The pressure that only knows depth

A fluid at rest is the simplest thing in this collection that still has something to say. Nothing is moving, nothing is changing, and there is exactly one quantity to keep track of. What makes it worth an essay is that the one quantity behaves in a way almost nobody predicts correctly the first time.

That figure is the whole of hydrostatics compressed into one picture, and it is a picture most people refuse to believe. The obvious account — the base holds up the water, so the pressure on it is the weight divided by the area — gives three different answers and is wrong in all three cases. What is actually true is that pressure depends on depth, and that the vessel’s shape enters nowhere at all.

Why a still fluid has no shear, and what that forces #

Start with what makes a fluid a fluid. A solid can be pushed sideways along a surface and stay where it is: it develops a shear stress that opposes the push, and the material sits there under strain. A fluid cannot. Whatever sideways stress is applied, it flows, and it keeps flowing until the stress is gone. That is very nearly the definition of the word.

The consequence for a fluid at rest is immediate and severe. If the fluid is not moving, there is no shear stress anywhere in it — because any shear stress would have set it moving, and by assumption it has not. So the only force one part of a still fluid may exert on another is along the normal to the surface between them. Not sideways. Perpendicular, and pushing.

That single restriction is enough to force the next result, which is the one people find strange.

Consider a tiny wedge of fluid inside the bulk, with faces pointing in three different directions. The forces on those faces are all normal to them, with magnitudes equal to the pressure on each face times its area. Now shrink the wedge. Its volume falls as the cube of its size and its faces as the square, so the weight of the wedge becomes negligible compared with the surface forces long before the wedge disappears. In that limit the surface forces must balance among themselves.

Write down that balance and something falls out: the pressures on the three faces must be equal, whatever directions they point in. The geometry cancels. The pressure at a point in a still fluid is one number, the same in every direction, and it is not a vector.

This is worth pausing over because the electric field at a point is three numbers, the velocity of a flow is three numbers, and pressure looks superficially like the same kind of thing. It is not. It is closer to temperature: a single value attached to each point of space, which is what a scalar field means. The force is what has a direction, and that direction is supplied by the surface being pushed on rather than by the fluid doing the pushing.

Draw the pressure acting on each face of a submerged block and the arrows point in different directions — in on the sides, down on the top, up on the bottom — all of them produced by one scalar field. They differ in direction because the faces do, not because the pressure does. The sideways pairs cancel exactly, being at equal depths; the vertical pair does not, because the bottom face is deeper than the top one and the pressure there is larger by $\rho g h$. That difference, and nothing else, is buoyancy.

The law, in one line #

Now take a column of fluid of cross-section $A$ standing between depth $h$ and depth $h + \mathrm{d}h$. Three vertical forces act on it: pressure pushing up on the bottom face, pressure pushing down on the top, and its own weight. Balance them and everything but the essentials cancels:

$$\frac{\mathrm{d}p}{\mathrm{d}h} = \rho g.$$

For a fluid whose density does not change, integrating is a single step:

$$p = p_0 + \rho g h.$$

The pressure at depth $h$ is whatever it was at the surface, plus the weight of a column of fluid one unit in area and $h$ tall. Nothing about the vessel’s width appears, because $A$ cancelled. Nothing about its shape appears, because the argument was made on a column and every column gives the same answer.

The number is larger than it feels. Ten metres of water is about one atmosphere, which is why a diver at thirty metres is at four atmospheres absolute and why the pressure quoted for a car tyre — a couple of hundred kilopascals — is the same as a twenty-metre dive.

The paradox, which is not one #

Return to the three vessels. The base pressure is $\rho g h$ in each, so the total force on each base is $\rho g h A$ with $A$ the base area, which is equal in all three. But the weights differ. Where does the difference go?

Into the slanted walls. In the flaring vessel the walls lean outward, so the fluid pushes outward and upward on them — and by reaction they push down on the fluid, adding to what the base must hold. In the narrowing vessel the walls lean the other way and carry part of the load, so the base holds less than the contained weight. The vertical forces balance in all three; it is only the base that fails to see the difference.

This has a name, the hydrostatic paradox, and it stopped being a paradox as soon as somebody drew the wall forces. It also has a consequence that is genuinely useful: a very narrow tube of water, containing almost no water at all, can burst a large barrel, provided the tube is tall enough. Pascal is said to have done exactly that. Whether or not he did, the arithmetic permits it — a tube ten metres high adds an atmosphere to the barrel below regardless of how thin it is.

Which zero is being counted from #

There is a bookkeeping detail here that causes more confusion than it deserves, and it is worth clearing out before the loads are computed.

The $p_0$ in the integrated law is whatever pressure sits on the free surface, and for an open vessel that is the atmosphere: about 101 kilopascals, which is the weight of the whole column of air standing above it. A depth gauge that reads zero at the surface is therefore not measuring the pressure. It is measuring the amount by which the pressure exceeds the atmosphere’s, and that difference is called the gauge pressure. The full quantity, counted from vacuum, is the absolute pressure.

