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For a smooth passing encounter of a force-driven sphere with a fixed sphere, isotropy and reflection symmetry imply $\mathbf{U}=[m_{\perp }(R)I+(m_{\parallel }(R)-m_{\perp }(R))\mathbf{nn}]\mathbf{F}$. Along a force axis $X$, $U_X$ is even and the transverse velocity is odd in $X$. Uniqueness then makes the trajectory fore-aft symmetric and restores the incoming impact parameter downstream. This consequence of kinematic reversibility of Stokes flow remains valid with lubrication resistance, provided the particles remain smooth, noncontacting and subject only to the stated forces. Surface roughness and contact can invalidate these assumptions.

For $x=ϵa\cos \omega t$ and a tangent $(\sin \theta ,\cos \theta )$ with $\theta =ϵ\cos (\omega t+\varphi )$, the leading force on the fluid is $⟨F_y⟩=-(c_{\perp }-c_{\parallel })L{ϵ}^{2}a\omega \sin \varphi /2$. The sign changes under reversal of the tilt convention. Quadrature strokes maximize its magnitude; reciprocal strokes give zero, consistent with kinematic reversibility of Stokes flow.

The Reynolds number compares convective acceleration with viscous acceleration in the Navier-Stokes equation:At large Reynolds number, inertia dominates in the bulk; viscous boundary layers can still control wall shear stress, separation and drag. Turbulence is possible but does not follow from large Reynolds number alone. At small Reynolds number, viscous stress balances the pressure gradient and applied body forces. In the quasi-steady Stokes approximation, the velocity field responds instantaneously to the boundary motion and has kinematic reversibility of Stokes flow. Small Reynolds number alone does not suppress rapid local acceleration.

For a swimmer, take a characteristic length $L\sim 1\mathrm{m}$ and speed $U\sim 1\mathrm{m}{\mathrm{s}}^{-1}$. The supplied kinematic viscosity givesHuman swimming is therefore an inertia-dominated fluid flow, rather than Stokes flow.

Choose the axial origin at the sphere's centre. Reflection $x\mapsto -x$ leaves the straight tube, the spherical geometry and the transverse displacement of the centre unchanged. It reverses the prescribed axial Hagen-Poiseuille flow, while it leaves a hypothetical transverse particle velocity unchanged. On the other hand, kinematic reversibility of Stokes flow says that reversing the imposed flow reverses every particle velocity, including that transverse component. The force-free and torque-free constraints are preserved by both operations. Uniqueness of the coupled Stokes flow and rigid-body motion therefore makes the transverse velocity equal to its own negative. It must vanish.

Consequently $\dot{c}=0$ as the particle is advected. This is no lateral migration of a single sphere in a uniform tube in Stokes flow. The argument uses the fore-aft symmetry of a rigid sphere and a uniform tube; it does not extend automatically to inertial migration or deformable particles.

Work in the laboratory frame, with fluid at rest at infinity and the instantaneous sphere centre at the origin. In the Unscaled Papkovich–Neuber representation, take the harmonic potentialsDifferentiating the two terms gives the translating sphere in Stokes flow:Here $\mathbf{n}=\mathbf{x}/r$. The potentials are harmonic outside the sphere, so they satisfy the Stokes equations. At $r=a$, the two coefficients of $\mathbf{nn}$ cancel and the coefficient of $I$ is one, proving the no-slip boundary condition $\mathbf{u}=\mathbf{V}$; the field decays at infinity. Its leading part is a Stokeslet of force $6\pi \mu a\mathbf{V}$ exerted on the fluid.

Let $\hat{\mathbf{z}}$ point upwards, $\delta =V_1-V_2$, and $D=a_2{V}_2-a_1{V}_1$. Each isolated sphere moves with velocity $-V_i\hat{\mathbf{z}}$. To leading order its correction is the other's incident Stokeslet velocity:The forces generating these fields are the fixed sedimenting loads, so their leading strengths are set by the isolated speeds. Subtraction yields the leading interaction of two sedimenting spheres:Since $\hat{\mathbf{z}}=\cos \theta \hat{\mathbf{r}}-\sin \theta \hat{\mathbf{\theta }}$, this reduces toThe stipulated higher-order errors can be appended to both equations; they do not enter this leading calculation.

