A thinning gas flows least at one free path across, and then more without end

Where a fluid stops being one set the two exact descriptions of a rarefied gas side by side — Navier–Stokes with a slipping wall, good when the gas is dense, and the free-molecular flux, good when it is not — and left the region between them to measurement. The one number in that essay that came from a laboratory rather than an expression was Knudsen’s minimum: the observation, made in 1909 with gas pushed through glass capillaries, that as the pressure falls the flow for a given pressure gradient first falls and then rises again. Neither closed form can produce that, since one of them rises steadily with density and the other does not depend on it.

Slip is a memory of one mean free path explained the first correction: a molecule arriving at a wall brings the velocity of the place it last collided, a free path away, so the gas at the wall slides. That is the dense end of the story. This essay computes the rest of it — the flow between two plates at every density, from the kinetic equation itself — and finds that the minimum is not a hole between two theories. It is a crossing between two kinds of transport, both of which the kinetic equation contains and each of which one of the closed forms keeps alone.

An equation with the minimum in it #

The gas is at rest, at uniform temperature, between two plates a height $H$ apart, and a small pressure gradient along the channel pushes it. Treating it as a continuum gives up exactly what is needed here: the state of a gas is not a velocity at each point but a distribution of molecular velocities at each point, and the Boltzmann equation says how that distribution changes as molecules fly and collide. Its collision term is the hard part. The model used here, due to Bhatnagar, Gross and Krook in 1954 and known by their initials, replaces it with the simplest thing that conserves what collisions conserve: at each point the distribution relaxes towards the local equilibrium at a single rate, the collision frequency. The disturbance is small, so the equation can be linearised about rest, and the walls are taken to re-emit every molecule that strikes them in equilibrium at the wall’s own temperature and velocity, the diffuse condition.

Two numbers describe the problem. The Knudsen number $\mathrm{Kn} = \lambda/H$ compares the free path with the height, and the kinetic literature uses its inverse in a convenient form, the rarefaction parameter

$$\delta = \frac{Hp}{\mu v_0} = \frac{\sqrt\pi}{2\,\mathrm{Kn}}, \qquad v_0 = \sqrt{2RT},$$

so large $\delta$ is a dense gas. The answer is a reduced flow rate $G$, the mass flow per unit width scaled by $H^2(-\mathrm{d}p/\mathrm{d}x)/v_0$, chosen so that it stays finite at both ends of the range.

The speeds along the channel and across its width can be integrated out, leaving a distribution of the one speed that matters, $c$, the speed at which a molecule crosses the gap. A molecule crossing at $c$ travels in a straight line from the wall it last left, and the probability that it has not collided after crossing a distance $s$ of the gap is $e^{-\delta s/c}$ in units of $H$. Adding up what reaches each point from every earlier point along every such path turns the kinetic equation into one equation for the bulk velocity $u(x)$ alone:

$$u(x) = \frac{\delta}{\sqrt\pi}\int_{-1/2}^{1/2} T_{-1}\!\big(\delta|x-x'|\big)\,u(x')\,dx' \;-\; \frac{1}{2\delta\sqrt\pi}\Big[\sqrt\pi - T_0\big(\delta(\tfrac12 - x)\big) - T_0\big(\delta(\tfrac12 + x)\big)\Big],$$

where $T_n(z) = \int_0^\infty c^n e^{-c^2 - z/c}\,dc$ are Abramowitz’s functions, which carry the averaging over crossing speeds. The first term is the velocity a point receives from molecules that collided somewhere else in the gas; the second is what the pressure gradient gives to molecules on their way from a wall. The flow rate is $G = -2\int u\,dx$.

The kernel $T_{-1}$ is infinite where $x' = x$, logarithmically, because molecules crossing very slowly linger at the height they are at. That is handled by holding the velocity constant on each of a few hundred cells, crowded towards the walls, and integrating the kernel over each cell exactly — it is the derivative of $T_0$, which is smooth — so the singular point is never evaluated. The result is a dense linear system, solved directly at two cell counts and extrapolated.

