The spin that feeds itself

The vorticity equation for an incompressible fluid is short, and one term in it has no counterpart in two dimensions:

$$\frac{\mathrm{D}\boldsymbol{\omega}}{\mathrm{D}t} = \underbrace{(\boldsymbol{\omega}\cdot\nabla)\mathbf{u}}_{\text{stretching and tilting}} + \nu\nabla^2\boldsymbol{\omega}$$

The first term on the right is why turbulence exists in three dimensions and not in two, why the small scales of a real flow are so much smaller than the large ones, and why nobody can compute a flow at the Reynolds number of an aircraft. It says that a vortex tube being pulled along its own axis spins faster.

This essay does three things with that: derives the amplification from a conservation law rather than from the equation, asks what stops it, and answers with the one exact solution where the balance closes.

The amplification, from two things that cannot change #

The derivation needs no differential equation at all.

Take a tube of fluid enclosing a circulation $\Gamma$, of cross-sectional area $A$ and length $L$. Kelvin’s theorem says $\Gamma$ round a material circuit cannot change in an inviscid fluid with conservative body forces. Incompressibility says the volume $AL$ cannot change either.

Now stretch it, so $L \to \lambda L$. Then $A \to A/\lambda$, and since vorticity is circulation per unit area,

$$\omega = \frac{\Gamma}{A} \;\longrightarrow\; \frac{\Gamma}{A/\lambda} = \lambda\,\omega$$

The spin rises by exactly the stretch ratio. No approximation entered, no model of the fluid was needed beyond incompressibility, and the same argument works for an ice skater pulling in their arms — except that for the skater the conserved quantity is angular momentum and for the tube it is circulation, which is a different quantity that happens to give the same answer here.

The solver states the two conservations and computes the third quantity, and the assertion checks that they agree: assertStretchingConserves recomputes the vorticity as circulation over area and compares it with the amplified value. A stretched tube that gained circulation is refused, which is the failure the argument would not survive.

Where the extra energy comes from #

The spin went up, so the kinetic energy went up. It is worth being explicit about what paid for it, because “conservation of circulation” can sound as though something was had for nothing.

The energy per unit length of a core of circulation $\Gamma$ and radius $r$ goes as $\rho\Gamma^2$ with a logarithm in it, and the length of the tube has grown by $\lambda$, so the total energy of the tube rises. What supplied it is the straining flow that did the stretching: pulling on a tube that resists being pulled is work, and the work goes into the tube.

That is the whole mechanism of the energy cascade, stated once and for all. A large eddy strains the fluid around it; a smaller vortex embedded in that strain is stretched by it; the stretching transfers energy from the large eddy to the small one and makes the small one smaller still. Nothing in the argument is statistical and nothing in it is a model. The cascade is this paragraph, repeated across a range of scales.

The question the argument does not answer #

If stretching multiplies vorticity and the strain field is always there, why is the vorticity in a real fluid finite?

The tempting answer — that the stretching stops — is not the right one. Something else happens, and it can be computed exactly.

Thinning a tube brings its own brake. Viscous diffusion spreads vorticity outwards on a timescale $r^2/\nu$, so a tube of radius $r$ smears itself out at a rate proportional to $\nu/r^2$. Halving the radius quadruples that rate. Meanwhile the stretching only multiplies the vorticity by $\lambda$. The brake strengthens as the square of what the drive strengthens as, so the two must cross, and where they cross is a length.

Burgers’ vortex, and the length it settles at #

The balance can be written down exactly, and it is one of a handful of exact solutions of the Navier–Stokes equations this site has.

Impose an axisymmetric straining field $u_r = -\tfrac{1}{2}\alpha r$, $u_z = \alpha z$: fluid drawn in from the sides and pushed out along the axis, at a constant rate $\alpha$. Look for a steady, axisymmetric vorticity distribution $\omega(r)$ in it. The steady vorticity equation becomes

$$-\frac{\alpha r}{2}\,\frac{\mathrm{d}\omega}{\mathrm{d}r} = \alpha\omega + \nu\left(\frac{\mathrm{d}^2\omega}{\mathrm{d}r^2} + \frac{1}{r}\frac{\mathrm{d}\omega}{\mathrm{d}r}\right)$$

and it is solved by a Gaussian:

$$\omega(r) = \frac{\Gamma\alpha}{4\pi\nu}\exp\!\left(-\frac{\alpha r^2}{4\nu}\right), \qquad r_{\text{core}} = \sqrt{\frac{4\nu}{\alpha}}$$

The left-hand side is inward transport by the strain, the first term on the right is the amplification, and the last is diffusion outwards. All three are the same size at the core radius, which is what “balance” means here.

What the solver computed, and how it was checked #

Three things, and the arrangement of the third is the point.

The core radius and the peak. At $\nu = 0.005$ and $\alpha = 2$ the core comes out at exactly $0.1$, with a peak vorticity of $31.83 = \Gamma\alpha/4\pi\nu$. Both are closed forms and neither is interesting on its own.

