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The creep compliance is the strain per unit step stress: if $\sigma (t)={\sigma }_{0}H(t)$, then $\gamma (t)={\sigma }_{0}J(t)$ for $t>0$. A Newtonian fluid obeys $\sigma =\eta \dot{\gamma }$, so under constant stress its initially zero strain grows as $\gamma ={\sigma }_{0}t/\eta$. Therefore

A viscoelastic material stores part of the deformation energy and dissipates part of it. In linear viscoelasticity, a small imposed shear strain produces a stress determined by its history, $\tau (t)={\int }_{-\infty }^{t}G(t-t^{\prime })\dot{\gamma }(t^{\prime })d{t}^{\prime }$. A shear step probes stress relaxation; a stress step probes creep compliance. A purely elastic solid has persistent stress after a held strain, while a Newtonian fluid relaxes that stress and creeps continuously under constant stress.

For a Maxwell model, a spring of shear modulus $G_0$ and a dashpot of viscosity $\eta$ are in series:For the convention $\gamma \propto e^{i\omega t}$,The storage modulus dominates at short times or high frequencies; the loss modulus dominates at long times or low frequencies. The energy dissipated per cycle per volume is $\pi G^{\prime \prime }{\gamma }_0^2$. A Kelvin-Voigt model places spring and dashpot in parallel and instead gives bounded creep $J(t)=(1-e^{-G_{0}t/\eta })/G_0$: a simple illustration of a viscoelastic solid. Real materials can require a spectrum of relaxation times rather than one element.

A small-amplitude oscillatory rheometer measures stress amplitude and phase relative to imposed strain, giving $G^{\prime }=|{\tau }_0/{\gamma }_0|\cos \delta$ and $G^{\prime \prime }=|{\tau }_0/{\gamma }_0|\sin \delta$. First sweep strain amplitude to identify the linear range; control temperature, sample ageing, wall slip and instrument inertia. Frequency-dependent phase lag alone is insufficient if these apparatus effects are unresolved.

The Stokes–Einstein relation connects thermal diffusion with dissipative mobility. For a dilute suspension of isolated rigid spheres of hydrodynamic radius $a$ in an unbounded Newtonian fluid of viscosity $\eta$,where $\zeta$ is translational drag coefficient, $D$ diffusion coefficient and $T$ absolute temperature. Stokes drag supplies $\zeta =6\pi \eta a$ under no-slip, continuum, low-Reynolds-number conditions. The Einstein diffusion relation $D=k_{B}T/\zeta$ is more general than that particular spherical drag law.

A short equilibrium argument explains the connection. Let a weak external potential $U$ exert force $\mathbf{F}=-\nabla U$, so drift velocity is $\mathbf{F}/\zeta$. The number flux is $\mathbf{j}=n\mathbf{F}/\zeta -D\nabla n$. A Boltzmann distribution $n\propto e^{-U/(k_{B}T)}$ has $\nabla n=n\mathbf{F}/(k_{B}T)$; requiring zero equilibrium flux fixes $D=k_{B}T/\zeta$. thermal fluctuations and viscous dissipation are therefore linked, rather than independent adjustable effects. The original theory also relates diffusion to measured random displacements.^[ref]

On times long compared with momentum relaxation, an isotropic diffusing sphere hassingle-particle tracking can measure the slope. Alternatively, dilute single-scattering dynamic light scattering measures a field correlation $g_1(q,t)=e^{-D{q}^{2}t}$; the corresponding intensity-correlation excess is proportional to $e^{-2D{q}^{2}t}$. The scattering wavenumber and the distinction between field and intensity decay must be retained.

