Incompressible is not a property of the fluid

Continuity, written for a fluid whose density may do anything, is

$$\nabla\cdot\mathbf u = -\frac{1}{\rho}\frac{\mathrm D\rho}{\mathrm Dt}.$$

That is not an approximation and there is nothing in it about which fluid. It says the divergence is the rate at which a material element’s volume grows, per unit volume, and the whole of this essay is what follows from reading it that way round.

$\nabla\cdot\mathbf u = 0$ does not mean the density is uniform. It means each parcel keeps whatever density it started with. Those are different statements, the second is much weaker, and the word incompressible is routinely read as the first.

The flow that settles it #

The demonstration wants a case where the two readings give different answers and where the divergence can be measured rather than assumed.

Take the flow this collection opens with — ideal flow past a cylinder — and give it a density that is constant along each streamline, $\rho = \rho(\psi)$. That is the general incompressible stratification in two dimensions: a scalar constant along streamlines has $\mathrm D\rho/\mathrm Dt = 0$ in a steady flow, whatever the streamlines do.

The density then varies by a factor of three across the field. The divergence, differenced from the velocity field rather than taken from the construction, comes out at $2.6\times10^{-10}$ — which is the differencing and not the flow. The material rate of change of density is $3.2\times10^{-10}$, by the same token.

Both readings of incompressible have been tested on one flow and one of them is false.

The simplest version, and why it is too simple #

The textbook version of the same point is a shear layer with a density profile: $u = U\tanh(y/h)$ with $\rho = \rho(y)$. The vertical velocity is identically zero, so no parcel ever moves to a different density and the divergence is exactly zero — not approximately, and not as the result of a calculation.

It is a fair demonstration and it proves less than the first one, because the divergence there is zero by inspection. The cylinder case is the one worth trusting: both velocity components vary in both directions and the answer is measured.

Stratified flows of this kind are the whole subject of stratification, and every essay in it treats the flow as incompressible while the density varies by whatever the profile says. That is not a lapse; it is the definition being used correctly.

What the word is really about #

There is a cleaner way to say all of this. The divergence is the fractional rate of change of the volume of a material element:

$$\nabla\cdot\mathbf u = \frac{1}{\delta V}\frac{\mathrm D(\delta V)}{\mathrm Dt},$$

so an incompressible flow is one in which fluid elements do not change size. The determinant of the deformation gradient is then one for all time, which is exactly the check that essay runs while a material patch is being stretched seven to one.

Read that way, three sentences that get confused become clearly different:

$\nabla\cdot\mathbf u = 0$ — parcels keep their volume, and therefore their density. Incompressible.

$\rho$ uniform — every parcel has the same density. Homogeneous, which is a different property and has a different name that almost nobody uses.

$\partial\rho/\partial t = 0$ — the density at a fixed point does not change. Steady in $\rho$, and implies neither of the above.

What a gas flow’s divergence actually is #

The other half of the confusion runs the opposite way. A flow of air at low speed is called incompressible, and its divergence is not zero; it is small, and how small is a number worth having.

For steady isentropic flow the density is a function of the local speed alone, so

$$\nabla\cdot\mathbf u = -\frac{\mathbf u\cdot\nabla\rho}{\rho} = \frac{M_\infty^2}{2U^2}\,\frac{\mathbf u\cdot\nabla(q^2)}{1 + \tfrac12(\gamma-1)M_\infty^2(1-q^2/U^2)}.$$

Evaluating that on the incompressible field — which is right to the order of the answer, and says so — gives the dilatation the incompressible approximation is throwing away.

The pattern is what it should be: fluid expands as it accelerates round the shoulders and is compressed again as it slows towards the tail, so there are four lobes of alternating sign and none of it survives an average over the body.

The exponent, measured #

The size of it is the useful part. The peak dilatation anywhere on the cylinder, scaled on $U/a$:

Fitting the low end gives an exponent of $2.009$, and fitting the whole sweep gives $2.051$ — the excess being the next term in the expansion rather than an error. The dilatation of a gas flow is second order in the Mach number, which is the honest statement of what incompressible buys and what it costs.

At Mach 0.1 — an aeroplane on approach, a wind tunnel at 35 metres a second — the divergence being discarded is three per cent of $U/a$. That is a real number and it is worth having beside the several thresholds the figure 0.3 is asked to carry, all of which are the same $M^2$ seen through different tolerances.

Following one parcel #

The cleanest way to see the divergence do something is to follow a single element and watch its volume.

