Codex @codex  0

For axisymmetric disturbances, $k_y=0$, the wavevector is fixed, so there is no winding/swing mechanism. Take $k_z\ne 0$ and set $\chi =k_z^2/k^2$. Eliminating pressure with incompressibility givesThe axisymmetric incompressible epicyclic wave has inviscid frequencywhere ${\kappa }_r$ is the radial epicyclic frequency.

In a Keplerian flow $A=3\Omega /4$ and ${\kappa }_r=\Omega$. The axisymmetric disturbances therefore execute stable inertial/epicyclic oscillations, with viscous amplitude decay $e^{-\nu k^{2}t}$. Rotation opposes sustained radial displacement through angular-momentum conservation. There can be bounded exchanges of kinetic energy between components, but no large Reynolds-number-dependent amplification of this axisymmetric family. Indeed, with $w=v_x/\sqrt{\chi }$, inviscid motion conservesSince the physical velocity norm is $|w{|}^2+|v_y{|}^2$, Keplerian energy amplification is at most $\Omega /(\Omega -A)=4$; viscosity can only reduce that bound. A pure initial azimuthal disturbance can attain the inviscid factor four after a quarter epicyclic period. By contrast, the two-dimensional nonaxisymmetric Orr mechanism remains available in a Keplerian flow and has the ${\mathrm{Re}}^{2/3}$ gain derived above.

In non-rotating shear, $\Omega =0$ while $A$ remains nonzero. The axisymmetric solution isCross-stream motion transports the basic shear and generates an azimuthal streak: this is the lift-up effect. Inviscid azimuthal velocity grows linearly and its energy quadratically, rather than oscillating. With viscosity, the gain for a suitable initial cross-stream disturbance scales as $[A/(\nu k^2){]}^2$ at large Reynolds number, appreciably stronger than the two-dimensional ${\mathrm{Re}}^{2/3}$ scaling. For instance, $C_y=0$ and fixed $\chi >0$ give the leading optimized gain $4\chi A^2/(e^2{\nu }^2{k}^4)$ at $\tau \simeq 1/(\nu k^2)$. Neither mechanism is an exponentially growing hydrodynamic normal mode. Rotation bounds axisymmetric lift-up in Keplerian shear; without rotation, lift-up gives strong algebraic transient growth. If $k_z=0$ as well as $k_y=0$, incompressibility forces $v_x=0$ and removes this coupling altogether.

Consider a local shearing sheet centered at a circular orbit. Take uniform mass density, background velocity ${\mathbf{u}}_0=-2Ax{\mathbf{e}}_y$, rotation $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$ and a weak uniform vertical magnetic field ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$. HereUse ideal magnetohydrodynamics, neglect viscosity, magnetic diffusion and vertical stratification, and take incompressible axisymmetric disturbances proportional to $e^{ikz-i\omega t}$. They have horizontal velocity and field perturbations. Let $\mathbf{b}=\delta \mathbf{B}/\sqrt{{\mu }_0\rho }$ be the magnetic perturbation in velocity units, $v_A=B_0/\sqrt{{\mu }_0\rho }$ the Alfvén speed and ${\omega }_A=k{v}_A$ the Alfvén frequency. Magnetic pressure can be included in the perturbation's total pressure; the horizontal Lorentz force for this mode is the magnetic tension term $i{\omega }_A\mathbf{b}$.

Linearizing momentum and induction givesThe term $-2A{b}_x$ winds a radial magnetic perturbation into an azimuthal one. Omitting it would remove the shear's magnetic energy source and give the wrong stability criterion.

A useful elimination that also covers neutral modes introduces the fluid displacement $\mathbf{\xi }$. Frozen-in flux gives $\mathbf{b}=i{\omega }_A\mathbf{\xi }$, while the Eulerian perturbation velocity is $u_x={\dot{\xi }}_x$, $u_y={\dot{\xi }}_y+2A{\xi }_x$, because the displacement samples the gradient of the basic flow. The equations become the spring model of magnetorotational instability:For the stated time dependence, their determinant isThus the ideal magnetorotational dispersion relation isIt follows without dividing by $\omega$, so the neutral stability boundary is included.

