= Solution
For body-force-free Stokes fields with the same <viscosity>,
$$
\boxed{\int_{\partial V}u^{(1)}\cdot\sigma^{(2)}n\,dS
=\int_{\partial V}u^{(2)}\cdot\sigma^{(1)}n\,dS.}
$$
To prove this <Lorentz reciprocal theorem>, integrate the divergence of the difference of the two cross-work fluxes. Both <stress> divergences are zero; <incompressibility> and symmetry of the Newtonian <stress> reduce each cross-gradient contraction to $2\mu e^{(1)}:e^{(2)}$. These are equal, so the divergence integrates to zero. The argument applies to an exterior domain by first cutting it off and taking the outer radius to infinity; the decaying <velocity> and <stress> make the outer boundary contribution vanish.
Define the generalized <force> as <force> and <torque> exerted by the rigid body on the fluid. With a resting far field and <no-slip boundary conditions>, $u=U+\Omega\times x$ on the body, and linearity gives
$$
\binom{F}{G}=\mathsf R\binom{U}{\Omega}.
$$
For two such motions the reciprocal theorem becomes
$$
U^{(1)}\cdot F^{(2)}+\Omega^{(1)}\cdot G^{(2)}
=U^{(2)}\cdot F^{(1)}+\Omega^{(2)}\cdot G^{(1)}.
$$
Since the six-component velocities can be chosen independently, $\boxed{\mathsf R=\mathsf R^T}$. The ordinary work identity gives
$$
\boxed{U\cdot F+\Omega\cdot G=2\mu\int_{V_{\rm fluid}}e:e\,dV>0}
$$
for every nonzero rigid-body motion. If the integral were zero, $e=0$ would make the connected exterior flow an infinitesimal rigid motion. Its decay at infinity requires that motion to be zero, which is incompatible with a nonzero no-slip body <velocity>. Thus $\mathsf R$ is positive definite. Fluid-on-body <forces> have the opposite sign; their map is $-\mathsf R$, not a positive-definite resistance <matrix>.
Write $s_\phi=\sin\phi$ and $c_\phi=\cos\phi$. The helix has $|dX/d\theta|=b\sqrt{1+\tan^2\phi}=b\sec\phi$, so
$$
\boxed{ds=b\sec\phi\,d\theta,\qquad L=N\pi b\sec\phi.}
$$
Its unit tangent is $X'=c_\phi e_\theta+s_\phi e_z$, where $e_\theta=(-\sin\theta,\cos\theta,0)$. The given <slender-body force density> is the <force> per arc length exerted on the fluid. For pure axial translation, $V=Ue_z$, it gives
$$
f_z=CU(1-s_\phi^2/2),\qquad f_\theta=-CU s_\phi c_\phi/2.
$$
For pure rotation, $V=b\Omega e_\theta$, it gives
$$
f_z=-Cb\Omega s_\phi c_\phi/2,\qquad f_\theta=Cb\Omega(1-c_\phi^2/2).
$$
The axial <torque> per unit arc length is $bf_\theta$. Integrating along the wire gives the <axial resistance matrix of a slender helix>
$$
\boxed{\binom{F_z}{G_z}=\begin{pmatrix}\mathcal A&\mathcal B\\\mathcal B&\mathcal D\end{pmatrix}\binom{U}{\Omega},}
$$
$$
\boxed{\mathcal A=\frac{CL}{2}(1+c_\phi^2),\qquad
\mathcal B=-\frac{CLb}{2}s_\phi c_\phi,\qquad
\mathcal D=\frac{CLb^2}{2}(1+s_\phi^2).}
$$
Thus translation gives $(F_z,G_z)=(\mathcal A U,\mathcal B U)$, and rotation gives $(\mathcal B\Omega,\mathcal D\Omega)$. These are the requested axial components; the axial swimmer approximation does not require proving statements about the other components. The equality of the two off-diagonal coefficients checks reciprocity, and
$$
\mathcal A\mathcal D-\mathcal B^2=\frac{C^2L^2b^2}{2}>0.
$$
For the organism, retain the specified entrainment approximation: the head moves with the local large-scale fluid translation, and its angular <velocity> $\Omega-\omega$ differs from the local fluid rotation $\Omega$ by $-\omega$. Put $D_0=8\pi\mu a^3$. Its <torque> on the fluid is therefore $-D_0\omega$, and its <force> is neglected. Neutral buoyancy and absence of external <torque> give
$$
\mathcal A U+\mathcal B\Omega=0,\qquad \mathcal B U+\mathcal D\Omega=D_0\omega.
$$
Solving,
$$
U=-\frac{\mathcal B D_0\omega}{\mathcal A\mathcal D-\mathcal B^2}
=\frac{D_0\omega s_\phi c_\phi}{CLb}
=\boxed{\frac{\omega a^3|\log\epsilon|\sin2\phi}{bL}}.
$$
This is the <entrained-head approximation for a helical microswimmer>, which differs from treating the head as an isolated sphere in otherwise stationary fluid. The flagellum's angular speed is also $\Omega=D_0\omega(1+c_\phi^2)/(CLb^2)$.
At fixed $b,L,a,\omega,\epsilon$, the speed is maximal at $\phi=\pi/4$ and tends to zero at either limiting pitch. At fixed other parameters it decreases as $1/L$: the longer flagellum has more resistance at the fixed <torque> supplied in this model. Holding the number of turns fixed instead would make $L$ change with pitch, so that is a different comparison.
The motor exerts equal and opposite <torques> of magnitude $D_0\omega$ on flagellum and head, whose relative angular speed is $\omega$. Therefore its rate of working is
$$
\boxed{\mathcal P=D_0\omega^2=8\pi\mu a^3\omega^2.}
$$
Equivalently, the work delivered to the fluid is $G_{\rm flag}\Omega+G_{\rm head}(\Omega-\omega)=D_0\omega\Omega-D_0\omega(\Omega-\omega)$. Translation contributes no net work because the total <force> vanishes. This explains the independence from $U$ and $\Omega$, without assigning an isolated-fluid drag law to the entrained head.
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