The shock in a river

Below a weir the water is shallow and moving fast. A few metres further on it is deep and moving slowly. Between the two is a stationary band of white, churning, noisy water that does not move downstream and does not go away.

That band is a shock wave. Not like a shock wave: a shock wave, in a fluid whose ratio of specific heats happens to be two, with the depth playing the density’s part and the Froude number playing the Mach number’s. And like a shock, it is a discontinuity across which momentum is conserved exactly, energy is destroyed, and only one direction is allowed.

Shallow water is a gas #

The correspondence is not a resemblance. It is a change of variables.

Take water of depth $h$ in a channel, moving with a depth-averaged velocity $V$, and assume the depth is small compared with the length over which things change — so the pressure is hydrostatic and the vertical velocity is negligible. The equations for $h$ and $V$ are then:

$$\frac{\partial h}{\partial t} + \frac{\partial (hV)}{\partial x} = 0 \qquad \frac{\partial V}{\partial t} + V\frac{\partial V}{\partial x} + g\frac{\partial h}{\partial x} = 0$$

Set those beside the one-dimensional equations of a compressible gas and they match term for term, with $h \leftrightarrow \rho$ and the pressure law $p \propto \rho^{2}$. That last is the identification: a gas with $p \propto \rho^{\gamma}$ and $\gamma = 2$.

Every consequence follows without further work.

• The wave speed in a gas is $\sqrt{\gamma p/\rho}$, which for $\gamma = 2$ and $p \propto \rho^2$ is proportional to $\sqrt{\rho}$. In shallow water it is $\sqrt{gh}$. Long waves in shallow water travel faster in deeper water for the same reason sound travels faster in a denser gas at the same $\gamma$.

• The Froude number $V/\sqrt{gh}$ is the Mach number. Below one, information can travel upstream and the flow knows what is coming; above one it cannot, and the warning cannot arrive.

• Subcritical and supercritical are subsonic and supersonic, with the same consequences: a disturbance in a subcritical stream spreads both ways, and one in a supercritical stream is swept into a wedge with a half-angle whose sine is $1/\mathrm{Fr}$ — a Mach cone, drawn on a river, visible any time water runs fast over a shallow apron and meets a pebble.

• The characteristics carry the same information in the same way. A disturbance travels at $V \pm \sqrt{gh}$, exactly as one in a gas travels at $u \pm c$, so a supercritical stream has both families pointing downstream and cannot be influenced from below — which is why a supercritical channel is controlled from its upstream end and a subcritical one from its downstream end. Every statement about domains of dependence that the compressible field makes is true here with the words changed.

• Waves steepen. A crest travels faster than a trough because it is deeper, so a long wave in shallow water leans forward and eventually breaks. That is the same nonlinear steepening that turns a compression wave in a gas into a shock, and the breaking wave on a beach and the shock ahead of a bullet are the same phenomenon.

The reason this is teachable rather than merely true is that it is visible. Everything in the compressible field happens in a transparent gas at speeds nobody can see; the same physics in a channel happens in slow motion, at walking pace, in something a person can put a hand in.

The jump, from momentum alone #

Draw a box round the jump. Steady flow, negligible bed friction over the short distance involved, hydrostatic pressure at each end. What crosses the ends is mass and momentum, and the pressure force on each end is the depth-integrated hydrostatic pressure, $\tfrac{1}{2}\rho g h^2$ per unit width.

The conserved combination is the specific force

$$M = \frac{q^2}{gh} + \frac{h^2}{2}$$

with $q = Vh$ the discharge per unit width. Setting $M_1 = M_2$ and discarding the trivial root gives the conjugate-depth relation:

$$\frac{h_2}{h_1} = \tfrac{1}{2}\left(\sqrt{1 + 8\,\mathrm{Fr}_1^{2}} - 1\right)$$

Nothing about the interior of the jump entered that. There is a violent roller in there, air is being entrained, the flow is three-dimensional and unsteady and nobody has solved it; the two depths either side follow anyway, because the box does not need to know what is inside it.

The energy, which is where the white water comes from #

Now compute the specific energy — the head, measured from the bed —

$$E = h + \frac{q^2}{2gh^2}$$

on each side, using the two depths the momentum balance produced. It falls. It has to fall, and the amount has a closed form:

$$\Delta E = \frac{(h_2 - h_1)^3}{4\,h_1 h_2}$$

which the solver checks against the difference of the two computed energies to nine significant figures.

