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Write $r=|\mathbf{x}|$ and take $\mu$ to be the dynamic viscosity. The Papkovich–Neuber representation of a homogeneous Stokes flow uses a harmonic vector potential $\mathbf{\Phi }$ and harmonic scalar potential $\chi$:These formulas give incompressibility because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and give $\mu {\nabla }^2\mathbf{u}=\nabla p$, the Stokes equation away from forcing.

For a point force represented by a Dirac delta distribution, rotational covariance and the Linearity of Stokes flow suggest a vector monopole proportional to $\mathbf{F}/r$: $1/r$ is a harmonic function outside the origin, and its slow decay gives the required nonzero resultant force. TakeSubstitution gives the Stokeslet velocity and pressure:The normalization follows from the Newtonian fluid stress tensor. Differentiating the velocity gives the rate of strain and vorticity of a Stokeslet:Here $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$. Thus ${\sigma }_{ij}=-p{\delta }_{ij}+2\mu e_{ij}=-3(\mathbf{F}\cdot \mathbf{x})x_i{x}_j/(4\pi r^5)$. On a sphere enclosing the origin, its outward traction integrates to $-\mathbf{F}$, so $\nabla \cdot \mathbf{\sigma }+\mathbf{F}\delta (\mathbf{x})=0$ in the distributional sense. All displayed fields away from the point force have $r>0$.

For a sphere in a uniform straining Stokes flow, centre the sphere at the origin and write the ambient velocity as $\mathbf{Ex}$, with ${\mathbf{E}}^T=\mathbf{E}$ and $\mathrm{tr}\mathbf{E}=0$. Faxén's first law and Faxén's rotational law give zero translation and rotation. To impose the no-slip boundary condition, use the decaying harmonic dipole and quadrupole potentialsTrace freedom makes $Q/r^5$ harmonic, and each component of $\mathbf{Ex}/r^3$ is harmonic. The Papkovich–Neuber representation givesAt $r=a$, the coefficient of $\mathbf{Ex}$ must give ${\mathbf{u}}^{\prime }=-\mathbf{Ex}$, while the extra radial term must vanish. Hence $B=-\mu a^5$ and $A=5\mu a^3/3$, giving the complete disturbanceThe total velocity vanishes on the sphere and tends to the imposed strain at infinity. This field has no Stokeslet or rotlet contribution, consistent with a force-free, torque-free sphere. Its leading far field is the radial stresslet $-5a^{3}Q\mathbf{x}/(2r^5)$.

Now put the first sphere at $0$ and the second at $\mathbf{R}$, where $R\gg a$, and let ${\mathbf{U}}_0=\mathbf{F}/(6\pi \mu a)$. The first sphere's leading disturbance is its Stokeslet. The rate-of-strain tensor and vorticity incident on the second sphere areThe second sphere translates with the local flow and rotates with half its local vorticity. Its force-free and torque-free conditions therefore remove any reflected force or torque monopole. The incident strain produces the stresslet just obtained. Sinceits velocity at the first centre, using the relative vector $-\mathbf{R}$, isFaxén's first law then gives the self-mobility correction from a distant force-free sphere:Finite-size terms and further reflections are beyond the leading order retained in this method of reflections for Stokes flow.

Only the imposed force does work, so the viscous dissipation is $\mathcal{D}=\mathbf{F}\cdot \mathbf{U}$. With $\zeta =6\pi \mu a$, define ${\epsilon }_R=a/R$ and $\gamma =15a^4/(4R^6)$. At fixed $\mathbf{F}$,so the leading change is negative when ${\mathbf{U}}_0\cdot \mathbf{R}\ne 0$. The direction is held fixed in taking ${\epsilon }_R\to 0$.

The fixed-force comparison of minimum viscous dissipation distinguishes the prescribed force from prescribed boundary velocity. Indeed,and henceAt the actual reduced velocity, adding the second sphere increases the dissipation relative to an isolated translating sphere at that same velocity, exactly as required by the Minimum-dissipation theorem for Stokes flow. One can extend the two-sphere velocity rigidly into the second sphere, adding zero strain and producing an admissible trial field in the domain without it. The reflected angular velocity of the first sphere contributes only a higher-order squared rotational power to this comparison.

Finally hold the second sphere's angular velocity at zero. Faxén's rotational law requires the applied coupleThe resulting rotlet adds at the first centreThis is the additional rotational constraint correction to sphere mobility. Including the strain reflection, the total leading change relative to the isolated velocity isFor motion parallel to the line of centres the constraint adds nothing at this order; for transverse motion it supplies an extra resisting correction.

