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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.

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.

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.