Inviscid does not mean irrotational

Two words are used interchangeably through most of this subject, including in the first two rungs of this collection’s own ideal-flow ladder, and they are not the same hypothesis.

Inviscid means the viscous term has been dropped. What is left is Euler’s equations: three components of momentum plus continuity, nonlinear, with vorticity in them.

Irrotational means the additional assumption that $\nabla\times\mathbf{u} = 0$. That collapses the whole velocity field into the gradient of one scalar, turns the momentum equation into a formula for the pressure, and leaves Laplace’s equation — linear, with a unique answer once the boundary is given, and with closed-form solutions for everything.

The step between them is enormous and it is not free.

What licenses the second step #

The licence is Kelvin’s circulation theorem: in a fluid with no viscosity, constant density and conservative body forces, the circulation round a loop of marked particles does not change. So a flow that started irrotational stays irrotational, forever.

Read the conditional in that sentence. The irrotational theory is a theory about flows with a particular history — flows that began at rest, or in a uniform stream, and have been disturbed only by bodies moving through them. That covers a great deal: an aeroplane accelerating down a runway through still air is exactly that case, and it is why potential flow is as useful as it is.

It does not cover a flow that arrives with vorticity already in it, and a great many do. The wind has a boundary layer hundreds of metres deep, so every wing near the ground is flying in a shear. A propeller works in the wake of a fuselage. A turbine blade works in the wake of the row in front. A second aeroplane in formation is in the trailing vortices of the first.

What steady Euler actually says #

For a steady plane flow the vorticity equation collapses to one statement:

$$\mathbf{u}\cdot\nabla\omega = 0,$$

the vorticity is constant along a streamline. Since $\psi$ is also constant along a streamline, that means $\omega$ is a function of $\psi$ alone:

$$\omega = F(\psi),\qquad \nabla^2\psi = -F(\psi).$$

That is the whole of steady two-dimensional Euler, and the important thing about it is what it leaves open. $F$ is not determined by the equations. It is carried in from upstream on every streamline that comes from upstream, and it is the flow’s memory of where it has been.

Potential flow is the case $F \equiv 0$. Every other $F$ is a different theory of the same fluid, and each of them is exact.

The case that is exactly soluble #

A uniform shear $U + Ky$ has vorticity $-K$ everywhere, so $F$ is a constant and the problem is $\nabla^2\psi = K$ — linear, with a particular solution $Kr^2/4$ that happens to be constant on any circle centred at the origin. So the rest of the problem is the harmonic one solved in this collection’s earliest essays, and the answer is closed-form:

$$\psi = U\left(r - \frac{a^2}{r}\right)\sin\theta \;+\; \frac{Kr^2}{4} \;-\; \frac{K}{4}\left(r^2 - \frac{a^4}{r^2}\right)\cos 2\theta.$$

Three terms. The first is the ordinary cylinder solution. The third is the image of the shear’s quadrupole part, reflected in the circle by the circle theorem. The second is the entire difference between this and a potential flow, and it is the only term with a non-zero Laplacian.

The vorticity is measured rather than asserted: differencing the streamfunction twice over five hundred points returns $-K$ to seven parts in a million, everywhere outside the body.

Bernoulli’s constant is not one constant #

The pressure is where the difference bites, and it bites in the way that produces most of the errors in this part of the subject.

Bernoulli’s equation is an integral of the momentum equation along a streamline. In an irrotational flow the constant of integration happens to be the same on every streamline, which is why it is normally written as a single number and applied between any two points in the flow. In a rotational flow it is not.

Measure it upstream, where the flow is the undisturbed shear and the transverse momentum balance makes the pressure uniform across the stream. Then $H(y) = p_\infty/\rho + \tfrac12(U+Ky)^2$, which varies across the streamlines before the body is anywhere near. Over the streamlines drawn in the figures here it varies by a factor of twenty.

Crocco’s theorem is the general statement: $\nabla H = \mathbf{u}\times\boldsymbol\omega$ for a steady flow. The gradient of the Bernoulli constant is perpendicular to both the velocity and the vorticity, so it vanishes when either does — and only then.

Two ways to get vorticity into a flow that has none #

If Kelvin’s theorem forbids the generation of vorticity, where does any of it come from? Three answers, and each of them is a hypothesis of the theorem being broken on purpose.

A wall. No-slip is a viscous boundary condition and viscosity is the term Kelvin’s theorem assumed away. Every boundary layer is a sheet of vorticity manufactured at a surface, and every wake is that sheet after it has left.

