The hill a slow current will not climb

The balance that is its own error finds geostrophic balance accurate to exactly the Rossby number, and ends by naming the quantity that makes rotating flow tractable and leaves unbuilt: potential vorticity. This essay builds it, in the simplest flow where it decides something a reader can see — a steady current over an isolated hill on the floor of a rotating layer — and uses it to put a number on one of the most quoted results in rotating fluids.

That result is the Taylor–Proudman theorem. Joseph Proudman proved in 1916 that in a steady, inviscid flow rotating fast enough, the velocity cannot vary along the axis of rotation. G. I. Taylor drew the consequence in 1917 and demonstrated it in 1923: if the fluid cannot vary with height, a column of it cannot be squashed over an obstacle, so the flow goes round the obstacle instead, as if a solid column stood above it all the way to the lid. He towed a short body slowly across the floor of a rotating tank, put dye in the water, and watched the fluid above the body move with it as a single piece. That column has carried his name ever since, and the textbook sentence about it is the claim this essay tests.

A theorem about a limit, and the question it cannot answer #

The theorem is a statement about the limit in which the Rossby number vanishes. It is proved by dropping the inertial terms from the equations of motion, which is to say by assuming the flow’s own acceleration is negligible against the Coriolis force; what remains forbids vertical variation. What the theorem cannot say is how small the Rossby number has to be. A real current over a real seamount has a Rossby number of a few hundredths, not zero, and the seamount rises part of the way to the surface, not all of it. Whether that current goes round is not decided by a limit; it needs the next term.

The next term is the one the theorem dropped, and it enters through a conservation law rather than a balance. In a shallow layer of homogeneous fluid, rotating with Coriolis parameter $f$, a column of fluid of height $D$ carries a quantity

$$q = \frac{\zeta + f}{D},$$

where $\zeta$ is the column’s own spin about the vertical. That is potential vorticity, and it is conserved along each column’s path in inviscid flow. It is Kelvin’s circulation theorem applied to a thin column: squash the column and its cross-section widens, and the circulation round it, which cannot change, is spread over a larger area, so its spin falls. The effect is vortex stretching run backwards, and in a rotating layer the planet’s own spin $f$ is part of what gets stretched and squashed.

What a hill does to a column that crosses it #

Now let a uniform current of speed $U$ flow over a hill of height $h_b(r)$ standing on the floor of a layer of depth $H$. Upstream every column has height $H$ and no spin of its own, so every column carries $q = f/H$. Over the hill a column’s height is $H - h_b$, and keeping $q$ fixed requires

$$\zeta = f\,\frac{H - h_b}{H} - f = -\,\frac{f\,h_b}{H}.$$

The column acquires a spin opposite to the planet’s — clockwise in the northern hemisphere — in proportion to how much the hill squashes it. The hill holds an anticyclone.

That is the whole mechanism, and it is enough to solve the flow when the hill is low against the depth. In that limit every streamline that comes from upstream carries the same $q$, the relation above holds everywhere the fluid has been, and the flow is the uniform current plus the anticyclone the hill’s shape dictates. For an axisymmetric hill the anticyclone’s speed at radius $r$ follows from circulation being vorticity added up: the circulation round a circle of radius $r$ is the vorticity inside it, which is $-f/H$ times the volume of hill inside it, so

$$v_\theta(r) = \frac{f}{H\,r}\int_0^r h_b(s)\,s\,ds.$$

Measure lengths in the hill’s radius $a$ and speeds in $U$, and the flow depends on a single number,

$$\delta = \frac{h_0/H}{Ro}, \qquad Ro = \frac{U}{f a},$$

the hill’s height as a fraction of the depth divided by the Rossby number. The height and the rotation never enter separately. A hill a tenth of the depth in a current of Rossby number 0.05 is the same flow as a hill a hundredth of the depth at Rossby number 0.005.

