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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.

In a thin constant-density disk with free surfaces $z=\pm H$, neglect slow radial gradients and let perturbations vary as $e^{ikr-i\omega t}$. With $h=p^{\prime }/\rho$, the local inviscid equations are $-i\omega u-2\Omega v=-ikh$, $-i\omega v+{\kappa }^{2}u/(2\Omega )=0$, $-i\omega w=-h_z$ and $iku+w_z=0$. Eliminating velocity gives $h_{zz}=k^2{\omega }^{2}h/({\omega }^2-{\kappa }^2)$. This local WKB approximation assumes $H/r\ll 1$ and $|kr|\gg 1$ and has both surface and inertial wave branches.

// Target: astrophysics.bigb

For a clean inviscid bubble a distance $d\gg a$ below a gravity-restored approximately flat free surface, the surface displacement scales as $F/(\rho g{d}^2)=O(a^3/d^2)$. Zero normal velocity and zero tangential traction are imposed to leading order by an opposite normal image Stokeslet at distance $2d$ from the bubble. Its incident velocity is $-F/(8\pi \mu d)$, which gives the displayed correction at fixed buoyancy. The image can be represented by a same-radius inviscid drop of density $2\rho$ in liquid of density $\rho$. Its deformation-producing strain is $O(F/(\mu d^2))$, so the leading fractional capillary deformation is $O(\mathrm{Bo}(a/d{)}^2)$; small Bond number is the conventional stronger sufficient criterion.

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 uniform-mass density self-gravitating cylinder has ${\Phi }_R=2\pi G{\rho }_{0}R$. Radial balance with the displayed rotation law gives $p_R={\rho }_0(k^2{R}^3-2\pi G{\rho }_{0}R)$. At central pressure ${\pi }^2{G}^2{\rho }_0^3/k^2$, pressure becomes ${\rho }_0{k}^2(R^2-R_0^2{)}^2/4$ and the free surface has zero effective gravity. The positive radius uses $|k|$ if either rotation direction is allowed.

Take a reservoir with constant horizontal cross-sectional area $A$, one small orifice of area $a$, no inflow, and constant stipulated discharge coefficient $C_d>0$. Let the initial head be $H>0$. For constant mass density $\rho$, the stored mass above the outlet level is $\rho Ah$ and outgoing mass flow rate is $\rho Q$. Mass conservation cancels $\rho$ and gives the initial value problemThe downward speed of the free surface is $Q/A$, smaller than the ideal jet speed by the factor $C_{d}a/A$. Thus a small opening relative to the reservoir supports the negligible-surface-velocity assumption.

For $h>0$, separation of variables or the chain rule gives $d\sqrt{h}/dt=-k/2$. Integrating yieldsThe model reaches zero head in finite time even though volumetric flow rate tends to zero. Extending the squared expression beyond $T$ would make head grow again and violate the original ordinary differential equation. The zero-head continuation enforces no further outflow. For an elevated side opening, zero head means the surface has reached that opening, leaving fluid below it; $T$ is not the time to remove every drop from such a tank.

For an original synthetic example, stipulateThe initial volume above outlet level is $AH=0.0250{\mathrm{m}}^3=25.0\mathrm{L}$. The initial ideal jet speed is $2.215\mathrm{m}{\mathrm{s}}^{-1}$, the initial volumetric flow rate is $Q_0=0.1373\mathrm{L}{\mathrm{s}}^{-1}$, and the predicted zero-head time is $T=364.1\mathrm{s}$. These are consequences of stipulated parameters, not experimental measurements. At $t=T/2=182.1\mathrm{s}$, head is $H/4=0.0625\mathrm{m}$, stored volume is $6.25\mathrm{L}$, and volumetric flow rate is $Q_0/2=0.06866\mathrm{L}{\mathrm{s}}^{-1}$. Half the initial volume has already left at $t=T(1-1/\sqrt{2})=106.7\mathrm{s}$.

