The streamfunction says which relaxed state

An equilibrium a three-dimensional flow cannot have takes a gas of point vortices in a box and shows that at high enough energy its temperature goes negative and the vortices cluster by sign: the statistical reason the energy of a cascade that runs backwards ends up in the largest structures the box allows. It closes on the gap between that argument and a real flow. A point-vortex gas predicts clustering. A real condensate is a pair of large smooth vortices with a definite profile, and three continuous theories predict three different profiles.

Each prediction is a relation between the vorticity $\zeta$ and the streamfunction $\psi$, since any steady two-dimensional flow has $\zeta = F(\psi)$ for some function $F$. Montgomery and Joyce’s mean-field theory of the point-vortex gas gives $F = \lambda\sinh\beta\psi$. The Miller–Robert–Sommeria theory, which keeps every invariant of the continuous vorticity field rather than only the energy, gives a family whose simplest member, for a flow that began as two levels of vorticity, is a tanh. And the minimum-enstrophy argument of selective decay gives a straight line. The essay asks for a statistic that separates them by more than a real flow’s scatter. This one computes the three states and looks for it.

How a flow gets to a dipole #

The question is worth asking because the end state is reached by a process that looks nothing like the theories that predict it. Decaying turbulence in a periodic box does not diffuse towards its equilibrium; it gets there by a long sequence of mergers. Two vortices of the same sign orbit each other, as vortices move each other describes, and if they are close enough they merge — and a vortex pair carries eddies no snapshot sees showed how much fluid such a pair carries with it as it goes round. Each merger makes a larger vortex and throws off filaments of vorticity that are sheared out and dissipated. At the end there is one vortex of each sign, or on a rotating plane a set of stripes, and whatever relation they obey was written by that history.

The three theories are three claims about what the history forgets. The mean field forgets everything except the energy and the circulation, as a gas of point vortices would. The maximum-entropy theory remembers how much fluid began at each level of vorticity, since two-dimensional flow only rearranges vorticity and never changes the area occupied by each value. Selective decay forgets even the energy’s distribution, and keeps only the fact that enstrophy is lost faster than energy. The relation the dipole obeys is therefore a measurement of what the turbulence remembered, which is why it is worth a statistic that can be trusted.

Three states in one box #

On the doubly periodic square, a steady state with $\zeta = F(\psi)$ solves

$$-\nabla^2\phi = \mu\,f(\phi), \qquad \phi = \beta\psi, \qquad f = \sinh,\ \tanh \text{ or the identity}.$$

For every $f$, the relaxed pair — the dipole, one vortex of each sign — starts from the lowest mode of the box, $\phi = a(\sin x + \sin y)/2$, at $\mu = 1$, and grows from it as the amplitude $a$ rises. It was followed here by a fixed-point iteration: invert the Laplacian spectrally on $64^2$ points, apply it to $f(\phi)$, and rescale so that the largest value of $\phi$ stays at $a$, repeating until nothing changes. Each state converges in a dozen or two sweeps, to a residual in the equation of three parts in $10^{12}$ of its largest vorticity.

Two things in the iteration mattered. A strongly peaked sinh state started from the sine can land on a different solution, so the branch is followed by continuation, each state starting from its neighbour. And the tanh dipole is not an attractor of the iteration at all: left alone, round-off grows into the one-dimensional bar state, a pair of opposite jets, which is. The dipole’s symmetry between $x$ and $y$ is imposed each sweep to hold it. The bar states were then used as a check: their equation is an ordinary differential equation, and solving it again by shooting gives the same $\mu$ to five parts in $10^{12}$.

The scale of $\beta$ drops out of everything compared here. The ratio of enstrophy to energy $Z/E$ and the flatness of any field are the same whatever multiple of $\psi$ is drawn, so each theory is one curve, parametrised by its amplitude, and the three can be compared state against state.

Same energy, same enstrophy, different vortices #

The first figure puts two of them side by side at the same ratio of enstrophy to energy, 1.1 — the kind of value a decayed flow in a periodic box ends with — cut along the diagonal that passes through both vortex centres. The sinh state has concentrated its vorticity into sharp cores reaching more than three times the rms value. The tanh state has spread it into flat-topped patches with steep edges, never more than 1.3 times the rms. The linear state, a plain sine, can only exist at $Z/E = 1$ exactly, since any straight-line relation in a box is one of the box’s own modes, and it sits between them.