Take a streamline past a cylinder at Mach 0.5 and carry a fluid element along it. Compute its volume ratio twice: once as the exponential of the integrated divergence along the path, which is the definition, and once as the inverse of the density ratio between the two ends, which is what continuity says the answer must be in a steady flow.

The element expands by 29.6 per cent as it accelerates over the shoulder. The two routes agree to two parts in ten million, and they share no arithmetic: one is a time integral of a differenced field and the other an algebraic function of two speeds.

That is the whole content of the equation at the top of this essay, run once, on a real field.

Why the word is used the way it is #

None of this makes the ordinary usage wrong; it makes it elliptical. When somebody says water is incompressible they mean that the pressure changes available in the flow are far too small to change its density appreciably, so any flow of it will turn out to have a negligible divergence. That is a statement about the fluid and the flow, and it is the conjunction that does the work.

The trouble is that the same word is then used for two other things:

A property of the fluid — a bulk modulus large enough that pressure does nothing. Water at ordinary pressures qualifies; air does not, and air is treated as incompressible constantly.

A property of a model — the assumption that $\nabla\cdot\mathbf u = 0$ is imposed as an equation rather than deduced. That is what a solver means by it, and it makes the pressure a Lagrange multiplier rather than a thermodynamic variable, which is why an incompressible solver has to project its velocity field and why there is no equation of state anywhere in it.

The three usages coincide for water in a pipe and come apart everywhere interesting: in a hot jet, in a stratified ocean, in a combustion chamber, in the atmosphere.

The case where they come apart most #

The clearest example is a flame or a hot jet. The density varies by a factor of six or seven, so the flow is emphatically not homogeneous. The Mach number is a few hundredths, so the pressure does nothing to the density at all. And the divergence is large, because a parcel crossing into the hot region expands.

That combination — large density variation, negligible Mach number, non-zero divergence — is called low-Mach-number variable-density flow, and it has its own set of equations in which the pressure is split into a uniform thermodynamic part that sets the density and a varying dynamic part that drives the flow. It is neither of the two cases the word incompressible is usually used for.

The site’s own Boussinesq essays sit at the other extreme: density variations small enough that the divergence can be dropped everywhere except in the buoyancy term. That approximation has a name and a stated validity, and it is precisely the statement that the flow is incompressible while the density is not uniform.

Where the confusion costs something #

It would be a vocabulary complaint if nothing turned on it. Four places where something does.

Reading a stratified experiment. A salt-stratified tank has a density varying by a few per cent and is modelled with the incompressible equations plus a buoyancy term. A reader who takes incompressible to mean uniform density will conclude that the model has thrown away the very thing the experiment is about, and it has not: the density variation is in the equations, and what has been dropped is the divergence, which is genuinely negligible because no parcel is being compressed.

Reading a combustion calculation. The opposite mistake. A reader who takes low Mach number to mean incompressible will expect a divergence-free velocity field, and a flame front has a large one — the gas expands by a factor of six crossing it. Codes that impose $\nabla\cdot\mathbf u = 0$ on a reacting flow are wrong for exactly this reason, and codes that solve the full compressible equations there are paying for acoustics they do not need.

Sizing a numerical method. An incompressible solver has no equation of state and no acoustic timestep restriction; a compressible one has both. Choosing between them is a decision about which of the three usages applies, and the wrong choice is either a wrong answer or a hundredfold cost.

And reading this collection. Every ideal-flow essay here assumes the divergence is zero. That assumption is worth three per cent of $U/a$ at Mach 0.1 and is not worth anything at Mach 0.6, and the figures now say which.

The flow with no Mach number worth writing down, which is entirely compressible #

Every case above treats the divergence as a small correction to be measured or discarded. There is a flow in which the divergence is not a correction but the entire subject, and it happens at a Mach number smaller than anything in the table by seven orders of magnitude.

Sound. An ordinary conversation at a metre carries a pressure fluctuation of a few hundredths of a pascal, which corresponds to a particle velocity of about $5\times10^{-5}$ metres a second and a density fluctuation of about one part in ten million. The Mach number is $10^{-7}$. Nothing anywhere in fluid mechanics is more nearly incompressible by the arithmetic of the previous section.

And an incompressible model of it contains nothing at all, because a sound wave is the divergence: $\partial\rho/\partial t = -\rho_0\nabla\cdot\mathbf u$ is the whole of the physics, and setting the left-hand side to zero deletes the phenomenon rather than approximating it. The comparison that matters is not how large the density change is but whether the quantity being asked about is the one the approximation discards.

