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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.

Let $\mathbf{t}=(x_s,y_s)$ be the unit tangent of the inextensible filament. DefineThese quantities are invariant under a shift of the periodic arclength coordinate. Since $x_s^2+y_s^2=1$, the average tangent tensor is $M=\left(\begin{array}{cc}\beta & \delta \\ \delta & 1-\beta \end{array}\right)$. Also $\int \mathbf{t}ds=\lambda {\mathbf{e}}_x$.

For finite-amplitude propulsion of an inextensible periodic filament, distinguish axial phase speed from tangential material speed. The $V$ in the displayed formula is the axial speed of the travelling pattern relative to the mean filament motion. In the wave frame an inextensible material line slides backwards tangentially at the constant speed $Q=V\Lambda /\lambda$: its mean axial sliding is $Q\lambda /\Lambda =V$. If the filament translates at $-U{\mathbf{e}}_x$ in the laboratory, its local fluid-relative velocity isThis relation is the axial and arclength travelling-wave speeds conversion. It ensures the arclength-averaged material velocity is $-U{\mathbf{e}}_x$, rather than confusing wave propagation with material transport.

Integrating the resistive-force theory hydrodynamic force over a period and imposing zero axial force givesSubstitution of $Q\lambda =V\Lambda$ yieldsFor $0<\rho <1$ and $\beta <1$ it is positive: anisotropic drag drives swimming opposite to the wave. A straight waveform has $\beta =1$ and gives no propulsion; isotropic drag $\rho =1$ also gives zero. The denominator $(1-\beta )+\rho \beta$ is positive. If instead $V$ denotes arclength propagation speed, the left-hand ratio acquires the extra factor $\lambda /\Lambda$; the two speeds are not interchangeable.

There is a geometric qualification to the asserted direction. For free swimming strictly along $x$, zero transverse force also requires $\delta =0$, which holds for the usual reflection-symmetric waveforms. The periodicity assumptions alone do not imply that symmetry. A general waveform has cross-resistance of an asymmetric planar waveform and can require a transverse translation. If the mean laboratory velocity is $(-U_x,U_y)$, the full force-free condition isWriting $A_0=1+(\rho -1)\beta$, $B_0=1+(\rho -1)(1-\beta )$ and $\mathcal{D}=A_0{B}_0-(\rho -1{)}^2{\delta }^2$, one obtainsThe printed expression is recovered when $\delta =0$, or as the axial force-balance result if transverse motion is externally constrained. For an explicit permitted asymmetric shape, concatenate tangents $(\sqrt{3}/2,1/2)$ and $(1/2,-\sqrt{3}/2)$ with arclength fractions $p=\sqrt{3}/(1+\sqrt{3})$ and $1-p$. Their mean vertical tangent is zero, but $\delta =\sqrt{3}(2p-1)/4\ne 0$. Repeating the segments gives a periodic graph; the corners can be smoothed without removing the nonzero cross integral. Thus pure opposite-to-wave free swimming also needs a zero cross-resistance assumption.

A freely swimming body has no net external force or torque in Stokes flow. Its force-free condition removes the $r^{-1}$ Stokeslet monopole, and its torque-free condition removes the antisymmetric force-dipole contribution. The leading generic far field is therefore a symmetric traceless rank-two tensor stresslet, decaying as $r^{-2}$. Its axial strength may vanish for special swimmers, in which case higher-order singularities dominate.

The axisymmetric stresslet from a Stokeslet pair provides an explicit construction and fixes the sign convention. Place forces $+F\mathbf{e}$ and $-F\mathbf{e}$ on the fluid at $+(a/2)\mathbf{e}$ and $-(a/2)\mathbf{e}$. The Stokeslet tensor isExpanding their sum for $r\gg a$ givesSincethe dipole strength is $S=Fa$ and the resulting stresslet is the given one. More general surface-traction distributions produce the same leading form through their symmetric traceless first force moment.

Writing $\cos \alpha =\mathbf{e}\cdot \hat{\mathbf{r}}$, its radial velocity isConsequentlyThis classifies pusher microswimmers and puller microswimmers in the force-on-fluid convention used by the given formula.

For the circular swimmer let $\phi =\omega t$, $\mathbf{p}=(\cos \phi ,\sin \phi ,0)$, ${\mathbf{x}}^{\prime }=R\mathbf{p}$, and choose the tangent director $\mathbf{e}=(-\sin \phi ,\cos \phi ,0)$. Its sign is immaterial to the stresslet. Write $\mathbf{r}=\mathbf{x}$ for the observation vector from the circle centre, and assume $r\gg R$. The leading far-field orbit average of a tangent stresslet is obtained by replacing $\mathbf{x}-{\mathbf{x}}^{\prime }$ by $\mathbf{r}$ while retaining the time-dependent director.

The orientation averaging of an axisymmetric stresslet over a full revolution givesSubstitution givesThuswith $\tilde{\mathbf{e}}$ equivalently $-\hat{\mathbf{z}}$. The effective axis is normal to the orbit plane. A positive original strength becomes negative, and a negative strength becomes positive: a pusher averages to a puller, and a puller averages to a pusher, at leading far-field order. The result is independent of $R$ and $\omega$ at this order.

