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Keep the background magnetic field, thermal gradient and wavevector fixed as $|\Omega |\to \infty$. In particular, $k_z/K$ is nonzero and of order one. Put $H=g\alpha \beta$ and $X={\omega }^2$. Multiplying the nonzero-frequency dispersion relation by ${\omega }^2$ givesthat is,The sum and product of its two roots areFor large rotation, the sum is of order ${\Omega }^2$ while the product is independent of $\Omega$. The quadratic therefore has one large root and one small root. More explicitly,The fast frequencies are consequentlywhich are inertial waves at leading order. For nonzero $a$ away from the zero-frequency boundary, the rapid-rotation slow magneto-Coriolis branch hasIts frequency is of order $|\Omega {|}^{-1}$, much smaller than the fast inertial wave frequency. Coriolis force balances most of the magnetic and buoyant forces in this slow motion, with inertia supplying only a small correction.

For $a^2>0$, the slow wave has real nonzero frequency whenStable thermal stratification has $H<0$ and automatically satisfies this condition. With adverse stratification, $H>0$, sufficiently strong magnetic tension along the wavevector is needed. If the inequality is reversed, $X_s<0$ and the slow branch has imaginary frequencies, including an exponentially growing mode. Equality gives a stationary limit, rather than a nonzero oscillation. If $\mathbf{B}\cdot \mathbf{k}=0$, the proposed slow wave degenerates into zero-frequency disturbances: there is no magnetic restoring coupling. The remaining nonzero branch has ${\omega }^2=(4{\Omega }^2{k}_z^2-H{h}^2)/K^2$. These qualifications also explain why zero roots introduced by clearing denominators must not be interpreted as ordinary propagating waves.

Here $\alpha$ is the thermal expansion coefficient and $\rho$ is the mass density. Let $K=|\mathbf{k}|$, $h^2=K^2-k_z^2$, $q=\mathbf{B}\cdot \mathbf{k}$, and $a^2=q^2/({\mu }_0\rho )$, the squared Alfvén frequency. We initially consider a nonzero frequency. The temperature and magnetic equations for a Fourier mode giveIn particular, the minus sign in the magnetic relation follows from $-i\omega \hat{\mathbf{b}}=iq\hat{\mathbf{u}}$. Substitution into the magnetohydrodynamic momentum equation, with $P=p/\rho$, givesFor incompressible flow, $\nabla \times (\hat{\mathbf{z}}\times \mathbf{u})=-{\partial }_z\mathbf{u}$. Taking the vertical component of the curl eliminates pressure and gives, with $\hat{\zeta }=(i\mathbf{k}\times \hat{\mathbf{u}}{)}_z$,Taking another curl and its vertical component uses $\nabla \times \nabla \times \mathbf{u}=-{\nabla }^2\mathbf{u}$ and $(\nabla \times \nabla \times (\theta \hat{\mathbf{z}}){)}_z=h^2\hat{\theta }$. Thusor, after eliminating temperature,The two equations for vertical velocity and vertical vorticity have a nontrivial solution only when their determinant vanishes. This proves the stratified magneto-Coriolis wave dispersion relationNo division by $\Delta$ was necessary, so a possible $\Delta =0$ is retained. Division by $\omega$, however, was necessary. Stationary disturbances must be checked directly in the original equations: for example, with $q\ne 0$, $\mathbf{u}=0$ and an appropriate solenoidal magnetic perturbation can balance temperature-induced buoyancy and pressure. Such static balances are separate from the nonzero-frequency wave branches.

Linearization removes the Poisson bracket from ${\nabla }_{\parallel }$. For Fourier perturbations proportional to $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$, the two reduced equations becomeEliminating either amplitude givesSince $k=(k_{\perp }^2+k_{\parallel }^2{)}^{1/2}=k_{\perp }[1+O({ϵ}^2)]$, this is the specified dispersion relation to the accuracy of the anisotropic reduction. The reduced equations cannot retain the subleading correction replacing $k_{\perp }$ by the full $k$.

The full electron magnetohydrodynamics equation checks that correction directly. Its linearization is ${\partial }_t\delta \mathbf{B}=-\alpha B_0{\partial }_z\nabla \times \delta \mathbf{B}$. HenceApplying the cross-product operator twice and using $\mathbf{k}\cdot \delta \mathbf{B}=0$ givesThus the full model gives the stated relation exactly and the reduced model gives its leading anisotropic limit, with all signs and normalization factors consistent.