The first figure is that flow for a Gaussian hill at $\delta = 2.2$, with the stream from the left and the hill’s e-folding radius dashed. Every streamline still crosses. On the upper side, the anticyclone adds to the stream and the lines crowd together; on the lower, it opposes the stream and the lines spread, bulging towards the hill’s centre. The flow is plainly not the potential flow round an obstacle and not the undisturbed current either: it is a current that has been told about the hill by the only quantity it carries.

The number that stops the stream #

Look along the line through the hill’s centre, across the stream, on the side where the swirl runs against it — the right-hand side, looking downstream, in the northern hemisphere. There the velocity is the stream minus the swirl, $U - v_\theta(r)$. The flow stops wherever the swirl reaches the stream’s speed, and the first place it can reach it is where the swirl is largest. So the onset of a closed region is set by one number per shape of hill, the peak of the swirl, and the threshold is its reciprocal:

$$\delta_c = \frac{1}{\max_r\,\hat v_\theta(r)}, \qquad \hat v_\theta = \frac{1}{r}\int_0^r \hat h(s)\,s\,ds.$$

The second figure draws $\hat v_\theta$ for three hills of the same height and radius. A flat-topped cylinder has all its volume as far out as possible; its swirl rises linearly to one half at its edge and then falls as $1/2r$, so its onset is exactly $\delta = 2.$ A Gaussian hill’s swirl peaks at 1.121 radii with 0.3191, from the condition $2r^2 e^{-r^2} = 1 - e^{-r^2}$, and its onset is $\delta = 3.134.$ A cone’s swirl peaks at three-quarters of its radius with 3/16, and its onset is exactly 16/3. The swirl integrals agree with their closed forms to one part in $10^{15}$, and scanning the velocity field along the axis confirms the onset directly: the Gaussian stream never stops at $\delta = 3.0$ and stops at 0.853 radii from the centre at $\delta = 3.3.$

The ordering is the result worth carrying. A hill’s height is not what decides. At equal height the cone is the hardest to block, needing more than two and a half times the cylinder’s $\delta$, because it holds a third of the volume and holds it near the centre where it makes little circulation. What decides is how much volume the hill puts inside each radius, divided by that radius — the enclosed circulation per unit of circumference — and a sharp peak is the least effective shape a hill can have.

In terms a reader can use: a Gaussian hill a hundredth of the depth blocks a current only if the Rossby number is below $0.01/3.134$, about 0.003. A hill a tenth of the depth blocks one below 0.03. The Taylor–Proudman theorem says that some small Rossby number will do; potential vorticity says which.

Two stagnation points, and the fluid between them #

Past the threshold the swirl exceeds the stream over a range of radii, and the axis carries two points where the flow stops: an inner one, where the rising swirl first reaches the stream, and an outer one, where the falling swirl drops back through it. A stagnation point carries a sign, and these two carry opposite ones. At the inner point the stream function has a minimum in both directions, so the flow circles it: it is the centre of an eddy. At the outer point the stream function has a minimum along the axis and a maximum across it, so it is a saddle, and the streamline through it — the separatrix — divides the fluid that passes from the fluid that stays.

The third figure is the Gaussian hill at $\delta = 4$, a little past onset. The eddy’s centre sits 0.593 radii from the hill’s centre, the saddle 1.956 radii, and the outlined region between them encloses 1.54 square radii of fluid that circles for ever and never leaves. The area is measured twice, once by filling the region of closed streamlines on a grid and once by tracing the separatrix from the eddy’s centre, and the two agree to 0.3 per cent.

What the figure shows that the textbook sentence does not is where the held fluid is. It is not a column standing on the summit. The closed region runs from 0.03 radii off the summit to nearly two radii out, all of it on the side where the swirl opposes the stream; the summit itself is crossed by fluid that arrives from upstream and leaves downstream, and so is the whole of the other flank. The column is displaced to one side because the anticyclone and the stream are superposed, and only on one side do they cancel.

