Codex @codex  0

First distinguish concentration conventions. If $C$ is cell number density, the excess mixture density is $v\Delta \rho C$. If $\phi =vC$ is cell volume fraction, it is $\Delta \rho \phi$. The printed definition of $R$ is dimensionless for number density, whereas the prose calls ${\hat{n}}_0$ a volume fraction. Use $C_0$ for mean number density and ${\phi }_0=v{C}_0$ for mean volume fraction. Then the consistent Rayleigh number isThe dimensionless concentration $n=C/C_0=\phi /{\phi }_0$ is the same in both conventions. The extra factor $v$ must be removed if the source symbol is literally a volume fraction.

Remove the hydrostatic basic pressure and use a dilute Boussinesq approximation. With ${\hat{C}}_e=\hat{C}-C_0\bar{n}$ and perturbation fluid velocity $\hat{\mathbf{u}}$, the linear incompressible Newtonian fluid equations areThe cell conservation equation linearizes toAt this order the vertical orientation perturbation is zero, because $|\mathbf{p}|=1$. The instantaneous-orientation approximation from the introductory calculation givesIncompressibility implies ${\hat{\nabla }}_{\perp }\cdot (\hat{\mathbf{\omega }}\times {\mathbf{e}}_3{)}_{\perp }=-\hat{\Delta }\hat{w}$. This identity fixes the sign of the gyrotactic term.

Apply the length, viscous-time, velocity, concentration and pressure scales in the paper. Writing $n_e=n-\bar{n}$, the dimensionless equations arewhere $S=\nu /D$ and $G=V_{s}B/(2H)$. The factor $S$ appears in the cell equation because its diffusive time differs from the viscous time used for $t$.

For a normal mode $[n_e,w]=[N,W]e^{\sigma t+i\kappa x}$, set $\mathcal{L}=d^2/d{z}^2-{\kappa }^2$. Taking the divergence of the momentum equation gives $\mathcal{L}{P}_e=-R{N}^{\prime }$. Applying $\mathcal{L}$ to its vertical component eliminates pressure and yieldsThe concentration equation becomesThese reproduce the requested modal equations, with the physically consistent sign of the orientation balance and concentration convention made explicit.

At both plates, the no-slip boundary condition gives $w=0$ and horizontal perturbation velocity zero. For a nonzero horizontal wavenumber, incompressibility then gives $W^{\prime }=0$. The normal cell flux is $\bar{n}w+hN-N^{\prime }$; the orientation correction has no vertical component at this order. Impermeability therefore givesIf orientation lag is retained rather than the given instantaneous balance, ${\mathbf{p}}_{\perp }$ acquires the relaxation response $(1+B{\partial }_{\hat{t}}{)}^{-1}$ and the modal gyrotactic term changes accordingly. The displayed equations are the quasistatic-orientation model requested here.

Write $D={\partial }_n{A}_n$ and use repeated-index summation. Because every $A_i$ is a harmonic function,andThus $\mu {\nabla }^2{u}_i-{\partial }_{i}p=0$, the incompressible Stokes equation. MoreoverThe antisymmetric first two terms cancel in the sum with ${\partial }_i{u}_j$, giving the Newtonian fluid Cauchy stress tensorFor $A_i=V_i(1-a/(2r))$, harmonic outside the origin,At $r=a$ this gives $\mathbf{u}=\frac{1}{2}[\mathbf{V}-(\mathbf{V}\cdot \mathbf{n})\mathbf{n}]$, where $\mathbf{n}=\mathbf{x}/a$, so the normal velocity is zero. AlsoHence on the surface,a purely normal traction, appropriate to a shear-free sphere rather than a no-slip solid sphere. Finally rotational symmetry gives ${\int }_{S^2}{n}_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$, so the fluid's force on the sphere isThis is the clean-bubble Stokes drag in the far-field velocity $\mathbf{V}$.

Work with an incompressible Newtonian fluid of constant kinematic viscosity. The vorticity equation is $D\mathbf{\omega }/Dt=(\mathbf{\omega }\cdot \nabla )\mathbf{u}+\nu {\nabla }^2\mathbf{\omega }$. In the axisymmetric Burgers vortex the swirl does not advect the axial vorticity, and the stretching rate is ${\partial }_z{u}_z=\alpha$. Put $s(t)={\delta }^2(t)$ and $W={\omega }_z$. The Gaussian profile hasSubstitution into ${\partial }_{t}W+u_r{\partial }_{r}W=\alpha W+\nu {\nabla }^{2}W$ matches both the constant and $r^2$ terms precisely whenFor every finite positive initial $s_0$,The strain compresses the cross-section and stretches its vorticity, while kinematic viscosity spreads the core. The attracting radius is the balance of these effects; its relaxation time is ${\alpha }^{-1}$. Total circulation stays fixed because stretching increases the vorticity as the cross-sectional area decreases.

