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Let ${ϵ}_{\mathrm{loc}}(\mathbf{x})=2\nu S_{ij}{S}_{ij}$ and define a coarse-grained energy dissipation at scale $r$, for example by a ball average,This is the random variable denoted ${ϵ}_{\mathrm{AV}}(r)$. The refined similarity hypothesis uses the local supply of energy available to motions of size $r$, rather than replacing it everywhere by a global mean. In its inertial range form,At high local Reynolds number, $V$ has a universal conditional distribution independent of ${ϵ}_r$, forcing details and $r$ within the inertial range. This givesFor signed odd orders the coefficients carry the appropriate sign; absolute moments use their own coefficients. Universality is retained in the normalized conditional velocity statistics, while the distribution of ${ϵ}_r$ can retain integral-scale information. This distinction resolves the particular averaging objection: its nonuniversal viscous dissipation moments are now explicitly present.

The quoted second viscous dissipation moment does not by itself determine every longitudinal structure function exponent. The extra lognormal intermittency model assumes $Y_r=\log ({ϵ}_r/ϵ)$ is Gaussian. Write its variance as ${\sigma }_r^2$. Normalization of the mean requires $⟨Y_r⟩=-{\sigma }_r^2/2$, soThe second-moment estimate gives $\exp ({\sigma }_r^2)=B(\ell /r{)}^{\mu }$. Hence, with $q=p/3$,In particular ${\zeta }_3=1$, ${\zeta }_2=2/3+\mu /9$ and ${\zeta }_6=2-\mu$.

The most problematic step is the extrapolation of a lognormal viscous dissipation law to rare events and all moment orders. It is not entailed by a measured second moment. Its quadratic exponents eventually decrease with $p$ and become negative for $p>3+6/\mu$. If the velocity has finite uniformly bounded moments, the bound $⟨|\Delta v{|}^p⟩\le 2^p⟨|u_x{|}^p⟩$ prohibits such divergence as $r\to 0$. Thus the model cannot be an unrestricted high-order inertial range law. This is a criticism of the lognormal closure, rather than an algebraic contradiction in refined similarity itself.

For the Landau intermittency counterexample, combine two locally homogeneous turbulent populations, with the same sufficiently wide inertial interval but different viscous dissipation rates ${ϵ}_1$ and ${ϵ}_2$. This can mean an ensemble randomly selected between two intensities, or large turbulent patches whose small-scale increments are sampled away from the patch boundaries. Suppose each population individually obeys the proposed universal coefficient ${\beta }_p$. If their proportions are $w$ and $1-w$, the combined longitudinal structure function isIts effective coefficient referenced to the global mean viscous dissipation is thereforeFor a concrete example let $w=1/2$ and ${ϵ}_{1,2}=ϵ(1\pm d)$, with $0<d<1$. ThenFor $p=6$ this is $1+d^2$, explicitly dependent on the strength of the large-scale modulation. For $p=2$ it is less than one by concavity. Thus coefficients based solely on mean viscous dissipation cannot be universal under arbitrary integral-scale intermittency. The common inertial interval can be maintained by taking both local Reynolds numbers sufficiently large.

At $p=3$ the ratio is exactly one. This exception is consistent with the signed third-order energy-flux law and must not be used to rescue universality of all other orders. For noninteger orders one uses absolute increments; the even-order examples already establish the counterexample without a sign convention ambiguity. The issue is the difference between a moment of the local viscous dissipation and the same power of its mean, not a failure of averaging itself.

Let $\mathcal{E}=⟨|\mathbf{u}{|}^2⟩/2$ be the turbulent kinetic energy per unit mass. In addition to $L>0$ and high Reynolds number, use the usual self-preserving large-eddy decay assumption: the normalized large-scale correlation has a fixed shape, sowhere $C_L>0$ is constant. For freely decaying high-Reynolds-number turbulence, the cascade viscous dissipation law is $ϵ=C_{ϵ}{\mathcal{E}}^{3/2}/\ell$. The kinetic-energy balance therefore givesPut $A=C_{ϵ}{C}_L^{1/3}{L}^{-1/3}$. Integrating,The constant $t_{\ast }$ is a virtual time origin. This is the Saffman decay law. The energy decreases while the integral scale of turbulence grows so that the conserved large-volume momentum variance remains constant.