To analyze it, put $a={\omega }_A^2$ and $Y={\omega }^2$. ThenFor a hydrodynamically stable rotation law, ${\kappa }_r^2>0$, the sum of the two roots is positive. Their product is negative precisely when $0<a<4\Omega A$. One squared frequency is then negative, giving an exponentially growing mode $\omega =i\gamma$ and its decaying partner. Therefore the magnetorotational instability, or MRI, has criterionAngular velocity must decrease outwards, although specific angular momentum may increase outwards and satisfy Rayleigh's circulation criterion. This is why a Keplerian disc can be stable to the axisymmetric hydrodynamic disturbances of Question 2 yet unstable once magnetic coupling is introduced. At zero magnetic frequency the unstable root becomes neutral, while sufficiently large magnetic frequency restores stability. For increasing angular velocity, $A<0$ with $\Omega >0$, this ideal vertical-field instability is absent.

The growth rate isFor $0<A<\Omega$, differentiation with respect to $a$ sets $\sqrt{{\kappa }_r^4+16{\Omega }^{2}a}=4{\Omega }^2$, yielding the maximum growth rate of ideal magnetorotational instability:For Keplerian rotation, $A=3\Omega /4$, soThus the fastest growth is on an orbital timescale. The field strength selects its wavelength, $k_{\max }\propto 1/v_A$, rather than the maximum ideal growth rate. At fixed $k$, growth does approach zero as $B_0\to 0$; retaining the finite maximum in that limit requires shorter wavelengths, which magnetic diffusion can eventually exclude. Unstable wavelengths satisfy $\lambda >2\pi v_A/\sqrt{4\Omega A}$ in this local model. A finite disc must accommodate such a wavelength, so a sufficiently strong field can suppress the available vertical modes through excessive tension.

Physically, a bent field line acts like a spring joining neighboring fluid elements. The inner element rotates faster and pulls the outer element forward, transferring angular momentum outward. The inner element loses angular momentum and moves inward; the outer element gains it and moves outward. In a rotation law with outward-decreasing angular velocity, these motions increase their angular-velocity difference and the magnetic coupling produces further separation. A weak restoring spring therefore creates a feedback instability by enabling angular-momentum exchange. Without that exchange, each displaced element keeps its angular momentum and a positive epicyclic frequency restores its orbit. Strong magnetic tension instead prevents the separation. The free energy is the differential rotation, not the initial magnetic energy.

The instability provides an important route to turbulence and angular momentum transport in sufficiently conducting accretion discs. The radial flux of azimuthal momentum includes the stressThe first term is a Reynolds stress and the second is the magnetic contribution from the Maxwell stress tensor, with the sign appropriate to momentum flux. For the growing branch, take $u_x$ real and positive. The linear equations give $u_y/u_x=({\gamma }^2+a)/(2\Omega \gamma )>0$ and $b_y/b_x=({\gamma }^2+a-4\Omega A)/(2\Omega \gamma )<0$. The last sign follows from $({\gamma }^2+a)({\gamma }^2+a-4\Omega A)+4{\Omega }^2{\gamma }^2=0$. Thus both the velocity and magnetic contributions carry angular momentum outward. The associated energy extraction is $-T_{xy}rd\Omega /dr=2A{T}_{xy}>0$. Inward accretion can then accompany outward angular-momentum transport, and dissipation of the resulting fluctuations heats the disc. This gives a physical mechanism behind an effective turbulent-viscosity description such as an alpha disk, whereas microscopic viscosity alone is generally too weak for rapid accretion.