For $\omega >N$, the dispersion relation gives $k^2=-m^2{\omega }^2/({\omega }^2-N^2)$. Put $\kappa =m\omega /\sqrt{{\omega }^2-N^2}>0$. Boundedness as $x\to +\infty$ selects $k=i\kappa$, rather than the exponentially growing root. The internal-wave polarization relations now yieldAgain take real parts and set $v=0$. This is an evanescent wave, decaying over distance ${\kappa }^{-1}$; there is no propagating far-field internal gravity wave at this frequency. The pressure amplitude $P=-i\omega u/\kappa$ is in quadrature with the horizontal velocity, so the time-averaged horizontal pressure-work flux vanishes. The equality $\omega =N$ is a singular limiting case and is outside the two frequency ranges considered here.

For $0<\omega <N$, put $\alpha =\omega /\sqrt{N^2-{\omega }^2}$. The dispersion relation permits $k=\pm \alpha m$. On the positive-frequency branch, for $k>0$,The radiation condition therefore selects $k=\alpha m$, since energy must travel into the fluid. Substitution into the internal-wave polarization relations gives the complex fieldsThe physical fields are their real parts; $v=0$. The vertical component of group velocity is negative, so this positive-$m$ plane internal gravity wave carries energy down and right, although its phase propagates in a different direction.

In the Boussinesq approximation, density departures from a constant reference density ${\rho }_0$ are small enough to neglect in inertia and mass continuity, but their gravitational effect is retained. Write the total buoyancy acceleration as $b=-g(\rho -{\rho }_0)/{\rho }_0$, choose a hydrostatic background $\bar{b}(z)$, and set $\sigma =b-\bar{b}$. Thus $\sigma$ is the buoyancy perturbation, positive for a parcel lighter than the background at its present height. The squared buoyancy frequency isStable stratification has $N^2>0$. A small vertical displacement $\xi$ of a parcel conserving its buoyancy gives $\sigma =-N^2\xi$, explaining the restoring acceleration and the name of $N$.

Let $P=p^{\prime }/{\rho }_0$ be pressure with the background hydrostatic pressure removed, $\mathbf{u}=(u,v,w)$, and $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ the material derivative. The ideal, nonrotating Boussinesq equations areThe last equation is equivalent to $Db/Dt=0$. There is neither viscosity nor buoyancy diffusion. The reference density occurs in the pressure normalization, while the small density anomaly occurs only through buoyancy.

For the two-dimensional disturbance, linearization about rest givesA plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$ therefore satisfiesThese internal-wave polarization relations and the dispersion relation determine the individual responses in parts i and ii. A radiating response has no incoming internal gravity wave from $x=+\infty$; an evanescent response is bounded there. The harmonic solutions describe the response after excluding arbitrary initial transients.

For the localized oscillatory boundary, define $\alpha =\omega /\sqrt{N^2-{\omega }^2}$ and $I(z)=ϵ{\int }_0^{\infty }{\hat{u}}_0(m)e^{imz}dm=f(z)+ig(z)$. Both signs of vertical wavenumber must radiate into $x>0$. Their horizontal wavenumber is $k=\alpha |m|$, so the positive-$m$ contribution involves $Z_1=z+\alpha x$ and the negative-$m$ contribution involves $Z_2=z-\alpha x$. Consequently the complex horizontal velocity isTaking its real part provesThe positive-$m$ internal-wave polarization has $w=-\alpha u$ and the negative-$m$ one has $w=+\alpha u$. HenceIn particular, $2f(z)=ϵu_0(z)$, so the boundary condition is recovered exactly. This is two-beam radiation from a localized oscillating boundary.

Take a smooth localized profile, for example $u_0(z)=e^{-z^2/H^2}$. Its effective support is a neighbourhood of width a few $H$; increasing that neighbourhood makes the tails arbitrarily small. The boundary displacement is $ϵu_0(z)\sin \omega t/\omega$ to first order, so at $t=0$ it is flat and moving at maximum speed. At that instant,There are two narrow oblique beams centered on $z=-\alpha x$ and $z=+\alpha x$. Away from their overlap, the velocity is parallel to the relevant beam, in the directions $(1,-\alpha )$ and $(1,+\alpha )$. The figure uses $\alpha =3/4$ and measures velocities in units of $ϵ$.