A density gradient that is not aligned with the pressure gradient. The theorem assumes a barotropic fluid; in a stratified one the baroclinic term generates circulation directly, which is what drives a sea breeze and what tilts the vorticity behind a curved shock.

And a body force that is not conservative, which in practice means a rotating frame, where the Coriolis term does the work.

So the vorticity in the shear above was made somewhere — in the ground’s boundary layer, most likely, tens of kilometres upstream — and has been carried since by a fluid that could neither create nor destroy it. That is what “carried in from upstream” means, and it is why $F(\psi)$ is a memory rather than a parameter.

There is a third hypothesis, and it is not either of the first two #

The essay opened by separating two words that are used as one. Crocco’s theorem separates a third, and it is the one that makes the plane case above look simpler than the subject is.

In two dimensions the vorticity is perpendicular to the velocity everywhere, because it has nowhere else to point. So $\mathbf{u}\times\boldsymbol{\omega}$ vanishes only when one of the two does, and a uniform Bernoulli constant is equivalent to irrotational flow — which is why the two are routinely treated as the same statement, and why the varying constant in the figure above reads as the signature of vorticity.

In three dimensions that equivalence fails. The cross product vanishes whenever the vorticity is parallel to the velocity, and a flow arranged that way has $\nabla H = 0$ everywhere: one Bernoulli constant through the whole field, applicable between any two points, with vorticity of any strength whatever.

Such flows have a name — Beltrami flows — and they are not contrived. The condition $\boldsymbol{\omega} = \lambda\mathbf{u}$ makes the nonlinear term in Euler’s equations a pure gradient, which can be absorbed into the pressure, so every Beltrami field is automatically a steady solution. A rigid-body rotation combined with an axial flow is one. So is the family of periodic fields whose streamlines are, notoriously, chaotic in a flow that is perfectly steady and written down in three lines. A tornado’s core is approximately one, and so is the swirling flow down the axis of a vortex tube in a turbulent field.

So there are three hypotheses where the textbooks usually offer two. No viscosity. No vorticity. And no component of vorticity across the velocity — which is weaker than the second, is enough for every algebraic convenience Bernoulli’s equation offers, and is available to flows with as much rotation in them as anyone could want.

What it does not buy back is the rest of the apparatus. A Beltrami flow still has no velocity potential, so the conformal maps, the image systems and the superposition are still gone; what it restores is only the single constant. That is a useful thing to know when reading a derivation: an author who applies Bernoulli globally is not necessarily assuming irrotational flow, and an author who proves a flow rotational has not thereby proved that Bernoulli’s constant varies.

The alignment also has a name as a quantity. The integral of $\mathbf{u}\cdot\boldsymbol{\omega}$ is the helicity, it is conserved by ideal flow for the same reasons circulation is, and a Beltrami flow is the extreme case — the field with the most helicity its energy and enstrophy allow. This collection has an essay on what that conservation forbids, and the connection is worth carrying: the quantity whose vanishing makes the plane theory simple is the same quantity whose maximum makes the three-dimensional theory tractable, and the ordinary case is neither.

The plane flows in this essay have helicity exactly zero, necessarily, since a vortex line in a plane flow is perpendicular to every velocity in it. Two dimensions is not the general case with one direction removed; it is the special case in which one of the three hypotheses collapses onto another, and every intuition trained there about what vorticity does to Bernoulli’s equation is an intuition about a coincidence.

The plane and the general case therefore need different words, and this collection uses them carefully from here: irrotational for the vanishing of the vorticity, homentropic or Bernoulli-uniform for the vanishing of its gradient of $H$, and Beltrami for the alignment. Three conditions, three names, and only in two dimensions do the first two coincide.

A lift with no circulation in it #

The force is where this stops being a technicality.

On the body’s own surface Bernoulli does hold, because the body is one streamline. So the surface pressure is $C - \tfrac12\rho q^2$ up to an additive constant, and an additive constant contributes nothing to a force round a closed curve. Integrating gives

$$L = 2\pi\rho U K a^2,$$

directed towards the fast side of the shear, on a body with no circulation round it whatever. The drag is zero to twelve decimal places: d’Alembert survives the vorticity intact.

The second route is worth dwelling on, because it is the only real check available. Getting the pressure field of a rotational flow requires inverting $\psi$ to find where each streamline came from, looking up its Bernoulli constant, and subtracting the local dynamic pressure. That the answer then matches a surface integral which never needed any of it is the evidence that both are right.