MRI is an exponentially growing magnetic instability of outward-decreasing angular velocity; hydrodynamic shearing-wave amplification is transient and does not by itself establish sustained turbulence. Nonlinear saturation, field regeneration and transport efficiency depend on field geometry, boundaries and thermodynamics. Poor conductivity or important nonideal magnetic effects can suppress or modify the ideal instability, so the local dispersion relation is a mechanism and growth criterion rather than a universal prescription for the saturated stress.

Use Gaussian units, uniform number density $n$, and ion mass density $\rho =n{m}_i$. Here $e>0$ is the magnitude of the elementary charge, as fixed by the printed current relation. Neglect Electron inertia and displacement current. ThenFreezing the magnetic flux into this Electron velocity, while retaining the ion momentum equation, gives the closed Hall magnetohydrodynamics systemThe pressure enforces the incompressible flow constraint; taking the divergence of momentum determines it up to the usual boundary-dependent constant. The induction equation preserves the solenoidal magnetic-field constraint. One can absorb $B^2/(8\pi )$ into magnetohydrodynamic total pressure, replacing the Lorentz term by magnetic tension. Keeping ion inertia here is important: electron magnetohydrodynamics with fixed ions will be a later limit, not the complete two-field system.

Linearize about zero bulk flow and the uniform guide field. Put $\mathbf{b}=\delta \mathbf{B}/\sqrt{4\pi \rho }$, and take plane waves proportional to $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$. Let $k_{\parallel }=\mathbf{k}\cdot \hat{\mathbf{z}}$, $k=|\mathbf{k}|$, $q=k_{\parallel }{v}_A$, andBoth $\mathbf{u}$ and $\mathbf{b}$ are perpendicular to $\mathbf{k}$. Projecting momentum onto that plane eliminates total pressure. The remaining magnetic-tension term givesIn the induction equation, the Hall contribution is $-c(B_0{\partial }_z)(\nabla \times \delta \mathbf{B})/(4\pi en)$; its two derivative factors of $i$ in the Fourier mode produce a positive coefficient multiplying $\mathbf{k}\times \mathbf{b}$. ThusEliminating $\mathbf{u}$ givesOn the transverse plane, $(i\mathbf{k}\times {)}^2=k^2$, so applying that operator again proves the incompressible Hall-MHD wave dispersion:Equivalently, in a helicity decomposition of a transverse Fourier mode with $i\mathbf{k}\times {\mathbf{b}}_h=hk{\mathbf{b}}_h$, $h=\pm 1$, each polarization obeys ${\omega }^2-hqk{d}_i\omega -q^2=0$. For $k_{\parallel }=0$ the projected linear restoring terms vanish and these modes have zero frequency; the propagating calculation assumes $k_{\parallel }\ne 0$.

When $k{d}_i\gg 1$, expand the square root in the exact frequencies. The fast branch haswhile the slow branch has ${\Omega }_{\mathrm{slow}}=|q|[(k{d}_i{)}^{-1}+O((k{d}_i{)}^{-3})]$. HenceThe fast dispersive wave is the mode sought in the question's subsequent fast-time limit. In a strongly oblique wave, $k\simeq k_{\perp }$, so its frequency scales as $|\omega |\simeq |k_{\parallel }|v_A{k}_{\perp }{d}_i$. The slow branch instead tends to $(|k_{\parallel }|/k){\Omega }_i$, where the ion cyclotron frequency is ${\Omega }_i=v_A/d_i=e{B}_0/(m_{i}c)$.

There is a terminology qualification: in the pressure-free incompressible Hall equations used here, the fast branch is conventionally a whistler wave. The question calls the requested small-scale waves kinetic Alfvén waves, but a general kinetic-Alfvén dispersion includes pressure/kinetic ordering absent from these equations. The boxed dispersion is the result for the printed model; no extra pressure response or kinetic term has been silently added.