The minimum is two transports crossing #

On the dense side the solve follows the slip formula $G = \delta/6 + 1.016$, rising in proportion to the density. That rise is ordinary viscous flow. The mass flow through a channel under a fixed pressure gradient is the density times a velocity, and the velocity is the gradient over the viscosity times the height squared; a gas’s viscosity does not depend on its density — the same kinetic arithmetic that puts every gas’s Prandtl number near three-quarters — so halving the pressure halves the mass flow. The constant $1.016$ is the slip of the previous essay, the wall’s contribution, worth one free path’s memory, a shorter cousin of the memory by which liquid above a moving wall answers what the wall did seconds before.

On the rarefied side the solve rises again, and without limit. Between infinite plates a molecule that leaves a wall at a grazing angle, crossing the gap at a small speed $c$, flies a distance along the channel proportional to $1/c$ before it reaches the other wall, and the pressure gradient accelerates it the whole way. The fraction of molecules leaving at small $c$ falls only linearly in $c$, so the momentum they carry grows like $\int dc/c$ — a logarithm with nothing to stop it. Nothing, that is, except the other molecules. A grazing molecule’s flight ends when it collides with another one, after about a free path, so the free path cuts the logarithm off and

$$G \approx -\frac{\ln\delta}{\sqrt\pi} + \text{const}, \qquad \delta \to 0.$$

The solve bears this out. From $\delta = 0.01$ to $0.001$ it rises from 3.049 to 4.449, against $\ln 10/\sqrt\pi = 1.30$ a decade in the limit. Thinning the gas lengthens the flights of exactly the molecules that do the carrying, and so it increases the flow.

So two transports are at work. One is collisional: momentum exchanged between neighbouring layers of gas, rising with density, and the whole of the Navier–Stokes description. The other is ballistic: momentum carried along the channel by molecules between collisions, falling with density, since collisions are what end the flights. The minimum is where the falling one hands over to the rising one. It sits at $\delta = 1.07$, a Knudsen number of $0.827$, where the reduced flow rate is $1.538$ — the height of the channel is a little more than one free path. It is not a coincidence of numbers that the crossing falls near one: it is the density at which a molecule crossing the gap at an ordinary angle is as likely to collide on the way as not, so that neither mechanism has the gas to itself.

Above both closed forms, not below them #

The usual way of describing the transition regime — and the way the essay on where a fluid stops being one described it — is as a gap between two exact theories with the truth somewhere outside both. Between plates, the solve says where the truth is, and it is above both.

The closed form for the rarefied end is not, between plates, the free-molecular flux: with no collisions at all the grazing molecules fly for ever and that flux is infinite. Its honest counterpart is the uncollided flux, the second term of the integral equation alone — the molecules that have come straight from a wall without meeting another one, cut off by the collision probability $e^{-\delta s/c}$ but given nothing back by the collisions that remove them. It is the first term of the series the integral equation generates when each collision is counted in turn.

The two closed forms cross at $\delta = 0.335$, where both give 1.07. The solve there gives 1.68, fifty-seven per cent above both. What neither closed form contains is the molecule that has collided once or a few times and is still carrying momentum along the channel — not enough collisions for viscosity to be the right description, too many for the uncollided count. That is the whole of the transition regime’s flow, and in a channel it adds rather than subtracts.

The width of that band is the practical result. A designer with a slip formula and a free-molecular formula, and the rule that the transition regime runs from Kn = 0.1 to 10, would expect each formula to be roughly right at its own end of that range. The slip formula at Kn = 0.1 ($\delta = 8.9$) is 4 per cent low; at Kn = 1 it is 25 per cent low. The uncollided flux at Kn = 10 ($\delta = 0.089$) is 16 per cent low. Both closed forms fail in the same direction everywhere in the band, which is what a missing transport would do and what a gap between two theories would not.

The profile goes flat before the formula goes wrong #

The bulk velocity across the channel says the same thing in a different way. In a dense gas it is a parabola standing on a small slip at each wall. As the gas thins, the profile flattens: more of the flow is carried by molecules that have recently left a wall and remember it, and the momentum they carry is spread across the gap by flights longer than the gap itself.

