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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.

The sheet elastocapillary energy ratio compares capillary and bending energy scales for an elastic sheet:Here $L$ is a lateral size, $\gamma$ is surface tension, $B$ is plate bending stiffness and ${\ell }_{\mathrm{ec}}$ is the elastocapillary length. The ratio is a dimensionless quantity: $\gamma L^2$ and $B$ both have units of energy.

The estimate assumes bending through an angle of order one over size $L$. Then curvature scales as $1/L$ and integrating the bending energy density over area $L^2$ gives a scale $B$. A comparable change in interfacial area gives a surface energy scale $\gamma L^2$. The same balance underlies the capillary origami study's definition of elastocapillary length.^[ref]

For an original calculation, take $B=2\times 10^{-8}\mathrm{N}\mathrm{m}$, $\gamma =0.05N/m$ and $L=1\mathrm{mm}$. Then $\eta =2.5$. Doubling $L$ alone raises it to 10. Halving thickness at fixed $L$, Young's modulus and Poisson's ratio divides $B$ by eight and raises $\eta$ to 20.

Changing all dimensions gives another trend. Write $s=L/L_0$ and keep material properties and surface tension fixed. At fixed thickness, $\eta /{\eta }_0=s^2$; at fixed aspect ratio $h/L$, $B/B_0=s^3$ and $\eta /{\eta }_0=s^{-1}$. Thus halving $L$ alone gives one quarter of the initial ratio, while halving both $L$ and $h$ doubles it.

A larger ratio makes capillary energy more competitive with bending, but does not establish complete closure. Actual wetting, contact angle, droplet volume, contact-line pinning and accessible deformation paths also matter. The ratio compares scales rather than computing the full change in energy along a folding path.

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.

A diffuse-interface free energy balances the bulk preference for coexisting compositions against a positive penalty for sharp composition gradients. For dimensionless composition $\varphi$, $f_b$ is a bulk energy density and $K$ has units of energy per length. Subtract the bulk coexistence common tangent to define $\Delta \omega (\varphi )$, with equal zero minima at the two compositions. A planar equilibrium profile obeys $K{\varphi }^{\prime \prime }=\Delta {\omega }^{\prime }(\varphi )$ and $K({\varphi }^{\prime }{)}^2/2=\Delta \omega (\varphi )$. Its surface tension is $\int K({\varphi }^{\prime }{)}^{2}dz$. As a thermodynamic critical point is approached, the composition contrast and bulk curvature vanish, the correlation length grows, and the intrinsic interface broadens. A scalar quartic domain wall gives the mean-field hyperbolic-tangent profile. Thermally excited capillary waves can broaden a measured profile further.

For viscous fluid inside a circular fluid free boundary between parallel plates of separation $h$, the depth-averaged Hele-Shaw flow satisfies $\bar{u}=-K\nabla p$, $K=h^2/(12\mu )$. With volume injection rate $Q$, ${\bar{u}}_r=Q/(2\pi hr)$ and $R^2=R_0^2+Qt/(\pi h)$. Neglecting surface tension and exterior viscosity, pressure is constant at the interface. Perturb its radius by $a_n\cos n\theta$ and solve the interior Laplace equation with a regular pressure term proportional to $r^n\cos n\theta$. The linearized pressure boundary condition gives $\delta p(R)=a_{n}Q/(2\pi hKR)$, hence velocity perturbation $-n{a}_{n}Q/(2\pi h{R}^2)$. The displaced base velocity adds another $-a_{n}Q/(2\pi h{R}^2)$, proving the displayed growth law. Injection stabilizes the circular blob; suction destabilizes it. An exact radial solution can therefore exist while giving a fragile experimental prediction. These signs concern viscous fluid inside the blob, rather than the opposite geometry of a low-viscosity injected bubble.

