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Archimedes' principle states that buoyancy equals the downward gravitational weight, or weight, of the displaced fluid. In a resting fluid of constant mass density ${\rho }_f$ under uniform gravitational acceleration $g$, its magnitude isHere $V_{\mathrm{displaced}}$ is the volume from which the body excludes the fluid: the whole external volume for a fully immersed solid body, and just the immersed volume for a partly immersed one. The body's own mass does not enter this buoyancy formula. ^[ref]

A short proof outline for a fully immersed body is to imagine replacing it by exactly the fluid it displaces. The surrounding hydrostatic pressure is unchanged. That replacement fluid is at rest, so its upward pressure force must balance its downward force from gravity. The original body receives the same inward pressure forces on the same closed surface, hence the same upward buoyancy.

Equivalently, let $\Omega$ be the displaced region, $\partial \Omega$ its closed surface and ${\mathbf{e}}_z$ the upward unit direction. The divergence theorem, applied component by component to the surface integral, and hydrostatic pressure giveThis also explains why horizontal forces cancel even for an irregular shape. A spatially constant surface pressure contributes zero to the closed-surface integral. If the mass density varies with depth, replace ${\rho }_f{V}_{\mathrm{displaced}}$ by the mass ${\int }_{\Omega }{\rho }_{f}dV$ of the displaced fluid.

The closed-surface argument assumes that the fluid applies pressure over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no fluid beneath it; then the lower-face pressure force used in the proof is absent, and a supporting contact force must be treated separately. For partial immersion, the usual buoyancy formula assumes the exposed surface receives the common atmospheric pressure, neglects the much smaller buoyancy from the air, and neglects surface tension.

For the $0.10\mathrm{m}$ cube in the stipulated water, full immersion displaces $0.001{\mathrm{m}}^3$, so $F_b=1000\times 9.81\times 0.001=9.81\mathrm{N}$. If its uniform mass density is $600\mathrm{kg}{\mathrm{m}}^{-3}$, its mass is $m=600\times 0.001=0.600\mathrm{kg}$ and its downward force from gravity is $mg=5.886\mathrm{N}$. To hold it fully immersed at rest requires an additional downward forceOn release from rest it initially moves upward; its eventual floating equilibrium has a smaller displaced volume. A same-size cube of uniform mass density $1200\mathrm{kg}{\mathrm{m}}^{-3}$ has mass $1.200\mathrm{kg}$ and downward force from gravity $11.772\mathrm{N}$. Full immersion still gives $9.81\mathrm{N}$ of buoyancy, leaving an initial net downward force of $11.772-9.81=1.962\mathrm{N}$. Subsequent motion also involves drag.

Let fluid 1 lie above fluid 2, with mass densities ${\rho }_1<{\rho }_2$ and dynamic viscosities ${\mu }_1<{\mu }_2$. Define $\Delta \rho ={\rho }_2-{\rho }_1>0$, $\Delta \mu ={\mu }_2-{\mu }_1>0$, porosity $\varphi$, permeability of a porous medium $k$, and downward interface speed $W$. With Darcy flux $\varphi W$, no capillarity, and two unbounded layers, Darcy's law givesFor a Fourier mode of interface displacement, the two perturbation pressures decay exponentially into the respective layers. The kinematic boundary condition makes their amplitudes proportional to $\varphi {\mu }_1\sigma /(k|\alpha |)$ and $-\varphi {\mu }_2\sigma /(k|\alpha |)$. Linearized pressure continuity adds the base-gradient difference $\varphi W\Delta \mu /k-g\Delta \rho$. Eliminating the amplitudes proves the formula. The critical interface speed is $kg\Delta \rho /(\varphi \Delta \mu )$; the critical Darcy velocity is $kg\Delta \rho /\Delta \mu$. Without surface tension this model has unbounded short-wave growth rate on its unstable side.

For a drop with internal viscosity $\lambda \mu$, exterior viscosity $\mu$, continuous velocity and a decaying exterior Stokes flow, use the drop-outward normal $n$ and curvature $\kappa ={\nabla }_s\cdot n$, twice the mean curvature. Constant surface tension gives $({\sigma }^{+}-{\sigma }^{-})n=\gamma \kappa n$. Multiply the interior boundary representation by $\lambda$ and add it to the exterior one: the traction difference yields the single-layer term shown, while the viscosity-independent stress kernel yields $(\lambda -1)D[u]$. Here $S,D$ denote the kernels in boundary integral representation of Stokes flow. For equal viscosities the double layer vanishes.

