Fill a plastic syringe with air, close the nozzle with a finger and push the plunger. It goes in easily, and springs back when you let go. Now fill the syringe with water and try again. The plunger will not move at all, however hard you push. Water seems impossible to squeeze.
Pressure from every side
A body placed in a liquid or a gas is pushed on every face, at right angles to the face. The normal force per unit area that a fluid exerts is its pressure \(P\), measured in pascals. The chapter on Pressure in Fluids studies it in detail. Here we need one fact: raise the pressure all around a body by \(\Delta P\), and every face is pushed harder by \(\Delta P\) per unit area.
The body then keeps its shape but shrinks in size. The extra pressure \(\Delta P\) is the stress, called the volume stress. The fractional change \(\Delta V/V\) is the volume strain, and it is negative, since the volume falls. Within the elastic range the two are proportional, and their ratio defines the bulk modulus \(B\).
Raise the pressure on every side of a body, and its volume falls by a small fraction. The bulk modulus of the material is the ratio of the rise in pressure to that fraction: \(B = -\Delta P/(\Delta V/V)\).
The minus sign is there to make \(B\) positive. A rise in pressure, \(\Delta P > 0\), always makes the volume fall, \(\Delta V < 0\). So \(-\Delta P/(\Delta V/V)\) is a positive number. For water \(B\) is about \(2 \times 10^9\,\text{Pa}\), for steel about \(1.6 \times 10^{11}\,\text{Pa}\).
In the sea the pressure grows by about \(10^7\,\text{Pa}\) for every kilometre of depth, taking the density of water as \(1000\,\text{kg/m}^3\) and \(g = 10\,\text{m/s}^2\). Move the slider to \(1\,\text{km}\). The volume strain is \(\Delta P/B = 10^7/(2 \times 10^9) = 5 \times 10^{-3}\), so the litre loses \(5\,\text{mL}\), about a teaspoon. At \(4\,\text{km}\) it has lost \(20\,\text{mL}\). The loss grows in step with the depth and stays small.
A litre of a liquid is put under an extra pressure of \(2 \times 10^7\,\text{Pa}\), and its volume falls by \(10\,\text{mL}\). Find the bulk modulus of the liquid.
The volume strain is \(\Delta V/V = -10/1000 = -0.01\).
\(B = -\Delta P/(\Delta V/V) = -(2 \times 10^7)/(-0.01) = 2 \times 10^9\,\text{Pa}\). The liquid behaves like water.
One litre of water, with \(B = 2 \times 10^9\,\text{Pa}\), is put under an extra pressure of \(4 \times 10^6\,\text{Pa}\). By how many millilitres does its volume decrease?
Why does the definition of the bulk modulus carry a minus sign?
Easy to squeeze, hard to squeeze
The inverse of the bulk modulus, \(1/B\), is called the compressibility. It is the fractional loss of volume for each pascal of extra pressure. For water it is \(1/(2 \times 10^9) = 5 \times 10^{-10}\,\text{Pa}^{-1}\). A material with a large bulk modulus has a small compressibility.
Fig 4.1 shows why the syringe of air gave way. Air at atmospheric pressure, squeezed slowly at a steady temperature, halves its volume when the pressure doubles. Its bulk modulus is only about \(10^5\,\text{Pa}\), equal to its pressure. Water's is \(2 \times 10^9\,\text{Pa}\), twenty thousand times larger. Steel's is eighty times larger again.
The reason lies in the spacing of the molecules. In a gas they are far apart, and squeezing only brings them closer through empty space. In a liquid or a solid they already touch, and squeezing must push against the strong repulsion between atoms. A solid is usually the least compressible, because its atoms are locked in place.
Rank these by compressibility, the most compressible first.
"Water is incompressible" is an approximation, not a law. It is excellent for a tank or a pipe, where the pressure changes by a few atmospheres. In the deep sea, or in a hydraulic press, the small change of volume is real and sometimes matters.
Denser under pressure
A squeezed body keeps its mass but loses volume, so its density rises. The derivation below finds by how much. It uses one rule for small changes. When \(V\) changes by a small fraction, \(1/V\) changes by the same fraction with the opposite sign.
\[ \rho = \dfrac{m}{V} \]
Why. the definition of density; the mass \(m\) does not change when the body is squeezed.
\[ \dfrac{\Delta\rho}{\rho} = -\dfrac{\Delta V}{V} \]
Why. for a small change, a fraction \(\Delta V/V\) more volume gives the same fraction less density.
\[ \dfrac{\Delta V}{V} = -\dfrac{\Delta P}{B} \]
Why. the definition of the bulk modulus.
\[ \dfrac{\Delta\rho}{\rho} = \dfrac{\Delta P}{B} \]
Read it. substitute step 3 into step 2. The two minus signs cancel.
Water at the surface has a density of \(1000\,\text{kg/m}^3\) and \(B = 2 \times 10^9\,\text{Pa}\). About \(1\,\text{km}\) down the pressure is greater by \(10^7\,\text{Pa}\). Find the density of the water there.
\(\Delta\rho/\rho = \Delta P/B = 10^7/(2 \times 10^9) = 5 \times 10^{-3}\), half a percent.
\(\rho' = 1000 \times (1 + 0.005) = 1005\,\text{kg/m}^3\). This matches the sim, where the litre lost \(5\,\text{mL}\) but kept its mass.
The pressure on water, with \(B = 2 \times 10^9\,\text{Pa}\) and a density of \(1000\,\text{kg/m}^3\), is raised by \(2 \times 10^7\,\text{Pa}\). What is its new density, in kg/m³?
The deepest point of the ocean is about \(11\,\text{km}\) down. Take \(B = 2 \times 10^9\,\text{Pa}\), a density of \(1000\,\text{kg/m}^3\) and \(g = 10\,\text{m/s}^2\). Estimate how many millilitres a litre of water loses when it is carried down there.
Now return to the syringe. Your thumb can add perhaps \(10^6\,\text{Pa}\) to the water, ten atmospheres. With \(B = 2 \times 10^9\,\text{Pa}\), that shrinks the water by only \(5 \times 10^{-4}\), five parts in ten thousand. The plunger does move, by a fraction of a millimetre, too little to feel. The air, with a bulk modulus twenty thousand times smaller, gives way at once.
The driver of a bus presses the brake pedal, and oil in thin pipes carries the push to every wheel. This works because the oil, like water, has a bulk modulus near \(10^9\,\text{Pa}\), so it passes the push on without shrinking. If a bubble of air gets into a brake pipe, the pedal goes soft and sinks. The push is spent in squeezing the air, which is why mechanics "bleed" the air out of the brake lines.
- Dropping the minus sign and getting a negative bulk modulus.\(\Delta V\) is negative when \(\Delta P\) is positive. \(B = -\Delta P/(\Delta V/V)\) is always positive.
- Using the total pressure instead of the change in pressure.\(B\) relates the change \(\Delta P\) to the change \(\Delta V\). Atmospheric pressure already acting does not squeeze further.
- Saying that density falls under pressure.The mass is fixed and the volume falls, so the density rises: \(\Delta\rho/\rho = \Delta P/B\).