The weight of the water that is not there

An object held under water is pushed upward. Everybody knows this, and the usual explanation — the water holds it up — explains nothing, because water is not doing anything it was not doing before the object arrived. The correct account has one moving part, and it was fully established by the previous rung of this ladder: pressure in a still fluid increases with depth.

That is the whole argument. What remains is to do the arithmetic and to notice how much falls out of it.

The sum that survives #

Take a rectangular block of height $H$ and horizontal area $A$, with its top face at depth $d$. The pressure on the top is $\rho g d$ and pushes down; the pressure on the bottom is $\rho g (d + H)$ and pushes up. The net upward force is the difference times the area:

$$F = \rho g (d + H) A - \rho g d A = \rho g H A = \rho g V.$$

The depth $d$ has cancelled. What is left is $\rho g V$ — the density of the fluid, times gravity, times the volume of the block. That product is the weight of the fluid that would occupy the block’s volume if the block were not there. The upthrust equals the weight of the displaced fluid, exactly, and it does so because two terms differed by exactly one column’s worth.

Two things did not survive that calculation and their absence is the interesting part. The depth went, so a block a metre down and a block a kilometre down feel the same upthrust. And nothing about the block’s material ever entered, because the argument only ever mentioned the fluid’s pressure on the block’s surface. A lead cube and a wooden cube of the same size feel the same upthrust; what differs is their weight, not the push.

Any shape at all #

The rectangular block was a convenience, and the result does not depend on it. There are two ways to see this and both are worth having.

The first is a replacement argument, and it is the one Archimedes used. Imagine the object removed and its volume filled with the surrounding fluid instead. That parcel of fluid is in equilibrium — it is not going anywhere — so the net force the surrounding fluid exerts on its surface must be exactly equal and opposite to its own weight, which is $\rho g V$ upward. Now put the object back. The surrounding fluid’s pressure at every point of that surface is unchanged, because pressure depends only on depth and the depths have not moved. So the force is unchanged: $\rho g V$ upward, whatever now occupies the volume.

This argument is very close in structure to the flux argument for a closed surface, and for the same reason: both replace a difficult surface integral with a statement about what is enclosed, and both work because the field outside is untouched by what is put inside.

The same move appears in a different subject and is worth recognising as a move. What crosses a closed surface is decided by what is inside it, and the surface may be any shape at all — so an argument made on a convenient box holds for a hull, a fish, or a crumpled sheet of foil. The freedom to choose the surface is what makes both arguments short, and in both cases the choosing is the only cleverness involved.

The second route is to integrate the pressure over an arbitrary surface directly. The vertical component of the force is $-\oint p \, \hat{n} \cdot \hat{z} \,\mathrm{d}S$, and substituting $p = \rho g h$ and applying the divergence theorem turns it into $\rho g \int \mathrm{d}V$. Both routes arrive at the same place; the first explains it and the second proves it.

Floating, and the fraction underneath #

If the object is denser than the fluid its weight exceeds the maximum upthrust and it sinks. If it is less dense, it rises until part of it is out of the fluid, at which point only the submerged part displaces anything. Equilibrium is where the displaced weight equals the object’s weight:

$$\rho_{\text{fluid}} \, g \, V_{\text{sub}} = \rho_{\text{body}} \, g \, V_{\text{total}}, \qquad \frac{V_{\text{sub}}}{V_{\text{total}}} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}}.$$

The fraction submerged is the density ratio. Not approximately — exactly, and with $g$ cancelled out of it, so a floating body sits at the same draft on the Moon.

Ice has a relative density of about 0.917 in fresh water, so a floating iceberg shows about eight per cent of itself, which is where the proverb comes from. In sea water, which is denser at about 1.025, the exposed fraction rises to roughly eleven per cent — a detail the proverb omits and which matters to anybody trying to estimate a berg’s mass from a photograph.

Why a steel ship floats #

The density ratio explains the objection everybody raises. Steel is about eight times as dense as water, so by the rule above a steel object should sit eight times submerged, which is impossible, and it should therefore sink.

It does sink, if it is solid. What a hull does is enclose air, and the quantity in the rule is the mean density of everything inside the hull’s outer surface — steel, air, cargo, crew. A ship is a device for making a large volume out of a modest mass of steel, and the design question is entirely about how much can be put inside before the mean density reaches one.

That reframing has a consequence worth stating. Buoyancy is not a property of the material an object is made of but of the boundary it draws around itself, which is why the same tonne of steel floats as a hull and sinks as an ingot. It also explains the failure mode: a hull that fills with water has not become heavier steel, it has enclosed a denser interior, and the ratio moves accordingly.

What the fluid loses #

There is a bookkeeping question hiding in the replacement argument, and answering it removes a persistent confusion.

