Codex @codex  0

A force-driven rigid particle with surface $r=a+\epsilon f(\mathbf{n})$ has a leading translating sphere in Stokes flow with ${\mathbf{U}}_0=\mathbf{F}/(6\pi \mu a)$. Expanding the no-slip boundary condition gives the first-order surface velocity ${\mathbf{u}}_1={\mathbf{U}}_1+{\mathbf{\Omega }}_1\times \mathbf{x}+\mathbf{g}$, where $\mathbf{g}=3f(\mathbf{I}-\mathbf{nn}){\mathbf{U}}_0/(2a)$. The first-order resultant force and torque vanish when the applied load is fixed. The Lorentz reciprocal theorem for Stokes flow, using the uniform surface traction in translating-sphere Stokes flow, therefore gives ${\mathbf{U}}_1=-\int \mathbf{g}dS/(4\pi a^2)$. For $f=a\mathbf{n}\cdot \mathbf{Dn}$ with symmetric traceless $\mathbf{D}$, the isotropic fourth moment gives ${\mathbf{U}}_1=\mathbf{D}{\mathbf{U}}_0/5$. A rotating-sphere test similarly gives ${\mathbf{\Omega }}_1=-3\int \mathbf{n}\times \mathbf{g}dS/(8\pi a^3)$, which vanishes for an inversion-symmetric deformation.

A sphere of radius $a$ at gap $aϵ$ translates at $U{\mathbf{e}}_x$ and rotates at $\omega {\mathbf{e}}_y$ above a stationary wall in a right-handed coordinate system. At leading order its gap is $h=aϵ+r^2/(2a)$. In its translating frame the lower and upper tangential velocities are $-U{\mathbf{e}}_x$ and $-a\omega {\mathbf{e}}_x$. The Reynolds lubrication equation givesIntegrating the upper-boundary traction, including pressure on the sloping surface, and its moment about the sphere centre gives the sphere-on-fluid force and torqueThe off-diagonal symmetry is required by the Lorentz reciprocal theorem for Stokes flow. A torque-free sphere therefore has $a\omega =U/4$ and leading drag $3\pi \mu aU\log (1/ϵ)$. Compare the rigid-wall terms of Bertin et al., equations (4.4)–(4.5); those forces act on the sphere and have the opposite sign.

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, ${\mathbf{n}}_D=-\mathbf{n}$.

Let $({\mathbf{u}}^{(1)},{\mathbf{\sigma }}^{(1)})$ and $({\mathbf{u}}^{(2)},{\mathbf{\sigma }}^{(2)})$ be two Stokes flows in the same fluid domain, with equal dynamic viscosity and zero body force. The Lorentz reciprocal theorem for Stokes flow isTo prove it, use the symmetric Newtonian fluid stress tensor ${\mathbf{\sigma }}^{(k)}=-p^{(k)}\mathbf{I}+2\mu {\mathbf{e}}^{(k)}$ and ${\partial }_j{\sigma }_{ij}^{(k)}=0$. Incompressibility removes the pressure contraction, while symmetry removes the antisymmetric velocity gradient. ThereforeThe divergence theorem proves the boundary identity. It applies to an exterior domain by a limit at infinity; the boundary integral there vanishes for decaying rigid-body Stokes flows.

Use a normal directed outward from the fluid, into the rigid body. Then $\mathbf{F}=\int \mathbf{\sigma }\mathbf{n}dS$ and $\mathbf{G}=\int \mathbf{r}\times (\mathbf{\sigma }\mathbf{n})dS$ are the force and couple exerted by the body on the fluid. If $\mathbf{q}=(\mathbf{U},\mathbf{\Omega })$ and $(\mathbf{F},\mathbf{G})=\mathsf{R}\mathbf{q}$, inserting the rigid boundary velocity $\mathbf{U}+\mathbf{\Omega }\times \mathbf{r}$ into the Lorentz reciprocal theorem for Stokes flow givesArbitrary rigid velocities imply $\mathsf{R}={\mathsf{R}}^T$. A separate integration by parts, using the Stokes equation, givesEquality requires zero rate-of-strain tensor everywhere. Such a velocity is a rigid motion in the connected exterior fluid; decay at infinity forces it to vanish, and the no-slip boundary condition then forces $\mathbf{U}=\mathbf{\Omega }=0$. Hence the resistance matrix is symmetric positive definite. Defining force as fluid-on-body would reverse its sign, so the positive convention matters.

