= Axial coupling of an elliptic helix
{title2=$F_z/\Omega=G_z/W=-cABb\int_{-\pi}^{\pi}(s')^{-1}d\theta$}
For a slender centreline $\mathbf X(\theta)=(A\cos\theta,B\sin\theta,b\theta)$, $-\pi\leq\theta\leq\pi$, let $s'(\theta)=(A^2\sin^2\theta+B^2\cos^2\theta+b^2)^{1/2}$ and use local <force> density $\mathbf f=c(2I-\mathbf t\mathbf t)\mathbf v$, $\mathbf t=d\mathbf X/ds$, $c=2\pi\mu/\ln(L/R)$. Rotation with angular speed $\Omega$ gives $\mathbf t\cdot\mathbf v=\Omega AB/s'$, hence
$$
F_z=-c\Omega ABb\int_{-\pi}^{\pi}\frac{d\theta}{s'(\theta)}.
$$
Axial translation with speed $W$ gives exactly the same coefficient for the <torque>: $G_z=-cWABb\int d\theta/s'$. Indeed $(\mathbf X\times\mathbf t)_z=AB/s'$. Both <forces> and <torques> here are exerted on the fluid. In the circular case $A=B=a$, this reduces to $F_z/\Omega=G_z/W=-2\pi c a^2b/\sqrt{a^2+b^2}$. The equality is also a consequence of the <Lorentz reciprocal theorem>, since the local resistance <tensor> is symmetric.
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