What a parcel does in the first instant

Put a small circle of dye into a moving fluid and watch it for a moment. It goes somewhere, which is dull. It also changes shape, and the way it changes shape in the first instant is the whole local description of the flow — everything a fluid does to itself at a point is in that one deformation, and there is nothing else.

The claim this essay makes is that the deformation splits into a stretch and a spin, that the split is unique rather than conventional, and that the two halves have completely different jobs. One of them is what viscosity acts on. The other cannot produce a stress at all, however violent it gets.

Four things, and the list is complete #

A parcel in a flow can do exactly four things, and it is worth being explicit that the list has an end.

It can translate, which is the velocity at its centre and is not interesting here — a flow looks the same to an observer who moves with it. It can rotate as a rigid body. It can stretch along some direction and squash along another. And it can shear, which turns a square into a parallelogram.

That is all. The reason the list stops there is arithmetic rather than physics. The velocity a small distance $\delta\mathbf{x}$ from the parcel’s centre differs from the centre’s velocity by

$$\delta \mathbf{u} = \mathbf{L}\,\delta\mathbf{x} + O(|\delta\mathbf{x}|^2)$$

where $\mathbf{L}$ is the matrix of velocity derivatives $L_{ij} = \partial u_i/\partial x_j$. Four numbers in two dimensions, nine in three. A parcel small enough for the quadratic term to be negligible sees a linear velocity field, and a linear map is the whole of what can happen to it.

The interesting part is what those four numbers are for.

The split, and why there is only one #

Any square matrix is the sum of a symmetric part and a skew-symmetric part:

$$\mathbf{L} = \underbrace{\tfrac{1}{2}\left(\mathbf{L} + \mathbf{L}^{\mathsf{T}}\right)}_{\mathbf{S},\ \text{the strain rate}} + \underbrace{\tfrac{1}{2}\left(\mathbf{L} - \mathbf{L}^{\mathsf{T}}\right)}_{\boldsymbol{\Omega},\ \text{the spin}}$$

and — this is the part that makes it a theorem rather than a definition — the decomposition is unique. If $\mathbf{L} = \mathbf{S}' + \boldsymbol{\Omega}'$ with $\mathbf{S}'$ symmetric and $\boldsymbol{\Omega}'$ skew, then transposing gives $\mathbf{L}^{\mathsf{T}} = \mathbf{S}' - \boldsymbol{\Omega}'$, and adding and subtracting the two recovers $\mathbf{S}$ and $\boldsymbol{\Omega}$ exactly. There is no other way to do it.

Each half has a physical name that the rest of this site already uses.

The skew half is the vorticity, halved. In two dimensions $\boldsymbol{\Omega}$ has one independent entry, and it is $\tfrac{1}{2}(\partial v/\partial x - \partial u/\partial y) = \omega/2$. So a parcel in a flow rotates at half the vorticity, which is the factor of two that makes vorticity so easy to mis-state: the vorticity is twice the angular velocity of the fluid, not equal to it.

The trace of the symmetric half is the divergence. $S_{11} + S_{22} = \partial u/\partial x + \partial v/\partial y$, which is the quantity continuity sets to zero in an incompressible flow. So incompressibility is not a statement about the whole gradient; it is a statement about one number in one half of it.

Those two identifications are what the solver checks. assertDecomposition refuses a split whose halves do not add back to the gradient, whose first half is not symmetric, whose second is not skew, whose symmetric trace is not the divergence, or whose skew entry is not half the vorticity. The fourth of those is the one worth having: a decomposition can satisfy every matrix identity and still be attached to the wrong physical quantity, and the label on the figure would be the only thing wrong.

What the solver computed, and how it was checked #

The pictures are not sketches of a distortion. In a uniform gradient the map from a parcel’s starting shape to its current one is the matrix exponential $\exp(\mathbf{L}t)$, computed here by scaling and squaring a truncated series — so there is no time step to argue about and no accumulated error to hide behind.

Two things are then measured off the transported polygon rather than read off the algebra.

The area. Liouville’s theorem says a material parcel’s area multiplies by $\exp(t\,\mathrm{tr}\, \mathbf{L})$, which for an incompressible flow is exactly one. The area of the deformed ring is computed by the shoelace formula over its own vertices and compared with that exponential; for the shear figure above it agrees to fifteen decimal places. The comparison matters because the claim the picture makes is the area claim: the sentence “stretching in one direction is squashing in the other” is not a description of the drawing, it is the statement that the area did not change.

The axes. The ellipse’s principal directions come from the eigenvectors of $\mathbf{F}\mathbf{F}^{\mathsf{T}}$, and for a small time they approach the eigenvectors of $\mathbf{S}$. In simple shear those sit at exactly 45° to the flow, which is why a blob of dye in a shear layer stretches along a diagonal and not along the streamlines.

Simple shear is half of each, exactly #

The most useful special case is the one every boundary layer is made of.