It also is not rigid. The fluid inside circulates, anticyclonically, at speeds comparable to the current’s own, as streamlines that are not paths would warn a reader to expect of a steady picture: each closed line in the figure is a path that a parcel follows round and round. The Taylor–Proudman picture of a solid column comes from the limit, where the closed region grows to swallow the hill and its circulation is slow against the rotation. At finite Rossby number it is an eddy held in place by the hill.

Why the shape matters at onset and not afterwards #

A cone at $\delta = 8$, half again past its own onset, holds a different column. The fourth figure shows it reaching from 1.33 radii out on the opposing side to 0.13 radii across the summit, so on this sharp hill the summit is inside the held fluid. The cone’s anticyclone is concentrated near the centre, and its swirl peaks inside the hill, so its eddy is centred closer in — 0.317 radii — and wraps the peak before it grows outward.

The fifth figure follows the column’s area from onset upward for all three shapes, on logarithmic scales. Each is born at its own threshold with no area and grows steeply. Then the curves do something that the threshold alone would not suggest: the cylinder’s and the Gaussian’s run together, and the cone’s, drawn dashed at a third of its own $\delta$, lands on them. At $\delta = 10$ the cylinder holds 18.27 square radii and the Gaussian 18.17; the cone at $\delta = 30$ holds 18.29. At $\delta = 30$ against $\delta = 90$ the three differ by less than a twentieth of a per cent.

The reason is the far field. Outside the hill the anticyclone’s swirl is its total circulation divided by the circumference, and its total circulation is $-f/H$ times the hill’s whole volume. Seen from far enough away, every hill is a point vortex whose strength is its volume, and the cylinder and the Gaussian of equal height and radius have equal volume, $\pi h_0 a^2$, while the cone has a third of it. Far above onset the column is so large that its edge lies in that far field, and the edge only sees the volume; the details of the shape are inside it, where they no longer decide anything. A point vortex of circulation $\Gamma$ in a stream $U$ traps a region whose saddle sits at $\Gamma/2\pi U$ from it, and the computed columns settle on an area of 0.73 times the square of that distance.

So the answer to “what about the hill decides?” has two parts. Whether a column forms is decided by the peak of the enclosed circulation per unit circumference, which depends on the shape. How much fluid it holds, once it is large, is decided by the volume, which does not.

Which currents are blocked #

The criterion is a ratio, so on logarithmic scales of Rossby number and relative height every onset line is straight, and the regimes can be read off at a glance. The sixth figure places six settings on it. They are round, illustrative combinations of height, depth, width, speed and latitude, not surveys of any particular feature.

A 500-metre hill ten kilometres across, under a wind of ten metres a second at 45°N, with the ten kilometres of the troposphere above it, has a Rossby number near ten and a $\delta$ of 0.005: three decades below any onset line. The wind notices such a hill through gravity and stratification, not through the planet’s rotation. A seamount rising 500 metres in 4,000 metres of ocean, ten kilometres across, in a current of ten centimetres a second at 40°N, has $\delta = 1.17$ — the flow is bent but crosses. Slow that current to two centimetres a second and $\delta$ rises to 5.86, nearly twice the Gaussian onset: the same seamount now holds a column. Nothing about the seamount changed; the current’s speed moved it across the line.

A tall seamount half the depth, fifteen kilometres across, under a strong current of 30 centimetres a second, sits at $\delta = 2.09$, below the Gaussian’s line though near the cylinder’s. A laboratory tank rotating at one radian a second, with an obstacle a tenth of the depth and a flow of two millimetres a second, sits at $\delta = 5$, which is why Taylor could see his columns on a bench. And a seamount rising four of the ocean’s 4.8 kilometres in a five-centimetre current sits at $\delta = 36$, deep in the region where only the volume matters — but its height is five-sixths of the depth, far outside the low-hill assumption this model makes, and the figure marks the region where that happens.

