A stratified sea blocks a current sooner and traps less

The hill a slow current will not climb put a hill on the floor of a rotating layer of water and let a slow current cross it. A column of water squashed by the hill must spin against the planet to keep its potential vorticity, so the hill holds an anticyclone, and on one side that anticyclone runs against the current. Once the hill’s height over the depth, divided by the Rossby number, passes a threshold — exactly 2 for a flat-topped cylinder, 3.134 for a Gaussian hill — the current stops there, and fluid over the hill circulates for ever: a Taylor column, running from the floor to the surface.

That essay’s layer was homogeneous, and it ended on the sea as it actually is. The ocean is stratified, its density rising with depth, and in a stratified fluid potential vorticity is not about the column’s height but about the spacing of the density surfaces the column lies between. A hill squashes those surfaces near the floor and much less far above it, so its anticyclone should decay upward, over a height $fa/N$ set by the rotation $f$, the hill’s radius $a$ and the buoyancy frequency $N$. The questions are what the threshold becomes, how tall the trapped water is, and whether the circulations measured over real seamounts are the caps this predicts.

What a column is, when the sea is layered #

The homogeneous essay’s column was literal: a vertical tube of water from floor to surface, whose height the hill changed. In a stratified sea there is no such tube. A parcel of water sits between two density surfaces, and what the hill changes is the spacing between them — squeezing them together over its summit near the floor, and hardly at all a few hundred metres up. The potential vorticity each parcel carries is its spin plus the planet’s, divided by that local spacing, and it is conserved along the parcel’s path as long as the water is not mixed. A squeezed layer must spin against the planet, as the squeezed column did; a layer far above the hill is barely squeezed and barely spins.

That single change carries everything below. It is the same conservation law that makes two-dimensional turbulence send energy to large scales in a rotating layer and makes a current follow the depth contours of a sloping floor, and the same one whose absence explains why the bath does not know the hemisphere: in a bath the Rossby number is in the thousands and nothing on its floor comes near any threshold here. At the other extreme, a river is slow and wide enough for the planet to bend it a little, and a deep ocean current over a seamount is slow and wide enough for the planet to stop it.

A Laplace equation with the vertical stretched #

The model is the stratified form of the earlier essay’s. A uniform current $U$ flows over a Gaussian hill $h_0 e^{-r^2/a^2}$ on the floor of a layer of depth $H$ with a rigid lid, in water of uniform buoyancy frequency $N$, rotating at $f$. The Rossby number is small and the hill low against the depth, the quasi-geostrophic setting. Upstream the potential vorticity is uniform, so it is uniform everywhere the water has come from, and the perturbation streamfunction obeys

$$\nabla^2\psi + \frac{f^2}{N^2}\,\frac{\partial^2\psi}{\partial z^2} = 0,$$

a Laplace equation in which heights are stretched by $N/f$. Rotation makes a stratified fluid treat a vertical distance $z$ as a horizontal one $Nz/f$ — which is why the hill’s influence reaches up a height $fa/N$, the vertical distance equivalent to the hill’s radius.

The hill enters through the floor. A current climbing the hill lifts the density surfaces there, and carried along streamlines from upstream, the density equation at the floor sets the vertical shear of the streamfunction to $-(N^2/f)$ times the hill’s height. At the lid there is no vertical motion. For an axisymmetric hill a Hankel transform in the radius solves this exactly: each radial wavenumber $k$ decays upward as $\cosh(\mu k(H - z))/\sinh(\mu kH)$ with $\mu = N/f$. Everything depends on one number besides the earlier essay’s $\delta$,

$$B = \frac{NH}{fa},$$

the depth over the height $fa/N$ to which the hill is felt.

Three checks fix the solution. As $B \to 0$ the anticyclone is the homogeneous layer’s at every height and its threshold 3.134, to four parts in $10^7$. The vertical shear the solution produces at the floor matches $-(N^2/f)$ times the hill’s slope within 0.02 per cent at two stratifications and two radii, which checks the floor condition by differencing the solution rather than by construction. And as $B$ grows the threshold times $B$ converges, to 2.243 at $B = 20$ and $40$ alike.

The threshold falls, and becomes a Froude number #

The first figure is the main result, and it runs against the obvious guess. Stratification is usually said to insulate — to suppress vertical motion and stop the floor being felt above it — so it would be natural to expect a stratified sea to be harder to block. It is easier. The threshold falls from 3.134 as the stratification grows: to 2.95 at $B = 0.3$, 1.92 at $B = 1$, 0.74 at $B = 3$ and 0.22 at $B = 10$.