At $\delta = 1$ the wall’s gas moves at 0.667 of the mean velocity and the formula puts it at 0.859; at $\delta = 0.1$ the numbers are 0.846 and 0.984. In both cases the formula is wrong at the wall in the same sense — it has the gas sliding faster than it does — and right enough in the middle. That is the clue to where the slip formula’s error lives, and it is near the wall.

Inside the first free path #

A slip condition is a statement about the gas a few free paths from the wall, extrapolated to the wall. Within the first free path the distribution of molecules is not the near-equilibrium one Navier–Stokes assumes, because half of the molecules there have just left the wall and carry its velocity, whatever the gas above is doing. That region is the Knudsen layer, and the slip formula does not describe it; it describes the flow outside it as though the layer were not there.

Measured in free paths, and as a share of the slip, the layer is the same at every density: the gas at the wall moves about 0.71 of the slip velocity the formula assigns it, and the deficit vanishes within a free path and a half. Replacing a thin layer the equations cannot describe by one length at the wall is the same move made for a porous wall and for a viscous nanometre of liquid read as slip, and it fails in the same way, inside the layer it replaces. That is why the slip formula works for the flow rate long before it works for the velocity at the wall. The flow rate is an integral across the whole channel and the layer is a fixed number of free paths thick, so its share of the integral falls as the channel grows in free paths. At $\delta = 40$ the channel is 45 free paths high and the layer is three per cent of it; at the minimum the channel is about one free path high and the layer is all of it.

The convention behind every number here matters. The Knudsen number is formed on the full height $H$, not the half-height, and the free path is the one defined from the viscosity, $\lambda = (\mu/p)\sqrt{\pi RT/2}$. On the half-height the minimum would be quoted at $\mathrm{Kn} = 1.65$, and on a hard-sphere free path at a different number again, so a minimum “at Kn ≈ 1” is only meaningful with the convention attached — the usual fate of a threshold quoted as “of order one”. The rarefaction $\delta$ does not carry that ambiguity, which is why the kinetic literature uses it.

What a channel does when it is pumped down #

In dimensional terms the result is a statement about conductance. Fix the pressure difference across a channel and lower the mean pressure: in the continuum the mass flow falls in proportion, since the density does and the velocity does not. A narrow channel leaves the continuum early, and its flow stops falling, levels off and rises.

For air at room temperature the free path is 68 nanometres at atmospheric pressure, so a channel one micrometre high is at $\mathrm{Kn} = 0.07$ before anyone pumps it, and its no-slip conductance is already a third low. Its minimum falls at 8.2 kPa, about a twelfth of an atmosphere; a ten-micrometre channel reaches it at 0.82 kPa. At 20 pascals the one-micrometre channel passes 2.4 times what it did at its minimum. Each of those numbers is an inversion of the same curve, because the flow rate depends on the pressure, the height and the gas only through $\delta$.

A hard-disc head flies a few nanometres above its platter at atmospheric pressure, far out on the rarefied side, where its bearing’s load comes from grazing molecules rather than viscosity. The pores of a gas-separation membrane and the passages of a microfabricated valve or pump, a micrometre or two high and run between a few hundred pascals and an atmosphere, sit across the minimum in normal operation, as do the gas films of a squeeze-film damper under a light vacuum. It is also why a leak test performed by pumping down is not a scaled version of the same test at atmosphere: below the minimum a leak path conducts better the harder it is pumped.

A tube is not a channel #

The minimum Knudsen measured was in tubes, and a tube is a different case for exactly the reason the channel’s flow rate diverges. A grazing molecule in a tube still leaves the wall at a small angle, but there is nowhere for it to skim along: every chord of a circle meets the wall again within the tube’s diameter, however close to the wall it starts. The longest flight is set by the geometry and not by the collisions, so the free-molecular flux through a tube is finite — $8/(3\sqrt\pi) = 1.505$ in the same reduced units — and the rarefied end of the tube’s curve flattens onto that constant rather than climbing for ever.

