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The occupied cross-sectional area and interface width areThe incoming volume flux per unit channel length is $h{w}_e$. Volume conservation therefore gives $A_t+(Au{)}_x=h{w}_e$. Ambient fluid carries no excess mass density, so integrating the excess-mass density balance gives $(A{g}^{\prime }{)}_t+(Au{g}^{\prime }{)}_x=0$. These two equations imply dilution of the reduced gravity, even though total integrated buoyancy is conserved.

The integrated horizontal hydrostatic pressure force, divided by the reference mass density, isThe prismatic walls contribute no axial component of pressure force. Since ambient fluid enters with zero horizontal velocity, it supplies no horizontal momentum flux. Thus the entraining shallow-water current in a triangular channel obeys the conservative equationsTo put them in primitive form, let $D={\partial }_t+u{\partial }_x$ be the material derivative. The volume equation gives $Dh+(h/2)u_x=w_e$. Subtracting $u$ times the volume equation from the momentum equation, and $g^{\prime }$ times it from the buoyancy equation, yieldsThe momentum source is the acceleration of newly entrained ambient fluid, not bed drag. The source in the reduced gravity equation describes mixing with ambient fluid of zero excess mass density. In particular, the rectangular-channel coefficients cannot be used: the triangular area produces both $h/2$ in the depth equation and $h/3$ in the buoyancy-gradient force.

For the state vector $q=(h,u,g^{\prime }{)}^T$, the primitive equations are $q_t+\mathsf{A}{q}_x=S$, withThe characteristic polynomial isConsequently the three characteristic speeds areFor $h>0$ and $g^{\prime }>0$, the hyperbolic system has three distinct real eigenvalues. The long-wave speed relative to the current is $c$, while $u$ transports a buoyancy/contact disturbance.

A left eigenvector for the material characteristic is $(0,0,1)$, givingFor $u\pm c$, left eigenvectors can be chosen asMultiplying $q_t+\mathsf{A}{q}_x=S$ by these rows gives the characteristic compatibility for an entraining triangular-channel current. With all differentials evaluated along the indicated characteristic,The source follows directly from ${\ell }_{\pm }S=-2u{w}_e/h\pm 2c{w}_e/h\mp 2g^{\prime }{w}_e/(3c)$ and $g^{\prime }/c=2c/h$.

Since $4dc=(2c/h)dh+(h/c)d{g}^{\prime }$, an equivalent form isCombining it with the material buoyancy equation also givesOnly when entrainment vanishes and $g^{\prime }$ is spatially constant do $u\pm 4c$ become constant Riemann invariants on these characteristic curves. Dropping the buoyancy differential in the general problem would lose one of the required transport couplings. The characteristic formulas with $1/c$ apply to the nondegenerate region $g^{\prime }h>0$; dry or neutrally buoyant limits require their conservative form instead.

The entrainment coefficient $\alpha$ is the ratio of the inward ambient-fluid velocity normal to a buoyant plume edge to the characteristic buoyant plume velocity. For an upward buoyant plume, take $u_e=\alpha w>0$. There are two edges at half-width $b$, so the full-section volume source is $2u_e$. Dividing every integral balance by two gives consistent half-section equations.

For top-hat plume profiles, volume conservation and mass conservation areThe mass entering the buoyant plume is ambient mass with local mass density ${\rho }_0(z)$. The Boussinesq approximation uses a common reference mass density in inertia while retaining the mass density deficit in buoyancy. The mass equation may be kept to the first order in that deficit to derive its transport.

Multiply the volume equation by the time-independent ambient mass density. The product rule givesSubtract the mass equation and multiply by $g$. The entrainment sources cancel exactly, leavingwhereFor stable stratification, ${\rho }_0^{\prime }<0$ and the buoyancy frequency $N$ is real. Rising fluid moves into lighter ambient fluid and loses its mass density deficit relative to that local ambient. Thus environmental stratification reduces the buoyant plume's buoyancy flux, even though the ambient fluid entering by entrainment contributes no mass density deficit at its own height.

Keep the half-section convention of part (a). Define the mass density-weighted mass flux, buoyancy flux and momentum flux byThe full-section fluxes are $2Q,2F,2M$. These factors must be changed consistently if full fluxes are used; mixing a full-width flux with a half-width entrainment source would introduce a spurious factor two.