Why rotation should drop out is worth a paragraph, since the whole subject began with rotation making hills block currents. In a deep stratified sea the hill lifts the density surfaces near the floor by an amount set by its height, and the anticyclone at the floor is the geostrophic flow those lifted surfaces demand. The height of the disturbance scales as $fa/N$ and its horizontal size as $a$, and in the velocity that results the $f$ from the geostrophic balance and the $f$ in the decay height cancel: the floor swirl scales as $Nh_0$. The rotation sets the shape of the disturbance and the Coriolis force sets its direction — anticyclonic — but its strength is the stratification’s.

The reason the threshold falls is where the anticyclone is. Homogeneous, the hill’s squashing is shared by the whole column from floor to surface, and its anticyclone is the same at every height. Stratified, the squashing of density surfaces is concentrated near the floor, and so is the anticyclone — it is weaker aloft and stronger at the floor, where the current first stops. The stratification does insulate the upper water; it does it by concentrating the hill’s effect in the lower water, which makes the blocking there easier.

In deep water the threshold falls as $2.243/B$, and $\delta B$ is $(h_0/H)(fa/U)(NH/fa) = Nh_0/U$. The rotation drops out: a current over a hill in a deep stratified sea is blocked when

$$\frac{N h_0}{U} > 2.243,$$

an inverse Froude number based on the hill’s height. It is the stratified cousin of the shallow-water Froude number that really is one at its threshold: there a disturbance can no longer travel upstream, here the water near the floor no longer has the kinetic energy to lift itself up the density gradient the hill imposes, and goes round. The waves that carry the hill’s influence upward in a stratified fluid are the internal waves whose frequency is set by N, and the height fa/N is where rotation cuts them off. Rotation still decides how the blocked water behaves — it circulates in an anticyclone rather than simply stagnating — but not when blocking starts.

At the floor the anticyclone tightens #

The second figure looks at the floor itself, where the current stops. Homogeneous, the anticyclone’s speed at radius $r$ is the circulation of all the hill’s volume inside $r$ spread round a circle of that radius — circulation being vorticity added up — so it peaks at 1.12 hill radii and falls slowly outside, still at half its peak three radii out. Strongly stratified, the floor no longer feels the hill’s whole volume at once. It feels the hill’s slope nearby much more than its bulk far away, since a distant part of the hill acts on the floor only through water higher up, where the influence has already decayed. The swirl peaks nearer the centre, at 0.84 radii, and falls to a seventh of its peak by three radii.

That tightening is the stratified sea’s version of a familiar fact about a balance that is its own error: geostrophic flow is decided by a field — potential vorticity here — that is set non-locally by an inversion, and what stratification changes is how far the inversion reaches. Homogeneous, it reaches the whole layer. Stratified, it reaches $fa/N$ upward and a comparable distance sideways, and a hill several radii away is invisible at the floor.

The anticyclone fades upward over fa/N #

The third figure shows the confinement. Weakly stratified, at $B = 0.3$, the anticyclone hardly changes with height: the earlier essay’s column. At $B = 1$ it falls to half its floor value at the surface. At $B = 3$ and $B = 10$ it falls by a factor $e$ within 0.62 of fa/N above the floor — a fifth of the depth and a sixteenth of it — and the same 0.62 at every stratification strong enough for the lid not to matter. The peak also moves inward as it gathers at the floor, from 1.12 hill radii homogeneous to 0.84.

That decay is what makes the next number meaningful. The trapped water is only as tall as the height at which the anticyclone still outruns the current, and that height is set by $fa/N$, not by the depth.

The cap, and how tall it grows #

The fourth figure measures the cap. Past the threshold the current stops at the floor over a ring of radii, and the trapped region extends upward as long as the anticyclone’s peak still exceeds the current. In deep water the result is universal when measured in $fa/N$: the cap reaches 0.23 of it at one and a half times the threshold, 0.69 at three times and 1.8 at ten times, and the curves for $B = 3$ and $B = 10$ lie on each other. In a weakly stratified or shallow sea the cap reaches the lid well past the threshold and becomes the homogeneous essay’s full column; at $B = 1$ that happens by three times the threshold.

The growth has a one-line reading. The anticyclone’s peak decays upward roughly exponentially, falling by $e$ every 0.62 of $fa/N$, and the cap ends where the decayed peak has fallen to the current’s speed, so its height should be about $0.62\,\ln(\delta/\delta_c)$ in units of $fa/N$: 0.25, 0.68 and 1.43 at one and a half, three and ten times the threshold, against the 0.23, 0.69 and 1.8 computed. The last is the furthest off because high above the floor the decay slows — the longest-wavelength part of the hill’s field decays most slowly — and a strongly overdriven cap reaches into it.