At other phases the $g$ terms contribute. For the positive-frequency convention above, $g$ is the Hilbert transform of $f$; it generally has longer tails than $f$. Thus even when the boundary is nearly stationary, the interior need not be stationary or confined to the bright cores shown here. Half a period later all velocities reverse. The beam directions remain the same, but the instantaneous velocity profile and its quadrature component change.

A short pulse has a broad frequency spectrum. Frequencies below $N$ produce outgoing internal gravity wave packets with frequency-dependent ray slopes $\pm \omega /\sqrt{N^2-{\omega }^2}$; frequencies above $N$ contribute an evanescent near-boundary response. Moreover the group velocity depends on wavelength as well as direction. The radiated packet therefore spreads into a dispersive fan or wake, with different frequencies traveling at different angles and different scales separating along those directions. After the forcing ends the propagating packet travels away, rather than maintaining the two permanently driven monochromatic beams.

To first order, $|1+u+iv{|}^2=1+2u$ and $(1-|\psi {|}^2)\psi =-2u$. Equating real and imaginary parts of the Gross–Pitaevskii equation therefore givesDifferentiate the second equation in time and use the first to eliminate $v$:For a Fourier mode with $k=|\mathbf{k}|$, the dispersion relation isThis is the normalized Bogoliubov spectrum. At small wavenumber the positive branch has sound speed ${\lim }_{k\to 0}{\omega }_{+}(k)/k=1/\sqrt{2}$; at large wavenumber it approaches the free-particle quadratic dispersion. The linear frequencies are real, confirming stability of the repulsive uniform bulk.

Take shear-horizontal waves with displacement $\mathbf{u}=(0,0,w(x,y,t))$. This is divergence-free, so the elastic equation reduces to $\rho w_{tt}=\mu (w_{xx}+w_{yy})$. Rigid boundaries require $w=0$ at $y=0,h$, giving modes $w=A\sin (n\pi y/h)\cos (kx-\omega t)$, $n\ge 1$. With shear speed $c_s=\sqrt{\mu /\rho }$, the dispersion relation isFor $k>0$, the phase velocity is $v_p=\omega /k>c_s$ and the group velocity is $v_g=d\omega /dk=c_s^{2}k/\omega <c_s$, with $v_p{v}_g=c_s^2$.

The kinetic energy density is $T=\rho w_t^2/2$. The only strains are $e_{xz}=w_x/2$ and $e_{yz}=w_y/2$, together with their symmetric counterparts; thus the isotropic linear-elastic energy density is $W=\mu (w_x^2+w_y^2)/2$. Average over one temporal period and integrate across the layer. The sine- and cosine-squared transverse integrals are both $h/2$, givingThe guided shear-horizontal mode has average equipartition after both operations; a time average at a fixed transverse point alone need not give equality.

Let the rapidly varying phase be $\Theta$, so $\mathbf{k}=\nabla \Theta$ and $\omega =-{\Theta }_t$. The dispersion law is the Hamilton-Jacobi equation ${\Theta }_t+\Omega (\nabla \Theta ,\mathbf{x},t)=0$. Differentiating it and following a path with group velocity gives the ray tracing equationsThe final relation is obtained by the chain rule along that path; its two spatial terms cancel.

The medium is stationary and independent of $y$, so $\omega ={\omega }_{\infty }$ and $k_2=K$ are conserved. For deep water far from shore, ${\kappa }_{\infty }={\omega }_{\infty }^2/g$. Choose the incident branch with $k_1<0$ and $K=({\omega }_{\infty }^2/g)\sin {\theta }_{\infty }>0$; choosing the opposite alongshore direction changes its sign. Since the dispersion relation depends on direction only through $\kappa$, the group-velocity component ratio is $k_2/k_1$. Hence the required implicit ray equation isNear shore, $\kappa h\to 0$ and $\kappa \sim {\omega }_{\infty }/\sqrt{g\alpha }{x}^{-p/2}$. Thus $y^{\prime }\sim -K\sqrt{g\alpha }{x}^{p/2}/{\omega }_{\infty }$, whose integral givesThe rays approach the shoreline normally. A crest is a level set of $\int k_{1}dx+Ky-\omega t$. Its slope is $dy/dx=-k_1/K\sim {\omega }_{\infty }{x}^{-p/2}/(K\sqrt{g\alpha })$, so crests become parallel to the shoreline. Their separation in the shore-normal direction is $2\pi /|k_1|\sim 2\pi \sqrt{g\alpha }{x}^{p/2}/{\omega }_{\infty }$ and decreases to zero. These are formal ray asymptotics; for $p<2$, the slow-variation assumption eventually fails arbitrarily close to the shoreline. The sketch shows a representative shallow-water case $p=1$; it illustrates these limiting directions rather than assuming that value in the derivation.

