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With a consistent force convention, the mobility matrix maps applied generalized forces to rigid-body velocities; it is the inverse of the hydrodynamic resistance matrix. For a sphere with force on the fluid $\mathbf{F}$, its isolated mobility is $1/(6\pi \mu a)$. Coupling between bodies represents hydrodynamic interactions.

For two Stokes flows $(u^{(1)},{\sigma }^{(1)})$ and $(u^{(2)},{\sigma }^{(2)})$ of the same dynamic viscosity in the same region, with no volume force, the Lorentz reciprocal theorem statesIndeed, the divergence of their cross-work difference isThe pressure terms vanish by incompressible flow, and the derivatives of the Cauchy stress tensors vanish by the Stokes equation. The divergence theorem proves the result.

Use the convention that $F,G$ are the force and torque exerted by the body on the fluid. By Linearity of Stokes flow, $(F,G)=\mathsf{R}(U,\Omega )$. Applying the Lorentz reciprocal theorem to two independent rigid motions gives $q_1^T\mathsf{R}{q}_2=q_2^T\mathsf{R}{q}_1$, so the hydrodynamic resistance matrix is a symmetric matrix. The boundary-power identity givesThis is strictly positive for nonzero rigid motion: equality would imply $e=0$, hence a rigid motion throughout the connected fluid, which must vanish since the fluid is at rest at infinity. The no-slip boundary condition would then force $q=0$. Therefore $\mathsf{R}$ is a positive-definite matrix. Reversing to the fluid-on-body force changes the sign of the force law, not the positive resistance coefficients.

For the two rods, take the torque about $O$ and use body axes. On the $x$-rod, $X=(s,0,0)$, $0\le s\le 2L$, and its slender-body force density isOn the $y$-rod it isIntegrate $f$ and $X\times f$ along both rods, using $\int ds=2L$, $\int sds=2L^2$ and $\int s^{2}ds=8L^3/3$. A convenient dimensionally uniform statement of the complete right-angle two-rod resistance matrix isIn physical coordinates this means translation–translation entries scale as $CL$, the two translation–rotation blocks as $C{L}^2$, and rotation–rotation entries as $C{L}^3$. The planar block has inverseThis verifies the printed hydrodynamic mobility matrix with its third velocity component $L{\Omega }_z$; the TeX aid's ${\Omega }_x$ is an OCR error.

Take laboratory vertical velocity positive upwards. The body axes are $e_1=(\cos \theta ,\sin \theta )$ and $e_2=(-\sin \theta ,\cos \theta )$. In quasistatic sedimentation the force and torque exerted on the fluid equal the gravitational resultants on the body:The out-of-plane block is unforced, so its positive hydrodynamic resistance matrix gives $U_z={\Omega }_x={\Omega }_y=0$. Substitution in the planar hydrodynamic mobility matrix givesandFor $0\le \lambda <1$, this ordinary differential equation has positive right side on $0\le \theta <\pi /4$, with a stable zero at $\pi /4$. Uniqueness prevents crossing that equilibrium in finite time. For $\lambda >1$ the initial angular velocity is negative, and the stable equilibrium reached from zero is $-3\pi /4$. The heavier-end body turns clockwise through $3\pi /4$; it does not settle at $\pi /4$. More explicitly, with $K=(1-\lambda )mg/(4C{L}^2)$,where the continuous branch has $\theta -\pi /4\in (-\pi ,0)$. At $\lambda =1$, $\theta$ remains zero.

Transforming the translational velocity back to laboratory axes givesFor $\lambda \ne 1$, eliminate time between $U_h$ and the angular velocity:At either limiting orientation $\sin \theta =\cos \theta$. Thus both cases have the same net horizontal displacement,For $\lambda <1$ the drift is monotonically rightwards. For $\lambda >1$ it first moves left, reaching $x_O-x_O(0)=(2L/3)(1-\sqrt{2})$ at $\theta =-\pi /4$, then reverses. The total horizontal path length in this second case is $(2L/3)(2\sqrt{2}-1)$, whereas its net displacement is $2L/3$ rightwards. At $\lambda =1$ it falls vertically without rotation or drift; taking the infinite-time limit before $\lambda \to 1$ is consequently singular.