The exponent does not follow from the invariant alone: it also uses unforced decay, one evolving large-eddy length and an asymptotically constant cascade viscous dissipation coefficient. A changing normalized spectral shape or a confining boundary can invalidate that closure. Moreover $\mathrm{Re}\sim {\mathcal{E}}^{1/2}\ell /\nu \propto (t-t_{\ast }{)}^{-1/5}$, so at fixed kinematic viscosity the high-Reynolds-number decay regime eventually ends; the $-6/5$ law is not a claim about the ultimate viscous limit.

Let the fluid approach a uniform, quiescent state $({\rho }_0,p_0,c_0)$ at infinity. Use the outward unit normal vector $N=\nabla f/|\nabla f|$, pointing from the object into the fluid. Write $H=H(f)$ for the Heaviside step function, $q=\rho -{\rho }_0$, and ${\delta }_S=\delta (f)|\nabla f|$. With the viscous stress tensor ${\tau }_{ij}$, defineThus $T_{ij}$ is the Lighthill stress tensor, and $L_i$ is the surface loading exerted on the fluid. The total force on the object has the opposite sign. Extend $q$ by zero inside the object. Since the boundary is fixed and impermeable, $u\cdot N=0$, and the distributional mass conservation and momentum conservation equations areFor example, the surface contribution to the momentum equation is $(\rho u_i{u}_j+P_{ij})N_j{\delta }_S=L_i{\delta }_S$. Differentiating the first equation in time and subtracting the divergence of the second gives the fixed-body Ffowcs Williams-Hawkings equation:There is no acoustic thickness noise: a fixed impermeable boundary injects no mass. Apply the retarded acoustic Green function $\delta (t-R/c_0)/(4\pi c_0^{2}R)$, where $R=|x-y|$. The supplied surface-delta identity converts the loading term to a surface integral. For an exterior observer, the required integral equation isThis acoustic analogy is an exact rearrangement of the fluid equations; source approximations enter later. In the distant linear acoustic region, $p^{\prime }=c_0^{2}q$.

The acoustic far field requires observer distance $r$ much larger than the source extent $\ell$ and $k_{0}r\gg 1$, so derivatives of the retarded phase dominate derivatives of $1/R$. The acoustic compact-source approximation requires $k_0\ell \ll 1$, so the source's differential propagation delays can be neglected. These conditions concern different lengths and neither implies the other. Put $n=x/r$, $t_r=t-r/c_0$, and define the loading and stress momentsThe leading compact acoustic far field isThe surface term is an acoustic dipole; the volume term is an acoustic quadrupole. The plus sign of the loading term follows from the minus sign of its spatial divergence and the derivative of retarded time.

For the power comparison, assume low Mach number $M=U/c_0$, high Reynolds number, eddy size $\ell$, characteristic time $\tau =\ell /U$, and stress magnitude ${\rho }_0{U}^2$. Keep the prescribed turbulent region and its leading volume stress comparable when inserting the object; neglect changes of source statistics and exceptional cancellations of the leading moments. Compactness follows from $\ell /(c_0\tau )=M\ll 1$. Since $Q\sim {\rho }_0{U}^2{\ell }^3$ and the outgoing acoustic intensity is $⟨p^{\prime 2}⟩/({\rho }_0{c}_0)$, the volume power isIf the object's size is comparable to $\ell$ and its fluctuating resultant loading is $F\sim {\rho }_0{U}^2{\ell }^2$, thenAngular factors and dimensionless loading coefficients have been omitted. The acoustic dipole and acoustic quadrupole interference vanishes in the leading power integrated over a full sphere, by odd angular parity; individual directions can show interference.