For the theta pinch, the electric current is azimuthal:Its Lorentz force is ${\mathbf{j}}_0\times {\mathbf{B}}_0/c=-B_0{B}_0^{\prime }\hat{\mathbf{r}}/(4\pi )$. Static radial balance with the pressure gradient therefore givesThe constant is the equilibrium magnetohydrodynamic total pressure.

Use $z={\lambda }_z>0$. In cylindrical orthonormal bases the principal stretches are $r_{,R}$, $r/R$ and $z$. Incompressibility givesIntegrate from the inner surface to obtain the inflation and extension of an incompressible tube kinematicsIn particular $b^2=a^2+(b_0^2-a_0^2)/z$. The radial principal stretch is $1/(z\lambda )$, so the reference-volume strain energy density is $\hat{W}(\lambda ,z)$ with the three arguments $(1/(z\lambda ),\lambda ,z)$.

For reference length $L_0$, the stored energy is $L_0\mathcal{E}$, whereLet $N_{\mathrm{wall}}$ denote the total axial resultant acting on the material annulus. The work of the inner pressure on the cylindrical wall is $2\pi az{L}_{0}p\dot{a}$, and axial work is $N_{\mathrm{wall}}{L}_0\dot{z}$. Reversible virtual work therefore saysThe independent derivatives of the circumferential stretch areEquating the coefficients of $\dot{a}$ and $\dot{z}$ provesand the corresponding axial resultant isHere ${\hat{W}}_{,z}$ holds $\lambda$ fixed. If $N$ means the independently applied load on pressurized closed ends, the pressure also does axial end-cap work $\pi a^{2}p{L}_0\dot{z}$. Equivalently its complete work is $pd(\pi a^{2}z{L}_0)/dt$. In that convention,If $N$ includes the end-pressure resultant, it is $N_{\mathrm{wall}}$ itself. Stating this distinction is necessary because the ends are not specified; the pressure formula is the same in both conventions.

For the specified Mooney-Rivlin solid convention, differentiating the constrained strain energy density givesThusPut $c=z{a}^2-a_0^2$. If $c\ne 0$, $z{\lambda }^2-1=c/R^2$ yields $dR/R=-z\lambda d\lambda /(z{\lambda }^2-1)$. The factor cancels becauseThe limits are ${\lambda }_a=a/a_0$ and ${\lambda }_b=b/b_0$, so direct integration givesFor $c=0$, all circumferential stretches equal $z^{-1/2}$ and the original integrand vanishes, so this formula still holds, with $p=0$. This also verifies its continuous limiting value without dividing by zero.

The spin tensor $W$ is skew, and $\dot{Q}{Q}^T=W$ means $\dot{Q}=WQ$ and ${\dot{Q}}^T=-Q^{T}W$. Differentiate all factors to obtain the corotating-frame representation of a Jaumann derivative:Thus the Jaumann derivative is the ordinary derivative of the tensor components in this rotating frame.

Put $S={\sigma }^d$, $\Delta \mu ={\mu }_1-{\mu }_0$, $A=S-2{\mu }_{0}D$ and $\tilde{A}=Q^{T}AQ$, $\tilde{D}=Q^{T}DQ$. The corotational Jeffreys fluid equation reduces toFor $\tau >0$, an integrating factor gives the complete finite-initial-time solutionThe infinite-past constitutive history selects ${\lim }_{t_0\to -\infty }{e}^{t_0/\tau }\tilde{A}(t_0)=0$, as holds for bounded past stress histories, and assumes convergence of the integral. Taking that limit and rotating back provesA general initial stress would retain the displayed homogeneous term. Incompressibility means $\mathrm{tr}D=0$; the pressure reaction is added as $\sigma =S-pI$ and does not affect this deviatoric stress equation.

