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For radius $b$, arc length $L$, tangent angle $\varphi$ to the horizontal plane, and local drag coefficient $C$, resistive-force theory givesThe sign of $B$ selects the handedness convention. With the force exerted on the fluid taken positive, $AD-B^2=C^2{L}^2{b}^2/2>0$. Translation-rotation coupling supplies propulsion for a rotating helical microswimmer with a spherical head.

Choose the helix handedness so its unit tangent is $\mathbf{t}=\cos \alpha \mathbf{i}+\sin \alpha {\mathbf{e}}_{\varphi }$. Let $q={\Omega }_0-\Omega$; the body rotates at $+\Omega \mathbf{i}$, while the flagellum rotates at $-q\mathbf{i}$ in the laboratory, since its relative motor rotation is $-{\Omega }_0\mathbf{i}$. Both translate at $-U\mathbf{i}$. Thus the relative fluid velocity on a flagellar element is $\mathbf{w}=U\mathbf{i}+bq{\mathbf{e}}_{\varphi }$.

Low-Reynolds-number resistive-force theory gives force per unit length on the flagellumIts axial and azimuthal components areDefine the integrated axial resistance matrix of a slender helix coefficientsThe sphere has translational resistance $A_b=6\pi \mu a$ by the Stokes drag law and rotational resistance $H_b=8\pi \mu a^3$. Its hydrodynamic force is $+A_{b}U\mathbf{i}$ and its torque $-H_b\Omega \mathbf{i}$. Integration of $f_x$ and $b{f}_{\varphi }$ gives the zero total force and torque conditionsDividing the first by $L$ and expanding the coefficients gives exactly the two requested balances. The internal motor torque cancels in the total torque balance; it is not an external applied torque.

Eliminate $U$ and use $q+\Omega ={\Omega }_0$. The effective flagellar rotational resistance is $H_e=H-C^2/(A+A_b)$, soThe determinant of helical resistance in resistive-force theory is $AH-C^2=b^2{L}^2{K}_N{K}_T>0$, hence the denominator is positive. Opposite handedness changes the sign of the translation-rotation coupling and reverses the swimming direction.

For $K_T=K$, $K_N=2K$, $\alpha =\pi /4$ and $a=b$, the coefficients are $A=3KL/2$, $C=KaL/2$, $H=3K{a}^{2}L/2$. ThereforeEquivalently, with $h=KL/(\pi \mu a)$, $U=a{\Omega }_{0}4h/(2h^2+21h+48)$. These are positive speed magnitudes for the selected handedness; the translational velocity is $-U\mathbf{i}$.

This helical microswimmer with a spherical head approximation neglects head-tail hydrodynamic interactions as permitted, treats the flagellum as slender and rigid, and averages away transverse forces associated with finite helical ends. The flow is quasistatic Stokes flow, so neither body inertia nor fluid inertia enters the balance.

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 a microscopic swimmer, the Reynolds number $\mathrm{Re}=\rho UL/\eta$ is usually small. Taking $L=2\mu \mathrm{m}$, $U=20\mu \mathrm{m}{\mathrm{s}}^{-1}$ and water-like properties gives $\mathrm{Re}\simeq 4\times 10^{-5}$. The Stokes equations then balance viscous stress and pressure almost instantaneously, so a swimmer cannot coast appreciably when its propulsive stroke stops.

The scallop theorem rules out a reciprocal opening-and-closing stroke in an otherwise unchanged Newtonian environment: reversing the sequence of shapes reverses the motion, and changing the stroke speed does not prevent cancellation. Organisms therefore use nonreciprocal motion. A sperm-like bending wave travels along a flagellum; a bacterium can rotate a helical flagellar bundle. The unequal parallel and perpendicular drag of a slender filament converts those motions into a net thrust. A freely suspended swimmer is force-free and torque-free overall, so body drag balances thrust and the body counter-rotates against the rotating bundle. Tracking a body and its flagellum together can reveal these opposite rotations. The helical microswimmer with a spherical head makes that force and torque balance quantitative.

A force-free, torque-free microswimmer in a stationary Newtonian fluid, with negligible fluid and body inertia and steady Stokes flow at each instant, gains no net displacement from a reciprocal stroke: one whose sequence of shapes is exactly retraced in reverse. The environment and boundary conditions must also be unchanged during that retracing. Opening a two-shell swimmer and then closing it follows the same sequence backwards; doing the closure faster does not help, because Linearity of Stokes flow changes the instantaneous velocity by the same factor as the stroke rate.

A single real shape coordinate that travels out and back along an interval gives this cancellation, as computed in the force-free two-sphere stroke. Two independently controlled shape coordinates can instead execute a loop that is not retraced, avoiding this obstruction. Continuous rotation also need not retrace an interval: the helical microswimmer with a spherical head provides another route to propulsion. The scallop theorem is a restriction on a stroke under stated conditions, not a claim that small organisms cannot swim. Purcell presented the opening-and-closing scallop and a swimmer with two hinges in Life at low Reynolds number (1977), figures 6–7.