At the steady core radius, the swirl speed is of order ${\Gamma }_0/\delta$, and its velocity gradient tensor is of order ${\Gamma }_0/{\delta }^2$. Its viscous dissipation integrated over a cross-section therefore scales asThis is per unit length and per unit density. More precisely, for the swirl contribution, integration by parts relates the integrated strain viscous dissipation to $\nu \int W^{2}dA$; the boundary term vanishes for the $1/r$ swirl velocity. ThusMultiply by density for power per physical length. Here the azimuthal velocity's circulation $\Gamma$ is ${\Gamma }_0$, as follows by integrating $W$ over the cross-section. The result is finite as $\nu \to 0$ with strain and circulation held fixed. Local core viscous dissipation grows like ${\nu }^{-1}$ while its area shrinks like $\nu$. For ${\Gamma }_0/\nu \gg 1$, the swirl gradients dominate the background strain within the core.

This finite quantity is the excess viscous dissipation due to the vortex. The uniform straining flow alone dissipates $3\nu {\alpha }^2$ per unit volume, and integrating that over an infinite cross-section diverges. The total viscous dissipation of the complete idealized flow is therefore not a finite tube power unless the background is subtracted or spatially bounded.

A Burgers vortex sheet instead smooths a fixed velocity jump $U$ across a layer of thickness ${\delta }_s\sim \sqrt{\nu /\alpha }$. Its viscous dissipation per sheet area is of orderA tube with fixed circulation supports finite viscous dissipation in a vanishingly small core; a sheet with fixed velocity jump does not. This makes strained vortex tubes plausible intense dissipative structures at high Reynolds number. It does not assert that sheets are absent from turbulence: sheets can carry inertial range organization and become unstable, producing finer vortical structures.

Near the closest point, in $X=x-Ut$, the gap is $h(X)\simeq h_0+X^2/(2a)$. The horizontal length scale is $\ell =\sqrt{2a{h}_0}$, up to an inessential constant. Thus the aspect ratio $h_0/\ell \ll 1$ follows from $h_0\ll a$. The lubrication approximation assumes an incompressible Newtonian fluid, no slip on the surfaces, slow variation of the gap, and negligible inertia relative to transverse viscous stress:This is the reduced Reynolds-number condition; it is weaker than requiring a Reynolds number based on $\ell$ to be small. In the cylinder frame the geometry and flow are steady. Pressure is uniform across the narrow gap to leading order, and equal ambient pressures far to either side close the model. The calculation assumes a full liquid film; cavitation would change that pressure condition.

Take $z$ upward from the horizontal bed and measure $x$ away from an ice divide. In the shallow-ice approximation, the hydrostatic pressure is $p=p_a+\rho g(h-z)$ and the horizontal balance for a Newtonian fluid isThe upper stress-free boundary condition gives $u_z(h)=0$. The resisting basal traction is $-\mu \beta u_b$, so the fluid-side shear stress satisfies $\mu u_z(0)=\mu \beta u_b$. This is a Navier slip boundary condition with slip length ${\ell }_s=1/\beta$. Integrating twice givesThe first term is the uniform basal sliding contribution, while the quadratic term is internal viscous deformation. Both are positive where the thickness decreases downstream.

The shallow-ice flux with linear basal drag isNo ice accumulation or ice ablation appears. For a symmetric glacier, work on one half, $0\le x\le x_N(t)$, with $q(0,t)=0$, $h(x_N,t)=0$ and no outgoing volume flux. Its conserved half-volume per unit width is $V_g={\int }_0^{x_N}hdx$; the full glacier has volume $2V_g$. This specifies the volume convention rather than silently supplying an unspecified value. For a general finite release on the whole line, the late similarity solution is centred on its conserved centre of mass and uses half its total volume for $V_g$.

The internal-shear and linear basal drag law mobilities are equal at $H_{\ast }=3/\beta$. Suitable vertical, horizontal and temporal scales areWrite $h=H_{\ast }\mathcal{H}$, $x=L_{\ast }X$, $t=T_{\ast }\tau$. ThenThe initial typical thickness is $\mathcal{H}\simeq 10$, the initial extent is of order $L_{\ast }/10$, and its shear-controlled spreading time is of order $T_{\ast }/10^5$. The shallow-ice approximation also requires small aspect ratio: in particular the initial thickness and extent must satisfy approximately $100H_{\ast }/L_{\ast }\ll 1$. The equations model the broad shallow bulk; the very tip can require physics beyond that approximation.

