A ring moves because it is bent

A smoke ring crosses a room. A straight line vortex, of the same circulation and in the same fluid, sits exactly where it was put and never moves at all.

The difference between them is curvature, and the whole of this essay is that word made quantitative: what the curvature does, how fast it does it, and why the answer contains a logarithm that nobody has ever been able to remove.

The mechanism is a cross product #

The velocity a filament induces at a point is the Biot–Savart integral, borrowed unchanged from magnetostatics because the two fields obey the same equations:

$$\mathbf{u}(\mathbf{x}) = \frac{\Gamma}{4\pi}\oint \frac{\mathrm{d}\boldsymbol{\ell}\times(\mathbf{x}-\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|^3}$$

Now evaluate it on the filament itself. For a straight filament along $z$, the element $\mathrm{d}\boldsymbol{\ell}$ points along $z$ and so does $\mathbf{x}-\mathbf{x}'$, because both points are on the same line. The cross product of two parallel vectors is zero. Every term is zero, so the sum is zero, and no amount of length or proximity changes it.

For a ring, the elements on the far side of the circle are not parallel to the line joining them to the point of interest, and their contributions do not cancel. Reducing the integral for a circular filament of radius $R$, evaluated at a point on itself, gives

$$u_z = \frac{\Gamma}{4\pi}\int \frac{R^2(1-\cos\varphi)}{\left(2R\sin\tfrac{\varphi}{2}\right)^3}\,\mathrm{d}\varphi$$

and every element of that integrand is positive: the whole ring pushes the point forward.

The integral does not converge, and that is the result #

The integrand behaves as $1/\varphi$ near $\varphi = 0$, so the integral diverges logarithmically as the neighbouring elements are approached. A filament of zero thickness moves infinitely fast.

That is not a defect of the method; it is a statement about the model. A filament is a line of infinite vorticity, and a real vortex has a core of finite size in which the vorticity is finite and spread out. Removing an arc of length $\delta$ centred on the evaluation point — a cutoff, standing in for the core — gives a finite answer, and doing the integral properly gives

$$u_{\text{ring}} = \frac{\Gamma}{4\pi R}\ln\frac{8R}{\delta}$$

So every number anybody has ever quoted for the speed of a vortex ring contains a decision about the core, and the honest way to present it is as a curve against the cutoff rather than as a number.

What the solver computed, and how it was checked #

Four things, and the third is the one that makes the essay’s claim testable.

The quadrature lands on the closed form. The integral is computed in $\ln\varphi$ rather than in $\varphi$, because the integrand goes as $1/\varphi$ at the cutoff end and a uniform mesh would put almost no points where almost all of the integral is — the same trap, and the same repair, as the rotor integral in the applied field. The result agrees with $(\Gamma/4\pi R)\ln(8R/\delta)$ to eight significant figures.

The two routes agree. The segment sum and the quadrature are different code with different approximations, and they land within a part in a thousand at a cutoff of a fiftieth of the radius. It was worth doing because the first attempt did not agree, by a factor whose logarithm was exactly $\ln 2$: the quadrature’s $\delta$ is the total arc removed and the segment sum’s cutoff is a radius about the point, so the same hole is $\delta/2$ there. A twelve per cent disagreement reads as a discretisation problem and was a definition.

The straight filament comes out at zero through the same code. Magnitude below $10^{-14}$, and the same integrator gives the ring $0.42\,\Gamma/R$ at the same cutoff — a check that is included because an integrator returning zero for everything would pass the first half of it.

The slope survives the core model. Fitting the speed against $\ln(1/\delta)$ over four decades gives $0.0795775$ against $\Gamma/4\pi R = 0.0795775$, to a part in ten million. That is the assertion the family rests on: the value depends on a modelling decision and the slope does not, so the slope is what may be asserted.

Kelvin’s quarter #

The standard formula for a ring with a solid-body core of radius $a$ is Kelvin’s:

$$U = \frac{\Gamma}{4\pi R}\left(\ln\frac{8R}{a} - \frac14\right)$$

and the difference between it and the bare cutoff result is that $-\tfrac14$, which is the only place the internal structure of the core appears. A hollow core gives $-\tfrac12$ instead; a core with a different vorticity distribution gives something else again. The logarithm swamps it — at $a/R = 0.05$ the logarithm is $5.08$ and the quarter is five per cent of it — which is why the formula is useful and why nobody should quote a ring speed to three figures without saying what the core was assumed to be.

This is the standing shape of the subject: the leading behaviour is a theorem and the constant is a model. The site draws Kelvin’s line on the figure and asserts the slope.