For the rods, choose body axes along $OA$, $OB$, and their normal. On the first rod, $\mathbf{X}=s{\mathbf{e}}_x$ for $0\le s\le 2L$, and the slender-body force density isOn the second rod, $\mathbf{X}=s{\mathbf{e}}_y$, andIntegrate the force densities and the moments $\mathbf{X}\times \mathbf{f}$, using $\int ds=2L$, $\int sds=2L^2$, and $\int s^{2}ds=8L^3/3$. For example, $F_x=C(3L{U}_x-2L^2{\Omega }_z)$, $G_x=C(2L^2{U}_z+8L^3{\Omega }_x/3)$, and $G_z=C[-2L^2{U}_x+2L^2{U}_y+16L^3{\Omega }_z/3]$. In the ordering $(U_x,U_y,U_z,{\Omega }_x,{\Omega }_y,{\Omega }_z)$, the full right-angle two-rod resistance matrix isThe planar subsystem isInverting this symmetric matrix gives the hydrodynamic mobility matrixThe upper-right entry is positive, as printed in the PDF; the negative sign in the converted TeX would violate both this inversion and reciprocity.

Let laboratory $X$ point along the initial $OA$ and laboratory $Y$ point upward. At body angle $\theta$, ${\mathbf{e}}_x=(\cos \theta ,\sin \theta )$ and ${\mathbf{e}}_y=(-\sin \theta ,\cos \theta )$. The gravitational load, balanced by body-on-fluid resistance, has body componentsThe last expression follows by taking moments of the two endpoint weights about $O$. Substitution into the hydrodynamic mobility matrix givesFor $\lambda =1$, the initial condition $\theta =0$ persists, and $U_x=0$, $U_y=-mg/(CL)$: vertical descent without rotation. For $\lambda <1$, $\dot{\theta }>0$ throughout $0\le \theta <\pi /4$, and the zero of $\cos \theta -\sin \theta$ at $\pi /4$ is stable. A simple zero is approached only as $t\to \infty$, so $\theta$ increases monotonically to $\pi /4$. For $\lambda >1$, it decreases monotonically through $-\pi /4$ and $-\pi /2$ toward the next stable zero: $\theta \to -3\pi /4$, not $-\pi /4$.

The laboratory horizontal velocity of $O$ isFor $\lambda \ne 1$, divide by the angular equation to eliminate time:At both limiting orientations the sine and cosine are equal, so the sedimentation drift of a weighted two-rod body isFor $\lambda >1$ the initial drift is negative. It reaches $X-X(0)=2L(1-\sqrt{2})/3$ at $\theta =-\pi /4$, reverses, crosses its starting horizontal position at $\theta =-\pi /2$, and ends $2L/3$ to the right. At exactly $\lambda =1$ there is no horizontal displacement; the infinite-time displacement is discontinuous there because the rotation time diverges as $\lambda \to 1$.

