The layer a parcel cannot leave

A parcel of air is lifted a hundred metres and let go. Whether it comes back is the question the whole of atmospheric stability is built on, and the answer contains no density, no pressure and no temperature — only two gradients, and which of them is steeper.

Why the comparison is between two slopes #

A parcel that moves does two things at once. It finds itself at a new pressure — one set by the weight of everything above it — so it expands or is compressed; and it does that faster than heat can leak in or out, which is the condition that makes a sound wave adiabatic too, so the change is adiabatic. Its temperature therefore falls at a rate fixed by thermodynamics and by nothing about the weather:

$$\Gamma_d = \frac{g}{c_p} = \frac{9.80665}{1005} = 9.76 \text{ K per kilometre}.$$

The environment’s temperature also falls with height, at whatever rate it happens to have. Nothing forces the two to agree, and the whole of the classification is the difference between them.

An isothermal column thins exponentially with a scale height $kT/mg$, and a real atmosphere is not isothermal — which is where the environment’s profile comes from and why the comparison in this essay is between two profiles rather than against a fixed background. The parcel’s own temperature falls at one rate as it rises and the surroundings’ falls at another, and stability is entirely a question of which falls faster.

Writing the restoring force out gives the frequency directly. A parcel displaced a height z has a temperature excess $(\Gamma - \Gamma_d)z$ over its surroundings, so a buoyant acceleration $g(\Gamma-\Gamma_d)z/T$, so

$$N^2 = \frac{g}{T}(\Gamma_d - \Gamma).$$

Positive means stable and $N$ is a frequency; negative means unstable and $|N|^{-1}$ is a growth time. The same expression covers both, and the sign flips exactly at the adiabat.

Two things are worth separating in that expression. The factor $g/T$ carries the units and varies hardly at all — it is 0.034 per kelvin per second squared at 288 K and 0.033 at 300. Everything interesting is in the bracket, which is a difference of two lapse rates in kelvin per kilometre, and it runs from about +15 for a strong inversion to about −3 for a superadiabatic layer near a hot surface. So N² spans a range of about six between the extremes, N a range of about two and a half, and the period a range of about two and a half in the other direction — from four minutes to ten. The atmosphere’s clock is remarkably uniform, and the interesting cases are the two ends where the bracket approaches zero and changes sign.

The ocean does the same arithmetic with a different variable, and buoyancy there is the same upthrust. There the adiabatic reference is tiny — compressing seawater warms it by about 0.1 K per kilometre — so the criterion is very nearly the plain density gradient, and N is set by temperature and salinity together. Thermocline values run to 0.01 per second, a period of about ten minutes, which is the same number the atmosphere gives for a completely different reason.

One gradient instead of two #

Comparing two slopes is awkward to do by eye on a sounding, and meteorology long ago replaced it with a single variable that does the subtraction in advance.

Take a parcel at height $z$ and ask what temperature it would have if it were brought adiabatically down to a reference pressure — sea level, by convention. That is its potential temperature,

$$\theta = T\left(\frac{p_0}{p}\right)^{R/c_p},$$

with the exponent 0.286 for dry air. A parcel moved adiabatically does not change its $\theta$, because $\theta$ was defined as the thing that does not change under exactly that operation. So a column’s stability becomes a question about one profile rather than two: if $\theta$ increases with height, a lifted parcel finds itself among air of higher $\theta$ and returns; if $\theta$ decreases with height, it keeps going. The buoyancy frequency is then

$$N^2 = \frac{g}{\theta}\frac{\mathrm{d}\theta}{\mathrm{d}z},$$

which is the expression above with the subtraction already performed.

Nothing has been added — the two forms are algebraically the same statement — but the picture is different, and it is the one worth carrying. A stable atmosphere is one whose potential temperature increases upward, which means the air is already sorted, lightest on top, in exactly the way a settled column of liquids is. Convection is then the ordinary business of a fluid that has been stacked in the wrong order, and the dry adiabat’s 9.76 K per kilometre is not a rule about air but the coordinate change that makes the sorting visible.

What the parcel does #

The overshoot is the point. A restoring force does not produce a return to equilibrium; it produces an oscillation about it. Damping is what produces a return — the three regimes of it — and a parcel of air in a clear atmosphere has almost none — mixing with its surroundings is slow compared with the period, and radiation is slower still. So a displaced parcel rings.

That divergence at the adiabat is worth pausing on, because it is where the two behaviours meet. Approaching it from the stable side the period runs to infinity; from the unstable side, the growth time does. A neutral column is the only one in which a displaced parcel neither returns nor runs away, and it is the state a vigorously convecting column is driven toward — convection removes the very instability that drives it, and stops when the profile has been beaten flat onto the adiabat.

And a stable column is not a still one. Stability means a displaced parcel comes back; it says nothing about whether parcels are being displaced. A stably stratified layer with something stirring it — a mountain below, a jet above, a front moving through — is full of oscillations at frequencies up to N, all of them going nowhere on average and all of them mixing a little. Calling such a layer “stable” is a statement about its response and not about its state.