A Stokes-Einstein experimental consistency test independently measures temperature and solvent viscosity, uses a narrow distribution of inert spheres, checks decay rate proportional to $q^2$ or displacement variance proportional to time, and extrapolates toward infinite dilution. Subtract drift, account for wall-enhanced drag, and verify that aggregation is absent. Tracking additionally needs spatial calibration and correction for finite exposure and localization precision. At finite concentration, collective diffusion is not automatically the single-particle self-diffusion in the equation. A hydrodynamic radius is not generally identical to a polymer radius of gyration; nonspherical particles, slip or viscoelastic solvents require a different mobility model.

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.

Assume steady fully developed laminar flow of a Newtonian fluid in a long circular tube, with no slip at a rigid wall. The Hagen-Poiseuille equation gives the volume flow ratewhere $R$ is the lumen radius, $\ell$ the tube length, $\eta$ the dynamic viscosity and $\Delta p$ the pressure drop. If $\Delta p$, $\ell$ and $\eta$ remain fixed, thenThus doubling the flow requiresThis estimate neglects pulsation, arterial compliance, entrance losses, the detailed shape of a stenosis and changes in the rest of the circulatory resistance. It treats blood as a homogeneous Newtonian fluid and assumes that enlarging the artery does not change its applied pressure drop. These assumptions supply the intended fourth-power scaling; an actual circulation need not keep that pressure drop fixed.

Diffusion and mobility. For a dilute isolated sphere in a Newtonian fluid, Stokes's law gives the drag magnitude $F_{\mathrm{drag}}=6\pi \eta Rv=\xi v$, where $\eta$ is the dynamic viscosity and $R$ is the hydrodynamic radius. The mechanical mobility is therefore $1/\xi$; the translational hydrodynamic mobility matrix is ${\xi }^{-1}I$.

To relate mobility to diffusion, put the particles in a weak external potential energy $U(x)$. Their drift speed is $-U^{\prime }(x)/\xi$. For number density $n(x)$, the sum of drift and Fick's first law fluxes isAt thermal equilibrium, the Boltzmann distribution gives $n\propto e^{-U/(k_{B}T)}$, and the net flux is zero. Substituting $n^{\prime }=-n{U}^{\prime }/(k_{B}T)$ into $J=0$ yields $D=k_{B}T/\xi$. Thus the Stokes–Einstein relation isHere $k_B$ is the Boltzmann constant and $T$ is absolute temperature. The assumptions include a no-slip approximately spherical particle, a continuum fluid, a small particle Reynolds number, and negligible hydrodynamic interactions with other particles or boundaries.

Gravitational sedimentation. The buoyant force must be included, since the mass of displaced water is comparable to the protein mass. Using $1\mathrm{Da}=1.66054\times 10^{-27}\mathrm{kg}$ givesso the buoyant mass in the stated spherical model isThe drag coefficient is $\xi =6\pi (10^{-3})(2.16\times 10^{-9})=4.0715\times 10^{-11}\mathrm{kg}{\mathrm{s}}^{-1}$. At terminal speed, $\xi v_g=m_{b}g$, soIgnoring buoyancy would give $1.47\times 10^{-11}\mathrm{m}{\mathrm{s}}^{-1}$ and would not use the supplied water density. This calculation treats the stated radius as both the hydrodynamic radius and the radius determining displaced volume; a real hydrated protein can require a separately measured partial specific volume.

Electrical drift. In the specified simple electrophoresis model, terminal force balance is $Ze\mathbf{E}=\xi \mathbf{v}$. Thereforewhere $\mu$ is the electrophoretic mobility and $e$ the elementary electric charge. For the given positive beta-lactoglobulin charge,in the direction of the electric field. This exceeds the gravitational sedimentation speed by about $8.6\times 10^6$.

Separation time. The two populations have equal drag coefficients because their radii are equal. Their difference in drift speed is therefore set by one elementary charge:Starting with coincident band centres, their separation is $\Delta vt$. HenceAt that time the original band has travelled $20\mathrm{mm}$ and the mutant $21\mathrm{mm}$. Their masses do not enter this ideal electrical force balance: equal-radius particles with different masses but the same charge would have equal electrophoretic mobility in this model.