For the same-sign case, interchange labels if necessary to arrange $\delta >0,D>0$, and put $c=3D/4$. The vertical separation $z=r\cos \theta$ satisfiesIf sphere one is initially above sphere two, its vertical order reverses; in every case $z\to -\infty$, implying $r\to \infty$. For a noncollinear trajectory there is no collision in the point-force equations. Indeed, divide the two polar equations and integrate:At side-by-side passage, $\theta =\pi /2$, the closest separation is the positive root of $\delta r^2+2cr=C$. Before passage $\dot{r}<0$ and after passage $\dot{r}>0$, proving overtaking and unbounded subsequent separation. This passing invariant for unequal point-force spheres describes a physical far-field passage only when the closest separation also greatly exceeds the sphere radii. Exactly collinear approach instead reaches the near-contact regime, where this leading approximation cannot describe two finite spheres passing through one another.

For opposite signs, choose $\delta >0,D<0$ and consider the invariant configuration with sphere one vertically above sphere two, $\theta =0$. The radial equation isIf $r>r_{\ast }$, separation decreases; if $r<r_{\ast }$, it increases. Thus it approaches this equilibrium along the vertical line. Linearization gives $\dot{\delta r}=-(\delta /r_{\ast })\delta r$. But a small tilt obeys $\dot{\theta }=(\delta /(2r_{\ast }))\theta$, so the vertical bound pair in the point-force sedimentation model is stable only against vertical perturbations and is a saddle in the full separation dynamics. The opposite vertical orientation has the reversed stability. Reversing gravity reverses both $\delta$ and $D$, leaves $r_{\ast }$ unchanged, and reverses every trajectory; radial attraction becomes repulsion. This is fully consistent with kinematic reversibility of Stokes flow.

There is an important restriction in the printed data. Its hierarchy $|D|\ll (a_1+a_2)|\delta |$ implies $r_{\ast }\ll a_1+a_2$, so the proposed equilibrium would overlap the physical spheres and lie outside the well-separated approximation. With that hierarchy there is no justified physical far-field bound pair. The intended far-field equilibrium instead requires the first inequality reversed, $|D|\gg (a_1+a_2)|\delta |$. The formal equilibrium and its vertical stability above show the intended mechanism, while also identifying the missing validity and transverse-stability qualifications.

Choose time $t_c$ when the passing sphere centres are at the same height, and let $S$ reflect positions in that horizontal plane. For each instantaneous configuration, Stokes flow determines particle velocities uniquely and linearly from the fixed external forces. Time reversal reverses all velocities and the dense sphere's applied force; reflecting in the horizontal plane reverses that force again. The neutral sphere remains force-free under both operations. Thus the reflected time-reversed motion obeys the same original forcing.

At $t_c$, reflection fixes both particle centres. Since the spheres have no orientational shape variables and there is no contact, uniqueness of the particle-trajectory initial-value problem makes the reflected backward motion coincide with the forward motion:This reflection symmetry of a two-sphere passing trajectory fixes the transverse coordinates and reverses the relative axial separation. Consequently, at matching large separations above and below, both spheres return to their original distances from the centreline. Their intermediate lateral excursions are undone. This uses kinematic reversibility of Stokes flow, geometric reflection symmetry and the noncontact result, not a dissipation-minimizing trajectory. It assumes the spheres pass one another; a perfectly collinear approach would not supply the side-by-side symmetry event.

For the usual anisotropy $c_{\perp }>c_{\parallel }$ and $a>0$, the maximum positive mean transverse force from a rocking rod occurs atThe maximum magnitude has either quadrature phase $\varphi =\pm \pi /2$. If positive tilt is defined towards $-x$, the positive-force optimum is instead $+\pi /2$. At our optimum $\theta =ϵ\sin \omega t$ is opposite in sign to $\dot{x}$: the rod tilts so that the weaker tangential drag and stronger normal drag combine into a positive $y$ force on both strokes.

For $\varphi =0$ or $\pi$, orientation is a single-valued function of the centre displacement. The configuration retraces its path under time reversal, and kinematic reversibility of Stokes flow makes the force impulse cancel. Equivalently, $⟨\theta \dot{x}⟩=0$. The enclosed loop in the displacement-orientation plane is largest in quadrature and vanishes for a reciprocal stroke, illustrating the principle behind the scallop theorem.

The mobility correction from a fixed distant sphere givesTaking their ratio and keeping the first nonzero transverse correction yieldsto the stated leading order. Set $b=Y_{\infty }\gg a$. The deflection is $O(a^2/b)$, so replacing $Y$ by $b$ on the right introduces only higher-order errors. Integrating from $X=-\infty$ givesThe maximum occurs at $X=0$, and the deflection and spin in a distant sphere encounter areFor the rotation, use $dt=dX/V$ and $Y=b$ to leading order in the angular velocity from part (b). Its signed angle about the positive $z$ axis isThus the rotation is clockwise when viewed from positive $z$, with the magnitude of the displayed leading term.

