= Young–Laplace equation
{c}
{title2=$\Delta p=\gamma\kappa$}
{wiki}
The <Young–Laplace equation> relates the <pressure> jump across a stationary interface to its <curvature> and <surface tension>. Choose a unit <normal vector> $\mathbf n$ pointing from the inside region to the outside region, and define $\kappa=\nabla_s\cdot\mathbf n$ to be positive for a convex <sphere>. In mechanical <equilibrium>, with constant <surface tension> and no additional normal <viscous stresses>,
$$
\Delta p=p_{\mathrm{in}}-p_{\mathrm{out}}
=\gamma\kappa=\gamma(k_1+k_2).
$$
Here $k_1,k_2$ are the signed <principal curvatures> in this convention. Their sum is twice the corresponding signed <mean curvature>; stating the convention prevents a factor-of-two ambiguity. The <Young–Laplace equation> is a local <normal stress> balance for one interface. https://arxiv.org/html/2011.11403v2[Tomas Bohr and Bernhard Scheichl, Surface Tension and Energy Conservation in a Moving Fluid, equations (8)–(9)]{ref}
A spherical <liquid> droplet of <radius> $R$ in a <gas> has one interface, with $k_1=k_2=1/R$. An ideal thin soap bubble has two <liquid>–<gas> interfaces. Adding their <pressure> jumps gives
$$
\Delta p_{\mathrm{droplet}}=\frac{2\gamma}{R},\qquad
\Delta p_{\mathrm{soap\ bubble}}\simeq\frac{4\gamma}{R}.
$$
For the soap bubble, $\Delta p$ now means the inner <gas> <pressure> minus the outer <gas> <pressure>. The second formula assumes equal <surface tensions> on both faces and film thickness negligible relative to $R$, so their radii can be identified. Retaining distinct radii gives $2\gamma/R_{\mathrm{inner}}+2\gamma/R_{\mathrm{outer}}$. A <gas> bubble inside bulk <liquid> has only one interface and therefore uses $2\gamma/R$. https://openstax.org/books/college-physics-2e/pages/11-8-cohesion-and-adhesion-in-liquids-surface-tension-and-capillary-action[OpenStax, College Physics 2e, §11.8: pressure in a soap bubble]{ref}
With the assumed constant toy <surface tension> $\gamma=0.072\,\mathrm{N\,m^{-1}}$, a droplet with $R=1\,\mathrm{mm}$ has excess <pressure> $144\,\mathrm{Pa}$; at $R=0.5\,\mathrm{mm}$ it has $288\,\mathrm{Pa}$. A thin soap bubble assigned the same $\gamma$ and $R=1\,\mathrm{mm}$ has excess <pressure> $288\,\mathrm{Pa}$. These are differences from the surrounding <pressure>, rather than absolute <pressures>. Halving the <radius> doubles the <capillary pressure>; doubling the interface count also doubles it at fixed <radius> and <surface tension>.
\Image[/surface-tension-droplet-pressure.png]
{title=Smaller droplets need larger excess pressure; two bubble interfaces double it}
{description=Left: a virtual outward displacement $dR$ of a <sphere> balances <pressure> <work> against increased <surface energy>. Right: excess <pressure> in pascals versus positive <radius> in millimetres for one and two interfaces, using the assumed constant toy $\gamma=0.072\,\mathrm{N\,m^{-1}}$.}
The spherical case has a short <work> argument. A virtual increase in <radius> gives $dV=4\pi R^2\,dR$ and $dA=8\pi R\,dR$. At mechanical <equilibrium>, the net <pressure> <work> balances the increase in <surface energy>:
$$
\underbrace{\Delta p\,4\pi R^2\,dR}_{\text{pressure work}}
=\underbrace{\gamma\,8\pi R\,dR}_{\text{surface energy increase}}
\quad\Longrightarrow\quad \Delta p=\frac{2\gamma}{R}.
$$
For a thin soap bubble, both interfaces gain approximately the same <surface area>, doubling the right-hand side and giving $4\gamma/R$. This is a virtual <work> balance between neighboring spherical configurations, rather than an assumption that a fixed-<volume> droplet actually expands. The general <Young–Laplace equation> replaces the spherical area-to-<volume> ratio by local <curvature>, making geometry directly measurable through <pressure>.
Back to article page