Most of the time the distinction cancels and can be ignored, because the atmosphere presses on both sides of whatever is being considered — on the water in a dam and on the dry face of its wall, on the inside of a tyre and on the outside. What survives is the difference, so working in gauge pressure is not a simplification but the correct accounting.

It stops cancelling in two places. The first is any calculation involving a gas, because a gas’s density depends on its absolute pressure and not on the excess over its surroundings; doubling the gauge pressure in a nearly-empty tyre does not double the amount of air in it. The second is anywhere a fluid is asked to go below zero absolute, which is where a liquid can be pulled apart rather than boiled, and where the ordinary intuition about pressure being a push stops applying at all.

The instrument that made all this legible is the manometer, and it is nothing but this page’s law read backwards. Connect a pressure to one arm of a U-tube of liquid and the other arm to a reference, and the fluid settles with a height difference of $\Delta p / \rho g$. The pressure is then a length — which is why blood pressure is quoted in millimetres of mercury and rainfall pressure in millibars that were once inches of it. A unit of pressure named after a distance is a fossil of this figure.

What the wall of a dam has to hold #

The pressure grows with depth, so the load on a vertical wall is not uniform. It is a triangle: zero at the surface, greatest at the bottom, and linear in between. Two questions follow, and the second is the one that matters structurally.

The total force per unit width is $\int_0^h \rho g z\,\mathrm{d}z = \tfrac12 \rho g h^2$. It grows as the square of the depth, which is why dams get so much thicker toward the bottom and why doubling a reservoir’s depth quadruples what the wall must resist.

The line of action is the more interesting half. It sits where the load’s centroid is, at

$$\bar{z} = \frac{\int_0^h z \cdot \rho g z\,\mathrm{d}z}{\int_0^h \rho g z\,\mathrm{d}z} = \frac{2h}{3},$$

two thirds of the way down, not halfway. A designer who assumes the resultant acts at mid-depth underestimates the overturning moment about the base by a third. The figure computes both integrals rather than printing the answer, so the two-thirds is something the picture found rather than something it was told.

Where the constant density went #

Every result above assumed $\rho$ does not vary with depth, and that assumption is doing more work than it looks. It is excellent for water and terrible for air, and the difference is worth stating precisely because it is the same equation in both cases.

Water’s bulk modulus is about 2.2 gigapascals, so a pressure of one atmosphere compresses it by roughly one part in twenty thousand. At the bottom of the deepest ocean trench, at about 110 megapascals, water is around five per cent denser than at the surface — enough to matter for an ocean model and irrelevant for a swimming pool.

Air is a different case entirely, because its density is proportional to its pressure. Substituting that into $\mathrm{d}p/\mathrm{d}h = \rho g$ turns a linear law into an exponential one, and the result is the barometric formula, which is the same hydrostatic balance solved for a compressible fluid. The two essays are about one equation with two constitutive laws stuffed into it.

Where the density is not constant the same balance gives a different curve. In air, density tracks pressure, so the equation $dp/dz = -\rho g$ becomes $dp/dz = -pg/RT$ and the solution is exponential rather than linear: pressure falls by a fixed factor per unit height instead of by a fixed amount. The scale height — about 8.4 km at 288 K, 7.3 km at 250 K — replaces the depth as the number that matters, and the two curves for those two temperatures diverge by a factor of two by 20 km. Water is a thousand times denser and a hundred thousand times less compressible, which is the whole reason the linear law is exact enough to build dams on.

What the picture is standing on #

There is a deeper assumption underneath all of this, and it is not about density. Every statement here treats the fluid as a continuous substance with a pressure defined at each point. Water is not continuous; it is molecules, and the pressure it exerts is the momentum they deliver per second as they arrive at a surface and bounce.

That reconciliation matters because it says when the continuum picture may be used. A pressure defined “at a point” is really an average over a region containing enough molecules for the fluctuations to vanish. In a litre of water that is a spectacularly good approximation — the numbers involved are of order $10^{25}$ — and in a rarefied enough gas it fails outright, because the region needed for a stable average becomes as large as the apparatus.

Underneath any of this is a rate of arrival. Molecules strike the wall, each delivers its momentum, and the pressure is the sum per unit area per unit time. The steady number the continuum picture uses is the average of that count, and it is steady only because the count is enormous: a square micrometre of wall in water at room temperature is struck something like $10^{20}$ times a second, so the fractional fluctuation is $10^{-10}$ and no instrument sees it. Shrink the area far enough and the pressure stops being a number and becomes a distribution.

The siphon, and the height it cannot exceed #

A tube filled with liquid, running from a full vessel up over a rim and down to a lower one, moves liquid uphill and keeps doing it without a pump. The usual explanation is a chain of liquid being pulled over the top by the longer side, and the usual explanation is wrong in a way this page can settle.