A parallel shear flow $\mathbf{u} = (\dot\gamma y,\, 0)$ has

$$\mathbf{L} = \begin{pmatrix} 0 & \dot\gamma \\ 0 & 0\end{pmatrix}, \qquad \mathbf{S} = \frac{\dot\gamma}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}, \qquad \boldsymbol{\Omega} = \frac{\dot\gamma}{2}\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$$

Simple shear is exactly half straining and half rotation. The two halves have the same magnitude, at every shear rate, in every fluid. That is not an approximation and not a coincidence of units: the skew and symmetric parts of a matrix with one non-zero entry are always equal in size.

Two consequences follow, and they are the reason this decomposition is worth an essay.

The first is that a paddle wheel in a shear flow turns at $\dot\gamma/2$ while the fluid is being stretched at $\dot\gamma/2$ along the diagonal — the same number doing two different jobs. The second is that in a shear layer the material lines that stretch are at 45° to the wall, which is where the intense small-scale structure of a turbulent shear flow comes from and why the energy cascade is a story about stretching rather than about shearing.

Nine numbers instead of four #

Two dimensions make the arithmetic readable and hide one thing worth stating before the essay leaves.

In three dimensions the gradient has nine components. The symmetric half keeps six of them and the skew half three, and those three are precisely the components of the vorticity vector — which is why vorticity is a vector in three dimensions and a scalar in two. The count is the same statement as before: a general linear map of space is a rotation about some axis, followed by a stretch along three mutually perpendicular directions.

Those three directions are the eigenvectors of $\mathbf{S}$ and they are always perpendicular, because a symmetric matrix has an orthogonal eigenbasis. So the local picture of any flow, however complicated, is a sphere of fluid becoming an ellipsoid with three principal axes, while the whole thing turns about one axis at half the vorticity. The three stretch rates sum to the divergence, so in an incompressible flow at least one of them is negative: something is always being squashed, and it is the price of whatever is being stretched.

The half that viscosity can see #

Here is the reason this is more than bookkeeping.

A Newtonian fluid’s viscous stress is proportional to the strain rate and to nothing else:

$$\boldsymbol{\tau} = 2\mu\,\mathbf{S}$$

The skew half does not appear. It is not that its contribution is small or usually negligible — it is that a term in $\boldsymbol{\Omega}$ is forbidden.

The argument is one sentence and it is the strongest kind. Take a tank of fluid on a turntable, rotating rigidly. Nothing in it is sliding past anything: every parcel keeps its shape, every distance between two particles is fixed, and no fluid anywhere is being deformed. A stress law containing $\boldsymbol{\Omega}$ would predict a viscous stress in that tank — a fluid that heats itself up by being carried round a corner — and no such thing happens. So the stress can depend only on the part of the gradient that vanishes for a rigid motion, and that part is $\mathbf{S}$ exactly.

This is why the split earns its keep. The gradient is what is easy to measure; the strain rate is what the physics acts through. The whole apparatus of the boundary layer is about one component of $\mathbf{S}$ becoming enormous in a thin film, and the Reynolds stresses that averaging leaves behind are modelled, in every closure this site has examined, as something proportional to the mean $\mathbf{S}$ and never to the mean $\boldsymbol{\Omega}$.

The turntable argument, made exact #

The rotating-tank argument above is decisive and informal. Its formal version is worth having, because it says precisely which quantities a constitutive law may contain, and it settles a question that a whole branch of rheology is organised around.

Let a second observer be related to the first by a time-dependent rotation $\mathbf Q(t)$. Working through, the velocity gradient transforms as

$$\mathbf L^* = \mathbf Q\mathbf L\mathbf Q^{\mathsf T} + \dot{\mathbf Q}\mathbf Q^{\mathsf T},$$

and the last term is skew. So it lands entirely in the spin: the strain rate transforms as $\mathbf S^* = \mathbf Q\mathbf S\mathbf Q^{\mathsf T}$, which is what a tensor should do, while $\boldsymbol\Omega^* = \mathbf Q\boldsymbol\Omega\mathbf Q^{\mathsf T} + \dot{\mathbf Q}\mathbf Q^{\mathsf T}$ picks up the observer’s own rotation.

A quantity that transforms the first way is called objective, and the principle of material frame indifference says a constitutive law may be written in objective quantities and nothing else. That rules $\boldsymbol\Omega$ out by an argument that needs no experiment: a term containing it would make the stress depend on how the laboratory is turning, and a material has no way of knowing.

And it has a consequence that catches everybody the first time. The ordinary time derivative of an objective tensor is not objective, because differentiating $\mathbf Q\mathbf T\mathbf Q^{\mathsf T}$ produces terms in $\dot{\mathbf Q}$. So a viscoelastic model with $\dot{\boldsymbol\sigma}$ in it — the natural way to write a stress that relaxes — is not frame indifferent as written and predicts a stress in a fluid on a turntable.

