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The expression in braces in the first equation is the net axial force. The term $-\pi a^2\gamma /a$ is the compressive force from the capillary pressure $\gamma /a$, $3\pi \mu a^2{w}_z$ is the Newtonian extensional tension with Trouton ratio three, and $2\pi a\gamma$ is the axial pull of surface tension around the circumference. Its $z$-derivative vanishes because axial force is conserved.

The final equation is conservation of an insoluble surfactant. The surface divergence $w_z+u/a$ is the sum of axial and circumferential extension rates. Positive surface divergence increases interfacial area and dilutes $C$; negative divergence concentrates it. The absence of a diffusion term expresses the assumption of negligible surface diffusion.

Linearizing the area and surfactant equations givesThusand hencewhere $\psi$ is fixed by the initial data.

Linearizing the displayed net axial force and using ${\gamma }^{\prime }=-A{C}^{\prime }$ givesEliminating $w_z$ and $C^{\prime }$ therefore gives directlyThus the three displayed evolution equations implyThe target formula printed later in the paper contains an additional factor of $\pi$ in both denominators. That factor does not follow from the displayed equations because every term in the axial-force balance contains the same factor $\pi$. If the target formula is adopted as the intended normalization, its corresponding value is $s=({\gamma }_0-A{C}_0)/(6\pi \mu a_0)$.

Initially $\eta =0$ and $C^{\prime }>0$, so $\psi >0$ and ${\eta }_t<0$: a surfactant-rich, low-tension region begins to neck as neighboring higher tension pulls fluid away. The accompanying axial extension dilutes the surfactant.

If $A<{\gamma }_0/C_0$, then $s>0$. The Rayleigh–Plateau instability overwhelms the weak surface-elastic response: the necking perturbation grows in the linear model while the original concentration excess is diluted and eventually changes sign.

If $A>{\gamma }_0/C_0$, then $s<0$. Strong surface elasticity arrests the disturbance atThe concentration perturbation becomes negative, raising the local surface tension until its axial force balances that of the wider regions. This is stabilization by a surfactant-induced Marangoni stress.

The condition $a\gamma =a_0{\gamma }_0$ makes the leading extensional axial force uniform, but it does not make the capillary pressure uniform. Sincea nonuniform radius gives a nonuniform pressure. Its axial gradient must drive flow, so the prediction $w=0$ is inconsistent.

For $a=a_0+\eta$ varying over the axial scale $L$,An axial Hagen-Poiseuille flow in a cylinder has speed scaleCross-sectional mass conservation gives ${\eta }_t\sim a_{0}w/L$, and consequently

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.

Suppose the apparent capillary pressure is $p_1-p_2=\gamma (1+\beta x)({\kappa }_0+\nabla \cdot \mathbf{n})$, with normal towards the displaced fluid and positive apparent tension at the flat front $x=X$. Linearizing the Darcy law interface problem gives the local growth rateThe pressure increase experienced by a forward protrusion is stabilizing when ${\kappa }_0\beta >0$. It suppresses every positive wavenumber when $U\le k\gamma {\kappa }_0\beta /({\mu }_2-{\mu }_1)$. For a fixed spatial gradient, $X=Ut/\varphi$ makes the rate time dependent; constant exponential growth is a frozen-position approximation.

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.

Immiscible fluids share pore space, with each phase's Darcy flux determined by its pressure, viscosity and relative permeability. Fluid saturation specifies how much pore space each phase occupies. A capillary pressure couples their pressures, while their individual mass conservation laws determine displacement and saturation evolution.

A U-tube differential manometer contains a connected column of liquid with one gas connection above each liquid meniscus. Write $p_L,p_R$ for the left and right pressures, and $z_L,z_R$ for the meniscus heights on an upward vertical coordinate. Assume constant mass density $\rho$ and downward gravitational acceleration of magnitude $g$; neglect the gas column's mass density and differences in capillary pressure. The signed vertical length difference $\Delta h=z_R-z_L$ then givesLarger left pressure means a lower left liquid meniscus: $p_L>p_R$ implies $z_L<z_R$. Equal pressures give equal meniscus heights.

The balance follows directly from hydrostatic pressure. At a common height $z_0$ below both surfaces, the two points in the connected homogeneous liquid have equal pressure:Subtracting cancels the common bottom depth and leaves $p_L-p_R=\rho g(z_R-z_L)$. The same-height rule follows from hydrostatic equilibrium, $\nabla p=-\rho g{\mathbf{e}}_z$: a horizontal displacement changes no pressure, and for constant $\rho$ the quantity $p+\rho gz$ is constant throughout the connected liquid. Equal height alone is insufficient for unrelated vessels or for fluid flow with significant acceleration.

For an original constructed example, choose water of stipulated mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and vertical length difference $\Delta h=0.060\mathrm{m}$. ThenThe SI pressure unit is the pascal: the product has units $\mathrm{kg}{\mathrm{m}}^{-3}\times \mathrm{m}{\mathrm{s}}^{-2}\times \mathrm{m}=\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}=\mathrm{Pa}$. Swapping the two connections gives $-588.6\mathrm{Pa}$ and reverses which liquid meniscus is lower. In equal-bore arms, volume conservation makes each surface move $30\mathrm{mm}$ from its initial equal-height position; using only one surface's displacement would halve the inferred pressure. With unequal arm areas $A_L,A_R$, a left fall $a$ and right rise $b$ instead obey $A_{L}a=A_{R}b$, while the vertical length difference is still $a+b$.