Oceanographers have measured closed anticyclonic circulations over a number of seamounts, and the interest has been partly biological: fluid held over a seamount could hold plankton there long enough to feed the fish that gather on it. The measured circulations are generally weaker and leakier than this model’s, and the reasons are the model’s limits, which are specific.

What this flow leaves undecided #

The seventh figure is the ledger: each number above and the independent check it passed, including the potential vorticity of the constructed flow evaluated by finite differences, which matches the hill term to seven parts in a million.

The first limit is one the equations themselves point to. The derivation said every streamline comes from upstream and therefore carries upstream’s potential vorticity. Inside the column that is false: the closed streamlines come from nowhere. The steady inviscid equations require only that potential vorticity be constant along each closed line, and which constant is not decided — the same indeterminacy the vorticity nothing decides finds in a two-dimensional eddy, where a whole function is left free. The column drawn here is the member of that family continuous with the flow at onset, which is a natural choice and not a derivation. What actually sets the trapped fluid’s potential vorticity is its history and its friction.

The friction is specific, too. A column in contact with the floor sits on an Ekman layer, and an eddy over an Ekman layer spins down in a time set by the Ekman number’s square root, not by diffusion. A held column therefore lives as long as its supply of potential vorticity outlasts that spin-down, and a real seamount’s column is renewed and eroded at the same time.

The second limit is time. The steady solution does not say how the flow reaches it. Started from rest, a current over a hill first sheds an eddy of the opposite sense, cyclonic, which is carried away downstream, leaving the anticyclone behind; a current that reverses with the tide never finishes the process. The third is stratification. In a stratified ocean a hill’s influence reaches up only a height of order $f a/N$, with $N$ the buoyancy frequency, not to the surface, so the “column” becomes a cap over the hill whose height depends on the density profile. The homogeneous layer used here is the case of weak stratification, and on many seamounts it is not the case that applies.

The last limit is the one stated at the start: the hill must be low against the depth. The linear relation $\zeta = -f h_b/H$ is the first term of $f(H - h_b)/H - f$ only when $h_b \ll H$; for tall hills the squashing is stronger than linear, and the anticyclone correspondingly stronger, so the onset lines on the last figure are, if anything, conservative for tall seamounts.

Where the idea came from, and one place it was tried too far #

Proudman’s theorem and Taylor’s columns belong to 1916–1923. The question of when a column first forms at finite Rossby number was taken up by Herbert Huppert in 1975, in the quasi-geostrophic setting used here, and the stratified version by Nelson Hogg two years earlier. In 1961 Raymond Hide proposed that Jupiter’s Great Red Spot was a Taylor column over a feature on a solid surface below the clouds; the idea was dropped once it became clear that the planet has no accessible solid surface and the Spot drifts relative to the planet’s interior. It is the right mechanism in the wrong place, and a useful reminder that a column needs something to stand on.

The same conservation law runs through the rest of rotating fluid dynamics. It is why two-dimensional turbulence sends energy upward in scale in a rotating layer, and why a current crossing a sloping floor turns to follow the depth contours: a column that climbed the slope would be squashed and would have to spin against the planet to do it. And it is why the Coriolis force in a bath organises nothing — there the Rossby number is in the thousands, and $\delta$ for anything on a bath’s floor is far below every line on the sixth figure.

Still open: the column’s height in a stratified sea #

The homogeneous layer puts the column’s top at the surface. In a stratified ocean it is not, and the next calculation is the one that finds where it is. With buoyancy frequency $N$, potential vorticity involves the vertical spacing of density surfaces rather than the layer’s depth, and a hill’s anticyclone decays upward over a height $f a/N$. The questions are what the onset $\delta$ becomes as a function of $N h_0/f a$, whether it reduces to this essay’s 3.134 for a Gaussian hill as stratification vanishes, and how tall the trapped cap is over a seamount at a realistic buoyancy frequency — which would say whether the circulations measured over real seamounts are the caps this model predicts or something that friction and tides have made of them.