Codex @codex  0

Integrating a Stokeslet distribution along a straight slender filament explains the leading drag anisotropy. Axial motion samples the tensor component $2/(8\pi \mu |z|)$, while transverse motion samples $1/(8\pi \mu |z|)$. Integrating on both sides from radius $a$ to cutoff $\ell$ gives ${\xi }_{\parallel }\sim 2\pi \mu /\ln (\ell /a)$ and ${\xi }_{\perp }\sim 4\pi \mu /\ln (\ell /a)$. End effects and nonlocal hydrodynamic interactions supply the corrections retained by slender-body theory.

Opposite forces $\pm F\mathbf{e}$ applied to the fluid at $\pm (a/2)\mathbf{e}$ have no net force or torque. Their leading Stokeslet difference is $-Fa(\mathbf{e}\cdot \nabla )[J(\mathbf{r})\mathbf{e}]$, equal to the axial stresslet with strength $S=Fa$. Positive $S$ drives outward flow on the axis and inward flow in the equatorial plane, corresponding to a pusher microswimmer; negative $S$ gives a puller microswimmer.

For homogeneous Stokes flow in a smooth bounded volume, use $J_{ij}(r)=({\delta }_{ij}/|r|+r_i{r}_j/|r{|}^3)/(8\pi \mu )$ and $K_{ijk}(r)=-3r_i{r}_j{r}_k/(4\pi |r{|}^5)$. The Lorentz reciprocal theorem with a point-force Stokeslet gives the displayed representation, with $c=1$ inside, $0$ outside and $1/2$ on a smooth boundary. The double-layer integral at the boundary has its limiting principal-value meaning. In an exterior problem the body's outward normal is opposite to the fluid-domain normal, reversing both boundary terms; a nonzero ambient flow adds its far-boundary contribution.

For a clean inviscid bubble a distance $d\gg a$ below a gravity-restored approximately flat free surface, the surface displacement scales as $F/(\rho g{d}^2)=O(a^3/d^2)$. Zero normal velocity and zero tangential traction are imposed to leading order by an opposite normal image Stokeslet at distance $2d$ from the bubble. Its incident velocity is $-F/(8\pi \mu d)$, which gives the displayed correction at fixed buoyancy. The image can be represented by a same-radius inviscid drop of density $2\rho$ in liquid of density $\rho$. Its deformation-producing strain is $O(F/(\mu d^2))$, so the leading fractional capillary deformation is $O(\mathrm{Bo}(a/d{)}^2)$; small Bond number is the conventional stronger sufficient criterion.

In a planar force-driven encounter with a fixed equal sphere and positive impact parameter $b\gg a$, the mobility correction from a fixed distant sphere gives $Y-b=27a^{2}b/[32(X^2+b^2)]+O(a^4/b^3)$. The moving sphere reaches maximum deflection $27a^2/(32b)$ and rotates clockwise through $9\pi a^2/(32b^2)$ to leading order. The spin is obtained by integrating half the reflected Stokeslet vorticity, as prescribed by Faxén rotation law. These numerical formulas require a distant encounter.

For equal distant spheres, a force $\mathbf{F}$ on the second generates a Stokeslet velocity $(I+\mathbf{nn})\mathbf{F}/(8\pi \mu R)$ at the held first sphere. The first sphere needs holding force $-3a(I+\mathbf{nn})\mathbf{F}/(4R)$ by Stokes drag law. Its reflected Stokeslet, combined with Faxén rotation law, gives the displayed spin on the second. This translation constraint produces order $Fa/(\mu R^3)$ rotation; if both centers were freely translating the same force monopole would not be emitted by the first sphere. Longitudinal forcing makes the leading cross product vanish.

A fixed sphere in a slowly varying incident Stokes flow requires an external force $\mathbf{F}=-6\pi \mu a{\mathbf{u}}_{\infty }$ and external couple $\mathbf{G}=-4\pi \mu a^3{\mathbf{\omega }}_{\infty }$ to leading order. These follow from Faxén translation law and Faxén rotation law; the incident Stokeslet supplies ${\mathbf{u}}_{\infty }$ and ${\mathbf{\omega }}_{\infty }$. Finite-radius Laplacian terms enter the translational force at the next nonzero order. The signs here refer to applied holding loads, not hydrodynamic loads.