Practical discrimination. A one-charge difference is enough for the stipulated separation, so the method is sensitive to mutations that change effective charge or friction. It cannot distinguish every mutation: substitutions that preserve charge and hydrodynamic size need not separate. Buffer pH changes protein ionization, and realistic electrophoresis also involves the ionic atmosphere, electro-osmosis, walls or gel interactions, and finite band widths.

For example, the Stokes–Einstein relation at $293.15\mathrm{K}$ gives $D=9.94\times 10^{-11}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. For an initially narrow band, diffusive broadening of electrophoretic bands givesat the calculated separation time. Thus the centres are about $3.15$ diffusion standard deviations apart, rather than forming infinitely sharp bands. The stated $1\mathrm{mm}$ centre-separation criterion is achievable in the model, but actual resolution also depends on the loading width and other dispersion mechanisms.

Equivalent centrifugal acceleration. For the original protein to attain its electrical drift speed by sedimentation, its buoyant centrifugal force must equal $20eE$:ConsequentlyThis is the acceleration at the protein's position in the rotating solution; no rotation rate can be assigned without a centrifuge radius. If the intended comparison is instead the relative separation speed of the two bands, use $\Delta m_b=600\mathrm{Da}$, since their displaced volumes are equal. Matching $|\Delta m_b|a_{\mathrm{eff}}=eE$ then gives $8.04\times 10^7\mathrm{m}{\mathrm{s}}^{-2}$. In that separation experiment the heavier original protein moves faster centrifugally, whereas the more highly charged mutant moves faster electrically.

Let $z$ increase down the pipe, $a=0.015\mathrm{m}$ be its radius and $r$ the distance from its axis. For steady, fully developed laminar flow of a Newtonian fluid, the axial Navier-Stokes equations reduce toWrite $K=-dp/dz+\rho g\sin 20^{\circ }$. Integrating twice, regularity at $r=0$ excludes a logarithmic term, and the no-slip boundary condition $u(a)=0$ gives the Hagen-Poiseuille flow profileThe volumetric flow rate follows by integrating over the circular cross-section:For the intended gravity-driven case with no imposed axial pressure gradient, $dp/dz=0$. Substitution givesandThe corresponding mean speed is about $0.1047\mathrm{m}{\mathrm{s}}^{-1}$ and the diameter-based Reynolds number is $\rho \overline{u}(2a)/\eta \simeq 3.49$, consistent with laminar flow. The original PDF does not specify inlet and outlet pressures. If they impose an additional pressure gradient, the general result with $K$ above applies; gravity alone does not determine that extra contribution.

The dynamic modulus, or complex viscoelastic modulus, is the ratio of complex stress amplitude to complex strain amplitude in the periodic steady response:This defines the sign convention; for passive response the loss modulus is nonnegative at positive frequency. A Newtonian fluid obeys $\sigma =\eta \dot{ϵ}$, while a Hooke's law solid obeys $\sigma =G_0ϵ$. HenceThe first is purely dissipative and the second purely elastic. Here $ϵ$ denotes the scalar deformation used in the problem, with any shear-versus-extension geometric factors already included in its modulus and viscosity.

Take $z$ positive away from the wall into the liquid, and write $p_a$ for atmospheric pressure. For a constant-mass density Newtonian fluid, the two-dimensional Navier-Stokes equation and continuity equation areThe stationary wall imposes the no-slip boundary condition and no-penetration boundary condition, $u=w=0$ at $z=0$. For a flat free surface with negligible air shear stress, the stress boundary condition is $u_z=0$, $p=p_a$ at $z=h_0$; the kinematic boundary condition is $w=0$ there.

For steady flow independent of $x$, these equations give $w=0$, $\mu u_{zz}=-\rho g\sin \alpha$ and $p_z=-\rho g\cos \alpha$. Two integrations and the boundary conditions give the uniform gravity-driven Newtonian film:Its surface velocity is $U_0=\rho g{h}_0^2\sin \alpha /(2\mu )$, while its mean velocity is $2U_0/3$.

