= Solution
Write $\boldsymbol\Omega=\mathbf G_1/(8\pi\mu a^3)$. The incident <rotlet> of sphere 1 at sphere 2 is
$$
\mathbf u_\infty(\mathbf R)=a^3\frac{\boldsymbol\Omega\times\mathbf R}{R^3},
$$
and it is harmonic away from sphere 1. Since sphere 2 is <force-free>, <Faxén's first law> therefore gives
$$
\boxed{\mathbf U_2=a^3\frac{\boldsymbol\Omega\times\mathbf R}{R^3}.}
$$
The <vorticity> of the rotlet is
$$
\boldsymbol\omega_\infty(\mathbf R)
=\frac{a^3}{R^3}\left[
3\frac{(\boldsymbol\Omega\cdot\mathbf R)\mathbf R}{R^2}
-\boldsymbol\Omega
\right].
$$
Because $\mathbf G_2=0$, <Faxén's rotational law> gives
$$
\boxed{\boldsymbol\Omega_2
=\frac{a^3}{2R^3}\left[
3\frac{(\boldsymbol\Omega\cdot\mathbf R)\mathbf R}{R^2}
-\boldsymbol\Omega
\right].}
$$
The symmetric <rate-of-strain tensor> of the incident rotlet at sphere 2 is
$$
\mathbf E=-\frac{3a^3}{2R^5}\left[
(\boldsymbol\Omega\times\mathbf R)\mathbf R
+\mathbf R(\boldsymbol\Omega\times\mathbf R)
\right].
$$
After translation and rotation have matched the uniform and antisymmetric parts of the incident flow, the leading perturbation from sphere 2 is the <stresslet> part of the supplied straining-sphere solution:
$$
\boxed{\mathbf u_2'(\mathbf x)
\sim-\frac{5a^3}{2}\frac{(\mathbf x\cdot\mathbf E\cdot\mathbf x)\mathbf x}{r^5}.}
$$
Part b gives its vorticity as
$$
\boldsymbol\omega_2'=5a^3\frac{\mathbf x\times(\mathbf E\cdot\mathbf x)}{r^5}.
$$
At the centre of sphere 1, $\mathbf x=-\mathbf R$, so
$$
\boldsymbol\omega_2'(-\mathbf R)
=-\frac{15a^6}{2R^6}\left[
\boldsymbol\Omega-\frac{(\boldsymbol\Omega\cdot\mathbf R)\mathbf R}{R^2}
\right].
$$
Applying <Faxén's rotational law> to sphere 1 produces half this ambient vorticity and proves
$$
\boxed{
\boldsymbol\Omega_1-\boldsymbol\Omega
=-\frac{15a^6}{4R^6}\left[
\boldsymbol\Omega-\frac{(\boldsymbol\Omega\cdot\mathbf R)\mathbf R}{R^2}
\right].}
$$
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