The deflection tends back to zero downstream: $Y(+\infty )=Y_{\infty }$. More generally, kinematic reversibility of Stokes flow combined with reflection in the plane $X=0$ makes a passing trajectory fore-aft symmetric. One can see this without using the distant-sphere approximation: the relevant translational hydrodynamic mobility matrix has the form $m_{\perp }(R)I+[m_{\parallel }(R)-m_{\perp }(R)]\mathbf{nn}$. Hence $\dot{X}$ is even in $X$ and $\dot{Y}$ is odd in $X$. Uniqueness of the trajectory through $X=0$ then gives $Y(X)=Y(-X)$.

For $Y_{\infty }=a/100$, the numerical deflection and spin approximations above are invalid: the encounter enters a narrow gap and requires lubrication theory. However, the same return to the incoming offset holds for an ideal passing encounter of perfectly smooth spheres in Stokes flow. Lubrication resistance prevents finite-time contact under a bounded force, and does not itself destroy kinematic reversibility of Stokes flow. Contact, surface roughness or nonhydrodynamic forces could change that conclusion; they are additional physics, not part of the ideal model. This distinction is the fore-aft symmetry of a sedimenting-sphere encounter.

For each incompressible Stokes flow without body force, ${\partial }_j{\sigma }_{ij}=0$ andThe divergence theorem givesIncompressibility removes the pressure term, antisymmetric velocity-gradient parts do not contract with the symmetric stress, and the remaining integrand is $2\mu e_{ij}^{(1)}{e}_{ij}^{(2)}$, symmetric under $1\leftrightarrow 2$. This proves the Lorentz reciprocal theorem for Stokes flow:

By Linearity of Stokes flow, the drag on a translating body is linear in its velocity, so $F_i=A_{ij}{U}_j$. Apply the reciprocal theorem to two translation solutions with velocities ${\mathbf{U}}^{(1)}$ and ${\mathbf{U}}^{(2)}$. Their surface velocities are constant and the far-field contribution vanishes, yieldingSince the two vectors are arbitrary,Thus the hydrodynamic resistance matrix is symmetric and, for fixed fluid viscosity, depends only on body geometry.

A straight rod has fore-aft symmetry about its centre. Gravity acts through that centre, so the translational-rotational coupling and applied torque vanish. Kinematic reversibility of Stokes flow and Uniqueness of Stokes flow then rule out spontaneous rotation: the inclination $\theta$ remains constant.

Let the speed for perpendicular translation be $U_0$. The given factor of two makes the parallel speed $2U_0$ for the same force component. Resolving gravity parallel and perpendicular to the rod givesResolving these velocities horizontally and vertically givesTherefore

For an incompressible Newtonian fluid with negligible inertia and no body force, the Stokes equations areThey are linear because any linear combination of solutions corresponds to the same linear combination of boundary data and applied forces. They are reversible because replacing all imposed velocities and forces by their negatives sendsFluid particles therefore retrace their paths under reversal. This is kinematic reversibility of Stokes flow.

Reflect the configuration in a plane perpendicular to the tube axis. The geometry and radial displacement are unchanged, whereas the imposed pressure gradient and every axial velocity reverse. By kinematic reversibility of Stokes flow, a radial migration velocity would also have to reverse; the spatial reflection leaves that component unchanged. Uniqueness of Stokes flow therefore forces it to vanish. Reflection in the meridional plane through the two axes similarly excludes azimuthal drift, so $c$ remains constant.

The Reynolds number compares inertial and viscous terms in the momentum equation:Here $\rho$ is fluid mass density, $U$ a characteristic speed, $L$ a swimmer or flow length, and $\eta$ the dynamic viscosity. The advective inertia scale is $\rho U^2/L$ and the viscous force-density scale is $\eta U/L^2$, whose ratio is Re. For an illustrative micron-scale swimmer with $L=2\mu \mathrm{m}$ and $U=20\mu \mathrm{m}{\mathrm{s}}^{-1}$ in water, taking $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$ and $\eta =10^{-3}\mathrm{Pa}\mathrm{s}$ gives $\mathrm{Re}\sim 4\times 10^{-5}$.

At such low Re, Stokes flow is the useful approximation. For an incompressible Newtonian fluid it obeys $-\nabla p+\eta {\nabla }^2\mathbf{u}=0$ and $\nabla \cdot \mathbf{u}=0$, with negligible inertia. Swimming is not a sequence of propulsive impulses followed by coasting: drag balances propulsive forces almost instantaneously. For example, the drag on a slowly translating sphere is $6\pi \eta aU$. A freely swimming organism has approximately zero total external force and torque; its body motion balances the forces and torques generated by its appendages.

