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The steady radial equation first gives the base-pressure gradientFor the stated normal mode, putRetaining terms linear in the disturbance in the Euler equations for an inviscid fluid givestogether with incompressible flowThe $-2\Omega v^{\prime }$ and $(2\Omega +r{\Omega }^{\prime })u^{\prime }$ terms are the linearized centrifugal and angular-momentum couplings of the swirling base flow.

At latitude ${\theta }_0$, resolve the planetary angular velocity asFor velocity $(u,v,w)$, the Coriolis acceleration isThe traditional approximation drops the terms involving the horizontal rotation component $\Omega \cos {\theta }_0$. The retained horizontal force is thereforeIts direct velocity-scale requirement isFor an incompressible flow, $W/U=O(H/L)$, so a sufficient condition istogether with the small aspect ratio that makes hydrostatic pressure the leading vertical momentum balance. The approximation consequently becomes delicate near the equator.

The beta plane is the local Taylor expansionwhereand $a$ is the planetary radius. It requires a local Cartesian region,and $H/a\ll 1$. Away from the equator, treating the variation as a perturbation of an f-plane also requires $\beta L/|f_0|\ll 1$; near the equator one instead retains the linear term as the leading Coriolis parameter.

Let $D=H-h$ be the constant interior depth. Assume a homogeneous hydrostatic interior with depth-independent horizontal velocity, negligible friction, steady small-Rossby number flow, and no normal flow through the flat bottom. The vertical velocity varies from $w(-H)=0$ to $w(-h)=w_E$, so incompressible flow givesThe leading horizontal momentum balance is geostrophic balance:Taking its vertical curl on a beta plane givesCombining the last two equations yields Sverdrup balanceThe zonal velocity is then fixed, up to its value on one side boundary, byEquivalently, substituting part a gives the depth-integrated formA lateral boundary condition, normally supplied by matching to a boundary current, determines the remaining zonally uniform part of $u$.

Differentiate horizontal momentum with respect to $z$ and vertical momentum with respect to $x$. When the two equations are subtracted, the terms proportional to ${\overline{u}}_z(u_x^{\prime }+w_z^{\prime })$ vanish by incompressible flow. The pressure derivatives also cancel, leaving

Differentiate this equation with respect to $x$. Since continuity givesone obtainsApply $D_t$ and use the buoyancy equation $D_t{\sigma }^{\prime }=-N^2{w}^{\prime }$. Multiplication by $-1$ then gives

The thermodynamic order parameter is the conserved scalar composition difference $\varphi$ of the binary fluid mixture. The momentum density $\rho \mathbf{v}$ is an additional conserved hydrodynamic mode; it is not a symmetry-breaking order parameter, but it must be retained because momentum relaxes only through spatial transport.

Composition conservation gives a continuity equation,The current $\mathbf{J}=-M\nabla \mu$ is the leading isotropic, dissipative constitutive law: it drives material down gradients of the chemical potential. For the Landau-Ginzburg theorythe functional derivative isThese statements give the advective Cahn--Hilliard equation.

Constant mass density and incompressible flow require $\nabla \cdot \mathbf{v}=0$. Momentum conservation gives the Navier-Stokes equation: material acceleration equals the sum of the Newtonian viscous force $\eta {\nabla }^2\mathbf{v}$, the pressure force $-\nabla P$, and the Korteweg force $-\varphi \nabla \mu$. The pressure is the Lagrange multiplier enforcing incompressibility. Together these equations are Model H dynamics; isothermality removes the need for a separate energy equation.

Assume late-stage bicontinuous phase separation has one characteristic domain size $L(t)$ and statistically self-similar morphology. The dynamical scaling of binary-fluid coarsening then estimatesThe interface has surface tension $\sigma$. Its curvature pressure is of order $\sigma /L$, so its coarse-grained gradient, equivalently the transverse part of $-\varphi \nabla \mu$, scales as $\sigma /L^2$. Diffusive composition transport is neglected because hydrodynamic advection controls the late stage, and $\rho ,\eta ,\sigma$ are treated as constants.

