The sign a stagnation point carries in space

The count a pattern cannot break gave every stagnation point of a plane flow an index, +1 for a centre and −1 for a saddle, and showed that the indices inside a closed curve add up to the number of times the velocity turns as the curve is walked. The count computed on a body ran the same rule on the skin-friction lines of a closed surface, where the total is the surface’s Euler characteristic. And Two kinds is a plane flow’s privilege moved the classification into space, found four kinds of point where the plane allows two, and ended by declining the one calculation that would connect it back to the counting: the index of a critical point in three dimensions.

This essay makes that calculation. The answer is shorter than the classification it completes. A three-dimensional critical point’s index is the sign of a single determinant, that determinant is already one of the two numbers the classification plots, and so the diagram that sorts points into four kinds sorts them into two indices by halving itself down the middle. The rest of the essay is what that buys on a flow where the answer can be computed exactly — including a pair of results about that flow that neither the index nor the classification would have suggested alone.

An index is how many times the direction wraps a sphere #

In the plane the index of an isolated zero of a vector field is the number of turns its direction makes round a small circle about the point. In space the circle becomes a sphere and a turn becomes a covering: take a small sphere round the point, map each point of it to the direction of the velocity there, and count how many times, with sign, the image covers the unit sphere. For a zero where the velocity gradient is non-singular that count is the sign of the gradient’s determinant, +1 or −1, and nothing else is possible.

Two conventions fix the signs. The gradient is $J_{ij} = \partial u_i/\partial x_j$, and the invariants are those of Chong, Perry and Cantwell used in the classification essay: $Q = -\tfrac12\operatorname{tr}J^2$ and $R = -\det J$. With those, a flow that is incompressible has a trace-free gradient, its eigenvalues sum to zero, and

$$\text{index} = \operatorname{sign}\det J = -\operatorname{sign} R.$$

There is a more physical way to say the same thing. The determinant is the product of the eigenvalues; complex pairs contribute a positive factor; so its sign is fixed by how many real eigenvalues are negative. For a trace-free gradient in three dimensions that leaves two cases. A point with one outgoing direction and two incoming ones has index +1; a point with two outgoing and one incoming has index −1. The index is the parity of the dimension of the point’s stable manifold, and in an incompressible flow that dimension is one or two.

The left half and the right half are the two indices #

The four kinds of point the classification found are a focus being stretched, a focus being squeezed, and a node with two saddle directions in each of the same two flavours. The stretching kinds, whose single odd direction points outward, have a positive determinant and negative R and sit on the left of the diagram; the compressing kinds sit on the right. So the four kinds are two of each index, and the division that decides the index is not the discriminant curve the classification essay was about but the vertical axis through its cusp.

That has an immediate consequence which is easy to state wrongly. Crossing the discriminant curve turns a node into a focus and leaves the index where it was, because the determinant does not pass through zero there. Only crossing the line R = 0 can change an index, and on that line the determinant vanishes, the zero is degenerate, and it is not isolated in the generic sense. An index can change only where a critical point is being created or destroyed, and the diagram shows exactly where that can happen.

The plane result sits on that line too, and it is worth saying why it is a different count. A plane flow embedded in space has an identically zero third row, so its determinant is zero and R = 0 — but that is not a degenerate three-dimensional zero, it is not an isolated zero at all. A plane saddle is a whole line of stagnation points along the ignored direction. The plane’s ±1 counts turns of a circle in that plane; the space index counts coverings of a sphere; and a flow that is truly three-dimensional is needed before the second means anything.

A flow with no outside, and its eight points #

The count needs a flow whose stagnation points can be found exactly. The ABC flow is the natural one: $u = (A\sin z + C\cos y,\; B\sin x + A\cos z,\; C\sin y + B\cos x)$ on a box of side 2π that repeats in every direction. It is an exact steady solution of Euler’s equations, it is the flow whose streamlines Steady, three-dimensional, and mixing anyway showed wandering chaotically through a volume, and it is a Beltrami flow — its vorticity equals its velocity — which is the property behind The knot a flow cannot untie.

The repeating box is what makes the count sharp. A periodic box has no outside: each face is the same face as the one opposite, traversed the other way, so the degree of the velocity’s direction over the box’s boundary cancels face against face. The indices of all the stagnation points inside therefore sum to zero, which is the Euler characteristic of the three-dimensional torus the box really is. Whatever stagnation points the flow has, they come in equal numbers of each sign.