The linear equations have kinematic reversibility of Stokes flow: reversing the boundary velocities reverses the fluid motion. A reciprocal stroke that retraces exactly the same sequence of shapes therefore retraces its displacement. This is the scallop theorem. A single hinge opening and closing along the same shape path cannot propel an ideal swimmer through an otherwise quiescent Newtonian Stokes fluid, regardless of merely changing how quickly it opens relative to closing.

A useful design has at least two shape coordinates driven out of phase, so the stroke follows a loop rather than a retraced line in shape space. In the three-bead example the two link lengths change in a cycle; hydrodynamic interactions differ at different points of the cycle, allowing net translation. The drawing illustrates this mechanism without assigning a particular swimming speed.

Biological routes include continuous rotation of a chiral helical flagellum, nonreciprocal ciliary strokes, and traveling bending waves along a flagellum. An elongated filament generally has greater drag for motion normal to itself than along itself; this drag anisotropy helps convert those motions into propulsion. The body may counterrotate to balance appendage torque. Micro-organisms must exploit nonreciprocal shape changes, rotation or another mechanism consistent with viscous force balance; inertia-based swimming intuition fails. Viscoelastic surroundings or appreciable inertia can invalidate the particular Newtonian assumptions behind the simple scallop result.

Stokes flow is the limit of fluid mechanics in which inertia is negligible: pressure gradients, viscosity and any body forces balance at each instant. A small Reynolds number can arise from a small object, a slow velocity, or a large dynamic viscosity. A microscopic sphere moving through water is a useful example: the surrounding fluid continually dissipates the supplied power through viscous dissipation.

The picture is a meridional slice of the three-dimensional axisymmetric flow around a sphere, with the sphere held fixed and the distant fluid moving rightward at speed $U$. The no-slip boundary condition makes both components of velocity vanish on the surface. The streamlines have fore-aft reflection symmetry; reversing the incident velocity reverses their arrows. The underlying field is the translating sphere in Stokes flow solution viewed in the stationary-sphere frame.

For an incompressible flow of a Newtonian fluid with constant dynamic viscosity $\mu$, the steady Stokes equation isHere $p$ is pressure, $\mathbf{u}$ is velocity, and $\mathbf{f}$ is body force per unit volume. The pressure gradient enforces incompressible flow; it need not be small just because inertia is small. At speed $U$ and length $L$, the motion-induced pressure gradient has typical scale $\mu U/L^2$, corresponding to a disturbance pressure difference of order $\mu U/L$. See Shreyas Mandre's fluid dynamics notes, sections 9.1.1–9.1.2.

The controlling Reynolds number iswhere $\rho$ is density and $\nu$ is kinematic viscosity. The condition ${\mathrm{Re}}_L\ll 1$ suppresses advection of momentum on the length $L$. Rapid changes require an additional check of the viscous diffusion time: small Reynolds number alone does not eliminate the time derivative in the Unsteady Stokes equation.

Two rewards of this limit are the exact Stokes's law for the drag on a sphere and kinematic reversibility of Stokes flow. The first predicts settling velocities and connects to the Stokes–Einstein relation; the second explains why a microswimmer needs a nonreciprocal stroke, as expressed by the scallop theorem. Despite viscous dissipation, reversing the driving can retrace the motion: an especially striking consequence of Linearity of Stokes flow.

George Gabriel Stokes obtained the sphere drag and its terminal velocity in his 1851 investigation of how viscosity changes physical pendulum motion. He also used the slow fall of small water droplets to discuss clouds. The paper was read on 9 December 1850; 1851 is its publication date. See Stokes, On the effect of the internal friction of fluids on the motion of pendulums, original 1851 printing, especially articles 42–43; the formulas are also accessible in the collected-paper reprint, articles 42–43.

If $\delta D<0$, the leading interaction of two sedimenting spheres has a formal vertical equilibrium at $r_{\ast }=-3D/(2\delta )$. Choose labels with $\delta >0,D<0$ and put sphere one above sphere two. On that invariant vertical line, $\dot{r}=-\delta +\delta r_{\ast }/r$, so both sides approach $r_{\ast }$ and the radial eigenvalue is $-\delta /r_{\ast }$. A small tilt instead grows at rate $\delta /(2r_{\ast })$: the full configuration-space equilibrium is a saddle, not an attracting node. A physically valid far-field bound pair requires $r_{\ast }\gg a_1+a_2$, or $|D|\gg (a_1+a_2)|\delta |$. Reversing both driving forces reverses the vector field, interchanging radial attraction and repulsion, consistent with kinematic reversibility of Stokes flow.