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Using only this upper-phase flux in $\dot{h}=v_n=-J_n/(2{\varphi }_B)$ givesPositive surface tension raises the chemical potential at a crest, drives material away from it, and therefore smooths the interface, which fixes the negative sign of the decay rate.

The dynamics is nonlocal because the conserved composition must diffuse through the bulk. Solving Laplace's equation maps a boundary Fourier amplitude to its normal derivative by multiplication by $|q|$. Combining this Dirichlet-to-Neumann factor with the curvature factor $q^2$ produces the nonanalytic $|q{|}^3$ rate described by diffusion-limited relaxation of an interface.

At the final state of the slump test for yield stress, an axisymmetric deposit of height $h(r)$ is just able to support itself. Hydrostatic pressure gives radial pressure gradient $\rho g{h}^{\prime }(r)$, while lubrication theory makes the magnitude of the basal shear stressThe marginally yielded final profile therefore satisfiesIntegration givesUsing mass conservation, the known volume isSolving for the yield stress produces the estimateThus one measures the final radius $R$ and inserts it with $V$ and $\rho$. The estimate assumes a thin deposit, negligible surface tension, complete initial yielding, and a spatially uniform yield stress.

Dry sand is a granular material governed primarily by frictional stability. Its final free surface reaches the angle of repose ${\theta }_r$, so the deposit is approximately a cone,This constant-slope profile differs from the square-root edge of the yield-stress-fluid model.

The phase velocity is the motion of a constant-wave phase surface. For a one-dimensional plane wave $u=a\cos (kx-\omega t+{\varphi }_0)$ with real $k\ne 0$ and real $\omega$, fixing $kx-\omega t+{\varphi }_0$ givesEach wave crest of this monochromatic travelling wave advances at that signed phase velocity. Its magnitude is the phase speed $|c_p|=\lambda /T$, where $\lambda =2\pi /|k|$ is the wavelength and $T=2\pi /|\omega |$ is the temporal period when $\omega \ne 0$. Choosing $\omega >0$ makes positive $k$ mean rightward phase velocity and negative $k$ mean leftward phase velocity. Reversing both signs leaves the real plane wave and its phase velocity unchanged. At $k=0$ there are no spatial crests to track, so $\omega /k$ is undefined; a finite limit as $k\to 0$ is a limiting phase speed, not the speed of a spatially uniform oscillation.

In several dimensions, a plane wave has wave phase $\mathbf{k}\cdot \mathbf{x}-\omega t+{\varphi }_0$. Constant-wave phase planes obey $\mathbf{k}\cdot \mathbf{v}=\omega$. Their normal phase velocity isTangential motion within a plane does not move that wavefront, which is why the normal direction fixes the vector convention. In an anisotropic medium the group velocity need not point along this wavevector. Likewise the horizontal phase velocity of a tilted wavefront is its crossing speed along a horizontal line, rather than a component of the normal phase velocity.

For the nondispersive wave branch $\omega =ck$, a wave crest and a wave envelope both move at $c$. For deep-water surface gravity waves, $\omega =\sqrt{gk}$ gives $c_p=\sqrt{g/k}$ and group velocity $c_g=c_p/2$. With the illustrative choices $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\lambda =1\mathrm{m}$, the model predicts $c_p\simeq 1.25\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.625\mathrm{m}{\mathrm{s}}^{-1}$. These are calculated values for negligible surface tension and sufficiently large $kh$, not measurements. A wave crest moves forward through the wave packet, so timing a crest does not time the packet's journey.

There is a direct laboratory meaning to this distinction. MIT's 2005 surface-wave lab uses a long water tank, a controllable hydraulic paddle, and an absorbing beach. Resistance probes convert submerged length into voltage; calibration converts their readouts into free surface elevation versus time. Two probes a known distance apart, backed by a ruler and stopwatch, let students time a wave crest and separately time a feature of the wave packet. The lab records depth and varies frequency and wave amplitude.^[ref]

For a clean measurement of phase velocity, fit the periodic portion of both records or follow one identifiable wave crest. A phase difference is only known modulo $2\pi$: probe spacing below half the shortest relevant wavelength, or continuity while sweeping frequency, avoids assigning the wrong whole-cycle delay. Then compare the timed phase velocity with $\lambda /T$. Tracking pepper grains instead measures fluid-particle motion, which has a different meaning from either phase velocity or group velocity.

