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For body-force-free Stokes fields with the same viscosity,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 givesFor two such motions the reciprocal theorem becomesSince the six-component velocities can be chosen independently, $\boxed{\mathsf{R}={\mathsf{R}}^T}$. The ordinary work identity givesfor 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 $s_{\varphi }=\sin \varphi$ and $c_{\varphi }=\cos \varphi$. The helix has $|dX/d\theta |=b\sqrt{1+{\tan }^2\varphi }=b\sec \varphi$, soIts unit tangent is $X^{\prime }=c_{\varphi }{e}_{\theta }+s_{\varphi }{e}_z$, 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, $V=U{e}_z$, it givesFor pure rotation, $V=b\Omega e_{\theta }$, it givesThe axial torque per unit arc length is $b{f}_{\theta }$. Integrating along the wire gives the axial resistance matrix of a slender helixThus 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, andFor 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 giveSolving,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 $\Omega =D_0\omega (1+c_{\varphi }^2)/(CL{b}^2)$.

At fixed $b,L,a,\omega ,ϵ$, the speed is maximal at $\varphi =\pi /4$ 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 isEquivalently, the work delivered to the fluid is $G_{\mathrm{flag}}\Omega +G_{\mathrm{head}}(\Omega -\omega )=D_0\omega \Omega -D_0\omega (\Omega -\omega )$. 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.