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The first term is the local viscous force per unit length from resistive-force theory. With unit tangent vector $\mathbf{t}={\mathbf{r}}_s$, the drag tensor gives $-c_{\parallel }{\mathbf{v}}_{\parallel }-c_{\perp }{\mathbf{v}}_{\perp }$, where $\mathbf{v}={\mathbf{r}}_t$ is velocity relative to the fluid.

The second term is the bending force of an inextensible filament. Varying its energy $E_b=(B/2)\int |{\mathbf{r}}_{ss}{|}^{2}ds$ gives bulk force density $-B{\mathbf{r}}_{ssss}$. The final term, ${\partial }_s(T{\mathbf{r}}_s)$, is the force density from filament tension; $T$ is the Lagrange multiplier enforcing $|{\mathbf{r}}_s|=1$. At low Reynolds number, these forces balance without a filament acceleration term.

With mass, length, and time dimensions $M,\mathsf{L},\mathsf{T}$, the coefficients haveThus the drag coefficients have the dimensions of dynamic viscosity, the filament bending modulus has dimensions force times length squared, and tension has dimensions force.

In the Giesekus model, $\lambda$ is the viscoelastic relaxation time: after deformation stops, polymeric stress relaxes on that timescale. The parameter $\eta$ has dimensions of dynamic viscosity and is the model's zero-rate viscosity scale. Since $\lambda {\mathbf{\tau }}^2/\eta$ already has dimensions of stress, $\alpha$ is dimensionless.

The symbol $\stackrel{▽}{\tau }$ is the upper-convected derivativeThe convective and velocity-gradient terms account for translation, rotation, and affine stretching of material elements. They make the constitutive law objective under time-dependent rigid changes of observer.

The Navier-Stokes equation for constant mass density $\rho$, dynamic viscosity $\mu$, and body force per unit mass $\mathbf{f}$ iswhere the Newtonian fluid stress tensor isTaking the dot product with $\mathbf{u}$ givesIncompressibility impliesand converts the left side into a local time derivative plus the divergence of kinetic-energy flux. The divergence theorem therefore gives the kinetic-energy balance for an incompressible Newtonian fluidThe four terms are respectively outward advective transport of kinetic energy, mechanical power supplied by surface traction, power supplied by the body force, and irreversible viscous dissipation into heat.

For a no-slip spherical particle in a Newtonian fluid at low Reynolds number, Stokes law gives $\zeta =6\pi \eta a$. The fluctuation-dissipation theorem then gives the Stokes–Einstein relation $D=k_{B}T/\zeta$. Using the conventional radial Brownian displacement in three dimensions,Here $k_B$ is the Boltzmann constant, $T$ absolute temperature, $\eta$ dynamic viscosity, $a$ particle radius and $t$ elapsed time. Larger thermal forcing raises diffusion; larger viscous drag reduces it. The formula requires times long relative to momentum relaxation and neglects walls, interactions and external drift.

The word “average” needs a statistic: the vector mean is zero, while the mean radial displacement for Gaussian three-dimensional diffusion is $⟨|\Delta \mathbf{r}|⟩=4\sqrt{Dt/\pi }$, not $\sqrt{6Dt}$. Both scale as $t^{1/2}$. Neither is the total length of a microscopic random path.

A fluid is described on a continuum scale by density, pressure and velocity fields. An Eulerian field is measured at fixed position; a material element changes at rate $D/Dt={\partial }_t+\mathbf{v}\cdot \nabla$. mass conservation requiresFor incompressible flow this becomes $\nabla \cdot \mathbf{v}=0$. A streamline is instantaneously tangent to velocity, whereas a pathline follows a particle through time; they coincide in a steady flow but need not in an unsteady one.

