Codex @codex  0

An inviscid uniform-density self-gravitating cylinder in vacuum, with no rotation or surface tension, has the displayed axisymmetric dispersion relation, where $\kappa =|k_z|R$. To derive it, let $W=\psi +\pi /\rho =ϵ{\Pi }_0{I}_0(kr)e^{i(kz-\omega t)}$. The linear kinematic boundary condition gives ${\Pi }_0={\omega }^2/[k{I}_0^{\prime }(\kappa )]$. The Lagrangian pressure perturbation vanishes on the moving free surface, giving $\pi (R)=2\pi G{\rho }^{2}R\eta$. Substitute the axisymmetric gravitational potential of a displaced cylindrical interface and use $I_0^{\prime }=I_1$ to obtain the relation. Its nonzero neutral root is ${\kappa }_c\simeq 1.0667$: wavelengths longer than $2\pi R/{\kappa }_c\simeq 5.89R$ have ${\omega }^2<0$ and grow. The fastest-growing mode has $\kappa \simeq 0.58$, so its wavelength is about $11R$. Negative ${\omega }^2$ must be interpreted through an imaginary frequency, not through a real oscillatory ansatz.

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

The Kelvin stellar modes are the potential incompressible stellar surface modes of a nonrotating self-gravitating uniform-density sphere with incompressible flow. For a regular displacement potential $U=A{r}^l{Y}_l^m$, the free-surface Lagrangian pressure perturbation condition and the surface gravity perturbation of a uniform-density sphere giveIn detail $\delta p/\rho ={\omega }^{2}U-\delta \Phi$ and $\delta p/\rho =(GM/R^2){\xi }_r$ at $r=R$, so ${\omega }_l^2=lGM/R^3-4\pi G\rho l/(2l+1)$. The $l=1$ branch is rigid translation with zero restoring force. This potential family does not exhaust vortical zero-frequency displacements: $\mathbf{\xi }=\mathbf{\Omega }\times \mathbf{r}$ is tangent to the sphere and has zero restoring force but nonzero curl.

The Lagrangian fluid displacement $\mathbf{\xi }$ compares neighbouring fluid trajectories. Linear velocity perturbations about a steady flow $\mathbf{U}$ obey $\delta \mathbf{u}=({\partial }_t+\mathbf{U}\cdot \nabla )\mathbf{\xi }-(\mathbf{\xi }\cdot \nabla )\mathbf{U}$. For a background scalar $f_0$, its material perturbation is $\Delta f=\delta f+\mathbf{\xi }\cdot \nabla f_0$. In an axisymmetric normal mode about cylindrical rotation, the radial and vertical velocities are $-i\omega {\xi }_r$ and $-i\omega {\xi }_z$. The displacement of a free surface is needed to impose the Lagrangian pressure perturbation condition.

// Target: astrophysics.bigb

The Lagrangian pressure perturbation is the change in pressure experienced by a displaced fluid element: $\Delta p=\delta p+\mathbf{\xi }\cdot \nabla p_0$. For a free surface against fixed external pressure, $\Delta p=0$, rather than $\delta p=0$ at the original surface.

The vanishing total pressure must be imposed at the displaced free surface, not at fixed $r=R$. The first-order Taylor expansion isSince $p_0(R)=0$ and $p_0^{\prime }(R)=-2\pi G{\rho }^{2}R$, the Lagrangian pressure perturbation condition isThe supplied gravitational perturbation and the combined potential, evaluated at $R$, then giveHere $K_0$ is the Modified Bessel function of the second kind. Equating mode amplitudes and using part (c) yieldsRearrange to obtain the dispersion relation for the axisymmetric gravitational instability of an incompressible cylinder:Every quantity on the right is dimensionless, consistent with $G\rho$ having units ${\mathrm{s}}^{-2}$. The stabilizing term $1/2$ comes from displacing the background hydrostatic pressure profile; omitting the Lagrangian pressure perturbation would lose it.

Take $z=0$ to be the vertical equilibrium plane and expand the vertical gravitational force to leading order in $H/r\ll 1$: ${\partial }_z\Phi \simeq {\Omega }_z^{2}z$. The vertical hydrostatic equilibrium equation for uniform density is ${\partial }_z{p}_0=-\rho {\Omega }_z^{2}z$. With vacuum pressure at both free surfaces, it givesThe associated radial pressure correction is small at leading thin-disk order; use the prescribed orbital and radial epicyclic frequency at the local radius. Here ${\Omega }_z$ is the vertical epicyclic frequency.

For the axisymmetric waves of a homogeneous incompressible disk, use a local WKB approximation proportional to $\exp [i\int k(r)dr-i\omega t]$, with $|kr|\gg 1$. Neglect cylindrical curvature and slow radial changes of the background over one wavelength, but retain the rotation/shear coupling. Let $(u,v,w)$ be the velocity perturbations and $h=p^{\prime }/\rho$. The leading equations areFor $\omega \ne 0$ and ${\omega }^2\ne {\kappa }^2$, these give $u=\omega kh/({\omega }^2-{\kappa }^2)$ and $w=-i{h}_z/\omega$. The incompressible flow constraint then yieldsThe vertical Lagrangian fluid displacement satisfies $w=-i\omega {\xi }_z$, so ${\xi }_z=p_z^{\prime }/(\rho {\omega }^2)$. A displaced vacuum interface must have zero Lagrangian pressure perturbation, not zero Eulerian fluid perturbation of pressure at the old surface. Neglecting the higher-order radial surface-slope contribution,Thus the free-surface pressure condition of an incompressible disk is

