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The original PDF has absolute-value bars around both gradients; they are missing from the transcription. The stability argument must respect the sign of the upward coordinate.

Displace a small parcel upward by $\xi$. Assume it remains in pressure equilibrium with its surroundings but exchanges negligible heat, so it follows an adiabatic process. For a perfect gas with adiabatic index $\gamma$, $T\propto p^{(\gamma -1)/\gamma }$ along that parcel's adiabat. Its temperature gradient is thereforeThe last step uses hydrostatic equilibrium, and $c_p$ is the specific heat capacity at constant pressure. At its new height, the temperature difference between the parcel and the environment is $(T_{\mathrm{ad}}^{\prime }-T^{\prime })\xi$. At equal pressure, its relative density excess is the negative of its relative temperature excess. The upward buoyancy acceleration is thusThis is the squared Brunt–Väisälä frequency. Convective stability requires $N^2>0$, or, in signed-gradient form,For a temperature decreasing upward, this is exactly the PDF's Schwarzschild criterionThe absolute-value condition is also sufficient more generally, although a sufficiently steep temperature inversion remains stable even when that stronger condition fails. Reversing the signed inequality, as the bar-free transcription does, would instead describe an unstable temperature decrease.

The buoyancy frequency, also called the Brunt–Väisälä frequency, is the angular frequency of a small vertical displacement restored by buoyancy in a stably stratified fluid. Take $z$ upward, gravitational acceleration $-g\hat{\mathbf{z}}$, and a background in hydrostatic equilibrium,Displace a small gas element upward by $\xi$, sufficiently slowly for its pressure to match its surroundings but without heat exchange. For this adiabatic process, $p/{\rho }^{\gamma }$ is constant along the element's motion, so its mass density changes by $\rho {\gamma }^{-1}(d\ln p/dz)\xi$. The surrounding mass density changes by $(d\rho /dz)\xi$. The element's excess mass density is consequentlyIts acceleration is $\ddot{\xi }=-g\Delta \rho /\rho =-N^2\xi$. ThusFor a gas of fixed composition obeying the ideal gas law, $\rho \propto p/T$. The adiabatic buoyancy frequency of an ideal gas is thereforewhere $c_p=\gamma \mathcal{R}/(\gamma -1)$ is the specific heat capacity at constant pressure, and $\mathcal{R}=p/(\rho T)$ is the specific gas constant. Here $N^2$ has units ${\mathrm{s}}^{-2}$; $N$ is an angular frequency, not a cyclic frequency. Positive $N^2$ gives simple harmonic motion, negative $N^2$ gives convective instability, and $N^2=0$ is marginal. The temperature form assumes fixed composition; the preceding density form is more general.

Write the background mass density as $\bar{\rho }(z)={\rho }_0+{\rho }_0^{\prime }z$ and its hydrostatic pressure as $\bar{p}(z)$, whereApply an Eulerian perturbation of a fluid variable to the background fields:Thus $\pi$ is the Eulerian pressure perturbation, in pascals; it is not the background hydrostatic pressure.

For incompressible flow, mass conservation becomes $D{\rho }_{\mathrm{tot}}/Dt=0$. Linearizing about rest givesThe Euler equations are ${\rho }_{\mathrm{tot}}D\mathbf{u}/Dt=-\nabla p_{\mathrm{tot}}-{\rho }_{\mathrm{tot}}g\hat{\mathbf{z}}$. The background pressure gradient and weight cancel. Dropping products of perturbations givesUnder $|{\rho }_0^{\prime }z|\ll {\rho }_0$, replace $\bar{\rho }$ by ${\rho }_0$ in inertia but retain the background mass density gradient in the displacement-induced buoyancy. This is the Boussinesq approximation. Its linear Boussinesq equations here areThe absolute value makes explicit that the small-density-variation condition concerns magnitude, including the usual negative ${\rho }_0^{\prime }$ of stable density stratification.