Faster as it shrinks, and the invariant that makes it so #

The speed goes as $\Gamma/R$, so a ring that shrinks speeds up. That single fact produces the most-filmed phenomenon in vortex dynamics.

Two coaxial rings, one behind the other, each sit in the other’s field. The rear one is drawn inwards by the front one’s induced flow, so it narrows and accelerates; the front one is pushed outwards, widens and slows; the rear passes through the front; and the roles exchange. They do it repeatedly, and the exchange is conserved by something computable.

The hydrodynamic impulse of a coaxial ring system, $\sum \Gamma \pi R^2$, cannot change: it is the momentum the vortex system imparts to the fluid, and there is nothing to change it. So one ring’s radius can only grow at the expense of the other’s, which is exactly what the figure shows.

What travels with the ring #

A ring in still fluid carries a closed body of fluid along with it, and this is the difference between a vortex ring and a jet.

That closed region is why a smoke ring keeps its smoke. In a jet the fluid streams through and disperses; in a ring the marked fluid is trapped, and travels with the vortex until viscosity and instability let it go.

A ring is a parcel of momentum with no jet attached #

The impulse that keeps appearing in the leapfrog has a physical reading worth making explicit.

$\rho\Gamma\pi R^2$ is the momentum the ring has given the fluid, and it is the momentum anything would have to absorb to stop it. A ring can therefore carry momentum across a room without carrying any mass across it — the fluid inside the bubble travels along, but it is the same fluid the whole way, and no stream of air crosses from one end of the room to the other.

That is a genuinely odd object, and it is what makes the ring the cleanest demonstration of the momentum theorem available. Draw a control volume round the ring at any instant and the momentum flux through its faces is exactly $\rho\Gamma\pi R^2$; draw it a second later, somewhere else entirely, and the answer is the same number, because nothing has dissipated. The momentum was delivered to the fluid once, when the ring was made, and it has been travelling ever since.

The connection to the disc that produces thrust is direct: a puff from a nozzle is an impulsive version of the same accounting, and the ring is what the vorticity in that impulse rolls up into. A jellyfish and a squid propel themselves by making rings, and the momentum they gain is this number.

A vortex line cannot end in the fluid #

Helmholtz’s second law says a vortex tube must either close on itself, end on a boundary, or go to infinity — it cannot simply stop. It is one of the few statements in the exact theory that survives into real flows unchanged, because it is a consequence of a vector identity rather than of any assumption about the fluid. The reason is the same as for a magnetic field line: the vorticity field is the curl of something and therefore has zero divergence, so its tubes have no ends.

That is a strong constraint and it is why rings are so common. A finite blob of vorticity created in open fluid — by a puff from a nozzle, a paddle, a collapsing bubble, a starting wing — has nowhere for its vortex lines to terminate, so they close, and what results is a ring.

The same law is what forces the wing’s horseshoe: the bound vortex on a wing cannot end at the tips, so it turns downstream as the trailing pair, and the far end is closed by the starting vortex left on the runway. An aircraft’s wake is a single vortex loop, thousands of times longer than it is wide, and it obeys Helmholtz’s law exactly.

Numbers, for something a reader can picture #

A smoke ring from a mouth is about $R = 4$ cm across the core centres with a core of perhaps $a = 6$ mm, and it crosses a room at something like half a metre a second. Reading the formula backwards, that requires a circulation of

$$\Gamma = \frac{4\pi R U}{\ln(8R/a) - \tfrac14} = \frac{4\pi (0.04)(0.5)}{\ln(53.3) - 0.25} \approx 0.068\ \mathrm{m^2/s}$$

which is a swirl velocity of about $1.8$ m/s at the edge of the core — brisk, and entirely plausible for air pushed out of a mouth. The ratio $\Gamma/\nu$ is then about $4500$, so the ring is firmly in the regime where viscosity does not dominate but is not negligible either, which is why a smoke ring lasts seconds rather than minutes.

The same arithmetic run on a ring of $R = 1$ m with the same core ratio gives a speed twenty-five times smaller for the same circulation, because the speed goes as $1/R$: large rings are slow. That is the same $1/R$ that drives the leapfrog, and it means a ring cannot be scaled up without becoming sluggish — a fact that governs everything from the design of a pulse-jet thruster to why the vortex shed by a jumbo’s wingtip descends at a metre a second rather than at ten.

What the picture cannot show #

A real core is not a line and does not stay the same size. Viscosity spreads it as $\sqrt{\nu t}$, so a ring slows as it ages, and the model here has no mechanism for that at all. The rings in the leapfrog figure would do it forever, and real ones manage two or three exchanges before the cores overlap and merge.

