Three vortices can meet at a point in a finite time, and one per cent stops them

Vortices move each other set out what a set of point vortices is: each one carried by the velocity all the others induce and by nothing of its own. Three is the most that can be predicted counted the invariants and found that three vortices have just enough of them to be integrable, with quasi-periodic motion in which neighbouring trajectories separate at worst linearly. One clause in that essay mentioned that a three-vortex system can have “self-similar collapse in finite time when the strengths satisfy” a stated relation, and went on to chaos, which was the point of it.

The clause deserves a good deal more attention, because of what it says. A set of vortices in an ideal, incompressible fluid, with no boundaries and no forcing, reaches a state in which all of them occupy one point, and it does so not as a limit at infinite time but at a particular moment on the clock. The velocity they induce on each other goes to infinity as it happens, and their kinetic energy, whatever that means for a point vortex, stays unchanged.

The two-vortex answer makes this look impossible. A pair of like sign keeps its distance and circles for ever; a pair of unlike sign keeps its distance and travels. Distance is a conserved quantity for two, and for the sum of strengths of three it looks as if it should be nearly so. But three vortices have a loophole, and it depends on a piece of arithmetic so precise that it fails at the third figure.

The three things that would have to stop it #

If anything forbids the meeting, it is one of the conservation laws, and there are three of them that matter.

The centre of strength, $\sum\Gamma_i z_i$ over $\sum\Gamma_i$, does not move, so there is a fixed point for them to meet at. That helps rather than hinders.

The energy, for point vortices the logarithmic interaction

$$H = -\frac{1}{2\pi}\sum_{i<j}\Gamma_i\Gamma_j\ln r_{ij},$$

is conserved, and each pair has a logarithm that diverges as the pair closes. This is the term that seems to forbid everything. For a like-signed pair alone, $H$ conserved means $r$ conserved. Yet the sum runs over three pairs of mixed signs, so scale the whole triangle by a factor $\lambda$ and the logarithms each pick up the same $\ln\lambda$:

$$H \to H - \frac{\ln\lambda}{2\pi}\sum_{i<j}\Gamma_i\Gamma_j .$$

If that sum of products vanishes, a uniform scaling does not change the energy at all. And $\Gamma_1\Gamma_2+\Gamma_2\Gamma_3+\Gamma_3\Gamma_1=0$ is the same statement as $\sum 1/\Gamma_i=0$, Gröbli’s condition. For strengths 1 and 1 it fixes the third at exactly $-1/2$. The energy is not violated by the collapse; at these strengths it simply cannot see the triangle’s size. It sees only its shape.

The second moment is the last. The combination

$$L = \sum_{i<j}\Gamma_i\Gamma_j\,r_{ij}^2$$

is also conserved, and it scales with the square of the size, so if all the sides are to go to zero, this conserved number must be zero already. For strengths 1, 1 and $-1/2$, $L = r_{12}^2 - \tfrac12(r_{23}^2+r_{13}^2)$, and a triangle with $r_{12}=1$ needs $r_{23}^2+r_{13}^2=2$. That is one condition on the shape, and it leaves a family: sides 1, 0.75 and 1.2 satisfy it, and so do many others. The equilateral triangle is in the family too, but it is a stationary shape that only rotates, so it is the one member that goes nowhere.

So nothing stops it. Three invariants, three exemptions. What remains is to see what actually happens, and the marching result is unusually simple.

The squared sides fall in straight lines #

The shape stays fixed. The triangle turns about the centre of strength and shrinks without changing its proportions, and because every velocity scales as one over the size and each side’s rate of change is velocity times side, the rate of change of a squared side does not depend on size at all. It is a constant throughout the motion.

That makes the meeting a straight-line extrapolation. Read the first side’s rate of change from the initial velocities, $d(r_{12}^2)/dt = -0.1296$, divide into the initial value of one, and the meeting is at $t=7.7189$. Then march the equations with a step that shrinks as the smallest side squared, the motion’s only time scale, and the size reaches a ten-thousandth of its starting value at that same moment, to eight figures. The check is a rather pretty one: the march is entirely unaware of the shortcut, and it arrives on the clock.

Nothing about this needs the triangle to be small. There is no asymptotic regime, and the first step of the motion is already a piece of the singular solution. The shape chooses the time. A triangle only a little off equilateral has velocities that almost exactly turn it and that hardly shrink it: 17.0 at a longest side of 1.1, and 35.3 at 1.05. Further from equilateral, the meeting comes sooner, and reaches about 4.56 as the longest side nears 1.3. The same mirror-image triangle does the reverse: reflected, it expands from a point rather than shrinks towards one. That is simply the equations’ time reversibility, the same one reversible and unusable tested, and it means that the same strengths can appear as an explosion out of nothing at some earlier moment, if the mirror triangle is what is given.