A spherical harmonic of angular degree $\ell$ has horizontal wavenumber $k_h=L/r$, with $L=\sqrt{\ell (\ell +1)}$. The local acoustic dispersion relation therefore gives the radial wavenumberIn the high-order regime a travelling stellar acoustic mode accumulates phase $\int k_{r}dr$ across its radial cavity. Reflection at the inner turning point and at the near-surface boundary fixes additional phases. A round trip must return the same complex amplitude, so WKB quantization givesHere $n$ counts radial oscillations, up to the choice of origin for radial order, and $\alpha$ incorporates both reflection phases. The outer acoustic reflection need not coincide exactly with the geometric stellar radius; replacing it by $R$ and absorbing its correction in $\alpha$ is the leading asymptotic approximation. The model neglects buoyancy, perturbations of self-gravity and rapid background variations on a wavelength.

Resolved observations of the Sun measure line-of-sight surface velocity through the Doppler effect, or brightness fluctuations. Projection on spherical harmonics identifies $\ell$ and azimuthal order $m$; a temporal Fourier transform identifies the mode angular frequencies $\omega$. The sequences of peaks of successive radial order form ridges in the frequency-degree diagram. Their ordering and comparison with a reference stellar oscillation model identify $n$; the surface image alone does not count inaccessible radial nodes. Fit the residual phase of the radial quantization relation across the observed mode set to estimate $\alpha$. A simultaneous fit with the sound-speed profile, or a calibrated reference model, is needed: $\alpha$ is not determined independently by one mode, and an integer shift of the numbering of $n$ shifts $\alpha$ oppositely. In a more accurate fit the surface phase depends on frequency. These steps constitute helioseismology.

Set $a(r)=c(r)/r$ and $w=\omega /L$. Assume $a$ decreases outwards in the region to be inverted. With the surface approximation above, the Duvall law isThus a suitably scaled mode set samples one function of $w$. Smooth the measured values before differentiating; inversion amplifies observational noise. At a finite outer reference point $r_s$, the already known contribution exterior to $r_s$ must be subtracted. For the remaining cavity, let $w_s=a(r_s)$ and $h(b)=-d\ln r/db$. Changing integration variable from radius to $b=a(r)$ givesThe boundary term vanishes because the square root vanishes at $b=w$. Now multiply the derivative by $w/\sqrt{a^2-w^2}$ and interchange the integrals. The kernel identityfollows by setting $w^2=b^2+(a^2-b^2){\sin }^2\theta$ and then integrating $1/[b^2{\cos }^2\theta +a^2{\sin }^2\theta ]$. It yields the Abel inversion of stellar acoustic travel times:Knowing the monotone relation $r(a)$ then gives $c=ar$. Observed turning points delimit the region recoverable without extrapolation; an unknown surface contribution or nonmonotone $a$ prevents this simple unique inversion.

There are genuine inconsistencies in the printed numerical example. Since $w$ and $w_0$ have dimensions of angular frequency, $F$ must have dimensions of time, but the displayed empirical right-hand side is dimensionless. Moreover its derivative supplies a factor $1/w_0$, whereas the requested inversion supplies $1/w_0^2$. The dimensionally repaired relation that produces the requested inversion integral isSubstitution in the boxed Abel transform inversion gives precisely the exponent coefficient $2a^2/(\pi w_0^2)$, with the square-root denominator inside the integral. This repairs the first normalization but does not repair the later algebra.

Indeed, put $x_s=w_s/a$ and $b=\sqrt{1-x_s^2}$. The required elementary integrals evaluate toThe finite-reference-point answer following from $F_c$ is thereforeIt still depends on $w_s$. The later printed profile suppresses that dependence, which requires the additional surface limit $w_s\to 0$, $r_s\to R$. Even in that limit it is not the result of this integral. To see this exactly, write $\mathcal{L}=\ln (R/r)$ and $y=a^2/(\pi w_0^2)$. The integral givesThe minus branch is selected by $a\to 0$ at the surface. Its decreasing-radius branch extends only to $\ln (R/r)=1/6$, where $a^2=\pi w_0^2/6$; it cannot describe $r=R/e$. The stated lower bound on the largest sampled $w$ does not fix this algebraic obstruction. In contrast, squaring and rearranging the printed closed profile would require $\mathcal{L}=18y-81y^2$. This differs identically from $2y-6y^2$, so the two printed steps cannot both follow from the same data.

