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For a flagellum with axial resistance matrix of a slender helix and a head with resistances $A_0=6\pi \mu a$, $D_0=8\pi \mu a^3$, neglect interactions between the parts. A motor with relative angular velocity $\omega$ givesThe head rotates at $\Omega -\omega$. A large head creates excessive translational resistance; a small head supplies too little rotational resistance, allowing the head to counterrotate while the flagellum barely moves relative to the fluid.

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.