For an ideal inviscid fluid, pressure and body forces give the Euler equation. In the constant-density gravitational case,A stationary impermeable boundary requires zero normal velocity, but inviscid theory does not impose no slip. Taking its steady projection along a streamline gives $p/\rho +v^2/2+gz=\mathrm{constant}$. If the flow is also irrotational in the connected region, that constant is shared between streamlines. Inviscid does not mean irrotational: rotating ideal-fluid flows are possible. A velocity potential exists locally for an irrotational field, and in incompressible flow it obeys Laplace's equation. The Bernoulli assumptions must be checked before converting a pressure difference into a speed.^[ref]

Real fluids transmit shear stress. An incompressible Newtonian fluid with constant dynamic viscosity $\eta$ obeysThe constitutive shear stress is proportional to strain rate, and stationary solid boundaries usually impose no slip. viscosity diffuses vorticity and converts organized mechanical energy into heat; its local dissipation $2\eta e_{ij}{e}_{ij}$ is nonnegative. It cannot generally be included merely by changing the constant in a lossless Bernoulli equation.

In a long straight circular tube, steady fully developed laminar flow hasThis Hagen-Poiseuille equation follows by balancing the axial pressure gradient with viscous shear and setting $v_z(a)=0$. The parabolic profile, the fourth-power radius sensitivity and the pumping power $\Delta pQ$ are experimentally distinct consequences. Entrance regions, rough walls, non-Newtonian response or a changing temperature require additional modelling.

The Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares inertia with viscosity. Small values favor creeping-flow balances; high values permit thin boundary layers, separation and turbulence, although transition depends on geometry and perturbations, not one universal critical number. At high Reynolds number an outer flow can be approximately inviscid while near-wall viscosity still determines drag. This resolves why a zero-drag ideal potential-flow model cannot predict the resistance of an ordinary moving body. In turbulent flow, fluctuating Reynolds stresses redistribute momentum and increase the required pressure drop; averaging the equations introduces closure requirements.

A controlled tube experiment measures pressure difference and collected volume per time while changing forcing. Use tracers or a velocity probe to test profile shape, allow sufficient entrance length, and monitor radius and temperature independently. Repeat with reversed flow and at several rates to test offsets and linearity. Dye mixing alone does not measure viscosity or establish a fully developed velocity profile. The original figure below separates a specified laminar velocity profile from an independent pressure/discharge measurement.

Dynamic viscosity measures resistance to a velocity gradient. In Newtonian simple shear, ${\tau }_{xy}=\eta \partial u_x/\partial y$; positive viscosity dissipates mechanical energy into heat. Kinematic viscosity is $\nu =\eta /\rho$. For an incompressible fluid of constant viscosity,Scaling velocity by $U$ and length by $L$ gives the Reynolds number $\mathrm{Re}=\rho UL/\eta =UL/\nu$, the ratio of inertial to viscous terms. At small Reynolds number, creeping-flow approximations neglect inertia; at large Reynolds number, inertia dominates much of the bulk but viscosity can remain crucial in boundary layers. A large Reynolds number permits instability and turbulence; it does not by itself specify a universal transition threshold. Geometry and disturbances also matter.

The dynamic viscosity measures the transport of tangential momentum between adjacent layers. The kinetic estimates of gas transport coefficients givewhere $n$ is number density, $m$ the molecular mass, $\lambda$ the mean free path, and ${\sigma }_c$ a collision cross section. At fixed temperature, raising the pressure increases the number of momentum carriers but shortens their mean free path by the same factor. Consequently the dynamic viscosity of a dilute gas is approximately independent of pressure.

For a temperature-independent collision cross section, the same estimate gives $\eta \propto \sqrt{m{k}_{B}T}/{\sigma }_c$, so $\eta \propto T^{1/2}$ in the hard-sphere model. Faster molecules carry momentum across layers more rapidly. Real intermolecular interactions make the effective collision cross section temperature-dependent, changing this simple exponent. The pressure-independence also fails when wall collisions, rather than molecule–molecule collisions, set the transport distance; the continuum estimate assumes a small Knudsen number.

For a Newtonian fluid with velocity $u(z)\hat{\mathbf{x}}$, $\tau =\eta {\partial }_{z}u$ is the $x$ component of traction on a plane whose normal points along $+z$. The opposite face has the opposite traction. The coefficient $\eta$ is dynamic viscosity, with units $\mathrm{Pa}\mathrm{s}$. Faster layers transfer positive $x$ momentum toward slower neighbours; the internal shear stress resists relative sliding.