Substitute the monochromatic plane waves into the two divergence equations of Maxwell's equations. They become $ik{E}_{0z}=0$ and $ik{B}_{0z}=0$. For nonzero wavenumber,Thus both fields are transverse to the direction of propagation. The wave equation also gives the vacuum dispersion relation ${\omega }^2=c^2{k}^2$.

Let the basic temperature be $\overline{T}=T_{\ast }-\beta z$ with $\beta >0$, the density be ${\rho }_0$, gravitational acceleration be $-g\hat{\mathbf{z}}$, and thermal expansion coefficient be $a_T$. Denote kinematic viscosity, thermal diffusivity and magnetic diffusivity by $\nu ,\kappa ,\eta$. Write the dimensional magnetic perturbation as ${\mathbf{b}}_d$, the temperature perturbation as ${\theta }_d$, and use constant permeability ${\mu }_0$. Linearizing the Boussinesq equations and the resistive induction equation about the static state givesThe Lorentz force identity is $(\nabla \times {\mathbf{b}}_d)\times B_0\mathbf{c}=B_0(\mathbf{c}\cdot \nabla ){\mathbf{b}}_d-\nabla (B_0\mathbf{c}\cdot {\mathbf{b}}_d)$. Its gradient term is absorbed into $P_d$, the perturbation magnetohydrodynamic total pressure. The temperature source follows from ${\mathbf{u}}_d\cdot \nabla \overline{T}=-\beta u_{d,z}$.

Choose any reference length $d$, since no vertical boundaries are prescribed. Scale length by $d$, time by $d^2/\kappa$, velocity by $\kappa /d$, temperature perturbation by $\beta d$, magnetic perturbation by $B_0$, and total pressure by ${\rho }_0\nu \kappa /d^2$. The dimensionless parameters areThey are the Prandtl number, magnetic-to-thermal diffusivity ratio, Rayleigh number, and Chandrasekhar number, respectively. Division by the chosen scales givesIn particular $Q\zeta =B_0^2{d}^2/({\mu }_0{\rho }_0\nu \kappa )$, so the diffusivity convention agrees with the momentum coefficient.

For a plane wave put $\mathbf{l}=(k,0,m)$, $q=|\mathbf{l}{|}^2=k^2+m^2$, and $d_{\parallel }=\mathbf{c}\cdot \mathbf{l}=k\sin \alpha +m\cos \alpha$. The induction and temperature equations imply $\hat{\mathbf{b}}=i{d}_{\parallel }\hat{\mathbf{u}}/(s+\zeta q)$ and $\hat{\theta }=\hat{w}/(s+q)$. The projection $P=I-\mathbf{l}{\mathbf{l}}^{\mathsf{T}}/q$ removes pressure. Because $(P\hat{\mathbf{z}}{)}_z=k^2/q$, its vertical momentum equation isThus the oblique-field plane-wave magnetoconvection dispersion relation isIf out-of-plane velocity is also allowed, its decoupled magnetic branch satisfies $(s/\sigma +q)(s+\zeta q)+Q\zeta d_{\parallel }^2=0$; it has no buoyant driving. At stationary marginality, for $k\ne 0$,These are stationary thresholds; an oscillatory marginal mode, if present for particular diffusivity ratios, is a different onset calculation.