A size threshold also requires a loading model. For a small nonseparating object of size $a\ll \ell$ in an approximately inviscid, smoothly varying incident eddy, a spatially uniform pressure has zero resultant on its closed surface. The first surviving loading is the pressure-gradient or added mass force, $F\sim {\rho }_0{a}^{3}U/\tau ={\rho }_0{U}^2{a}^3/\ell$. Under these explicitly stated assumptions,The two mechanisms become comparable at $a\sim \ell M^{1/3}$; for smaller objects the cases with and without the object have the same leading order of power. This is the small-body loading-noise threshold. It is not a geometry-independent size law: a separated drag force $F\sim {\rho }_0{U}^2{a}^2$ fluctuating on the original eddy time instead gives ${\mathcal{P}}_D/{\mathcal{P}}_Q\sim (a/\ell {)}^4/M^2$ and $a\sim \ell \sqrt{M}$. If both force and time are set by body-scale eddies, $\tau =a/U$, their loading power is ${\rho }_0{U}^6{a}^2/c_0^3$ and comparison with the original volume source gives $a\sim M\ell$. Each estimate states which source time and force it uses.

The physical distinction is momentum conservation. Unforced turbulence in a fluid supplies internal stresses, whose net-force contribution cancels and leaves acoustic quadrupole radiation. The supported object exchanges momentum with the fluid and the external support, permitting a time-dependent resultant force and acoustic dipole radiation. One fewer retarded derivative makes the loading radiation more efficient at low Mach number.

Take $\Delta \rho >0$ and $\tau >0$, with $x$ pointing along the imposed shear stress, and put $G=\Delta \rho g$. Relative to the upper fluid's hydrostatic reference, the hydrostatic pressure in the current is $p=G(h-z)$. Thus the horizontal pressure gradient is $G\nabla h$. The lubrication theory momentum balance, lower no-slip boundary condition, and imposed upper shear stress giveIts depth-integrated volume flux isDefine $D=G/(3\mu )$ and $A=\tau /(2\mu )$. The continuity equation gives the shear-driven viscous gravity current equationThe two terms have opposite roles: imposed shear stress carries fluid downstream, whereas the hydrostatic pressure gradient spreads it from thick regions towards thin ones.

For the approximation, a representative horizontal velocity is $U\sim \tau H/\mu +G{H}^3/(\mu L)$, with vertical velocity $UH/L$. Small slopes require $H/L\ll 1$. Horizontal fluid inertia relative to vertical viscous resistance is ${\rho }_{c}U{H}^2/(\mu L)$, where ${\rho }_c=\rho +\Delta \rho$. Hence sufficient small parameters, expressed without an unknown velocity, areFor the hydrostatic pressure and normal-stress approximation also require $\mu U/(GHL)\ll 1$. Its gravity-driven part is already $(H/L{)}^2$; its shear-driven part addsThis condition controls viscous normal stress, vertical viscous corrections and the normal projection of shear at a slightly tilted interface relative to $GH$. With it and the previous inertia bounds, vertical inertia is small as well. Time variations here are on the transport timescale $L/U$; separately imposed rapid forcing would need its own unsteady inertia bound. A small ordinary Reynolds number is a stronger sufficient restriction, but the reduced inertia ratios above are what the thin-layer momentum balance directly requires.

In the steady far-downstream regime, cross-stream derivatives dominate gravity-driven spreading. Dropping the smaller downstream gravity flux leavesWrite $Y=y_N(x)$ and let $h_c$ be a typical central thickness. Balance gives $Y^2\sim (D/A)h_c^{2}x$, while conservation of the source volume flux gives $Q\sim A{h}_c^{2}Y$. Therefore the downstream similarity of a shear-driven gravity current hasThe omitted downstream gravity flux relative to the imposed shear flux is $D{h}_c^2/(Ax)\sim (Y/x{)}^2\ll 1$, so this approximation is self-consistent in the stated regime.

For the full profile put $w=h^2$. Then $h^3{h}_y=(w^2{)}_y/4$ andThis is a porous medium equation with downstream distance as its evolution coordinate. Conservation of $\int wdy=Q/A$ and a similarity solution $w=x^{-1/3}f(\eta )$, $\eta =y/x^{1/3}$, givewhere symmetry gives zero integration constant at $\eta =0$. Inside the positive region, $f^{\prime }=-2A\eta /(3D)$, so $f=(A/(3D))({\eta }_N^2-{\eta }^2)$. Requiring the total downstream volume flux to be $Q$ fixes the coefficient:Consequentlyfor $|y|<y_N$, with $h=0$ outside. The squared height is a parabolic Barenblatt solution; the height cross-section is a semicircular profile after rescaling its axes. The cross-stream volume flux vanishes at the edges even though the height slope becomes singular there. The profile describes the outer lubrication theory region, not a resolved microscopic front.

