A wave on the wall is a pump

The number on a streamline is a flow rate showed that the streamfunction is not a label: the difference between its values on two streamlines is the volume passing between them every second, by any route. Mass has nowhere to go, before it, said that squeezing a stream speeds it up because the same amount must get through a smaller gap. Both were about walls that stay put. This essay moves the wall.

Take a channel and make its walls ripple — not slide along, only move in and out — in a wave that travels down the channel. That is how the ureter moves urine, how the gut moves its contents, and how a roller pump moves blood without touching it. The question is how much it pumps, and the answer comes in two parts of very different character. One part is continuity and holds whatever the fluid and whatever the speed. The other needs the momentum equation, and it is a single number. What that number is decides not only the output but whether some of the fluid is carried along with the wave in a closed pocket, and whether the fluid near the walls goes forwards or backwards.

In the wave’s frame the walls are streamlines #

The conventions are these. The channel is two-dimensional and symmetric, with half-width $h = a(1 + \varphi\sin\theta)$ and phase $\theta = 2\pi(X - ct)/\lambda$: a wave of amplitude ratio $\varphi$ travelling at speed $c$. Flow rates are per half-channel, in units of $ca$; pressures are in units of $\mu c\lambda/a^2$; distances across the channel are in units of the mean half-width $a$. For the momentum part the wavelength is long against the width and inertia is negligible — the lubrication limit. Continuity needs neither assumption.

In the laboratory this flow is unsteady: at any fixed station the wall moves in and out and the velocity changes through each period. Move with the wave, at speed $c$, and everything stops changing. The walls become fixed curves. A flow that is steady in some frame has streamlines that are particle paths in that frame, which is exactly the condition Streamlines are not the paths particles take said is needed, and the walls, which no fluid crosses, are streamlines themselves. So the flow rate between the centreline and the wall in the wave frame, call it $q$, is one number: the difference between the streamfunction on the wall and on the centreline, the same at every section along the channel. The walls here behave exactly as a body made of a level set does in steady flow.

That frame is where the pattern in the first figure is drawn. At an amplitude ratio of 0.7 and the flow at which the channel works against no pressure, $q = -0.410$: in the wave frame the fluid mostly runs backwards, as it must, since the walls in that frame are sliding backwards at $c$. Under each crest the centreline velocity changes sign at two points, and the streamline through them closes round a pocket of fluid moving forwards relative to the rest. That pocket is the bolus, and it is the first thing the momentum number decides.

Continuity alone turns the wave frame into the laboratory #

Now go back to a fixed station. Every particle there has the wave-frame velocity plus $c$, so the laboratory flow rate across the section is the wave-frame flow rate plus $c$ times the width the fluid occupies:

$$Q = q + c\,h.$$

Averaged over a period the mean of $h$ is $a$, so the pump’s time-mean output is $\Theta = q/(ca) + 1$ in the units here. Nothing in that argument used the momentum equation, the viscosity, the Reynolds number or the shape of the velocity profile. It holds for any flow that is steady in the frame of the wave. The one thing it cannot supply is $q$.

The second figure is that statement tested. With a mean flow of 0.5 at amplitude ratio 0.7, integrating the laboratory velocity profile across the channel at 24 moments reproduces $q + h$ at every one, to better than 10⁻⁸. The flow past the station swings from −0.200 as the narrowest section goes by to 1.200 under the widest. For part of every period the fluid at a fixed station runs backwards, and the pump still delivers a steady forward mean — which is the difference between a flow rate at a point in time and a flow rate averaged over a cycle, drawn as a sine wave with its axis lifted off zero. What a fixed observer and a moving one see are different fields, as The picture belongs to the watcher insisted; here the change of frame is what makes the problem solvable at all.

What the momentum equation has to supply #

Everything left is the value of $q$, and for a long wave at negligible Reynolds number it follows from lubrication. Locally the channel looks straight, so the wave-frame profile is a parabola whose walls slide backwards at $c$: $u/c = -1 + \tfrac32(\hat q + H)(H^2 - Y^2)/H^3$, with $H = h/a$, $Y = y/a$ and $\hat q = q/ca$. Its flow rate is $\hat q$ by construction, and the pressure gradient that drives it is

$$\frac{dp}{dX} = -\frac{3\mu c\,(\hat q + H)}{a^2 H^3}.$$

This is the same mechanism Nothing but the shape of the gap found in a bearing: a gap that narrows forces a fixed flow rate through less space and pays for it in pressure. Here the narrowing travels. Integrating over one wavelength, the pressure rise is $-3(\hat q I_3 + I_2)$ in units of $\mu c\lambda/a^2$, where $I_2 = (1 - \varphi^2)^{-3/2}$ and $I_3 = (1 + \varphi^2/2)(1 - \varphi^2)^{-5/2}$ are the averages of $H^{-2}$ and $H^{-3}$ over a wavelength.

