Two eddies can stand behind a cylinder, but not for long

The mirror that is a circle builds the image of a point vortex in a circular cylinder: an opposite vortex inside at the inverse point, and a compensating one at the centre, so that the cylinder’s surface is a streamline whatever the vortex outside does. It leaves the vortex drifting round the cylinder, pushed by its own images.

Add a uniform stream past the cylinder, and a second, opposite vortex below the axis, and a new question appears. Behind a real cylinder at a Reynolds number of a few tens there are two eddies, one either side of the axis, standing still in the flow: the stream coming round the cylinder is balanced against the eddies’ own tendency to move. Can ideal flow — no viscosity, no boundary layer, no wake — hold such a pair still? Ludwig Föppl answered in 1913 that it can, and that it can do so at any distance behind the cylinder.

A pair held still by its images and the stream #

Each vortex is moved by four things: the stream, deflected round the cylinder; its own image inside the cylinder and the compensating one at the centre; its partner across the axis; and the partner’s images. The stream pushes it downstream. The partner, of opposite sign, pushes it upstream — a vortex pair moves perpendicular to the line joining it, and this pair’s line is across the stream. The images push it towards or away from the cylinder depending on where it is. For the vortex to stand still, the two components of its velocity must vanish together.

That is two conditions and, for a given position, one free number, the pair’s strength. So a pair cannot stand anywhere: the standing positions form a curve. Solved by Newton’s method from the conditions themselves, the curve comes out as

$$r - \frac1r = 2y,$$

with $r$ the vortex’s distance from the cylinder’s centre and $y$ its distance from the axis, both in cylinder radii, and the strength needed at each point as

$$\Gamma = 2\pi Ua\left(r - \frac1r\right)\left(1 - \frac1{r^4}\right).$$

These are Föppl’s closed forms. The Newton solution lands on both to three parts in $10^{16}$ at five positions from 1.2 to 5 radii behind the cylinder, and the vortex’s leftover velocity there is at the level of rounding.

A pair just behind the cylinder is weak and sits close to the axis; at 1.2 radii from the centre its strength is 1.34 in units of the stream speed times the radius. Two radii behind, it sits 0.86 either side and needs a strength of 10.3. At five radii it sits 2.77 out and needs 34.8. Every point of the curve is a different steady flow, and nothing in the equations prefers one.

The two ends of the curve #

The curve has two limits, and each says something about what holds the pair. Close behind the cylinder, as $r \to 1$, the standing position slides onto the axis and the strength falls to zero: an infinitesimally weak pair can sit just off the rear stagnation point, where the stream has almost stopped and there is almost nothing to balance. The angle the vortex makes with the axis, seen from the cylinder’s centre, grows from zero there to 23 degrees two radii behind and 29 degrees five radii behind.

Far behind, the curve straightens. The ratio $y/r$ tends to exactly one half — 0.496 at ten radii, 0.4998 at fifty — so the pair ends up on two lines at thirty degrees to the axis, and its strength tends to $2\pi U r$. That strength is the one at which a free pair of opposite vortices, a distance $2y = r$ apart, would drive itself upstream at the stream’s own speed: far from the cylinder the pair stands because it swims against the stream at exactly the stream’s speed, and the cylinder’s only remaining job, through the small velocities it induces there, is to fix the angle.

The body the stream sees #

The dividing streamline that leaves the cylinder’s rear and rejoins the axis behind the pair encloses the eddies, and to the stream outside it the cylinder and its eddies are a single, longer body. For the pair two radii behind, the rear of that body — the point where the flow along the axis stops and turns back — is 3.46 radii from the cylinder’s centre, so the eddies extend 1.23 diameters behind the cylinder. For the pair 1.5 radii behind, 0.63 diameters; for the pair three radii behind, 2.3.

Real standing eddies have lengths in the same range and grow with the Reynolds number, roughly in proportion to it: about one diameter at a Reynolds number near twenty, two near forty. So the ideal family contains a member of the right length for every Reynolds number at which real eddies stand, and says nothing about which member a given Reynolds number picks. That choice belongs to viscosity, which sets how much vorticity the shear layers deliver into the eddies and so how strong, and therefore where, they are.