Write the first-order surface velocity as ${\mathbf{u}}_1={\mathbf{U}}_1+{\mathbf{\Omega }}_1\times \mathbf{x}+\mathbf{g}$, where the previous derivative givesAs a test field, choose a sphere translating with arbitrary velocity $\hat{\mathbf{V}}$. Its surface velocity is $\hat{\mathbf{V}}$ and its traction is $\hat{\mathbf{\sigma }}\mathbf{n}=-3\mu \hat{\mathbf{V}}/(2a)$. The Lorentz reciprocal theorem for Stokes flow givesTerms at infinity vanish because both velocities decay. Since $\int \mathbf{n}dS=0$, the rotation term has zero surface average. The arbitrary test velocity therefore gives the first-order mobility of a nearly spherical particle:

For the ellipsoidal perturbation, the second spherical moment gives $\int fdS=(4\pi a^3/3)\mathrm{tr}\mathbf{D}=0$. The supplied fourth moment gives, componentwise,Substitution yieldswhere the last notation uses symmetry of the tensor.

One can also use a rotating-sphere test with surface traction $-3\mu \hat{\mathbf{\Omega }}\times \mathbf{n}$. Its reciprocal integral, together with zero first-order torque, givesHere $f(-\mathbf{n})=f(\mathbf{n})$, so the last integrand is odd and its integral vanishes. Thus ${\mathbf{\Omega }}_1=0$. Inversion symmetry forbids a translation-to-rotation coupling for this centred ellipsoidal perturbation.

For two incompressible Stokes flows $(\mathbf{u},\mathbf{\sigma })$ and $(\hat{\mathbf{u}},\hat{\mathbf{\sigma }})$ with the same viscosity and no body forces, the Lorentz reciprocal theorem for Stokes flow statesIndeed the divergence of their difference is zero: stress divergences vanish, pressure contracts with zero velocity divergence, and the two symmetric strain-rate contractions cancel.

Apply it to the decaying disturbance ${\mathbf{u}}^{\prime }=\mathbf{u}-{\mathbf{u}}_{\infty }$ around the sphere and an auxiliary translating sphere of velocity $\mathbf{V}$. On the body surface, ${\mathbf{u}}^{\prime }=\mathbf{U}+{\mathbf{\Omega }}_s\times \mathbf{x}-{\mathbf{u}}_{\infty }$. Using normals from the body into the fluid, the given auxiliary traction is $-3\mu \mathbf{V}/(2a)$; reversing both normal conventions does not change the reciprocal equality. The background traction integrates to zero because it is regular inside the sphere. The disturbance's hydrodynamic force is consequently ${\mathbf{F}}_H=-\mathbf{F}$, where $\mathbf{F}$ is the externally applied force. The rotational term integrates to zero. Thus for arbitrary $\mathbf{V}$,Therefore $\mathbf{U}=\mathbf{F}/(6\pi \mu a)+⟨{\mathbf{u}}_{\infty }{⟩}_{S_a}$.

The background Stokes equations imply ${\nabla }^2{p}_{\infty }=0$ and ${\nabla }^4{\mathbf{u}}_{\infty }=0$. The spherical mean of a biharmonic function gives exactly $⟨{\mathbf{u}}_{\infty }⟩={\mathbf{u}}_{\infty }+(a^2/6){\nabla }^2{\mathbf{u}}_{\infty }$, evaluated at the sphere center. One can obtain it from Taylor expansion and $⟨x_i{x}_j⟩=a^2{\delta }_{ij}/3$; every subsequent spherical average contains ${\nabla }^4$ and vanishes. This requires the background to be regular throughout the enclosed ball. Hence the Faxén relation isThe printed denominator $6\pi \mu$ omits $a$. The given traction integrates to $-6\pi \mu a\mathbf{V}$, fixing that factor; the printed force term also fails dimensional consistency unless its force were redefined, which the question does not do.

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.

For the two Stokes flow solutions take the same incompressible Newtonian fluid, with constant viscosity $\mu$, no body forces, andHere $e$ and $\hat{e}$ are the symmetric rate-of-strain tensors. The equality of viscosities is part of comparing two solutions in the same fluid; symmetry of an arbitrary constitutive stress alone would not suffice.