Rings are unstable to azimuthal waves. A ring supports bending waves round its own circumference, and in a strained environment some of them grow — which is why a real smoke ring eventually develops a wavy shape and then breaks up. Nothing in a thin-filament model with a fixed circular shape can show it.

A real ring entrains fluid. The bubble in the figure is a fixed body of fluid; a real one exchanges fluid with its surroundings continuously across a shear layer, gaining mass and losing sharpness. That is what makes a ring visible for as long as it is and what eventually stops it being a ring.

Where the model stops #

Everything here is inviscid and everything here is thin-core. The filament model is at its best when $a/R$ is small, which for a well-formed smoke ring is about $0.1$ — and the logarithm of the core ratio is then only about four, so the “small parameter” is not very small and the corrections are larger than they look.

The model also cannot compute its own core. Where the core radius comes from — how a puff of fluid rolls up into a ring of a particular thickness — is a viscous and unsteady problem that requires a solver this site does not have.

And the speed formula assumes the ring is circular and planar. A ring passing near a wall, or through a shear, deforms, and then the self-induction has to be computed along a curve of varying curvature, which is the general filament problem: local induction plus everything else, and the “everything else” is a divergent integral again.

The general curve, and the equation it turns out to obey #

The essay closes on the general filament problem — a curve of varying curvature, whose self-induction is a divergent integral again — and it is worth going one step further, because the divergence regularised produces one of the tidiest results in the subject.

Look at where the divergence comes from. The integrand blows up only near the evaluation point, so the divergent part of the self-induced velocity is entirely local: it depends on the shape of the filament in a small neighbourhood, which to leading order means its curvature. Carrying that through gives the localised induction approximation,

$$\mathbf{u}_{\text{self}} \approx \frac{\Gamma}{4\pi}\ln\!\left(\frac{L}{a}\right)\,\kappa\,\hat{\mathbf{b}},$$

with $\kappa$ the local curvature and $\hat{\mathbf{b}}$ the binormal. A filament moves perpendicular to both itself and its own bending, at a rate proportional to how sharply it is bent. The ring’s $1/R$ falls straight out of it, since a circle’s curvature is $1/R$ and its binormal is its axis, and so does the straight filament’s zero.

What is remarkable is what happens when that rule is written as an evolution equation for the curve. Hasimoto showed in 1972 that combining the curvature and the torsion into a single complex function turns the filament equation into the cubic nonlinear Schrödinger equation — one of the small handful of integrable nonlinear equations, with an exact solution method and an infinite family of conserved quantities.

The consequence is that a vortex filament supports solitons. A localised loop of bending can travel along an otherwise straight vortex at constant speed without changing its shape, and two of them can pass through one another and emerge unaltered but for a shift in position. That is a startling thing to find in a fluid, it was derived from the same divergent integral this essay had to cut off, and such waves have been seen — on vortices in rotating tanks, and as Kelvin waves on the quantised vortices of superfluid helium, where the core radius is an atomic length and the logarithm is genuinely large.

The approximation’s limits are as instructive as its successes, and they follow from what was thrown away. Keeping only the local part means keeping only the leading term in $\ln(R/a)$, so the accuracy is logarithmic — poor, and improving very slowly as the core is thinned. It preserves the filament’s length exactly, so it can never describe a vortex being stretched. And it contains no interaction between distant parts of the curve at all, so two filaments approaching each other feel nothing, and reconnection — the event that decides what a real tangle does — is invisible to it. The approximation that makes the problem integrable is the one that removes everything a real vortex spends its life doing.

Who found it, and when #

Helmholtz’s 1858 paper contains all of it: the vortex laws, the impossibility of a line ending in the fluid, the leapfrog, and the observation that a ring propels itself. Kelvin was so taken by the permanence of the vortex ring in an inviscid fluid that he proposed in 1867 that atoms were knotted vortex rings in the aether — a theory that was wrong, that produced the mathematical classification of knots, and that is the reason knot theory exists.

Kelvin’s speed formula with the quarter dates from 1867; Hicks and Lamb refined the core treatment, and Saffman’s 1970 work put the whole thing on a footing where the arbitrary cutoff could be related to a physically meaningful core structure. Maxworthy’s experiments in the 1970s measured the entrainment and the decay that the inviscid theory cannot supply.

Where the ladder goes next #

Vorticity has been followed from a point, through a pair, to a closed loop that moves under its own influence. What is still missing from every one of these pictures is pressure: the fields have been computed and the forces have not. The pressure field of an incompressible flow turns out to have no propagation speed at all, which is the next thing worth knowing — and is the assumption the compressible field is built to break.