What averaging costs

Nobody wants the instantaneous velocity field of a turbulent flow. It has structure down to scales nobody can compute, it never repeats, and almost every question anybody asks of it — how much drag, how much mixing, how much heat — is a question about an average.

So the natural move is to average the equations rather than solve them, and ask what the mean flow obeys.

The move is legitimate, the algebra is short, and the result is exact. It is also the single most important unsolved problem in classical physics, and this essay is about why those two sentences are both true.

The decomposition, and the one term that survives #

Write every field as a mean plus a fluctuation about it:

$$u_i = \langle u_i\rangle + u_i', \qquad \langle u_i'\rangle = 0$$

The averaging is a linear operation and it commutes with differentiation, so almost every term in the Navier–Stokes equations passes through it unchanged. The time derivative averages to the derivative of the mean. The pressure gradient averages to the gradient of the mean pressure. The viscous term, being linear in the velocity, averages to the same operator applied to the mean.

One term does not: the convective term u_j ∂u_i/∂x_j, which is quadratic. Averaging a product is not the product of the averages, and what comes out is

$$\langle u_j \frac{\partial u_i}{\partial x_j}\rangle = \langle u_j\rangle \frac{\partial \langle u_i\rangle}{\partial x_j} + \frac{\partial \langle u_i' u_j'\rangle}{\partial x_j}$$

The first piece is the mean flow carrying itself, which is what one would expect and want. The second is new. It involves the correlation between two fluctuating components — a quantity about which the mean-flow equations say nothing at all.

Moved to the other side and multiplied by the density, it is conventionally called the Reynolds stress, −ρ⟨u′ᵢu′ⱼ⟩, and it sits in the averaged momentum equation exactly where a stress would sit.

Counting, which is the whole argument #

The temptation at this point is to describe the difficulty. It is much more effective to count it.

Unknowns. Three components of mean velocity. One mean pressure. Six independent components of ⟨u′ᵢu′ⱼ⟩ — six rather than nine, because the tensor is symmetric. Ten.

Equations. Three momentum components. One continuity equation. Four.

The system is short by six, and no amount of care will change that, because nothing was approximated anywhere in the derivation. The averaged equations are exact. They are exact and they do not determine their own solution.

That combination is unusual enough to be worth dwelling on. This is not a model that is inaccurate, nor an equation that is hard to solve, nor a numerical method that does not converge. It is a set of equations, correct in every respect, with more unknowns in it than equations.

Why symmetric, and why six #

The tensor ⟨u′ᵢu′ⱼ⟩ is symmetric because multiplication is commutative: ⟨u′v′⟩ and ⟨v′u′⟩ are the average of the same product. So of nine components, three pairs coincide and six values are independent — the three variances ⟨u′²⟩, ⟨v′²⟩, ⟨w′²⟩ and the three covariances ⟨u′v′⟩, ⟨u′w′⟩, ⟨v′w′⟩.

The variances are never zero in a turbulent flow: they are the mean-square fluctuations, and their sum is twice the turbulent kinetic energy. The covariances may be zero, and whether they are is what decides whether the turbulence does anything to the mean flow.

That is the one useful simplification available for free. In a flow with a single mean shear direction, only ⟨u′v′⟩ appears in the mean momentum equation, so the problem reduces from six unknowns to one. A great deal of turbulence modelling for boundary layers and pipes lives inside that reduction, and it stops being available the moment the flow has more than one shear direction — which is to say, on any real three-dimensional geometry.

What the Reynolds stress actually is #

The name is a historical accident that has cost a great deal of clarity, so it is worth being exact about the object.

Consider a plane in a shear flow, with a mean velocity that increases upward. A parcel that moves upward through the plane — v′ > 0 — has come from below, where the mean velocity is lower, so it tends to arrive carrying a velocity deficit: u′ < 0. A parcel moving downward brings a surplus. Either way the product u′v′ is negative on average.

That product, averaged, is the mean flux of streamwise momentum in the vertical direction, carried by the fluctuating motion. It has the dimensions of a stress because a momentum flux per unit area is a stress dimensionally, and it enters the momentum equation exactly where a stress does because a momentum flux and a force are the same thing to a control volume.

But the analogy stops there, and the difference is the reason for the whole difficulty.

A viscous stress has a constitutive relation. For a Newtonian fluid it is μ times the rate of strain, μ is a property of the substance, and it can be measured once in a viscometer and used everywhere thereafter. Nothing about the flow enters.