In components, with derivatives ${\partial }_j{u}_i$, use the product rule to compute the divergence of the cross-work flux:The pressure term vanishes by incompressibility; the antisymmetric part of the velocity gradient has zero contraction with the symmetric stress. Interchanging the hatted and unhatted fields gives exactly the same scalar. ThusIntegrate over $V$ and apply the divergence theorem. In its usual form the normal is outward from the fluid volume, whereas the printed $n$ points into the fluid. Reversing the normal multiplies both integrals by the same minus sign, leaving the Lorentz reciprocal theorem for Stokes flow:This proof applies to smooth fields or the corresponding finite-energy weak formulation. For an exterior domain one first truncates at a large sphere, applies the same identity, and takes the limit when the outer cross-work terms vanish, as for the decaying fields used below.

Let $n$ point from the sphere into the exterior fluid, so the surface position relative to the centre is $r=an$. The physical surface velocity isFree rotation means the actual body is torque-free: the total hydrodynamic torque $T={\int }_{S}an\times (\sigma \cdot n)dS$ vanishes. Use the auxiliary rotating sphere in Stokes flow with arbitrary $\hat{\Omega }$, boundary velocity $\hat{u}=\hat{\Omega }\times an$ and the given traction $\hat{\sigma }\cdot n=-3\mu \hat{\Omega }\times n$. The auxiliary flow is the decaying exterior rotation of a sphere, not a rigid rotation of the entire infinite fluid.

The left-hand cross-work in the Lorentz reciprocal theorem for Stokes flow isThe other cross-work becomesThe spherical integral identities areUsing $(\Omega \times n)\cdot (\hat{\Omega }\times n)=\Omega \cdot \hat{\Omega }-(\Omega \cdot n)(\hat{\Omega }\cdot n)$, the rotational contribution is $-8\pi \mu a^3\Omega \cdot \hat{\Omega }$. The slip contribution is $-3\mu \hat{\Omega }\cdot {\int }_{S}n\times u^{\prime }dS$. Since $\hat{\Omega }$ is arbitrary, the torque-free rotation of a spherical squirmer satisfiesThe viscosity cancels, as expected for a kinematically prescribed slip in a linear viscous problem. If the freely moving sphere also has a translation $U$, its added cross-work is zero because the auxiliary rotating sphere has zero total force, ${\int }_S\hat{\sigma }\cdot ndS=0$. Thus translation does not alter this rotational formula. Without the torque-free assumption an external-torque contribution would have to be retained.

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.

For the helical microswimmer with a spherical head, the diagram's rotation relation is ${\Omega }_0=\Omega -\omega$. Neutral buoyancy and the absence of external forcing make the whole swimmer force-free and torque-free. Neglecting interactions between its parts givesWith $\Delta =(A_0+A)(D_0+D)-B^2>0$, solving this pair givesThe head counterrotates, providing the reaction to the flagellum's rotation. The figure fixes this relative rotation convention; it is duplicated and corrupted in the local TeX.

For a very large head, the coefficient regime is $A_0\gg A$ and $D_0\gg D$. ConsequentlyThe large translational resistance $A_0$ makes the microswimmer slow even though the flagellum rotates almost at the motor rate.

For a very small head, $A_0\ll A$ and $D_0\ll D$. The appropriate denominator retains the translation-rotation coupling:Both tend to zero with $D_0\propto a^3$, while ${\Omega }_0\simeq -\omega$. The motor mainly rotates the low-resistance head, and produces little flagellar motion relative to the fluid.

In the intermediate coefficient regime $A_0\ll A$ and $D_0\gg D$, the head supplies a strong rotational reaction with little translational penalty. ThenThe head radius has dropped out. To find the optimal pitch of a helical microswimmer, put $z=\tan \varphi$. The dimensionless speed becomes $z/(z^2+2)$ and has derivative $(2-z^2)/(z^2+2{)}^2$. Hence, for $\omega >0$ with the chosen handedness,The angle is measured from the horizontal plane. An intermediate pitch combines axial and transverse tangent directions, allowing anisotropic drag to convert rotation into translation; either extreme removes that coupling.