Take a thin fluid element with horizontal area $A$ and thickness $dz$. The difference between its upper and lower viscous tractions is $A\eta u_{zz}dz$. The pressure gradient supplies $-A{p}_{x}dz$. Incompressibility is satisfied because $u$ has no $x$ dependence, and the convective acceleration vanishes: $(\mathbf{u}\cdot \nabla )\mathbf{u}=u{\partial }_x\mathbf{u}=0$. Equating net force to mass times acceleration therefore gives the relevant component of the Navier-Stokes equation:For steady flow, integrate $\eta u^{\prime \prime }=p^{\prime }$ twice and impose the no-slip boundary condition at each plate, $u(0)=0$ and $u(a)=u_0$. The Couette-Poiseuille flow profile isThe linear term is Couette flow and the parabolic term is plane Poiseuille flow. A negative $p^{\prime }$ drives flow in $+x$. Integrating the profile across the gap gives the volumetric flow per unit width,It has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying by the plate width gives the actual volume per time.

With zero pressure gradient, substituting $u=\mathrm{Re}[A{e}^{i(kz-\omega t)}]$ gives $-i\rho \omega =-\eta k^2$. Select the root that decays into fluid occupying $z>0$:The other sign applies to decay in the opposite half-space. With a real wall velocity amplitude $U_0$, the established periodic solution of Stokes' second problem isThe oscillation's amplitude decays exponentially and its phase lags with distance from the plate. This is diffusion of tangential momentum, not a freely propagating undamped elastic wave. Initial transients must have decayed before this purely periodic expression applies.

The fluid-displacement amplitude and the wall-displacement amplitude have the same ratio as their velocity amplitudes, since both are divided by the common $\omega$. Thus for the specified amplitudes,Using the standard approximate water density $\rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, explicitly an approximation rather than another value printed on the paper, gives $\delta =0.564\mathrm{mm}$ and $z\simeq 2.21\mathrm{mm}$ from either face of a sufficiently large plate. The viscous penetration depth varies as ${\rho }^{-1/2}$ if a more precise water density is used. The ideal parallel-flow solution ignores finite plate edges and tank walls.

A Newtonian fluid exerts viscous shear stress proportional to the rate of strain. For an incompressible fluid of constant dynamic viscosity $\eta$,The local momentum balance is the Navier-Stokes equation:The left side is density times material acceleration. A steady velocity field can still have convective acceleration when a moving parcel changes speed or direction; setting $\partial \mathbf{u}/\partial t=0$ does not eliminate it.

The Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares inertia with viscous effects. At low Reynolds number, the fluid's own inertia can often be neglected, giving Stokes flow. This does not say that every immersed body has zero acceleration: its mass and the net pressure, viscous and body forces still enter its own equation of motion.

For a small rigid sphere moving through a slowly varying viscous background, Stokes drag is $-6\pi \eta a(\mathbf{v}-\mathbf{u})$. In still fluid, a simple settling model includes weight, buoyancy and drag:Ignoring fluid added mass and unsteady history forces, this gives relaxation time $m/(6\pi \eta a)$ and terminal velocityAt the terminal velocity the net force and acceleration vanish, while viscous drag and gravity remain individually nonzero. The dissipated power $6\pi \eta a{v}_t^2$ is supplied by the loss of gravitational potential energy relative to buoyant support. The creeping-flow and quasisteady-drag assumptions must be checked before applying this expression at large speed or immediately after a rapid start.

Following the moving fluid. The Eulerian velocity field $\mathbf{u}(\mathbf{x},t)$ assigns a velocity to each position. A fluid particle satisfies $\dot{\mathbf{x}}=\mathbf{u}$, so the chain rule gives the material derivative, also called the convective derivative,Even in steady flow a particle can accelerate because it moves between regions with different velocity. For example, flow around a curved obstacle changes direction despite ${\partial }_t\mathbf{u}=0$. Mass conservation isFor a constant-density incompressible fluid, $\nabla \cdot \mathbf{u}=0$.