For the requested sedimentation drift of a weighted two-rod body sketch, a full parametric trajectory follows by also integrating $U_v/\dot{\theta }$. Put $\psi =\theta -\pi /4$ and set the initial height to zero:It tends to $-\infty$ in either case while $x_O$ tends to $2L/3$. At $\lambda <1$ both rods end pointing upwards from $O$ symmetrically; at $\lambda >1$ they end pointing downwards symmetrically. The asymptotic downward speed is $(1+2\lambda )mg/(3CL)$.

Use the same resistive-force theory convention as before: the force density is the fluid's force on the filament. Let $\phi$ be the usual cylindrical azimuth. Choose the orientation of a left-handed helix so that, as $z$ increases, $\phi$ decreases. With $s=\sin \theta$, $c=\cos \theta$,Reversing the orientation of the tangent leaves the local drag tensor unchanged. Define $k={\xi }_{\perp }-{\xi }_{\parallel }$. From resistive-force theory, the relevant force-density components areThese scalar axial and azimuthal components are uniform along the helix. Integration over its contour length $\ell$, and use of the axial torque density $a{f}_{\phi }$, give the axial resistance matrix of a slender helix:Thus $B=C$, as also required by the Lorentz reciprocal theorem. The signs here belong to $(F,L{)}^T=-\left(\begin{array}{cc}A& B\\ B& D\end{array}\right)(U,\Omega {)}^T$, with $F,L$ forces and moments on the helix. The handedness reversal of helical hydrodynamic resistance reverses $B,C$ but leaves $A,D$ unchanged. If the positive rotation direction or handedness convention is reversed, the coupling sign reverses with it.

As a check on physical admissibility, the determinant of helical resistance in resistive-force theory isTogether with $A,D>0$, this makes the hydrodynamic resistance matrix positive definite and the viscous power loss $A{U}^2+2BU\Omega +D{\Omega }^2$ positive. Isotropic local drag would have $B=0$, so rotating a helix would not propel it in this approximation.

For the pair, use a local additive resistive-force theory model: both helices have the same drag coefficients, and inter-helix hydrodynamic interactions and unresolved end effects are omitted. The right-handed helix has resistance entries $(nA,-nB,-nB,nD)$. Define the signed motor motion byThe printed relative rotation specifies a magnitude rather than which helix rotates relative to which. Choosing ${\Omega }_2-{\Omega }_1=\omega$ instead reverses all signed speeds below.

The common translation speed $U$ and zero total force and torque obeyInternal motor forces and torques cancel in these totals. IntroduceSolving the three linear equations yields the opposite-handed counterrotating helical swimmer:For $n>0$, ${\xi }_{\perp }>{\xi }_{\parallel }$ and $\omega >0$, $B>0$, so $U<0$, ${\Omega }_1>0$ and ${\Omega }_2<0$. The opposite-handed helices counterrotate but contribute thrust in the same axial direction. The formula satisfies the specified relative rotation without identifying either laboratory rotation rate with the motor rate.

At $n=0$, the second helix supplies no hydrodynamic resistance. The remaining helix must have $F=L=0$, and invertibility of its hydrodynamic resistance matrix forces $U={\Omega }_1=0$. Formally ${\Omega }_2=-\omega$ is the rotation of a zero-resistance motor shaft or vanishing second rotor; there is no physical finite second helix to propel or to provide a reaction torque. This is the vanishing reaction rotor in a helical swimmer.

At $n=1$, equal-length opposite-handed helices haveThe translation-generated axial torques cancel between the two helices, and equal counterrotation balances the rotational torques. Their propulsive forces add, rather than cancel, because handedness and rotation both reverse. These are equal-length opposite-handed helices.

For $n\to \infty$, the large reaction helix limit isThe increasingly long second helix nearly anchors the whole assembly: its very small translation and rotation suffice to balance the finite force and torque generated by the first helix. The first helix rotates at almost the full motor rate, but moving the large resistive second helix makes the common translation tend to zero. These are idealized limits of the additive local model, rather than a resolution of short-helix end effects or infinitely extended interacting filaments.

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.

The force exerted on the fluid through the upper gap boundary includes pressure acting on its slope:The pressure cancels the $x^2/h^2$ term, givingPressure exerts no torque about the centre of a sphere because its traction is radial. To logarithmic order, the shear has lever arm $-a{\mathbf{e}}_z$, soThe other components vanish by parity: $G_x$ involves an integral of the $xy$ term, and $G_z$ has an integrand odd in $y$. The combined hydrodynamic resistance matrix isIts equal off-diagonal coefficients are required by the Lorentz reciprocal theorem for Stokes flow. It is a positive-definite matrix, as required by viscous dissipation.