Standing eddies cost nothing #

The pair changes the pressure round the cylinder completely: behind it the stream no longer recovers its pressure over the rear half, because the eddies occupy the space it would have flowed into. It is natural to expect a drag, and the usual informal account of bluff-body drag — low pressure in the eddies sucking the body backwards — predicts one.

Blasius’ theorem gives the force on a body in plane potential flow from an integral of the square of the complex velocity round its surface, which is computed here with four thousand points, a spectrally accurate rule for a periodic integrand. For a standing pair at 1.3, 2 and 4 radii behind, the drag and the lift are both zero to five parts in $10^{16}$. Standing eddies exert no drag. It is d’Alembert’s paradox, surviving the eddies.

The reason is a momentum balance. The whole flow is steady; the vortices, standing still, feel no force from the fluid, because a free vortex’s force is its strength times its velocity relative to the fluid, which here is zero; so the total force on body and vortices together is the momentum carried in from infinity, and a steady flow whose disturbance falls off like a pair of opposite vortices carries none. The only thing left to push on is the cylinder, and it is pushed on by nothing.

The integral can fail, and the figure shows it failing. Give the same pair a strength a fifth too weak or too strong, and it can only be kept in place by a force from outside; the cylinder then feels the reaction, a drag of order $\rho U^2 a$ with the sign of the imbalance. Drag in a steady inviscid flow needs something to be held against the stream. In a real flow that something is not held but thrown away: the vortices a real cylinder sheds are carried downstream, and the drag is the momentum they take with them. Föppl’s pair takes nothing anywhere. The contrast with the free-streamline flow past a plate is exact: there an ideal flow does have drag, because its wake is an open cavity reaching to infinity and the momentum deficit goes with it. Close the wake into a pair of standing eddies and the drag goes to zero.

Neutral one way, unstable the other #

An equilibrium is a claim that must survive being disturbed, and the pair’s symmetry splits its disturbances in two. A displacement that keeps the mirror symmetry — both vortices moved downstream together, or both moved away from the axis together — is one family; a displacement that breaks it — both moved across the stream in the same direction — is the other. By symmetry each is a two-by-two linear problem.

Symmetric displacements oscillate and do not grow: the linearised eigenvalues are purely imaginary, with a frequency of 0.45 in units of the stream speed over the radius at two radii behind. The pair, pushed symmetrically, rocks about its station. Antisymmetric displacements grow, everywhere on the curve: at a rate of 1.12 when the pair sits 1.3 radii from the centre, 0.40 at two radii, 0.16 at three and 0.054 at five. Whatever strength the standing pair has, and however far behind the cylinder it stands, a sideways push that treats its two members differently grows exponentially.

Leaving #

Followed beyond the linear stage, the displaced pair shows what the instability does. Moved a thousandth of a radius across the stream, the two vortices sit apparently still for about ten time units while the displacement grows at 0.40 per unit time, too small to see. By fifteen it is a tenth of a radius. Then the pair lifts away from the axis, the upper vortex rising faster than the lower, and the stream, faster further from the cylinder, carries the whole pair off downstream; by thirty time units it is five radii behind where it stood, still bound together as a travelling pair.

Two things in the result are worth separating. The instability is real and its mode is the right one: a displacement across the stream is exactly how a real wake’s symmetry first breaks, with one eddy growing at the other’s expense before shedding begins. What the point vortices then do is not what a real wake does. A real cylinder does not lose its eddies as a pair; the shear layers leaving its surface feed new vorticity into the eddies continuously, and the growing asymmetry turns into an alternating release of one eddy and then the other. The model has no source of new vorticity and can only let its fixed pair go.

What holds real eddies still #

Behind a real cylinder the standing eddies are stable from a Reynolds number of about six, where they first appear, to about forty-seven, where the wake starts to oscillate and the Kármán street begins. Föppl’s pair, the ideal version of the same arrangement, is unstable at every position. So something present in the real flow and absent from the ideal one must hold the eddies still below forty-seven, and the only candidate is viscosity: diffusion spreads the eddies’ vorticity so that they are not points, and the attached shear layers anchor them to the cylinder’s surface so that they cannot drift freely.