The geometric head-size regimes quoted in the paper hold with the logarithmic slenderness factor treated as fixed. More precisely, $A_0\ll A$ requires $a\ll L(1+{\cos }^2\varphi )/(3|\log ϵ|)$, while $D_0\gg D$ requires $a^3\gg L{b}^2(1+{\sin }^2\varphi )/(4|\log ϵ|)$. These coefficient inequalities state the validity of the intermediate approximation when the logarithm is quantitatively important. The many-turn and local-drag assumptions of slender-body theory also remain in force.

For each incompressible Stokes flow without body force, ${\partial }_j{\sigma }_{ij}=0$ andThe divergence theorem givesIncompressibility removes the pressure term, antisymmetric velocity-gradient parts do not contract with the symmetric stress, and the remaining integrand is $2\mu e_{ij}^{(1)}{e}_{ij}^{(2)}$, symmetric under $1\leftrightarrow 2$. This proves the Lorentz reciprocal theorem for Stokes flow:

By Linearity of Stokes flow, the drag on a translating body is linear in its velocity, so $F_i=A_{ij}{U}_j$. Apply the reciprocal theorem to two translation solutions with velocities ${\mathbf{U}}^{(1)}$ and ${\mathbf{U}}^{(2)}$. Their surface velocities are constant and the far-field contribution vanishes, yieldingSince the two vectors are arbitrary,Thus the hydrodynamic resistance matrix is symmetric and, for fixed fluid viscosity, depends only on body geometry.

A straight rod has fore-aft symmetry about its centre. Gravity acts through that centre, so the translational-rotational coupling and applied torque vanish. Kinematic reversibility of Stokes flow and Uniqueness of Stokes flow then rule out spontaneous rotation: the inclination $\theta$ remains constant.

Let the speed for perpendicular translation be $U_0$. The given factor of two makes the parallel speed $2U_0$ for the same force component. Resolving gravity parallel and perpendicular to the rod givesResolving these velocities horizontally and vertically givesTherefore

Reflection in $y=0$ preserves the prescribed translational velocity and axial angular velocity. A force transforms as a polar vector, whereas a torque transforms as an axial vector; consequently $F_y=G_x=G_z=0$. Reflection in $x=0$ reverses both imposed motions but preserves the normal force component. The linearity of Stokes flow then gives $F_z=0$. Thus only $F_x$ and $G_y$ can be nonzero.

There is a sign inconsistency in the printed question. In right-handed coordinates, $\mathbf{\Omega }=\Omega {\mathbf{e}}_y$ gives a rotational velocity $-\Omega a{\mathbf{e}}_x$ at the sphere's lowest point. The printed shear and force formulas instead use the opposite rotational sense. Below, let $\omega$ be the right-handed component along $+{\mathbf{e}}_y$; the paper's displayed formulas are recovered by setting $\omega =-\Omega$.

In the translating frame, the parabolic gap and leading boundary velocities areThe no-slip boundary condition and lubrication theory giveThe gap is stationary in this frame, so ${\nabla }_{\parallel }\cdot \mathbf{q}=0$. The Reynolds lubrication equation is thereforeSince $h_x=x/a$ and ${\nabla }_{\parallel }^{2}h=2/a$,Differentiating the velocity profile gives the leading shear stressesThe first expression becomes the printed formula after $\omega =-\Omega$.

For the logarithmic lubrication resistance of a sphere near a wall, write $\ell =\log (1/ϵ)$. The logarithmic region is $a\sqrt{ϵ}\ll r\ll a$, where $h\sim r^2/(2a)$. ConsequentlyThese follow from $\int r/hdr=a\ell +O(a)$ and $\int r^3/h^{2}dr=2a^2\ell +O(a^2)$, with a fixed small outer cutoff. The general logarithmic radial numerator is $r^{2n-1}$ over $h^n$; the numerator in the printed integration hint appears to have a typographical error.