An independent mass conservation check integrates the outgoing volumetric flow rate:Multiply both sides by mass density to obtain discharged mass. A dimensional analysis check gives $[k]={\mathrm{m}}^{1/2}{\mathrm{s}}^{-1}$, $[Q]={\mathrm{m}}^3{\mathrm{s}}^{-1}$ and $[T]=\mathrm{s}$. The triangular area under $Q(t)$ explains why the initial-rate estimate $AH/Q_0$ is only half the actual model time. Doubling $A$ doubles $T$; doubling $a$ or $C_d$ halves $T$, within the same assumptions.

This finite-time result is an idealized prediction. The constant $C_d$ and quasi-steady approximation need not survive very small head or the initial startup. Appreciable viscosity in a narrow outlet tube requires a different pressure balance, as in Poiseuille flow; appreciable surface tension can prevent zero head altogether.

An external gravity wave displaces a fluid's free surface and moves nearly uniformly through the depth in the long-wave limit. In a layer of depth $H$, its leading shallow-water approximation speed is $\sqrt{gH}$.

For a deep inviscid liquid above a gas of negligible inertia, an inverted free surface has dispersion relation ${\omega }^2=q(\sigma q^2/\rho -g)$, $q=|k|$. The Rayleigh-Taylor instability affects $0<q<k_c=\sqrt{\rho g/\sigma }$. A bounded aperture can suppress it if its smallest allowed wavenumber is at least $k_c$. The capillary length is ${\ell }_c=k_c^{-1}$ and the marginal wavelength is $2\pi {\ell }_c$. A pinned one-dimensional span of width $L$ with a half-sine lowest mode instead has $k_{min}=\pi /L$, giving $L\le \pi {\ell }_c$. A numerical maximum container size therefore also requires aperture geometry and contact-line boundary conditions; the marginal wavelength alone is not a universal diameter.

A free surface is a fluid interface whose position evolves as part of the solution. A water-air boundary provides a familiar example when the air's mechanical effect is approximated by a prescribed external pressure. Its motion obeys a kinematic boundary condition, while balance of traction supplies a dynamical condition involving pressure, viscous stress and, where relevant, surface tension.

For a surface $z=\eta (x,t)$ in two dimensions, the kinematic condition is ${\eta }_t+u{\eta }_x=w$ at the surface. A material particle already on the interface stays on it. The free-surface height is therefore coupled to the underlying velocity field, rather than being a fixed wall. Small disturbances generate surface gravity waves or capillary effects, depending on the relevant restoring force. Measuring both surface shape and flow tests different parts of this coupled problem.

For a steady Rankine vortex in a fluid of constant density $\rho$ with gravity $-gk$, the Euler equations give $p_r=\rho u_{\theta }^2/r$ and $p_z=-\rho g$. Integrating on the two sides and matching pressure at $r=a$ givesSetting a constant surface pressure yieldsThus the free surface and its radial slope match at the core edge. The central depression is ${\Omega }^2{a}^2/g$, twice the depression at $r=a$. Bernoulli's theorem applies throughout the exterior; the rotational core instead uses the radial momentum balance above.

The front velocity is the propagation speed of the earliest boundary of a newly introduced disturbance's support. It asks when an initially unaffected observer can first respond, whereas group velocity asks how a narrow wave envelope translates. A detector threshold defines a practical arrival time, but can miss arbitrarily small earlier responses; a packet peak is another distinct timing convention. In a causal dielectric model, a high-angular frequency contour argument gives exactly zero field for $t<x/c$, even when phase velocity exceeds $c$. This is a statement about that model's relativistic causality, not a universal identification of group velocity with front velocity.^[ref]

For the stipulated Klein-Gordon equation with compactly supported real Cauchy data, front velocity is bounded by $c$, and $c$ is the model's maximal speed of influence. A short energy method outline usesIntegration outside an expanding interval shows that zero initial energy there remains zero. Thus initial support in $[-a,a]$ cannot influence $x>a$ before $t=(x-a)/c$. The positive energy density includes the restoring term ${\omega }_0^2{u}^2/2$: this adds oscillation behind the front without extending its domain of dependence. A particular disturbance may fail to excite the leading boundary, so the maximal speed is not a promise of a detectable response for every source.