Euler equations for an inviscid fluid and Bernoulli's equation. Neglect viscosity and take the body force per unit mass to be $-\nabla \Phi$. Newton's second law for a fluid element givesFor steady flow, dot with displacement along a streamline. The tangential component of $\mathbf{u}\cdot \nabla \mathbf{u}$ is the derivative of $u^2/2$ along that line. For constant density, integration givesIn uniform gravity, $\Phi =gz$. Thus pressure, kinetic energy per unit mass and gravitational potential energy per unit mass exchange along a streamline. For a barotropic compressible fluid, replace $p/\rho$ by $\int dp/\rho (p)$. Different streamlines need not have the same Bernoulli constant.

Vorticity and potential flow. Vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. A small fluid element rotates locally with angular velocity $\mathbf{\omega }/2$ after its deformational part is separated. The vector identitygives $\nabla B=\mathbf{u}\times \mathbf{\omega }$ for steady constant-density inviscid flow. Thus an irrotational region has one common Bernoulli constant on each connected region. In a simply connected irrotational domain, introduce a velocity potential $\mathbf{u}=\nabla \varphi$. Incompressibility then gives Laplace's equation, ${\nabla }^2\varphi =0$. For unsteady irrotational flow,An impermeable stationary wall fixes ${\partial }_n\varphi =0$, but ideal potential flow does not enforce no slip. In a multiply connected domain, irrotational flow may still have nonzero circulation around an excluded vortex core, so a globally single-valued potential requires care.

For example, uniform two-dimensional flow past a circular cylinder has $\varphi =U(r+a^2/r)\cos \theta$ outside $r=a$. Its normal velocity vanishes at the cylinder, and its surface tangential velocity is $-2U\sin \theta$. Bernoulli's equation predicts a symmetric pressure distribution and zero drag. This is D'Alembert's paradox: the assumptions omit the viscous boundary layers, separation and wake responsible for real drag.

For an incompressible fluid of constant kinematic viscosity $\nu$, the curl of the motion equation givesThe first term stretches and tilts vortex lines; the second diffuses vorticity. In two-dimensional incompressible inviscid flow the stretching term is absent and vorticity is carried with the fluid. In a three-dimensional inviscid flow it can be amplified by stretching. Density gradients not parallel to pressure gradients provide an additional source in a general compressible fluid.

Viscous flow. A Newtonian fluid's shear stress is proportional to rate of strain. For constant viscosity and incompressibility the Navier-Stokes equation iswith dynamic viscosity $\eta$ and kinematic viscosity $\nu =\eta /\rho$. At an ordinary solid wall, the no-slip boundary condition sets fluid velocity equal to the wall velocity. Viscosity transfers momentum between neighbouring layers and converts mechanical energy into heat; unlike the pressure force, its shear contribution depends on velocity gradients.

For plane Couette flow between plates $y=0,h$, with the upper plate moving at $U$, the steady solution is $u_x=Uy/h$, with shear stress $\eta U/h$. For pressure-driven flow through a pipe of radius $a$, a constant pressure gradient gives the parabolic Poiseuille profileThe Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares convective inertia with viscous force. Small Reynolds number often permits the Stokes-flow approximation; large Reynolds number can produce thin boundary layers and, under suitable conditions, turbulence. It does not justify neglecting the wall conditions. Bernoulli's inviscid constant is generally not conserved along a viscous streamline, and pipe flow instead shows a pressure loss corresponding to dissipation.

Molecular transport across gradients. A gas can transport momentum, thermal energy and species abundance even when its average bulk velocity is zero. Molecules arriving at a point carry properties acquired near their preceding collisions. A small gradient makes the populations arriving from opposite sides slightly different, producing transport from larger to smaller values of the carried property. A continuum transport law requires a mean free path $\lambda$ small compared with the gradient length, and relaxation times short compared with the observation time.

For a dilute hard-sphere gas,where $d$ is molecular collision diameter. Both speed and collision distance matter: high speed alone does not determine a transport coefficient.