The surface polytrope test can nevertheless be performed exactly, and clarifies the final discrepancy. For a plane-parallel polytropic atmosphere of index $m=3$, take $g=GM/R^2$ constant and $z=R-r$. The stellar hydrostatic equation and $p=K{\rho }^{4/3}$ give $4p/\rho =gz$, hence the surface sound speed of a polytropic star isThe exponent in the equilibrium polytropic equation of state need not equal the perturbation adiabatic exponent $\gamma$. For the corrected inversion above, expansion at $z/R\ll 1$ gives $c^2\simeq \pi w_0^{2}Rz/2$, so matching the surface polytrope fixes $w_0^2=\gamma GM/(2\pi R^3)$.

Alternatively, if the later printed closed profile is adopted as a separate imposed model, its surface expansion gives $c^2\simeq \pi w_0^{2}Rz/18$. It also has the correct square-root depth dependence, but requiresAt $r=R/e$ that imposed profile then givesThe claimed coefficient $\pi /3$ is inconsistent with this substitution; $1/e\ne \pi /3$. Thus the surface scaling is demonstrable, but the empirical normalization, the requested closed inversion and the final numerical coefficient cannot all be asserted as printed. These discrepancies are visible in the original PDF, not introduced by its TeX conversion.

Take a dense lower layer and write $g^{\prime }=g({\rho }_1-{\rho }_0)/{\rho }_0>0$. In the Boussinesq approximation the fractional density difference is small; use the reference density in inertia and retain the difference in buoyancy. The one-layer model additionally treats the ambient as much deeper, with negligible distributed horizontal inertia away from the head. The shallow water approximation requires depth small relative to the horizontal variation length, small vertical accelerations, nearly hydrostatic pressure, and an approximately depth-uniform horizontal velocity. Neglect mixing, viscosity, bed friction and rotation. A slowly varying rectangular channel allows cross-sectional averages without large lateral separation.

A slice of width $b(x)$ contains volume $bhdx$. Its outgoing volume flux is $bhu$, so volume conservation gives $(bh{)}_t+(bhu{)}_x=0$. The integrated hydrostatic excess pressure is ${\rho }_0{g}^{\prime }b{h}^2/2$. Pressure on the sloping side walls contributes ${\rho }_0{g}^{\prime }{h}^2{b}^{\prime }/2$ to the longitudinal force. The conservative momentum equation is thereforeUse the volume equation to simplify it. The resulting shallow water equations areOmitting the wall force would incorrectly introduce a width term into the material acceleration equation.

For $(h,u)$ the principal matrix is $\left(\begin{array}{cc}u& h\\ g^{\prime }& u\end{array}\right)$. Its eigenvalues are $u\pm c$, where $c=\sqrt{g^{\prime }h}$. For $h>0$ they are real and distinct, so the system is strictly hyperbolic. It degenerates at a dry front. From the volume equation, $c_t+u{c}_x+(c/2)u_x=-(cu/2)b^{\prime }/b$. Combine this with the momentum equation to obtain the variable-width shallow-water characteristics:In a constant-width channel $u\pm 2c$ are Riemann invariants along the corresponding characteristic curves; with varying width they satisfy compatibility equations with a source term.

For the upward incident internal gravity wave, $P={\rho }_0\omega m_{0}W/k^2$. Using real physical fields, its time-averaged vertical energy flux per unit area isThe assumed mixing power is $\mathcal{I}/4$. A deepening law also requires the gravitational potential energy of the mixed layer. The wave problem specifies a well-mixed half-space above the interface, but supplies neither a finite upper mixed-layer depth nor a replacement buoyancy/energy boundary condition. There is therefore no unique finite-layer $d(t)$ from the printed information alone.

The missing parameter can be displayed explicitly. Let the initially mixed upper region have finite depth $a$, with a fixed upper boundary at $z=a$, and conserve its integrated buoyancy while homogenizing down to $z=-d$. Its new buoyancy isIntegrate $-{\rho }_{0}zb$ over the initial and final profiles in $-d<z<a$. The potential energy of mixed-layer deepening with an initial buoyancy jump isWith $d(0)=0$ and a maintained constant incident flux, the resulting law isThe right-hand side is positive and determines a unique depth once $a$ is supplied. Initially $\dot{d}=\omega m_0|w_0{|}^2/(4k^2\Delta ba)$ for $a>0$, already showing why the upper depth matters. If the added assumptions instead specify a zero initial mixed depth, the cubic term gives $d=[3\omega m_0|w_0{|}^{2}t/(2k^2{N}_0^2){]}^{1/3}$; the initial jump then has no finite volume above it to mix with. Neither assumption is contained in the original half-space wave statement. Literal homogenization of the entire upper half-space is the $a\to \infty$ limit and costs infinite energy for any fixed positive $d$. This limitation should not be concealed by a guessed entrainment depth or by keeping the upper buoyancy fixed while claiming conservation.