For an isotropic Newtonian fluid in incompressible flow, the constitutive equation isHere $p$ is pressure, ${\delta }_{ij}$ is the Kronecker delta, $\mu$ is the constant dynamic viscosity, and $E$ is the strain-rate tensor. The more general compressible constitutive equation can include a term proportional to $(\nabla \cdot \mathbf{u}){\delta }_{ij}$; that term vanishes here.

Differentiate the Cauchy stress tensor, retaining the constant dynamic viscosity outside the derivatives:The last equality uses incompressibility. Substitution into the Cauchy momentum equation gives the Navier-Stokes equationThe pressure gradient and the Laplacian of the velocity field therefore give the two contact force densities in this model.

Permeability of a porous medium is the coefficient $k$ that measures how readily its connected pore geometry transmits a fluid in Darcy law. Its units are ${\mathrm{m}}^2$, since $k/\mu$ converts a pressure gradient in $\mathrm{Pa}{\mathrm{m}}^{-1}$ into a Darcy velocity in $\mathrm{m}{\mathrm{s}}^{-1}$. Large open, well-connected passages generally give larger permeability of a porous medium than narrow or winding passages. Porosity alone cannot determine it: two samples can contain the same void volume but have different passage sizes and connectivity.

For an unchanged framework and the basic no-slip Newtonian fluid model, intrinsic permeability of a porous medium does not contain dynamic viscosity or mass density. These properties instead enter hydraulic conductivity $K=k\rho g/\mu$. At the same horizontal pressure gradient, replacing a liquid with another of twice the dynamic viscosity halves Darcy flux even though $k$ stays fixed. Structural changes, such as swelling or clogging, can change $k$; the independence claim assumes the pore geometry stays fixed.^[ref]

An isotropic sample has scalar permeability of a porous medium. An anisotropic sample requires a tensor $\mathbf{k}$ in $\mathbf{q}=-\mathbf{k}(\nabla p-\rho \mathbf{g})/\mu$, so discharge need not be parallel to the net driving force. Layered porous permeability gives a concrete example: arranging individually isotropic materials in horizontal layers produces different vertical and horizontal effective permeabilities.

Now both plates at $z=0$ and $z=H$ are stationary and both obey the Navier slip boundary condition with the same slip length $b$. Drive steady flow of a Newtonian fluid in incompressible flow by the constant pressure gradient $G=-dp/dx>0$, with constant dynamic viscosity $\mu >0$ and no other streamwise body force. The planar Poiseuille flow profile and its volumetric flow rate per unit width areThe full volumetric flow rate for channel width $W$ is $Q=Wq$ when side-wall effects are negligible. Thus $q$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, while $Q$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-1}$. Denote the value at $b=0$ by $q_0=G{H}^3/(12\mu )$.

The two wall checks expose the normal signs:Both Navier slip boundary conditions therefore hold exactly. The two walls exert backward tangential tractions of magnitude $GH/2$ on the fluid. Increasing $b$ at fixed $G$ adds a uniform velocity offset, leaving the velocity gradient and wall shear stress unchanged. This differs from Couette flow with a slipping lower wall, where upper-wall velocity is fixed and shear stress decreases.

For $H=10\mu \mathrm{m}$, $b=2\mu \mathrm{m}$, $G=10^3\mathrm{Pa}{\mathrm{m}}^{-1}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$, the wall velocity is $10\mu \mathrm{m}{\mathrm{s}}^{-1}$, the centre velocity is $22.5\mu \mathrm{m}{\mathrm{s}}^{-1}$, and $q=1.83\times 10^{-10}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. The enhancement is $q/q_0=2.2$. Equivalently, a prescribed volumetric flow rate needs only $1/2.2$ of the original pressure gradient. This makes slip length useful for estimating pumping requirements in small channels.