A simple kinetic estimate makes the common structure explicit. For a quantity $a(y)$ carried per particle, a molecule moving at angle $\theta$ to $+y$ brings approximately $a(y-\lambda \cos \theta )$. Expanding to first order and averaging an isotropic velocity distribution givessince $⟨\cos \theta ⟩=0$ and $⟨{\cos }^2\theta ⟩=1/3$. This mean-free-path argument estimates the coefficients; a detailed collision calculation changes their numerical prefactors and distinguishes different relaxation rates.

Momentum transport and viscosity. For a shear flow with mean velocity $U_x(y)$, the transported directed momentum per molecule is $m{U}_x$. The net flux of $x$ momentum in the $+y$ direction isHere $\eta$ is dynamic viscosity in $\mathrm{Pa}\mathrm{s}$. The associated shear stress on a material surface has the corresponding traction sign convention; the flux carries momentum from the faster-moving layer towards the slower-moving one. The kinematic viscosity is $\nu =\eta /(nm)$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Viscosity smooths velocity gradients, and the lost ordered mechanical energy becomes heat.

Heat transport. Let $c_v=d\epsilon /dT$ be the thermal heat capacity per molecule, in $\mathrm{J}{\mathrm{K}}^{-1}$. Substituting $a=\epsilon (T)$ gives Fourier's law,Thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a monatomic gas, translational equipartition gives $c_v=3k_B/2$. Rotational or vibrational excitations in other gases can increase the energy carried per molecule, while collisions govern how those modes exchange energy. A temperature gradient can therefore be transported differently from directed momentum even when the dimensional estimates look similar.

Species transport. For a dilute tracer or a mixture with prescribed total density and negligible bulk flow, molecules diffuse down concentration gradients. The first-gradient number flux is Fick's first law,The diffusivity $D$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying number flux by the species mass gives mass flux. More generally one uses diffusion flux relative to the mixture's bulk velocity and specifies mass or mole fractions. The approximation is not a claim that an arbitrary pure-gas pressure gradient can remain at rest while only a Fick flux acts. In a mixture, collision cross-sections and relative molecular speeds, including the reduced mass, determine the species diffusivities.

Dependence on density and temperature. The kinetic estimates of gas transport coefficients predict, for fixed collision diameter and molecular heat capacity,Thus dynamic viscosity and thermal conductivity are approximately density independent in the dilute continuum regime. Reducing density reduces the number of carriers but increases their mean free path in the same proportion. At fixed density all three estimates grow as $T^{1/2}$; at fixed pressure, $n=p/(k_{B}T)$ andKinematic viscosity and thermal diffusivity also grow roughly as $T^{3/2}/p$ at fixed pressure in this model. Real interaction potentials make effective collision cross-sections temperature dependent, and internal excitations change $c_v$. Dense-gas correlations and boundaries invalidate the simple cancellation. In particular, when the mean free path approaches the apparatus scale, transport becomes free molecular; it cannot be described by extending the same local, pressure-independent Fourier and viscosity coefficients indefinitely.

Heat diffusion equation. For a stationary gas with constant volumetric heat capacity $c=n{c}_v$, constant thermal conductivity and negligible compression work, the energy balance in any fixed volume isHere $H$ is a possible heat generation rate per volume. The divergence theorem and Fourier's law yieldThermal diffusivity $\alpha$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is of order $\overline{v}\lambda /3$. In a moving or compressing gas, the material derivative, pressure-volume work and viscous heating must also enter the internal-energy equation. The displayed pure diffusion equation describes the approximation in which those effects are absent or negligible.

Representative solutions. With no source, a steady one-dimensional slab between temperatures $T_a,T_b$ at $x=0,L$ has $T=T_a+(T_b-T_a)x/L$ and uniform heat flux $-\kappa (T_b-T_a)/L$. Thus a constant spatial gradient transports heat without changing local temperature in time when the boundaries continually supply and remove it.

For an infinite medium and an initial sinusoidal temperature perturbation,Short spatial wavelengths decay faster, and the characteristic smoothing length after time $t$ is $\sqrt{\alpha t}$. A finite slab with both ends held at $T_b$ has a sine expansion with factors $e^{-\alpha (n\pi /L{)}^{2}t}$; its slowest relaxation time is $L^2/({\pi }^2\alpha )$.