The pressure along a wavelength shows where the work is done. At zero net flow the pressure is nearly flat until the throat and then climbs steeply, to 11.87 at the end of the wavelength: the pump holding back a column of fluid. At the free-pumping flow of 0.590 it dips to −0.874 approaching the throat, rises to 0.337 past it, and returns exactly to where it started. At 0.9 the flow outruns the wave’s ability to pump and has to be driven, and the pressure falls by 6.23 over the wavelength.

Free pumping, and what a deeper wave buys #

Setting the pressure rise to zero gives the flow the channel delivers against nothing:

$$\Theta_0 = \frac{3\varphi^2}{2 + \varphi^2}.$$

It is 0.0588 at an amplitude ratio of 0.2, 0.222 at 0.4, 0.333 at 0.5, 0.590 at 0.7 and 0.865 at 0.9, approaching 1 as the throat closes completely — the limit of a roller pump, which occludes the tube and simply carries forward the fluid ahead of each roller. For a shallow wave it is $\tfrac32\varphi^2$: second order in the amplitude, and so absent from any linear description of the wall motion. That is the same arithmetic as The drift in a wave that has none, where a wave with zero mean velocity at every point carries mass at the square of its steepness, and as A swimmer that cannot go backwards, whose waving sheet swims at a speed proportional to the square of its amplitude. A peristaltic channel is that swimmer held still: the wave that would have pushed the sheet forward pushes the fluid back past it instead.

The other end of the pump curve is the pressure the channel can hold with no flow at all, $3(I_3 - I_2)$: 0.199 at an amplitude ratio of 0.2, 1.11 at 0.4, 2.31 at 0.5, 11.87 at 0.7, 37.0 at 0.8 and 232 at 0.9. It grows without bound as the throat closes, as $(1 - \varphi)^{-5/2}$. Between the two ends the pressure rise falls linearly with the mean flow, because the lubrication equation is linear in $q$.

A bolus rides with the wave #

The centreline velocity in the wave frame is $\tfrac12 + \tfrac32\hat q/H$. It changes sign wherever the half-width equals $-3\hat q$, and if that happens somewhere inside the channel — where $H$ lies between $1 - \varphi$ and $1 + \varphi$ — the fluid under the crest runs forwards with the wave while the rest runs back, and the streamline between them closes. In terms of the mean flow, a bolus exists when $(2 - \varphi)/3 < \Theta < (2 + \varphi)/3$.

The free-pumping curve enters that band where $3\varphi^2/(2 + \varphi^2) = (2 - \varphi)/3$, which is $\varphi = \sqrt{13} - 3 = 0.6056$. A channel pumping against no pressure traps fluid only if its wall wave is at least that deep. Below it, every streamline in the wave frame runs through and no fluid travels with the wave.

At an amplitude ratio of 0.7 the centreline velocity at a mean flow of 0.4 is −0.029 under the crest and −2.50 at the throat: backwards everywhere, just, and no bolus. At the free-pumping flow of 0.590 it is +0.139 under the crest and −1.55 at the throat, changing sign at 0.053 and 0.447 of the wavelength, and the bolus between those points holds 30.5 per cent of the fluid in each wavelength. At a mean flow of 0.5 the bolus holds 13.3 per cent. At 0.8 the forward region under the crest has spread across most of the wavelength and only the throat still runs back. A bolus is fluid that moves at the wave speed indefinitely: whatever is in it — a particle, a bubble, a stone — is carried along the whole channel, which is the reason the trapping threshold matters to anyone designing or diagnosing a peristaltic tube.

The speed matters as much as the trapping. The bolus travels at $c$, while the fluid delivered by the pump advances on average at the mean flow divided by the mean half-width — 0.590 of $c$ at this amplitude. Whatever rides in a bolus therefore crosses the channel in 1/0.590 = 1.69 times less time than the average fluid, and a residence-time estimate made from the flow rate alone overstates how long its contents spend in the tube by that factor.

A photograph in the laboratory shows no bolus #

The bolus is a closed streamline in the wave frame. It is worth asking what an instantaneous picture of streamlines taken in the laboratory would show, and the answer is short enough to do exactly. The laboratory velocity at every point is the wave-frame velocity plus $c$, so the laboratory streamfunction is the wave-frame one plus $cy$, and its derivative across the channel is $u + c$, which in the lubrication profile is $\tfrac32(\hat q + H)(H^2 - Y^2)/H^3$ in these units. That has one sign across the whole section — the sign of $\hat q + H$ — and vanishes only on the walls. A function whose slope across a section never changes sign has no maximum inside it, so no laboratory streamline can close: at every instant the laboratory picture is a set of open curves running along the channel, forwards wherever the section is wider than $-\hat q$ and backwards across the whole width wherever it is narrower.