Reading it as a ceiling #

That linear relation between speed and height is the reason a plume in a stable layer flattens into a sheet at a definite level rather than thinning away. Everything arriving with the same buoyancy reaches the same height, overshoots, comes back, and spreads sideways — because sideways is the one direction with no restoring force in it.

The other reading is what happens when the ceiling is a bargain. Pollution released at the ground into a stable layer is confined to a depth set by the same w/N — the vertical speed of whatever is carrying it, divided by the buoyancy frequency. A morning inversion has a large N and a small w, so the mixing depth is a few tens of metres and the concentration is whatever the source divided by that depth gives; by mid-morning the sun has warmed the ground, the profile has been driven toward the adiabat, N has fallen through zero and the depth is a kilometre. The concentration falls by a factor of thirty in an hour with no change in the source at all.

Shear can overturn a layer the parcel argument calls stable #

Everything above is done in a column at rest, and the atmosphere is not at rest. Once there is a wind that changes with height, stability stops being decided by $N$ alone.

The reason is an energy account rather than a force one. Overturning a stable layer costs work, because it means lifting dense air over light — and a sheared flow has kinetic energy available to pay for it, an amount set by how fast the wind changes with height. The comparison between the two is a dimensionless ratio,

$$\mathrm{Ri} = \frac{N^2}{(\mathrm{d}u/\mathrm{d}z)^2},$$

the stratification’s stiffness against the shear’s supply. Where it is large the layer is safe; where it falls below a quarter, small disturbances can grow and the layer can break into billows and mix. A quarter rather than one is the result of a proper stability calculation rather than a dimensional estimate, and it is a necessary condition rather than a sufficient one — a layer with $\mathrm{Ri}$ below a quarter somewhere may overturn, and one above it everywhere cannot.

That is the ordinary cause of clear-air turbulence, and it is why an aircraft can be thrown about in a layer whose temperature profile is stable by every measure in this essay. It is also why the sharp inversions that trap pollution are simultaneously the most likely places to find mixing: an inversion has a large $N$, and a strong inversion usually caps a layer with a large wind change across it, so both terms of the ratio are large and which one wins is not decided by either alone.

The other integral: what a whole column has stored #

The ceiling $w/N$ answers what a given push buys in a stable layer. The opposite question — what an unstable column will do on its own — has an answer of the same kind, and it is the number a forecaster actually looks at.

Integrate the buoyant acceleration over every height at which a lifted parcel is warmer than its surroundings, and the result is an energy per unit mass: the work the column will do on a parcel that gets started. Converting it to a speed gives $w_{\max} = \sqrt{2E}$, so a column holding two thousand joules per kilogram would drive an updraught at sixty metres per second.

Measured updraughts reach perhaps half of that, and the gap is the two omissions this essay already owns. Entrainment dilutes the parcel with the air it passes through, and the condensed water it is carrying has weight of its own that the buoyancy calculation ignores. So the integral is an upper bound, in the same direction and for the same reasons as the ceiling — which is the useful property for a forecast, where the question is whether a column can do something rather than exactly what it will do.

What the model has in it and what it does not #

The parcel is assumed not to mix. Everything above treats the parcel as a sealed bag exchanging pressure with its surroundings and nothing else. A real thermal entrains the air it passes through, which dilutes its buoyancy and makes it rise less far than w/N says. The parcel argument is therefore an upper bound on the ceiling and a lower bound on the stability, and both errors point the same way.

And there is no water in it. A rising parcel that reaches saturation begins to condense, at the temperature the vapour-pressure curve fixes, the latent heat released warms it, and it then cools at a smaller rate — around 5 K/km rather than 9.76, depending on temperature. A column can therefore be stable to dry displacements and unstable to saturated ones, which is a condition with no counterpart in the figures above and is the whole of why a cloud can exist in an atmosphere that is not convecting.

Boiling a kilogram of water costs 2,260 kJ against the 4.2 kJ per kelvin needed to warm it, so condensing a gram of water into a kilogram of air warms that air by about half a kelvin. That is the latent heat doing the work: a rising parcel that begins to condense stops cooling at the dry rate and cools at a slower one, so it can remain warmer than its surroundings and keep rising — which is the difference between a fair-weather cumulus and a thunderstorm.

The dry adiabat itself is an idealisation with a domain. g/c_p assumes the parcel exchanges no heat, which is good for the minutes an oscillation takes and poor over a day; it assumes the parcel’s pressure equals its surroundings’ at every instant, which is good because pressure equilibrates at the speed of sound and the parcel moves at metres per second; and it assumes c_p is constant, which it is to a fraction of a per cent through the troposphere. Of those three, only the first ever fails on the timescales the figures draw, and it fails by radiation on a scale of days.

And the environment is assumed not to move while the parcel does. A displaced parcel is treated as an infinitesimal intruder in a column that carries on as though nothing had happened. Once enough parcels are moving at once — which is what convection is — the profile they are oscillating in is the profile they are changing, and the frequency computed from a sounding taken before the event is not the frequency of anything afterwards.