A stress boundary condition specifies the traction acting on a boundary, rather than directly fixing its velocity. If $\mathbf{\sigma }$ is the Cauchy stress tensor and $\mathbf{n}$ a chosen unit normal, the traction on the corresponding oriented surface is $\mathbf{\sigma }\mathbf{n}$. Its component along $\mathbf{n}$ is normal traction; the remaining components are tangential.

At a clean liquid-gas interface with negligible gas viscosity, tangential traction is approximately zero, giving a stress-free boundary condition. The normal stress still balances external pressure and the curvature-dependent surface tension contribution. If surface tension varies along the interface, a tangential Marangoni stress must also be balanced.

These distinctions matter for a falling liquid film: a stationary solid wall usually imposes no slip, whereas its exposed surface can move while satisfying a prescribed traction. Calling the free surface stress-free does not set every component of its total stress tensor to zero.

Consider a long-wave film-rupture model whose strong-Marangoni stress asymptotic growth law isHere $h_0$ is the base film thickness, $\mu$ is dynamic viscosity, $\gamma >0$ is the base surface tension, $V>0$ is the model's destabilizing interaction coefficient with dimensions of energy, and $K=k{h}_0$ is dimensionless wavenumber. The precise relation of $V$ to a microscopic interaction depends on the chosen film model.

Set $y=K^2$. Differentiating $y(1-\Gamma y)$ gives $1-2\Gamma y$, so the unique unstable-band maximum isThe second derivative is negative. Since $\Gamma \gg 1$, the selected $K_m\ll 1$ is consistent with the long-wave assumption. The growth timescale is not itself a rupture time or bubble lifetime: a perturbation starting at amplitude $\delta h_0$ must grow to a specified nonlinear scale, giving an additional logarithmic dependence on the initial disturbance, and the base thickness may evolve before that happens.

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.

Outward interfacial flow from a thinning region dilutes surfactant there and raises its surface tension. The resulting Marangoni stress pulls fluid back toward the thinning region, opposes further stretching, and slows rupture.

For a clean spherical gas bubble with negligible internal dynamic viscosity and thermal conductivity, a weak temperature gradient produces the displayed migration velocity, where ${\gamma }_T=d\gamma /dT$. The exterior Papkovich–Neuber representation can use only the scalar potential $\chi =-a^3\mathbf{U}\cdot \mathbf{r}/(2r^3)$. Its surface normal velocity equals $\mathbf{U}\cdot \mathbf{n}$, and its tangential traction balances minus the surface surface tension gradient from the temperature dipole around an insulating sphere. For the usual ${\gamma }_T<0$ the bubble moves toward warmer liquid. Small Reynolds number, Péclet number, capillary number, and fractional surface-tension variation justify the approximation; surfactants can oppose the temperature-driven Marangoni stress.

For a liquid film heated below and cooled above, the quasistatic temperature of a conductively cooled film makes a thicker region colder at the surface. If surface tension decreases with temperature, Marangoni stress draws fluid towards thicker regions, competing with smoothing by capillary pressure. Writing ${\gamma }_s={\gamma }_0+bh$ gives $a_c={\gamma }_0/(3\mu )$, $a_m=b/(2\mu )$. A balanced long-wave scaling gives $H_{\tau }+(H^2{H}_X{)}_X+(H^3{H}_{XXX}{)}_X=0$. The dispersion relation $s=k^2-k^4$ has unstable band $0<|k|<1$ and fastest normal mode at $|k|=1/\sqrt{2}$. This thermal mechanism does not require surfactant.

Tillandsia usneoides, commonly called Spanish moss, is an epiphytic angiosperm in Bromeliaceae, the taxonomic family that also includes the pineapple. Despite its common name, Spanish moss is not a moss. Its slender shoots hang from supports, and conspicuous surface trichomes contribute to its silvery appearance and water relations.^[ref]

A 2021 study made surface water transport visible. Fluorescent droplets spread along shoots with intact trichome wings but much less effectively after those wings were removed. Imaging showed neighboring flexible wings flattening in succession as the wetting front advanced: a coupling between surface tension and deformable plant structures.^[ref]

This example separates three stages that are easy to collapse into one: bringing water to a shoot, spreading it over the surface, and absorbing it into living tissue. Seeing the wet front advance establishes transport along the surface; it does not by itself measure the rate at which every living cell takes up that water.