For positive $Q,M$, $w=M/Q$, $\rho b=Q^2/M$, and $({\rho }_0-\rho )gb=FQ/M$. The mass equation in part (a), the momentum equation, and the buoyancy equation therefore becomeThese are direct algebraic rewritings with the mass density factors retained. In particular, $FQ/M$ is the buoyancy inventory per unit height as well as the integrated buoyancy force. If desired, $\rho$ can be reconstructed from $F/(gQ)={\rho }_0/\rho -1$.

At leading Boussinesq order use a constant reference mass density $R$ in the inertial fluxes, $Q=Rbw$, $M=Rb{w}^2$, and write $g^{\prime }=g({\rho }_0-\rho )/R$ so $F=R{g}^{\prime }bw$. Set ${\rho }_0/\rho =1$ in the stratification source and use $N^2\simeq -g{\rho }_0^{\prime }/R$. The unsteady top-hat line-plume balances are thenEquivalently the kinematic fluxes $q=Q/R=bw$, $m=M/R=b{w}^2$ and $f=F/R=g^{\prime }bw$ obey the same equations with $R$ removed from the first source. This explicit distinction between mass density-weighted and kinematic fluxes fixes the normalization for the power-law solution.

In a homogeneous ambient fluid, $N=0$ and $R={\rho }_0$ is constant. Use kinematic fluxes or, equivalently, the primitive form of the unsteady top-hat line-plume balances:Here $g^{\prime }=g({\rho }_0-\rho )/R>0$ is the positive buoyant reduced gravity. Seek a nonsteady separable power-law ansatz with both time differentiation and vertical transport retained. Matching powers in the volume equation forces $b\propto z$ with no time dependence. Matching the two momentum terms then gives $w\propto z/\tau$, and the buoyancy force has $g^{\prime }\propto z/{\tau }^2$, where $\tau =t+t_{\ast }>0$ is a shifted age. Thus writeSubstitution determines the constants rather than just the exponents. The volume equation gives $2BA=\alpha A$, so for a nonzero buoyant plume $B=\alpha /2$. The momentum equation gives $C=-A+3A^2$, while the buoyancy equation gives $C(-2+3A)=0$. The buoyant nontrivial branch has $A=2/3$ and $C=2/3$. Consequently the separable decaying top-hat line plume isThe Boussinesq approximation requires $2z/(3g{\tau }^2)\ll 1$. The corresponding mass density-weighted fluxes, to this order, areFor a direct check, $Q^2/M=R\alpha z/2$ is time independent and $FQ/M=R\alpha z^2/(3{\tau }^2)$. Hence $(Q^2/M{)}_t+Q_z=2R\alpha z/(3\tau )=R\alpha M/Q$, $Q_t+M_z=R\alpha z^2/(3{\tau }^2)=FQ/M$, and $(FQ/M{)}_t+F_z=0$. This verifies all three balances and the required width coefficient.

This nonsteady branch is not the steady, continuously supplied, constant-flux line plume. The latter has constant $w$ and half-width $b=\alpha z$ in the same top-hat convention. Also, the unshifted power law has $Q,F\to 0$ as $z\to 0$ and is singular at $\tau =0$. It is therefore an interior separable solution, not by itself a solution with a nonzero continuously operating point-source flux at the origin. A finite-width source boundary, a virtual origin, or a matching region is needed to connect it to physical source and initial data. For example, replacing $z$ by $z+z_{\ast }$ supplies a finite base width and time-varying positive base fluxes when $z_{\ast }>0$. No source or initial flux history is specified in the question to select that matching; the requested power-law width follows from the local balances above.

In a thin disk dominated by a central stellar mass, the vertical gravitational potential is harmonic to leading order, so its vertical acceleration is $-{\Omega }_0^{2}z$. For a locally isothermal background, $p=c_s^2\rho$ with $c_s$ independent of height. Vertical hydrostatic equilibrium therefore givesThis is the isothermal vertical structure in stellar gravity. On the upper half of the disk the downward gravity magnitude is $g={\Omega }_0^{2}z>0$; the density profile extends evenly through the midplane.