The linearisation is good and it is not exact. The acceleration is $g(T_p - T_e)/T_e$, which is not proportional to the displacement, because the denominator changes with height too. The integrations above use the full expression and the periods agree with 2π/N to about a per cent at 300 metres of displacement — which is the honest size of the error and the reason the figures integrate rather than assume.

The same stability question is easier to see for a body than for a parcel. An object in a stratified fluid finds the level where its density matches, and whether it returns after being displaced depends on whether its density changes with depth faster or slower than the fluid’s. A parcel is that problem with the object made of the same stuff as the surroundings, which is what makes the comparison a comparison of rates rather than of values.

The history, which is one measurement and one argument #

The adiabatic lapse rate was worked out before anybody had been up to check it. Kelvin derived g/c_p in 1862 from the same thermodynamics as above; Reye and Hann established the parcel argument in the years after; and Väisälä and Brunt arrived at the frequency independently in the 1920s, which is why it carries both names.

What settled it was not the derivation but the balloon. Systematic soundings from the 1890s onward returned a lapse rate that was consistently below the adiabatic one through the troposphere — about 6.5 K/km rather than 9.8 — and consistently zero or negative above about eleven kilometres. The first of those says the troposphere is stably stratified nearly everywhere, which is why weather is confined to it. The second was completely unexpected and is the discovery of the stratosphere: a layer so stable that vertical motion in it is negligible, which is exactly what N² being large means.

The frequency as an instrument #

The buoyancy frequency is the natural clock of a stratified fluid, and everything that oscillates in one has a period longer than 2π/N. That bound is not obvious and it is exact: a wave restored by buoyancy has a frequency $N\cos\phi$ where φ is the angle its motion makes with the horizontal, so the fastest possible oscillation is purely vertical and everything else is slower.

Internal waves have a dispersion relation stranger than any surface wave’s: the frequency depends on the direction of travel and not on the wavelength at all. So a disturbance at a given frequency propagates along a fixed angle to the horizontal, whatever its size — which is why internal waves in the ocean and the atmosphere form beams rather than spreading fronts, and why they can be seen as clean diagonal stripes in a laboratory tank.

There is a neat consequence of that bound. Since nothing restored by buoyancy can oscillate faster than N, a stratified layer is a low-pass filter for its own motion: a disturbance forced at a frequency above N cannot propagate as a wave and decays with distance instead, exactly as a wave below a cutoff does in a pipe or a plasma. The atmosphere therefore has a maximum internal-wave frequency of about one cycle per ten minutes, and everything faster than that is sound.

The measurement runs backwards. N is hard to measure directly and easy to infer, because a temperature profile is easy to take. Every radiosonde ascent is a measurement of Γ against height, on a column whose pressure profile is an exponential with a scale height, and N² follows from it point by point — so the oscillation period of every layer of the atmosphere is known routinely, and the layers that trap pollution are the ones with the shortest periods.

One more thing the two-gradient form makes obvious. The criterion has no length scale in it. A parcel displaced a metre and a parcel displaced a kilometre are both governed by the same N, and both oscillate at the same period — which is the signature of a linear restoring force and the reason the period does not depend on the amplitude. Everything else about the atmosphere has a scale height in it; this does not, and it is why one number classifies a layer of any thickness.

What a very stable layer is worth #

The stratosphere is the extreme case of this page’s criterion, and its consequences are worth stating because they are what a large N² actually buys.

Above the tropopause the temperature stops falling and begins to rise, so $\Gamma$ is negative while the parcel still cools at 9.76 K/km on being lifted. The difference between the two slopes is therefore larger than anywhere in the troposphere, N² is correspondingly large, and vertical displacement is opposed harder than it is anywhere else in the atmosphere.

Nothing mixes across it. Aerosol injected into the troposphere is rained out in about a week; the same material placed in the stratosphere stays for one to three years, because there is no vertical motion to carry it down and no precipitation up there to scavenge it. That single contrast is why a volcano that reaches the stratosphere cools the planet measurably for two summers and one that does not reach it does nothing at all, and it is the whole reason the aerosol geoengineering proposals specify an injection altitude rather than a quantity.

And it is why airliners cruise where they do. Vertical motion is what turbulence consists of, so a layer that suppresses vertical motion is smooth. The cruise altitudes of long-haul aircraft sit at or just above the mid-latitude tropopause, which is a fuel-efficiency choice and a comfort choice at once — the aircraft climbs out of the layer where the two gradients are close together and into the one where they are furthest apart.

Where this ladder goes next #

This rung has one parcel moving vertically in a column at rest. The next puts many of them together and lets the motion be a wave: an internal gravity wave, which travels through the interior of a stratified fluid rather than along a surface, carries energy at right angles to its own crests, and is the mechanism by which a mountain range hundreds of kilometres upwind shows up as a row of lens-shaped clouds standing still in a moving airstream.