The cell conservation equation has the form ${\bar{n}}_{\bar{t}}=-\bar{\nabla }\cdot \bar{\mathbf{J}}$, withThe first term is advection by the suspension's fluid velocity, the second is active swimming at fixed speed $V_c$ in direction $\mathbf{p}$, and the last is an isotropic diffusive flux due to random swimming. Here $D$ has units length squared per time, $\bar{n}$ is cell number per volume, and the divergence measures local loss by transport. No birth or death term is included.

For a spherical cell of radius $a_c$, its excess buoyancy force is $(4\pi a_c^3/3)\Delta \rho g$. The Stokes drag law gives passive settling speed $V_{\mathrm{set}}=2a_c^2\Delta \rho g/(9\mu )$. Neglecting individual settling requires $V_{\mathrm{set}}\ll V_c$, not exact neutral buoyancy. A dilute suspension can still generate appreciable collective buoyancy because many small excess masses contribute to the bulk density.

Use the half-rate convention for the gyrotactic reorientation time $B$, required by the coefficients printed in the question:The first torque tends to point the bottom-heavy cell upward; the second rotates a sphere with half the fluid vorticity. This is a quasistatic torque balance, assuming orientation relaxation is fast enough to follow the local flow. Keeping orientation dynamics instead would put $d\mathbf{p}/d\bar{t}$ on the left. The convention with aligning rate $1/B$ uses a reorientation parameter twice this $B$. Thus the factor of two is a definition, not a physical discrepancy. Close to the upward state the quasistatic gyrotactic balance with half-rate reorientation gives ${\mathbf{p}}_{\perp }=B\bar{\mathbf{\omega }}\times {\mathbf{e}}_z$.

For a dilute Newtonian suspension with the Boussinesq approximation, the other equations areThe background-water hydrostatic pressure is absorbed in $\bar{p}$. The cell volume is $\upsilon$, density is $\rho$, cell density is $\rho +\Delta \rho$, dynamic viscosity is $\mu =\rho \nu$, and $\upsilon \bar{n}\ll 1$ is the volume fraction. Extra cell mass produces the downward buoyancy term. Constant viscosity and no explicit swimming stress are part of this closure. Rigid impermeable plates impose no-slip boundary condition and zero normal total cell flux.

At rest, the stable cell direction is upward and the steady vertical flux is $V_c\bar{n}-D{\bar{n}}_{\bar{z}}$. It is constant in depth by conservation and zero at an impermeable plate, soThe pressure gradient balances its excess weight. The mean-concentration formula follows by averaging over $0\le \bar{z}\le h$ and is continuous at $\beta =0$.

Use $\mathbf{x}=\bar{\mathbf{x}}/h$, $t=D\bar{t}/h^2$, $\mathbf{u}=h\bar{\mathbf{u}}/D$, $n=\bar{n}/{\bar{N}}_0$, and $p=h^2\bar{p}/(\rho \nu D)$. ThenwhereThe dimensionless base concentration is $n_0=e^{\beta z}$.

Linearize with the horizontal Fourier mode $e^{\sigma t+ilx+imy}$ and $k^2=l^2+m^2$. Write the first-order orientation perturbation as $\mathbf{p}={\mathbf{e}}_z+ϵ{\mathbf{p}}_1+\cdots$. Unit length gives $p_{1z}=0$, while torque balance gives ${\mathbf{p}}_{1\perp }=G{\mathbf{\omega }}_1\times {\mathbf{e}}_z$. Incompressibility impliesLet $\mathcal{L}=d^2/d{z}^2-k^2$. The perturbation of fluid advection is $n_0^{\prime }W=\beta n_{0}W$, and swimming contributes $\beta N^{\prime }+\beta n_0\nabla \cdot {\mathbf{p}}_1$. Cell conservation therefore givesorThis explains the destabilizing concentration-advection term and the gyrotactic redistribution term with their signs.

