incompressible Stokes flow $u$ with every admissible incompressible velocity field $v$ having the same boundary values. The trial fields need not themselves satisfy the momentum equation. With $e(u)=(\nabla u+\nabla u^T)/2$ and constant dynamic viscosity $\mu$, define the viscous dissipationAssume no nonconservative body force, or absorb a conservative force into the pressure. Set $w=v-u$, so $\nabla \cdot w=0$ and $w=0$ on the boundary. The Stokes stress tensor $\sigma =-pI+2\mu e(u)$ obeys $\nabla \cdot \sigma =0$. Because pressure has zero contraction with $e(w)$,Expanding the square provesEquality requires $e(w)=0$, so $w$ is a rigid motion; the homogeneous boundary values force it to vanish. The theorem concerns fixed boundary velocities, not fixed applied forces or arbitrary comparisons between different fluid domains.

For the concentric-sphere motion, seek $u=F(r)\mathbf{\Omega }\times x$. Incompressibility is automatic. The choice $F=A+B/r^3$ gives a harmonic velocity: the linear term is harmonic, and each component of $x/r^3$ is a derivative of the harmonic function $1/r$. The no-slip conditions $F(a)=1$, $F(b)=0$ giveThus the Rotational Stokes flow between concentric spheres isIt satisfies the Stokes equation with constant pressure, and the uniqueness implied by the theorem identifies it as the required flow.

In the Unscaled Papkovich–Neuber representation, take $\Phi =-u/2$ and $\chi =0$. Both potentials are harmonic and $x\cdot \Phi =0$, so the representation gives $u=-2\Phi$. Alsoand therefore $p=0$, after choosing the arbitrary pressure constant.

This pressure conclusion also follows from symmetry and Linearity of Stokes flow. Pressure is a scalar linear in the axial vector $\mathbf{\Omega }$, and isotropy leaves only a possible term proportional to $\mathbf{\Omega }\cdot x$. That is a pseudoscalar, forbidden by reflection symmetry of the concentric spherical domain. Hence the pressure perturbation vanishes. Equivalently, axisymmetry and reflection with flow reversal make the motion purely azimuthal, while its meridional momentum balance has no pressure gradient.

Let $v=\mathbf{\Omega }\times x$ in the following stress calculation. The symmetric velocity gradient isThe full stress tensor, in the zero-pressure gauge, is thereforeFor $n=x/r$, the traction on a radial surface is $\sigma n=-3\mu B(\mathbf{\Omega }\times n)/r^3$. This completely specifies the stress field; with the polar axis along $\mathbf{\Omega }$, its only nonzero spherical shear component is ${\sigma }_{r\varphi }=-3\mu B\Omega \sin \theta /r^3$.

The fluid couple on the inner sphere is the integral of $x\times \sigma n$ with the normal pointing from the solid into the fluid. Using $\int n_i{n}_{j}d\omega =(4\pi /3){\delta }_{ij}$,Hence the sustaining applied Torque in rotational Stokes flow between concentric spheres isFor $a\ll b$, it approaches the isolated rotating sphere in Stokes flow result $8\pi \mu a^3\mathbf{\Omega }$. For a narrow gap $d=b-a\ll a$, it becomes $8\pi \mu a^4\mathbf{\Omega }/(3d)$: the divergent resistance is that of a thin Couette shear layer, with speed scale $a\Omega$ and gradient scale $a\Omega /d$.

Now add the force-free, couple-free rigid particles and denote the resulting velocity by $u_p$. Extend $u_p$ through each particle as its rigid translational and rotational velocity. No slip makes this extension continuous across every particle surface; it is incompressible and has zero rate-of-strain tensor inside each particle. It therefore gives an admissible trial velocity in the original particle-free annulus, with exactly the same sphere boundary velocities. Its full-annulus dissipation is the actual fluid dissipation ${\mathcal{D}}_p$, since the added interior integrals are zero. The Minimum-dissipation theorem for Stokes flow consequently gives ${\mathcal{D}}_p\ge {\mathcal{D}}_0$.

The boundary-work identity follows by integrating $\nabla \cdot (\sigma u)$: dissipation equals the mechanical power supplied at all rigid boundaries. The outer sphere is stationary; the inner sphere has no translation, so any force needed to hold its centre does no work; and each added particle contributes zero power because its net force and couple are zero. Thus ${\mathcal{D}}_p={\mathbf{G}}_p\cdot \mathbf{\Omega }$, while ${\mathcal{D}}_0={\mathbf{G}}_0\cdot \mathbf{\Omega }$. ThereforeFor $\mathbf{\Omega }\ne 0$ and at least one particle of positive volume in the open annulus, the inequality is strict. Equality would make the extended velocity identical to the particle-free solution, yet that solution has nonzero strain on every open region away from the rotation axis and cannot be rigid throughout any particle interior. Hence the applied couple component along the imposed angular velocity increases. This is the fixed-velocity extra dissipation due to a rigid inclusion argument; it does not assert an increase in every transverse torque component.