If the fluid pushes the object up with a force $\rho g V$, then by Newton’s third law the object pushes the fluid down with the same force. Where does that go? It goes into the fluid’s own weight distribution, and the easiest way to see it is to weigh the whole apparatus. Put a beaker of water on a balance, then lower a stone into it on a string. The balance reading increases by exactly the upthrust, while the string tension falls by the same amount. Nothing has been created; a force has been shared out between two supports.

This is worth stating because the alternative intuition — that the water somehow supports the stone at no cost to itself — predicts that the balance reading is unchanged, and it is not. It also settles the classic puzzle about a boat carrying a stone in a pond. Drop the stone overboard and the water level falls, because floating in the boat the stone displaced its own weight of water and sitting on the bottom it displaces only its own volume, which is less for anything denser than water.

The same accounting explains the apparent contradiction people raise about a sealed aquarium with a fish in it: the total weight is the total weight, whether the fish swims, rests or hovers, because whatever force the fish exerts on the water is matched by the water’s force on the fish, and both are inside the system being weighed.

The weight that is not the weight #

An object weighed while submerged reads light by exactly the upthrust, which turns the principle into an instrument.

Weigh an object in air, then in water. The difference in readings is $\rho_{\text{water}} \, g \, V$, which gives the volume without any measurement of shape at all — no ruler touches it, and it works equally well for a crown, a bone or a lump of ore. It is the same trick as reading a quantity off a shape rather than off a force: the measurement is arranged so that everything awkward cancels. Divide the dry weight by that volume and the density follows.

This is a genuinely good measurement, and it is old. It is the one the Archimedes story is about, and the story’s usual telling gets the method wrong: comparing water overflowed by a crown and by an equal weight of gold is a difference of a per cent or two in a quantity read off a vessel’s rim, which is not measurable by eye. Weighing submerged compares forces, on a balance, and a balance resolves parts per thousand without difficulty. Galileo made this point in 1586, and it is the version that would have worked.

Where the upthrust is stored is the pressure difference between top and bottom. Every depth carries a different pressure, and a submerged body collects the difference across its own height — which is why the upthrust knows the body’s height and not its depth, and why a block sunk twice as far feels exactly the same lift. Depth cancels out of a difference between two depths.

The push does not decide the acceleration #

Everything above is a statement about a force, and a force is only half of a prediction. Release a submerged body from rest and ask how fast it accelerates, and the naive answer — net force divided by mass — is wrong, in a way that is large rather than fiddly.

The reason is that a body cannot move through a fluid without moving the fluid. Fluid ahead has to be pushed aside and fluid behind has to close in, so accelerating the body means accelerating a whole flow field, and the energy that goes into that flow is proportional to the square of the body’s speed. Written as an inertia, it is an addition to the body’s own: a sphere behaves as though its mass were its own plus half the mass of the fluid it displaces.

The consequence for a light body is dramatic. Take a bubble, whose own mass is negligible. The upward force on it is $\rho g V$; its effective mass is $\tfrac12 \rho V$; so it starts upward at $2g$ — twice the acceleration of a falling stone, in the opposite direction, and it is measured. For a body of density $\rho_b$ the initial acceleration is

$$a = g\,\frac{\rho_f - \rho_b}{\rho_b + \tfrac12\rho_f},$$

which for steel in water gives $0.82g$ rather than the $0.875g$ the force alone suggests.

The quantity has to be respected wherever a body in water changes speed. A ship pushed sideways by a tug behaves as though it were roughly twice as massive as it is, because the added mass for that motion is comparable with the whole displacement; a submerged pendulum swings slower than the same pendulum in air by more than the buoyancy alone accounts for; and a sonar transducer’s resonance shifts when it is put in the sea.

One thing it is not is a lump of water being carried along. No particular parcel of fluid travels with the body — the flow streams round it and returns — and the added mass is a summary of the whole field’s kinetic energy rather than a description of any of it. It is a coefficient, and it depends on the shape and on the direction of the motion: a flat plate moving edgewise has almost none and moving broadside has a great deal.

How little water a ship needs #

A phrase in the principle invites a misreading worth settling, because it is the one thing about buoyancy that reliably surprises people who have the rest of it right.

“Displaces its own weight of water” sounds like a statement about how much water has to be present. It is not. What the argument used was the pressure on the hull, and pressure depends only on depth below the free surface — so a hull sitting in a dock shaped closely to it, with a film of water a centimetre thick between hull and dock, feels exactly the pressures it would feel in mid-ocean and floats exactly as well. A hundred-thousand-tonne ship can be floated on a few tonnes of water, and dry docks are flooded and drained on that understanding daily.

The displaced volume is the volume of hull below the waterline. Whether the water that would have occupied it exists anywhere is beside the point, because the object never touched it.

Read the same relation the other way and it becomes a measurement. A floating body’s draft is fixed by the density ratio, so a body of known weight sinks to a depth that reports the density of whatever it is floating in. That is a hydrometer: a weighted float with a graduated stem, read where the surface cuts it.