To eliminate pressure from the fluid equations, take their divergence, giving $\mathcal{L}{p}_1=-(R/\beta )N^{\prime }$. Apply $\mathcal{L}$ to the vertical equation $S{c}^{-1}\sigma W=-p_1^{\prime }+\mathcal{L}W-(R/\beta )N$. The concentration second derivatives cancel, leavingNo-slip conditions give $W=0$ and $W^{\prime }=0$ at each plate; for nonzero $k$, the latter follows from the vanishing horizontal velocity and incompressibility. Zero perturbed normal cell flux givesThe normal orientation correction is zero at this order, and the fluid velocity is zero at the plate, so neither adds another term to this concentration condition.

Now use the stated distinguished scalings $k=\beta \alpha$, $\sigma ={\beta }^2{\sigma }_2$, $G={\beta }^{-1}{G}_{-1}$, $N=N_0+\beta N_1+{\beta }^2{N}_2+\cdots$, $W=\beta W_1+{\beta }^2{W}_2+\cdots$, and $R=R_0+\beta R_1+\cdots$. Here $\alpha$ is a scaled wavenumber, unrelated to Question 2's shear rate. At order one, the concentration equation and boundary conditions give $N_0^{\prime \prime }=0$, $N_0^{\prime }=0$ at both endpoints, so $N_0$ is constant. Absorb its nonzero value into the perturbation amplitude and chooseAt order $\beta$, the fluid equation is $W_1^{(4)}=-R_0{\alpha }^2$ with $W_1=W_1^{\prime }=0$ at both ends. Its unique solution isA homogeneous difference is cubic and the four clamped conditions make it zero.

The concentration equation at order $\beta$ givesso $N_1=z-G_{-1}{W}_1+C_1$, where $C_1$ fixes the amplitude normalization. At order ${\beta }^2$ it givesThe boundary conditions expand to $N_2^{\prime }=N_1$ at each endpoint. Integrating this equation from zero to one cancels $[N_2^{\prime }{]}_0^1$ against $[N_1{]}_0^1$. Also $[W_2^{\prime }{]}_0^1=0$ andThus the required solvability condition isSince ${\int }_0^1{z}^2(1-z{)}^{2}dz=1/30$, the clamped-quartic solvability in shallow bioconvection gives

For nonzero long horizontal waves, $R_0<720$ gives decay and $R_0>720$ gives growth; the leading neutral threshold is $R_0=720$. The leading instability is stationary and independent of $G_{-1}$ and $Sc$. The exactly uniform horizontal mode $\alpha =0$ has zero growth at this order because total cell number is conserved, so it does not determine the threshold. Upward swimming produces a dense upper layer; the unstable disturbance is its buoyancy-driven overturning, or bioconvection. Higher orders are needed for wavelength selection and finite-depth threshold corrections.

With fixed material parameters and mean concentration, $R\sim h^4{V}_c\upsilon \Delta \rho g{\bar{N}}_{00}/(\rho \nu D^2)$ for small $\beta$, so a sufficiently thin chamber is stable to this branch. Formally its threshold corresponds to $h_c\sim [720\rho \nu D^2/(V_c\upsilon \Delta \rho g{\bar{N}}_{00}){]}^{1/4}$ when the weak-stratification and distinguished-gyrotaxis assumptions apply. In particular, the assumed $G=O({\beta }^{-1})$ is a specified asymptotic analysis; at fixed $B,D,V_c$ and decreasing $h$, one instead has $G=O({\beta }^{-2})$. The leading result should not be asserted unchanged in that different ordering without redoing the expansion.

The quasi-geostrophic approximation isolates slowly evolving, nearly balanced motions in a rotating, stably stratified fluid. Ocean eddies and mid-latitude atmospheric disturbances often have this character: a leading Coriolis force–pressure balance suppresses rapid horizontal acceleration, while hydrostatic balance suppresses rapid vertical acceleration. Small ageostrophic flow nevertheless supplies the vertical motion that changes the leading vorticity. The approximation filters fast inertia-gravity waves while retaining nonlinear transport, vertical shear and Rossby waves.

Use a reference density ${\rho }_0$ and remove the background hydrostatic pressure. Let $\pi =p^{\prime }/{\rho }_0$ and let the upward buoyancy perturbation be $b=-g{\rho }^{\prime }/{\rho }_0$. A vertically varying reference density stratification has $N^2(z)>0$. Under the Boussinesq approximation, the inviscid and adiabatic equations areHere $D/Dt={\partial }_t+{\mathbf{u}}_H\cdot {\nabla }_H+w{\partial }_z$. Density variations are small compared with ${\rho }_0$ but retained in the vertical buoyancy force. This is a useful local, incompressible description; a compressible atmosphere needs the corresponding weighted or pressure-coordinate version.

