Surface tension acts tangentially along an interface. When the interface is curved, those tangential forces have a net normal effect that must be balanced by a pressure difference across the interface.

The resulting relation is the Young-Laplace equation:

$$\boxed{\Delta p=\gamma\left(\frac1{R_1}+\frac1{R_2}\right)},$$

where

• $\Delta p$ is the pressure on the concave side minus the pressure on the convex side under the chosen curvature convention;
• $\gamma$ is the surface tension;
• $R_1$ and $R_2$ are the two principal radii of curvature of the interface.

The key physical message is independent of sign convention: greater curvature requires a larger pressure jump to balance the same surface tension.

Spherical droplet

For a spherical liquid droplet of radius $R$, both principal radii are $R$. Therefore

$$\boxed{p_{\rm inside}-p_{\rm outside}=\frac{2\gamma}{R}}.$$

This result can also be derived directly from force balance.

Imagine cutting the droplet into two hemispheres. The excess internal pressure acts over the circular cross-sectional area $\pi R^2$, producing force

$$F_p=\Delta p,\pi R^2.$$

Surface tension acts around the circular rim of circumference $2\pi R$, producing the opposing force

$$F_\gamma=\gamma(2\pi R).$$

Equilibrium requires

$$\Delta p,\pi R^2=2\pi R\gamma,$$

so

$$\boxed{\Delta p=\frac{2\gamma}{R}}.$$

A smaller droplet therefore has a larger internal excess pressure.

Cylindrical interface

For an ideal cylindrical interface of radius $R$, one principal radius is $R$ and the other is infinite because the surface is not curved along the cylinder axis. Hence

$$\boxed{\Delta p=\frac{\gamma}{R}}.$$

This distinction shows why curvature must be considered in two independent surface directions.

Soap bubble: two interfaces

A thin soap bubble has an inner liquid-gas interface and an outer liquid-gas interface. Each interface contributes approximately

$$\frac{2\gamma}{R}$$

when the film is thin compared with the bubble radius. The total pressure excess is therefore

$$\boxed{p_{\rm inside}-p_{\rm outside}\approx\frac{4\gamma}{R}}.$$

The factor of four is not the Young-Laplace formula for one spherical interface; it comes from two interfaces, each contributing $2\gamma/R$.

Worked example

A water droplet of radius

$$R=1.0,\mathrm{mm}=1.0\times10^{-3},\mathrm m$$

has surface tension approximately

$$\gamma=0.072,\mathrm{N/m}.$$

Its internal excess pressure is

$$\Delta p=\frac{2(0.072)}{1.0\times10^{-3}} =144,\mathrm{Pa}.$$

Thus

$$\boxed{p_{\rm inside}=p_{\rm outside}+144,\mathrm{Pa}}.$$

If the radius were reduced by a factor of ten, the pressure jump would increase by a factor of ten.

Curvature rather than surface tension alone

A flat interface has

$$R_1,R_2\to\infty,$$

so the curvature contribution tends to zero and there need not be a surface-tension-induced pressure jump across a static flat interface.

Surface tension therefore does not create one fixed pressure difference. The pressure jump depends on surface tension multiplied by curvature. This relation underlies droplets, bubbles, menisci, capillary rise, wetting phenomena, and many small-scale fluid systems.