For $m=1$, write $R_{mathrmstat}=F(k)+QG(k)$, where $F=(1+k^2{)}^3/k^2$ and $G=(1+k^2)(\cos \alpha +k\sin \alpha {)}^2/k^2$. Direct differentiation gives $F^{\prime }(k)=-2k^{-3}+6k+4k^3$, so its two minima are $\pm 1/\sqrt{2}$, with $F=27/4$ and $F^{\prime \prime }=36$. For $Q>0$ the negative sign has the smaller magnetic penalty: at every positive $k$, $G(-k)<G(k)$ because both sine and cosine of the inclination are positive. The wavenumber selection in oblique-field magnetoconvection therefore starts from $a=-1/\sqrt{2}$. Expanding $F^{\prime }(k_c)+Q{G}^{\prime }(k_c)=0$ gives $k_c=a+C_{1}Q+O(Q^2)$, where $C_1=-G^{\prime }(a)/F^{\prime \prime }(a)$. Taylor expansion of the value then givesThis proves the small-field forms, including the order of the correction.

For large $Q$, $G$ is nonnegative and vanishes only at $b=-\cot \alpha$. The candidate $k=b$ has finite threshold $F(b)$, so a global minimizer must satisfy $QG(k_c)\le F(b)$ and hence tend to $b$. Here $G^{\prime \prime }(b)=2(1+b^2){\sin }^2\alpha /b^2>0$. Expanding the stationarity condition at $b$ yieldsThe value expansion isThe leading coefficient follows from $1+{\cot }^2\alpha ={\csc }^2\alpha$. The expansion holds at fixed $0<\alpha <\pi /2$.

Finally choose $\cot \alpha =1/\sqrt{2}$. Then the nonmagnetic minimizing wavevector also has $d_{\parallel }=0$, and its magnetic penalty vanishes. Since $F(k)\ge 27/4$ and $QG(k)\ge 0$, this is the global minimum for every nonnegative field strength:This field-independent stationary threshold in oblique magnetoconvection has a simple physical explanation: the disturbance is constant along the imposed field, so it does not bend field lines.

In the Boussinesq approximation, replace density by a constant reference density ${\rho }_0$ in inertia and incompressible flow, retaining the small density departure only in the gravitational force. Write $\rho =\bar{\rho }(z)+{\rho }^{\prime }$ and subtract the resting hydrostatic pressure. The upward buoyancy perturbation acceleration and squared buoyancy frequency areThus stable density stratification has $N^2>0$. Let $p$ henceforth denote the pressure anomaly divided by ${\rho }_0$. The ideal, nonrotating Boussinesq equations areThe buoyancy equation follows by materially conserving the total buoyancy $b_0(z)+\sigma$, where $b_0^{\prime }=N^2$; it does not assume constant $N$.

For an $x$-periodic flow of period $P$, define $\bar{a}=P^{-1}{\int }_0^{P}adx$. An infinite-domain average with vanishing end contributions works as well. The necessary pressure condition is no mean zonal pressure gradient, $\overline{p_x}=0$: periodic pressure satisfies it, whereas a pressure containing a term proportional to $x$ does not. Use incompressible flow to write the zonal momentum equation in conservative form,Averaging its zonal derivative terms yields the exact Reynolds stress relation

For two-dimensional small disturbances about rest, the Linearized Boussinesq equations are $u_t=-p_x$, $w_t=-p_z+\sigma$, ${\sigma }_t=-N^{2}w$, and $u_x+w_z=0$. Differentiate the two momentum equations to eliminate pressure: $(u_z-w_x{)}_t=-{\sigma }_x$. Differentiating in $x$ and using $u_{xz}=-w_{zz}$ gives ${\nabla }^2{w}_t={\sigma }_{xx}$. One further time derivative givesThe equation remains valid for $N=N(z)$. No vertical derivative of $N$ was taken: ${\partial }_x^2(N^{2}w)=N^2{w}_{xx}$.

For constant $N$ and $k\ne 0$, a plane internal gravity wave hasBoth signs give the prescribed lower-boundary velocity. The radiation condition excludes energy arriving from infinity. At positive frequency, the vertical group velocity isFor a source below the fluid, energy must propagate upward, so the outgoing solution has $m<0$, even though its vertical phase velocity $\omega /m$ is downward.