Near the point source the two horizontal dimensions are comparable, say $\ell$. Upstream spreading arrests where outward gravity-driven volume flux balances downstream shear-driven volume flux:Eliminating $h_{\ast }$ givesThis is a scaling estimate, not a determination of a numerical prefactor. The associated depth is $h_{\ast }\sim (\mu Q/(\Delta \rho g){)}^{1/4}$. The same horizontal scale is obtained by setting $y_N\sim x$ in the downstream similarity solution, confirming where that solution fails.

For the line source the steady volume flux per unit width is $q=A{h}^2-D{h}^3{h}_x$. On the downstream constant-height branch, $q=Q_{2d}$ gives $h_d^2=Q_{2d}/A$. Upstream there is no net flux through the finite nose, so $q=0$. Thus $h{h}_x=A/D$, and continuity of height at the source givesThe line-source upstream reach under imposed shear is thereforeIn the upstream part, shear-driven and gravity-driven volume fluxes cancel. The volume flux per unit width jumps by exactly $Q_{2d}$ at the source. As in the point-source profile, the square-root nose is a formal outer solution with an unresolved steep front; its finite upstream reach follows from flux balance, not from assuming a small slope all the way to the nose.

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.

In a steady axisymmetric flow without swirl, time derivatives and azimuthal derivatives vanish. For a slender high-Reynolds number jet of width $\delta \ll z$, radial viscous derivatives dominate axial ones by $(z/\delta {)}^2$. The transverse momentum balance makes pressure nearly constant across the jet, matching the constant ambient pressure, so the leading axial pressure gradient vanishes. Retaining axial convection and radial diffusion in the Navier-Stokes equation gives the boundary-layer equationThe radial velocity is small but its convective term is of the same order as axial convection because radial gradients are larger.

The hydrostatic approximation gives $p=\rho g(h-z)$ above the horizontal floor, up to atmospheric pressure. The horizontal Stokes flow equation is $\mu {\mathbf{u}}_{zz}=\rho g\nabla h$, with the no-slip boundary condition at $z=0$ and zero tangential stress on the free surface. Integrating twice gives $\mathbf{u}=(\rho g/\mu )\nabla h(z^2/2-hz)$ and the lubrication gravity-current fluxFor the squeegee lubrication model, use $h_0$ vertically, $\ell =\rho g{h}_0^3/(\mu U)$ horizontally, and $\ell /U$ in time. If $X$ is the laboratory coordinate in the direction of travel, set $x=(X-Ut)/\ell$, $y=Y/\ell$, $t=U{t}_{\mathrm{lab}}/\ell$, and $h=h_{\mathrm{lab}}/h_0$. The blade half-length is $\alpha =L/\ell$. The moving-frame flux is $\mathbf{q}=-h{\mathbf{e}}_x-h^3\nabla h/3$. Conservation of volume gives the dimensionless lubrication equationUsing these scales as representative of the actual flow, lubrication theory requires a small aspect ratio and negligible inertia relative to vertical viscous resistance:The second is the Reynolds number based on $h_0$, multiplied by $h_0/\ell$. Surface tension is neglected as assumed. Large piles, rounded gaps, and narrow end regions need their own local slope and inertia checks if these representative scales cease to describe them.

Both eigenvalues of the linear generator, $-1/R$ and $-1/(4R)$, are negative. Nevertheless the non-normal matrix transfers the slowly decaying second component into the first before viscous decay wins. Modal decay coexists with energy amplification of order $R^2$ over a time of order $R$.

This illustrates transient growth from non-normal modes: large Reynolds number permits much greater disturbance amplification even though the linear stability analysis finds no unstable eigenvalue. The energy-neutral rotational nonlinearity can redirect the amplified state and affect its subsequent evolution. The linear gain alone establishes neither sustained nonlinear instability nor a nonlinear transition threshold.