At A = B = C = 1 there are eight, and they can be written down. The zero conditions force $x = \pi/4 + k\pi/2$, then two values of $y$ for each, then one value of $z$. The four with index +1 are at $(\tfrac14, \tfrac54, \tfrac34)\pi$, $(\tfrac34, \tfrac14, \tfrac54)\pi$, $(\tfrac54, \tfrac34, \tfrac14)\pi$ and $(\tfrac74, \tfrac74, \tfrac74)\pi$; the four with index −1 at $(\tfrac14, \tfrac74, \tfrac54)\pi$, $(\tfrac34, \tfrac34, \tfrac34)\pi$, $(\tfrac54, \tfrac14, \tfrac74)\pi$ and $(\tfrac74, \tfrac54, \tfrac14)\pi$. Every one has Q = −1.5; the first four have R = −0.7071 and the second four R = +0.7071. A Newton search from 512 seeds scattered through the box finds these eight and no others, and each one agrees with the closed form to 10⁻⁸.

The index counted twice, and the sum counted a third way #

Reading an index off a determinant is algebra about one tensor. The definition is a count of how a direction wraps a sphere, and the two are equal by a theorem rather than by construction, so the count is worth making directly.

For each point, a sphere of radius 10⁻³ is cut into 4,608 quadrilaterals, each split into two triangles, and the velocity’s direction at the corners maps each triangle to a spherical triangle on the unit sphere. Adding up the signed solid angles of all 9,216 and dividing by 4π gives the degree. It comes out +1.0000000000000002 at the first point and −0.9999999999999986 at the second, and matches the sign of the determinant at all eight. The mesh is fine enough that neighbouring directions never differ by more than 10.5°, far inside the right angle beyond which a triangle’s orientation stops being unambiguous.

The sum can be checked a third way, over boxes. A box from 0.05 to 3.2 in each direction contains one stagnation point, of index −1, and the degree of the direction over its six faces is −1.000000; a box containing two points of opposite index gives 0.000000; a third containing a single point of index −1 gives −1.000000. The boundary of a whole periodic cell gives 3 × 10⁻¹⁵. The index theorem, which the plane essays used as a rule, is here a measurement that can fail, and it does not.

The points exist only where the squares could be a triangle #

The count says how many of each sign there must be. It does not say whether there are any. Squaring the three zero conditions and adding them in pairs gives a linear system for $\sin^2 x$, $\sin^2 y$ and $\sin^2 z$, and those have to lie between zero and one. Working that through, a solution exists only when each of $A^2$, $B^2$ and $C^2$ is no larger than the sum of the other two — when the three squares could be the sides of a triangle.

The search agrees with the inequality at all 256 lattice sites, and wherever there are points there are exactly eight. Two marks on the map are worth reading. The line up the middle is the sweep the next sections follow, A = B = 1 with C rising, and it leaves the region at $C = \sqrt{2}$, where $C^2 = A^2 + B^2$. The green point is the ABC flow that is usually used to show that these streamlines are chaotic — $A = \sqrt{3}$, $B = \sqrt{2}$, $C = 1$, the values the chaos essay advected — and it lies exactly on the circle $B^2 + C^2 = A^2$. With A reduced by one part in ten thousand it has eight stagnation points whose determinants are all already smaller than 0.1; with A increased by the same amount it has none. The textbook example of a chaotic steady flow sits precisely on the boundary where its stagnation points are born, with each of its eight points merged into a partner of the opposite sign.

A Beltrami flow’s stagnation points are pure strains #

The classification essay’s most interesting kind of point was the stretched focus, a vortex pulled out along its own axis — the structure it called the one turbulence is made from. The ABC flow is steady, exact and chaotic, and it has none.

The reason is one line. In a Beltrami flow the vorticity is the velocity, so at a stagnation point the vorticity is zero, and a velocity gradient with no vorticity is symmetric. A symmetric matrix has real eigenvalues, so the point is a node with saddle directions and can never be a focus: the discriminant is never positive. Across 224 stagnation points sampled over the existence region the largest discriminant is 4 × 10⁻¹⁶ and the largest antisymmetric part of any gradient 10⁻¹². The same argument gives Q in closed form, $Q = -(A^2\cos^2 z + B^2\cos^2 x + C^2\cos^2 y)$, which is never positive — so the Q-criterion that What a parcel does first’s decomposition underlies would never mark one of these points as a vortex, correctly.

That is the surprising connection this flow offers. The chaos of its streamlines lives in the regions between its stagnation points, where the vorticity is large and aligned with the velocity; the points themselves are where the vorticity is exactly absent, and there the flow is pure strain. A flow can mix chaotically with every one of its stagnation points being the least rotational kind of point there is, and Inviscid does not mean irrotational is the reminder that the vorticity elsewhere in it is what the Euler equations are carrying.