The Klein-Gordon packet velocities example makes the distinction especially sharp: its phase velocity exceeds $c$, yet its narrow wave envelope moves below $c$. A carrier wave labels successive oscillations; its speed alone does not establish information velocity. The front velocity depends on the full causal model, including angular frequencies far outside the carrier's narrow range.

A symmetric thin viscous sheet with shear-free faces has leading longitudinal velocity independent of thickness coordinate $z$, and incompressibility gives $w=-z{u}_x$. Young–Laplace equation and the Newtonian fluid stress tensor give ${\sigma }_{zz}=-p_a+\gamma h_{xx}/2$ and ${\sigma }_{xx}=-p_a+\gamma h_{xx}/2+4\mu u_x$. The effective longitudinal tension per unit transverse width is $\mathcal{T}=4\mu h{u}_x+\gamma h{h}_{xx}/2-\gamma h_x^2/4+2\gamma$. The last two terms include the horizontal projections of the two surface tension pulls. Force balance gives ${\mathcal{T}}_x=0$, yielding the displayed longitudinal equation; the face kinematic conditions give the displayed conservation of mass equation.

For an inviscid interface with no surface tension, continuity of normal traction reduces to continuity of pressure. The same condition is used in a sharp Darcy law interface model that neglects capillary pressure and resolved interfacial viscous normal stress. For a general viscous interface, continuity of normal traction balances the full stress tensor; it does not by itself imply continuity of pressure.

For spherical expansion in an unbounded incompressible flow, continuity gives $u_r=R^2\dot{R}/r^2$. Integrating the kinetic energy outside the bubble gives $K=2\pi \rho R^3{\dot{R}}^2$. The Rayleigh equation for an inviscid spherical bubble then yields $\dot{K}=(p_b-p_{\infty })\dot{V}$, where $V=4\pi R^3/3$. With $p_b=p_{\infty }{V}_0/V$ and constant $p_{\infty }$, integration gives $K-K_0=p_{\infty }[V_0\log (V/V_0)-V+V_0]$. This neglects viscosity, surface tension and external body-force differences, which contribute additional terms in other bubble models.

The Rayleigh-Plesset equation balances the radial inertia of an incompressible liquid around a spherical bubble against internal and ambient pressure, surface tension and viscous stress. Removing the last two effects and setting the bubble pressure to zero gives the following vacuum-collapse case.

The Saffman–Taylor instability occurs when a more mobile fluid displaces a less mobile fluid in a porous medium or Hele–Shaw cell. Viscosity contrast amplifies interface corrugations, while surface tension suppresses sufficiently short wavelengths.

A slender viscous thread has a cross-sectional scale much smaller than its axial variation scale. For an axisymmetric Newtonian thread of area $A$ and nearly uniform axial velocity $w$, mass conservation and the Trouton ratio givewhen inertia and surface tension are negligible and $z$ points downward.

For an ideal soap film with equal constant surface tension $\gamma$ on both faces, the effective tension is $\tau =2\gamma$. Choose a unit normal $N$ from the minus side to the plus side, and $\Delta p=p_{-}-p_{+}$. Translating the Young–Laplace equation to $S=-dN$ givesA single interface uses $\tau =\gamma$. An outward-oriented spherical soap bubble has $H=-1/R$, giving $\Delta p=4\gamma /R$. The positive-convex curvature sum in the Young–Laplace equation article is $-2H$. ^[ref]

An open film with the same resting gas on both sides has zero pressure jump. Neglecting film weight, soap film pressure balance gives $H=0$: a minimal surface. A pressurized bubble instead has nonzero mean curvature. These statements assume equilibrium, constant surface tension, and negligible normal viscous stress. Surfactant gradients or motion require the fuller normal stress balance. ^[ref]

The variational outline is short. With fixed boundary, a normal displacement $fN$ gives $\delta A=-2{\int }_{M}HfdA$ by the first variation of area formula. For a closed interface oriented outward, $\delta V={\int }_{M}fdA$, so stationarity of $\tau A-\Delta pV$ gives $-2\tau H-\Delta p=0$. This is a first variation calculation; testing for a stable minimal surface needs the second variation, and deciding a global minimum needs a comparison of competitors.

