The price of a gradient

A fluid in solid-body rotation — a bucket of water spun up and left alone — has a velocity that varies from place to place, a velocity gradient everywhere in it, and a viscosity that is not zero. It destroys no energy at all. Not a little; none.

That is the fact this essay is built on, and it is not an edge case. It is the definition of what viscosity charges for, and every number about drag, lubrication, heating and loss in the rest of this collection is a consequence of it.

What the stress actually does work against #

Take a small parcel of fluid and ask what its neighbours are doing to it. They pull on its faces with a stress, and the rate at which they do work on it is the stress contracted with the velocity gradient. That is the starting point, and it looks as though the whole velocity gradient ought to appear in the answer.

It does not, and the reason is one line of algebra. The viscous stress tensor is symmetric. The velocity gradient is not: it splits into a symmetric part, the rate of strain, and an antisymmetric part, which is the local rotation and is the vorticity in disguise. Contracting a symmetric tensor with an antisymmetric one gives zero, identically. So the antisymmetric half of the velocity gradient does no work against a viscous stress, ever, and what is left is

$$\Phi = 2\mu\,e_{ij}e_{ij}, \qquad e_{ij} = \tfrac12\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right).$$

This is the dissipation function: the rate at which mechanical energy is being destroyed, per unit volume, at a point.

Two properties fall straight out of the form. It is a sum of squares, so it is never negative — a viscous fluid cannot give energy back, which is the only place an arrow of time enters the equations of motion at all. And it is zero exactly when every component of the rate of strain is zero, which happens exactly on the rigid motions: a translation, a rotation, and any sum of the two. The decomposition a parcel undergoes in its first instant is therefore also a decomposition of the bill.

Why a picture of it proves nothing #

Every other field on this site can be checked by looking at it, at least a little. A streamline that runs through a solid body is wrong and looks wrong. A pressure field with a maximum where the flow is fastest is wrong and looks wrong.

A dissipation map has no such property. It is a scalar field, drawn as bands, and any smooth velocity field produces a plausible one — bright where the picture looks sheared, dark where it looks calm. There is nothing in the drawing to be wrong about. The site’s standing worry that a wrong flow field is beautiful is at its sharpest here.

So the dissipation function needs a test that is not a picture, and it has one that is better than most: the same number can be computed a completely different way.

The second route #

Integrate the mechanical-energy equation over a fixed region V with surface S. For a steady incompressible flow the result is

$$\int_V \Phi\,dV \;=\; \oint_S \mathbf{u}\cdot(\boldsymbol{\sigma}\cdot\mathbf{n})\,dS \;-\; \oint_S \rho\tfrac12|\mathbf{u}|^2\,(\mathbf{u}\cdot\mathbf{n})\,dS.$$

The first term on the right is the work the surface tractions do on the boundary; the second is the kinetic energy carried out through the same surface by the flow itself. Neither integral knows anything about the other. One is a volume integral of a squared derivative in the interior; the other is a pair of surface integrals of a stress and a flux. A field that is not a solution of the equations of motion will not make them agree, and a drawing of that field would not show which of them was wrong.

Where a pipe’s heat is actually made #

The first flow in that table is worth stopping on, because it contains a result that contradicts almost everybody’s first guess.

Plane Poiseuille flow between stationary walls has velocity $u = u_{\max}(1 - y^2/h^2)$. The walls do no work, because they do not move. So every joule the flow destroys is paid for at the ends, by whatever is maintaining the pressure gradient, and the boundary route is simply $(-dp/dx)\,Q$ — the pressure drop times the flow rate, which is what a pump reads.

The volume route is $\int \mu (du/dy)^2\,dy$. And $du/dy$ is zero on the centreline and largest at the wall.

The reason this matters beyond the arithmetic is that it is the same statement as the boundary-layer one. The claim that everything happens in a thin layer is usually presented as a fact about where the vorticity is. It is equally a fact about where the bill is, and the second version is the one that explains why a body’s drag is so insensitive to what the outer flow is doing.