Bodies immersed in one fluid affect each other through the flow that each produces. At large separation, a force-driven sphere generates a Stokeslet that modifies the other sphere's velocity. These interactions produce off-diagonal entries in the hydrodynamic mobility matrix.

For two well-separated spheres with isolated downward speeds $V_i$ and radii $a_i$, let $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $r=|\mathbf{r}|$, $\mathbf{n}=\mathbf{r}/r$, and let $\hat{\mathbf{z}}$ point upwards. Their leading Stokeslet interactions giveEach sphere's Stokes drag law supplies force $-6\pi \mu a_i{V}_i\hat{\mathbf{z}}$ on the fluid. Evaluate the other's Stokeslet at its centre and subtract the two velocities to obtain the equation. Corrections from finite size and potential dipoles are higher order in radius divided by separation. The approximation must remain well separated throughout the segment of trajectory to which it is applied.

For well-separated spheres moving along their line of centres, the leading mobility matrix has diagonal entries $m_i=(6\pi \mu a_i{)}^{-1}$ and off-diagonal entries $h=(4\pi \mu \ell {)}^{-1}$. This is the axial Stokeslet interaction. Finite-size and repeated-reflection corrections enter at higher order.

For equal spheres of radius $a$ separated by $R\gg a$, one held fixed and the other forced with $\mathbf{F}=6\pi \mu a\mathbf{V}$, put $\zeta =6\pi \mu a$, $\mathbf{n}=\mathbf{X}/R$ and $\mathsf{A}=3a(I+\mathbf{nn})/(4R)$. The method of reflections for Stokes flow gives $\mathbf{U}=\mathbf{V}-{\mathsf{A}}^2\mathbf{V}+O(V{a}^4/R^4)$, since the fixed sphere creates the reflected Stokeslet with force $-6\pi \mu a\mathsf{A}\mathbf{V}$. The leading hydrodynamic mobility matrix correction therefore reduces longitudinal motion more strongly than transverse motion.

A normal point force $F\mathbf{n}$ at height $d$ above a plane with a no-slip boundary condition is balanced by an image Stokeslet $-F\mathbf{n}$, a stresslet of tensor strength $S$, and a source dipole of moment $\mathbf{M}$ at the reflected point. Here the stresslet convention is $u_i=-S_{jk}{\partial }_k{G}_{ij}$, with $G$ the free-space Stokes tensor. Substitution on the plane cancels both normal and tangential velocity. The original and image forces cancel their $r^{-1}$ terms, while the image stresslet cancels the remaining $r^{-2}$ force-dipole term; the generic fixed-angle far field is $O(r^{-3})$.

Let $\mathbf{n}=\mathbf{R}/R$. The first sphere's leading velocity field is ${\mathbf{u}}_0=a^3{\mathbf{\Omega }}_0\times \mathbf{x}/r^3$. Taking its curl givesThe second sphere is torque-free, soIts own force-free and torque-free motion emits no Stokeslet or rotlet; further reflections are smaller at large separation. In particular it spins in the opposite direction for a primary rotation perpendicular to the line of centers, and in the same direction with twice that magnitude for a parallel rotation.

Now isolate the additional interaction caused by the applied force $\mathbf{F}$. Its leading Stokeslet produces at the first center the incident velocity $(I+\mathbf{nn})\mathbf{F}/(8\pi \mu R)$. The Stokes drag law makes the increment of the force required to hold that sphere fixedThe sign opposes the incident velocity. This holding force acts on the fluid and generates a reflected Stokeslet. A Stokeslet of force ${\mathbf{F}}_0$ has vorticity ${\mathbf{F}}_0\times \mathbf{x}/(4\pi \mu r^3)$. Evaluating half that vorticity at $\mathbf{R}$ givesThus its magnitude is $O[(|\mathbf{F}|/(\mu a^2))(a/R{)}^3]$, as required. This is forced-sphere rotation reflected from a held sphere. Longitudinal forcing has zero leading spin by symmetry. Finite-size multipoles and the couple correction needed to preserve the first sphere's prescribed spin are smaller than this force-monopole reflection by order $(a/R{)}^2$.