The epiphyte's supporting branch therefore supplies a place to grow, while the shoot surface solves part of the water-collection problem. Physical attachment and physiological resource acquisition need not occur at the same site.

An ideal fluid issuing from a small hole a vertical depth $h$ below a free surface has speed

Here $h$ is the vertical distance from the free surface to the centre of the orifice, and $g$ is the positive magnitude of gravitational acceleration. The ideal prediction assumes incompressible flow, the same ambient pressure at the free surface and emerging jet, negligible velocity of the reservoir's free surface, and negligible viscous dissipation and surface tension effects. For a falling free surface, a quasi-steady approximation additionally assumes that the local outflow adjusts much faster than $h$ changes. It applies the Bernoulli equation at each instant, rather than claiming that the whole drainage is steady flow.^[ref]

A short proof uses a streamline connecting the reservoir's free surface at elevation $z_s$ to the emerging jet at elevation $z_o$. With constant mass density $\rho$, negligible surface speed and no mechanical energy losses, the Bernoulli equation givesEqual pressures cancel, leaving $v^2/2=gh$: a unit mass gains the same kinetic energy as a body in free fall through height $h$. This makes Torricelli's law a useful first estimate before the detailed velocity field is known. At one quarter of the original head, ideal jet speed is one half as large; it is not proportional to head.

Far from the accelerating outlet region, the reservoir has approximately hydrostatic pressure $p_{\mathrm{ambient}}+\rho gh$ at the outlet's elevation. The emerging jet instead has approximately ambient pressure when surface tension is negligible. Do not assign that bulk hydrostatic pressure to the emerging jet as well as adding the height difference: that would count the available mechanical energy twice. A sealed reservoir, a submerged outlet or significant surface tension changes the pressure conditions and requires a different balance.

Actual volumetric flow rate is commonly represented by a discharge coefficient:where $a$ is the geometric area of the orifice. This definition compares volumetric flow rates, not just speeds. If a nearly uniform jet at its vena contracta has area $a_j$ and speed $v_j$, define the two separate ratios $C_c=a_j/a$ and $C_v=v_j/\sqrt{2gh}$. Then $Q=a_j{v}_j$ gives $C_d=C_c{C}_v$. A contracted jet can consequently have a speed close to $\sqrt{2gh}$ while $C_d$ is appreciably below one. No particular $C_d$ is universal: calibrate it for the opening and operating range.

Consider a nearly spherical bubble of radius $a$ translating with velocity $\mathbf{U}$ through a liquid of dynamic viscosity $\mu$. Neglect internal gas viscosity and use exterior Stokes flow. Let the surfactant concentration be $C=C_0+C^{\prime }$ and linearize surface tension as $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$ with ${\gamma }_1>0$. Surface transport has surface diffusion coefficient $D_s$ and exchange-relaxation rate $k$.

In the bubble's translating frame, write its tangential surface velocity aswhere $\mathbf{n}$ points out of the bubble. The clean-bubble value is $A=-1/2$; a completely immobile interface has $A=0$. The dipolar surfactant distribution is $C^{\prime }=B\mathbf{U}\cdot \mathbf{n}$, withCombining that transport solution with the tangential Marangoni stress balance givesIncreasing $\lambda$ suppresses interfacial motion, whereas stronger diffusion or exchange reduces the concentration contrast and restores mobility. The linearization separately requires $|C^{\prime }|/C_0\ll 1$, or $2|A||\mathbf{U}|a/(k{a}^2+2D_s)\ll 1$. Small Reynolds number and capillary number are also needed for Stokes flow and a nearly spherical shape; a large immobilization parameter alone does not establish those assumptions.

