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Write the angular dependence as $f(\varphi )={\sum }_{n\ge 1}{a}_n\sin n\varphi +{\sum }_{n\ge 0}{b}_n\cos n\varphi$, so $u^{\prime }=\sin \theta f(\varphi )e_{\varphi }$. In the spherical coordinate system, $n=e_r$ and $e_r\times e_{\varphi }=-e_{\theta }$, withThus the integral in the torque-free rotation of a spherical squirmer has componentsThe vanishing follows from north-south cancellation, including the potentially relevant first azimuthal harmonics. In the $z$ direction, orthogonality of complex exponentials givesConsequently every rotational component is determined:This assumes an integrable prescribed angular function, with the Fourier series convergent sufficiently to justify the surface integrals; a finite Fourier sum is enough for the construction below. Other harmonics can generate exterior flows even though they contribute no net rotation.

For flow-free rotation of a spherical squirmer, choose $b_0\ne 0$ and set every $a_n$ and every $b_n$ with $n\ge 1$ to zero. Then $u^{\prime }=b_0\sin \theta e_{\varphi }$, while the rigid-rotation velocity isThe total laboratory-frame boundary velocity is therefore zero everywhere. With fluid at rest at infinity, Uniqueness of Stokes flow givesA constant pressure supplies zero resultant force and torque, so this is consistent with free motion. The imposed surface actuation counter-rotates relative to the material sphere; no claim is being made that ordinary no-slip rigid rotation produces zero flow.