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A rotlet dipole is the derivative of a rotlet along the separation of two opposed torques. A torque-free bacterium can have this far field because its rotating flagella and counterrotating body exert spatially separated internal torque reactions. The displayed axial formula uses the observation-coordinate derivative convention and signed torque moment $D$.

For a two-mode tensorial squirmer flow of radius $a$ and a passive equal sphere held at $X=Re$, with $a/R\ll 1$, let $s=e\cdot Be$. Faxén's first law gives holding force $F_h=-9\pi \mu a^{3}se/R^2+O(\mu a^4|A|/R^3+\mu a^5\Vert B\Vert /R^4)$. Its reflected Stokeslet changes swimming velocity by $-9a^{3}se/(4R^3)$. Since this leading force is radial, its vorticity at the swimmer vanishes. The next holding force, generated by the potential dipole, yields rotation $a^{4}A\times X/(4R^6)$. Terms from the passive sphere's finite radius, reflected stresslet and rotlet are needed at higher order. These formulas assume fixed slip and no externally applied swimmer force or torque.

For a Keplerian disk with evolution ${\partial }_t\Sigma =(2\pi r{)}^{-1}{\partial }_r\mathcal{F}+S$ and $\mathcal{F}=6\pi r^{1/2}{\partial }_r(r^{1/2}\bar{\nu }\Sigma )$, take zero viscous torque at $r_1>0$ and zero flux outside the source region at $r_2>r_1$. The steady state gives ${\mathcal{F}}^{\prime }=-2\pi rS$, and its outer condition gives the displayed flux. Integrating once more from the zero-torque inner boundary condition yieldsThe equation determines the viscosity-density product, so a thermodynamic viscosity closure can be imposed afterward. For nonnegative integrable $S$, it is nonnegative and the inner accretion rate is exactly the total supplied mass per unit time.

For an axial rotlet dipole parallel to a flat free surface, its reflected image gives the displayed normal vorticity at height $h$. A spherical torque-free body responds with angular velocity ${\omega }_z/2$ and an elongated body also responds to strain. At constant height, self-propulsion plus yaw gives a circular path; the signed moment sets its handedness.

For a sphere of radius $a$ with prescribed surface slip $u^{\prime }$ and zero external torque, the Lorentz reciprocal theorem for Stokes flow giveswhere $n$ points from the sphere into the fluid. Use an auxiliary rotating sphere in Stokes flow with arbitrary angular velocity $\hat{\Omega }$ and traction $-3\mu \hat{\Omega }\times n$. The actual torque-free cross-work vanishes. Substituting the boundary velocity $\Omega \times an+u^{\prime }$ into the other cross-work leaves $-8\pi \mu a^3\hat{\Omega }\cdot \Omega -3\mu \hat{\Omega }\cdot \int n\times u^{\prime }=0$. A possible translation contributes zero because the auxiliary net force is zero.

In an axisymmetric thin disc with no vertical mass or torque flux, integrate mass and conservation of angular momentum over height. With fixed height-independent $l=r^2\Omega$, the equations are ${\partial }_t\Sigma +r^{-1}{\partial }_r(r\Sigma v_r)=0$ and ${\partial }_t(\Sigma l)+r^{-1}{\partial }_r(r\Sigma v_{r}l-r^3\bar{\nu }\Sigma {\Omega }^{\prime })=0$. Their difference gives $r\Sigma v_r{l}^{\prime }=(r^3\bar{\nu }\Sigma {\Omega }^{\prime }{)}^{\prime }$. The density-weighted viscosity of a disk supplies the integrated stress.