These are two kinds of vortex met separately elsewhere. The sinh core is what a gas of like-signed point vortices becomes when it clusters, and the tanh patch is the uniform vortex patch whose edges a strained vortex keeps until it is torn. Energy and enstrophy do not choose between them.

The scatter plot shows only curvature #

The second figure is the plot everyone draws. The sinh relation bends away from the straight line — it adds vorticity where the streamfunction is largest — and the tanh bends towards it and flattens. A straight line fitted to either misses by 0.27 of the rms vorticity. The tanh form, fitted to the sinh state, does no better than the straight line, because a tanh cannot bend that way at all; its best fit is its own linear limit.

That sounds decisive, and in a computed state it is. A measured state is not a curve but a cloud, because a decayed flow still carries small eddies that have not relaxed, and those scatter each point vertically by an amount comparable to the curvature being sought. The cloud from a sinh state and the cloud from a tanh state overlap most where there are most points, near the middle of the plot, and differ at the ends, where there are fewest. That is the observation that left the question open.

The count of points makes it concrete. In the sinh state at $Z/E = 1.1$, seven grid points in ten have a streamfunction less than half its largest value, where the three curves are within a fraction of the rms vorticity of each other, and only nine in a hundred lie beyond four-fifths of it, where the sinh curve climbs away. The tanh state puts more of its area near its extremes — seventeen in a hundred — but there its curve is flat and close to the line’s. A cloud of points is dominated by its middle, and its middle is where the theories agree. A fit that weights every point equally is therefore mostly a fit to the part of the plot that cannot tell the answer, and the part that can is a few per cent of the points, the ones nearest the vortex centres, which are also the ones most disturbed by a small eddy passing through a core.

A number instead of a shape #

A shape is hard to compare; a number is not. The vorticity’s flatness, $\langle\zeta^4\rangle/\langle\zeta^2\rangle^2$, measures how much of the field sits in extremes, and the third figure follows it along both branches against $Z/E$. They start together at the linear state’s 9/4 and part at once. At $Z/E = 1.1$ the sinh state’s flatness is 4.94 and the tanh state’s 1.35, nearly a factor of four apart. The tanh branch cannot even reach a ratio above about 1.3: its vorticity saturates, and two uniform patches are as far as it can go.

That would settle it if $Z/E$ could be measured cleanly, since the comparison has to be made at the measured ratio. It cannot.

What unrelaxed eddies do to each number #

The fourth figure adds to each state a field of small eddies, at wavenumbers 8 to 16 with random phases, whose rms vorticity is a fraction of the state’s, and recomputes three statistics. The enstrophy-to-energy ratio moves most: at 30 per cent it has risen by nine per cent for both states, and at 40 per cent by sixteen, because small eddies carry enstrophy and almost no energy. Any comparison made at the measured $Z/E$ is therefore made at the wrong point on each branch. The vorticity’s flatness moves too — the tanh state’s rises towards the Gaussian value of 3, the sinh state’s falls towards it — by up to a third.

The streamfunction’s flatness does not move at all. An eddy of wavenumber $k$ contributes to the streamfunction its vorticity divided by $k^2$, so eddies eight to sixteen times smaller than the box add a few parts in ten thousand to it. Even eddies at wavenumbers 2 to 4 — not small at all, and the kind a flow that has not quite finished its inverse cascade would still carry — move the streamfunction’s flatness by 0.01 at twenty per cent of the vorticity, against a gap between the sinh and tanh states of 0.6.

The sign rule #

The fifth figure is the statistic the question asked for. It divides the streamfunction’s flatness by that of the linear state of the same shape — 9/4 for the dipole, 3/2 for the bar — and follows it along all four branches. Every sinh state lies above one, and rises with amplitude; every tanh state lies below one, and falls. At every amplitude computed, in both shapes, the side of one that the ratio falls on says which kind of relation made the state.

The reason is not hidden. A relation that concentrates vorticity where the streamfunction is extreme — sinh, or anything convex like it — makes the streamfunction itself more peaked than a sine, and a peaked field has a high flatness. A relation that saturates — tanh, or any concave relation — spreads the vorticity into patches, and the streamfunction over a uniform patch is rounder-topped than a sine. The sign test does not say that a relation is exactly sinh rather than some other convex function, and the Miller–Robert–Sommeria family contains convex members too. It says which side of the straight line the relation bends, which is the question the scatter plots were asked, and it answers it from a quantity the unrelaxed eddies cannot reach.