For $f(u)=C{u}^p$, the dimensional expression is again $L=C{\eta }^{2-3p}{\rho }^{p-1}{\sigma }^{2p-1}{t}^p$. Independence of dynamic viscosity requires $2-3p=0$, giving inertial hydrodynamic coarsening:For this law, both inertial terms and the capillary term scale as $u^{-4/3}$, whereas the viscous term scales as $u^{-5/3}$. It is consequently subleading at large $u$. Direct substitution gives the leading amplitude conditionA growing branch requires the indicated ratio to be positive. The inertial combination must be nonzero: if $2\beta =\alpha$ its leading acceleration and convective contributions cancel, so this particular schematic equation cannot realize the generic inertial-capillary balance with that pure power law. The requested scaling assumes no such geometrical degeneracy. This also shows why the coefficients cannot determine a universal numerical prefactor without a morphology calculation. The relative viscous contribution decays as $u^{-1/3}$, consistently with an increasingly large domain-scale Reynolds number.

For a power law $f(u)=C{u}^p$, restoring dimensions givesIndependence of mass density requires $p-1=0$. Hence viscous hydrodynamic coarsening hasAt small $u$, the capillary and viscous terms are both of order $u^{-2}$, while the convective inertial term is of order $u^{-1}$ and the acceleration of the leading linear law is zero. Thus the leading force balance is viscous resistance against capillary driving, giving $C_v=c_s/c_v=-\delta /\gamma >0$ in the signed convention of the preceding part.

Equivalently, the domain-scale Reynolds number is $\rho L\dot{L}/\eta \sim C_v^{2}u$, so inertia is negligible as $u\to 0$. “Small” here refers to time compared with $t_0$ within a late-stage hydrodynamic window: it does not extend the model through the microscopic initial stage, where diffusion or interface formation can dominate. An initial-time offset changes the intercept, not this asymptotic growth exponent.

Choose a length $\ell$, velocity $U_s$, time $\ell /U_s$, and pressure $\rho U_s^2$, giving Reynolds number $Re=U_s\ell /\nu$. Write perturbation velocity as $(u,v,w)$ and let $\mathcal{L}={\partial }_t+U{\partial }_x-R{e}^{-1}{\nabla }^2$. Linearizing the incompressible Navier-Stokes equations givesTaking the divergence gives ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying ${\nabla }^2$ to the wall-normal momentum equation and usingthen cancels the pressure derivatives and yields $\mathcal{L}{\nabla }^{2}v-U^{\prime \prime }{v}_x=0$. Define wall-normal vorticity $\eta =u_z-w_x$. Applying ${\partial }_z$ to the streamwise equation minus ${\partial }_x$ to the spanwise equation gives $\mathcal{L}\eta =-U^{\prime }{v}_z$.

For the assumed normal modes, set $D=d/dy$ and ${\kappa }^2={\alpha }^2+{\beta }^2$. The two equations become the Orr-Sommerfeld equation and Squire equation:At rigid no-slip walls the conditions are $\hat{v}=D\hat{v}=\hat{\eta }=0$ for nonzero horizontal wavenumber.

Put $q=10/Re>0$. The frequency formula becomesFor large positive Reynolds number, $\sqrt{q-q^2/4}=\sqrt{q}[1-q/8+O(q^2)]$. ThereforeMore precisely the real correction is $-5/Re$, and the next imaginary correction is $O(R{e}^{-3/2})$. The growing branch has a rate tending to zero as $Re\to \infty$, despite being unstable at every sufficiently large finite $Re$.

Let $w$ be downward Darcy flux, measured per unit total horizontal area. Neglect the resistance of the displaced phase, capillary pressure, and horizontal subsurface flow. Taking atmospheric pressure as zero, the upper end of the saturated column has pressure $\rho gh$ and its advancing lower end has pressure zero. Across a column of depth $l$, Darcy law therefore givesThe pressure head from the overlying layer supplies $h/l$, and gravity supplies one. Because the new liquid fills a pore volume $\varphi dl$ per unit area, the vertical imbibition under a draining liquid layer obeys $\varphi \dot{l}=w$. Volume conservation gives $h+\varphi l=h_0$, so, until the surface layer is exhausted,The initial condition is understood as a limiting solution of this singular equation. For $0<\varphi <1$, separation of variables givesAt small penetration, expansion of the logarithm yields $t\sim \varphi \nu l^2/(2gk{h}_0)$, henceA sufficient dimensional early-time condition is $t\ll \varphi \nu h_0/(gk)$, for which $l\ll h_0$. The printed $t\ll 1$ needs a chosen time unit or nondimensionalization. In the limiting case $\varphi =1$, $l^2=2gk{h}_{0}t/\nu$ holds exactly until $h=0$. The divergent initial Darcy flux is an idealization of sudden contact with a dry continuum; it cannot describe penetration below the pore scale.

