= Solution
Put $r=|\mathbf x|$ and $s=\mathbf F\cdot\mathbf x$. In the <Unscaled Papkovich–Neuber representation>, take the harmonic potentials $\boldsymbol\Phi=-\mathbf F/(8\pi\mu r)$ and $\chi=0$. Differentiating $s/r$ gives the <Stokeslet> <velocity> and its <pressure>:
$$
\boxed{\mathbf u=\frac1{8\pi\mu}\left(\frac{\mathbf F}{r}+\frac{s\mathbf x}{r^3}\right)},\qquad p=\frac{s}{4\pi r^3}.
$$
Indeed, writing $K=1/(8\pi\mu)$,
$$
\partial_j u_i=K\left[-\frac{F_i x_j}{r^3}+\frac{F_j x_i}{r^3}+\frac{s\delta_{ij}}{r^3}-\frac{3s x_i x_j}{r^5}\right].
$$
The antisymmetric first two terms cancel in the <rate-of-strain tensor>, whereas their <curl> supplies the <vorticity>. Thus
$$
\boxed{e_{ij}=\frac{s}{8\pi\mu r^3}\left(\delta_{ij}-\frac{3x_i x_j}{r^2}\right),\qquad \boldsymbol\omega=\frac{\mathbf F\times\mathbf x}{4\pi\mu r^3}}.
$$
The <rate-of-strain tensor> has zero trace, as required by <incompressibility>. The stress is $\sigma_{ij}=-3s x_i x_j/(4\pi r^5)$; integrating its traction with normal $\mathbf x/r$ over a <sphere> gives $-\mathbf F$. This fixes the <force> normalization of the <Stokeslet>.
For a <sphere in a uniform straining Stokes flow>, write the ambient <velocity> as $\mathbf E\mathbf x$, where $\mathbf E$ is symmetric and trace-free, and set $q=\mathbf x\cdot\mathbf E\mathbf x$. Symmetry gives zero translation and rotation of the centred <force-free>, <torque-free> <sphere>. Seek a decaying disturbance using the <Unscaled Papkovich–Neuber representation> with
$$
\boldsymbol\Phi=C a^3\frac{\mathbf E\mathbf x}{r^3},\qquad \chi=D a^5\frac{q}{r^5}.
$$
The components of $\mathbf E\mathbf x/r^3$ are derivatives of $1/r$, and $q/r^5$ is a trace-free harmonic quadrupole, so both potentials satisfy <Laplace's equation> outside the <sphere>. Direct differentiation yields
$$
\mathbf u'=-3Ca^3\frac{q\mathbf x}{r^5}+2Da^5\frac{\mathbf E\mathbf x}{r^5}-5Da^5\frac{q\mathbf x}{r^7}.
$$
The <no-slip boundary condition> at $r=a$ requires the coefficient of $\mathbf E\mathbf x$ to be $-1$ and that of $q\mathbf x$ to vanish: $2D=-1$, $-3C-5D=0$. Hence $D=-1/2$, $C=5/6$, and
$$
\boxed{\mathbf u'=-\frac{a^5}{r^5}\mathbf E\mathbf x-\frac{5a^3}{2r^5}\left(1-\frac{a^2}{r^2}\right)q\mathbf x},\qquad p'=-\frac{5\mu a^3q}{r^5}.
$$
This disturbance cancels the ambient <velocity> at $r=a$, decays at infinity, and has a leading <stresslet> proportional to $a^3\mathbf E$.
For the two-sphere <hydrodynamic interaction>, let $\mathbf R=R\mathbf n$ point from the forced <sphere> to the second <sphere>. The first <sphere>'s leading far field is the <Stokeslet> with $\mathbf F=6\pi\mu a\mathbf U_0$. Its incident <rate-of-strain tensor> at the second <sphere> is
$$
\mathbf E_2=\frac{3a}{4R^3}(\mathbf U_0\cdot\mathbf R)(\mathbf I-3\mathbf n\mathbf n),\qquad \mathbf E_2:\mathbf R\mathbf R=-\frac{3a}{2R}(\mathbf U_0\cdot\mathbf R).
$$
The <force-free> <sphere> translates with the local uniform incident flow, and the <torque-free> <sphere> rotates with half the incident <vorticity>. Those parts cause no leading <Stokeslet> or <rotlet> disturbance. The remaining incident strain produces the <stresslet> just calculated. At displacement $-\mathbf R$ from the second <sphere> its <velocity> is
$$
\Delta\mathbf U=\frac{5a^3}{2R^5}(\mathbf E_2:\mathbf R\mathbf R)\mathbf R=-\frac{15a^4}{4R^6}(\mathbf U_0\cdot\mathbf R)\mathbf R.
$$
The first <sphere> translates in this returned incident flow. Finite-size sampling and higher incident multipoles enter only beyond the retained reflection order. Thus the <stresslet reflection between two force-free and forced spheres> gives
$$
\boxed{\mathbf U=\mathbf U_0-\frac{15a^4}{4R^6}(\mathbf U_0\cdot\mathbf R)\mathbf R+O\!\left(|\mathbf U_0|(a/R)^5\right)}.
$$
The factor $|\mathbf U_0|$ makes explicit the <velocity> scale implicit in the printed remainder.
The resulting input power, equal to the <viscous dissipation>, is
$$
\mathbf F\cdot\mathbf U=6\pi\mu a\left[|\mathbf U_0|^2-\frac{15a^4}{4R^6}(\mathbf U_0\cdot\mathbf R)^2\right]+\text{higher orders}.
$$
It decreases at fixed <force>. The <Minimum-dissipation theorem for Stokes flow> instead compares admissible fields at fixed first-sphere <velocity>: extending a two-sphere field rigidly into the second <sphere> gives an admissible one-sphere field, so the extra rigid inclusion can only increase the minimum resistance. Allowing the second <sphere> to choose its translation and rotation minimizes over its rigid motions, producing zero <force> and couple there. An increase of this effective <hydrodynamic resistance matrix> decreases its inverse <hydrodynamic mobility matrix>, and hence decreases fixed-force power. \b[The negative mobility correction is therefore consistent with minimum dissipation.]
If the second <sphere> is prevented from rotating, its <angular velocity> relative to the incident fluid becomes
$$
\boldsymbol\Omega_{\mathrm{rel}}=-\frac12\boldsymbol\omega_1(\mathbf R)=-\frac{3a}{4R^3}\mathbf U_0\times\mathbf R.
$$
Use the <rotating sphere in Stokes flow> at $-\mathbf R$. The additional reflected <rotlet> gives
$$
\Delta\mathbf U_{\mathrm{rot}}=\frac{3a^4}{4R^6}(\mathbf U_0\times\mathbf R)\times\mathbf R=-\frac{3a^4}{4R^4}\left[\mathbf U_0-(\mathbf U_0\cdot\mathbf n)\mathbf n\right].
$$
Adding this <rotation-clamped reflection between two spheres> to the strain reflection,
$$
\boxed{\mathbf U=\mathbf U_0-\frac{a^4}{R^4}\left[\frac{15}{4}(\mathbf U_0\cdot\mathbf n)\mathbf n+\frac34\left(\mathbf U_0-(\mathbf U_0\cdot\mathbf n)\mathbf n\right)\right]+O\!\left(|\mathbf U_0|(a/R)^5\right)}.
$$
The holding couple supplies no power because the held <sphere>'s <angular velocity> is zero.
Back to article page
