Slender-body force density

At leading logarithmic order, a straight slender rod with unit tangent $\mathbf{t}$ and local velocity $\mathbf{v}$ exerts force per unit lengthwhere $C$ depends logarithmically on the rod aspect ratio.

For two perpendicular rods of length $2L$ joined at one endpoint, the slender-body force density gives translation resistance $CL\mathrm{diag}(3,3,4)$, rotation resistance $C{L}^3\mathrm{diag}(8/3,8/3,16/3)$, and translation–rotation block $C{L}^2\left(\begin{array}{ccc}0& 0& -2\\ 0& 0& 2\\ 2& -2& 0\end{array}\right)$. The opposite block is its transpose. Here $G$ is the torque about the joint and $F$ the force on the fluid. Integrating along each rod proves the entries and makes the symmetry and positivity of a rigid-body resistance matrix explicit.

Attach weights $mg$ at the joint and $\lambda mg$ at each outer endpoint of the right-angle two-rod resistance matrix geometry. With the first rod initially horizontal and the second vertically upwards, quasistatic Stokes flow gives $\dot{\theta }=(1-\lambda )mg(\cos \theta -\sin \theta )/(4C{L}^2)$. For $\lambda <1$ the orientation tends to $\pi /4$; for $\lambda >1$ it tends to $-3\pi /4$. Eliminating time gives $x-x_0=(2L/3)(1+\sin \theta -\cos \theta )$ and the same rightward net displacement $2L/3$ in both cases. For $\lambda >1$ the initial drift is leftwards before reversing. At $\lambda =1$ there is no rotation or lateral drift, illustrating noncommuting infinite-time and parameter limits.

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.

For a slender centreline $\mathbf{X}(\theta )=(A\cos \theta ,B\sin \theta ,b\theta )$, $-\pi \le \theta \le \pi$, let $s^{\prime }(\theta )=(A^2{\sin }^2\theta +B^2{\cos }^2\theta +b^2{)}^{1/2}$ and use local force density $\mathbf{f}=c(2I-\mathbf{tt})\mathbf{v}$, $\mathbf{t}=d\mathbf{X}/ds$, $c=2\pi \mu /\ln (L/R)$. Rotation with angular speed $\Omega$ gives $\mathbf{t}\cdot \mathbf{v}=\Omega AB/s^{\prime }$, henceAxial translation with speed $W$ gives exactly the same coefficient for the torque: $G_z=-cWABb\int d\theta /s^{\prime }$. Indeed $(\mathbf{X}\times \mathbf{t}{)}_z=AB/s^{\prime }$. Both forces and torques here are exerted on the fluid. In the circular case $A=B=a$, this reduces to $F_z/\Omega =G_z/W=-2\pi c{a}^{2}b/\sqrt{a^2+b^2}$. The equality is also a consequence of the Lorentz reciprocal theorem, since the local resistance tensor is symmetric.

Local resistive-force theory for a helix gives $AD-B^2={\ell }^2{a}^2{\xi }_{\parallel }{\xi }_{\perp }$. Positive local drag therefore guarantees a positive-definite axial resistance matrix and positive viscous power loss. The identity also keeps the denominator in counterrotating two-helix balances positive.

Mirroring the handedness of a helix reverses its axial translation-rotation coupling coefficients while preserving its pure translation and rotation resistances. With $\mathbf{t}=\cos \theta {\mathbf{e}}_z-\sin \theta {\mathbf{e}}_{\phi }$ for a left-handed helix and the force-on-body convention, $B=C=\ell a({\xi }_{\perp }-{\xi }_{\parallel })\sin \theta \cos \theta$.