The Minimum-dissipation theorem for Stokes flow compares fields with fixed geometry and boundary velocities. Moving the sphere changes the geometry, while sedimentation prescribes its effective weight rather than its velocity. It therefore supplies no rule for migrating toward a position of least dissipation. Indeed, at fixed force $W$, the power is $WV$, and a lower drag resistance gives a larger settling speed and larger power, rather than a smaller power. This is a fixed-force comparison of minimum viscous dissipation issue.

The proposed migration is itself false in ideal Stokes flow. Reflect the uniform vertical tube in a horizontal plane through the sphere centre. The sphere's lateral position is unchanged, but the applied axial force reverses. Reflection leaves a transverse translational velocity unchanged. By Linearity of Stokes flow, however, reversing the force reverses that transverse velocity. It must therefore be zero. This proves no lateral migration of a single sphere in a uniform tube in Stokes flow:Rotation and a wall-dependent vertical speed are allowed. Inertial lift, particle deformation or tube-end effects belong to different problems and do not follow from the dissipation theorem.

For a fixed fluid domain, consider an incompressible Newtonian Stokes flow with prescribed velocities on its boundaries and prescribed far-field behaviour. Conservative body forces, such as gravity, may be absorbed into the pressure. Among all sufficiently regular divergence-free trial velocity fields with exactly those boundary velocities and far-field behaviour, the Stokes solution minimizes the viscous dissipationTo see this Minimum-dissipation theorem for Stokes flow, write a trial field as $\mathbf{v}=\mathbf{u}+\mathbf{w}$, with $\nabla \cdot \mathbf{w}=0$ and homogeneous velocity boundary data. The Newtonian fluid stress tensor satisfies $\nabla \cdot \mathbf{\sigma }=0$, so integration by parts givesThereforeEquality requires the difference to be a rigid motion, eliminated by the fixed wall and far-field data. The theorem varies the velocity field in one fixed geometry; it does not minimize over particle positions. If a nonconservative body force is prescribed, the appropriate variational functional also includes its work, rather than being the dissipation alone.

Let downward velocity be positive in this part, and let $W>0$ be the dense sphere's weight minus buoyancy. For the original centreline configuration, write its six-component translation-and-rotation velocity as $q$, and its diagonal positive hydrodynamic resistance matrix as $R_0$. Its single-sphere dissipation isWith only axial force and no torque, the original velocity is $V_0=W/R_{zz}$.

Now add the neutral sphere. In its actual Stokes flow, extend the velocity rigidly through the new sphere's interior. This extension is a valid incompressible trial field for the domain without that sphere, with exactly the dense sphere's actual six boundary velocities; its strain and dissipation inside the inserted ball are zero. The Minimum-dissipation theorem for Stokes flow and extra dissipation due to a rigid inclusion therefore giveThe neutral sphere has zero net external force and torque, and the dense sphere has only the axial force $W$. The dissipation equals total external power, so $D=W{V}_z$. Positivity gives $V_z>0$, and the comparison yields $V_z\le W/R_{zz}$ without assuming that either sphere translates only vertically or that either angular velocity vanishes.

For a finite nonoverlapping placement, the first inequality is strict. Equality would make the extended field the old Stokes solution and thus make that solution rigid throughout the added ball. Analyticity and connectedness then force its strain to vanish throughout the original fluid domain; the stationary tube wall and quiescent far field would force the whole motion to vanish, contradicting $V_z>0$. Hence the force-free inclusion reduces axial mobility of a centred settling sphere:The difference tends to zero as the added sphere recedes infinitely far away. The lateral and rotational contributions are nonnegative in the resistance comparison, which is precisely why they cannot invalidate the bound.

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.

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.

Let $\mathbf{u}$ be the Stokes flow and let $\mathbf{v}$ be any other admissible incompressible flow with the same no-slip boundary condition on the sphere and the same decay at infinity. PutThen $\nabla \cdot \mathbf{w}=0$ and $\mathbf{w}=0$ on the sphere. The viscous dissipation satisfiesFor the Stokes stressincompressibility of $\mathbf{w}$ givesThe divergence theorem, the homogeneous boundary data, decay at infinity, and $\nabla \cdot \mathbf{\sigma }=0$ giveThereforeThis proves the Minimum-dissipation theorem for Stokes flow.