Young-Laplace Equation

Equivalent forms

\Delta P = \gamma \left( \frac{1}{r_1} + \frac{1}{r_2} \right)

\Delta P_{soap} = \frac{4 \gamma}{r}

A curved interface generates pressure — geometry alone determines the squeeze.

Symbol	Name	SI unit	Dimension	Range
$\gamma$	Surface Tension Energy per unit interfacial area	N/m	M·T⁻²	0.001 – 1
$r$	Radius of Curvature Radius of the curved interface	m	L	0.00001 – 0.1
$\Delta P$	Excess Pressureoutput Pressure difference across the interface	Pa	M·L⁻¹·T⁻²	0 – 100000

Unit systems

Symbol	SI	CGS	Imperial
$\gamma$	N/m	dyn/cm	lbf/ft
$r$	m	cm	ft
$\Delta P$	Pa	dyn/cm²	psi

Where it holds

Quasi-static curved interface between two fluid phases. Assumes

\gamma is

constant (no Marangoni effects), gravity is negligible compared to curvature pressure, and the interface is in mechanical equilibrium. Excludes dynamic situations and very thin films where disjoining pressure matters.

Dimensional analysis

$[\gamma ]/[r] = (M\cdot T^{-2})/(L) = M\cdot L^{-1}\cdot T^{-2} \checkmark$ (pressure)

Discovery

Thomas Young & Pierre-Simon Laplace · 1805

Thomas Young proposed the qualitative law in 1804; Laplace independently derived the equation in 1805 from molecular force considerations, giving rise to modern capillarity theory.

Try this

Why are smaller soap bubbles harder to inflate than larger ones?

A water droplet has radius 0.001 m. Water surface tension is 0.072 N/m. What is the excess pressure inside the droplet?

Research status: stable

Real-world applications

Aerosol and droplet formation in inkjet printing
Pulmonary surfactant in lung alveoli (Laplace's law for breathing)
Capillary rise in soils and plants
Microfluidic device design

Common misconceptions

For a soap bubble (TWO interfaces) the factor is 4, not 2
$\Delta P$ is the JUMP across the interface, not absolute internal pressure
$\gamma$ depends on temperature and surfactant concentration — it's not a fixed property of a liquid

Experimental verification

Direct measurement using pendant drop, sessile drop, and Wilhelmy plate tensiometry consistently confirms the law. The maximum-bubble-pressure method, used in industrial surfactant testing, is a direct application of Young-Laplace.

Derivation

Consider a spherical droplet of radius r.

The surface tension

\gamma

pulls along the equator with circumference

2\pi r

, giving an inward force

\gamma \cdot 2\pi r

.

This must be balanced by the excess pressure

\Delta P

acting on the cross-sectional area

\pi r^{2}

,

so \Delta P\cdot \pi r^{2} = 2\pi r\gamma

, giving

\Delta P = 2\gamma /r

.

Generalizing to a non-spherical surface gives

\Delta P = \gamma (1/r_{1} + 1/r_{2})

where

r_{1} and r_{2} are

the principal radii of curvature.

Limiting cases

r \to \infty

⟶

\Delta P \to 0

A flat interface has no excess pressure (Laplace pressure vanishes).

r \to 0

⟶

\Delta P \to \infty

Tiny droplets have enormous internal pressures — explains nanoparticle stability and nucleation barriers.

Soap bubble (two surfaces)⟶

\Delta P = 4\gamma /r

Soap bubbles have inner AND outer surfaces, doubling the pressure jump.

What if…

What if the droplet shrinks

10\times

?

The excess pressure rises $10\times$ . This is why nanoparticles are so chemically active — internal pressures reach megapascals.

What if surfactant is added?

$\gamma$ drops, $so \Delta P$ drops too. Lung surfactant uses this trick to prevent small alveoli from collapsing into larger ones.

What if you have a non-spherical interface?

$Use \Delta P = \gamma (1/r_{1} + 1/r_{2})$ . For a saddle (negative curvature) the pressures can partially cancel — the basis of minimal-surface theory.

1

Water droplet pressure

Given ·

\gamma:: 0.072
r:: 0.001

Find · \Delta P

Steps

Apply $\Delta P = 2\gamma /r$
$\Delta P = 2 \times 0.072 / 0.001$
$\Delta P = 144\,\mathrm{Pa} (\approx 0.14$ % of atmospheric)

Result ·

\Delta P = 2 \times 0.072 / 0.001 = 144\,\mathrm{Pa}

2

Soap bubble (double surface)

Given ·

\gamma:: 0.025
r:: 0.01

Find · \Delta P

Steps

Soap bubble has inner and outer surfaces
$Use \Delta P = 4\gamma /r$
$\Delta P = 10\,\mathrm{Pa}$ — tiny but enough to keep the bubble round

Result ·

\Delta P = 4\gamma /r = 4 \times 0.025 / 0.01 = 10\,\mathrm{Pa}

Jurin's Law (Capillary Rise)→Hydrostatic Pressure→