The case that does not close, and why that is the good one #

Three flows agreeing to machine precision is reassuring and is not, on its own, evidence of much: they are all flows in which the fluid is bounded and the accounting is easy. The interesting test is a body moving through unbounded fluid, and there the balance appears to fail.

Take Stokes’ solution for a sphere translating at speed $U$ through fluid at rest. The power needed to tow it is the drag times the speed, $6\pi\mu a U^2$. Integrate $\Phi$ over the fluid out to a radius $R$ and the answer is less than that — at every finite $R$, by a fraction that does not go away as the resolution is improved.

That shortfall is not error. It is the second term in the balance — the work still being done by the viscous stress across the surface at $R$ — and the closed form for it is exactly $3a/2R$ of the total. A calculation that treats it as a numerical failure will conclude that Stokes’ solution does not conserve energy, which is a strong claim to arrive at by dropping a term.

The physical statement is the surprising one. A body creeping through a fluid heats a region far larger than itself, and the heating falls off so slowly that no finite region contains most of it. That is the energetic face of the same long-range disturbance that makes creeping flow past a cylinder have no solution at all: the 1/r decay of a Stokeslet is too slow for anything to be local.

The thing it is not: enstrophy #

There is a second quantity that looks like the dissipation and is quoted as though it were one, and the difference between them is worth being precise about because it is the commonest slip in the literature on vortices.

In two dimensions, with the fluid extending to infinity and everything decaying there, the total dissipation equals $\mu$ times the integral of the squared vorticity — the enstrophy. That identity is exact and it is genuinely useful, because vorticity is often the thing a calculation carries.

It is also only true over the whole plane. Over a finite region the two differ by a boundary term, and the boundary term is not small. A decaying vortex is the clean case: outside its core the fluid is irrotational, so it contributes nothing whatever to the enstrophy — and it is being sheared, so it contributes plenty to the dissipation. Inside a circle of radius $R$ the two differ by exactly $\mu\Gamma^2/\pi R^2$, which is measured in the essay on what a vortex loses.

The general rule is the one this whole essay is about: a region’s books do not balance until every way a joule can cross its boundary has been counted, and there are three of them — work by the tractions, kinetic energy carried by the flow, and, if the region is not fixed, the energy the moving boundary sweeps. Naming the dissipation without naming the region is naming half of a subtraction.

The term that is the Reynolds number’s fingerprint #

In all four cases above the kinetic-energy flux term is zero, and it is zero for a reason: creeping flows have no inertia, so nothing is carrying energy anywhere. The instant a flow has a Reynolds number worth naming, that term is not zero, and its size is a direct measure of how much of the power delivered to a region is leaving it as motion rather than as heat.

The map makes a second point that the profile figure only hinted at. The dissipation is not where the body is. It is on a surface round the body and in two lines behind it, and at higher Reynolds number the second of those dominates — which is taken up in the next essay, where the fraction still being made five diameters downstream is measured.

What a joule of dissipation is actually worth #

The function says how much mechanical energy is destroyed. It does not say how much is lost, and the two are different quantities as soon as the fluid has a temperature.

Dissipation does not annihilate energy — it converts it to heat, at the place where the straining happens. What has gone is the energy’s availability, and the measure of that is entropy. The entropy produced per unit volume by the straining is

$$\dot{s}_\Phi = \frac{\Phi}{T},$$

with $T$ the local absolute temperature. So the same $\Phi$, deposited in a compressor inlet at 250 K and in a turbine passage at 1,500 K, produces six times as much entropy in the first as in the second — and the work that could have been recovered and was not is proportional to the entropy, not to the heat. A joule dissipated cold is worth six of a joule dissipated hot, which is why a loss audit for a gas turbine is done in entropy and not in pressure drop, and why the same aerodynamic inefficiency is a different-sized problem at different stations of the same machine.

There is a second entropy source in any flow that carries heat, and it competes with the first. Conducting heat across a finite temperature difference is irreversible on its own, at a rate

$$\dot{s}_q = \frac{k\,|\nabla T|^2}{T^2},$$

and the two together are the whole of the entropy a flow produces. What makes them worth writing side by side is that they respond to a design change in opposite directions.