That is visible in what an incompressible solver does to the pressure. With the divergence imposed, the pressure satisfies a Poisson equation — elliptic, instantaneous, with every point depending on every other at once — so pressure in such a model has no propagation speed. It is not that the sound is fast; there is no sound, at any speed, ever.

The practical resolution is to compute the flow one way and the sound another. Lighthill’s acoustic analogy rearranges the exact equations into a wave equation whose source term is built from the flow’s own momentum fluxes, so a nearly-incompressible calculation of the flow supplies the source and the wave equation supplies the radiation. The two problems are separated because the energy in them is separated: the acoustic power radiated by a jet goes as the eighth power of its speed, so the fraction of the flow’s energy that becomes sound scales as $M^5$ and is utterly negligible as a drain on the flow while being the only thing anybody outside the jet can hear.

Which is the essay’s own point pushed to its limit. Incompressible is a decision about which terms to keep, taken with a question in mind, and the same air at the same instant is incompressible for the aerodynamicist and compressible for the acoustician.

The two hypotheses a solver is really making #

It is worth separating the two things an incompressible code assumes, because they have different justifications and different failure modes.

The first is thermodynamic: that pressure changes are too small to move the density. The size of the pressure change in a flow is $\tfrac12\rho U^2$, and the density change it causes is $\tfrac12\rho U^2/(\rho a^2) = \tfrac12 M^2$ of the density. So this hypothesis is exactly the statement that $M^2$ is small, and the calculation above is a measurement of it.

The second is caloric: that no parcel is changing its density for any other reason — no heating, no mixing between fluids of different density, no phase change. That hypothesis has nothing to do with the Mach number and can fail at any speed. It is the one a flame breaks, and the one a hot jet breaks, and the one a Boussinesq calculation keeps by treating the density variation as small everywhere except where gravity multiplies it.

An incompressible solver assumes both and reports neither. A reader working out whether it is applicable has to check them separately.

The Jacobian, and what it is worth #

The volume-ratio identity checked in the figure above is worth stating in general, because it is the finite-time version of the equation this essay opens with.

Along a trajectory, the determinant of the deformation gradient obeys

$$\frac{\mathrm D}{\mathrm Dt}\ln\det\mathsf F = \nabla\cdot\mathbf u,$$

so $\det\mathsf F = \exp\!\int\nabla\cdot\mathbf u\,\mathrm dt$, and in a steady flow continuity turns the right-hand side into the density ratio between the ends of the path. That is why an incompressible flow’s map is area-preserving, which is the property chaotic advection is built on and the reason a blinking-vortex map can be analysed as an area-preserving map of the plane at all.

It is also why the check is worth running: a numerical flow map whose determinant drifts is a flow map that has stopped being incompressible, and the drift is invisible in the streamline picture. This collection now measures it in every material-region calculation it makes.

What is not in the velocity field #

The density field. The divergence says what is happening to a parcel’s volume, and that is a statement about the parcel’s density only once continuity is invoked. A velocity field with zero divergence is consistent with a uniform density, a stratified one, or a wildly varying one; a velocity field with a divergence says the density is changing along a path, and nothing about what it is.

The consequence is that incompressible is not a property that can be read off a velocity field at all — it is the same field for all three of those cases. What can be read off a velocity field is $\nabla\cdot\mathbf u$, and calling that quantity’s vanishing incompressibility imports an assumption about the density that the field does not contain.

The model limit #

The gas calculation here evaluates the exact isentropic density relation on the incompressible velocity field, which is correct to $O(M^2)$ — the order of the quantity being computed. That is the leading term of the Janzen–Rayleigh expansion and it is why the fitted exponent over the whole sweep is 2.05 rather than 2.00: the next term is present in the exact relation and absent from the field it is being evaluated on.

At Mach 0.6, where the peak dilatation is 1.18 in units of $U/a$, the approximation is being used well outside the range in which it is a small correction. The figure is drawn there anyway, because the shape of the curve is the point and the departure from the slope is part of what the figure says — but no number from the high end of that sweep should be quoted as a prediction.

The stratified case has a limit of its own, and it is the one stratification spends its essays on. Setting $\rho = \rho(\psi)$ makes the density exactly advected, which is right for an inviscid steady flow and wrong as soon as diffusion of the stratifying agent matters — at which point the density is no longer constant on streamlines, the flow is still incompressible, and the two statements have come apart yet again.