The wall that shakes

Take a flat wall in a fluid at rest and slide it back and forth in its own plane, a hundred times a second. How far up does the fluid know?

The answer is a fifth of a millimetre in air, it can be computed exactly, and the computation is one of the handful of cases where the Navier–Stokes equations can be solved in closed form without approximating anything at all.

Why this one is exact #

The Navier–Stokes equations are hard because of one term: $(\mathbf{u}\cdot\nabla)\mathbf{u}$, which is quadratic in the unknown and couples everything to everything.

In a parallel flow that term is identically zero. If the velocity is $u(y,t)$ in the $x$ direction and nothing else, then $\mathbf{u}\cdot\nabla u = u\,\partial u/\partial x = 0$, because $u$ does not depend on $x$. The nonlinearity does not need to be small; it is absent.

What is left is the diffusion equation:

$$\frac{\partial u}{\partial t} = \nu\frac{\partial^2 u}{\partial y^2}$$

with the wall’s motion $u(0,t) = U_0\cos\omega t$ as its boundary condition. That is a linear equation with constant coefficients, and it can be solved.

The whole family of exact Navier–Stokes solutions is essentially this observation used in different geometries: flow along a pipe, flow between plates, flow above an impulsively started wall, and this one. The equations are solvable exactly in the cases where they are not the equations everybody complains about.

The solution, and the length it contains #

Seeking a solution of the form $e^{i\omega t}$ times a function of $y$ gives

$$u(y,t) = U_0\,e^{-y/\delta}\cos\!\left(\omega t - \frac{y}{\delta}\right), \qquad \delta = \sqrt{\frac{2\nu}{\omega}}$$

and the striking part is that $\delta$ appears twice, doing two different jobs. It is the depth over which the amplitude falls by a factor of $e$, and it is also the depth over which the motion falls one radian behind the wall. Amplitude and phase are controlled by the same length, so a wave that has travelled far enough to be a quarter cycle behind has already lost 80 per cent of its amplitude.

That is what “dies within a wavelength” means quantitatively. A full wavelength is $2\pi\delta$, and $e^{-2\pi} = 0.0019$: the second crest is two parts in a thousand of the first, which is why nobody ever draws it.

What the solver computed, and how it was checked #

Three things, and the arrangement of the first is the site’s standard discipline.

The diffusion equation, differenced off the profile. $\partial u/\partial t$ and $\nu\,\partial^2 u/\partial y^2$ are evaluated at a lattice of depths and phases using five-point stencils in both variables, on the function the figures draw. The worst relative residual is $3\times10^{-9}$. A residual computed from the algebra that produced the answer would be a check on the typing; this one is a check on the drawing.

The decay and the lag, measured. Both come out at one per $\delta$ to five decimal places, by a procedure that would work equally well on laboratory data.

A rejection that looks like nothing. The same solution with the decay length made 30 per cent larger than the phase length is still a smooth, decaying, plausible-looking wave — and its residual is eight orders of magnitude worse. Nothing about the picture would give it away. That is the characteristic failure this site’s assertions exist for.

The numbers, which are small #

The depth is $\sqrt{2\nu/\omega}$, and since $\nu$ for air is $1.5\times10^{-5}\ \mathrm{m^2/s}$ the answers are millimetres and fractions of millimetres.

Water’s layer is thinner than air’s, which is worth pausing on. Water is far more viscous than air in the ordinary sense — its dynamic viscosity is fifty times greater — but what governs diffusion is the kinematic viscosity $\nu = \mu/\rho$, and water is so much denser that its $\nu$ comes out fifteen times smaller. Momentum diffuses through air more readily than through water, and this figure is that fact measured in millimetres.

Reading a viscosity off the lag #

The two roles of $\delta$ make the solution into an instrument.

Measure the fluid’s velocity at a known height $y$ above an oscillating wall and record how far behind the wall it runs. The lag is $y/\delta$ radians, so

$$\nu = \frac{\omega y^2}{2\,(\text{lag in radians})^2}$$

and the viscosity has been obtained from a phase measurement, a height and a frequency — with no force measured anywhere, and no calibration of anything.

The amplitude gives a second, independent estimate from the same experiment: the ratio of the local amplitude to the wall’s is $e^{-y/\delta}$, so its logarithm gives $y/\delta$ again. Two routes to the same length, sharing no arithmetic, which is the arrangement this site trusts most. If the two disagree, something is wrong with the assumption of a parallel flow — an edge is being felt, or the layer has gone turbulent — and the disagreement is the diagnostic.

That is not a hypothetical use. Oscillating-plate and oscillating-cylinder viscometry works exactly this way, and the reason the method is good is that a phase is easy to measure precisely and a small force is not.

