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An adiabatic fluid perturbation preserves the specific entropy of each displaced parcel. For a perfect gas, mass conservation gives $\Delta \rho =-\rho \nabla \cdot \mathbf{\xi }$ and the adiabatic equation of state gives $\Delta p=\gamma p\Delta \rho /\rho$. The corresponding Eulerian perturbation of a fluid variable is $\delta p=-\gamma p\nabla \cdot \mathbf{\xi }-\mathbf{\xi }\cdot \nabla p$. An isothermal equilibrium may still have adiabatic perturbations, because the thermodynamic constraint on displaced parcels differs from the equilibrium temperature profile.

Vertical hydrostatic equilibrium gives $p_0=\rho {\Omega }_z^2(H^2-z^2)/2$. At a vacuum interface, the Lagrangian pressure perturbation vanishes: $p^{\prime }+{\xi }_z{\partial }_z{p}_0=0$. The vertical equation gives ${\xi }_z=p_z^{\prime }/(\rho {\omega }^2)$, henceThe moving interface makes this different from setting the Eulerian perturbation of a fluid variable $p^{\prime }$ to zero on the original surface.

// Target: astrophysics.bigb

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.

Write total pressure and gravitational potential asHere $\pi$ is the Eulerian perturbation of a fluid variable applied to pressure, while $\psi$ is the corresponding gravitational potential perturbation. They have units $\mathrm{Pa}$ and ${\mathrm{m}}^2{\mathrm{s}}^{-2}$, respectively. The constant interior mass density is $\rho$.

The Euler equations areSubtract the hydrostatic equilibrium relation $\nabla p_0/\rho +\nabla {\Phi }_0=0$ and discard terms quadratic in the perturbation. Mass conservation gives incompressible flow, and the gravitational Poisson equation has no bulk density perturbation. Thus, strictly inside the fluid,The last equation is Laplace equation, not a statement that self-gravity has disappeared. The displaced surface carries a first-order surface mass density $\rho \eta$, where $\eta$ is its radial displacement, and this supplies the source and matching conditions for $\psi$ at the moving interface. The axisymmetric gravitational potential of a displaced cylindrical interface, expressed using modified Bessel functions, incorporates that boundary source.

It is useful to define $W=\psi +\pi /\rho$. Taking the divergence of the momentum equation gives ${\nabla }^{2}W=0$. For an axisymmetric Fourier mode with axial wavenumber $k>0$, regularity at $r=0$ selects radial dependence $I_0(kr)$, the Modified Bessel function of the first kind, since $K_0(kr)$ is singular there. In what follows $k=|k_z|$ can always be chosen nonnegative.

Let $p_1(z)$ denote the Eulerian perturbation of a fluid variable that is the pressure amplitude, and keep $\rho$ constant. The linearized Euler equations and incompressible flow constraint areThe last equation gives $u=i{w}^{\prime }/k$. Substituting it into the first yieldsDifferentiating this formula and using the vertical momentum equation produces the Rayleigh equation for inviscid shear flow:For a growing normal mode, $c$ is nonreal, so $U-c$ never vanishes on the real $z$ axis. In each open region the base velocity is constant or linear and $U^{\prime \prime }=0$; therefore $w^{\prime \prime }-k^{2}w=0$. Decay away from the layer selects $e^{-kz}$ above and $e^{kz}$ below, while both exponentials are retained inside. Conveniently normalize these asThese are the spatially localized discrete modes relevant to instability, rather than all possible generalized neutral disturbances at a critical layer in a shear flow.

At each corner $U$ is continuous and $U^{\prime }$ jumps. The two sides have the same normal velocity and normal stress: there is no membrane, density jump or singular external force supplying a pressure jump. Thus $w$ and $p_1$ must be continuous. The pressure formula proves the required continuity of $(c-U)w^{\prime }+U^{\prime }w$. This is pressure matching at a piecewise-linear shear interface; it does not require $w^{\prime }$ to be continuous. Equivalently integrating the Rayleigh equation for inviscid shear flow through a corner gives $(U-c)[w^{\prime }]=[U^{\prime }]w$.

Write $\alpha =kd$, $y=c/U_0$, and $E=e^{2\alpha }$. Continuity of normal velocity givesAt $z=d$, the outside derivatives are $w_{+}^{\prime }=-kA$, $U_{+}^{\prime }=0$, while the inside derivatives are $w_{-}^{\prime }=-kB+kEC$, $U_{-}^{\prime }=U_0/d$. Hence pressure matching reduces toSimilarly at $z=-d$, the outside derivative is $kD$ and the inside derivative is $-kEB+kC$. Matching givesThe two equations can be written asA nonzero mode therefore requiresThis is the dispersion relation for the unbounded piecewise-linear shear layer.

With the phase convention $e^{ik(x-ct)}$, positive $\mathrm{Im}c$ gives exponential temporal growth. Since the displayed $c^2$ is real, a growing/decaying pair occurs precisely whenFor $0<\alpha \le 1/2$, $e^{-2\alpha }>1-2\alpha$ by the strict exponential tangent inequality, so every such mode is unstable. For $\alpha \ge 1/2$, define $f(\alpha )=2\alpha -1-e^{-2\alpha }$. Its derivative $2+2e^{-2\alpha }$ is positive, while $f(1/2)<0$ and $f(1)>0$. There is consequently exactly one root in that interval, andAt the cutoff the discrete roots merge at $c=0$; above it the roots are real and have no exponential growth. The physical wavelengths in the unstable band satisfy ${\lambda }_x=2\pi /k>2\pi d/{\alpha }_s\simeq 9.83d$.

This is a finite-thickness Kelvin-Helmholtz instability: long disturbances roll up the layer and draw kinetic energy from its shear. The off-diagonal $e^{-2kd}$ terms describe coupling of the two vorticity discontinuities. Short disturbances decay across the layer before strongly coupling its edges and are neutral. Indeed at $kd\to \infty$ the two phase speeds approach the velocities $\pm U_0$ of the separated edges. At $kd\to 0$, expansion of the numerator gives $c^2/U_0^2\to -1$, recovering the vortex sheet result $c=\pm i{U}_0$; the growth rate then behaves as $k|U_0|$.