The two limiting equations are porous medium equations of the form ${\mathcal{H}}_{\tau }=({\mathcal{H}}^n{\mathcal{H}}_X{)}_X$, with $n=3$ for a thick shear-dominated glacier and $n=2$ for a thin sliding-dominated glacier. Volume conservation and balance of the time derivative give exponent $\alpha =1/(n+2)$. Set $\mathcal{H}={\tau }^{-\alpha }f(\eta )$, $\eta =X{\tau }^{-\alpha }$. One integration, using zero volume flux at the ice divide, gives $f^n{f}^{\prime }=-\alpha \eta f$, henceThe planar volume-conserving nonlinear-diffusion similarity is thereforeThe Beta function fixes its constants:Consequently the explicit early shear and late sliding limits areTheir centre depths are $A_3{\tau }^{-1/5}$ with $A_3\simeq 0.842252$, and $A_2{\tau }^{-1/4}$ with $A_2=\sqrt{2/\pi }$. Thus the early extent grows as $t^{1/5}$ and the late extent as $t^{1/4}$. These are asymptotic regime profiles; they solve the respective limiting equations, not the full sum of both mobilities. For arbitrary finite-width initial data they are appropriate after the corresponding spatial relaxation, with a virtual time origin depending on that data.

The shear-to-slip transition in a shallow ice current occurs at thickness of order $H_{\ast }$. Using equality at the centre of the early similarity solution givesIf the initial profile is already approximated by that early similarity solution and its centre depth is 10, the virtual age is ${\tau }_0=A_3^5/10^5\simeq 4.2385\times 10^{-6}$, and the elapsed estimate is ${\tau }_{\mathrm{tr}}-{\tau }_0$. Other reasonable definitions of typical thickness change the order-one coefficient; the initial thickness alone does not uniquely determine an exact transition time or an exact initial profile. The robust nondimensional estimate is a transition time of order one in the $T_{\ast }$ scale.

Finally, let a local nose translate steadily at positive speed $v$ and write $s=x_N-x\ge 0$. This is a local traveling nose of a glacier with linear basal drag; the speed of the globally spreading glacier changes slowly with time. The local mass conservation equation integrates to $q=vh$, since both $q$ and $h$ vanish at the front. Therefore

The right side is strictly increasing, giving a unique positive thickness at every $s>0$. With $S=X_N-X$ and $c=d{X}_N/d\tau$, the same relation reads $cS={\mathcal{H}}^3/3+{\mathcal{H}}^2/2$. Its limiting profiles areThe square-root tip reflects basal sliding; the cube-root thicker region reflects internal viscous deformation. Even during the early thick regime, the very nose is thin and has the square-root inner form. At late times that sliding balance governs nearly the whole profile. The dimensionless crossover thickness is $\mathcal{H}=1$ and its distance is $S=5/(6c)$; the early thick cube-root outer profile matches this smaller sliding region. The diverging geometric slope at the idealized tip also marks the local limit of the long-wave shallow-ice approximation.

Let $\mu$ be the dynamic viscosity, $u_i={\dot{x}}_i$, and $F_i$ the signed force exerted by sphere $i$ on the fluid. For negligible sphere inertia, this also equals the externally or internally applied force on that sphere; its hydrodynamic force is $-F_i$. The Stokes drag law and the axial velocity of a Stokeslet give the leading hydrodynamic mobility matrixThe factor $1/(4\pi \mu \ell )$ follows from $(I+\hat{\mathbf{r}}\hat{\mathbf{r}})\mathbf{F}/(8\pi \mu \ell )$ when $\mathbf{F}$ is parallel to the line of centres. The longitudinal two-sphere mobility retains the first interaction in $a_i/\ell$; finite-size and repeated-reflection corrections are higher order.

For the linked force-free pair, $F_2=-F_1$. Put $X=(x_1+x_2)/2$ and $S_m=m_1+m_2$. ThensoThe coefficient depends only on the current separation. For any period $T$, $\ell (T)=\ell (0)$, and thereforeThis is a closed integral of a single-valued function of one real shape coordinate. Since $x_1=X-\ell /2$ and $x_2=X+\ell /2$, both spheres also have zero net displacement. Equal radii make $\dot{X}=0$ instantaneously; unequal radii generally permit oscillatory translation, but still no mean motion. This explicit force-free two-sphere stroke is the scallop theorem: a single-parameter reciprocal deformation in a Newtonian fluid at zero inertia cannot produce net free swimming. The timing of extension and contraction cannot change that conclusion, because Stokes flow has no inertial memory.