A Gaussian function is nonzero arbitrarily far away, so the Gaussian envelope in a Klein-Gordon medium has no sharp initial support boundary. Its small distant tails are already in the initial data. Likewise, keeping only a narrow interval of the frequency spectrum cannot reproduce an exact abrupt onset. Cutting off a Gaussian tail changes its Fourier transform and the accompanying initial velocity; it is a different initial-value problem. The exponential initial-velocity tail for the Klein-Gordon equation is another concrete example of a pre-existing tail outside the main disturbance.

The water-tank measurement in phase velocity records free surface elevation with calibrated resistance probes and uses a ruler and stopwatch for an independent check. The inspected protocol calls the front or back of a group a timing feature. That visible edge establishes an envelope transit estimate, not the first nonzero disturbance in the sense of front velocity.^[ref] A proposed additional control would repeat the transit fit at several detection thresholds and train durations, using matched probe calibration and the same sampling rate. Dependence on the threshold diagnoses wave envelope reshaping or limited sensitivity; records should be compared before reflections return. Such a test could expose limitations of a group velocity estimate, but would not establish an exact causal front or validate the stipulated Klein-Gordon equation for water waves.

Geophysical fluid dynamics studies fluid flow on rotating, stratified planets, including oceans, atmospheres and planetary interiors. The competition among rotation, buoyancy, pressure gradients and inertia organizes its useful approximations.

For speed $U$, horizontal length $L$ and nonzero Coriolis parameter $f$, the Rossby numbercompares advective acceleration $U^2/L$ with Coriolis acceleration $|f|U$. Small $\mathrm{Ro}$ motivates geostrophic balance between horizontal pressure gradients and Coriolis force, provided rapid transients and other leading forces are absent. Stratification introduces a buoyancy frequency, while vertical hydrostatic balance can be appropriate when vertical acceleration is small.

A rotating-tank experiment makes these balances visible. In the MIT balanced-vortex apparatus, water spirals toward a drain; a camera rotating with the tank measures motion relative to it. The free surface slope supplies a pressure gradient. Comparing relative speed, tank rotation and surface slope tests the balance and its Rossby number correction. ^[ref]

This shared mathematics connects laboratory rotating fluids to planetary circulation without assuming that every atmospheric vortex is geostrophic.

A constant-depth layer initially at rest with a step in its free surface adjusts by radiating inertia-gravity waves. Conservation of linearized shallow-water potential vorticity leaves the bounded balanced profile ${\eta }_s={\eta }_0\mathrm{sgn}(x)(1-e^{-|x|/R_D})$, where $R_D=\sqrt{gH}/|f|$ is the Rossby deformation radius. The released potential energy per unit transverse length is $3\rho g{\eta }_0^2{R}_D/2$; the final kinetic energy is $\rho g{\eta }_0^2{R}_D/2$, leaving $\rho g{\eta }_0^2{R}_D$ in radiated waves.

A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.

A gravity wave is a fluid disturbance restored by gravity or buoyancy. A surface gravity wave displaces a free surface, while an internal gravity wave displaces density surfaces within a stratified fluid. Rotation can additionally contribute to restoration, giving an inertia-gravity wave.

A steady incompressible jet at atmospheric pressure conserves Bernoulli's equation and volume flux. Its radius contracts as $R=a[u/u_0{]}^{-1/2}$. After impact, a thin nearly horizontal axisymmetric film has $2\pi rhv=\pi a^2{u}_0$. Evaluating Bernoulli's equation on its free surface gives $v^2=u_0^2+2g(H-h)$, hence $a^4/(4r^2{h}^2)=1+2g(H-h)/u_0^2$. This inviscid thin-film description applies away from the impact region and does not describe a dissipative hydraulic jump.