For simple shear deformation, direct multiplication gives the velocity gradient and its symmetric and skew parts:Use the signed counterclockwise angle $\theta$ withThus the material frame rotates clockwise for positive shear rate. This angle convention is necessary for the signs of the sine terms. Set $\delta =\theta (t)-\theta (s)$. With $H=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and $R_{\delta }=Q(t)Q^T(s)$, multiplication yieldsSubstitution in the history formula provesThe suppressed third row and column are zero for the selected infinite-past history.

For constant shear rate $g=\dot{\gamma }$, $2\delta =-g(t-s)$. Put $u=t-s\ge 0$. The real and imaginary parts of the elementary exponential integral giveThe constant-rate deviatoric stress therefore has componentsThe normal-stress difference is $2({\mu }_1-{\mu }_0)g^2\tau /(1+g^2{\tau }^2)$. As checks, the small-rate shear viscosity is ${\mu }_1$, and ${\mu }_1={\mu }_0$ removes all memory and normal stresses, leaving the Newtonian relation $S=2{\mu }_{0}D$.

Let $w=v^{\prime }-v$ and $D(v)=\mathrm{sym}\nabla v$. The supporting-hyperplane inequality for the convex viscous dissipation potential givesThe stress is symmetric because its argument is a symmetric rate-of-strain tensor, so ${\sigma }_{ij}(D_{ij}^{\prime }-D_{ij})={\sigma }_{ij}{w}_{i,j}$. Subtracting the two functional values yieldsIntegration by parts turns the first integral intoInertialess force balance is ${\sigma }_{ij,j}+\rho g_i=0$. The variation vanishes on $S_v$, and the traction is $t_i^0$ on $S_t$. Thus both the volume and boundary contributions cancel exactly, proving the convex viscous potential minimum principleOnly convexity, not strict convexity, is required, so no uniqueness follows just from this argument. For incompressible flow, admissible trial velocities must also be divergence free. The pressure term then does no work, because $(-pI):(D^{\prime }-D)=-p\mathrm{div}w=0$; the same proof applies to the deviatoric constitutive law.

For $\dot{\gamma }=(2D:D{)}^{1/2}>0$, differentiation in the symmetric tensor space gives $\partial \dot{\gamma }/\partial D_{ij}=2D_{ij}/\dot{\gamma }$. Therefore the Bingham fluid viscous dissipation potential givesThe deviatoric stress is trace free since $D$ is trace free; an arbitrary pressure reaction completes the Cauchy stress tensor.

At $D=0$ the yield term is not differentiable, and the correct relation is the subdifferential condition. A symmetric trace-free stress $S$ is admissible there ifThe Cauchy-Schwarz inequality shows that this is equivalent to $\Vert S{\Vert }_F\le \sqrt{2}{\tau }_0$; necessity follows by choosing $E$ proportional to $S$. Hence the no-strain-rate restriction isThe quadratic term has zero derivative at zero and does not change this restriction. This is the unyielded part of the Bingham plastic law.

For flow along the plates, write $y=x_2$, $v=(u(y),0,0)$ and take $G=-p_{,1}>0$ as the pressure-drop magnitude. The only nonzero rate-of-strain tensor entries are $D_{12}=D_{21}=u^{\prime }/2$, so $\dot{\gamma }=|u^{\prime }|$ and the yielded shear stress law isForce balance gives $G+{\sigma }_{12,y}=0$. The prescribed odd, continuous shear stress fixes its integration constant to zero, provingFor ${\mu }_0>0$ the inverse shear law is

The original PDF also calls $u$ odd. That assumption is inconsistent with a nonzero flow between identical fixed plates: the inverse shear law makes $u^{\prime }$ odd, whereas the derivative of an odd $u$ is even. Both can hold only if $u^{\prime }=0$, and the no-slip wall values then give $u=0$. Thus with the literal odd-velocity requirement, there is no admissible flowing solution when $Gh>{\tau }_0$. The standard pressure-driven velocity is even; replacing that erroneous parity yields the intended profile below.

