= Solution
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 is
$$
u|_S=\Omega\times an+u'.
$$
Free rotation means the actual body is <torque-free>: the total <hydrodynamic torque> $T=\int_San\times(\sigma\cdot n)dS$ vanishes. Use the auxiliary <rotating sphere in Stokes flow> with arbitrary $\widehat\Omega$, boundary velocity $\widehat u=\widehat\Omega\times an$ and the given <traction> $\widehat\sigma\cdot n=-3\mu\widehat\Omega\times n$. 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> is
$$
\int_S\widehat u\cdot\sigma\cdot n\,dS
=\widehat\Omega\cdot T=0.
$$
The other cross-work becomes
$$
0=-3\mu\int_S(\Omega\times an+u')\cdot(\widehat\Omega\times n)dS.
$$
The spherical integral identities are
$$
\int_S n_i n_j\,dS=\frac{4\pi a^2}{3}\delta_{ij},\qquad
\int_SdS=4\pi a^2.
$$
Using $(\Omega\times n)\cdot(\widehat\Omega\times n)=\Omega\cdot\widehat\Omega-(\Omega\cdot n)(\widehat\Omega\cdot n)$, the rotational contribution is $-8\pi\mu a^3\Omega\cdot\widehat\Omega$. The slip contribution is $-3\mu\widehat\Omega\cdot\int_S n\times u'\,dS$. Since $\widehat\Omega$ is arbitrary, the <torque-free rotation of a spherical squirmer> satisfies
$$
\boxed{\Omega=-\frac{3}{8\pi a^3}\int_S n\times u'\,dS.}
$$
The 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, $\int_S\widehat\sigma\cdot n\,dS=0$. Thus translation does not alter this rotational formula. Without the torque-free assumption an external-torque contribution would have to be retained.
Back to article page