What it needs from a measured flow is the streamfunction, which any simulation computes and any measurement of the velocity field yields by one integration, and the geometry of the final state, dipole or bar, which is visible by eye. It does not need the energy, the enstrophy or their ratio.

What either answer would mean #

A ratio above one would say that the relaxed vortices are more concentrated than the box’s own mode — that the turbulence has gathered its vorticity towards the centres as a gas of like-signed elements would, forgetting how that vorticity was originally distributed. That is the mean-field picture’s claim, and it is a strong one: it says the relaxed state of any two flows with the same energy is the same, whatever they looked like at the start.

A ratio below one would say that the vorticity has been spread into patches whose levels are bounded — that the flow still remembers the largest vorticity it began with, and cannot concentrate beyond it. That is what a flow made of a few uniform patches must do, since area-preserving stirring can rearrange a patch but not raise its level, and it is the heart of the maximum-entropy theory. On this reading the tanh is not a rival to the sinh but its opposite limit: a flow that began as two levels relaxes to a tanh; a flow that began as a sprinkle of intense small vortices relaxes towards a sinh; and most flows are somewhere between, on the side set by where their initial vorticity was.

That makes the sign a test of the theories’ premises rather than a curve fit. It can be applied to one flow, it gives one bit of information, and the bit is the one that matters.

Why not simply look at the peaks #

There are other candidates, and the fourth figure’s argument ranks them. The largest vorticity in the field is the most obvious — the sinh core reaches 3.3 times the rms and the tanh patch 1.3 — but an unrelaxed eddy is exactly a local peak of vorticity, and the maximum is the statistic it moves most. The velocity is the streamfunction differentiated once, so an eddy of wavenumber $k$ contributes to it the vorticity divided by $k$ rather than $k^2$: the velocity’s flatness is better than the vorticity’s and worse than the streamfunction’s. Each integration makes the field smoother and pushes the small eddies further down, and the streamfunction, two integrations away from the vorticity, is the smoothest field the flow has. The statistic that separates the theories is the one that looks at the flow at the scale the theories are about.

What was checked #

The sixth figure is the ledger: residuals of three parts in $10^{12}$; the small-amplitude limit, where every relation becomes the linear one with $\mu = 1$, flatness 9/4 and $Z/E = 1$; the bar states against an independent shooting solution; and the two flatnesses at $Z/E = 1.1$. The states were also recomputed on $128^2$ points, and their statistics agreed with the $64^2$ values to eight figures.

What the picture cannot show #

Steady states only. The three theories predict where a flow ends; the calculation computes those end points and not the approach. A flow that has not finished relaxing is represented here only by the contamination of the fourth figure, which is added by hand rather than left over by a simulation.

One box. The doubly periodic square admits the dipole and the bar; a box with walls or a different aspect ratio has other end states, and the reference value of the linear state changes with them.

A two-member caricature of the maximum-entropy family. The Miller–Robert–Sommeria theory predicts a relation determined by the whole initial distribution of vorticity levels. The tanh is its two-level case; others are convex, and for those the sign rule would read “sinh-like”.

Inviscid states. Viscosity keeps decaying the final state slowly, and a slowly decaying dipole’s relation drifts towards the linear one as its cores spread.

The convention the numbers depend on #

The square has side $2\pi$, so the lowest modes have wavenumber one. $\phi = \beta\psi$ is scaled so that the relation reads $-\nabla^2\phi = \mu f(\phi)$, and the amplitude is the largest value of $\phi$. Flatness is the fourth moment over the square of the second, with the mean, which is zero, not subtracted.

Who found it, and when #

The sinh-Poisson relation is Joyce and Montgomery’s of 1973, and its comparison with decaying simulations Montgomery, Matthaeus and collaborators’ of the early 1990s. The maximum-entropy theory of the continuous field is Miller’s of 1990 and Robert and Sommeria’s of 1991, and selective decay, the minimum-enstrophy picture, Bretherton and Haidvogel’s of 1976. The mean-field limit of the point-vortex gas that the earlier essay sampled is Onsager’s idea of 1949 made quantitative.

Still open: the sign in a decaying simulation #

The sign rule is a prediction about data the calculation has not touched. The next step runs a decaying two-dimensional simulation in the same periodic box, from several initial conditions — random phases, two-level patches, a turbulent spectrum — follows the streamfunction’s flatness over the linear state’s through the decay, and asks on which side of one it settles, and whether that side depends on how the flow began, which is what the maximum-entropy theory says it should and what the mean-field theory says it should not.