For equal internal and external pressures, zero axial strain rate and positive surface tension, annular viscous extension gives constant $A=R_e^2-R_i^2$ and thickness $D(t)=R_e(0)-R_i(0)+\gamma t/(2\mu )$. Since $R_i=(A/D-D)/2$, the central hole closes atThe formula holds until closure and requires a genuine initial annulus; the subsequent topology has no inner cylindrical interface.

For a uniform annular sheet attached to a fixed outer rim $R_0$, two surface tension pulls give $h{\sigma }_{rr}=-2\gamma$ at the hole radius $R$. The zero-pressure radial equation gives $u=Ar+B/r$, with $B=-A{R}_0^2$ and $A=-\gamma R^2/[\mu h(R_0^2+3R^2)]$. The thickness remains uniform by conservation of mass. With negligible initial hole volume, $h=h_0/(1-x^2)$ andA positive seed hole grows initially exponentially. An exactly zero initial radius remains zero within this continuum evolution law.

A thin axisymmetric coating of physical thickness $d=ϵah$ on a rigid cylinder of radius $a$ has leading curvature $a^{-1}-d/a^2-d_{\bar{z}\bar{z}}$. With constant surface tension $\gamma$, negligible Newtonian gravity, no-slip boundary condition at the wall and negligible exterior stress, lubrication theory gives $q=-d^3{p}_{\bar{z}}/(3\mu )$. Set $z=\bar{z}/a$ and $t=\gamma {ϵ}^3\bar{t}/(3\mu a)$ to obtain the displayed thin-film equation. A perturbation of a uniform film $h_0$ proportional to $e^{st+ikz}$ hasThus the fastest Rayleigh–Plateau instability has $|k|=1/\sqrt{2}$, or physical wavelength $2\pi \sqrt{2}a$.

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]

The capillary number compares viscous stresses with surface tension. Its velocity and viscosity conventions must be specified; a porous-medium displacement can use the Darcy velocity and the viscosity of the displaced fluid.

When a nonwetting phase recedes from pores, surface tension and pore geometry can retain a residual saturation $s$. If $h$ is the mobile invaded thickness and $M$ its maximum past value, stored phase volume per plan area is $\varphi [(1-s)h+sM]$. Newly invading pores have $M=h$; during recession $M$ remains fixed. With mobile Darcy velocity $u$, conservation gives advance speed $u/\varphi$ and recession speed $u/[\varphi (1-s)]$. The latter is faster for $0<s<1$.

The height jump in linear similarity profiles for surfactant spreading is smoothed by surface tension. If $C_x\sim M/x_N^2$ through a region of width $\Delta$, the Marangoni stress flux is $AM{h}_0^2/(\mu x_N^2)$ and the capillary flux is ${\gamma }_0{h}_0^4/(\mu {\Delta }^3)$. Their balance gives $\Delta \sim ({\gamma }_0{h}_0^2{x}_N^2/(AM){)}^{1/3}\propto t^{2/9}$. This estimate assumes gravity is negligible and the smoothing region remains narrow compared with the pool.

For a hollow slender viscous thread, $A=R_e^2-R_i^2$ and both surfaces contribute axial surface tension. The liquid cut force relative to exterior pressure, including these surface terms, is $\pi [3\mu A{w}_z+(p_i-p_e)R_i^2+\gamma (R_e+R_i)]$. The axial traction on the sloping inner gas surface contributes $-\pi (p_i-p_e)(R_i^2{)}_z$ to force balance. For constant gas pressures the extra derivative cancels, giving $3\mu (A{w}_z{)}_z+\rho gA+\gamma (R_e+R_i{)}_z=0$. The displayed $T$ is this effective tension after accounting for gas-pressure work, not the unadjusted liquid cut force.

For an axisymmetric slender viscous thread, the Young–Laplace equation and radial extensional viscous stress give $p=p_e+\gamma /a-\mu w_z$. The axial liquid stress relative to ambient pressure is $3\mu w_z-\gamma /a$. Its cross-sectional integral plus the circumferential surface tension force $2\pi \gamma a$ gives the displayed tensile force. Thus the net capillary contribution is $\pi \gamma a$, and axial force balance with downward gravity is $T_z+\pi \rho g{a}^2=0$.

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.