The repair is to build a derivative that removes those terms, and there are several: the upper-convected rate $\dot{\boldsymbol\sigma} - \mathbf L\boldsymbol\sigma - \boldsymbol\sigma\mathbf L^{\mathsf T}$, the lower-convected one, and the co-rotational rate that subtracts $\boldsymbol\Omega$ instead. All are objective, so the principle does not choose between them — which is why the upper-convected Maxwell model has a hyphenated qualifier in its name at all, and why swapping the qualifier gives a different fluid with different normal stresses.

The principle says what is forbidden and leaves what is permitted open, and the choice among the permitted rates is a modelling decision rather than a theorem.

Irrotational does not mean undistorted #

Now the case that catches people, and it is the one the refutation above is built on.

The exact theory works with irrotational flows, which have $\boldsymbol{\Omega} = 0$ everywhere outside the body. It is tempting to read that as “nothing is happening to the fluid”, and the streamlines round a cylinder — curving smoothly over the top, closing up behind — encourage it.

The measurement says otherwise.

The parcels over the shoulder of the cylinder are being distorted violently while their angular velocity is exactly zero. A blob of dye there would be drawn into a long thin filament — which is what happens in a real experiment, and is often mis-read as evidence of rotation.

Nothing in the streamline picture reports this. The streamlines are the same smooth curves whether the strain is strong or weak, and it is the spacing of neighbouring streamlines rather than their shape that carries the information — which is exactly the property a drawing throws away.

What the picture cannot show #

Three things, and the third is a whole essay.

It is the first instant only. The matrix exponential is exact for a uniform gradient, and a real flow’s gradient changes as the parcel moves. Everything here is the leading term of a Taylor series in the parcel size and holds for as long as the parcel is small and the gradient it sits in is roughly constant. Follow the parcel far enough and the linear picture fails; that is where the material derivative and the acceleration field take over.

The parcel eventually stops being one. Viscosity diffuses momentum across the parcel’s boundary and molecular diffusion smears the dye. The picture of a material blob is exact for an inviscid fluid and a good approximation for as long as the deformation is faster than the diffusion — which is the same competition, written differently, as the Reynolds number.

Two dimensions have no vortex stretching. In two dimensions the vorticity vector points out of the plane and the stretching is all in the plane, so the two never interact. In three they do, and the interaction is the single most important term in turbulence. That is the next rung, and it is where the strain rate stops being something viscosity acts on and becomes something that amplifies the spin itself.

Where the model stops #

The decomposition itself does not stop anywhere: it is linear algebra, it holds for every flow of every fluid, and it needs no assumption about viscosity, compressibility or Reynolds number. What does have limits is the interpretation.

The identification of $2\mu\mathbf{S}$ as the stress needs a Newtonian fluid. A polymer solution’s stress depends on the strain-rate history and not only on its present value, and for those fluids the split is still exact and the stress law is not.

The identification of the parcel as a material object needs the continuum assumption. A “parcel” a few molecular mean free paths across is not a thing that deforms; it is a fluctuating population of molecules, and the velocity gradient at that scale is not defined.

And the split is frame-dependent in one half. An observer on a turntable measures a different $\boldsymbol{\Omega}$ from one standing still, because rigid rotation adds a constant to the spin. The strain rate $\mathbf{S}$ is the same for both, which is the deeper reason the stress law can only contain $\mathbf{S}$: a material cannot know how the laboratory is turning.

How to read a flow figure after this #

The practical residue of all of the above is a habit, and it is worth stating plainly because it changes what a reader looks for.

Streamline shape carries almost nothing about the local motion. A curved streamline may have no rotation in it and a straight one may have a great deal. What carries the information is how neighbouring streamlines converge, diverge and slide past each other — which is the gradient, and which is exactly the quantity a hand-drawn figure is least likely to get right.

So a figure on this site that wants to show what is happening to the fluid rather than where the fluid is going draws parcels, not paths — and the two pictures of the same flow can look entirely different without either being wrong.

Who found it, and when #

Cauchy had the decomposition of a general deformation into strain and rotation by the 1820s, in the theory of elastic solids, where it is a statement about displacement rather than velocity. Stokes carried it to fluids in 1845 in the memoir that also produced the Navier–Stokes equations, and it is Stokes’ argument above — that a rigid rotation cannot produce stress — that fixes the form of the viscous term.

Helmholtz gave the theorem its modern shape in 1858, in the paper that founded vortex dynamics, by noticing that the skew half of the gradient is the object that obeys its own conservation laws. That is the observation the rest of this field is built on: the spin is not merely a component of the motion, it is a substance with a history.

Where the ladder goes next #

The strain rate and the spin have been separated here and treated as independent. In three dimensions they are not: a strain field can stretch a vortex tube, and a stretched tube spins faster. That coupling is the next rung, it has one exact solution, and it is the reason turbulence is hard.