Laplace Pressure and Curved Interfaces

Blow a soap bubble and it snaps into a sphere. It never becomes a cube or a pancake. That shape is not an accident. It is the direct consequence of a single pressure law. This law governs every curved liquid surface, from raindrops to the alveoli in your lungs. The Young–Laplace equation describes this behavior. A curved interface can only be in mechanical equilibrium under one condition. The pressure inside must exceed the pressure outside. That excess is set by the surface tension and the curvature. This idea underlies capillary rise and bubble stability. It also governs emulsion droplet size. It connects to the thermodynamics of nucleation covered later. This article derives the Young–Laplace equation from a simple mechanical energy balance. You will see the general form with two radii of curvature. You will see it specialize to spheres and cylinders. Then you will apply it to bubbles, droplets, and capillaries. Fully worked numbers accompany each application.

Key takeaways

A curved liquid surface sustains a pressure difference: the inside of a convex surface is always at higher pressure.
The general law is $\Delta p = \gamma(1/R_1 + 1/R_2)$ , using the two principal radii of curvature.
For a sphere, $\Delta p = 2\gamma/R$ ; for a cylinder, $\Delta p = \gamma/R$ ; for a flat interface, $\Delta p = 0$ .
Soap films have two surfaces, so a soap bubble obeys $\Delta p = 4\gamma/R$ , doubling the droplet result.
Smaller radius means larger pressure difference — this is why small bubbles feed into large ones when connected.
Key idea: Surface tension pulls a curved interface inward like a stretched membrane, and only a pressure excess on the concave side can balance that inward pull.

The core insight

A curved interface is a stretched elastic skin; for it to sit still, the pressure pushing outward from the concave side must exactly balance the inward mechanical pull generated by surface tension acting over the interface's curvature, and the amount of that required pressure excess is $\Delta p = \gamma(1/R_1+1/R_2)$ .

The physical picture

Surface tension, $\gamma$ , is the free energy cost per unit area of creating interface (see Surface Tension and Surface Free Energy). A flat interface has no preferred direction to contract toward, so it exerts no net pressure. A curved interface is different: tension acts everywhere tangent to the surface, and when the surface bends, those tangential pulls acquire a net inward component. Picture a spherical droplet. Every element of surface is pulled toward its neighbors by tension, and because the surface closes on itself, all these pulls sum to a net inward force per unit area — a pressure. The only way the droplet can be static is if the liquid inside pushes back with a higher pressure. The smaller the sphere, the more sharply the surface bends per unit area, and the larger the required pressure excess. This is why fine mist droplets have enormously higher internal pressure than a raindrop, and why an over-inflated small balloon can feed air into a slack, larger one — the smaller radius of curvature demands the higher internal pressure. The general geometry needs two radii of curvature, $R_1$ and $R_2$ , measured in two perpendicular planes containing the local normal. A saddle-shaped surface (like the neck of a soap film between two rings) can have $R_1$ and $R_2$ of opposite sign, and the two curvature contributions can cancel, giving $\Delta p = 0$ even though the surface is curved everywhere.

Figure

Pressure jump across a spherical interface: outward push from inside balances inward pull of tension

Figure

Sphere versus cylinder versus soap bubble: how the pressure jump scales with geometry

The core relations

\Delta p \;=\; p_{\text{inside}} - p_{\text{outside}} \;=\; \gamma\left(\frac{1}{R_1} + \frac{1}{R_2}\right)

where $\gamma$ is the surface tension and $R_1, R_2$ are the two principal radii of curvature (general Young–Laplace equation).

\Delta p = \frac{2\gamma}{R}

where $R$ is the single radius of a sphere ( $R_1 = R_2 = R$ ), the case of a liquid droplet or gas bubble in bulk liquid.

\Delta p = \frac{\gamma}{R}

where $R$ is the radius of curvature of a cylindrical surface ( $R_1 = R$ , $R_2 \to \infty$ ), e.g. a liquid cylinder or jet.

\Delta p_{\text{soap bubble}} = \frac{4\gamma}{R}

where the factor of 4 arises because a soap bubble has two liquid–air surfaces (inner and outer film faces) each contributing $2\gamma/R$ .

h = \frac{2\gamma \cos\theta}{\rho g r}

where $h$ is capillary rise, $\theta$ the contact angle, $\rho$ the liquid density, $g$ gravitational acceleration, and $r$ the tube radius — the classic application of Young–Laplace to capillarity.

Derivation AThe Young–Laplace equation from a virtual-work energy balance

Goal. Show that mechanical equilibrium of a curved fluid interface under surface tension $\gamma$ requires

\Delta p = \gamma\left(\frac{1}{R_1}+\frac{1}{R_2}\right).