For an adiabatically displaced parcel kept in pressure equilibrium, the parcel's logarithmic density gradient is ${\gamma }^{-1}d\log p/dz$, whereas the environmental density gradient is $d\log \rho /dz$. A small upward displacement $\eta$ consequently produces a density contrast $\delta \rho /\rho =[{\gamma }^{-1}d\log p/dz-d\log \rho /dz]\eta$. Its buoyancy acceleration is $-g\delta \rho /\rho =-N^2\eta$. Thus the Brunt–Väisälä frequency isBoth background gradients equal $-z/H^2$ in the isothermal atmosphere, soAn isothermal background is therefore stably stratified for adiabatic perturbations with $\gamma >1$, but not buoyant for isothermal perturbations. In applying the adiabatic equation of state to a stratified atmosphere, its proportionality relates Lagrangian parcel changes: $\Delta p=(\gamma p/\rho )\Delta \rho$. Eulerian perturbations generally obey $p^{\prime }+\eta p_z=(\gamma p/\rho )({\rho }^{\prime }+\eta {\rho }_z)$ instead. Thus the printed simple proportionality must be interpreted as a parcel relation for the buoyancy calculation. For $\gamma =1$ in this isothermal background, the gradient terms cancel and the Eulerian proportionality holds too.

Now take isothermal perturbations, so $\gamma =1$ and $N^2=0$. The two supplied wave equations reduce toFor the nonsingular oscillatory case $\omega \ne 0$ and ${\omega }^2\ne {\Omega }_0^2$, substitute $u_z={\xi }_z/{\omega }^2$. With $x=z/H$, the resulting Hermite differential equation isThe exceptional frequencies must be treated through the original velocity-pressure equations rather than division by these factors.

The physical vertical condition is finite perturbation energy when integrated with the Gaussian density. In particular the pressure contribution is proportional to $\int e^{-x^2/2}|\xi {|}^{2}dx$ for nonzero frequency. Generic nonterminating solutions have an exponentially growing component at at least one vertical end and fail this condition; acceptable modes are the terminating Hermite polynomials. This is a finite-energy condition, not pointwise boundedness of the velocity in the vanishing-density atmosphere.

To prove the allowed values of $\alpha$, let the nonzero polynomial have degree $n$ and leading coefficient $a_n$. The highest power in ${\xi }_{xx}-x{\xi }_x+\alpha \xi$ has coefficient $(\alpha -n)a_n$. Hence $\alpha =n\in \{0,1,2,\dots \}$. Conversely, writing $\xi ={\sum }_j{a}_j{x}^j$ gives the recurrenceFor $\alpha =n$, choose the parity of $n$ and iterate until the series terminates at degree $n$. Thus all allowed Hermite vertical modes of an isothermal disk havewhere ${\mathrm{He}}_n$ is the Probabilists' Hermite polynomial. In particular, ${\mathrm{He}}_0=1$ and ${\mathrm{He}}_1=x$.

Let $K=kH$ and $s={\omega }^2/{\Omega }_0^2$. Eliminating the denominator for regular modes givesFor $n=0$, the nonzero-frequency solution isIts pressure variable is independent of height and $u_z=0$. These are inertial-acoustic waves, restored by epicyclic motion and pressure; at short radial wavelength they become ordinary acoustic waves. The multiplied dispersion polynomial also has $s=0$. This is not an extra oscillatory solution represented by nonzero $\xi$, since $\xi =i\omega p^{\prime }/\rho$ vanishes at zero frequency for finite pressure perturbation. The full horizontal equations have a zero-frequency balanced mode of an axisymmetric disk, with no radial velocity and azimuthal Coriolis acceleration balancing a radial pressure gradient. It must be recovered without dividing by $\omega$, rather than counted as a propagating wave. At $k=0$, the propagating branch limits to an epicyclic oscillation at $|\omega |={\Omega }_0$, another value at which the eliminated form must be interpreted by a limit.