At fixed $H$, $G>0$ and $\mu$, $b\to 0$ recovers ordinary planar Poiseuille flow. As $b\to \infty$, the velocity and $q$ grow without bound in this model, although the varying part of the profile stays the same. At exactly zero tangential wall traction, no steady solution can balance this nonzero pressure gradient: $u^{\prime }(0)=u^{\prime }(H)=0$ conflicts with $\mu [u^{\prime }(H)-u^{\prime }(0)]=-GH$. Unlimited enhancement is consequently not a prediction for a real channel; finite wall friction, finite driving capacity and the validity of the assumed steady flow must be assessed. At fixed finite $q$ instead, $G\to 0$ and the profile approaches uniform velocity $q/H$.

A short derivation integrates the reduced Navier-Stokes equation $\mu u^{\prime \prime }=-G$ twice. Equal slip lengths and the opposite inward unit normals give $u=G[z(H-z)+bH]/(2\mu )$. Its integral over the gap supplies the enhancement factor $1+6b/H$.

Porous-media flow is fluid motion through the interconnected voids of a porous medium, such as a sand bed, a rock or a filter. Individual passages can be tiny and tortuous, yet their combined discharge can be predicted without resolving every passage. Darcy law supplies this bulk description when viscous resistance dominates pore-scale inertia; porosity measures how much space stores fluid, while permeability of a porous medium measures how readily that space transmits it.

The basic fully saturated model assumes one Newtonian fluid fills the connected pore space of a stationary solid framework, with incompressible flow, a small pore-scale Reynolds number, and averaging volumes large compared with individual pores but small compared with macroscopic variations. These are assumptions of the model developed below, rather than a claim that Darcy law describes every porous-media flow. Two-phase porous-media flow and poroelastic aquifers require additional constitutive and storage information.

For a first calculation, see Darcy flow through a homogeneous slab: a pressure drop determines the discharge, but porosity determines the corresponding pore velocity. Layered porous permeability then shows how geometry alone can make the bulk response anisotropic.

Rectangular duct flow is steady flow of a Newtonian fluid down a straight rectangular pipe under a constant axial pressure gradient $-G$. For stationary walls and unidirectional flow, incompressibility eliminates axial dependence and the Navier-Stokes equation becomes a two-dimensional Poisson equation with zero Dirichlet boundary conditions. Subtracting a plane Poiseuille flow profile leaves a harmonic function that repairs the side-wall values using separation of variables. Uniqueness follows from Green's first identity: the difference of two solutions has zero boundary values and zero integral of the squared gradient, and hence vanishes.

For a sphere of radius $a$ in an otherwise stationary, unbounded Newtonian fluid with dynamic viscosity $\eta$, impose the no-slip boundary condition and the slow-flow assumptions of Stokes flow. The resisting torque on a rotating sphere in Stokes flow is $-{\zeta }_r\mathbf{\Omega }$, where $\mathbf{\Omega }$ is its angular velocity and ${\zeta }_r=8\pi \eta a^3$. Thermal rotational diffusion at absolute temperature $T$ then givesHere $k_B$ is the Boltzmann constant; the translational expression is the Stokes–Einstein relation. The ratio has units of inverse length squared, so comparing the numerical values of $D_r$ and $D_t$ without their units is meaningless. At fixed temperature and dynamic viscosity, decreasing radius tenfold increases $D_r$ a thousandfold but $D_t$ only tenfold. This strong size dependence explains why small particles can lose directional memory extremely quickly.

Choose $T=298\mathrm{K}$ and $\eta =0.001\mathrm{Pa}\mathrm{s}$ as illustrative model inputs. Using $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ gives $k_{B}T=4.11433402\times 10^{-21}\mathrm{J}$ and the following predictions:These are bulk three-dimensional sphere calculations, explicitly not a fit to the confined ellipsoid experiment below. Walls, nonspherical shape and departure from the no-slip boundary condition can change rotational resistance. A visible surface mark is still necessary to read the sphere's turning in an image, even when its rotational diffusion coefficient is well defined physically.