In the rotational convention of the printed force formula, set $\Omega =-\omega$ and measure the scalar couple along $-{\mathbf{e}}_y$, so $G=-G_y$. ThenFor sedimentation under the buoyancy-adjusted weight $F$, a uniform sphere is torque-free. Thus $\omega a=U/4$, and the force balance givesThe rotational scalar in the paper's displayed-force convention is $\Omega =-U/(4a)$. The sphere rolls in the right-handed $+{\mathbf{e}}_y$ sense but still slips: its lowest point has laboratory velocity $U-a\omega =3U/4$. The signs and coefficients of the right-handed resistance agree with the rigid-wall terms of Bertin et al., equations (4.4)–(4.5), after reversing the forces and torques there from fluid-on-sphere to sphere-on-fluid.

Choose the handedness for which the helix tangent is $\mathbf{t}=\cos \theta {\mathbf{e}}_z+\sin \theta {\mathbf{e}}_{\varphi }$. Its local rigid velocity is $\mathbf{u}=\tilde{U}{\mathbf{e}}_z+R\tilde{\Omega }{\mathbf{e}}_{\varphi }$, so $\mathbf{u}\cdot \mathbf{t}=\tilde{U}\cos \theta +R\tilde{\Omega }\sin \theta$. Integrating the axial force and torque from the drag law gives the symmetric hydrodynamic resistance matrix in the question, withReversing helical handedness reverses the sign of $B$ but leaves $A$ and $D$ unchanged.

Choose the vertical plane through the rod's axis. Reflection in that plane leaves the horizontal rod, the vertical gravitational force, and the Stokes equations unchanged, but reverses any velocity normal to the plane. A second reflection in the vertical plane perpendicular to the rod reverses velocity along the rod. By uniqueness of Stokes flow, both horizontal velocity components vanish. The rod therefore translates vertically.

Equivalently, the hydrodynamic resistance matrix of an axisymmetric rod has one coefficient parallel to its axis and another perpendicular to it. A vertical force on a horizontal rod lies entirely in the perpendicular eigenspace, so its velocity is vertical.

The uniform rod is invariant under inversion through its centre. This centrosymmetry makes the translation-rotation blocks of its generalized hydrodynamic resistance matrix vanish: a polar translation velocity changes sign under inversion, whereas an axial angular velocity does not, so an invariant linear coupling between them is impossible.

Gravity acts through the centre of mass and supplies no torque. The remaining rotational resistance is positive definite, so zero applied torque implies zero angular velocity. Thus an initially inclined rod keeps its orientation while settling.

At negligible Reynolds number, the Stokes equations and their no-slip boundary conditions are linear in velocity. Scaling or superposing a prescribed rigid translation therefore scales or superposes the resulting stress and total hydrodynamic force. Hence, for a fixed rod orientation,The minus sign records that viscous drag opposes motion, and $\mathbf{R}$ is the translational hydrodynamic resistance matrix.

For two body-force-free Stokes flows $({\mathbf{u}}^{(1)},{\mathbf{\sigma }}^{(1)})$ and $({\mathbf{u}}^{(2)},{\mathbf{\sigma }}^{(2)})$ in the same domain, the Lorentz reciprocal theorem for Stokes flow isIndeed, the difference of the two volume integrands isbecause both flows are incompressible and the Newtonian fluid stress tensor is symmetric. The divergence theorem proves the boundary identity.

On a rigid body, ${\mathbf{u}}^{(r)}={\mathbf{U}}^{(r)}+{\mathbf{\Omega }}^{(r)}\times \mathbf{x}$. The reciprocal theorem becomesWriting $(\mathbf{F},\mathbf{G}{)}^T=\mathsf{R}(\mathbf{U},\mathbf{\Omega }{)}^T$ and choosing arbitrary pairs of rigid velocities shows thatThus the hydrodynamic resistance matrix is symmetric.

Here $\mathbf{F},\mathbf{G}$ are forces and torques exerted by the body on the fluid, or the external force and torque needed to maintain its motion. The Lorentz reciprocal theorem for Stokes flow makes the six-dimensional hydrodynamic resistance matrix symmetric. The power identity $\mathbf{F}\cdot \mathbf{U}+\mathbf{G}\cdot \mathbf{\Omega }=2\mu \int e:edV$ makes it a positive-definite matrix for a body moving in otherwise stationary unbounded fluid.