The drag arrives before the motion #

The shear stress at the wall is $\mu\,\partial u/\partial y$ evaluated at $y = 0$, and it comes out as

$$\tau_w = -\mu\frac{U_0}{\delta}\left(\cos\omega t + \sin\omega t\right) = -\sqrt{2}\,\mu\frac{U_0}{\delta}\cos\!\left(\omega t - \frac{\pi}{4}\right)$$

The shear leads the wall’s velocity by exactly 45°, at every frequency, in every fluid.

The split is what makes that interesting. One of the two terms is in phase with the wall’s velocity — that is dissipation, and it is the part that does net work on the fluid over a cycle. The other is in phase with the wall’s acceleration, and does no net work at all: it is an added-mass term, the fluid in the layer being accelerated and decelerated along with the wall. Exactly half of each, and the half-and-half is why the phase is 45° rather than anything else.

The same clock as a boundary layer #

The result above looks unrelated to the boundary layer on a plate and it is the same physics with a different clock.

Momentum diffuses a distance $\sqrt{\nu t}$ in a time $t$. In the oscillating problem the available time is a fraction of a period, $t \sim 1/\omega$, so the depth is $\sqrt{\nu/\omega}$ — the formula above, up to the factor of $\sqrt2$. In Blasius’ problem the available time is how long a parcel has been next to the plate, $t = x/U$, so the thickness is $\sqrt{\nu x/U}$, which is the √x that every boundary-layer figure has in it.

So a boundary layer’s thickness is not set by the geometry. It is set by a competition — how fast momentum spreads against how long it has to spread — and the length of the plate enters only through the time. The oscillating wall is the clean demonstration because it has no geometry at all: an infinite wall, an infinite fluid, and a layer with a perfectly definite thickness.

Where this sits among the exact solutions #

It is worth knowing how short the list is that this belongs to.

The Navier–Stokes equations have been solved exactly perhaps a dozen times, and almost every one of those solutions is a flow in which the convective term switches itself off. Steady flow between sliding plates; steady flow along a pipe, whose parabolic profile and $64/\mathrm{Re}$ friction factor are exact at every Reynolds number; the impulsively started wall; the oscillating wall above; the suction layer of the next essay; and Burgers’ vortex, where the nonlinearity survives but arranges itself into a balance that can be written down.

What that list has in common is not that the flows are simple to look at. It is that in each of them the geometry removes the term that couples different parts of the flow to each other, so what is left is a linear equation — or, in Burgers’ case, a nonlinear one in a single variable.

Everything else on this site is either a closed-form solution of a different equation — the potential flows, which solve Laplace’s equation and are not solutions of Navier–Stokes at all — or a numerical solve, or an honest admission. Knowing which of the three a figure is showing is most of what it means to read this collection carefully.

The same solution, with heat instead of momentum #

Nothing in the derivation used the fact that the diffusing quantity was momentum. The equation solved was the diffusion equation with an oscillating boundary value, so the identical solution describes an oscillating temperature at a surface, with the thermal diffusivity in place of the kinematic viscosity:

$$\delta_T = \sqrt{\frac{2\alpha}{\omega}}, \qquad \frac{\delta}{\delta_T} = \sqrt{Pr}.$$

The ratio of the two depths is the square root of the Prandtl number, which is the same factor that separates the momentum and thermal layers in every other pairing on this site, arriving here without any new work.

The thermal version has the advantage of a case everybody has stood on. The ground is a semi-infinite solid with an oscillating surface temperature, twice over: once a day and once a year. For ordinary soil, with a thermal diffusivity near $5\times10^{-7}\ \mathrm{m^2/s}$, the daily wave has a depth of about 12 centimetres and the annual one about 2.2 metres.

Read the two roles of $\delta$ off that. At 2.2 metres the annual swing is 37 per cent of the surface’s and runs about two months behind it. At seven metres — $\pi\delta$ — the swing is four per cent and the lag is exactly half a cycle: the ground down there is at its warmest in midwinter and its coolest in midsummer, out of phase with the sky by six months, for the same reason the fluid a few $\delta$ above an oscillating wall moves backwards relative to it.

That single number decides several practical things without any further physics. It is why a cellar is cool in August and mild in February. It is why water mains are buried below the frost line, which is the depth at which the winter half of the daily-plus-annual wave no longer reaches freezing. It is why a ground-source heat pump takes its loop down a couple of metres and no further — below that the oscillation it is trying to exploit has died out, and the ground is simply at the annual mean. And it is why the annual wave is the one that matters for a building and the daily wave the one that matters for a road surface.