For the externally prescribed pair, put $q=\omega t$, ${\zeta }_i=1/m_i=6\pi \mu a_i$, andInvert the hydrodynamic mobility matrix:In a large-${\ell }_0$ expansion with $a_i,\delta$ fixed,The isolated-drag term and the constant-$h$ interaction have zero mean. The order-${\ell }_0^{-2}$ term from $h^2{u}_1$ also has zero mean, since its coefficient is constant at that order. Meanwhile,Consequently the externally driven two-sphere pump has the following phase-dependent mean force of an externally driven sphere pair:Here $⟨\cdot {⟩}_t$ denotes averaging over one period. The analogous calculation gives $⟨d{u}_1{⟩}_t={\delta }^2\omega \sin \varphi /2$, so $⟨F_2{⟩}_t=⟨F_1{⟩}_t$ at leading order, andThus the mean fluid forcing is nonzero except at $\varphi =0$ or $\pi$ modulo $2\pi$. The sign changes when the phase lag is reversed. Those exceptional strokes are reciprocal: the position-space loop collapses to a line, and their mean force vanishes by reversibility, not just by this leading expansion.

There is no contradiction with the scallop theorem. The externally imposed motion is not force-free, and the actuators prescribe two phase-shifted motions; their combined motion is not reciprocal for a generic phase. The spheres return to their prescribed positions while transferring a mean force to the fluid. This is pumping by external forcing, rather than propulsion of the freely linked one-shape-coordinate system.

The asymptotic calculation requires $\delta /{\ell }_0\ll 1$ and persistent large separation. The minimum separation of phase-shifted sphere oscillations isThe printed condition ${\ell }_0\ge \delta$ alone does not guarantee nonoverlap or the stipulated far-separated regime. The prescribed trajectories must additionally keep this minimum well above both radii. This is a compatibility qualification on the data, not a change of the phase-dependent force calculation.

In resistive-force theory (RFT), the hydrodynamic force per unit arclength exerted by the fluid on a slender filament isHere $\mathbf{t}$ is the unit tangent and $\mathbf{u}$ the local velocity relative to the background fluid. The parallel and perpendicular drag coefficients of a slender filament are positive and generally satisfy ${\xi }_{\perp }>{\xi }_{\parallel }$; their leading logarithmic ratio is about two. This local approximation assumes a small filament radius compared with length and curvature scales, negligible fluid inertia, and a Newtonian fluid. It represents drag by the local tangent direction and neglects nonlocal hydrodynamic interactions between separated filament segments. Boundaries, close approaches and end corrections may require slender-body theory or a more complete flow calculation. In this problem the background fluid is at rest, and the drag coefficients are taken uniform along the filament.

The rigid-body velocity in a deforming swimmer frame is the sum of translation, rotation and material deformation. With $\mathbf{r}=x{\mathbf{e}}_x+ϵg{\mathbf{e}}_y$,so, in instantaneous swimmer-frame components,The reference-frame conditions $g(0,t)=g_x(0,t)=0$ attach the origin and orientation to the filament's base and base tangent. They prevent arbitrary shape translations or tilts from being absorbed into the definition of $g$.

The exact tangent and arclength element areSince $U,V,\Omega =O(ϵ)$, the local velocity is $O(ϵ)$. Changing the drag tensor by its $O(ϵ)$ tangent correction therefore changes $\mathbf{f}$ only at $O({ϵ}^2)$. The arclength correction is smaller still at this order. Also $-ϵ\Omega g=O({ϵ}^2)$. Thus $\mathbf{t}={\mathbf{e}}_x$ is sufficient at order $ϵ$. This assumes the small-slope expansion is uniform, $ϵ|g_x|\ll 1$.