At the case drawn — an arriving Froude number of 5.05 — the jump destroys 49.5 per cent of the energy that arrived. Half. And the bed does not appear anywhere in the calculation.

That is the essay’s refutation, and the reason the folk explanation is so tempting is that the loss is obviously happening somewhere, the bed is obviously right there, and white water obviously looks like friction. It is friction, in the sense that it is ultimately viscous. It is not friction with the bed: it is the same internal dissipation the sudden expansion has, a shear layer feeding a cascade that ends at the smallest scales, and its size is fixed by the momentum balance before any of that happens.

There is a second thing that goes with it, and it is the same one a shock in a gas has. A jump does not merely lose energy — it loses energy and has to satisfy mass and momentum at the same time, and those three demands together are what fix the downstream state uniquely. Given the upstream depth and discharge, there is exactly one depth the flow can arrive at, and no freedom anywhere: not in how fast the transition happens, not in how rough the bed is, not in how the roller happens to be behaving that afternoon. That is the property that makes the calculation worth anything to a designer, and it is the property that Bernoulli’s equation alone could never deliver.

This is why stilling basins work. A dam spillway delivers water with enough energy to scour a river bed to bedrock, and the standard remedy is not to armour the bed — it is to force a hydraulic jump in a concrete basin at the foot of the spillway and let the jump destroy half the energy before the water reaches anything erodible. The design of such a basin is the design of a controlled shock wave, and the fraction it removes is read off the curve above.

Which way it is allowed to go #

The conjugate-depth relation is symmetric. Feed it a subcritical depth and it happily returns the supercritical partner: a jump upwards in Froude number, from deep and slow to shallow and fast, satisfying mass and momentum exactly.

It does not happen, and the reason is the same one that forbids the expansion shock. Running the arithmetic backwards, the energy of the stream would rise across the discontinuity — energy appearing from nowhere in a channel with nothing driving it. So the direction is fixed, and the solver refuses to compute a jump from a subcritical approach rather than returning the number:

A jump from subcritical to supercritical satisfies mass and momentum exactly and raises the energy of the stream, so it is forbidden for the same reason an expansion shock is. The conjugate-depth relation is symmetric and will happily return it; the flow will not.

That refusal is checked in the site’s gate, alongside the refusals about shocks in gases, because the two are the same statement about different fluids. What is worth noticing is that the mechanism of the prohibition differs. In a gas the forbidden branch lowers the entropy, and it is the second law that rules it out; the energy is conserved either way. In shallow water there is no entropy — the model has no thermodynamics in it at all — and what rules it out is that mechanical energy would have to be created. Two different arguments, the same conclusion, in fluids that satisfy the same equations.

Where the analogy stops, computed rather than asserted #

Here is the interesting part, and it is the part most treatments skip.

The shallow-water equations and the $\gamma = 2$ gas equations are the same for smooth flow. They are not the same across a discontinuity, and the reason is exactly the difference just described: a shock conserves energy and makes entropy, while a jump conserves momentum and destroys energy. Those are different jump conditions, so they must give different answers.

They do. But not immediately.

The two derivatives at $\mathrm{Fr} = M = 1$ come out at 1.333333333 and 1.333333333. That is asserted rather than admired: an analogy claimed to be exact would be a false claim, and an analogy whose curves were not even tangent would mean the correspondence had been set up wrongly. Both halves are checked, and the second half — that they are more than 15 per cent apart by Froude 3 — is checked too, because a version of the code that had accidentally used the same formula twice would pass the tangency test perfectly.

An analogy that is exact to first order and wrong at the second is the most useful kind, because it states precisely how far it may be trusted. Weak jumps and weak shocks are interchangeable. Strong ones are not, and the reason they are not is a real physical distinction rather than an approximation anybody could improve.

The same jump, walking #

Everything above is written for a jump that stands still, and standing still is a choice of observer rather than a property of the physics. Watch the same discontinuity from a frame moving with it and nothing in the momentum balance changes, which means the conjugate-depth relation describes a travelling front just as exactly as a stationary one.

Such a front is a bore, and its speed comes out of the relation with one line of algebra. Set the approach Froude number so that the water ahead is at rest in the laboratory, and

$$c = \sqrt{\frac{g h_2 (h_1 + h_2)}{2 h_1}}$$

for a bore of depth $h_2$ advancing into still water of depth $h_1$. It is worth checking the weak limit: as $h_2 \to h_1$ this becomes $\sqrt{g h_1}$, which is the long-wave speed. A bore is a wave that has grown until it is a discontinuity, and its speed exceeds the wave speed of the water it is running into — which is exactly the statement that a shock outruns sound, and the reason the water ahead has no warning.