The accuracy of the centre replacement can be made explicit. Half a revolution changes $\mathbf{p}$ and $\mathbf{e}$ to their negatives; the stresslet is even in $\mathbf{e}$. Pairing those times replaces the displaced field byHenceThe finite-radius average is not exactly a point stresslet at every location. For example, on the positive $z$ axis the exact average of the supplied model iswhich has the derived point-stresslet limit for $z\gg R$. This even displacement correction in an orbit-averaged stresslet is consistent with the far-field restriction.

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.

Let $\mu$ be the dynamic viscosity, $u_i={\dot{x}}_i$, and $F_i$ the signed force exerted by sphere $i$ on the fluid. For negligible sphere inertia, this also equals the externally or internally applied force on that sphere; its hydrodynamic force is $-F_i$. The Stokes drag law and the axial velocity of a Stokeslet give the leading hydrodynamic mobility matrixThe factor $1/(4\pi \mu \ell )$ follows from $(I+\hat{\mathbf{r}}\hat{\mathbf{r}})\mathbf{F}/(8\pi \mu \ell )$ when $\mathbf{F}$ is parallel to the line of centres. The longitudinal two-sphere mobility retains the first interaction in $a_i/\ell$; finite-size and repeated-reflection corrections are higher order.

For the linked force-free pair, $F_2=-F_1$. Put $X=(x_1+x_2)/2$ and $S_m=m_1+m_2$. ThensoThe coefficient depends only on the current separation. For any period $T$, $\ell (T)=\ell (0)$, and thereforeThis is a closed integral of a single-valued function of one real shape coordinate. Since $x_1=X-\ell /2$ and $x_2=X+\ell /2$, both spheres also have zero net displacement. Equal radii make $\dot{X}=0$ instantaneously; unequal radii generally permit oscillatory translation, but still no mean motion. This explicit force-free two-sphere stroke is the scallop theorem: a single-parameter reciprocal deformation in a Newtonian fluid at zero inertia cannot produce net free swimming. The timing of extension and contraction cannot change that conclusion, because Stokes flow has no inertial memory.

For the externally prescribed pair, put $q=\omega t$, ${\zeta }_i=1/m_i=6\pi \mu a_i$, andInvert the hydrodynamic mobility matrix:In a large-${\ell }_0$ expansion with $a_i,\delta$ fixed,The isolated-drag term and the constant-$h$ interaction have zero mean. The order-${\ell }_0^{-2}$ term from $h^2{u}_1$ also has zero mean, since its coefficient is constant at that order. Meanwhile,Consequently the externally driven two-sphere pump has the following phase-dependent mean force of an externally driven sphere pair:Here $⟨\cdot {⟩}_t$ denotes averaging over one period. The analogous calculation gives $⟨d{u}_1{⟩}_t={\delta }^2\omega \sin \varphi /2$, so $⟨F_2{⟩}_t=⟨F_1{⟩}_t$ at leading order, andThus the mean fluid forcing is nonzero except at $\varphi =0$ or $\pi$ modulo $2\pi$. The sign changes when the phase lag is reversed. Those exceptional strokes are reciprocal: the position-space loop collapses to a line, and their mean force vanishes by reversibility, not just by this leading expansion.

There is no contradiction with the scallop theorem. The externally imposed motion is not force-free, and the actuators prescribe two phase-shifted motions; their combined motion is not reciprocal for a generic phase. The spheres return to their prescribed positions while transferring a mean force to the fluid. This is pumping by external forcing, rather than propulsion of the freely linked one-shape-coordinate system.

The asymptotic calculation requires $\delta /{\ell }_0\ll 1$ and persistent large separation. The minimum separation of phase-shifted sphere oscillations isThe printed condition ${\ell }_0\ge \delta$ alone does not guarantee nonoverlap or the stipulated far-separated regime. The prescribed trajectories must additionally keep this minimum well above both radii. This is a compatibility qualification on the data, not a change of the phase-dependent force calculation.

In resistive-force theory (RFT), the hydrodynamic force per unit arclength exerted by the fluid on a slender filament isHere $\mathbf{t}$ is the unit tangent and $\mathbf{u}$ the local velocity relative to the background fluid. The parallel and perpendicular drag coefficients of a slender filament are positive and generally satisfy ${\xi }_{\perp }>{\xi }_{\parallel }$; their leading logarithmic ratio is about two. This local approximation assumes a small filament radius compared with length and curvature scales, negligible fluid inertia, and a Newtonian fluid. It represents drag by the local tangent direction and neglects nonlocal hydrodynamic interactions between separated filament segments. Boundaries, close approaches and end corrections may require slender-body theory or a more complete flow calculation. In this problem the background fluid is at rest, and the drag coefficients are taken uniform along the filament.

