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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.

Use the downward-positive displacement and the filament bending modulus $B$ and load density $w$ defined above. Two applications of integration by parts give the first variationThus the elastic-plus-gravitational force density is $-\delta E/\delta h=-B{h}_{xxxx}+w$. In resistive-force theory, transverse motion has drag per unit length $-{\zeta }_{\perp }{h}_t$. Neglecting filament and fluid inertia, instantaneous force balance gives the forced small-slope elastohydrodynamic filament equationClamping fixes position and tangent, while the free end has zero bending moment and zero transverse shear force. ThereforeThe last two conditions also follow as natural boundary conditions from the variation, since the free-end values of $\delta h$ and $\delta h_x$ are arbitrary. Distributed weight does not add a concentrated force or moment at the tip.

At steady state, $B{h}_s^{\prime \prime \prime \prime }=w$. Integrating first from the free end gives $h_s^{\prime \prime \prime }=(w/B)(x-L)$ and $h_s^{\prime \prime }=(w/2B)(x-L{)}^2$. Integrating twice more with the clamp conditions gives the uniform-load bending of a clamped filament:In particular the free-end displacement is $\delta =h_s(L)=w{L}^4/(8B)$ and the total buoyancy-corrected gravitational load is $F=wL$. Thus the tip stiffness of a uniformly loaded cantilever isThe last expression uses $A$ as Young's modulus; if $A$ itself is flexural rigidity, the same result reads $k_{\mathrm{eff}}=8A/L^3$. This is Hooke's law for the specified loading pattern. It is not the stiffness for a point force at the tip, which would be $3B/L^3$; replacing a distributed force by its resultant does not preserve its bending moment distribution.

For dimensional analysis, $[A]=$ force/length squared, $[I]=$ length to the fourth power, hence $[B]=$ force times length squared. Consequently $[B/L^3]=$ force/length, as required for a spring constant. Also $[{\zeta }_{\perp }]=$ force times time/length squared makes ${\zeta }_{\perp }{h}_t$, $B{h}_{xxxx}$ and $w$ all forces per unit length. Neither viscosity nor density enters the stiffness, although viscous drag controls the rate of approach to the steady shape. The small-slope approximation requires $w{L}^3/B\ll 1$, since $h_s^{\prime }(L)=w{L}^3/(6B)$.

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.

The first term is the local viscous force per unit length from resistive-force theory. With unit tangent vector $\mathbf{t}={\mathbf{r}}_s$, the drag tensor gives $-c_{\parallel }{\mathbf{v}}_{\parallel }-c_{\perp }{\mathbf{v}}_{\perp }$, where $\mathbf{v}={\mathbf{r}}_t$ is velocity relative to the fluid.

The second term is the bending force of an inextensible filament. Varying its energy $E_b=(B/2)\int |{\mathbf{r}}_{ss}{|}^{2}ds$ gives bulk force density $-B{\mathbf{r}}_{ssss}$. The final term, ${\partial }_s(T{\mathbf{r}}_s)$, is the force density from filament tension; $T$ is the Lagrange multiplier enforcing $|{\mathbf{r}}_s|=1$. At low Reynolds number, these forces balance without a filament acceleration term.

With mass, length, and time dimensions $M,\mathsf{L},\mathsf{T}$, the coefficients haveThus the drag coefficients have the dimensions of dynamic viscosity, the filament bending modulus has dimensions force times length squared, and tension has dimensions force.

In the small-slope Monge representation, $s=x+O(y_x^2)$, $\mathbf{t}=(1,y_x)+O(y_x^2)$, and the leading transverse velocity is $y_t$. With no imposed axial prestress, the induced filament tension is second order in transverse amplitude, so its transverse contribution is higher order. The transverse component of force balance reduces to the small-slope elastohydrodynamic filament equationAn externally imposed zeroth-order tension would instead add $T{y}_{xx}$.

For the transverse motion, the off-diagonal part of the resistive-force theory tensor gives axial propulsive force density $(c_{\perp }-c_{\parallel })y_x{y}_t$ to second order. Its integral isSubstitute the bending equation and integrate by parts:The boundary expression for transverse filament thrust is thereforeOnly endpoint slope, bending moment, and shear enter this expression. Exact inextensibility also generates second-order longitudinal material motion; if that motion is retained in instantaneous total axial drag, it adds $-c_{\parallel }\int v_{x}dx$. For a periodic deformation with fixed axial base position, that additional term has zero cycle average. Thus the displayed formula is the transverse propulsive contribution and also gives the cycle-averaged propulsive force.

In resistive-force theory, the force per unit length is local and linear in the filament velocity relative to the fluid:The distinct parallel and perpendicular drag coefficients encode the anisotropic resistance of a slender filament. Neglecting nonlocal hydrodynamic interactions makes this a leading logarithmic approximation to slender-body theory.