That changes how deep the minimum is. In a tube the curve falls from the free-molecular constant, through a shallow minimum, into the continuum’s rise; in a channel it falls from a logarithm that has no ceiling. The deep minimum and the unbounded rise are properties of a gap with no side walls, and any real rectangular channel has side walls: once the free path exceeds its width $W$, the grazing flights are cut off at $W$ rather than at $\lambda$, and the logarithm stops growing at roughly the value it had reached where the free path equalled the width. A channel ten times wider than it is high would level off near $\delta \approx 0.1$; one a thousand times wider, near $10^{-3}$.

Where a fluid stops being one described the tube’s measured minimum as a dip below both closed forms. Its own numbers say otherwise — at δ = 0.6 the slip line is 1.166 and the measured value 1.321 — and that essay’s caption has been corrected. The tube’s minimum lies above the slip line and below the free-molecular constant, and the channel’s lies above both. Neither is a dip under its asymptotes.

What the picture cannot show #

The model is a model. The BGK equation relaxes every molecular velocity at one rate, which a real gas does not do, and its slip coefficient and its flow rates differ from the full Boltzmann equation’s for hard spheres or for a realistic potential by amounts this calculation does not test. The diffuse wall forgets every incoming molecule completely; a smooth, clean surface reflects some of them specularly, which raises the slip at the dense end and the grazing flights at the rarefied end, deepening nothing and moving the whole curve up. The disturbance is linear: the pressure must change by a small fraction of itself over a free path, which holds in a long channel and not across a short orifice, nor in the entrance length where the profile is still forming. The temperature is uniform, so the thermal creep that drives gas along a wall with a temperature gradient — the other effect Maxwell found, and described in where a fluid stops being one — is absent. And the plates are infinite, which is what lets the logarithm run.

The figure of the carriers sorts molecules by their crossing speed at the midplane only. Near a wall the picture is different, since half the molecules there have just left it, and the share of the flux carried by grazing molecules is larger there still.

Who found it, and when #

Martin Knudsen measured the minimum in 1909, in a paper on the flow of gases through narrow tubes at low pressure, and it was a puzzle for half a century: the free-molecular theory he had developed alongside it gave a constant, and the continuum theory a straight line. The kinetic equation for the channel was solved by Cercignani and Daneri in 1963, with the BGK model, and they found the minimum at a reduced flow rate of about 1.54 near $\delta = 1$ — the number recomputed here. Sharipov and Seleznev’s review of 1998 collected the solutions for channels and tubes with every common wall condition, and it is where a reader wanting the tube’s numbers on the same footing should go.

How it was checked #

Abramowitz’s $T_0$ is tabulated and interpolated, and agrees with direct quadrature to two parts in a million; its slope matches $-T_{-1}$ computed independently to five parts in a million. The flow rate at $\delta = 1$ moves by seven parts in a hundred million when the cells are doubled from 100 to 200 after extrapolation. At the dense end, the remainder $G - \delta/6$ at $\delta$ = 10, 20 and 40 fitted as $\sigma + a/\delta + b/\delta^2$ gives $\sigma = 1.0161$, against the model’s published slip coefficient of 1.016 — an independent number, recovered from a calculation that never imposes a slip condition anywhere. At the rarefied end the solve exceeds the uncollided flux by 1.1, 3.0 and 7.1 per cent at $\delta$ = 0.003, 0.01 and 0.03, the share of collided molecules falling towards zero as it should. The minimum was located by a golden-section search and lies at $\delta = 1.072$; the published BGK value of the least flow rate, about 1.54, agrees. The solver refuses a channel with no gas, a negative rarefaction, fewer than eight cells, and a δ past four hundred where its cells no longer resolve the Knudsen layer.

Still open: a channel with sides #

Every number here is for plates of infinite width, and the rarefied end of the curve is decided by the width that is missing. A rectangular channel cuts the grazing flights off at its side walls, so the logarithm stops and the flow rate levels off at a value set by the aspect ratio. The next calculation solves the same kinetic problem in a rectangular cross-section — a two-dimensional integral equation, with the free path now competing with two lengths rather than one — and asks how the depth and position of the minimum move with the ratio of width to height: at what aspect ratio the channel’s minimum becomes as shallow as the tube’s, and whether there is a ratio at which it disappears.