So a stratified Taylor column is a cap: a lens of water over the hill whose height is set by the stratification and grows only slowly — roughly as the logarithm of the overdrive — as the current slows. The water above it flows over the hill barely disturbed.

A seamount #

The fifth figure puts numbers on a seamount of a common size: 800 metres high, ten kilometres in radius, on a floor 4.5 kilometres deep at mid-latitude. With no stratification it would block currents below 5.7 centimetres a second. In deep-ocean stratification, $N$ around $10^{-3}$ per second, it blocks currents up to about 36 centimetres a second — six times faster — and at $N = 2\times10^{-3}$, about 71. The deep ocean’s currents, a few centimetres a second, are blocked by such a seamount with a large margin.

In a current of 5 centimetres a second the cap is the full depth when the water is nearly unstratified, and falls as the stratification grows: about 1,490 metres above the floor at $N = 10^{-3}$ and 1,140 at $2\times10^{-3}$ — 690 and 340 metres above the 800-metre summit. That is a cap reaching some hundreds of metres above the seamount’s summit — the scale of the anticyclonic caps measured over real seamounts, which are commonly confined to the lower part of the water column and extend a few hundred metres above the summit.

What the cap does, and what the model cannot follow #

A trapped cap is water that stays over the seamount while the current passes, and it matters for the same reason the earlier essay’s column did: whatever is in it stays. Seamounts are known for high concentrations of plankton and fish above their summits, and one explanation is exactly this — retention in an anticyclonic cap that isolates water over the summit for weeks. The calculation says how tall that retained water is, which is the number that decides whether vertically migrating plankton can stay inside it.

What it cannot follow is how long the cap lasts. Tides slosh water over a seamount twice a day, often faster than the mean current; friction at the floor spins the cap down; and the quasi-geostrophic model needs the seamount to be low against the depth, where this one is a sixth of it. All three erode the idealised cap, and measured caps are both weaker and more transient than this steady solution.

What was checked #

The ledger holds three checks: the unstratified limit against the homogeneous layer’s swirl and threshold, to four parts in $10^7$; the floor’s density condition against the hill’s own slope, within 0.02 per cent; and the deep limit’s threshold times $B$ agreeing at two stratifications to three parts in $10^5$.

What the smooth cap leaves out #

Tides. The mean current is steady here; over most seamounts the tidal current is as large or larger, and a cap can be formed and flushed each tidal cycle.

Friction. A bottom Ekman layer spins the anticyclone down over a time $H/\sqrt{\nu f}$ in a homogeneous sea and over less in a stratified one, where the spin-down is confined to the cap.

Height. Quasi-geostrophy needs the hill low against the depth. A seamount rising a third of the way to the surface is outside it, and the numbers for it are the model’s trend rather than a prediction.

Variable N. The ocean’s stratification is strongest in the thermocline and weak at depth; a uniform $N$ is the simplest case, and a real profile puts more of the decay where the stratification is strongest.

The convention: δ and B #

$\delta = (h_0/H)/\mathrm{Ro}$ with $\mathrm{Ro} = U/fa$, as in the homogeneous essay. $B = NH/fa$ is the depth over the stratified decay height $fa/N$. Heights are above the floor; swirl speeds are in units of $fh_0a/H$, the homogeneous anticyclone’s scale; the hill is Gaussian with $e$-folding radius $a$. The seamount example takes $f = 10^{-4}$ per second, the Coriolis parameter near 43° of latitude, and a uniform buoyancy frequency over the whole depth, which a real ocean has only below its thermocline.

Who worked it out #

Hogg in 1973 solved the stratified Taylor column in quasi-geostrophic theory and introduced the stratification parameter; Huppert’s homogeneous threshold was two years later. Chapman and Haidvogel in the 1990s computed the formation of caps over tall seamounts numerically, including their spin-up and their erosion by tides; and the retention of plankton over seamount caps has been argued from observations since the work of Genin and Boehlert.

Still open: a cap that the tide flushes #

The steady cap here is the limit of a slow, steady current. The next calculation adds a tidal current oscillating at the semidiurnal frequency on top of the mean, follows the quasi-geostrophic potential vorticity through a tidal cycle, and asks how much of the trapped cap survives each cycle: whether there is a tidal amplitude, relative to the mean current, below which the cap is only rocked and above which it is swept off the summit twice a day — which would say which of the world’s seamounts can hold water long enough for the retention the ecology relies on.