= Solution
To the requested order, sphere 1 has velocity $\mathbf V$ and creates the translating-sphere field from part a. At the centre of sphere 2, a vector distance $\mathbf R$ away, this incident field is
$$
\mathbf u_\infty(\mathbf R)=
\left[
\frac{3a}{4R}(\mathbf I+\widehat{\mathbf R}\widehat{\mathbf R})
+\frac{a^3}{4R^3}(\mathbf I-3\widehat{\mathbf R}\widehat{\mathbf R})
\right]\mathbf V.
$$
Sphere 2 is force free, so <Faxén's first law> gives
$$
\mathbf U_2=\mathbf u_\infty(\mathbf R)
+\frac{a^2}{6}\nabla^2\mathbf u_\infty(\mathbf R).
$$
The source-dipole term is harmonic, while
$$
\nabla^2\left[
\frac1R(\mathbf I+\widehat{\mathbf R}\widehat{\mathbf R})
\right]\mathbf V
=\frac2{R^3}
(\mathbf I-3\widehat{\mathbf R}\widehat{\mathbf R})\mathbf V.
$$
Therefore
$$
\boxed{
\mathbf U_2=\left[
\frac{3a}{4R}(\mathbf I+\widehat{\mathbf R}\widehat{\mathbf R})
+\frac{a^3}{2R^3}(\mathbf I-3\widehat{\mathbf R}\widehat{\mathbf R})
\right]\mathbf V
+O\left(\frac{Va^7}{R^7}\right)}.
$$
This is the two-sphere <Rotne--Prager mobility> through order $a^3/R^3$.
The incident strain at sphere 2 is $O(Va/R^2)$. A force-free sphere in this strain creates a <stresslet> of size $O(\mu a^3Va/R^2)=O(\mu Va^4/R^2)$, whose velocity back at sphere 1 is $O(Va^4/R^4)$. Hence
$$
\boxed{\mathbf U_1-\mathbf V=O(Va^4/R^4)}.
$$
The nearly uniform returned flow merely advects sphere 1 and does not change its fixed Stokeslet strength, because its applied force remains fixed. The next scattered disturbance is therefore generated by the returned velocity gradient, of order $Va^4/R^5$. It induces a stresslet of size $O(\mu Va^7/R^5)$ at sphere 1 and hence velocity $O(Va^7/R^7)$ at sphere 2. This <method of reflections for Stokes flow> explains both the absence of an $O(Va^5/R^5)$ term and the next order $O(Va^7/R^7)$.
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