For the spreading surface layer, measure $z$ upwards from the substrate. A thin, slow gravity current has hydrostatic pressure $p=\rho g(h-z)$ and a leading horizontal viscous balance $\nu u_{zz}=g{h}_x$. A no-slip boundary condition at $z=0$ and a stress-free boundary condition at $z=h$ giveThis lubrication gravity-current flux drives horizontal spreading down the surface-pressure gradient. Surface volume conservation is $h_t+q_x=-w$, while conservation in the pores is $\varphi l_t=w$. Combining these balances gives the deep-substrate drainage of a gravity current modelDrainage transfers liquid from the surface layer into the substrate. Indeed $(h+\varphi l{)}_t+q_x=0$.

For definiteness take a one-sided current on $0<x<x_N(t)$, supplied at $x=0$. The imposed volume is per unit span, so $V_0$ has dimensions of area. The boundary conditions and volume conservation constraint areThe source starts from zero injected volume. Initially dry points acquire $l>0$ only when the advancing current reaches them; the formula $h/l$ is not applied outside that region. At the advancing nose these are limiting conditions for a singular drainage equation. For a source feeding two identical sides, each side has half the volume and inlet flux; only the corresponding numerical prefactors change. The integral and inlet-flux conditions are equivalent once the local balance and nose conditions hold.

The physical approximations are lubrication theory, negligible inertia, slowly varying surface height, small pore-scale Reynolds number, and negligible capillary entry pressure. Treating the substrate as vertically draining columns also requires its lateral flow to be negligible and its depth to exceed the penetration depth. The model is valid while surface liquid remains available above each draining column.

For a similarity solution, let $h$ and $l$ scale as $t^a$, and let the current length scale as $t^b$. Since $h/l$ then has no time dependence, drainage requires $a-1=0$. Balancing $h_t$ with the horizontal spreading term gives $a-1=4a-2b$, so $a=1$ and $b=2$. These exponents also give injected volume $t^{a+b}=t^3$, explaining the particular supply law.

Define scaleswhich satisfy $CD=V_0/{\tau }^3$ and $D^2=(g/\nu )C^3$. The cubic-input similarity for a draining gravity current isSubstitution into the two conservation equations giveswhereThus only $K$ and $\varphi$ remain in the dimensionless problem. Its constraints areIntegrating the sum of the two ordinary differential equations verifies that the inlet flux and integral normalization agree. The dimensionless endpoint ${\xi }_N={\xi }_N(K,\varphi )$ is the undetermined multiplicative constant in the length law; solving for it is not required.

The exact gap between the plane and the circle isIn the narrow region relevant to lubrication theory,so changes occur over the streamwise scale $x=O(\sqrt{a{h}_0})$; below use $x=\sqrt{2a{h}_0}X$.

Choose the positive $x$ direction to be the direction of the cylinder surface velocity $U=a\Omega$ in the gap, and write $\mu =\rho \nu$ for the dynamic viscosity. The lubrication velocity profile and volume flux per unit axial length areSince $Q$ is constant and the pressure returns to the same ambient value as $x\to \pm \infty$, ${\int }_{-\infty }^{\infty }{p}_{x}dx=0$. ThereforeUsing $h=h_0(1+X^2)$ and the supplied integrals givesin the direction of the cylinder surface motion.

The shear stress exerted by the fluid on the cylinder is opposite toEliminating $p_x$ and inserting $Q=2U{h}_0/3$ giveswhere the sign is relative to $U$. Sincethe resisting torque per unit axial length is

Finally, the ratio of streamwise fluid inertia $U^2/\sqrt{a{h}_0}$ to transverse viscous acceleration $\nu U/h_0^2$ is of orderThe stated restriction makes this lubrication-scale Reynolds number small, justifying neglect of inertia.

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.