The pressure cannot simply be discarded: in an incompressible flow it cancels the longitudinal part of the other forces and contains both capillary and dynamic contributions. Applying the divergence-free projection to the momentum equation eliminates $\nabla P$ while retaining a solenoidal force with the same scaling. Geometry, projection, signs, and correlations are absorbed into dimensionless order-one coefficients. The result is

Write the density as a hydrostatic reference profile $\hat{\rho }(z)$ plus a small perturbation ${\rho }^{\prime }$, and define buoyancy and buoyancy frequency byThe Boussinesq approximation to the Navier-Stokes equation, together with mass conservation, iswhere $D/Dt$ is the material derivative, $\nu$ is kinematic viscosity, and $\kappa$ is mass diffusivity. Linearizing about rest and eliminating pressure and buoyancy givesFor a plane wave proportional to $e^{i(kx+mz-\omega t)}$, with $K^2=k^2+m^2$, the viscous-diffusive dispersion relation isIn the inviscid limit this becomesFor weak diffusion the two oscillatory roots areso a freely evolving Fourier mode decays.

A single plane wave is also an exact solution of the nonlinear equations. Every field depends only on its phase $\Theta =\mathbf{k}\cdot \mathbf{x}-\omega t$, while incompressible flow gives $\mathbf{u}\cdot \mathbf{k}=0$. Therefore $\mathbf{u}\cdot \nabla$ annihilates both $\mathbf{u}(\Theta )$ and $b(\Theta )$, and all nonlinear advection terms vanish.

For a boundary-forced wave with real $\omega$, nonzero $\nu$ or $\kappa$ instead makes the bulk vertical wavenumber complex, attenuating the propagating beam. Because diffusion raises the spatial order of the equations, additional short vertical-wavenumber roots form viscous and scalar boundary layers; they allow a no-slip velocity condition and a scalar no-flux condition to accompany impermeability. These layers and bulk attenuation become essential near critical internal-wave reflection, where the inviscid reflected wavelength collapses.

The Boussinesq approximation replaces density by a common reference value ${\rho }_0$ in inertia, mass conservation, and pressure acceleration, while retaining small density differences in the gravitational buoyancy term. It gives incompressible flow and is appropriate here wheneven though those small differences drive the room-scale motion. It would fail for order-one thermal density contrasts or strongly compressible ventilation.

The Batchelor entrainment hypothesis sets the mean inflow speed across a turbulent plume edge to $\alpha$ times a representative axial plume speed, where $\alpha$ is the entrainment coefficient. It closes integral plume balances by relating plume growth to its speed. Applied here, it produces an entraining axisymmetric warm plume above the floor source and a one-sided cold wall line plume below the vent. Treating both as turbulent top-hat plume models neglects source regions, detailed profiles, wall friction, interaction between the two plumes, and the finite thickness of the density interface; these are the principal modelling assumptions.

Let $z$ measure distance across the narrow gap and let $\theta$ be polar angle about the tube axis. In lubrication theory, radial velocity and radial pressure variation are negligible. Axisymmetric incompressible flow is obtained frombecause ${\partial }_{\theta }(u_{\theta }\sin \theta )=0$. The tangential Stokes equation then separates:and hence, after choosing an irrelevant pressure constant,

In the sphere frame, $u_z=-u_{\theta }\sin \theta =-\varphi$. The no-slip boundary condition gives $\varphi (0)=0$ on the sphere and $\varphi (\delta )=U$ on the membrane translating backward relative to it. The sphere-frame volume flux inherited from the narrow remote tube is $-\pi r_0^{2}U$, soUnder the asymptotic condition $r_0^2\ll R\delta$, the right-hand side is negligible at leading order. Solving the quadratic profile subject to the two wall values and zero leading-order integral givesChanging the chosen positive tube direction reverses both signs but leaves the drag magnitude unchanged.