If the right port is open to the local atmosphere, $p_L-p_R$ is the left gauge pressure. Only with that reference and known atmospheric pressure can one obtain absolute pressure fromFor an explicitly stipulated atmospheric pressure of $100000\mathrm{Pa}$, the example gives $100588.6\mathrm{Pa}$ absolute pressure. If the right port is connected to another vessel or a pressure tapping, the same reading is a differential pressure and supplies no absolute pressure without knowing that reference.^[ref]

A liquid layer of depth $h$ above a dry deep porous substrate drives a Darcy flux $w=(gk/\nu )(1+h/l)$ through its saturated depth $l$, assuming negligible capillary pressure and negligible resistance from the displaced phase. The pressure head contributes $h/l$ and gravity contributes one. Volume conservation gives $\varphi \dot{l}=w$. For a finite column, $h+\varphi l=h_0$ and initiallyThe singular initial flux is an idealization; the continuum model requires penetration beyond the pore scale.

NASA's Wind Tunnel Experiment Details, slide 27, describes a practical manometer measurement: partially fill a Tygon tube with water and connect its two ends to total pressure and static pressure ports in a wind tunnel. The total pressure connection pushes its liquid meniscus downward; the vertical length difference supplies the pressure difference. The slide predicts increasing separation as the tunnel speed increases. It supplies an apparatus and method, rather than a table of observed readings.^[ref]

A control for this practical is to expose both connections to the same atmospheric pressure before the run: settled liquid meniscus heights should agree in a symmetric column. Read each liquid meniscus at eye level, reconnect sealed tubing to the identified ports, and wait for settling. Recheck the equal-pressure zero afterwards; reverse connections to check the sign. These checks establish neither mass density nor ruler accuracy.^[ref]

The inference is a differential pressure from hydrostatic pressure, with stagnation pressure on one side and static pressure on the other in NASA's practical. Converting that difference to a flow speed additionally needs the hypotheses discussed under dynamic pressure and Pitot-static tube. A changing pressure can leave the column oscillating; leaks, blocked pressure tappings or trapped liquid in a gas connection can alter or delay the readout. The column should be read after settling, and temperature should be recorded when choosing the measuring liquid's mass density.

Two corrections clarify the limits. If both gas ports are at the same height and carry the same gas of constant mass density ${\rho }_g$, keeping its hydrostatic pressure gives $p_L-p_R=(\rho -{\rho }_g)g\Delta h$. Different port heights require their separate gas column heights. Also, a curved liquid meniscus can sustain a capillary pressure jump through surface tension. Write $J_L,J_R$ for liquid pressure minus gas pressure at the two surfaces. Neglecting gas weight, the balance becomes $p_L-p_R=\rho g\Delta h+J_R-J_L$. Identical meniscus shapes in matching clean bores cancel those jumps; unequal bores or different wetting conditions need not. Thus the useful simplicity of $\rho g\Delta h$ rests on identifiable physical conditions.

Injecting water to displace oil through a porous reservoir. Spatial permeability contrasts cause preferential flow and early water breakthrough, while unswept slower regions retain oil. The recovery also depends on fractional flow, capillary pressure and phase mixing; a passive streamline model alone does not capture all two-phase effects.

The Young–Laplace equation relates the pressure jump across a stationary interface to its curvature and surface tension. Choose a unit normal vector $\mathbf{n}$ pointing from the inside region to the outside region, and define $\kappa ={\nabla }_s\cdot \mathbf{n}$ to be positive for a convex sphere. In mechanical equilibrium, with constant surface tension and no additional normal viscous stresses,Here $k_1,k_2$ are the signed principal curvatures in this convention. Their sum is twice the corresponding signed mean curvature; stating the convention prevents a factor-of-two ambiguity. The Young–Laplace equation is a local normal stress balance for one interface. ^[ref]

A spherical liquid droplet of radius $R$ in a gas has one interface, with $k_1=k_2=1/R$. An ideal thin soap bubble has two liquid–gas interfaces. Adding their pressure jumps givesFor the soap bubble, $\Delta p$ now means the inner gas pressure minus the outer gas pressure. The second formula assumes equal surface tensions on both faces and film thickness negligible relative to $R$, so their radii can be identified. Retaining distinct radii gives $2\gamma /R_{\mathrm{inner}}+2\gamma /R_{\mathrm{outer}}$. A gas bubble inside bulk liquid has only one interface and therefore uses $2\gamma /R$. ^[ref]

With the assumed constant toy surface tension $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, a droplet with $R=1\mathrm{mm}$ has excess pressure $144\mathrm{Pa}$; at $R=0.5\mathrm{mm}$ it has $288\mathrm{Pa}$. A thin soap bubble assigned the same $\gamma$ and $R=1\mathrm{mm}$ has excess pressure $288\mathrm{Pa}$. These are differences from the surrounding pressure, rather than absolute pressures. Halving the radius doubles the capillary pressure; doubling the interface count also doubles it at fixed radius and surface tension.

The spherical case has a short work argument. A virtual increase in radius gives $dV=4\pi R^{2}dR$ and $dA=8\pi RdR$. At mechanical equilibrium, the net pressure work balances the increase in surface energy:For a thin soap bubble, both interfaces gain approximately the same surface area, doubling the right-hand side and giving $4\gamma /R$. This is a virtual work balance between neighboring spherical configurations, rather than an assumption that a fixed-volume droplet actually expands. The general Young–Laplace equation replaces the spherical area-to-volume ratio by local curvature, making geometry directly measurable through pressure.