So the object that carries a third of the fluid along at the wave speed does not appear in any snapshot a fixed camera can take. It exists only in the frame that moves with the wave, and only as a statement about which fluid stays with which — exactly the kind of fact that a pattern belonging to whoever is watching predicts, and a sharper instance of it than the plane examples, because here the invisible structure is the one that decides what the pump transports.

Net flow forwards, and the fluid near the walls going back #

The bolus is one fact about where particles go that the flow rate does not show. The other is that particles on different streamlines have different mean speeds in the laboratory, and near the walls some of them go backwards.

In the wave frame a particle on an open streamline crosses a wavelength in some time $T$, so in the laboratory it advances $cT - \lambda$ each period; its mean velocity is $1 - 2\pi/T$ in these units. At an amplitude ratio of 0.4 and zero net flow, particles near the centreline drift forwards at 0.058 of the wave speed, while those at 0.58 of the half-width drift back at 0.012 and those at 0.81 at 0.047. At a small forward net flow of 0.1, the centreline drifts at 0.211 and particles at 0.97 of the half-width still go back at 0.0067. At 0.2 every streamline drifts forwards. The backward layer disappears at a mean flow of 0.160, where the pump is still working against a pressure rise of 0.313 — short of free pumping, which at this amplitude is 0.222.

That is reflux: a net forward flow with a backward-moving layer against the walls, and it is a statement about particle means, not about the flow rate. The flow rate is an Eulerian average taken at a fixed station; the particle drift is a Lagrangian one, taken following the fluid. They are different averages of the same field, exactly the distinction A drift made of two things that average to zero drew for a water wave, and a pump rated by its flow rate can carry fluid backwards along part of its cross-section while delivering a forward mean.

Every closed form against a route that does not use it #

The results rest on a handful of closed forms, and each has been checked by a calculation that shares none of its algebra.

The streamfunction agrees with a midpoint integration of the velocity profile across the channel to 2.3 × 10⁻⁹, and the velocity it implies is tangent to the wall to 5.7 × 10⁻¹¹. The laboratory flow rate integrated from the profile agrees with $q + h$ to 9.4 × 10⁻⁹ over fifty phases, and its average over a period with $\Theta$ to better than 10⁻⁷. The pressure rise found by integrating the local gradient over a wavelength matches $-3(\hat q I_3 + I_2)$ to 7 × 10⁻¹⁵ across amplitude ratios from 0.2 to 0.95. The drift of a streamline computed from its crossing time matches a single particle marched through the wave-frame field by fourth-order Runge–Kutta to 1.4 × 10⁻⁸. The trapping threshold is found twice, from the closed criterion and by scanning the centreline for sign changes, and the two agree a thousandth either side of $\sqrt{13} - 3$. And the calculation refuses a wave deep enough to close the channel, a negative amplitude, a drift through a trapped bolus and a streamline outside the channel.

What the long-wavelength channel cannot show #

Short waves. The lubrication profile assumes the channel is locally straight. When the wavelength is only a few times the width, the streamlines curve, the pressure varies across the channel, and the flow rate for a given pressure rise changes. The continuity part, $Q = q + ch$, does not.

Inertia. At a finite Reynolds number the wave-frame flow is no longer the lubrication parabola, and the bolus and reflux boundaries move. The frame change and the flow-rate identity are untouched, because they never mentioned the momentum equation.

A tube. An axisymmetric tube has the same argument with areas in place of widths, a factor of $2\pi r$ in the flow rate, and different averages of the radius, so its free-pumping curve and trapping threshold have different numbers.

A real contraction. The wave here is an infinite sine train, which is what makes the wave frame steady. A single muscular contraction travelling down a ureter is not periodic, the flow ahead of and behind it is not the same, and the frame in which it looks steady exists only near the contraction. The wall shape is also prescribed here rather than produced by muscle acting against the fluid pressure, and a soft wall would change shape under the very pressures the figures report.

Still open: the pump that closes the tube #

The next calculation takes the wall all the way in. As the amplitude ratio approaches one the free-pumping flow approaches the displacement of the wave and the pressure the pump can hold grows as $(1 - \varphi)^{-5/2}$; at exact occlusion the channel becomes a positive-displacement pump, and the lubrication solution breaks down in the throat where the gap goes to zero. Computing the approach to occlusion with the gap’s own lubrication layer resolved, and the leakage back past a nearly closed throat, would say how much of a roller pump’s delivery is the wave and how much is the seal.

Beside it is the energy account. The wall does work on the fluid every period, the pump delivers flow against a pressure, and the ratio is an efficiency with a maximum somewhere on the linear pump curve. The same bookkeeping for Taylor’s swimming sheet gave a cost of transport independent of amplitude; whether a peristaltic pump’s best efficiency has a similar independence, and where on the curve it sits, is a question the closed forms here are one integral away from answering.