Let a sphere of radius $a$ descend along the $z$ axis through a stationary unbounded Newtonian fluid. A second sphere with centre $(\ell ,0,0)$ is held against translation but has zero applied torque. In the prescribed isolated-sphere incident flow, Faxén's rotational law and the vorticity of translating-sphere Stokes flow giveThe speed $U$ cancels on integrating along the complete passage. Since ${\int }_{-\infty }^{\infty }dZ/({\ell }^2+Z^2{)}^{3/2}=2/{\ell }^2$, the result is the displayed angle about $-{\mathbf{e}}_y$. The target radius is absent from the rotational Faxén response to this harmonic incident vorticity. This is a one-way incident-flow calculation; nonoverlap alone does not justify ignoring mutual hydrodynamic interactions between finite spheres. A fully coupled encounter requires the method of reflections for Stokes flow or an exact two-body solution.

A force-free, torque-free microswimmer in a stationary Newtonian fluid, with negligible fluid and body inertia and steady Stokes flow at each instant, gains no net displacement from a reciprocal stroke: one whose sequence of shapes is exactly retraced in reverse. The environment and boundary conditions must also be unchanged during that retracing. Opening a two-shell swimmer and then closing it follows the same sequence backwards; doing the closure faster does not help, because Linearity of Stokes flow changes the instantaneous velocity by the same factor as the stroke rate.

A single real shape coordinate that travels out and back along an interval gives this cancellation, as computed in the force-free two-sphere stroke. Two independently controlled shape coordinates can instead execute a loop that is not retraced, avoiding this obstruction. Continuous rotation also need not retrace an interval: the helical microswimmer with a spherical head provides another route to propulsion. The scallop theorem is a restriction on a stroke under stated conditions, not a claim that small organisms cannot swim. Purcell presented the opening-and-closing scallop and a swimmer with two hinges in Life at low Reynolds number (1977), figures 6–7.

For a homogeneous Newtonian fluid with constant kinematic viscosity, a steady axisymmetric purely azimuthal flow obeys $v_{\varphi }^{\prime \prime }+v_{\varphi }^{\prime }/r-v_{\varphi }/r^2=0$. Writing $v_{\varphi }=r\Omega$ gives $(r^3{\Omega }^{\prime }{)}^{\prime }=0$, so $\Omega =A+B/r^2$. A domain containing the rotation axis requires $B=0$ for regularity; rotating walls in an annulus can sustain $B\ne 0$. An arbitrary time-dependent profile instead evolves by ${\partial }_t\Omega =\nu ({\Omega }^{\prime \prime }+3{\Omega }^{\prime }/r)$ within this ansatz.

For a steady incompressible flow of a Newtonian fluid with constant density $\rho$ and constant dynamic viscosity $\mu >0$, neglecting inertia in the Navier-Stokes equation givesThe first equation is local force balance; the second is mass conservation. The term $\mu {\nabla }^2\mathbf{u}$ is the divergence of the viscous part of the Newtonian fluid stress tensor. The body force $\mathbf{f}$ has units of force per unit volume, $\mathrm{N}{\mathrm{m}}^{-3}$, and equals $\rho \mathbf{g}$ for uniform gravity. Appropriate boundary data for velocity or traction, together with suitable distant conditions for an exterior flow, complete the problem; a rigid surface normally imposes the no-slip boundary condition.

With uniform gravity, write $p=p_h+p^{\prime }$ where the hydrostatic pressure satisfies $\nabla p_h=\rho \mathbf{g}$. Then the disturbance equation is $-\nabla p^{\prime }+\mu {\nabla }^2\mathbf{u}=0$. This removes gravity from the fluid disturbance equation, while the hydrostatic pressure still supplies buoyancy on an immersed body. Omitting it from the total force balance would give the wrong Stokes settling velocity.

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.