A buoyantly rising inviscid bubble in otherwise quiescent unbounded liquid has an exterior Stokeslet with $F=4\pi \mu aU=4\pi \rho g{a}^3/3$. Its tangential traction vanishes on the sphere. The dynamic normal traction is $-3F\cos \theta /(4\pi a^2)$, while the hydrostatic normal-traction variation is $\rho ga\cos \theta$. Their dipoles cancel at the terminal velocity, leaving a constant total normal stress compatible with uniform gas pressure even when surface tension is zero. This is an exact spherical solution in the stated Stokes model, not a proof of shape stability under arbitrary perturbations.

For a liquid interface with constant surface tension $\gamma$, the mechanical surface energy is the energy associated with its surface area $A$:The additive constant $C$ does not affect the work needed to change surface area. For several interfaces with the same surface tension, $A$ means their total surface area; the two faces of a thin soap film must both be counted. ^[ref]

With assumed $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, creating $1\mathrm{c}{\mathrm{m}}^2=10^{-4}{\mathrm{m}}^2$ of one interface requires $7.2\mu \mathrm{J}$ of reversible work. In the sliding-rim example illustrated under surface tension, moving a $20\mathrm{mm}$ rim outwards by $1\mathrm{mm}$ creates $4.0\times 10^{-5}{\mathrm{m}}^2$ across the two faces and requires $2.88\mu \mathrm{J}$. Independently, multiplying its $2.88\mathrm{mN}$ holding force by the displacement gives the same work.

This makes the preference for a round isolated droplet precise. At fixed volume $V$, the three-dimensional isoperimetric inequality giveswith equality for a sphere. Thus a sphere minimizes surface energy among sufficiently regular closed shapes enclosing the same volume, when surface tension is uniform and other energy contributions are negligible. This is an application of the isoperimetric inequality, rather than a proof of it. Walls, wetting, and gravity can change the shape by adding constraints or other energy contributions. ^[ref]

The differential rule $d{E}_{\mathrm{s}}=\gamma dA$ integrates directly to $E_{\mathrm{s}}=\gamma A+C$ because surface tension is constant. At fixed liquid volume, decreasing surface area can drive viscous fluid flow; the released surface energy can supply viscous dissipation. This mechanical bookkeeping is useful for predicting which deformations can release energy. If surface tension depends on temperature or composition, distinguish this mechanical area-work relation from the complete interfacial thermodynamic free energy and internal energy.

Surface tension $\gamma$ is the tensile force per unit length across a line drawn in a liquid interface. For a clean interface with isotropy, it has the same value in every tangential direction: each side pulls the other along the surface, perpendicular to the line. A straight segment of length $\ell$ therefore transmits a force of magnitude $\gamma \ell$ from one interface. Here we assume constant $\gamma >0$, with fixed temperature and composition. The same coefficient measures the reversible work required to create surface area:The equality of units follows from $1\mathrm{J}=1\mathrm{N}\mathrm{m}$; the equality of the mechanical coefficients follows from the moving-rim argument below. ^{[ref] [ref]}

For a concrete example, take an ideal flat soap film spanning a rectangular wire frame with a straight sliding rim of length $\ell$. Pull the rim outwards in the plane of the film, perpendicular to the rim. Each of the film's two liquid–gas interfaces pulls the rim back towards the film with force $\gamma \ell$, so the outward holding force is $F=2\gamma \ell$. The two faces are two interfaces, each with its own surface tension. With the assumed toy value $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $\ell =20\mathrm{mm}=0.020\mathrm{m}$, the required force is $2.88\mathrm{mN}$; one interface would contribute $1.44\mathrm{mN}$. This value of $\gamma$ is a stipulated model parameter, rather than a measured value for a particular soap solution.

Why does an interface resist being enlarged? Its molecules have a different environment from those in the bulk liquid; changing that environment has an energy cost. In the constant-surface tension model, reducing surface area lowers surface energy. This connects an easily measured rim force to the shape of a droplet, the Young–Laplace equation for its pressure, and the restoring mechanism of capillary waves. It is useful because one coefficient links forces, energy and geometry. ^[ref]

The moving-rim argument also explains the factor of two without memorizing a special formula. An outward displacement $dx$ creates surface area $\ell dx$ on each face, hence $d{A}_{\mathrm{total}}=2\ell dx$. Slow pulling without viscous dissipation requires work $Fdx=\gamma d{A}_{\mathrm{total}}$, giving $F=2\gamma \ell$. For one interface, the same argument gives $F=\gamma \ell$. A surface tension gradient instead produces Marangoni stress; the constant-$\gamma$ assumption excludes that additional effect.