The rigid-body velocity in a deforming swimmer frame is the sum of translation, rotation and material deformation. With $\mathbf{r}=x{\mathbf{e}}_x+ϵg{\mathbf{e}}_y$,so, in instantaneous swimmer-frame components,The reference-frame conditions $g(0,t)=g_x(0,t)=0$ attach the origin and orientation to the filament's base and base tangent. They prevent arbitrary shape translations or tilts from being absorbed into the definition of $g$.

The exact tangent and arclength element areSince $U,V,\Omega =O(ϵ)$, the local velocity is $O(ϵ)$. Changing the drag tensor by its $O(ϵ)$ tangent correction therefore changes $\mathbf{f}$ only at $O({ϵ}^2)$. The arclength correction is smaller still at this order. Also $-ϵ\Omega g=O({ϵ}^2)$. Thus $\mathbf{t}={\mathbf{e}}_x$ is sufficient at order $ϵ$. This assumes the small-slope expansion is uniform, $ϵ|g_x|\ll 1$.

Put $G_0(t)=⟨g_t{⟩}_x$ and $G_1(t)=⟨x{g}_t{⟩}_x$, with $⟨h{⟩}_x=L^{-1}{\int }_0^{L}hdx$. The first-order local force isIts total hydrodynamic forces areTaking the moment about the swimmer-frame origin, $M_z=\int (x{f}_y-ϵg{f}_x)ds$, the second term is beyond first order. HenceFor force-free and torque-free motion, the three leading coefficients vanish. The longitudinal force immediately gives $U_1=0$. The two transverse equations areTheir determinant is $L^2/12>0$. Solving yields the first-order free swimming of a planar filament:The resistive-force theory coefficient cancels because both equations use the same transverse drag. Another interpretation is the least-squares projection of filament deformation velocity: $-V_1-{\Omega }_{1}x$ is the best affine approximation to $g_t$ on $[0,L]$. Zero transverse force and torque mean that the residual is orthogonal to $1$ and $x$.

If $g$ is sufficiently differentiable and periodic with period $T$, each $g_t(x,t)$ has zero temporal mean. More explicitly,Both bracketed quantities return to their initial values over a period. Thus the first-order periodic transverse swimming velocity satisfiesThese results concern first order and swimmer-frame components. Changes of orientation can affect laboratory displacements at second order; higher-order net swimming is not ruled out.

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.

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.

Take $\mathbf{F},\mathbf{G}$ to be the force and couple exerted by the body on the fluid, equivalently the external force and couple needed to maintain its motion. This fixes the sign for a positive hydrodynamic resistance matrix. If one uses the fluid's force on the body, both resultants have the opposite sign.

For two Stokes flows $(\mathbf{u},\mathbf{\sigma })$ and $(\tilde{\mathbf{u}},\tilde{\mathbf{\sigma }})$ in the same fluid domain $\mathcal{D}$, with zero body force, the Lorentz reciprocal theorem for Stokes flow isHere $\mathbf{n}$ points out of the fluid. To see the identity, take the divergence of the difference of the two cross-work fluxes. The stress divergences vanish, while incompressibility and symmetry of the Newtonian fluid stress tensor reduce the remaining terms to $2\mu (\mathbf{e}:\tilde{\mathbf{e}}-\tilde{\mathbf{e}}:\mathbf{e})=0$. The divergence theorem proves the result.

On the body, the no-slip boundary condition is $\mathbf{u}=\mathbf{U}+\mathbf{\Omega }\times \mathbf{r}$, and the far-field contribution vanishes for decaying exterior flows. Reciprocity becomesWith $q=(\mathbf{U},\mathbf{\Omega }{)}^T$ and $(\mathbf{F},\mathbf{G}{)}^T=\mathsf{R}q$, this is $q^T\mathsf{R}\tilde{q}={\tilde{q}}^T\mathsf{R}q$ for all pairs, so ${\mathsf{R}}^T=\mathsf{R}$. The power identity givesEquality would force zero strain throughout the connected exterior domain, hence a rigid fluid motion. Decay at infinity makes that motion zero, and the no-slip boundary condition then gives $q=0$. This proves the symmetry and positivity of a rigid-body resistance matrix:

For the helix, assume $0<\varphi <\pi /2$ and use cylindrical unit vectors. Its arclength element and tangent areIntegrating to $\theta =L\cos \varphi /b$ gives $\boxed{\text{total arc length}=L}$. In a combined axial translation and rotation, $\mathbf{V}=U{\mathbf{e}}_z+b\Omega {\mathbf{e}}_{\theta }$. The slender-body force density, or local resistive-force theory, givesWriting $s_{\varphi }=\sin \varphi$, $c_{\varphi }=\cos \varphi$, its axial and azimuthal components areThe axial couple density is $b{f}_{\theta }$. Integration over arclength gives the axial resistance matrix of a slender helix:Thus pure translation gives $(F_z,G_z)=(AU,BU)$, and pure rotation gives $(F_z,G_z)=(B\Omega ,D\Omega )$. The mixed coefficients agree, as reciprocity requires, and $AD-B^2=C^2{L}^2{b}^2/2>0$. The sign of $B$ follows the handedness in the parametrization; reversing handedness reverses propulsion.