Nothing is pulling. Follow the pressure. At the surface of the upper vessel the liquid is at atmospheric. Going up the short arm to the crest, the pressure falls by $\rho g$ times the height climbed. Going down the long arm it rises again, by $\rho g$ times a larger drop, so the liquid arrives at the outlet at above atmospheric and flows out. The whole driving difference is $\rho g$ times the difference between the two liquid levels, and the crest’s height cancels out of it entirely — which is why a siphon’s flow rate depends on how far apart the two surfaces are and not on how high the tube goes over the rim.

What the crest’s height does decide is whether the siphon works at all. The pressure there is atmospheric minus $\rho g H$, so raising the crest lowers it, and a pressure cannot be lowered indefinitely: at somewhere near the liquid’s vapour pressure the liquid boils, the column breaks, and the siphon stops. For water at room temperature that puts the ceiling at about ten metres above the upper surface.

Ten metres is the same number that limits a suction pump, and it is a piece of hydrostatics that was known practically long before it was understood — Florentine well-diggers could not lift water more than about that far, whatever pump they built, and Torricelli’s explanation of why is the experiment that produced the first laboratory vacuum. The ceiling is not a property of the pump. It is the height of a column of water that one atmosphere can support.

The same law, inside an animal #

A vertical column of fluid produces $\rho g h$ whether it is in a vessel or in a vessel of the biological kind, and the numbers for a standing body are large enough to have shaped the anatomy.

Blood is a little denser than water, so every metre of height is about eleven kilopascals, or eighty millimetres of mercury in the units the clinic uses. A standing person’s feet are some 1.3 metres below the heart, so the arterial pressure there exceeds the pressure at the heart by around a hundred millimetres of mercury — roughly doubling it — and the veins in the legs must hold that head without ballooning. They do it with valves that break the column into segments and with the squeezing of the surrounding muscle, and when a person stands still long enough for both to fail, blood pools, the return to the heart falls, and they faint. Compression stockings and the practice of tensing the legs before standing are both interventions on one term in this page’s equation.

The extreme case is a giraffe. Its head is about two metres above its heart, so simply delivering blood to the brain costs an extra hundred and fifty millimetres of mercury before any is available to drive circulation — which is why a giraffe’s arterial pressure is roughly twice a human’s, the highest of any land animal, and why its heart wall is correspondingly thick.

The interesting part is what happens when it drinks. Lowering the head swings it from two metres above the heart to nearly two metres below, a change of some three hundred millimetres of mercury in the pressure the brain’s vessels would otherwise see. A giraffe has a dense network of small vessels at the base of the skull that buffers the swing, and valves in the jugular that stop the descending venous column running backwards. All of it is engineering against one linear term, which is the whole content of this page.

Where the model stops #

Four assumptions, each of which fails somewhere useful.

The fluid is at rest. This is the one that has been doing all the work. The moment the fluid moves, shear stresses appear, pressure is no longer the same in all directions in the same simple way, and the depth law acquires corrections. A body moving through a fluid, and the pressure field around it, is a whole subject and it belongs to the collection that owns flow around a shape rather than to this one.

The acceleration is $g$ and it points down. In a bucket of water on a turntable the effective gravity tilts, the free surface becomes a paraboloid, and the constant-pressure surfaces follow it. In a freely falling vessel there is no pressure gradient at all — which is why a drop of water in orbit is held together by the surface, and nothing else.

The surface is flat. It is not, at small scales, because surface tension curves it. In a tube a millimetre across the curvature adds tens of pascals and lifts the water measurably, which is capillary rise, and it is the first correction to every result on this page once the scale is small enough. What does the curving is the surface, which costs energy per unit area.

Density is uniform. Warm water sits on cold, salt water under fresh, and the resulting layering is stable rather than incidental. Stratified fluids support internal waves at their density interfaces, and the density difference is what makes those waves slow.

The history, and why it took so long #

The result on this page is one line of algebra, and it was not obtained for a very long time. Stevin had it around 1586, and stated the paradox as a theorem — the force on a base depends on the depth and the base, not on the vessel — which is exactly the claim that the picture at the top of this page makes.

Pascal’s contribution, half a century later, was to see that the pressure at a point is transmitted undiminished throughout a connected fluid, and to notice what that permits mechanically. That is the next rung on this ladder, and it turns a statement about depth into a machine.

The delay is not surprising once the argument is laid out, because it depends on a negative: a still fluid supports no shear. That is not something anybody observes directly. It is something inferred from the fact that fluids flow whenever they are pushed sideways, and inferring a property of the state at rest from the behaviour of the state in motion is a longer step than it looks from here.

The ladder from here #

Later rungs on this anchor: pressure transmitted through a connected fluid, and the machine that follows from it. The manometer, which turns a pressure difference into a length that can be read with a ruler. The free surface of a rotating fluid, where the constant-pressure surfaces are paraboloids and the shape can be used to grind a telescope mirror. Stratification and the internal waves it supports. And the case where the fluid is the atmosphere and the vessel is a planet, where hydrostatic balance is what decides how large a body has to be before its own gravity rounds it.

The neighbouring ladder is the one this page has just set up: the pressure on a submerged body’s top and bottom faces do not cancel, and what is left over is buoyancy.