An instantaneous heat input $Q_A$ per unit transverse area at $x=0$ in an otherwise uniform infinite medium gives the Gaussian solutionIts integral $\int c(T-T_b)dx=Q_A$ conserves the deposited heat. The peak falls as $t^{-1/2}$ while its width grows as $\sqrt{t}$; diffusion redistributes energy instead of transporting a sharp front at a single velocity.

Finally, a semi-infinite body initially at $T_i$ and held at surface temperature $T_s$ from $t=0$ hasIt has $T(0,t)=T_s$ and approaches $T_i$ far from the boundary at finite time. These solutions demonstrate steady conduction, mode decay, spreading of localized heat and boundary-driven diffusion. Their idealized behaviour at extremely short times or arbitrarily long distances must be interpreted within the continuum regime rather than as an assertion of instantaneous microscopic communication.

Meaning and microscopic origin. For a Newtonian fluid, the dynamic viscosity $\eta$ measures tangential momentum transfer between fluid layers moving at different speeds. For the Newtonian shear considered here, ${\tau }_{xz}=\eta d{v}_z/dx$, with units $\mathrm{Pa}\mathrm{s}$ for $\eta$. Molecular motion and collisions transport momentum across layers in a gas; intermolecular interactions contribute strongly in liquids. The resulting viscous forces oppose relative shear motion and convert organized kinetic energy to heat. The constitutive law is Newtonian and isotropic; incompressibility alone does not establish it.

Vertical force balance. Choose $z$ downward, as in the original diagram, and $x$ outward normal to the plate. In a uniform steady film, $\mathbf{v}=v_z(x)\hat{\mathbf{z}}$ has zero advective acceleration and no along-plate pressure gradient. On a slice of thickness $dx$, the net shear force per unit face area is $\tau (x+dx)-\tau (x)$, while its weight per unit area is $\rho gdx$. ThereforeHere $\tau =\eta d{v}_z/dx$ denotes the downward traction on a face with positive $x$ normal. The opposite face has the opposite traction orientation, accounting for the sign in the force difference.

Boundary conditions. At the fixed solid wall, the no-slip boundary condition is $v_z(0)=0$, with no normal penetration. At the free surface $x=d$, neglect air shear, so $\tau (d)=0$ and hence $v_z^{\prime }(d)=0$. The flat free surface has normal pressure $p_{\mathrm{atm}}$, neglecting surface-curvature effects. These are a velocity condition at the wall and a stress condition at the free surface, not zero velocity at both boundaries.

The parabolic speed profile. Integrating ${\tau }^{\prime }=-\rho g$ from the shear-free surface gives $\tau (x)=\rho g(d-x)$. Thusand the wall condition yieldsIt is zero at the wall and largest at the free surface, where its gradient vanishes. The wall shear $\rho gd$ supports the downslope weight of the film per unit plate area.

Volume flux per unit width. Integrate through the depth:This is volume flow per unit length in $y$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$.

Inclining the plate. Keep $x$ normal outward from the plate and use $s$ downslope. The gravitational components are $g\sin \theta$ downslope and $-g\cos \theta$ normal to the plate. For a uniform film of normal thickness $d$, the steady slice equations areThe free-surface pressure condition givesUniform $d$ makes ${\partial }_{s}p=0$. The same no-slip and zero-surface-shear conditions then give the uniform gravity-driven Newtonian film,The component normal to the plate changes the hydrostatic pressure, while the tangential component drives the flow. The thickness is measured normal to the tilted plate.

The inclined-film volume flux. Integrating the new profile givesIt recovers the vertical result at $\theta =\pi /2$ and vanishes on a horizontal plate at fixed uniform depth. A varying depth, imposed pressure gradient or significant air shear would require additional terms; the solution here uses the uniform film specified by the simple draining-plate model.