The pressure scale is $p\sim \mu UR/{\delta }^2$, whereas the viscous shear scale is $\tau \sim \mu U/\delta$. After multiplication by comparable areas, pressure drag exceeds shear drag by $R/\delta \gg 1$, an instance of lubrication pressure dominates shear stress. Put $c=\cos \Delta \theta$. The axial pressure force isAs $\Delta \theta \to 0$, the bracket tends to $-2$, and thereforeThe resulting confined-sphere drag coefficient isThus it exceeds the free Stokes drag law coefficient by $2R^2/{\delta }^2$. The Stokes–Einstein relation then gives

The broad faces have curvature zero to leading order, so their stress boundary condition gives ${\sigma }_{xx}=-p_e$. The semicircular edges have curvature $1/h$, giving ${\sigma }_{yy}=-p_e-\gamma /h$. Incompressible flow gives $u_x+v_y+w_z=0$; uniform transverse normal stresses imply uniform transverse extension rates, and eliminating them from the Newtonian fluid stress tensor yields

At an edge, the kinematic boundary condition balances axial advection of $b$, lateral strain, and capillary retraction. This givesThe equation $hbw=Q$ expresses conservation of volume flux, whilestates that the total axial tension is constant. DefineThen $[\Gamma ]=L^{-2}$ and $[T]=L^{-1}$, and substitution of $h=Q/(bw)$ givesSubtracting these logarithmic-derivative equations givesThe remaining width equation isThus, for $T\ne 0$,while the continuous $T=0$ limit is $1/b=1/b_0+\Gamma z/6$.

If $\Gamma =0$, then $b/b_0=e^{-Tz/6}$, $w/w(0)=e^{Tz/3}$, and mass conservation gives $h/h(0)=e^{-Tz/6}$. Therefore a prescribed thinning ratio $R=h(L)/h(0)$ requires

For a sufficiently smooth vector field on a volume $V$ and its closed boundary $S$, the divergence theorem stateswith the unit normal directed outwards. HereHence the net volume flux through every closed surface in the liquid is zero. This verifies incompressible flow directly. It does not require the velocity to vanish pointwise on an arbitrary closed surface; inward and outward contributions can cancel.

Since $\mathbf{V}=\nabla u$, its divergence is the Laplacian of the velocity potential. Using the Cauchy-Riemann equations and equality of mixed partial derivatives givesThus this is an incompressible flow, and $u$ is a harmonic function.

For a twice continuously differentiable stream function, equality of mixed derivatives givesThis is the incompressible flow condition in two dimensions. A streamline is a curve whose tangent at each regular point is parallel to the instantaneous velocity field. Parameterize it by ${\gamma }^{\prime }(s)=c(s)(u,v)$. The chain rule yieldsTherefore the stream function is constant on each streamline. This proof includes arbitrary regular parametrizations; at a stagnation point the velocity vanishes and the usual tangent direction is not determined.

For a planar incompressible flow, a stream function satisfies $u={\psi }_y$, $v=-{\psi }_x$. A velocity potential satisfies $u={\varphi }_x$, $v={\varphi }_y$ where the flow is irrotational; such a potential exists locally, and globally on a simply connected nonsingular region. The irrotational flow condition is $v_x-u_y=0$. Where both exist, the Cauchy-Riemann equations make the complex potential $w=\varphi +i\psi$ analytic with $w^{\prime }=u-iv$.

For the logarithmic potential, write $z-z_0=r{e}^{i\theta }$. ThenThe outward volume flux per unit span across a circle is $2\pi r{u}_r=m$. Hence it is a line source of strength $m$. The potential may use local logarithm branches; its velocity is single-valued away from the source.

By superposition, the full flow hasFor positive $m$ and $U$, the stagnation points areThey are real, one upstream of the source and one downstream of the sink. Away from the singularities, streamlines are level curves ofThe zero level through the two stagnation points contains an upper and a lower separating curve, symmetric about the real axis and about $x=1/2$. Together they bound a Rankine oval. Flow inside goes from source to sink; the external stream divides around this oval and tends to horizontal flow far away. This also specifies the sketch for every positive parameter ratio; the figure illustrates one ratio.

Orient the body boundary $C$ counterclockwise. Its outward normal vector satisfies $\mathbf{n}ds=(dy,-dx)$, so the pressure force on the body has complex componentFor steady incompressible flow and irrotational flow, Bernoulli's principle gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant-pressure integral vanishes. Write $q=w^{\prime }=u-iv$. The impermeability condition makes velocity tangent to $C$, implying $qdz=\bar{q}d\bar{z}$. Thus $|q{|}^{2}d\bar{z}=q^{2}dz$ along the boundary, andThis proves the Blasius force theorem, with force measured per unit length perpendicular to the flow plane.