Which strengths can do this #

The condition $\sum 1/\Gamma_i = 0$ rules out most of what might be tried, and what it rules out has a plain meaning. Three vortices of one sign can never meet: all their reciprocals are positive and cannot sum to zero. More than that, each of their products is positive, so the energy’s logarithms all pull the same way, and the energy would have to diverge. Like-signed vortices are kept apart by the energy absolutely, however many of them there are. That is why a cluster of same-signed vortices can spread and merge in a real flow, but as points cannot concentrate.

So the set needs mixed signs, and it needs a particular ratio. With two strengths $a$ and $b$ of one sign, the third must be

$$\Gamma_3 = -\frac{ab}{a+b},$$

opposite in sign and smaller than either. That is the formula for two resistors in parallel, and for the reduced mass of two bodies, which is not a coincidence of notation: it is the same reciprocal sum. A pair of strengths 1 and 1 needs $-1/2$. A pair of 1 and 3 needs $-3/4$. A pair of equal and opposite vortices, $1$ and $-1$, can never be joined by any third, because its reciprocals already sum to zero with the third’s reciprocal left over. That is also why no vortex dipole, however it is approached, can be drawn into a collapse. Three is also the smallest number for which this can happen at all: two vortices need $1/\Gamma_1+1/\Gamma_2=0$, which is the dipole, and the dipole’s distance is fixed by its own two invariants, so it travels for ever instead.

An infinite number of turns in a finite time #

Each vortex’s speed is of order one over the size, and the size goes as the square root of the time left, so the speed goes as one over the square root of the time left. That is an integrable singularity. The distance each vortex travels before the meeting, $\int dt/\sqrt{t_c-t}$, is finite, and so each path has a stated length.

But the angular speed goes as one over the square of the size, which is one over the time left. And $\int dt/(t_c-t)$ is a logarithm, and it diverges.

So the triangle makes infinitely many turns in a finite time along paths of finite length, which is the logarithmic spiral. Each tenfold reduction of the time left adds 0.73 turns. Going from a thousandth of the time left to a ten-thousandth takes exactly as many turns as going from a hundredth to a thousandth. The orientation of the triangle at the meeting has no limit, and so the solution cannot be continued past that point. It is not merely that it blows up. Even if the three vortices were to re-emerge, the direction in which they re-emerge has no defined value. Something from outside the equations would have to choose. That is the point vortices’ version of a sheet that cannot stay a sheet, where the solution also stops at a stated time and there too the mathematics simply declines to say what comes next.

One part in a hundred #

Now take that exact triangle and change the third strength a little. The positions are held; only the strength moves, from $-1/2$ to $-1/2+\delta$.

The motion starts almost exactly as before. Then, around the time the collapse would be finishing, the triangle stops shrinking, distorts, and comes back out, having turned round the centre a few times. The reason is that both exemptions have gone, and they fail together. The sum of products is now $1 + 2(-1/2+\delta) = 2\delta$, so the energy can see the size. And the second moment, instead of vanishing, is $L = 2\delta$ on the same starting positions, so a triangle cannot shrink to nothing while keeping any shape on which $L/r^2$ is finite. The conserved quantity forces $r^2 \approx 2\delta/f$, where $f$ is a function of the shape alone, and $f$ can be small but the triangle has to change its shape to make it so. Hold the shape and the smallest size would go as $\sqrt{\delta}$.

The slopes come out at 0.42 and 0.35, somewhat below that square root because the shape does adapt in the close approach and lets the vortices closer than a fixed shape would. So a thousandth of a strength still stops the triangle at a tenth of its size. A tenth of a strength barely shrinks it at all. There is a good case for reading the numbers the other way round: the exponents mean that getting within a hundredth of the collapse would take the strength right to something like one part in a hundred thousand. Physical vortices with strengths measured to three figures would never see it.

That is what makes the collapse a statement about a set of measure zero. The Hamiltonian system has a three-dimensional relative phase space and collapse occupies a thread of it. Almost every three-vortex motion is quasi-periodic, as the earlier essay said, and collapse is not a contradiction of that; it is the case where the torus has been squeezed into a line that reaches the origin. Nearby tori pass close and are thrown out, and the size a neighbour reaches is what the figure measures. That near-miss structure is also the nearest thing to the singularity that a real flow could visit, and how a vortex pair carries eddies is what its fluid is doing meanwhile. For a triad among many vortices, encounters of this kind are how close pairs form, and a family of them is what makes the four-vortex problem chaotic and not just complicated.