In the small-slope approximation the worm-like chain bending energy isThe clamped end obeys $h(0)=h_x(0)=0$, while a force-free and torque-free tip has the natural conditions $h_{xx}(L)=h_{xxx}(L)=0$. A static transverse tip force $f$ produces the cantilever compliance $h(L)/f=L^3/(3A)$. The equipartition theorem, or equivalently the static fluctuation--response relation, therefore gives the exact tip variance

For the dynamics, resistive-force theory gives the transverse drag per lengthat logarithmic accuracy and the stochastic beam equationLet ${\varphi }_n$ be clamped--free bending modes with $q_{n}L={\alpha }_n$, whereWriting $N_n={\int }_0^L{\varphi }_n^{2}dx$, independent thermally driven modes giveThe first mode contains almost all of the tip variance, so

For a microtubule, $A=k_{B}T{L}_p$. With $L=10\mu \mathrm{m}$ and $L_p=3\mathrm{mm}$,Taking water at room temperature and $a=0.5\mu \mathrm{m}$ gives ${\zeta }_{\perp }\simeq 3.6\times 10^{-3}\mathrm{Pa}\mathrm{s}$ andwith an order-one uncertainty from the logarithmic slender-body drag approximation.

On the upper half-filament, write $\mathbf{r}(s)=(h(s),s)$ with $0\le s\le L$. To first order in slope, resistive-force theory supplies the uniform transverse load $f_x={\zeta }_{\perp }U$, and Euler--Bernoulli balance givesMidpoint symmetry and clamping impose $h(0)=h^{\prime }(0)=0$; the free-end force and moment conditions are $h^{\prime \prime \prime }(L)=h^{\prime \prime }(L)=0$. With $H=h/L$ and $\xi =s/L$,Four integrations giveIn particular, $H(1)=\mathrm{Sp}/8$ and $H^{\prime }(1)=\mathrm{Sp}/6$. The small-slope approximation first fails when the end slope is order one, giving the estimate $\mathrm{Sp}\sim 6$; parametrically, breakdown occurs at Sperm number of order one.

The leading viscous dissipation due to a small planar filament stroke is $\dot{W}={\xi }_{\perp }{T}^{-1}{\int }_0^T{\int }_0^{\lambda }{y}_t^{2}dxdt$. It follows from the work against the local resistive-force theory force. The transverse velocity controls the leading power; the quadratic longitudinal velocity contributes only at higher order.

For a small-amplitude planar filament without mean axial drift, resistive-force theory gives the leading time-averaged force on the filament as $F_x=({\xi }_{\perp }-{\xi }_{\parallel })T^{-1}{\int }_0^T{\int }_0^{\lambda }{y}_x{y}_{t}dxdt$. The parallel and perpendicular drag coefficients of a slender filament must differ to obtain propulsion at this order. A right-going wave gives negative axial force; a left-going wave gives positive axial force.

For a heavy spherical head with a much thinner long rigid tail, let $P$ be its excess weight, $H=6\pi \mu a$ the head translation resistance, and $K=K_{N}L$ the tail normal translation resistance. Neglect tail mass relative to head mass, sphere rotation resistance relative to tail rotation resistance, and $a/L$ in tail lever arms. In shear $\alpha z{\mathbf{e}}_1$, the stable orientation satisfies $\sin \theta =\alpha /(\beta +\sqrt{{\beta }^2+{\alpha }^2})$, $\cos \theta >0$, with $\beta =3P/[L(K+4H)]$. Resolving resistive-force theory force and torque balance gives the displayed signed vertical velocity. Since $0<\gamma <1$, stronger shear increases $U_z$ towards zero and reduces the downward settling speed. The stable head is below the far tail end. Very large shear can make the neglected sphere-rotation contribution significant.

For a straight, unloaded inextensible filament with small transverse displacement, resistive-force theory gives viscous force density $-c_{\perp }{y}_t$. Variation of the bending energy gives $-B{y}_{xxxx}$, while the induced filament tension is higher order. Their instantaneous balance yields $c_{\perp }{y}_t=-B{y}_{xxxx}$. It is an overdamped fourth-order diffusion equation: short-wavelength bends relax much faster than long ones. Wiggins and Goldstein's flexive-propulsion paper develops the elastic-wave mechanism for low-Reynolds-number propulsion.

Let positive transverse displacement point along a uniform load per unit length $w$. The filament bending modulus $B$ gives energy ${\int }_0^L[B(h_{xx}{)}^2/2-wh]dx$. The variational derivative is $B{h}_{xxxx}-w$, so resistive-force theory gives ${\zeta }_{\perp }{h}_t=-B{h}_{xxxx}+w$. Clamping imposes $h(0)=h_x(0)=0$, while a free tip imposes $h_{xx}(L)=h_{xxx}(L)=0$. The steady solution isIts tip displacement is $w{L}^4/(8B)$. For a submerged cylindrical filament under Newtonian gravity, buoyancy gives $w=({\rho }_f-\rho )\pi a^{2}g$. This linear approximation requires small slope and negligible inertia.