For a circle of radius $a$, Milne-Thomson's circle theorem replaces an incident analytic potential $f(z)$ by $f(z)+\overline{f(a^2/\bar{z})}$, making the stream function constant on the circle. Its reflected singularities must lie inside the circle; here the physical source lies outside. With $c=Q/(2\pi )$ and $d=a^2/b$, this yields, up to additive constants,The interior image source and sink have equal and opposite strengths, so there is no extra net source inside the body. No independent vortex term is added. The potential may have logarithmic branches, but the velocity and pressure are single-valued.

Only $d$ and $0$ lie inside $C$. The residues of $(w^{\prime }{)}^2$ areTheir sum is $-2c^2{a}^2/[b(b^2-a^2)]$. The residue theorem and the preceding force formula therefore giveThe force points towards the external source. Reflection symmetry across the horizontal axis independently checks the vanishing vertical component.

The velocity is $\mathbf{u}=\nabla \varphi$ because the flow is irrotational. Incompressible flow requires $\nabla \cdot \mathbf{u}=0$, hence the velocity potential satisfies Laplace's equation:The exact kinematic boundary condition at the free surface is ${\eta }_t+{\varphi }_x{\eta }_x={\varphi }_y$. With constant atmospheric pressure, the dynamic condition from Bernoulli's principle is ${\varphi }_t+\frac{1}{2}|\nabla \varphi {|}^2+g\eta =0$, after choosing the time-dependent potential gauge. For small amplitudes, the products of perturbations are second order. Evaluating the remaining first-order terms at the undisturbed level $y=0$ givesSubstitution of $\varphi =f(y)\sin (kx-\omega t)$ into Laplace's equation gives $f^{\prime \prime }-k^{2}f=0$. For $k\ne 0$, decay as $y\to -\infty$ selects $f(y)=B{e}^{|k|y}$. The two boundary conditions for $\eta =A\cos (kx-\omega t)$ yieldFor a nonzero wave amplitude these are compatible precisely whenWith the conventional $k>0$ and positive frequency, the dispersion relation is $\omega (k)=\sqrt{gk}$. The exponential decay is the infinite-depth boundary condition; retaining a growing exponential would violate it.

Milne-Thomson's circle theorem inserts an impermeable circular boundary into a two-dimensional irrotational incompressible flow. If $f(z)$ is analytic in a neighborhood of the closed disc and its flow singularities lie outside, the new complex potential outside $|z|=a$ isReflection and coefficient conjugation make the second term analytic outside the circle, with its image singularities inside. On $|z|=a$, $a^2/\bar{z}=z$, so the two terms sum to $2\mathrm{Re}f(z)$: the streamfunction is constant there. Thus no velocity crosses the circle. At infinity the image term tends to the constant $\overline{f(0)}$, which has no effect on velocity, and the far-field flow is retained. This proves the construction, with an additive potential constant immaterial.

Put $A=\mu e^{i\alpha }$, $B=\mu e^{-i\alpha }$. For the dipole the reflected potential is $B/(a^2/z-b)=Bz/(a^2-bz)$, givingIts image pole is $c=a^2/b$, inside the cylinder. The complex velocity isNear $c$, the second term is $C(z-c{)}^{-2}$ with $C=B{a}^2/b^2$, while $v_0(z)=-A/(z-b{)}^2$ is analytic there. The square of the pole term has no residue, nor does $v_0^2$. The cross term has residue $2C{v}_0^{\prime }(c)$, soTake the contour counterclockwise around the cylinder, as in the supplied Blasius force theorem. The residue theorem gives $\frac{1}{2}i\rho \oint (w^{\prime }{)}^{2}dz=-\pi \rho \mathrm{Res}(w^{\prime }{)}^2$. ThereforeThe force points toward the dipole. Its independence of dipole orientation follows from the product $AB={\mu }^2$ in the sole residue contribution.

