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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.

An inviscid uniform-density self-gravitating cylinder in vacuum, with no rotation or surface tension, has the displayed axisymmetric dispersion relation, where $\kappa =|k_z|R$. To derive it, let $W=\psi +\pi /\rho =ϵ{\Pi }_0{I}_0(kr)e^{i(kz-\omega t)}$. The linear kinematic boundary condition gives ${\Pi }_0={\omega }^2/[k{I}_0^{\prime }(\kappa )]$. The Lagrangian pressure perturbation vanishes on the moving free surface, giving $\pi (R)=2\pi G{\rho }^{2}R\eta$. Substitute the axisymmetric gravitational potential of a displaced cylindrical interface and use $I_0^{\prime }=I_1$ to obtain the relation. Its nonzero neutral root is ${\kappa }_c\simeq 1.0667$: wavelengths longer than $2\pi R/{\kappa }_c\simeq 5.89R$ have ${\omega }^2<0$ and grow. The fastest-growing mode has $\kappa \simeq 0.58$, so its wavelength is about $11R$. Negative ${\omega }^2$ must be interpreted through an imaginary frequency, not through a real oscillatory ansatz.

Why can hot liquid wait for a charged particle rather than immediately turn into vapor? A small cavity costs surface energy. Consider an original macroscopic capillarity model at fixed temperature with surface tension $\sigma >0$, imposed liquid pressure $p_l$, and driving difference $\Delta p=p_{\mathrm{sat}}(T)-p_l>0$, where $p_{\mathrm{sat}}$ is the planar-interface vapor pressure. Neglect dissolved gas, interface thickness, heat transfer and curvature dependence of $\sigma$. The reversible formation-work approximation isThe first term charges for surface area; the second rewards the new vapor volume. This is a macroscopic free energy model, not the energy that an actual detector must receive as localized heat. It shares the capillarity balance used in the cloud-chamber expansion model, but the new phase is now a bubble.

Differentiating givesAt the critical nucleus, $d^2\Delta G/d{r}^2=-8\pi \sigma <0$. The barrier is a maximum: cavities smaller than $r_{\ast }$ tend to shrink, while larger ones can grow in this model. Doubling $\Delta p$ halves $r_{\ast }$ and quarters $\Delta G_{\ast }$ if $\sigma$ stays fixed. This explains the strong sensitivity to operating conditions without claiming a universal detector threshold.

Choose $\sigma =0.010\mathrm{J}{\mathrm{m}}^{-2}$ and $\Delta p=2.0\times 10^5\mathrm{Pa}$ solely for a teaching example. ThenFor the same chosen $\sigma$ and twice the driving pressure, $r_{\ast }=50\mathrm{nm}$ and $\Delta G_{\ast }\simeq 0.654\mathrm{keV}$. These are calculated formation-work scales, not measured bubble thresholds for diethyl ether or any other substance.

A simplified way to discuss localization is to stipulate deposited energy $E_{\mathrm{loc}}$ within a segment of length $\ell$ and a trial conversion fraction $0<\eta \le 1$. A toy criterion $\eta E_{\mathrm{loc}}\gtrsim \Delta G_{\ast }$ gives an energy per unit length scale $\Delta G_{\ast }/(\eta \ell )$. It is only a bookkeeping estimate: a real bubble also needs latent heat, must exchange heat, and is affected by the spatial pattern and time scale of energy deposition. Consequently neither a single stopping power value nor this barrier alone predicts detection efficiency or the number of bubbles. Two tracks with the same mean stopping power can distribute their energy differently on the scale of a critical nucleus.

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.

Motion of a liquid through a narrow space due to surface tension and contact with the walls; it draws the chromatographic mobile phase through a thin-layer chromatography coating.

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.

The threshold pressure difference required for a nonwetting fluid to invade a connected pore throat. In a simple cylindrical throat of radius $r$, the magnitude is approximately $2\gamma |\cos \theta |/r$, using surface tension $\gamma$ and contact angle $\theta$. A sealing caprock can fail to retain a buoyant phase if its pressure exceeds this threshold or if connected fractures bypass the fine pore network.

A force caused by an interface through its contact-line surface tension and the capillary pressure it produces. Both contributions must be counted with consistent geometry. For a narrow, completely wetting liquid bridge between flat plates of wetted radius $a$ and gap $d\ll a$, the leading attractive force is $2\pi a^2\gamma /d$.

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.

Capillary origami uses surface tension to fold an initially flat elastic plate around a liquid droplet. The fold trades a reduction in surface energy for an increase in bending energy; the initial outline helps select the resulting enclosure.^[ref]

In the 2007 experiments of Charlotte Py and collaborators, 40–80 micrometre polydimethylsiloxane sheets rested on a water-repellent support. A water droplet reached the sheet's corners; time-lapse images followed its evaporation. Squares changed from four-corner bending to a tubular shape; triangles formed a tetrahedron. Smaller or thicker sheets could bend and then reopen without enclosing the droplet.^[ref]

The elastocapillary length makes the thickness dependence quantitative. For fixed Young's modulus, Poisson's ratio and surface tension, the bending stiffness obeys $B\propto h^3$ and ${\ell }_{\mathrm{ec}}\propto h^{3/2}$. Halving thickness therefore reduces ${\ell }_{\mathrm{ec}}$ by $2^{3/2}$, about $2.83$. At the same lateral size $L$, the sheet elastocapillary energy ratio $\eta =\gamma L^2/B$ increases eightfold. This calculation explains why making the sheet thinner can enable wrapping even without changing its outline.

The sheet elastocapillary energy ratio compares available energy scales; it does not determine a unique final shape. The contact angle, initial liquid volume and outline also matter. A cut chosen to permit several panels to meet is a mechanical design choice, while the surface tension supplies the driving force. Predicting the complete folding path requires more than the scalar energy comparison. The same group reported cross-shaped templates forming a cube and flower-shaped templates forming nearly spherical enclosures.^[ref]

For downslope length scale $L$, film thickness scale $h_0$ and surface velocity scale $U_0=\rho g{h}_0^2\sin \alpha /(2\mu )$, this normalization makes the capillary part of the Benney equation flux $\epsilon S{h}^3{h}_{xxx}$. The dimensionless capillary pressure is $-3S{h}_{xx}/2$. Keeping $S=O(1)$ specifies a distinguished long-wave approximation in which surface tension remains relevant.

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 tube is a tube with a small internal bore. Its large wall-area-to-fluid-volume ratio makes interfaces important: surface tension and the contact angle at the wall can produce capillary action, while viscosity strongly limits flow through a narrow passage.

For an ideal circular tube of radius $a$ in contact with a liquid reservoir, a static meniscus model gives $h=2\gamma \cos \theta /(\rho ga)$, where $\gamma$ is surface tension, $\theta$ the contact angle, $\rho$ the liquid mass density and $g$ gravitational acceleration. This follows by balancing the meniscus pressure jump with hydrostatic pressure. The sign permits a depression for a nonwetting liquid. Comparing tubes of different radii tests the predicted inverse-radius scaling within the model's assumptions.