A Reynolds stress has none. It is not a property of the fluid. It is a property of this flow, at this point, at this moment in its history — and its value at a point can depend on what the flow was doing some distance upstream, because the eddies carrying the momentum flux were made there.

This is why the name misleads. A student who reads “stress” reasonably expects a constitutive relation to exist, and spends time looking for the one that turbulence has. It does not have one.

The equation for the stress, and where it leads #

The obvious response to being short of six equations is to derive six more. It can be done: take the Navier–Stokes equation for u′ᵢ, multiply by u′ⱼ, add the same with the indices exchanged, and average. The result is a transport equation for ⟨u′ᵢu′ⱼ⟩, exact, with terms that have physical names — production, dissipation, pressure–strain redistribution, and transport.

It also contains ⟨u′ᵢu′ⱼu′ₖ⟩, the triple correlation, which has ten independent components in three dimensions and about which the new equations say nothing.

Deriving equations for those produces the quadruple correlation, with fifteen. The count of new unknowns exceeds the count of new equations at every level, and the gap widens rather than closing. The ladder that never closes takes this seriously and is about the shape of the divergence.

The general statement is that the moment hierarchy of a nonlinear stochastic system does not close, and the reason is not fluid mechanical: it is that averaging a nonlinear operator loses information which no amount of further averaging recovers.

Where the information went #

It is worth asking, plainly, what was thrown away, because the answer explains why the loss is irrecoverable.

An average is a projection. Applying it to a field discards everything about the field except its first moment, and the discarded part is exactly what the nonlinear term needed in order to be evaluated. Nothing was approximated; something was deleted, and then a quantity that depended on the deleted part was still required.

That framing makes two things clear at once.

Why it is not a fluid problem. The same difficulty arises in any nonlinear system with fluctuations — in statistical mechanics as the BBGKY hierarchy, in field theory as the Schwinger–Dyson equations, in population dynamics as the moment-closure problem. Fluid turbulence is the case where the hierarchy matters most in engineering, not the case where it is worst behaved.

Why a closure is a decision. Every practical scheme supplies the missing information from somewhere outside the equations — dimensional analysis, an assumed analogy, a calibration experiment, or a resolved-scale computation. That is legitimate and it is not a derivation, and the honest form of any turbulence model states which of those it used.

What is still true, and it is not nothing #

The averaged equations are exact, and exactness has uses even without closure.

Integral constraints hold. Mass is still conserved by the mean flow. The momentum balance over a control volume still holds, with the Reynolds stress appearing as an extra flux through the faces. A great deal of practical turbomachinery and aerodynamics is done with such balances, using measured stresses rather than modelled ones.

Symmetries constrain the answer. In a flow with a homogeneous direction, every derivative along it vanishes on averaging, and several stress components drop out for reasons of symmetry rather than of modelling. This is how the classical free-shear-flow solutions — the jet, the wake, the mixing layer — are obtained: self-similarity plus symmetry removes enough unknowns for one modelled quantity to close the problem.

The energy budget is exact. The rate at which the mean flow loses energy to the fluctuations is −⟨u′ᵢu′ⱼ⟩ ∂⟨uᵢ⟩/∂xⱼ, the production term, and it appears with opposite signs in the two budgets. That is an exact statement about where the energy goes, and it is the first step of the cascade argument.

Two flows, one strain rate, two different stresses #

The clearest way to see that no constitutive relation exists is to name two flows in which the mean rate of strain is the same and the Reynolds stress is not.

Take a turbulent boundary layer at a station where ∂⟨u⟩/∂y has some value, and take the centre of a turbulent mixing layer where ∂⟨u⟩/∂y has the same value. An eddy viscosity that deserved the name would return the same stress for both. The measured stresses differ substantially, and they differ because the eddies in the two flows have different sizes: the boundary layer’s are limited by their distance from the wall, the mixing layer’s by the layer’s own thickness.

Worse for the analogy: there are flows where the Reynolds stress and the mean strain rate have opposite signs over part of the domain. An asymmetric channel, and the near-wake of a body, both show regions where the momentum flux runs up the gradient rather than down it. An eddy viscosity fitted there is negative, which is not a viscosity in any sense a physicist would accept.

None of that makes the eddy-viscosity idea useless. It makes it a correlation with a restricted domain of validity, like every other correlation on this site, and the honest practice is to say which flows it was calibrated on. What it is not is a derivation, and the word “stress” is what keeps suggesting that it might be.

The six are undetermined and they are not free #

The count says nothing determines the six unknowns, which is true and slightly overstates the position. Nothing determines them, and not every set of six values is possible, and the constraints are exact and cost nothing to check.