Use the Papkovich–Neuber representation in the formwhere $\mathbf{\Phi }$ and $\chi$ are harmonic functions. A torque is a axial vector, so rotational covariance and decay select the harmonic vector fieldHere $\mathbf{x}\cdot \mathbf{\Phi }=0$ and $\nabla \cdot \mathbf{\Phi }=0$, soThis rotlet equals the rotating sphere in Stokes flowwhen $\mathbf{\Omega }=\mathbf{G}/(8\pi \mu a^3)$. It satisfies the no-slip boundary condition on $r=a$, decays at infinity, and its Newtonian fluid stress tensor transmits the applied couple $\mathbf{G}$.

To hold sphere 2 fixed against the incident angular velocity from part ii, its applied couple must generate the bare rotationThe resulting rotlet advects the force-free sphere 1. Since the field is harmonic, Faxén's first law gives

The returned rotlet has velocity $O(\Omega a^6/R^5)$ and rate of strain $O(\Omega a^6/R^6)$ at sphere 1. That strain induces a stresslet of strength $O(\mu \Omega a^9/R^6)$, whose velocity at sphere 2 is $O(\Omega a^9/R^8)$. This method of reflections for Stokes flow gives the stated order of the next correction to ${\mathbf{U}}_2$.

Write $\mathbf{\Omega }={\mathbf{G}}_1/(8\pi \mu a^3)$. The incident rotlet of sphere 1 at sphere 2 isand it is harmonic away from sphere 1. Since sphere 2 is force-free, Faxén's first law therefore givesThe vorticity of the rotlet isBecause ${\mathbf{G}}_2=0$, Faxén's rotational law gives

The symmetric rate-of-strain tensor of the incident rotlet at sphere 2 isAfter 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:Part b gives its vorticity asAt the centre of sphere 1, $\mathbf{x}=-\mathbf{R}$, soApplying Faxén's rotational law to sphere 1 produces half this ambient vorticity and proves

In the preceding two-sphere geometry, holding the second sphere's angular velocity at zero needs a torque $\mathbf{G}=-4\pi \mu a^3{\mathbf{\omega }}_{\infty }$. Its rotlet addsto the first sphere's velocity, in addition to the self-mobility correction from a distant force-free sphere. This leading constraint correction is transverse to the line of centres.

If the otherwise force-free second sphere is held against rotation, its angular velocity relative to the incident fluid is $-3a({\mathbf{U}}_0\times \mathbf{n})/(4R^2)$. The rotlet reflected back to the first sphere addsThis follows by evaluating the rotating sphere in Stokes flow at displacement $-R\mathbf{n}$ and using the vector triple-product identity. The holding couple does no work because the held sphere's angular velocity is zero.

A rotlet dipole is the derivative of a rotlet along the separation of two opposed torques. A torque-free bacterium can have this far field because its rotating flagella and counterrotating body exert spatially separated internal torque reactions. The displayed axial formula uses the observation-coordinate derivative convention and signed torque moment $D$.

A torque-driven sphere advects and rotates a distant force-free sphere through its rotlet. The incident rate of strain makes the second sphere emit a stresslet; the vorticity of that reflected field supplies the first correction to the original sphere's angular velocity.

For a two-mode tensorial squirmer flow of radius $a$ and a passive equal sphere held at $X=Re$, with $a/R\ll 1$, let $s=e\cdot Be$. Faxén's first law gives holding force $F_h=-9\pi \mu a^{3}se/R^2+O(\mu a^4|A|/R^3+\mu a^5\Vert B\Vert /R^4)$. Its reflected Stokeslet changes swimming velocity by $-9a^{3}se/(4R^3)$. Since this leading force is radial, its vorticity at the swimmer vanishes. The next holding force, generated by the potential dipole, yields rotation $a^{4}A\times X/(4R^6)$. Terms from the passive sphere's finite radius, reflected stresslet and rotlet are needed at higher order. These formulas assume fixed slip and no externally applied swimmer force or torque.

With body-outward normal $n$, put $F=-\int \sigma ndS$, ${\Sigma }_{ij}=\int x_i(\sigma n{)}_{j}dS$, $U_{ij}=\int u_i{n}_{j}dS$, and $M=\Sigma -2\mu U$. Taylor expansion of the exterior boundary integral representation of Stokes flow gives a Stokeslet from $F$, a rotlet from $G_j=-{ϵ}_{jkl}{\Sigma }_{kl}$, a source from $Q=\mathrm{Tr}U$, and a stresslet from $S=(M+M^T)/2-I\mathrm{Tr}M/3$. Thus $F$ and $G$ are the force and torque exerted on the fluid, while $Q$ is the instantaneous displaced-volume rate. A volume-preserving deforming body has $Q=0$.