Step 1. Set up a virtual displacement. Consider a small patch of interface with principal radii of curvature $R_1$ and $R_2$ at that point, separating an inside fluid at pressure $p_{\text{in}}$ from an outside fluid at $p_{\text{out}}$ . Imagine displacing this patch outward, along its local normal, by an infinitesimal distance $dz$ . This is a virtual expansion of the enclosed volume.

Step 2. Compute the area change. For a patch of dimensions $x_1 \times x_2$ along the two principal directions, curvature means each side scales as the patch moves outward by $dz$ :

x_1' = x_1\left(1 + \frac{dz}{R_1}\right), \qquad x_2' = x_2\left(1 + \frac{dz}{R_2}\right).

The new area is the product $x_1' x_2'$ . Expanding to first order in $dz$ and dropping the $(dz)^2$ term gives

dA = x_1 x_2\left(\frac{dz}{R_1} + \frac{dz}{R_2}\right) = A\left(\frac{1}{R_1}+\frac{1}{R_2}\right)dz.

Step 3. Compute the volume change. Moving the patch of area $A$ outward by $dz$ sweeps out an extra volume

dV = A\, dz.

Step 4. Write the total work for the virtual displacement. Two things happen: the interface area grows, costing energy $\gamma\, dA$ ; and the inside fluid does work pushing the boundary outward against the outside fluid, net mechanical work $(p_{\text{in}} - p_{\text{out}})\, dV$ . At equilibrium, this virtual displacement must not lower the total energy — the system sits at a stationary point, so the work supplied by the pressure difference is exactly what pays for the new surface energy:

(p_{\text{in}} - p_{\text{out}})\, dV = \gamma\, dA.

Step 5. Substitute the geometric results. Using $dV = A\,dz$ and $dA = A(1/R_1 + 1/R_2)\,dz$ from Steps 2 and 3,

(p_{\text{in}} - p_{\text{out}})\, A\, dz = \gamma A\left(\frac{1}{R_1}+\frac{1}{R_2}\right) dz.

The area $A$ and displacement $dz$ cancel from both sides, leaving the local relation independent of the size of the chosen patch.

Step 6. Read off the boxed result.

\boxed{\Delta p = p_{\text{in}} - p_{\text{out}} = \gamma\left(\frac{1}{R_1} + \frac{1}{R_2}\right)}

For a sphere, $R_1 = R_2 = R$ gives $\Delta p = 2\gamma/R$ . For a cylindrical surface, one radius is infinite, giving $\Delta p = \gamma/R$ . For a flat interface, both radii are infinite and $\Delta p = 0$ , consistent with the everyday fact that flat liquid surfaces need no pressure jump to stay flat.

Worked examples

Example 1 — Pressure inside a small water droplet (foundational). Find the excess pressure inside a water droplet of radius $r = 1.0\ \mu\text{m} = 1.0\times10^{-6}\ \text{m}$ at 25 °C, where $\gamma_{\text{water}} = 72.0\times10^{-3}\ \text{N m}^{-1}$ . Because a droplet has a single liquid–vapor spherical surface,

\Delta p = \frac{2\gamma}{r} = \frac{2 \times 72.0\times10^{-3}\ \text{N m}^{-1}}{1.0\times10^{-6}\ \text{m}}.

\Delta p = 1.44\times10^{5}\ \text{Pa} \approx 1.42\ \text{atm}.

Sanity check: atmospheric pressure is $1.013\times10^5$ Pa, so this micron-sized droplet has an internal pressure roughly 1.4 times atmospheric added on top of ambient — a large but physically reasonable excess for a micron-scale radius, since $\Delta p$ scales as $1/r$ . For a millimeter raindrop ( $r = 1.0\times10^{-3}$ m) the same formula gives only $\Delta p = 144$ Pa, a thousand times smaller, confirming that curvature effects become negligible for macroscopic drops.

Example 2 — Two connected soap bubbles of different size (applied, exam-level). Two soap bubbles, radii $R_1 = 2.0\ \text{cm}$ and $R_2 = 5.0\ \text{cm}$ , are connected by a tube with a valve, both blown from a soap solution with $\gamma = 25.0\times10^{-3}\ \text{N m}^{-1}$ (soap solutions have lower $\gamma$ than pure water because surfactant adsorbs at the interface). Predict what happens when the valve opens, and compute the pressure excess of each bubble beforehand. Because a soap bubble has two surfaces (inner and outer film faces), the excess pressure is

\Delta p = \frac{4\gamma}{R}.

For the small bubble ( $R_1 = 2.0\times10^{-2}$ m):

\Delta p_1 = \frac{4 \times 25.0\times10^{-3}\ \text{N m}^{-1}}{2.0\times10^{-2}\ \text{m}} = 5.00\ \text{Pa}.

For the large bubble ( $R_2 = 5.0\times10^{-2}$ m):

\Delta p_2 = \frac{4 \times 25.0\times10^{-3}\ \text{N m}^{-1}}{5.0\times10^{-2}\ \text{m}} = 2.00\ \text{Pa}.

