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The capillary length measures when gravity and surface tension are comparably important in shaping a liquid interface. For a liquid of uniform mass density $\rho$, beneath a gas whose mass density is negligible, with uniform gravitational acceleration $g>0$ and constant surface tension $\gamma >0$,A deformation of height $a$ and horizontal length scale $L$ has hydrostatic pressure variation of order $\rho ga$ and small-slope capillary pressure of order $\gamma a/L^2$. Equating these gives $L\sim {\ell }_c$: surface tension smooths small features, whereas gravity matters for broad ones. For two resting fluids, replace $\rho$ by their positive mass density difference $\Delta \rho$ in this static balance. This change alone does not give the two-fluid wave dynamics, because both fluids can have inertia. ^[ref]

With the stipulated illustrative parameters $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the capillary length is ${\ell }_c=2.71\mathrm{mm}$. These parameters define the ideal calculations below; they are not a measurement of a particular sample.

For a sinusoid with wavenumber $k>0$, the relevant horizontal scale is $L=1/k$, while its wavelength is $\lambda =2\pi /k$. Thus the Bond number for this scale and the reciprocal restoring-term ratio areEquality is at $k_c=1/{\ell }_c$, hence ${\lambda }_c=2\pi {\ell }_c=17.02\mathrm{mm}$ for these parameters. Using $L=\lambda$ instead produces a different numerical Bond number; state the length convention. The capillary length itself is not the equality wavelength. This distinction helps choose whether a measured ripple should be modeled as a surface gravity wave, a nearly pure capillary wave, or a gravity-capillary wave. ^[ref]

A capillary wave is a wave along a fluid interface whose restoring stress is predominantly supplied by surface tension. A curved free surface requires a pressure jump by the Young–Laplace equation; increasing its corrugation also costs surface energy. In the ideal limit, inertia carries the interface through its flat position and the resulting oscillation propagates. When gravity also contributes appreciably, use the combined gravity-capillary wave model. Within the capillary wave regime, shortening a ripple makes it travel faster. ^[ref]

The laser-diffraction measurement of capillary waves below turns this mechanism into a test: set the frequency, measure the wavelength independently with light, and ask whether the dispersion relation follows. The theory is useful both for predicting how ripple wave packets separate and for inferring surface tension from their motion.

A dispersion relation specifies which angular frequencies and wavevectors a linear wave model admits. Fix the plane wave conventionHere $a$ is a constant complex amplitude, $\mathbf{k}$ is the wavevector, and $\omega$ is the angular frequency. Substitution into a constant-coefficient linear partial differential equation replaces ${\partial }_t$ by $-i\omega$ and $\nabla$ by $i\mathbf{k}$, turning the wave problem into an algebraic condition. A system can have several roots; $\omega =\omega (\mathbf{k})$ means one chosen branch, rather than a promise that the full dispersion relation is single-valued. For real $\mathbf{k}$, $\mathrm{Im}\omega >0$ means temporal growth in this convention, while $\mathrm{Im}\omega <0$ means damping. The real-branch discussion of phase velocity and group velocity below assumes negligible loss.

For the one-dimensional wave equation $u_{tt}=c^2{u}_{xx}$, the dispersion relation is ${\omega }^2=c^2{k}^2$. The branch $\omega =ck$ carries right-moving travelling waves for $c>0$; both its phase velocity and group velocity are $c$. One may instead use the positive-angular frequency branch $\omega =c|k|$, with right-moving components at $k>0$ and left-moving components at $k<0$. A real wave pairs each $(k,\omega )$ with $(-k,-\omega )$; that conjugate pair represents the same travelling pattern. Branch and sign choices therefore matter when interpreting a dispersion diagram.

A particularly tangible dispersive example is a small-amplitude surface gravity wave on still water of uniform depth $h$. Neglect surface tension and viscosity, assume incompressible flow and irrotational flow, and impose a flat impermeable bottom. With $k>0$ and positive angular frequency,Here $g$ is gravitational acceleration and $\tanh$ is the hyperbolic tangent. Deep water means $kh\gg 1$: $\omega \simeq \sqrt{gk}$ and longer wavelengths have larger phase speed. Shallow water means $kh\ll 1$: $\omega \simeq \sqrt{gh}k$, an approximately nondispersive wave. Thus depth relative to wavelength, rather than depth alone, determines whether crests and wave packets travel together.^[ref]

The dispersion diagram below also includes the ideal deep-water capillary wave limit $\omega =\sqrt{\gamma /\rho }{k}^{3/2}$, where $\gamma$ is surface tension and $\rho$ is mass density. This limit requires the surface tension restoring term to dominate gravity; it reverses the trend of the surface gravity wave, giving larger phase speed at shorter wavelength. The combined gravity-capillary wave has ${\omega }^2=gk+(\gamma /\rho )k^3$.^[ref]

The curves use dimensionless normalizations chosen separately for each model; they are analytical diagrams, not experimental data. Their beauty is that a single dispersion relation encodes several observable behaviours: its value gives the angular frequency, its origin-to-point slope gives the one-dimensional phase velocity, its tangent slope gives the group velocity, and variation of that tangent slope predicts distortion of a wave packet. A straight line through the origin describes a nondispersive wave; an affine branch $\omega =ck+b$ with $b\ne 0$ has constant group velocity but phase velocity $c+b/k$, so those properties must be distinguished.

A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.