For $n=1$, solve $(s-1{)}^2=K^{2}s$. The two branches areBoth signs of frequency occur on each squared-frequency branch. The upper branch is acoustic in the large-$|K|$ limit, $|{\omega }_{+}|\sim |k|c_s$; the lower branch is an inertial wave with $|{\omega }_{-}|\sim {\Omega }_0/|K|$. At $|K|\ll 1$, both frequencies approach ${\Omega }_0$, reflecting the equality of the radial and vertical epicyclic frequencies in a Keplerian disk. Here $\xi \propto z$, so $u_z={\xi }_z/{\omega }^2$ is independent of height: these modes have a coherent vertical displacement of the disk column, rather than a vertical breathing motion.

For a local non-axisymmetric disturbance proportional to $e^{i(kR+m\varphi -\omega t)}$, the material frequency is $\hat{\omega }=\omega -m{\Omega }_0$. In the same local radial-wave approximation, replace $\omega$ by $\hat{\omega }$. For $n=1,m=1$, this givesThe four frequencies can equivalently be written $\omega /{\Omega }_0=1\pm (\sqrt{K^2+4}\pm |K|)/2$, with independent choices of the two signs. This is a local dispersion relation, not a claim that Doppler substitution supplies an exact global non-axisymmetric solution in a differentially rotating disk. The radial-wave ordering neglects azimuthal gradients compared with radial gradients; $|k|R\gg 1$ can coexist with $|k|H\ll 1$ in a thin disk.

Write $w=\omega /{\Omega }_0$. For the two low inertial-frame frequency branches, $|w|\ll 1$, while the material frequency remains near $-{\Omega }_0$. Expanding the dispersion relation givesThereforeFor signed $k$, the two leading branches are conventionally labelled by $\pm kH/2$. These are bending waves of a Keplerian disk: an $m=1$ vertical displacement varies once around an annulus and corresponds to a small tilt of its orbital plane. A radial variation of the tilt is a warp. The waves transport the warp radially at group speeds $c_s/2$ in opposite directions. At zero radial wavenumber their low-frequency limit is a stationary rigid tilt; the apparent slow pattern does not mean that the orbiting gas has slow vertical motion in its own frame.

Set $S={\partial }_z\overline{u}>0$ and $N^2=-(g/{\rho }_0){\partial }_z\overline{\rho }>0$, using the constant reference mass density ${\rho }_0$ of the Boussinesq approximation. The mean-shear production is $\mathcal{P}=-\overline{u^{\prime }{w}^{\prime }}S>0$. The stable buoyancy destruction is $B=(g/{\rho }_0)\overline{{\rho }^{\prime }{w}^{\prime }}>0$. ThereforeThe Flux Richardson number compares stable buoyancy destruction with shear production; the gradient Richardson number compares restoring stratification with mean shear rate. They are generally different. The stationary energy budget gives $ϵ=\mathcal{P}(1-{\mathrm{Ri}}_f)$, so positive dissipation in this balance requires ${\mathrm{Ri}}_f<1$.

For $\mathrm{Ri}>{\mathrm{Ri}}_c$, the fixed-correlation equilibrium model for stratified turbulence has $R_{\min }>1$. Thus buoyancy destruction plus dissipation exceeds shear production for every $q>0$, and the local kinetic-energy budget predicts decay rather than any positive statistically steady turbulent equilibrium. The stable and unstable turbulent branches have disappeared together at the critical point.

The implied outcome within this closure is suppression or decay of sustained turbulence. It is not a proof that every physical stratified shear flow above this value must be permanently laminar: energy transport, external forcing, intermittent mixing or variable correlations and length-scale ratios can violate the model assumptions. As turbulence decays, the original fully turbulent high-Reynolds-number approximation itself ceases to apply. The critical Richardson number is consequently a prediction of the stated local closure, not a universal nonlinear stability bound.

Differentiating $R=a/q+bq$ gives $R^{\prime }=-a/q^2+b$, so for positive stratification its minimum has $q_m^2=a/b$. At that pointThus the two sinks are equal. The buoyancy fraction of total turbulent sinks is always $B/(B+ϵ)=1/2$ there. But the Flux Richardson number defined in part (i) is insteadIt equals $1/2$ only when the minimum is also an equilibrium, $R_{\min }=1$, namely at the critical gradient. This is the qualification needed for the printed assertion; an arbitrary minimum on a curve plotted away from equilibrium need not have ${\mathrm{Ri}}_f=1/2$.

