The Dominance of Viscosity at Low Reynolds Numbers — Epoche C2
Two drag laws, and the question of which one applies A fog droplet ten micrometres in radius, settling through still air, feels a drag force proportional to the first power of its speed — Stokes' law, $F_D = 6\pi\mu R v$, with $\mu$ the dynamic viscosity of the air, $R$ the droplet radius and $v$ its speed. A cricket ball feels a drag proportional to the square of its speed, $F_D = \tfrac{1}{2}\rho v^2 C_D A$, where $\rho$ is the fluid density, $A$ the frontal area and $C_D$ a dimensionless drag coefficient of order unity. These are not competing accounts of the same thing; they are the two ends of one continuum, and the object of this essay is to establish exactly what governs the crossover, what the linear law follows from, and where the two laws leave a gap between them that neither covers. The quadratic law comes from momentum transfer: a body sweeping out volume $Av$ per unit time gives the mass $\rho A v$ in that volume a velocity of order $v$, so the momentum shed per unit time is of order $\rho A v^2$, and the coefficient $C_D$ absorbs the geometry of how the fluid actually gets out of the way. Nothing in that argument mentions viscosity. The linear law is what remains when viscosity is the only thing left in the problem, and it is the more surprising of the two, because it makes the drag on a sphere independent of the density of the fluid it moves through. Where the Reynolds number comes from The criterion that decides between the two laws is not imposed from outside; it falls out of the equation of motion. For an incompressible Newtonian fluid, $$\rho\left(\frac{\partial \mathbf{v}}{\partial t} + (\mathbf{v}\cdot\nabla)\mathbf{v}\right) = -\nabla p + \mu\nabla^2\mathbf{v}, \qquad \nabla\cdot\mathbf{v} = 0 .$$ Compare the two terms that could dominate the balance. For a flow with characteristic speed $U$ varying over a characteristic length $L$, the inertial term $\rho(\mathbf{v}\cdot\nabla)\mathbf{v}$ is of order $\rho U^2/L$, and the viscous term $\mu\nabla^2\mathbf{v}$ is of order $\mu U/L^2$. Their ratio is $$\frac{\rho U^2/L}{\mu U/L^2} = \frac{\rho U L}{\mu} \equiv Re ,$$ which is the Reynolds number, and which is therefore not a definition but a result: it is the only dimensionless group that measures the relative weight of the two terms. The same conclusion arrives more formally on scaling lengths by $L$, velocities by $U$, time by $L/U$ and — this choice matters — pressure by the viscous scale $\mu U/L$ rather than the inertial scale $\rho U^2$. Writing tildes for the scaled variables, $$Re\left(\frac{\partial \tilde{\mathbf{v}}}{\partial \tilde{t}} + (\tilde{\mathbf{v}}\cdot\tilde\nabla)\tilde{\mathbf{v}}\right) = -\tilde\nabla \tilde{p} + \tilde\nabla^2 \tilde{\mathbf{v}} .$$ The whole inertial side of the equation carries the factor $Re$. Setting $Re$ to zero is not an approximation applied to one term among several; it deletes the only nonlinear term in the problem and leaves a linear system, the Stokes equations, $\nabla p = \mu\nabla^2\mathbf{v}$ with $\nabla\cdot\mathbf{v} = 0$. Note also that time has vanished except through the boundary conditions — a fact with consequences taken up at the end. Two cautions about the number itself. First, $L$ is a convention, and the convention must be stated: for a sphere one may use the radius or the diameter, and the resulting values differ by a factor of two. Everything below uses the diameter, $L = d = 2R$, which is the choice that makes the standard correlations come out right. Second, the critical value of $Re$ is not universal but geometry-specific and, in some geometries, not even sharp. Osborne Reynolds established the group experimentally in 1883 by injecting dye filaments into water flowing through glass tubes and observing where the filament broke up: transition from what he called direct to sinuous motion occurred at a value of the group near two thousand under ordinary inlet conditions, and could be pushed several times higher by suppressing disturbances at the entry. A number governing a transition can be well defined without the transition being. The range in question The following values are computed from $Re = \rho v d/\mu$ with $\rho = 1.2\ \mathrm{kg\,m^{-3}}$, $\mu = 1.8\times10^{-5}\ \mathrm{Pa\,s}$ for air and $\rho = 10^{3}\ \mathrm{kg\,m^{-3}}$, $\mu = 10^{-3}\ \mathrm{Pa\,s}$ for water. Object Size and speed Fluid $Re$ E. coli swimming 2 µm at 30 µm/s water $6\times10^{-5}$ fog droplet settling 20 µm at 1.2 cm/s air $1.6\times10^{-2}$ raindrop falling 2 mm at 6.5 m/s air $8.7\times10^{2}$ cricket ball 72 mm at 30 m/s air $1.4\times10^{5}$ swimmer 1.8 m at 1.5 m/s water $2.7\times10^{6}$ Close to eleven orders of magnitude separate the first row from the last — the ratio is $4.5\times10^{10}$ — and the two drag laws sit at its two ends. It is worth noticing already that the raindrop, the stock example of quadratic drag, sits at $Re \approx 870$ — below, not inside, the range in which the quadratic law is actually clean. That point is taken up in the conclusion. What Stokes actually solved Stokes' 1851 memoir on the effect of internal friction on the motion of pendulums contains, among much else, the solution for steady slow flow past a sphere. The problem is to solve the Stokes equations outside a sphere of radius $R$ subject to no slip at the surface — both velocity components vanishing on $r = R$, since the fluid adheres to the solid — and to a uniform stream $U$ at infinity. For axisymmetric flow the two velocity components can be replaced by a single stream function $\psi(r,\theta)$, and the Stokes equations reduce to $E^4\psi = 0$, whose general solution regular in $\theta$ is a combination of $r^{-1}$, $r$, $r^2$ and $r^4$ multiplying $\sin^2\theta$. The $r^4$ term is discarded because it grows faster than the uniform stream at infinity; the $r^2$ coefficient is fixed at $U/2$ by matching that stream; the remaining two coefficients are fixed by the two no-slip conditions. The re