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For a two-dimensional internal gravity wave with phase $kx+\int m(z)dz-\omega t$, a constant-phase line at fixed time has slope $dz/dx=-k/m(z)$. Its normal spacing from the next crest is $2\pi /(k^2+m^2{)}^{1/2}$; the wavevector $(k,m)$ is normal to the line. The intrinsic energy ray is tangent to these phase lines, while mean-flow advection changes the laboratory ray direction.

For ideal magnetohydrodynamics at constant mass density $\rho$, a shearing sheet with velocity $-q\Omega x{\mathbf{e}}_y+{\mathbf{v}}_h(z,t)$ and magnetic field ${\mathbf{B}}_h(z,t)+B_z{\mathbf{e}}_z$ obeysHorizontal invariance and vanishing vertical velocity eliminate nonlinear advection. These equations retain Coriolis acceleration, orbital shear and magnetic tension. The vertical pressure adjusts to balance gravity and magnetic pressure.

Expanding the kinetic helicity conservation law separates advection from the other flux:Thus material conservation of kinetic helicity density holds precisely whenAn especially useful sufficient pair of conditions is incompressible flow, $\nabla \cdot \mathbf{u}=0$, and constancy of $w-u^2/2$ along integral curves of the vorticity, $\mathbf{\omega }\cdot \nabla (w-u^2/2)=0$. Together these conditions make both terms vanish without cancellation. They are not necessary individually, because the two terms in the displayed balance can cancel. An irrotational flow is a trivial conserved case with $q=0$.

For contrast, combining the balance with mass conservation givesConsequently constancy of $w-u^2/2$ along integral curves of the vorticity makes $q/\rho$ an advected scalar even in compressible flow; it does not generally make $q$ itself constant.

In an adiabatic quasi-geostrophic approximation, hydrostatic approximation and geostrophic balance give $b^{\prime }=f_0{\psi }_z$. The leading buoyancy equation is $D_g{b}^{\prime }+N^{2}w=0$; hence $w=-D_g(f_0{\psi }_z/N^2)$. At the rigid surface $z=\alpha y$, the no-normal-flow condition is $D_g(z-\alpha y)=0$, so $w=\alpha v_g=\alpha {\psi }_x$. The small boundary slope allows the disturbance condition to be evaluated at the flattened boundary $z=0$ to the retained order.

For thermal wind $U=\Lambda z$, choose the basic quasi-geostrophic streamfunction $\overline{\psi }=-\Lambda zy$. Its buoyancy anomaly is $\overline{b}=-f_0\Lambda y$, and its interior quasi-geostrophic potential vorticity is constant. Therefore the linear disturbance equations areThe term $-\Lambda {\psi }_x^{\prime }$ is the advection of the basic meridional buoyancy gradient. Since $q^{\prime }$ is simply advected at each fixed height, $q^{\prime }=0$ initially implies $q^{\prime }=0$ forever. At the lower boundary define $a=\alpha N^2/f_0$; thenThe PDF uses $\alpha$ for the lower-boundary slope throughout; the TeX's isolated $z=cy$ is a transcription error.

For nonzero zonal wavenumber let $\mu =N|k|/|f_0|>0$. The zero-interior-potential vorticity condition gives ${\hat{\psi }}_{zz}-{\mu }^2\hat{\psi }=0$. In the semi-infinite domain choose decay as $z\to \infty$. Writing $B_0(t)={\hat{\psi }}_z(0,t)$, potential-vorticity inversion then givesThis sloping-boundary Eady edge wave has all its time dependence in the boundary buoyancy equation; the interior responds instantaneously through elliptic potential-vorticity inversion. A single boundary wave oscillates without exponential growth. Vertical shear and topography contribute oppositely to its propagation. The contributions cancel when $\alpha =f_0\Lambda /N^2$, the slope of a background isopycnal. For $f_0>0$, positive $\Lambda$ without slope gives eastward propagation, while a positive slope without shear gives westward propagation. The phase velocity varies as $1/|k|$, and the trapping depth is $1/\mu$. Its frequency is independent of $|k|$ on each signed branch, so its zonal group velocity is zero in this ideal semi-infinite model.

With a rigid horizontal upper boundary, put $B_D(t)={\hat{\psi }}_z(D,t)$ and $M=\mu D$. Solving the same second-order inversion problem with both derivative data givesDifferentiation at $0$ and $D$ checks the prescribed values. At the upper boundary $w^{\prime }=0$, so $({\partial }_t+ik\Lambda D)B_D=ik\Lambda \hat{\psi }(D)$. Combining this with the lower-boundary condition givesThus two Boundary Rossby waves interact across the layer. For normal modes proportional to $e^{-ikct}$ their dimensional wave-speed matrix isFor $\Lambda \ne 0$, define $C=c/(\Lambda D)$ and $\tilde{\alpha }=a/\Lambda$. Its characteristic polynomial is the Eady model with a sloping lower boundary dispersion relationThe source writes $\tilde{\mu }=NkD/f_0$; for $k,f_0>0$ this is $M$, and the dispersion relation is unchanged by replacing that signed quantity by its absolute value. Its two roots areIf $\tilde{\alpha }>1$, the second term is positive, so $\Delta >0$ for every $M>0$: all modes are spectrally stable. If $\tilde{\alpha }<1$, the increasing function $M\tanh M$ runs from zero to infinity, so there is a unique $M$ withThere the squared term is zero and $\Delta <0$, giving an exponentially growing member of the complex conjugate pair. By continuity it lies in an unstable band. This proves both the requested instability for $0\le \tilde{\alpha }<1$ and instability for negative slopes in the sloping-boundary Eady model. It does not assert that every wavelength is unstable. Strongly negative slopes move the resonant band to large $M$, where the coupling is exponentially weak.