As the Reynolds number rises both of those weaken, and the stabilisation fails at a definite value, which is where the antisymmetric mode found here takes over. The frequency the wake then chooses is set by more than this model contains — the Strouhal number of about 0.2 comes from the whole shear layer’s dynamics — but the growth rate of the antisymmetric mode, a few tenths of the stream speed over the radius, is the right size for the time a real wake takes to break its symmetry once it is unstable.

The standing pair is therefore a model that is wrong in the right way. It is stable in the way the real eddies are not — to symmetric pushes, which in a real wake are damped by viscosity — and unstable in the way they become, to the sideways push that starts shedding. That is why it has survived a century as the starting point for models of the near wake, and why every such model has had to add the one thing it lacks, a supply of vorticity from the body.

Why points are enough for this question #

Point vortices are a drastic idealisation, and what a point vortex is not lists what they leave out: a core with a size, a velocity that stays finite, a shape that can deform. The question here — whether the pair can stand, and which displacements it survives — depends on them less than it might, for a reason that can be stated. The standing condition and the linear stability both involve only the velocity each vortex receives from everything else: the stream, the cylinder’s images and the partner. A vortex’s own core induces no net velocity on its own centre if the core is symmetric, so a finite core changes the answer only through how it deforms in the strain of its surroundings, which is a correction of the order of the core’s area over the square of the distance to the nearest other vortex or image.

For a pair two radii behind a cylinder, whose nearest images sit inside the cylinder about a radius and a half away, a real eddy’s size is not small by that measure, and the model’s numbers are estimates rather than predictions. What survives the idealisation is the structure: a one-parameter family of standing states, neutral to symmetric disturbances and unstable to antisymmetric ones, with rates of the order of the stream speed over the radius. Finite-core versions of Föppl’s pair computed since have kept that structure and moved the numbers.

What the picture cannot show #

Point vortices. The eddies are concentrated in points, with no size. A real eddy is spread over a region comparable with the cylinder, and its induced velocity near its own centre is nothing like a point vortex’s.

No source of vorticity. The pair’s strength is fixed. The separating shear layers that make and feed a real wake’s eddies are absent, which is why the model can say how the symmetry breaks and not what follows.

Two dimensions and no viscosity. Everything is plane and inviscid; the real standing eddies’ stability below a Reynolds number of forty-seven is a viscous effect the model cannot contain.

The convention the numbers depend on #

Lengths are in cylinder radii and speeds in the stream speed, so time is in radii per stream speed. The upper vortex rotates clockwise and the lower anticlockwise, as the vorticity of the flow separating from the top and bottom of a cylinder does; strengths are quoted as positive magnitudes in units of stream speed times radius. Positions are measured from the cylinder’s centre, $x$ downstream and $y$ across. Drag is the force on the cylinder along the stream, in units of $\rho U^2 a$ per unit span.

Who found it, and when #

Ludwig Föppl published the standing pair in 1913. Its neutral response to symmetric displacements and its instability to antisymmetric ones were established by later analyses, and were settled in the second half of the twentieth century as part of the effort to understand the onset of vortex shedding. The circle theorem that gives the images is Milne-Thomson’s of 1940, written down in general long after Föppl had used its special case, and Blasius gave his force theorem in 1910.

The structure — a steady flow, found exactly, and then shown to be unstable to the one disturbance that matters — is the same as three vortices being the most that can be predicted, where an exact solution exists and cannot be reached.

Still open: a pair that is fed #

The model’s missing piece is identifiable, and adding it is the next calculation. Let each vortex’s strength grow at the rate a separating shear layer would feed it, from a separation point on the cylinder, with the vorticity travelling from the surface to the vortex along a cut in the plane — the construction known since the 1950s as the Brown–Michael model. The fed pair can then grow asymmetric and release one vortex at a time, and the question is whether it does so at a Strouhal number near the measured 0.2, and at what feeding rate — the stand-in for the Reynolds number — its symmetric state first loses stability. That is the point-vortex version of a wake told what to do, and it would say how much of vortex shedding a model with two moving points and two feeding sheets can reproduce.