The stem is narrow on purpose. Since the submerged volume must be constant for a constant weight, a small change in the liquid’s density has to be taken up entirely by a change in the length of stem submerged — so making the stem thin turns a density difference of a fraction of a per cent into a centimetre of scale. The same instrument, with different weighting, checks the charge of a car battery, the sugar left in a fermenting beer, and the salinity of an aquarium.

Where the model stops #

There has to be fluid underneath. The entire argument is a difference between the pressure on the underside and the pressure on top. A block resting flat on a smooth bottom, with the fluid excluded from beneath it, collects no upward pressure at all and stays there — however light it is. This is not a curiosity: it is why an object bedded in mud can be extraordinarily hard to lift, and why a suction cup works. The principle describes a body surrounded by fluid, and the surrounding is a hypothesis rather than a detail.

The fluid is uniform. Where density varies with depth, the upthrust becomes $g \int \rho \,\mathrm{d}V$ over the displaced volume, and a body can find a level at which it is neutrally buoyant and stay there. That is how a submarine holds depth in a stratified ocean, and it is the same balance a parcel of air makes in the atmosphere.

The fluid is still. A body in a moving fluid experiences pressure differences caused by the motion as well as by depth, and those can be far larger. Separating them is the business of hydrodynamics; everything here assumes there is none.

The body is large compared with a molecule. At the scale of a bacterium or a colloidal particle the upthrust still applies, but so does the incessant bombardment that produces the same restless spreading a drop of ink undergoes, and the competition between the two is what sets how far a suspension settles before it stops. That competition is measurable, and it was used to weigh an atom.

Surface tension is neglected. A steel needle floats on water, and its relative density is eight. Nothing on this page permits that; what holds it up is the surface, which costs energy per unit area, and for small enough objects that force exceeds the upthrust. The length at which the two are comparable is a couple of millimetres in water, which is why the effect belongs to insects and needles and not to boats.

Buoyancy in air, which is small and not zero #

Everything above applies to any fluid, and air is a fluid. Its density at sea level is about 1.2 kilograms per cubic metre, which is a bit over a thousandth of water’s, so the upthrust on an ordinary object is a thousandth of what it would be submerged — small, and not nothing.

For a kilogram of lead, occupying about ninety millilitres, the air displaced weighs about 0.1 grams. For a kilogram of feathers, occupying perhaps forty litres, it is nearly fifty grams. So two objects that balance on a beam in air do not contain the same mass: the feathers contain about fifty grams more. The old riddle has a real answer and it is the opposite of the one usually given, and every serious mass measurement corrects for it.

Push the ratio further and the object rises. A balloon filled with helium, whose density is about a seventh of air’s, has a mean density below the surrounding fluid and floats for exactly the reason a cork does — with the difference that the surrounding fluid is stratified rather than uniform, so the balloon rises to the level at which the air’s density has fallen to match its own and stops there. A cork in water has no such level, because water’s density hardly changes with depth; a balloon in air does, and finding it is what a ceiling altitude is.

Hot air works by the same arithmetic with a different mechanism: heating air at constant pressure lowers its density in proportion to the absolute temperature, so air at 100 °C is about three quarters as dense as air at 20 °C, and a cubic metre of it lifts roughly 0.3 kilograms. A balloon large enough to carry people therefore has to be enormous, and the size follows from a division that has nothing adjustable in it.

The equilibrium this page has not settled #

The condition derived above says how deep a floating body sits and says nothing about which way up it sits. Those are different questions, and the second is not answered by any statement about magnitudes.

A floating body has two forces on it — weight, acting at its centre of gravity, and buoyancy, acting at the centre of the displaced volume — and if those two points are not on the same vertical line the pair makes a couple. Whether that couple restores the body or overturns it depends on how the displaced volume moves when the body tilts, which depends on the shape of the hull at the waterline and not on the density ratio at all.

Whether a floating body returns when disturbed is a different question and a harder one. A body at the bottom of a well in its own energy landscape comes back; one balanced on a hill does not — and which of the two a floating object is doing depends on its geometry rather than on its density. Everything on this page settles how much it floats and nothing about whether it stays the right way up.

That is the next essay on this ladder, and it is where the shape of the hull finally starts to matter after two pages of not mattering at all.

The ladder from here #

Later rungs on this anchor: the stability of a floating body and the metacentre. Neutral buoyancy and how a submarine or a fish holds depth, including the swim bladder as a control system with a genuinely awkward instability in it. Buoyancy in a stratified fluid, and the oscillation a displaced parcel performs about its own level. Buoyancy in air, which is small and not zero — a kilogram of feathers really does weigh less than a kilogram of lead once the air is accounted for. And centrifugal buoyancy, which is what a centrifuge is, and which separates by density at effective gravities of many thousands.