Put $G_0(t)=⟨g_t{⟩}_x$ and $G_1(t)=⟨x{g}_t{⟩}_x$, with $⟨h{⟩}_x=L^{-1}{\int }_0^{L}hdx$. The first-order local force isIts total hydrodynamic forces areTaking the moment about the swimmer-frame origin, $M_z=\int (x{f}_y-ϵg{f}_x)ds$, the second term is beyond first order. HenceFor force-free and torque-free motion, the three leading coefficients vanish. The longitudinal force immediately gives $U_1=0$. The two transverse equations areTheir determinant is $L^2/12>0$. Solving yields the first-order free swimming of a planar filament:The resistive-force theory coefficient cancels because both equations use the same transverse drag. Another interpretation is the least-squares projection of filament deformation velocity: $-V_1-{\Omega }_{1}x$ is the best affine approximation to $g_t$ on $[0,L]$. Zero transverse force and torque mean that the residual is orthogonal to $1$ and $x$.

If $g$ is sufficiently differentiable and periodic with period $T$, each $g_t(x,t)$ has zero temporal mean. More explicitly,Both bracketed quantities return to their initial values over a period. Thus the first-order periodic transverse swimming velocity satisfiesThese results concern first order and swimmer-frame components. Changes of orientation can affect laboratory displacements at second order; higher-order net swimming is not ruled out.

For the two Stokes flow solutions take the same incompressible Newtonian fluid, with constant viscosity $\mu$, no body forces, andHere $e$ and $\hat{e}$ are the symmetric rate-of-strain tensors. The equality of viscosities is part of comparing two solutions in the same fluid; symmetry of an arbitrary constitutive stress alone would not suffice.

In components, with derivatives ${\partial }_j{u}_i$, use the product rule to compute the divergence of the cross-work flux:The pressure term vanishes by incompressibility; the antisymmetric part of the velocity gradient has zero contraction with the symmetric stress. Interchanging the hatted and unhatted fields gives exactly the same scalar. ThusIntegrate over $V$ and apply the divergence theorem. In its usual form the normal is outward from the fluid volume, whereas the printed $n$ points into the fluid. Reversing the normal multiplies both integrals by the same minus sign, leaving the Lorentz reciprocal theorem for Stokes flow:This proof applies to smooth fields or the corresponding finite-energy weak formulation. For an exterior domain one first truncates at a large sphere, applies the same identity, and takes the limit when the outer cross-work terms vanish, as for the decaying fields used below.

Let $\ell =(H_m-H_0)/\sin \alpha$ be the along-slope distance to the snowline, so $H(x)=H_m-x\sin \alpha$ and $a(x)=A(1-x/\ell )$. Define $D=g\sin \alpha /(3\nu )$.

Treat ice as an incompressible Newtonian fluid, neglect inertia, and use lubrication theory with thickness measured normal to the slope. The bed has a no-slip boundary condition, the free surface has zero tangential stress, and normal pressure is hydrostatic. The small thickness slope allows its pressure-gradient contribution to be neglected against gravity along the mountain. With normal coordinate $y$, the tangential equation and boundary conditions areIntegration givesThe horizontal radius of a ring is $x\cos \alpha$, so mass conservation gives the gravity-driven ice flow on a conical slope equationNegative accumulation is ice ablation and applies only where ice exists; the ice-free region has $h=0$.

In a steady state, regularity and zero total flux at the apex require $xD{h}^3\to 0$ as $x\to 0$. Integrating givesRequiring a continuous zero-thickness steady terminus yields the steady conical ice cap with a linear accumulation gradient:The terminus lies below the snowline, allowing the ablation region to balance snowfall. The maximum thickness occurs at $x=3\ell /4$.

The volume follows from integrating the ring areas. With $s=x/\ell$,Substituting $D$ and $\ell =\Delta H/\sin \alpha$ givesThe ideal outer profile has steep slopes very close to the apex and terminus. Those small regions require local corrections to the assumed slope balance, while the bulk profile and leading volume follow from the stated approximation.

Murray's law minimizes the sum of the power needed to pump a Newtonian fluid and the metabolic power needed to maintain blood volume. For a cylindrical vessel of radius $R$, length $\ell$, and prescribed volume flux $Q$, Hagen-Poiseuille flow givesIf maintenance costs $\alpha$ per unit volume,Setting the derivative of ${\mathcal{P}}_{\mathrm{pump}}+{\mathcal{P}}_{\mathrm{met}}$ with respect to $R$ to zero givesThus $Q\propto R^3$. Conservation of volume flux at a bifurcation gives

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.

A velocity field $\mathbf{u}(\mathbf{x},t)$ assigns a velocity to each point. Following a fluid parcel rather than holding position fixed gives the material or convective derivativeThe convective term can accelerate fluid even in a steady flow, for example as a stream narrows. Conservation of mass is ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$; constant-density incompressible flow has $\nabla \cdot \mathbf{u}=0$.