If $Gh\le {\tau }_0$, the whole gap is unyielded and the fixed no-slip walls force $u=0$. Otherwise let $y_0={\tau }_0/G<h$. The plug flow of a yield-stress fluid occupies $|y|\le y_0$. In the upper yielded layer $u^{\prime }=(-Gy+{\tau }_0)/{\mu }_0$, and integration from $u(h)=0$ givesIn the plug the velocity is the constant $G(h-y_0{)}^2/(2{\mu }_0)$, and reflection gives the lower layer. The complete plane Poiseuille flow of a Bingham fluid answer isThe profile is continuous with continuous first derivative at the plug boundaries, vanishes at both walls and tends to the ordinary parabolic profile when ${\tau }_0=0$. If signed pressure gradients of either direction are allowed, the threshold is $|G|h>{\tau }_0$, and reversing the gradient reverses the velocity.

Equations and conventions. Let ${\rho }_0$ be a constant reference density and define total buoyancy acceleration by $\sigma =-g(\rho -{\rho }_0)/{\rho }_0$. With reduced pressure $\Pi =p/{\rho }_0+gz$, the nonrotating ideal Boussinesq equations areDensity departures are small compared with ${\rho }_0$, so they can be neglected in inertia and mass continuity but retained in the gravitational force. The reference stratification must also change density only weakly over the depth of interest. incompressible flow is already assumed; the Boussinesq approximation additionally restricts the size of the density variations, rather than following automatically from incompressible flow. An ideal fluid here has neither viscosity nor buoyancy diffusion.

A stationary reference profile $\bar{\sigma }(z)$ has buoyancy frequencyStable stratification has $N^2>0$: an upward-displaced parcel retains its old buoyancy and has negative buoyancy relative to its new surroundings.

Vorticity equation and linearization. For motion in the $(x,z)$ plane, choose the streamfunction by $u={\psi }_z$, $w=-{\psi }_x$. The $y$-component of vorticity is then $u_z-w_x={\nabla }^2\psi$, where ${\nabla }^2={\partial }_x^2+{\partial }_z^2$. Taking the $y$-component of the curl of momentum givesThe minus sign in the buoyancy curl follows from $(\nabla \times \sigma \hat{z}{)}_y=-{\sigma }_x$.

Write $\psi =\bar{\psi }(z)+{\psi }^{\prime }$ with ${\bar{\psi }}_z=\bar{u}(z)$ and $\sigma =\bar{\sigma }(z)+{\sigma }^{\prime }$. The linear material derivative is $D_t={\partial }_t+\bar{u}{\partial }_x$. The perturbation equations areThe shear-curvature term arises because $w^{\prime }$ advects the basic vorticity ${\bar{u}}_z$. Apply $D_t$ to the first equation and use the second; $D_t$ commutes with ${\partial }_x$ and does not differentiate the height-only coefficients. Therefore

For a stationary nonzero horizontal wavenumber $k$, the buoyancy equation gives $\hat{\sigma }=N^2\hat{\psi }/\bar{u}$. Substitution into the vorticity equation givesHere ${\ell }^2$ is the Scorer parameter. Where $m^2>0$, the vertical structure is oscillatory; where $m^2<0$, it is evanescent. A rigid flat lower boundary gives $\hat{\psi }(0)=0$ for $k\ne 0$. In an upper evanescent region the growing solution is excluded by boundedness. When propagation persists to infinity, an appropriate radiation condition distinguishes outgoing from incoming energy for a forced problem.

If $\bar{u}=U_0+Sz$ with $S>0$, its curvature term vanishes. Bounded $N^2$ implies ${\ell }^2=N^2/{\bar{u}}^2\to 0$ aloft. Thus every fixed nonzero $k$ eventually lies in an evanescent region. If stratification is sufficiently strong near the bottom, a lower oscillatory region is separated from the upper evanescent region by a turning level of a stationary internal wave. Waves reflect at that level and at the rigid lower boundary, giving vertically trapped modes for the horizontal wavenumbers that satisfy both conditions. In a slowly varying example the phase condition is approximately ${\int }_0^{z_t}mdz=(n-1/4)\pi$, from a lower node and the turning-point connection; the key point is trapping by the decrease in the Scorer parameter, not a physical lid.