Since $\Delta p_1 = 5.00\ \text{Pa}$ exceeds $\Delta p_2 = 2.00\ \text{Pa}$ , the small bubble is at higher internal pressure. When the valve opens, air flows from the region of higher pressure to lower pressure: the small bubble empties into the large one. The small bubble shrinks further, which (counter-intuitively) raises its pressure even more, accelerating its own collapse, while the large bubble grows and its pressure drops further still. The process runs away until the small bubble vanishes and all its air joins the large one. Sanity check: this matches the everyday observation that when you connect two balloons of different sizes, the smaller one always deflates into the larger one — direct macroscopic evidence for the $1/R$ scaling of Laplace pressure, and a cautionary example that curved-surface equilibria can be unstable rather than stable.

Limiting cases & common traps

Case	Result	Note
Flat interface	$R_1, R_2 \to \infty \Rightarrow \Delta p = 0$	No pressure jump needed to hold a flat surface still
Sphere (droplet or bubble in liquid)	$\Delta p = 2\gamma/R$	Single surface, both radii equal
Cylinder (jet, fiber-wetting)	$\Delta p = \gamma/R$	One radius infinite
Soap bubble in air	$\Delta p = 4\gamma/R$	Two surfaces, doubles the droplet formula
Saddle (minimal surface, soap film neck)	$\Delta p = 0$ even though curved	$R_1$ and $R_2$ have opposite sign and cancel

The chemist behind it

The equation carries two names because two people derived it independently around the same time using different methods: Thomas Young reasoned about it physically in 1805, describing capillary action and surface tension in largely qualitative and geometric terms in his Essay on the Cohesion of Fluids. Pierre-Simon Laplace supplied the rigorous mathematical treatment in 1806, in the supplement to his Mécanique Céleste, treating the interface with a full analytical theory of molecular forces. Their combined result, still called the Young–Laplace equation today, became the foundation for everything from capillary rise measurements of surface tension to the modern theory of nucleation and colloidal stability.

Origins · 1 minPierre-Simon Laplace and how this was discovered

Pierre-Simon Laplace

1749–1827 · French mathematician and physicist who gave the Young–Laplace equation its rigorous analytical form in 1806, alongside foundational work in celestial mechanics and probability

Ask the tutor: How does this same Laplace pressure raise the vapor pressure of a small droplet, and lead to the Kelvin equation?

Why is the pressure inside a bubble always higher than outside?

Surface tension acts like a stretched elastic skin trying to minimize surface area. For a curved surface, that inward pull can only be balanced if the fluid on the concave (inside) has a higher pressure pushing outward. If the inside pressure equaled the outside pressure, the tension would have nothing to resist it, and the surface would collapse inward. This is true for any convex interface, from bubbles to droplets to the meniscus in a capillary tube.

Why does a soap bubble use 4γ/R rather than 2γ/R like a droplet?

A liquid droplet has exactly one interface: liquid on the inside, air outside. A soap bubble, however, is a thin liquid film with air on both faces — an inner air-film surface and an outer film-air surface. Each of these two surfaces independently obeys the sphere result $2\gamma/R$ , and since both act in the same direction (both pulling inward), their pressure contributions add, giving a total of $4\gamma/R$ .

How does the Young-Laplace equation connect to capillary rise?

In a narrow tube, the liquid meniscus forms a curved (roughly spherical-cap) surface with radius of curvature $R = r/\cos\theta$ , where $r$ is the tube radius and $\theta$ the contact angle. The Laplace pressure jump across this meniscus, $2\gamma\cos\theta/r$ , must be balanced by the hydrostatic pressure of the risen liquid column, $\rho g h$ . Setting these equal and solving for $h$ gives the capillary rise formula, making capillarity a direct macroscopic manifestation of the Young–Laplace law.

Does the Young-Laplace equation say anything about equilibrium vapor pressure over a curved surface?

Not directly — Young–Laplace is purely mechanical, relating pressure and curvature. But the extra internal pressure it predicts changes the chemical potential of the liquid inside a small droplet, which in turn raises its equilibrium vapor pressure above the flat-surface value. That thermodynamic consequence is quantified separately by the Kelvin equation, which builds directly on the Laplace pressure derived here.

What happens to Δp as the radius becomes very large or very small?

As $R \to \infty$ (a flat surface), $\Delta p \to 0$ : no pressure difference is needed to sustain a flat interface, consistent with everyday flat liquid pools sitting at ambient pressure. As $R \to 0$ (a vanishingly small droplet or bubble), $\Delta p \to \infty$ , meaning enormous pressures are needed to sustain extremely fine curvature — this divergence underlies why very small embryonic nuclei in nucleation are thermodynamically unstable and tend to redissolve unless they cross a critical size.