At C = 1 the index +1 point’s rates are 1.414, −0.707 and −0.707: two coincide, which is axisymmetric strain, and a repeated eigenvalue is exactly the condition for sitting on the discriminant curve. That is why all eight points in the first figure lie on the curve rather than below it. The index −1 point has the same rates with the signs reversed. Along the sweep the repeated rate splits again on either side of C = 1, and as $C$ approaches $\sqrt{2}$ the middle rate of each point runs to zero — 0.0038 at $\sqrt{2} - 10^{-5}$ — which is the determinant vanishing.

Pairs meet on R = 0, the one line an index can change on #

Following each point as C changes makes the rule about R = 0 visible.

At C = 0.3 the index +1 points sit at R = −0.088, Q = −1.045. As C rises they move out to the discriminant curve, touch it at C = 1 at R = −0.707, Q = −1.5, and turn back without entering the lobe, as a symmetric gradient must. As C approaches $\sqrt{2}$ they run in towards R = 0 at Q = −2, and the index −1 points arrive there from the other side. The sign of R does not change on any of the eight paths at any of the 113 values of C followed — the index is constant along each track — and at $C = \sqrt{2} - 10^{-4}$ each point of index +1 is within 0.05 of exactly one point of index −1. They meet in four pairs, on the planes x = 0 and x = π, at z = π/2 or 3π/2.

This is the statement that an index cannot change continuously, seen in motion. The eight points cannot leave the box one at a time, because each departure would change the total; they can only leave in pairs whose indices cancel, and a pair can only merge at a degenerate point, which is a point with R = 0. The paths are forced to meet on the one line in the diagram where that is allowed.

Where a pair meets, it closes as a square root #

The last figure asks how the pair closes.

Between $\sqrt{2} - 10^{-3}$ and $\sqrt{2} - 10^{-5}$ the magnitude of R falls with slope 0.500 and the gap between partners with slope 0.500: at the closest approach followed, $|R| \approx 2.378\sqrt{\sqrt2 - C}$ and the gap $\approx 3.364\sqrt{\sqrt2 - C}$. A square root is the signature of a fold, the generic way two zeros of a smooth field approach each other and disappear. The first essay on this subject drew a plane merger of the same shape, on a spinning cylinder, where two stagnation points on the surface meet and a single point leaves into the fluid carrying their combined index. Here the two indices are opposite, so the merged point carries nothing and nothing is left: a point with one outgoing direction and a point with two cancel, and the total index is unchanged at every value of the parameter, before, during and after.

What the picture cannot show #

Degenerate points. Everything here is about isolated zeros whose gradient is invertible, where the index is ±1. A degenerate zero can have index 0, or ±2, or more, and the flow at $C = \sqrt{2}$ exactly — or at $A = \sqrt{3}$, $B = \sqrt{2}$, $C = 1$ — has points of exactly that kind, which the Newton search finds only approximately and the determinant cannot classify. The figures stay a hair away from them and say so.

A body. The count on a periodic box is zero because the box has no outside. A flow round a body has one, and on a wall where the fluid does not slip the velocity is zero everywhere, so every point of the wall is a stagnation point and the index as defined here does not apply. The critical points of a separated flow are counted with half-saddles and half-nodes on the wall and full ones in the fluid, and that bookkeeping is a different theorem — the one the next section names.

Whose stagnation points they are. A stagnation point is where the velocity vanishes, and which places those are depends on the observer’s frame, as The picture belongs to the watcher showed for the plane. Q and R are unchanged by a steady translation of the observer and changed by a rotating one, so the index of a point is objective in one sense and not in the other. Every claim here is in the frame in which the ABC flow is steady.

Anything dynamical. An index says nothing about whether a stagnation point is stable in time, whether particles near it are trapped, or how fast they are carried away. The eigenvalues say something about the last of these, and The spin that feeds itself is about what the stretching they describe does to vorticity; neither is a statement about topology.

Still open: counting the points of a flow that has a wall #

The next calculation is the count for a three-dimensional separated flow over a body, the case the classification essay pointed at. There the critical points in the fluid carry the indices computed here, the points on the wall are counted as halves, and the rule relating them involves the topology of the body and of the section through the flow. Computing it on a flow with an exact wall — a three-dimensional stagnation flow onto a plate with a vortex added, whose wall pattern and interior points can both be found — would test whether half-points really behave as halves, and whether a pair annihilation in the fluid can be traded for a merger on the wall, which the plane count cannot see.

Beside it is the objectivity question, which this essay has answered only in part. The index is a sign of a determinant and the determinant is frame-indifferent under translation, so two observers moving steadily relative to each other agree on every index of every point they both see — but they do not see the same points, and in a rotating frame they do not agree on R. Constructing a flow in which a rotating observer and a fixed one count different numbers of stagnation points, with the totals still obeying their own theorem, would say exactly how much of this bookkeeping belongs to the flow.