The tensor $\langle u_i'u_j'\rangle$ is a covariance matrix, so it inherits every property a covariance matrix has. Its diagonal entries are mean squares and cannot be negative. Its off-diagonal entries are bounded by Cauchy–Schwarz — $\langle u'v'\rangle^2 \le \langle u'^2\rangle\langle v'^2\rangle$ — so a correlation coefficient cannot exceed one. And the whole matrix must be positive semi-definite, which is the general statement containing both.

Those conditions are called realisability, and a set of stresses that violates any of them did not come from any fluctuating field whatever. That makes them a free test on every closure. A linear eddy-viscosity model computes each normal stress as an isotropic part minus a multiple of the local strain rate, and where the strain rate is large enough the subtraction wins: the model returns a negative mean square, which is not a wrong number but an impossible one. It happens routinely at a stagnation point, where the strain is large and the turbulence is not, and the resulting excess of turbulent kinetic energy in front of a blunt body is a well-known and long-standing defect of the simplest models.

The geometry of the constraint is worth having because it is more informative than the inequalities. Strip out the trace — write the anisotropy $b_{ij} = \langle u_i'u_j'\rangle/2k - \delta_{ij}/3$ — and every realisable state corresponds to a point inside a definite region in the plane of that tensor’s two invariants: the Lumley triangle. Its interior is ordinary three-dimensional turbulence; its vertices are the three extreme states, isotropic, two-component axisymmetric, and one-component; and its edges are the axisymmetric and two-component limits.

Real flows travel across it in ways that are physically legible. Approaching a solid wall the wall-normal fluctuation is suppressed while the other two are not, so the state moves to the two-component edge — which is a strong statement about the near-wall region that no model containing a single scalar eddy viscosity can represent, since an isotropic coefficient cannot make one component vanish while the others survive.

So the honest form of the essay’s count has two halves. The averaged equations do not determine the Reynolds stress, and they do constrain it to a bounded region whose boundary means something. A model that leaves the triangle has produced a state no fluctuating field could have, and it has done so without any residual rising and without any conservation check failing. Testing for it is one eigenvalue computation per cell, it is not universally done, and it is the cheapest check available on a quantity nothing else in the equations can verify.

Where the model stops #

Nothing on this site computes a Reynolds stress, and the count above is why: the six unknowns are determined by the fluctuating field, and this site has no fluctuating field.

What is drawn here is arithmetic — a count of tensor components — and it is drawn as a count rather than as a flow for exactly that reason. The figures in this essay contain no velocity field at all, which is unusual for this site and is the honest presentation of the material.

There is also a positive statement to make about what the count does not forbid, because the pessimism is easy to overdo. Nothing above says the mean flow is unpredictable. Mean profiles in pipes and boundary layers are reproducible to within a per cent between laboratories, they collapse onto universal curves under the right scaling, and engineering predictions built on fitted closures are routinely good enough to design aircraft with. What the count says is that this predictability is not derivable from the averaged equations alone — it comes from those equations plus information obtained elsewhere, and the elsewhere is always, in the end, an experiment.

One further limit worth stating. The averaging above is written as though ⟨·⟩ were unambiguous. It is not: it may be a time average, an ensemble average, or a spatial average over a homogeneous direction, and these coincide only for a statistically stationary and ergodic flow. For a flow with a slow unsteadiness in it — a gust, a manoeuvre, a rotating machine — the choice matters, and the literature’s habit of writing ⟨·⟩ without saying which has caused real confusion about what a “turbulence model” is being asked to predict.

Who found it, and when #

Reynolds published the decomposition in 1895, twelve years after the pipe experiment, and derived the averaged equations that carry his name. He was clear that the new terms were momentum fluxes and he did not claim to have closed anything.

Boussinesq had already proposed the eddy-viscosity analogy in 1877 — before the decomposition existed — and it is worth noticing the order: the model that supplies the missing information was proposed eighteen years before the thing it supplies information about was written down.

Prandtl’s mixing length arrived in 1925 and Kármán’s similarity hypothesis in 1930, and the law of the wall is the best-known consequence of the first. Keller and Friedmann had set out the moment hierarchy and its non-closure in 1924, so the impossibility of the direct route was established before most of the models that route around it.

Where the ladder goes next #

The next rung follows the obvious remedy to its end: the ladder that never closes counts the moment hierarchy properly and shows the gap widening at every level.

The rung after that is the first honest closure, and what it costs: a guess with a constant in it takes Prandtl’s mixing length, integrates it, and finds that it does reproduce the law of the wall exactly — which is a fact about the model, not about a fluid.