The fastest swirl is not at the core radius. The swirl velocity is $u_\theta = \Gamma\left(1 - e^{-r^2/r_c^2}\right)/2\pi r$, and its maximum sits at $1.1209\,r_{\text{core}}$ — found here by search over the returned profile rather than by differentiating the formula. It matters because “the radius of a vortex” is quoted in two different senses in the literature and they differ by twelve per cent.

The residual. The steady vorticity equation is evaluated on the returned profile, with every derivative taken by five-point differences of the function the figures draw, and the worst relative residual over the whole core is under $10^{-9}$. That arrangement is deliberate: a residual computed from the algebra that produced the answer tests the typing. A residual computed from the drawn profile tests the drawing.

The check earns its keep in the rejection test. A profile whose core radius is one per cent wrong is still a Gaussian, still smooth, still indistinguishable from the right one by eye — and its residual is three orders of magnitude above the tolerance. The assertion catches it; nothing else here could.

The factor is not a limit and not an estimate, so it is worth drawing at a second value rather than asserting that it scales.

What the balance says about turbulence #

Two consequences, both quantitative.

A smallest scale exists, and this is it. Burgers’ vortex says a strain field of rate $\alpha$ supports vortex cores of size $\sqrt{4\nu/\alpha}$. Estimating $\alpha$ from the energy dissipation rate $\varepsilon$ as $\alpha \sim \sqrt{\varepsilon/\nu}$ gives a core of order $(\nu^3/\varepsilon)^{1/4}$, which is the Kolmogorov length exactly. The dimensional argument and the exact solution arrive at the same length by entirely different routes, and that agreement is the strongest reason to believe either.

The cost of computing a flow follows from it. If the smallest structure is set by the strain and the largest by the geometry, the ratio between them fixes how many grid points a simulation needs, and the count runs as the Reynolds number to the nine quarters. The stretching term is why that ratio is large.

The solution applied to a real vortex, and what it gets wrong #

It is worth putting numbers into the formula for something a reader can picture, because the answer is wrong by a factor of a thousand and the reason is the interesting part.

A tornado has a strain rate of order $0.1\ \mathrm{s^{-1}}$ — the inflow converges over a few hundred metres in a few tens of seconds. Air has $\nu = 1.5\times10^{-5}\ \mathrm{m^2/s}$. Burgers’ balance then puts the core at

$$r_{\text{core}} = \sqrt{\frac{4\nu}{\alpha}} = \sqrt{\frac{6\times10^{-4}}{1}} \approx 2.4\ \text{cm}$$

A real tornado’s core is tens of metres across. The formula is out by three orders of magnitude, and nothing is wrong with the formula.

What is wrong is the $\nu$. The balance is between stretching and whatever spreads vorticity outwards, and in a tornado that is not molecular diffusion — it is the turbulent mixing of the vortex itself, which is enormously more effective. Running the argument backwards, a fifty-metre core at that strain rate needs an effective diffusivity of $\alpha r^2/4 \approx 60\ \mathrm{m^2/s}$, which is four million times the molecular value.

That is a respectable use of an exact solution: it does not describe the tornado, and the size of its failure measures something real. The same substitution — a turbulent diffusivity where a molecular one belongs — is the whole content of an eddy-viscosity closure, and here it can be seen doing its job on a single vortex rather than on a statistical average.

The same question asked of a fluid with no viscosity at all #

The essay’s central question — what stops the amplification — was answered with viscosity, and the answer is complete for a real fluid. Asked of an ideal one it is the deepest open problem the subject has, and it is worth stating because it is exactly this mechanism taken seriously.

Remove the viscous term and nothing in the stretching argument sets a limit. A tube stretched by $\lambda$ spins $\lambda$ times faster, and the faster spin can stretch other tubes harder, which can stretch it harder in turn. Whether that feedback can run away to infinite vorticity in finite time, starting from perfectly smooth initial data, is not known.

What is known is a criterion. Beale, Kato and Majda showed in 1984 that a solution of the Euler equations stays smooth for as long as

$$\int_0^T \max_{\mathbf{x}} |\boldsymbol{\omega}|\;dt$$

is finite. So a singularity, if it exists, must announce itself in exactly the quantity this essay is about: the peak vorticity has to diverge, and to diverge fast enough that its integral does too. Every numerical hunt for blow-up since has been a hunt for that integral, and the searches have been delicate for a reason the essay’s own arithmetic supplies — a solution approaching a singularity concentrates into a thinner and thinner core, so the resolution required grows exactly as fast as the thing being resolved.

The current position is worth stating plainly, because it is unusual for a question this old. For axisymmetric flow with a boundary there is strong numerical evidence of genuine blow-up. For the general three-dimensional problem without boundaries it remains open, and the corresponding question for the Navier–Stokes equations is one of the Clay problems. The mechanism in this essay is the whole of what is at stake in it.

The event the tube argument forbids #

There is a second thing viscosity permits that the conservation argument rules out, and it is how a tangle of vortices actually gets rid of itself.