To obtain this selection causally, let $T=\mu t$ and $Z=\mu z$, and replace the constant amplitude by $a(Z,T)$. The method of multiple scales treats $(z,t,Z,T)$ as independent during differentiation: the carrier derivatives hold $Z,T$ fixed, while $a_T$ holds $Z$ fixed and $a_Z$ holds $T$ fixed. Only after differentiating do we restrict to $T=\mu t$, $Z=\mu z$. Acting on the complex wave, the operators becomeWith $K^2=k^2+m^2$, the wave equation has residualThe zeroth-order term is the dispersion relation; the first-order term vanishes precisely whenIf $a_0(T)$ is the switched-on boundary amplitude, with $a_0(T)=0$ for $T<0$, the outgoing envelope is $a(Z,T)=a_0(T-Z/b)$ for $b>0$. Its characteristics carry the boundary signal into initially undisturbed fluid. If $b<0$, characteristics instead arrive from the upper half-space; zero initial disturbances and no incoming signal cannot establish the proposed boundary-forced wavetrain. This is the internal-wave envelope radiation condition.

The leading incompressible flow polarization gives $u^{\prime }=-(m/k)w^{\prime }$. Hence, to second order in amplitude and leading order in the slow modulation,There is no meridional transport here. Envelope transport gives ${\partial }_t|a{|}^2=-b{\partial }_z|a{|}^2$, so an initially vanishing mean develops asThese are the leading expressions for internal-wave momentum deposition; higher slow-modulation corrections to the stress are beyond the displayed approximation. In particular, a spatially uniform fully established wavetrain has zero stress divergence, but the front has already produced its mean flow.

The mean wave energy density per unit mass follows from ${\sigma }^{\prime }=-i{N}^2{w}^{\prime }/\omega$:Thus $\bar{u}=(k/\omega )\mathcal{E}$. In the original frame the mean kinetic energy is ${\bar{u}}^2/2=O(a^4)$. In a frame translating at $c=\omega /k$, the undisturbed velocity is $-c$, and the change in mean kinetic energy is $-c\bar{u}+O(a^4)=O(a^2)$. Indeed $\mathcal{E}-c\bar{u}=0$ at this order behind the front. Mean-flow energy is therefore essential at the same order as wave energy in the translating frame. This follows also from the frame transformation of energy, $E\mapsto E-cP$ up to the fixed background constant, and is why conservation of momentum cannot be omitted from the moving-frame conservation of energy budget.

A precise stratification–rotation analogy uses constant $N$, ideal fluid, and independence of one horizontal coordinate. For the stratified system take ${\partial }_y=0$, $v=0$, and allow prescribed forces $(F_x,0,F_z)$. The possibly nonlinear Boussinesq equations areFor the unstratified system choose right-handed coordinates $(X,Y,Z)=(z,y,-x)$, rotation $2\mathbf{\Omega }=N{\mathbf{e}}_Z$, and ${\partial }_Y=0$. Its rotating Euler equations, with the static gravitational and centrifugal contributions absorbed into the background pressure, arewhere $D_R={\partial }_t+U{\partial }_X+W{\partial }_Z$. The identification isThen $D_R=D$, $P_X=p_z$, and $P_Z=-p_x$, proving the component-by-component equivalence. In particular, buoyancy perturbation maps to transverse rotating velocity, and stratified horizontal velocity maps to rotation-axis velocity. The mapping permits arbitrary prescribed forces in the two-dimensional plane, but no transverse force: a transverse force in the rotating problem would create a source in the stratified buoyancy equation. A transverse stratified force would also invalidate $v=0$. Constant $N$ is essential to this constant-rotation equivalence.

A physical free surface introduces extra structure beyond the interior equations. After subtracting hydrostatic pressure, its zero total pressure condition gives $p=g\eta$ to first order, together with the kinematic condition ${\eta }_t=w$ for the stratified vertical displacement. The mapping takes this normal velocity into $U$, whereas an actual rotating free surface normal to its physical vertical $Z$ has normal velocity $W$ and a displacement along $Z$. The image of the stratified horizontal surface is a vertical surface in the rotating coordinates. Consequently the physical free-surface problems are not analogous, although deliberately mapped rigid boundaries or mathematical boundary data can be.

For the first prescribed force, the steady linear stratified equations are satisfied byReal parts are understood. The horizontal pressure gradient balances the force; the vertical pressure gradient is balanced by buoyancy; and $w=0$ satisfies buoyancy conservation and incompressible flow. Thus neither $u(z)$ nor $b(z)$ is fixed by the steady equations alone.