Consider a heated duct carrying a fixed heat load. Speed the flow up and the heat-transfer coefficient rises, so the temperature differences needed to move the heat fall, so $\dot{s}_q$ falls — and the velocity gradients rise, so $\Phi$ and $\dot{s}_\Phi$ rise with them, roughly as the cube of the speed in turbulent flow. Slow it down and the trade reverses. One term falls as the other climbs, so their sum has a minimum, and the minimum is at a definite Reynolds number.

That is entropy-generation minimisation, and it is a genuinely different design criterion from either of the two it replaces. Minimising pressure drop alone says go slowly, which is wrong. Maximising heat transfer alone says go fast, which is also wrong. Minimising the total lost work says there is a speed, and the speed depends on the heat load, the fluid, the tube and the temperature level — because $T$ appears in the denominators and the two denominators are different powers of it.

The general lesson belongs beside the rest of this essay. $\Phi$ is a local, positive, exactly computable quantity, and it is not by itself an answer to any question a designer has. What a gradient costs is $\Phi$; what it is worth is $\Phi/T$; and what should be done about it depends on what else in the system is producing entropy for the same reason. The dissipation function supplies the first of those exactly and the other two only once a temperature field has been put beside it.

What this makes computable #

Three things become straightforward once the function is written down, and each is the subject of one of the essays this one is the base of.

A minimum. Because Φ is a positive-definite quadratic in the rate of strain, the total dissipation of a flow is a quadratic functional of its velocity field — and quadratic functionals have minima. The parabola in a pipe is not the shape the equations produce; it is the cheapest shape the walls allow, and the equations produce it for that reason.

A cost that is a product. In a bearing, in a coating film and in a heated seal, the dissipation is not a loss to be minimised — it is the mechanism. A film that carries a steel shaft on oil does so entirely by dissipating, and the load it supports is proportional to how much it destroys.

An arrow. Φ ≥ 0 is the only inequality in the equations of motion, and it is what makes a viscous flow irreversible while the shock that a compressible flow makes is forbidden in one direction by a completely separate argument about entropy. The two turn out to be the same statement, and the essay that connects them is the one about the coefficient this site had never used.

What the picture cannot show #

The grid solve is coarse, and its dissipation is the coarsest thing on it. Φ is a squared derivative of a bilinearly interpolated field, so it is one order less accurate than the velocity and two less than the streamfunction. The maps above are honest about where the dissipation is and are not a measurement of how much; the numbers quoted in this essay all come from closed forms.

Nothing here is compressible. The form Φ = 2μ e:e assumes the divergence of the velocity is zero. A gas whose density is changing has a second, independent coefficient multiplying the square of that divergence, and every figure on this site has quietly set it to zero. That is a separate essay and it is not a small correction.

And the fluid’s properties are constant. Real oils thin as they warm, and the warming is caused by the dissipation being computed. Where that feedback matters the calculation above is the first step of an iteration rather than an answer, and there is a value of the driving beyond which the iteration does not converge at all.

Who found it, and when #

Stokes wrote the function down in 1845, in the same paper in which he set the second viscosity to zero and said he had no argument for doing so. Rayleigh gave it the name it now has and used it to argue, in 1873, that a slow viscous flow arranges itself to dissipate as little as it can — which was the first appearance of the variational idea this collection’s next essay is about.

The surprising connection is with a result that has nothing to do with fluids. The condition Φ = 0 picks out exactly the velocity fields whose rate of strain vanishes, and those are exactly the rigid motions of three-dimensional space: three translations and three rotations, six parameters, no more. So the null space of the dissipation function is the Lie algebra of the Euclidean group — a statement about what a fluid cannot be charged for, which turns out to be a statement about what it means for a body not to deform at all.

Where the ladder goes next #

Above this rung is the accounting for a flow that has inertia, where the kinetic-energy flux is not zero and the question becomes where the heat of a drag is actually made. Beside it is the variational result — the cheapest shape a set of walls allows — which is what turns Φ from a bookkeeping device into a way of predicting a flow without solving for it.

Below it are the two facts it was built from: what a parcel does in its first instant, which supplies the rate of strain, and the thin layer, which is where almost all of it happens.