The free surface is material, so applying the material derivative to $y-\eta (x,t)$ gives the exact kinematic boundary conditionFor an incompressible fluid with constant density, the Euler equations areSince the flow is irrotational, $\mathbf{u}=\nabla \varphi$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\nabla \varphi {|}^2/2)$. Integrating the resulting gradient equation gives a function of time on the right of Bernoulli equation. Adding a time-dependent constant to $\varphi$ removes that function, without changing velocity, and choosing the atmospheric-pressure reference givesThe pressure immediately above the surface is atmospheric; the liquid-side pressure includes the stated surface tension jump.

At first order in wave amplitude, the products ${\varphi }_x{\eta }_x$ and $|\nabla \varphi {|}^2$, the slope-squared correction to curvature, and displacement corrections in evaluating $\varphi$ at $y=\eta$ are all higher order. Hence, on $y=0$,For $\eta =A\cos (kx-\omega t)$, a harmonic velocity potential decaying as $y\to -\infty$ and satisfying the first condition isIndeed $\Delta \varphi =0$ and ${\varphi }_y{|}_0=A\omega \sin (kx-\omega t)={\eta }_t$. Substituting in the second condition yields the dispersion relation for a gravity-capillary wave:Taking the positive-frequency branch, the phase velocity and group velocity areSubtracting givesSince $\omega >0$,The threshold is also the wavenumber at the minimum phase speed of a gravity-capillary wave.

The vorticity of a fluid velocity field is $\mathbf{\omega }=\nabla \times \mathbf{v}$. A vortex line is an integral curve tangent to that vector, not generally a particle path or streamline. Since $\nabla \cdot \mathbf{\omega }=0$, vorticity flux through successive cross-sections of a vortex tube is equal; a regular line cannot simply end inside the fluid. By Stokes's theorem, the tube strength is the circulationFor an inviscid barotropic fluid with conservative body force, Kelvin's circulation theorem conserves circulation around a material circuit. Under those assumptions vortex tubes move with the fluid. In three-dimensional incompressible ideal flow,stretching a material tube raises its vorticity as its cross-section shrinks. In two-dimensional incompressible flow there is no such stretching, so scalar vorticity is materially conserved. Constant viscosity adds a diffusion term $\nu {\nabla }^2\mathbf{\omega }$; baroclinicity and boundaries can also generate vorticity.

An ideal line vortex has $v_{\varphi }=\Gamma /(2\pi r)$ and zero ordinary vorticity for $r>0$, with its circulation concentrated in the singular core. Circular trajectories therefore do not imply nonzero local vorticity everywhere. A finite physical core removes the velocity singularity.

For steady inviscid incompressible flow subject to a conservative body force, Bernoulli's theorem states that $p/\rho +u^2/2+gz$ is constant along a streamline. The same constant applies across streamlines only with an additional condition, such as an irrotational connected flow with the same upstream head.

Neglect gravity and let the film's free-surface pressure be atmospheric. Its radial speed is then $v$ in the ideal model. Conservation of mass equates the incoming flux $\pi r_1^{2}v$ to $2\pi rt(r)v$, soThe latter value already warns that the thin-film approximation is poor near impact; the radial expression is most useful beyond that region. The incoming vertical momentum flux gives downward force $F_m=\rho \pi r_1^2{v}^2$ when the outgoing flow is horizontal.

In the stipulated stagnant-rear-layer idealization, the excess rear pressure is the stagnation value $\rho v^2/2$. Its upward force is $\rho v^2\pi r_2^2/2$. Thus, with downward positive and ignoring the water's weight,For a large plate, the radial-film volume is approximatelyIts weight therefore changes the model toCancellation occurs nearThe printed claim of negligible force for every radius above this scale does not follow from this model. Near the cancellation radius the force is small; for substantially larger radii the quadratic upward-pressure term dominates. The original PDF, not just its transcription, uses that inequality and wording. Moreover, the rear layer's mass and detailed gravity-dependent pressure field are unspecified. The calculation establishes a cancellation scale under the elementary stated pressure approximation, rather than a general large-radius zero-force theorem. A full free-surface flow model would be needed to extend it consistently.