Let horizontal and vertical lengths be $L,H$, speed be $U$, and take the slow time $L/U$. Take $f_0>0$ for definiteness; use $|f_0|$ in the scale estimates in the opposite hemisphere. The principal small parameter is the Rossby number $\mathrm{Ro}=U/(f_{0}L)\ll 1$ at a latitude where $f_0\ne 0$. For the hydrostatic, strongly stratified regime, $H/L\ll 1$ and $f_0/N\ll 1$. Retaining both horizontal vorticity and vortex stretching gives the Burger numberThe geostrophic balance pressure scale is $f_{0}UL$ and the buoyancy perturbation scale is $B=f_{0}UL/H$. Hence $B/(N^{2}H)=\mathrm{Ro}/\mathrm{Bu}\ll 1$: buoyancy perturbations produce small displacements relative to the background stratification scale. The ageostrophic flow is $O(\mathrm{Ro}U)$ and incompressibility then gives $w=O(\mathrm{Ro}UH/L)$. Vertical acceleration is smaller than hydrostatic terms by order ${\mathrm{Ro}}^2(f_0/N{)}^2$ in this scaling. Weak forcing or diffusion can be added at the slow order, but neither is present here.

Define the quasi-geostrophic streamfunction by $\psi =\pi /f_0$. Leading geostrophic balance and hydrostatic balance implyDifferentiating horizontally and vertically gives the thermal-wind balanceThus vertical shear is determined by horizontal buoyancy gradients. The homogeneous Taylor–Proudman theorem is recovered only when those gradients vanish. Stable stratification allows sloping density surfaces and different geostrophic flow at different levels, without abandoning leading balance.

The Prandtl ratio of scales makes that modification quantitative. Horizontal and vertical terms in balanced potential-vorticity inversion have sizes $\psi /L^2$ and $(f_0^2/N^2)\psi /H^2$. Equality givesFor constant $N$, $Z=(N/|f_0|)z$ makes the inversion operator an ordinary three-dimensional Laplacian. Equivalently, an interior balanced disturbance of horizontal size $L$ naturally penetrates a depth of order $|f_0|L/N$. With $N\gg |f_0|$ this is a thin, vertically sheared structure, rather than the arbitrarily tall homogeneous columns suggested by strict Taylor–Proudman theorem behaviour. This is an aspect-ratio statement, not the diffusivity Prandtl number.

To allow a weak meridional variation of the Coriolis parameter, use the beta plane $f=f_0+\beta y$, with $\beta L/f_0=O(\mathrm{Ro})$, or equivalently $\beta L^2/U=O(1)$. The small correction to the leading balance is dynamically important on the slow time. Let $\xi ={\nabla }_H^2\psi$ be vertical relative vorticity and let $D_g={\partial }_t+{\mathbf{u}}_g\cdot {\nabla }_H$. At the first nontrivial order, the vertical vorticity equation and buoyancy equation reduce toRelative-vorticity stretching and tilting are higher order here; planetary-vorticity stretching $f_0{w}_z$ must remain. Eliminate $w$ to expose a materially conserved balanced scalar. Because ${\mathbf{u}}_{g,z}=(-b_y/f_0,b_x/f_0)$, the commutator term ${\mathbf{u}}_{g,z}\cdot {\nabla }_{H}b$ is exactly zero. For $N=N(z)$ this givesAdding the two equations proves the quasi-geostrophic potential-vorticity equationAn additive constant such as $f_0$ changes no dynamics. The horizontal streamfunction advection bracket can express this as $Q_t+{\psi }_x{Q}_y-{\psi }_y{Q}_x=0$. Although small departures from balance generated this evolution equation, the leading geostrophic flow itself advects $Q$ nonlinearly. The stretching term couples horizontal motion to vertical displacement of density surfaces.