Use the pipe-diameter convention for the Reynolds number, with mean velocity $\bar{v}$, density $\rho$ and dynamic viscosity $\mu$:The Reynolds number compares inertial and viscous effects in the fluid flow. A conventional estimate for the onset of turbulence in a pipe is ${\mathrm{Re}}_{\mathrm{crit}}\simeq 2000$, givingHence the usual onset estimate is $\boxed{D\gtrsim 1.6\mathrm{mm}}$. Using the also common transition criterion ${\mathrm{Re}}_{\mathrm{crit}}=2300$ gives $1.84\mathrm{mm}$, whereas $\mathrm{Re}\gtrsim 4000$, used for established turbulence, gives $D\gtrsim 3.2\mathrm{mm}$. There is no universal sharp minimum: inlet disturbances and pipe conditions affect the transition from laminar flow, so the threshold convention must accompany the answer.

The independent fluid-mechanical relation iswhere $\rho$ is fluid mass density, $\eta$ dynamic viscosity and $\nu$ kinematic viscosity. For a generic rigid sphere, the convention must specify whether $L$ is its diameter or radius; their Reynolds numbers differ by a factor of two. The ratio compares advective inertia $\rho U^2/L$ with viscous stress gradients $\eta U/L^2$. Small Reynolds number supports a creeping-flow approximation when unsteady fluid inertia is also negligible. No organism-specific size, speed or propulsion calculation is made here.

The ordinary physics is Stokes settling velocity. For a generic isolated rigid sphere of radius $a$, particle mass density ${\rho }_p$ and surrounding-fluid mass density ${\rho }_f$, balance the buoyancy-corrected body force against Stokes law:Here $\eta$ is the dynamic viscosity. If the effective acceleration is approximately constant over the displacement and the terminal regime has been reached, $s=Ut$. In an actual centrifuge $g_{\mathrm{eff}}={\Omega }^{2}r$, so a sufficiently large radial change requires integrating $dr/dt\propto r$ instead of assuming constant $U$. This generic relation does not resolve the deferred biological application.

The Reynolds numbercompares advective inertia with viscous force on a specified length scale $L$ and velocity scale $U$. Here $\rho$ is fluid mass density, $\eta$ is dynamic viscosity and $\nu =\eta /\rho$ is kinematic viscosity. In the Navier-Stokes equation, advective acceleration scales as $U^2/L$, while the viscous term per unit mass scales as $\nu U/L^2$; their ratio is Reynolds number.

When $\mathrm{Re}\ll 1$ and unsteady fluid inertia is also negligible, the incompressible flow equations reduce to Stokes flow:Thus pressure, applied body forces and viscous stresses balance at each instant. For a translating rigid sphere, drag is proportional to velocity, $F_D=6\pi \eta aU$, rather than the inertial scale $\rho a^2{U}^2$. Given the same instantaneous geometry, reversing the imposed boundary velocities reverses the Stokes flow; there is no fluid-inertial coasting in this approximation. This statement concerns the fluid limit and does not discard a separately significant particle inertia.

A low steady Reynolds number alone does not justify dropping rapid unsteady acceleration. For forcing frequency $\omega$, the additional dimensionless ratio is $\rho \omega L^2/\eta$. No organism-specific sizes, velocities or propulsion mechanism are assumed here.

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.

The Reynolds number compares inertial and viscous effects:where $\rho$ is fluid mass density, $U$ a characteristic swimming speed, $L$ a characteristic body length, and $\eta$ the dynamic viscosity. Both swimmers are in the same fluid, so the fluid properties cancel from the ratio:For an order-of-magnitude estimate, take $L_b=2\mu \mathrm{m}$, $U_b=20\mu \mathrm{m}{\mathrm{s}}^{-1}$, $L_w=10\mathrm{m}$ and $U_w=5\mathrm{m}{\mathrm{s}}^{-1}$. With water values $\rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$ and $\eta \simeq 10^{-3}\mathrm{Pa}\mathrm{s}$, these giveand henceThe dimensions and swimming speeds are estimates, so this ratio is not a universal species-independent constant. Its physical significance is robust: viscous forces dominate the bacterium's motion, whereas fluid inertia is crucial for the whale.

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.