Kelvin’s theorem says the circulation round a material circuit cannot change, and a consequence is that vortex lines are material: they are carried by the flow and cannot break, cross or rejoin. Two linked vortex rings in an ideal fluid stay linked for ever, and a knotted vortex stays knotted — which is what conserved helicity forbids.

In a real fluid they do not. Bring two antiparallel tubes close together and they flatten against one another, the gap between them thins, and in that thin region the cross-diffusion of vorticity is fast — because diffusion goes as the inverse square of the separation, which is the same brake this essay has already used. The tubes reconnect: each cuts and rejoins to the other, the topology changes, and the circulation round a circuit that used to enclose one tube now encloses something else.

That has been watched happen, in simulations since the 1980s and in the laboratory since knotted vortices could be made in water and photographed. A knot tied in a vortex unties itself, in a fluid whose ideal limit forbids it, through a process confined to a thin sheet that lasts a moment.

So the two things this essay’s conservation law rules out are the two things that decide a real vortex’s fate — an unbounded concentration, and a change of topology — and both are permitted by the same term, acting in the same thin region, for the same reason. Viscosity is not merely the brake on the amplification. It is the only mechanism by which the arrangement of vortex lines can ever change at all.

Two dimensions, where none of this happens #

In two dimensions the vorticity vector points out of the plane and the velocity gradient lies in it, so $(\boldsymbol{\omega}\cdot\nabla)\mathbf{u}$ is identically zero. There is no stretching term at all.

The consequences are enormous and are the reason two-dimensional turbulence is a separate subject rather than an easier case of the same one. Without stretching, vorticity is merely carried about and diffused, so it cannot be concentrated; energy moves to larger scales rather than smaller ones; and long-lived vortices form and persist instead of being torn into filaments. A weather map and a wind tunnel are not the same phenomenon at different sizes.

It also means the flat pictures on this site cannot show the mechanism this essay is about. Every figure here is a cross-section through a three-dimensional argument, and the stretching itself happens along the axis that is not drawn.

The one number that does not move #

Through all of this, one quantity has been fixed: the circulation.

The tube’s vorticity rose by $\lambda$ and its area fell by $\lambda$; Burgers’ core thins as the strain rises and its peak vorticity rises in proportion. In every case the product — the circulation $\Gamma$ — is the same number it started as. Stretching does not create vorticity. It concentrates it, and concentration is what makes it visible, what makes it dissipate, and what makes it hard to compute.

That is the sense in which this whole field is one conservation law with different things done to it. Circulation is vorticity added up over an area, and every result in this essay follows from holding that sum fixed while the area changes.

What the picture cannot show #

A real vortex tube is not axisymmetric and not straight. Burgers’ solution is a straight tube in a uniform strain, and the vortices in a real turbulent flow are curved, tangled and stretched at rates that change along their own length. What survives is the balance and the scaling, not the shape.

The tilting half of the term is missing. $(\boldsymbol{\omega}\cdot\nabla)\mathbf{u}$ does two jobs: it stretches vorticity that is aligned with the strain and it tilts vorticity that is not, turning spin about one axis into spin about another. Nothing here draws the tilting, and in a real flow it is what feeds vorticity into the stretching directions in the first place.

Nothing here is a solution of turbulence. Burgers’ vortex is a single vortex in an imposed strain of somebody’s choosing. It says what a strained vortex settles at; it does not say where the strain came from, and in a turbulent flow the strain is produced by other vortices doing the same thing. This site’s standing limit applies here too.

Where the model stops #

The stretching argument needs the circulation to be conserved, which needs the flow to be inviscid round the circuit being followed. Over a viscous core the circulation is not conserved, and the argument only ever applies to a circuit drawn outside the core — which is exactly why the balance above has to be solved rather than argued.

Burgers’ solution needs the strain rate to be uniform, steady, and axisymmetric about the vortex. Real strain is none of these, and the solution’s honest status is as the local model of a core rather than as a description of a flow.

And the whole picture assumes incompressibility. In a compressible flow a tube can be stretched without thinning, by getting less dense instead, and the amplification is correspondingly weaker. That is a real difference in the compressible field and it is why the vorticity equation there carries an extra term.

Who found it, and when #

Helmholtz stated the vortex laws in 1858 and the stretching result is his: a vortex tube’s strength is constant along its length and in time, so a tube that is drawn out must spin faster. Kelvin gave the circulation theorem the tube argument rests on in 1869.

Burgers published the exact solution in 1948, looking for a model of turbulent dissipation that could be written down, and Rott extended it in 1958. Townsend measured vortex structures of about the right size in a wind tunnel in 1951, which is the earliest evidence that the solution describes something real. The modern reading of it — as the local structure of the fine scales rather than as a model of turbulence — arrived with the numerical simulations of the late 1980s, which found tubes of exactly this kind and of about this radius.

Where the ladder goes next #

Vorticity has now been followed from what it is, through the circuit integral that measures it, to the term that amplifies it. What has not been asked is what a vortex does once it exists: a single one sits still forever, and two of them move each other. That is the beginning of vortex dynamics, and it is one field along.