This gives the practical advection-and-inversion description. Start with interior three-dimensional quasi-geostrophic potential vorticity, advect it with ${\mathbf{u}}_g$, and recover $\psi$ fromThe stratified quasi-geostrophic inversion is elliptic for positive $N^2$. It must be supplied with lateral and vertical boundary conditions and any circulation or mean-flow constraints; interior potential vorticity alone is not sufficient. At rigid horizontal lids, $w=0$ makes boundary buoyancy satisfy $D_{g}b=0$, and $b=f_0{\psi }_z$ supplies the vertical derivative data for inversion. Boundary buoyancy perturbations can therefore support flow even when the interior anomaly is zero. Once $\psi$ is found, its horizontal derivatives give velocity, its vertical derivative gives buoyancy, and the thermodynamic equation diagnoses $w$. This balanced evolution excludes the independent fast-wave initial data present in the full equations.

Finally, a gradient in $f$ supplies a restoring mechanism even in an otherwise uniform fluid. Linearizing about rest with constant $N$, the quasi-geostrophic potential-vorticity equation becomesSubstitution of $\psi \propto e^{i(kx+ly+mz-\omega t)}$ gives the Rossby wave dispersionFor $\beta >0$, its zonal phase speed is westward relative to a resting background. A meridionally displaced parcel retains potential vorticity, so it acquires a relative-vorticity anomaly that induces the velocity tending to return the disturbance. Background shear and stratification modify the full potential-vorticity gradient; they can alter propagation or permit instability. On a constant f-plane the planetary contribution vanishes, but gradients of relative vorticity, boundary buoyancy or topography can still support balanced wave motion. Close enough to the equator that $f_0$ vanishes, this mid-latitude quasi-geostrophic approximation and its scaling must be replaced.

A density deficit in a flux tube is not by itself an instability calculation. For example, a straight magnetized tube in otherwise field-free gas at the same temperature has lateral balance $p_i+B_i^2/(2{\mu }_0)=p_e$. The perfect gas law then gives ${\rho }_i<{\rho }_e$, so the tube rises unless some other force balances its weight deficit. This configuration simply lacks magnetostatic equilibrium. A magnetic buoyancy instability, in contrast, starts from a force-balanced atmosphere and asks whether a small allowed displacement grows. A balanced horizontal magnetic layer can be stable or unstable; the presence of magnetic pressure alone does not decide this.

Assume a smooth equilibrium, with downward gravitational acceleration $g>0$, and write $\mathcal{R}$ for the gas constant per unit mass. The ambient total pressure and force balance areWe use ideal magnetic flux freezing on the displacement time scale and rapid lateral pressure equilibration. A bodily displacement perpendicular to a straight horizontal tube does not change its length or bend its field. If its area is $\mathcal{A}$, conservation of magnetic flux and mass per unit length gives $B\mathcal{A}=\mathrm{constant}$ and $\rho \mathcal{A}=\mathrm{constant}$. Thus the tube preservesEfficient heat exchange fixes its temperature to the ambient temperature at its new height, not to its original temperature. This is the thermal assumption that distinguishes the requested criterion from the adiabatic one.

Take a tube initially at $z_0$ and let $f_0=B_0(z_0)/\rho (z_0)$. After a small upward displacement $\xi$, let ${\rho }_i(z)$ be the density it would have in lateral balance at $z=z_0+\xi$. Its preserved $f_0$ and imposed local temperature giveAt the initial position ${\rho }_i=\rho$. Differentiate this relation at $z_0$:For the ambient atmosphere the same differentiation must also differentiate $f(z)=B_0(z)/\rho (z)$. It givesSubtracting isolates the actual displaced density contrast:The pressure-gradient force on the tube balances the weight of the displaced ambient gas, so its leading vertical buoyancy acceleration isThe coefficient multiplying $\xi$ is positive precisely when the tube becomes lighter on upward displacement. Thus the isothermal interchange criterion for magnetic buoyancy isWhere $B_0>0$, this is simply $d(B_0/\rho )/dz<0$, or $d\log B_0/dz<d\log \rho /dz$. The squared form is independent of the field's choice of direction. Equality is neutral at this order, and the reversed inequality is restoring for these displacements. A layer with an interval satisfying the strict inequality admits local growing interchange disturbances under the assumed pressure/thermal equilibration. The calculation does not address field-bending undular modes, finite heat-transfer times or magnetic diffusion; those can change the criterion. In particular, the localized interchange criterion for a magnetized atmosphere with an adiabatic gas response is not interchangeable with this fast-thermal-exchange result.