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At leading order in slope, the hydrodynamic torque on the distal filament is $-\mu K_N{\int }_0^{L}s(Y_t-V)ds$. The displayed opposite torque is required at its base. It is the total internal bending moment; identifying it with the active moment assumes no separate passive elastic moment. If a passive moment is specified, subtract that contribution to determine the required active moment.

For steady incompressible flow and irrotational flow of constant mass density, with no body force, this formula gives the hydrodynamic torque about the origin per unit span on an impermeable body. Orient its boundary counterclockwise, and let positive torque be counterclockwise. The proof uses $M=\mathrm{Re}{\oint }_{C}pzd\bar{z}$, the Bernoulli equation $p=P_{\ast }-\rho |w^{\prime }{|}^2/2$, and the tangency identity $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$. The real part of the constant-pressure term vanishes because $\mathrm{Re}(zd\bar{z})=d|z{|}^2/2$. The full complex integral of $zd\bar{z}$ need not vanish.

Two widely separated spheres rigidly joined by a hydrodynamically negligible connector obey the free-draining approximation. Their translational hydrodynamic resistance matrix is $6\pi \mu (a_1+a_2)I$. The displayed point is the centre about which every pure translation has zero hydrodynamic torque: ${\sum }_i{\zeta }_i({\mathbf{r}}_i-{\mathbf{r}}_c)=0$. A pure rotation about that point gives no resultant force at this order. This is an exact cancellation within the approximation, rather than an exact formula at arbitrary finite separation. It does not assert that every chiral rigid body has an origin eliminating its translation-rotation coupling.

Hold the two-sphere object fixed in a uniform stream $\mathbf{V}$. In the free-draining approximation, its drag forces are ${\mathbf{f}}_i={\zeta }_i\mathbf{V}$, with ${\zeta }_i=6\pi \mu a_i$. The hydrodynamic torque about ${\mathbf{r}}_c$ isTo make it zero for every incoming-stream direction, choose the hydrodynamic centre of a two-sphere dumbbell:It lies closer to the larger sphere, whose larger drag requires a shorter lever arm. Uniform translation does not spin an isolated sphere, so no additional isolated-sphere spin torque enters this leading approximation.

Finite-separation hydrodynamic interactions and connector drag can change the displayed location; the calculation has the same approximation as part c.

Apply the reciprocal identity with flow1 the actual body-force-driven flow and flow2 the rotlet auxiliary field. Its body force is zero. The large-sphere boundary terms vanish under the usual decay and convergence assumptions for this unbounded body-force problem.

On the sphere, the actual rigid velocity is $\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}$. The auxiliary force is zero and its torque with the outward fluid normal is $\mathbf{G}$. The actual torque with that normal is the negative of the hydrodynamic torque on the sphere, hence zero for a torque-free sphere. Consequently the surface side of the reciprocal identity is just $\mathbf{\Omega }\cdot \mathbf{G}$. The volume side isSince the auxiliary couple is arbitrary,An unknown translation contributes nothing because the auxiliary rotlet has zero net force. This is body-force-driven rotation of a torque-free sphere.

Use the Unscaled Papkovich–Neuber representationwith harmonic vector and scalar potentials. For a torque-driven sphere chooseIts components are harmonic for $r>a$, and both $\mathbf{x}\cdot \mathbf{\Phi }$ and $\nabla \cdot \mathbf{\Phi }$ vanish. HenceThe last equality follows from the no-slip boundary condition at $r=a$. To check that $\mathbf{G}$ is indeed the applied couple, the traction on the sphere, with $\mathbf{n}=\mathbf{x}/a$ directed into the fluid, is ${\mathbf{\sigma }}_G\mathbf{n}=-3\mu {\mathbf{\Omega }}_G\times \mathbf{n}$. Its hydrodynamic torque isHere $\int \mathbf{nn}dS=(4\pi a^2/3)I$. The applied couple balances this resisting torque. The resulting field is the rotlet of a rotating sphere in Stokes flow.

For the Lorentz reciprocal theorem for Stokes flow, let ${\mathbf{n}}_D$ point out of the fluid domain $D$ and take the same viscosity in both flows. With $\nabla \cdot {\mathbf{\sigma }}_i+{\mathbf{f}}_i=0$, the theorem isIt follows by taking the divergence of the boundary integrand: incompressibility removes pressure-work terms, the two Newtonian strain contractions are equal, and the remaining stress divergences give the displayed body-force terms. This convention matters: on the inner boundary of an exterior domain, ${\mathbf{n}}_D=-\mathbf{n}$.

Let $n$ point from the sphere into the exterior fluid, so the surface position relative to the centre is $r=an$. The physical surface velocity isFree rotation means the actual body is torque-free: the total hydrodynamic torque $T={\int }_{S}an\times (\sigma \cdot n)dS$ vanishes. Use the auxiliary rotating sphere in Stokes flow with arbitrary $\hat{\Omega }$, boundary velocity $\hat{u}=\hat{\Omega }\times an$ and the given traction $\hat{\sigma }\cdot n=-3\mu \hat{\Omega }\times n$. The auxiliary flow is the decaying exterior rotation of a sphere, not a rigid rotation of the entire infinite fluid.

The left-hand cross-work in the Lorentz reciprocal theorem for Stokes flow isThe other cross-work becomesThe spherical integral identities areUsing $(\Omega \times n)\cdot (\hat{\Omega }\times n)=\Omega \cdot \hat{\Omega }-(\Omega \cdot n)(\hat{\Omega }\cdot n)$, the rotational contribution is $-8\pi \mu a^3\Omega \cdot \hat{\Omega }$. The slip contribution is $-3\mu \hat{\Omega }\cdot {\int }_{S}n\times u^{\prime }dS$. Since $\hat{\Omega }$ is arbitrary, the torque-free rotation of a spherical squirmer satisfiesThe viscosity cancels, as expected for a kinematically prescribed slip in a linear viscous problem. If the freely moving sphere also has a translation $U$, its added cross-work is zero because the auxiliary rotating sphere has zero total force, ${\int }_S\hat{\sigma }\cdot ndS=0$. Thus translation does not alter this rotational formula. Without the torque-free assumption an external-torque contribution would have to be retained.