The same solution read as an instrument gives a measurement technique. Drive a thin metal line on a surface with an alternating current at frequency $\omega$ and it dissipates heat at $2\omega$; the resulting temperature wave penetrates $\delta_T$ into the material and returns to the line as a resistance oscillation at $3\omega$, whose amplitude and phase give the thermal conductivity. It is the thermal twin of reading a viscosity off a phase lag, it works for the same reason — one length doing two jobs — and it is the standard method for thin films precisely because the penetration depth can be tuned by choosing the frequency.

And where the two layers coexist, they do something neither does alone. A gas oscillating in a tube carries a viscous Stokes layer and a thermal one, of different thicknesses because $Pr \ne 1$, so a parcel’s displacement and its temperature change are out of phase by an amount the two depths set — which is exactly the offset a thermoacoustic engine or refrigerator exploits, pumping heat along a stack of plates with no moving parts. The device is built by placing the plates a fraction of $\delta_T$ apart, and the design variable is the number this essay computes.

Where the layer is felt #

Three places, and the first is why the solution is quoted so often.

Sound at a wall. A sound wave running along a surface has an oscillating velocity parallel to it, and the fluid must satisfy the no-slip condition, so a Stokes layer of exactly this thickness sits on every wall a sound wave touches. That layer dissipates energy, and it is why sound in a narrow tube attenuates far faster than sound in the open — the loss goes as the surface-to-volume ratio, and the layer’s thickness sets how much fluid is involved.

Oscillating bodies. A sphere or a wire vibrating in a fluid carries this layer with it, which is why the drag on an oscillating body has both a dissipative and an inertial part, and why a vibrating wire’s damping depends on the square root of the frequency.

Sand under waves. The oscillating flow a surface wave drives over a sea bed carries a layer of exactly this kind, a centimetre or so thick at wave frequencies, and it is the layer — not the wave — that moves the sediment. The same formula with the wave’s angular frequency in it is the first thing any account of that subject writes down.

Particles in a sound field. A small particle in an oscillating flow follows it or does not, depending on the same comparison of timescales that governs whether a droplet turns with a flow round a body. The Stokes layer’s thickness is what a particle inside it experiences instead of the free oscillation.

What the picture cannot show #

Steady streaming is missing. The solution above is exactly periodic, so the fluid returns to where it started at the end of every cycle. In a real oscillating flow with any curvature or any variation along the wall, the nonlinear term returns as a small rectified effect and drives a slow steady circulation — acoustic streaming — that the linear solution has no room for. It is second order and it can carry mass a long way.

Nothing here is turbulent. The layer stays laminar while a Reynolds number built on the layer itself, $U_0\delta/\nu$, is below a few hundred. Above that it goes turbulent, and a turbulent Stokes layer is thicker, dissipates more, and has no closed form.

The wall is infinite. A real oscillating surface has edges, and near an edge the flow is not parallel — which is exactly the assumption that made the problem solvable.

And the fluid is Newtonian. The stress was taken as proportional to the symmetric half of the velocity gradient with a constant of proportionality, and for a polymer solution or a suspension it is not. Those fluids have their own oscillating-plate solutions, they are the standard way of measuring what such a fluid does, and the layer in them is not a simple exponential.

Where the model stops #

The exactness has a price and it is worth stating plainly: this solution describes a fluid whose only motion is the one the wall imposes. Add a mean flow along the wall and the nonlinear term comes back, because the oscillation now has a mean shear to interact with; the result is the unsteady boundary layer, which has no general solution and is a research subject.

The solution also assumes the fluid is unbounded above. Put a second wall a few $\delta$ away and the two layers interact — which is a different exact solution, still tractable, and the basis of every measurement of viscosity by oscillating-plate viscometry.

Who found it, and when #

Stokes solved it in 1851, in the same memoir on pendulums that gave the world the drag law for a sphere at low Reynolds number. The problem he was actually attacking was practical: a pendulum swinging in air is damped by the fluid, and astronomers needed to know by how much before they could use pendulum clocks to measure gravity. The oscillating plate is the simplest case of that problem, and its layer has carried his name ever since.

The impulsively started version — Stokes’ first problem, sometimes attributed to Rayleigh — is the companion result, and its solution is an error function whose thickness grows as $\sqrt{\nu t}$ forever. That growing layer is what the next essay stops.

Where the ladder goes next #

Here is a layer whose thickness is set by a clock. The obvious question is whether anything can stop a layer growing at all — whether there is a boundary layer with no time and no distance in it — and there is exactly one, held still by sucking fluid through the wall at precisely the rate it diffuses outwards.