The Lagrangian fluid displacement $\mathbf{\xi }$ compares neighbouring fluid trajectories. Linear velocity perturbations about a steady flow $\mathbf{U}$ obey $\delta \mathbf{u}=({\partial }_t+\mathbf{U}\cdot \nabla )\mathbf{\xi }-(\mathbf{\xi }\cdot \nabla )\mathbf{U}$. For a background scalar $f_0$, its material perturbation is $\Delta f=\delta f+\mathbf{\xi }\cdot \nabla f_0$. In an axisymmetric normal mode about cylindrical rotation, the radial and vertical velocities are $-i\omega {\xi }_r$ and $-i\omega {\xi }_z$. The displacement of a free surface is needed to impose the Lagrangian pressure perturbation condition.

// Target: astrophysics.bigb

A liquid in rigid rotation about a vertical axis is at rest in its rotating reference frame. Constant surface pressure makes $gz-{\Omega }^2{r}^2/2$ constant along the free surface, giving the displayed paraboloid. Comparing with $z-z_0=r^2/(4f)$ gives the focal length $f$. The result assumes uniform gravitational acceleration, a vertical rotation axis and negligible surface-tension corrections. It explains how a rotating liquid can form a reflecting surface or a mould for a mirror.

Let $x$ measure distance down the incline and $y$ distance perpendicular to it, with the wall at $y=0$ and the free surface at $y=h$. A uniform steady falling film flow has $\mathbf{u}=u(y){\mathbf{e}}_x$. Its convective acceleration vanishes because $\mathbf{u}\cdot \nabla =u{\partial }_x$ and ${\partial }_{x}u=0$. A flat surface exposed to constant atmospheric pressure hasThus ${\partial }_{x}p=0$, and the downslope component of the Navier-Stokes equations reduces to $0=g\sin \alpha +(\eta /\rho )u^{\prime \prime }$. The printed pressure-free equation must be understood as this downslope projection: its full vector version would incorrectly omit the normal hydrostatic pressure gradient that balances $g\cos \alpha$.

The no-slip boundary condition gives $u(0)=0$. Neglecting air shear stress makes the surface shear-free, so $u^{\prime }(h)=0$. Two integrations give the uniform gravity-driven Newtonian film profileThe velocity increases up to the free surface, whereThis assumes a uniform laminar Newtonian fluid film, a stationary wall and negligible surface shear; a varying thickness or externally imposed downslope pressure gradient would alter the balance.

For an inverted free surface, gravity promotes a displacement instead of restoring it. The surface tension term still resists curvature. Hence the gravity-capillary wave relation becomes, with $q=|k|$,For $0<q<k_c$, where $k_c=\sqrt{\rho g/\sigma }$, ${\omega }^2<0$. Writing the displacement as $\eta \propto e^{st+ikx}$ gives the Rayleigh-Taylor instability growth rateShorter wavelengths, $q>k_c$, remain oscillatory because surface tension dominates. The marginal scale is determined by the capillary length,A sufficiently small aperture excludes the long unstable modes: this is finite-aperture stabilization of an inverted liquid surface. The estimate requested is therefore a container opening of order one centimetre, with $1.7\mathrm{cm}$ the marginal full wavelength rather than a geometry-independent maximum diameter.

The exact largest opening cannot be fixed from the stated infinite-surface dispersion relation alone. It requires the aperture shape and contact-line boundary conditions. For example, a one-dimensional span of width $L$ whose contact line is pinned at both ends has a half-sine lowest mode with $q_{\min }=\pi /L$. In that simple model,Different geometries change the numerical factor. This is a small-disturbance stability estimate with negligible air inertia and no leakage or contact-line failure; a guarantee that a real inverted container retains its water needs those additional boundary details.