For example choose closure parameters $C_u=C_{\rho }=1/4$, units with $L_u=L_{\rho }=S=1$, and $N^2=1/16$. Then $a=1/64$, $b=4$, $q_m=1/16$ and $R_{\min }=1/2$. The actual flux Richardson number at this minimum is $1/4$, while the buoyancy fraction of the total sinks is $1/2$. This is a concrete counterexample to the unqualified wording. With zero mass density gradient there is no positive interior minimum at all, and ${\mathrm{Ri}}_f=0$.

Let downward velocity be positive in this part, and let $W>0$ be the dense sphere's weight minus buoyancy. For the original centreline configuration, write its six-component translation-and-rotation velocity as $q$, and its diagonal positive hydrodynamic resistance matrix as $R_0$. Its single-sphere dissipation isWith only axial force and no torque, the original velocity is $V_0=W/R_{zz}$.

Now add the neutral sphere. In its actual Stokes flow, extend the velocity rigidly through the new sphere's interior. This extension is a valid incompressible trial field for the domain without that sphere, with exactly the dense sphere's actual six boundary velocities; its strain and dissipation inside the inserted ball are zero. The Minimum-dissipation theorem for Stokes flow and extra dissipation due to a rigid inclusion therefore giveThe neutral sphere has zero net external force and torque, and the dense sphere has only the axial force $W$. The dissipation equals total external power, so $D=W{V}_z$. Positivity gives $V_z>0$, and the comparison yields $V_z\le W/R_{zz}$ without assuming that either sphere translates only vertically or that either angular velocity vanishes.

For a finite nonoverlapping placement, the first inequality is strict. Equality would make the extended field the old Stokes solution and thus make that solution rigid throughout the added ball. Analyticity and connectedness then force its strain to vanish throughout the original fluid domain; the stationary tube wall and quiescent far field would force the whole motion to vanish, contradicting $V_z>0$. Hence the force-free inclusion reduces axial mobility of a centred settling sphere:The difference tends to zero as the added sphere recedes infinitely far away. The lateral and rotational contributions are nonnegative in the resistance comparison, which is precisely why they cannot invalidate the bound.

Take the upward direction as ${\mathbf{e}}_z$, with polar angle $\theta$ measured from it. The liquid is quiescent at infinity. On a translating clean inviscid bubble, the kinematic boundary condition is $(\mathbf{u}-U{\mathbf{e}}_z)\cdot \mathbf{n}=0$, and the stress-free boundary condition is zero tangential traction. The normal traction must balance uniform gas pressure and any constant-curvature surface-tension jump.

For a vertical Stokeslet of strength $F$, the radial velocity is $u_r=F\cos \theta /(4\pi \mu r)$. Its traction on a concentric sphere is purely normal, so zero tangential traction is automatic. The kinematic condition at $r=a$ givesThe singularity is only a device for generating the exterior solution; there is no liquid point force inside the actual bubble. The integrated dynamic traction on the bubble is $-F{\mathbf{e}}_z$, giving the clean-bubble Stokes drag. The upward buoyancy is $4\pi \rho g{a}^3/3$, hence force balance gives

The normal stress also shows that surface tension is not required for this particular spherical solution. Let the ambient hydrostatic pressure at the bubble centre be $p_c$, so on its surface $p_h=p_c-\rho ga\cos \theta$. Adding the Stokeslet's dynamic normal traction giveswhere the terminal buoyancy value of $F$ cancels the dipole. This is the spherical clean bubble without surface tension balance: an initially spherical bubble can remain spherical in this ideal steady Stokes solution even when surface tension is zero. With uniform surface tension $\gamma$, choose gas pressure $p_b=p_c+2\gamma /a$; with $\gamma =0$, $p_b=p_c$. This proves compatibility of the spherical shape, not stability of arbitrary deformations or validity when inertia matters.

The force scale is $F\sim \rho g{a}^3$. At the distant free surface, the Stokeslet's pressure and viscous normal-stress scales are $F/d^2$. Gravity restores a displaced surface by a pressure change $\rho gh$, soThus $h/d=O((a/d{)}^3)\ll 1$, consistent with treating the boundary as approximately flat and impermeable at leading order. This uses the stipulated gravity-dominated restoring balance.