Why this is not a catastrophe for the equations #

A singularity of an inviscid equation is usually a serious matter: the possibility of finite-time blow-up in three-dimensional Euler is an open problem of the first rank. Yet here is one in two dimensions, with a formula. The resolution is that it is not a singularity of the same equation.

Two-dimensional Euler with smooth vorticity carries each parcel’s vorticity along unchanged. The peak value of vorticity can never rise, since nothing can raise it. That is enough to prove, as Wolibner and Hölder did in 1933, that a smooth solution stays smooth for all time. Yudovich extended it in 1963 to any bounded vorticity, including the vortex patches of the shape a vortex keeps. No smooth blobs of vorticity, however arranged, can do what these points do.

A point vortex has already abandoned that bound. Its vorticity is infinite from the first moment, and all that “the equations” can mean for such an object is the rule that each is carried by the others, with its own contribution discarded. What a point vortex is not spelled out what that costs. So the blow-up is not in Euler; it is in the point-vortex system, a different dynamical system that agrees with Euler while the vortices are far apart. When the triangle’s size reaches any finite core size, three things happen that the points know nothing about. The cores overlap and deform each other, and fine filaments are thrown off. The opposite-signed vortex sits in a straining field that can tear it. And with a viscosity added, the cores merge and cancel by an amount that does not depend, in the limit, on how small that viscosity is taken. The last of these is an active research field: a finite amount of a vortex system’s small-scale enstrophy is destroyed however small the viscosity is, because the collapse delivers the scales to meet it.

The surprising connection is therefore with the cascade in two dimensions. Two-dimensional turbulence decays as a dilute gas of long-lived vortices, and encounters among mixed-sign neighbours in such a gas have been proposed as one route by which enstrophy reaches the viscous scale without any cascade to carry it. That part of the story cannot be computed from the three-point model; the three-point model says only that such an encounter has no stopping size of its own, so the core size decides.

What the picture cannot show #

The core. The picture shows points. Real vortices have cores, and the collapse must fail before the size reaches a core radius. Nothing in the figures can say what happens beyond that, or how much a finite core shifts the time. For a small core the time should be nearly the same, because most of the time is spent at large size, as a straight line in the square of the size has to: half of the remaining time is at sizes above 70% of the current one. The last doublings of a shrinking triangle take almost no time at all.

The fluid. A point-vortex computation tracks the vortices and has no field at all. Its picture of three spirals shows paths, and the surrounding fluid, also spiralling, is not drawn. The fluid in the triangle at the start is mostly not there at the end; it has been carried out between the vortices as they close, because the fluid is incompressible and three points have no volume.

Boundaries. A wall makes an image for each vortex, breaking the second moment’s conservation. Collapse in a bounded domain is a different question, and the exactness of the strengths’ arithmetic is gone in any case.

The convention #

Strengths are circulations in units of the first vortex’s; lengths are in units of the first side; time is in units of the side squared over the circulation, with the usual $2\pi$ in the velocity $\Gamma/2\pi r$. The triangle’s size is the root mean square of its three sides. “Off by $\delta$” changes the third strength with the positions held, so it changes both invariants at once; it is a statement about a near-miss, not a perturbation that keeps $L=0$. A step proportional to the smallest side squared is what lets the march continue through six decades of size at a small fraction of a turn per step.

How it was checked #

The second invariant, the energy and the shape’s ratio are each conserved through the collapse to round-off, or close to it, even though the velocities grow by a factor of ten thousand in the march. The time agrees to eight figures with the rate on the first step. The model refuses anything but three vortices, and refuses a triangle that no choice of its third side can make satisfy the invariant, which a longest side above $\sqrt2$ does.

Who found it, and when #

Walter Gröbli’s thesis of 1877 integrated the three-vortex problem and found the collapse. His thesis went unread for most of a century, and Synge found the condition again in 1949, as did Novikov in 1975. Aref’s work of 1979 mapped the three-vortex phase space and made the result standard; Tavantzis and Ting gave the systematic classification in 1988, and the account by Aref, Rott and Thomann of 1992, of Gröbli’s solution and his forgotten life, is the place to start.

Still open: what a core costs the meeting #

Every result here is for points, and the question that would make this useful to a real flow is what a finite core takes away. The next calculation would give each vortex a small core and a strength that it keeps, either as a contour-bounded patch or as a Gaussian with a fixed small viscosity, and march them from the same triangle. It would ask how small the triangle actually gets, how the closest size depends on the core radius, whether the third vortex tears before the first two reach it, and how much of the enstrophy goes to the viscous scale in that one encounter as the viscosity falls: whether the loss tends to a limit as the viscosity tends to zero, which would make one encounter a small dissipative anomaly in a flow with no direct cascade.