At $\tilde{\alpha }=1$ the speeds are real, but their crossing at $M=\coth M$ can give a defective neutral mode and algebraic growth; absence of exponential instability is weaker than boundedness of every initial disturbance. The nondimensional slope is undefined when $\Lambda =0$; the dimensional matrix remains valid and then has two real speeds. The instability for $\tilde{\alpha }<1$ is the counterpropagating wave instability mechanism: lower and upper boundary waves can match their laboratory phase speeds and exchange energy with the vertical shear.

Let $p$ denote the fluid pressure perturbation divided by the reference mass density, and let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation. The buoyancy frequency satisfies $N^2=-g{\hat{\rho }}_z/{\rho }_0$. Linearization requires small boundary slope $k|{\eta }_0|\ll 1$, small displacements compared with the vertical scales of the disturbance and background, and advection small compared with the oscillatory acceleration. For a propagating internal gravity wave this includes $|m{\eta }_0|\ll 1$; boundary slope alone is insufficient when $|m|\gg k$. The Boussinesq approximation also requires small relative mass density differences over the region of interest. These conditions must hold for the resulting disturbance, including any amplification by resonance.

The inviscid Linearized Boussinesq equations and linear kinematic boundary condition areWrite the velocity field as $(u,w)=\mathrm{Re}\{(U(z),W(z))e^{i(kx-\omega t)}\}$. Incompressibility and horizontal momentum balance giveThe vertical momentum balance therefore yieldsFor a real vertical wavenumber $m$, this gives the dispersion relationWhen $0<\omega <N$, define $\ell =k\sqrt{N^2/{\omega }^2-1}$. The radiation condition selects $m=-\ell$, because energy must leave the boundary upward. The boundary-forced internal gravity wave isIts wavevector is $\mathbf{K}=(k,m)$, its phase velocity is ${\mathbf{c}}_p=\omega \mathbf{K}/(k^2+m^2)$, and its group velocity isThus the phase velocity points down and right, while the group velocity points up and right. They are perpendicular. Internal-wave polarization makes the particle motion an oscillation along the group velocity direction, with zero first-order mean transport. The constant-phase lines of an internal gravity wave are also parallel to the group velocity. Their inclination $\vartheta$ above the horizontal satisfies $\sin \vartheta =\omega /N$.

When $\omega >N$, set $\kappa =k\sqrt{1-N^2/{\omega }^2}$. Boundedness at infinity selects the evanescent waveThe horizontal and vertical velocity components are in quadrature: fluid particles describe small ellipses, and the response decays over ${\kappa }^{-1}$. There is no upward time-averaged energy flux, because $P$ and $W$ are in quadrature. A real vertical group velocity is not defined for this evanescent wave. The pattern travels horizontally with phase velocity $\omega /k$.

For $N=0$ the response is the evanescent wave with $\kappa =k$; the ratios $\omega /N$ should not be used. At the cutoff $\omega =N>0$, the bounded harmonic solution has $W=-i\omega {\eta }_0$, $U=0$: it neither decays nor has nonzero upward group velocity. It is the limiting cutoff response, rather than a localized radiating disturbance. A uniform nonzero buoyancy frequency in an infinitely deep Boussinesq approximation is itself a local idealization of the background mass density.

The arrows distinguish the phase velocity, group velocity, and oscillatory particle motion of the internal gravity wave. The right panel shows the decay envelope and particle ellipses of the evanescent wave.

In a steady state, mass conservation makes $\mathcal{F}$ independent of radius. Taking the accretion rate $\dot{M}>0$ for inward flow gives $\mathcal{F}=-\dot{M}$. The conservation of angular momentum equation then makes $-\dot{M}h+\mathcal{G}$ constant. The zero-torque inner boundary condition fixes this constant to $-\dot{M}{h}_{\mathrm{in}}$, soThe viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^3\bar{\nu }\Sigma {\Omega }^{\prime }=2\pi qh\bar{\nu }\Sigma$. Combining these expressions gives the steady accretion disk with arbitrary rotation lawFor $x=r/R_g$, the Paczyński-Wiita circular orbit formulas giveConsequentlyIn particular, $f(6)=0$ and $f\to 1$ far from the black hole, recovering the outer Keplerian accretion disk.

Inside the innermost stable circular orbit, the gas enters the plunging region of a black-hole accretion disk. Its rapid inward motion leaves little time for stresses to exchange angular momentum, motivating the zero-torque inner boundary condition. This is a thin-disc approximation; a strong magnetic stress could change it.

Matter supplied from very large radius has negligible specific orbital energy, while matter crossing the inner edge carries ${\epsilon }_{\mathrm{in}}=-\eta c^2$. With zero inner torque, no energy is supplied by a stress at that edge. If the heat released outside it escapes by radiative transfer, rather than being lost through inward advection, the integrated conservation of energy balance gives