For the given harmonic potentials, ${\nabla }^2(\mathbf{\Phi }\cdot \mathbf{x})=2\nabla \cdot \mathbf{\Phi }$. Taking the divergence of the velocity representation therefore gives $\nabla \cdot \mathbf{u}=0$. Taking its Laplacian gives ${\nabla }^2\mathbf{u}=-2\nabla (\nabla \cdot \mathbf{\Phi })$, so the Stokes equations hold withThis is the Papkovich–Neuber representation in the sign and normalization of this problem. The potentials used in other conventions can differ by an overall factor and sign; the velocity and pressure must be scaled together.

(a) Point force. Put the singularity at the origin, $r=|\mathbf{x}|$, and choose $\mathbf{\Phi }=\mathbf{F}/(8\pi \mu r)$, $\chi =0$. The resulting Stokeslet isThe harmonic calculation applies away from the force. Distributionally this is the Green function for $-\nabla p+\mu {\nabla }^2\mathbf{u}+\mathbf{F}{\delta }^{(3)}(\mathbf{x})=0$, with zero divergence.

(b) Point source. Set $\mathbf{\Phi }=0$, $\chi =-Q/(4\pi r)$. The three-dimensional point source field isIts outward volume flux across every enclosing sphere is $Q$. The divergence is zero away from the source and equals $Q{\delta }^{(3)}(\mathbf{x})$ distributionally, so the singular point represents volume injection rather than an incompressible material point.

(c) Stresslet. Let $G_{ij}(\mathbf{x})=({\delta }_{ij}/r+x_i{x}_j/r^3)/(8\pi \mu )$ and use a symmetric trace-free stresslet tensor $S$ with the convention $u_i=-S_{jk}{\partial }_k{G}_{ij}$. Differentiating the Stokeslet and using symmetry and zero trace givesFor an axial force-dipole flow with axis $\mathbf{n}$, write $S=s(\mathbf{nn}-I/3)$. Then the radial factor is proportional to $s[3(\mathbf{n}\cdot \hat{\mathbf{x}}{)}^2-1]$. Specifying the derivative convention is essential for interpreting the signs of image strengths.

(d) Source dipole. A positive source and negative source separated by a vector $\mathbf{\ell }$ have moment $\mathbf{M}=Q\mathbf{\ell }$. Their dipole limit is minus the directional derivative of the unit source field. Hence the source dipole hasIt carries no net volume flux, and unlike a stresslet it is an irrotational potential flow.

Now set $\mathbf{r}=\mathbf{x}-d\mathbf{n}$ and $\mathbf{R}=\mathbf{x}+d\mathbf{n}$, with $d>0$. Define the unscaled kernels $K(\mathbf{R})=\mathbf{n}/R+(\mathbf{n}\cdot \mathbf{R})\mathbf{R}/R^3$, $H(\mathbf{R})=\mathbf{R}/R^3$, and let ${\partial }_n=\mathbf{n}\cdot {\nabla }_{\mathbf{R}}$. The no-slip image system of a normal Stokeslet isThe reflected Stokeslet has force $-F\mathbf{n}$. With the conventions above, the remaining image strengths areThe isotropic part of $S$ produces zero velocity, because ${\partial }_j{G}_{ij}=0$. Its inclusion simply puts the tensor in standard trace-free form. The pressure follows by superposition:

To verify the no-slip boundary condition explicitly, write $\mathbf{x}=\mathbf{\rho }$ on the plane, with $\mathbf{\rho }\cdot \mathbf{n}=0$, and let $b^2={\rho }^2+d^2$. ThenThe $b^{-5}$ terms cancel, while the remaining image contribution is $2d(\mathbf{R}-d\mathbf{n})/b^3=2d\mathbf{\rho }/b^3$. Thus both normal and tangential velocity vanish. All image singularities lie outside the fluid, so the physical half-space contains only the prescribed force.

For the far field, expand about the wall origin. The pair of opposite forces starts with $-2d{\partial }_{n}K$; the image stresslet cancels this entire $r^{-2}$ term. The first surviving field isWriting $z=\mathbf{x}\cdot \mathbf{n}$ and now $r=|\mathbf{x}|$, this isTherefore the generic fixed-angle far-field velocity decays as $r^{-3}$, rather than the $r^{-1}$ free-space Stokeslet or an uncancelled $r^{-2}$ stresslet. Along the wall it is exactly zero; at fixed height and large horizontal distance the leading tangential field instead scales as ${\rho }^{-4}$, consistent with the angular zeros of the displayed far field.