What it looks like when it is not a cylinder #

A wing in a shear does the same thing and the effect has a name in the trade: it is part of why a wing in ground effect on a gusty day behaves differently from the same wing at altitude, and why a propeller in a fuselage wake produces a periodic side force that shakes the airframe once per blade.

The mechanism is easy to state once the cylinder case is understood. The body sits in a stream that is faster on one side than the other. The stagnation streamline arrives off-centre. The flow round the fast side is faster still and the pressure lower, so there is a force towards the fast side — which sounds like Bernoulli’s fast-means-low-pressure and is not, because the two sides are on different streamlines and their constants differ. The right answer and the wrong argument agree in sign here, which is exactly the circumstance in which a wrong argument survives.

What the picture cannot show #

The upstream shear is uniform and no real one is. A uniform shear extends to infinity in both directions, so the velocity is negative below $y = -U/K$ and the streamfunction cannot be inverted there. Every figure here is drawn well inside that limit and the solution is a local model of a shear that must be bounded somewhere.

Nothing here is unstable, and a real shear is. A shear layer with an inflection in it is unstable to every wavelength, and a uniform shear with no inflection is stable in the inviscid theory and unstable in a viscous one at large enough Reynolds number. The steady solution drawn is a solution; it is not necessarily one a fluid would stay in.

And the vorticity is invisible. It is uniform, so there is nothing to shade: a contour map of it is one flat colour. What can be drawn is its consequence — the asymmetry of the streamlines, the varying Bernoulli constant, the force — and the quantity itself is a number printed in a caption.

What it costs to get this wrong #

The failure mode is specific and worth naming, because it is not “the answer is a bit off”.

Take a rotational flow and apply Bernoulli between two points on different streamlines. The result is wrong by the difference of their constants, which in the shear above is up to a factor of twenty in dynamic pressure. Then infer a pressure difference from a velocity measurement, or a velocity from a pressure measurement, and the error propagates into whatever the number was for.

The concrete version: a pitot-static probe in a shear reads a total pressure that belongs to the streamline the probe’s nose is on and a static pressure that belongs to the streamlines its side taps are on. In a uniform stream those are the same and the instrument works. In a strong shear they are not, and what the airspeed indicator believes becomes a question with a genuinely complicated answer.

What comes next in the ladder #

The obvious question is what happens when $F$ is not a constant, and the answer opens two rungs.

If $F$ is linear in $\psi$, the equation is still linear and there are still closed forms; Hill’s spherical vortex is the famous one, and it is a lump of rotating fluid travelling steadily through fluid at rest with no body in it anywhere.

And if the streamlines are closed — a recirculation bubble, an eddy in a cavity — then nothing comes in from upstream and $F$ is not determined by anything at all. That ambiguity is not one number but a whole function, and closing it needs an argument that takes the viscosity to zero rather than setting it to zero.

The list of what goes is worth having. No velocity potential, so no complex potential, no conformal mapping, no image systems, and no superposition — the equation is nonlinear in $\psi$ the moment $F$ is. No unique solution from the boundary data alone, because $F$ has to be supplied as well. And no single Bernoulli constant, so no algebraic route from a speed to a pressure. What survives is that streamlines exist and that vorticity is carried along them, which is enough to solve the one case above and not much more.

Who found it, and when #

Euler wrote the equations in 1757 and the irrotational specialisation is Lagrange’s, from 1781, with the circulation theorem that justifies it arriving from Kelvin ninety years later. Crocco’s theorem is from 1937, out of compressible aerodynamics, where the connection between entropy gradients and vorticity behind a curved shock made the distinction unavoidable. The shear-past-a-cylinder solution is in Lamb.

The surprising connection is with meteorology, where this is not a special case but the ordinary one. The atmosphere is stratified and rotating and its large-scale flow has vorticity everywhere; the quantity conserved along a streamline is potential vorticity rather than $\omega$, but the structure — a scalar carried by the flow, with the streamfunction satisfying a Poisson equation whose source is that scalar — is identical. Weather forecasting is done with the rotational form of the equations on this page, and the irrotational specialisation that dominates aerodynamics is not used there at all, because no atmospheric flow ever started from rest.

Where the ladder goes next #

Above this rung are the one rotational solution anybody can write down and, above that, the vorticity nothing decides.

Below it are the exact theory this is the general case of, and Kelvin’s theorem, which is the licence for specialising it.

Beside it is what vorticity actually is, which is the thing this essay spends its length insisting is not zero.