Tidal bores are this: a rising tide funnelled into a narrowing estuary until the leading edge steepens into a front and walks upriver against the current. The Severn’s is a metre or two high in water a few metres deep, and the formula puts it at several metres a second, which is roughly a brisk cyclist — one of the few places in this subject where a shock wave can be accompanied on a towpath.

And the moving version is what makes the dam-break problem the shock tube. Remove a barrier between deep water and shallow, and the shallow-water equations produce precisely what the gas equations produce when a diaphragm bursts: a rarefaction fan spreading backwards into the deep side, smoothly and reversibly, and a bore running forwards into the shallow side, abruptly and dissipatively. Two waves, opposite in character, from one initial discontinuity — and the asymmetry between them is the whole content of the direction rule above. Compression may steepen into a discontinuity; expansion may not.

If the downstream bed is dry rather than merely shallow there is no bore at all, because there is nothing for the front to jump onto. The solution is then a fan all the way to a tip that advances at $2\sqrt{gh_0}$, twice the wave speed of the reservoir — which is the shallow-water counterpart of a gas expanding into vacuum, and the reason a dam-break wave arrives faster than the still-water wave speed would suggest.

So the stationary jump below a weir is the special case, held in place by a stream running into it at exactly the right speed. Everything else the model can do with a discontinuity, it does on the move.

Where the model stops #

Three absences, each of which a real jump has and this calculation does not.

No length. The shallow-water model contains no viscosity, no turbulence model and no vertical structure, so there is no scale in it that could set how far the transition takes. The figure leaves that region blank on purpose and marks the empirical six-depth rule as borrowed. Anything drawn there would be an invention.

No roller. The recirculating, air-entraining region on the face of a strong jump is genuinely three-dimensional, genuinely unsteady, and outside everything this site can compute. Its existence is why the loss happens; its structure is not needed to compute the loss.

No classification. Real jumps are catalogued by strength — undular below about Froude 1.7, where the transition happens through a train of smooth waves with no white water at all; then weak, oscillating, steady and strong. The undular case is the one where this model is least honest, because there the transition is dispersive rather than dissipative and the shallow-water equations have no dispersion in them.

No air. A strong jump entrains a great deal of air — that is what makes it white — and the aerated water is measurably less dense than the water either side of it. The model has one fluid of one density, so the depth it computes is the depth of clear water and a real jump’s free surface sits somewhat higher. For a spillway that difference is a wall height, and it is estimated from experiment rather than derived from anything.

There is also a bed, and it does do something — over a long channel, bed friction is the dominant loss and the entire subject of gradually varied flow. What the essay’s refutation says is that it does not do this: over the few metres a jump occupies, the friction on the bed is small compared with the momentum flux, and dropping it is what makes the calculation exact.

Who found it, and when #

Giorgio Bidone observed and described the jump in Turin in 1820. Bélanger derived the conjugate-depth relation in 1828, and it carries his name. Both precede the gas-dynamics results they mirror by fifty years: Rankine’s shock conditions are 1870 and Hugoniot’s 1887.

That ordering is worth registering. The hydraulic jump is the older result, and when Riemann, Rankine and Hugoniot came to argue about whether a compression discontinuity in a gas was physically admissible — and about whether energy or entropy was the thing that had to be tracked — the river had already been settled for half a century. The analogy was used the other way round: as a visible, wadeable model of something invisible.

Ernst Mach, who spent the 1880s photographing shock waves, was working the same seam from the other side. The two subjects were one subject in the nineteenth century and separated only when the aeroplane made one of them urgent.

Where the ladder goes #

The jump is what a channel does when it must change regime. The next rung asks what decides which regime it is in at all, and the answer is a curve with a minimum on it.

For a given discharge there are two depths at every energy above a certain floor — one fast and shallow, one slow and deep — and exactly one at the floor itself. That depth is the critical depth, the Froude number there is exactly one, and the least energy is exactly three halves of it. All three are computed by search on the next rung and checked against their closed forms.

The floor has consequences a nozzle designer would recognise immediately. Raise the bed of a channel and the flow has less energy to spend; raise it too far and there is no depth at all that will carry the discharge, so the flow does not thin — it backs up. A weir chokes exactly as a nozzle chokes, and it is the same statement about not passing through the sonic point twice.