The liquid-air interface has negligible tangential traction. The stress-free image of a normal Stokeslet places an opposite point force at the reflection point, a distance $2d$ above the bubble centre. Across the flat surface the combined normal velocity is odd and the tangential velocity is even, so the normal velocity and tangential stress vanish there. A same-radius inviscid drop of density $2\rho$ has downward excess weight $4\pi \rho g{a}^3/3$, exactly opposite the bubble's upward buoyancy. Its exterior Stokeslet therefore supplies this image, giving the asserted full-space comparison to leading order in $a/d$.

The image's incident axial velocity at the bubble isTo first reflection, the bubble translates with this locally uniform incident velocity plus its isolated rise speed $F/(4\pi \mu a)$. Hence the bubble rise near a distant stress-free free surface isOnly the first relative correction is retained; the image's spatial variation and additional reflections affect higher terms.

The customary capillary-controlled small-deformation condition is a small capillary number,equivalently a small Bond number $\mathrm{Bo}=\rho g{a}^2/\gamma$. This is a sufficient condition, not a necessary condition for the special isolated spherical solution in part (b). More precisely, the image's strain is $O(F/(\mu d^2))$, so its deformation-producing normal stress compared with $\gamma /a$ is $O(\mathrm{Bo}(a/d{)}^2)$. When exploiting the exact isolated solution, that smaller combination controls the leading boundary-induced distortion. If capillarity at the distant free surface is also retained, the gravity-dominated deflection scaling requires $\gamma /(\rho g{d}^2)\ll 1$; $a\ll \sqrt{\gamma /(\rho g)}\ll d$ allows both the usual spherical-bubble criterion and gravity-dominated surface restoration.

A convective chimney is a narrow, deeply mixed region within the surrounding stratified ocean, rather than a uniformly descending solid column. Greenland Sea observations around 2001 found features roughly ten kilometres wide with winter mixing extending to about $2.5$ km. Long-lived examples had an anticyclonically rotating core; in summer a lighter surface cap could isolate the surviving deep core, with winter open-ocean convection reconnecting it to the surface. Wadhams's 2005 observations describe this structure. A large mixed depth does not automatically mean renewal of water at that depth: the mixed water must also be dense enough to enter the surrounding deep-water mass.

Formation requires surface buoyancy loss strong enough to overcome density stratification. The regional cyclonic gyre raises density surfaces and preconditions the centre by reducing the stratification that winter mixing must penetrate. Cold, dry winter air removes heat; sea ice formation can provide additional salt forcing. Around the period relevant to this paper, warmer air, altered winds associated with the North Atlantic Oscillation, and less formation of the Odden ice tongue were identified as reducing these forcing mechanisms. Fresh surface water also resists deep mixing. Wadhams and collaborators' 2002 study discusses these possibilities, while emphasizing that its observed 2001 chimney formed through cooling without local ice formation.

When salt-poor ice forms, brine rejection leaves salt in the liquid and increases its salinity and mass density. Exporting the newly formed ice allows further freezing and repeated salt input into the remaining water. Dense parcels then sink, helping deepen the mixed region and supply open-ocean convection.

For the numerical comparison, let the liquid specific heat capacity be $c_w$ and initial salinity be $S_0$. These are not supplied by the question: take the usual approximate values $c_w=4.0\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $S_0=35$ psu for a numerical estimate. The ice heat capacity cannot replace $c_w$ during liquid cooling. Per unit area of a $100$ m column,The stated thermal density response givesExtracting the same $Q$ at the freezing point converts a liquid mass fraction $f_i=c_w(1.8)/L_f\simeq 0.02143$ into salt-free ice. Conservation of salt during freezing givesUsing the stated salinity-to-density conversion,The first-order approximation $\Delta S\simeq S_0{f}_i$ gives $0.75$ psu and a factor $4.2$. Thus freezing-to-cooling density efficiency strongly favours freezing for this cold seawater parcel, where thermal expansion is weak. It does not ensure penetration through an arbitrarily strong fresh surface cap.