Hydrostatic Pressure

Equivalent forms

\Delta P = \rho g h

Container shape doesn't matter — only depth does.

Symbol	Name	SI unit	Dimension	Range
$P$	Pressure at Depthoutput Total pressure at depth h	Pa	M·L⁻¹·T⁻²	0 – 1000000
$P_0$	Surface Pressure Pressure at the surface	Pa	M·L⁻¹·T⁻²	0 – 200000
$ρ$	Density Fluid density	kg/m³	M·L⁻³	1 – 13600
$g$	Gravity Gravitational acceleration	m/s²	L·T⁻²	9.8 – 9.8
$h$	Depth Depth below surface	m	L	0 – 1000

Unit systems

Symbol	SI	CGS	Imperial
$P$	Pa	Ba	psi
$P_0$	Pa	Ba	psi
$ρ$	kg/m³	g/cm³	lb/ft³
$g$	m/s²	cm/s²	ft/s²
$h$	m	cm	ft

Where it holds

Static, incompressible fluid in a uniform gravitational field. Fails for compressible atmospheres over large altitude ranges — use the barometric formula instead.

Dimensional analysis

$[\rho gh] = (M\cdot L^{-3})(L\cdot T^{-2})(L) = M\cdot L^{-1}\cdot T^{-2} = [P] \checkmark$

Discovery

Simon Stevin · 1586

Stevin showed pressure depends only on depth, not container shape — the famous hydrostatic paradox illustrated with vessels of different shapes filled to the same height.

Try this

Why do your ears hurt at the bottom of a swimming pool?

Find the gauge pressure on your eardrum at 3 m below the surface of fresh water.

Research status: stable

Real-world applications

Submarine hull design and crush depth ratings
Dam wall thickness (thicker at the base where pressure is highest)
Scuba diving depth limits and decompression
IV drip bag height calculations in hospitals
Manometers for pressure measurement

Common misconceptions

The shape or width of the container does NOT affect the pressure — only depth matters
Pressure acts equally in all directions at a point, not just downward
Gauge pressure $(\rho gh)$ and absolute pressure $(P_{0} + \rho gh)$ are different — be careful which one you need

Experimental verification

Verified by Stevin's hydrostatic paradox demonstration with different-shaped vessels showing equal pressures at equal depths. Modern depth gauges and pressure transducers used in oceanography confirm to ppm precision.

Derivation

Consider a thin horizontal slab of fluid at depth h, area A, thickness dh.

Force balance: pressure on bottom

\times A

= pressure on

top \times A

+ weight of slab.

(P + dP)A = PA + \rho gA\cdot dh

, giving

dP/dh = \rho g

.

Integrating from surface

(P = P_{0})

to depth h yields

P = P_{0} + \rho gh

.

Limiting cases

h = 0

⟶

P = P_0

At the surface, pressure equals atmospheric (or whatever sits on top).

\rho \to 0

⟶

P \to P_0

A massless fluid carries no weight — no depth dependence.

h = 10\,\mathrm{m}

of water⟶

\Delta P \approx 1\,\mathrm{atm}

Every 10 m of water adds roughly one atmosphere — handy diving rule of thumb.

What if…

What if you use mercury instead of water?

Mercury is $13.6\times$ denser, so the same depth gives $13.6\times$ more pressure. This is why mercury barometers are short — only $\sim 76\,\mathrm{cm}$ to balance one atmosphere.

What if the container is shaped like an hourglass or wide cone?

The pressure at any depth depends only on h, not shape. This is Stevin's hydrostatic paradox — a tall narrow tube exerts the same pressure on the base as a wide vat at the same depth.

What if gravity changed (e.g., on the Moon)?

Pressure scales with g. On the Moon $(g \approx 1.6\,\mathrm{m/s}^{2})$ , the same depth of water would exert about 1/6 the pressure.

1

Eardrum pressure at 3 m depth

Given ·

ρ:: 1000
g:: 9.8
h:: 3

Find ·

\Delta P

Steps

Gauge pressure $\Delta P = \rho gh$
$\Delta P = (1000)(9.8)(3)$
$\Delta P = 29400\,\mathrm{Pa} \approx 0.29\,\mathrm{atm}$ above atmospheric

Result ·

\Delta P = \rho gh = 1000 \times 9.8 \times 3 = 29400\,\mathrm{Pa} \approx 29.4\,\mathrm{kPa}

2

Pressure at the bottom of the Mariana Trench

Given ·

ρ:: 1025
g:: 9.8
h:: 11000

Find ·

\Delta P

Steps

Use seawater density $\rho = 1025\,\mathrm{kg/m}^{3}$
$\Delta P = \rho gh = 1025 \times 9.8 \times 11000$
$\Delta P \approx 1.10 \times 10^{8} Pa$ — over 1000 atmospheres

Result ·

\Delta P = 1025 \times 9.8 \times 11000 \approx 1.10 \times 10^{8} Pa \approx 1090\,\mathrm{atm}

Bernoulli's Equation→Archimedes' Principle→