For an inviscid flow of constant mass density $\rho$ and depth $h$, impose excess pressure $p_1\cos (k(x-ct))$ with $k>0$ on a free surface. Neglect surface tension. A travelling solution of the linearized free-surface boundary conditions isThe velocity potential satisfies the Laplace equation and the bed no-penetration boundary condition. Its surface time derivative is $-A{c}^{2}k\coth (kh)\cos (k(x-ct))$, so the dynamic condition gives the displayed signed response. The nonnegative amplitude is $|A|$. It is finite except atthe squared phase speed of a free surface gravity wave at that wavenumber. Equality produces resonance. This describes a travelling forced component; free surface gravity waves must be added to meet general initial conditions. At exact resonance, a solution of the initial value problem generally grows with time rather than having a bounded periodic amplitude.

For two inviscid incompressible half-spaces, lower density/velocity ${\rho }_1,U_1$, upper density/velocity ${\rho }_2,U_2$, downward gravity $g$ and surface tension $\sigma$, a mode $e^{i(kx-\omega t)}$ with $k>0$ obeysA negative right-hand side produces exponential growth. The gravitational term recovers Rayleigh-Taylor instability for heavy-over-light ordering; the shear term can destabilize otherwise stable stratification. Finite layer thickness, viscosity, compressibility and magnetic tension lie outside this interface model.

Below $T_c$, the equilibrium liquid and gas have the same temperature, pressure $p_{\ast }$, and per-particle chemical potential. For macroscopic bulk thermodynamic phases, neglect interface contributions. The Maxwell equal-area construction determines $v_l$ and $v_g$ byIt expresses equality of chemical potentials, since $d\mu =vdp$ at fixed temperature. Equivalently, the equilibrium Helmholtz free energy per particle follows a common tangent line between the two endpoint states, lying below the homogeneous continuation between them. The construction uses that mathematical continuation to compare endpoints; it is not a physical compression path through negative isothermal compressibility states.^[ref]

Here is a reproducible calculated example at $t=0.9$. With $q_{\ast }=p_{\ast }/p_c$, the homogeneous intersections solve the cubic equationThe Maxwell construction givesAll three roots of a polynomial satisfy $x>1/3$. Only the smallest and largest are the phase coexistence endpoints; the middle one has negative isothermal compressibility. Equal pressures at three admissible values of volume therefore do not imply three stable thermodynamic phases. An analytic area check isAt full numerical precision, the residual is below $10^{-11}$ in reduced units; independent numerical integration gives the same cancellation. At $t=0.9$, the spinodal points are instead $x\simeq 0.718597189$ and $1.528504964$.

The equilibrium plateau changes the phase fraction rather than the endpoint densities. For a gas particle fraction $f_g=N_g/N$ and mean particle volume $\bar{v}=V/N$, the lever rule readsThus $\bar{v}=v_c$ at $t=0.9$ gives $f_g\simeq 0.227219$: this is a particle fraction, not the volume fraction occupied by gas. Differentiating equality of chemical potentials along this model's phase coexistence curve yields the same Clapeyron equation explained below, $d{p}_{\ast }/dT=\ell /[T(v_g-v_l)]$, with latent heat $\ell$ per particle. Both endpoint differences vanish as the Van der Waals critical point is approached; the equation's critical limit must be taken as a ratio.

The Maxwell construction presumes a pure bulk fluid at thermodynamic equilibrium. Its endpoint equality does not specify how rapidly a real compression reaches equilibrium. A homogeneous metastable sample can persist while nucleation is delayed; the spinodal marks loss of local mechanical stability in this mean-field theory, rather than the equilibrium onset of condensation. Surface tension, finite sample size and impurities can affect the observed transition. These qualifications matter when comparing a mathematical Van der Waals isotherm with a visible liquid meniscus.