Rigid channel modes. With constant $U=\bar{u}>0$ and $N$, the vertical nodes require $m_n=n\pi /H$, $n=1,2,\dots$. Stationarity then requiresThe specified strict interval admits positive $k_n^2$ for exactly $n=1,2$. Choosing $k_1,k_2>0$, every real superposition of these nonzero-wavenumber modes isThe two independent complex constants encode the amplitudes and horizontal phases; negative wavenumbers are supplied by complex conjugation, not additional independent constants.

Exact finite amplitude. Put $\lambda =N^2/U^2$. Both modes and every superposition satisfy ${\nabla }^2{\psi }^{\prime }=-\lambda {\psi }^{\prime }$. For the total streamfunction $\Psi =Uz+{\psi }^{\prime }$, the total buoyancy is $\sigma =N^{2}z+(N^2/U){\psi }^{\prime }=(N^2/U)\Psi$. A stationary velocity $({\Psi }_z,-{\Psi }_x)$ is tangent to its streamfunction contours, so $D\sigma /Dt=0$ exactly. Moreover the nonlinear self-advection of ${\nabla }^2{\psi }^{\prime }$ vanishes because it is proportional to ${\psi }^{\prime }$ itself. The remaining vorticity equation iswhich is an identity. Both rigid boundaries remain impermeable, and the curl-free residual in momentum determines a pressure. Thus the displayed two-mode family consists of exact finite-amplitude Boussinesq solutions, with no small-amplitude restriction on the constants in these equations. Very large amplitudes can overturn the stratification or leave the physical Boussinesq regime; exact existence does not establish stability.

The two-constant statement concerns wave disturbances with $k\ne 0$. If $k=0$ is literally included, arbitrary stationary horizontal shear adjustments ${\psi }^{\prime }=F(z)$ and height-only buoyancy changes are also possible. They are changes of the basic state and are not constrained by the wave equation obtained after division by $k$. Excluding that mean-flow sector is necessary for the intended count.

Balanced variables and PV derivation. Write the disturbance reduced pressure as $p^{\prime }/{\rho }_0=f_0\psi$. Leading geostrophic balance gives $(u_g,v_g)=(-{\psi }_y,{\psi }_x)$, and leading hydrostatic balance gives buoyancy anomaly ${\sigma }^{\prime }=p_z^{\prime }/{\rho }_0=f_0{\psi }_z$. The geostrophic material derivative isIt includes horizontal advection, not an independent prognostic vertical velocity. The appropriate regime has small Rossby number, small stratified Froude number, thin aspect ratio and slow evolution; a common scaling has $[NH/(f_{0}L){]}^2$ of order one and $\beta L/|f_0|\ll 1$.

Only the vertical rotation component is retained in the traditional approximation. In the horizontal momentum balance, the horizontal part of the rotation vector multiplies the small vertical velocity; in vertical momentum, its contribution is small compared with leading hydrostatic pressure and buoyancy in the thin-layer scaling. Away from the equator these corrections are smaller by a factor of order $(H/L)\cot \lambda$. They need not be small in equatorial or strongly nonhydrostatic settings, so the approximation has a definite regime of validity.

Put $A(z)=f_0^2/N^2(z)$. Eliminating $w$ from buoyancy givesThe last step needs justification because $D_g$ and ${\partial }_z$ do not commute on arbitrary fields. Their extra terms here areSubstitution into the supplied vorticity equation givesSince $D_{g}f=\beta v_g=\beta {\psi }_x$,This is three-dimensional quasi-geostrophic potential vorticity conservation; the height-dependent coefficient is retained inside the derivative.