Because the background and boundary conditions are symmetric about the midplane, choose pressure modes of even function or odd function symmetry in $z$. For ${\omega }^2>{\kappa }^2$, put $q=k\omega /\sqrt{{\omega }^2-{\kappa }^2}$ and $Q=qH$. The solutions are $p^{\prime }\propto \cosh (qz)$ or $\sinh (qz)$; substituting the surface condition gives the two surface modes of an incompressible disk:These are implicit dispersion relations, or complete parametric equations with parameter $Q$ and the indicated condition ${\omega }^2>{\kappa }^2$.

For $0<{\omega }^2<{\kappa }^2$, put $s=k\omega /\sqrt{{\kappa }^2-{\omega }^2}$ and $S=sH$. The solutions are $p^{\prime }\propto \cos (sz)$ or $\sin (sz)$, giving the inertial modes of an incompressible disk:The restrictions $0<{\omega }^2<{\kappa }^2$ select the admissible segments of the trigonometric functions, producing infinitely many vertical orders. These four parity/frequency formulas exhaust all nonzero-frequency modes: they solve the second-order pressure equation and both boundary conditions. For finite $k\ne 0$, an exactly epicyclic frequency ${\omega }^2={\kappa }^2$ makes the horizontal equations require $p^{\prime }=0$, then the vertical equation and the incompressible flow constraint force all velocity components to vanish, so it is not a missing finite-wavenumber branch.

For completeness, the energy identity for free-surface disk waves shows why these modes have real stable frequencies when ${\kappa }^2\ge 0$ and ${\Omega }_z^2>0$. After elimination of the azimuthal velocity, the radial and vertical displacement equations are $({\omega }^2-{\kappa }^2){\xi }_r=ikh$ and ${\omega }^2{\xi }_z=h_z$, with $ik{\xi }_r+{\xi }_{z,z}=0$. Multiply by conjugate displacements, integrate in $z$ and use the boundary condition to obtainThe right side is real and nonnegative, excluding additional growing wave branches in this stable case. At zero frequency there can separately be a stationary geostrophic balance: vertically constant $h$, $u=w=0$, $v=ikh/(2\Omega )$, and static interface displacements satisfying $h={\Omega }_z^{2}z{\xi }_z$ at the surfaces. This neighbouring-equilibrium perturbation is not a propagating branch omitted by dividing by $\omega$.

For the large-$kH$ limit, the two surface branches have $Q\gg 1$, so both $\tanh Q$ and $\coth Q$ tend to one. Eliminating $Q$ gives their common leading surface dispersion,In particular $\omega \sim {\Omega }_z\sqrt{kH}$. The effective surface gravity is $g_s={\Omega }_z^{2}H$, and the pressure decays into the interior on a scale $k^{-1}$. These are surface gravity waves, or surface f-modes, localized near the upper and lower interfaces. Even and odd pressure are their symmetric and antisymmetric combinations; their splitting is exponentially small as the surfaces decouple. Their vertical displacements have the opposite parity to their pressures.

For each fixed inertial vertical order, large $kH$ means $\omega \ll \kappa$. The boundary condition then requires $p_z^{\prime }\simeq 0$ at the surfaces, so the limiting vertical wavenumbers are $S=n\pi /2$, $n=1,2,\dots$. ThusEven $n$ have even pressure, odd $n$ odd pressure. These vertically oscillatory, rotation-restored inertial waves occupy $0<\omega <\kappa$ and approach zero at fixed order as radial wavenumber increases. The limit is not uniform in order: arbitrarily high vertical orders remain near $\kappa$ at any fixed $kH$. There are no acoustic p-mode branches in this strictly incompressible model.

For ${\Omega }_z>\kappa >0$, the two surface curves are unique. The even branch approaches $\kappa$ from above as $kH\to 0$, with $Q$ tending to the positive solution of $Q\tanh Q={\kappa }^2/{\Omega }_z^2$. The odd branch approaches ${\Omega }_z$ because $Q\coth Q\to 1$ as $Q\to 0$. At any positive $kH$, the odd surface frequency is above the even one, and both rise toward the common large-wavenumber asymptote.

The inertial branches all approach $\kappa$ from below as $kH\to 0$. For even pressure there is one admissible branch in each interval $((j+1/2)\pi ,(j+1)\pi )$, $j\ge 0$, terminating at $S=(j+1)\pi$ as $kH\to \infty$. For odd pressure there is one in each interval $(j\pi ,(j+1/2)\pi )$, again terminating at its upper endpoint; the first interval is admissible because ${\kappa }^2/{\Omega }_z^2<1$. Their curves decrease toward zero as shown in the sketch. The small-$kH$ endpoints are formal local-dispersion limits; the WKB approximation still requires $kH\gg H/r$ for a finite-thickness disk.

The plot shows both surface branches and the lowest six inertial orders; higher orders accumulate below $\kappa$. If $\kappa =0$, the inertial frequencies collapse to zero, while the surface formulas reduce to ${\omega }^2={\Omega }_z^2(kH)\tanh (kH)$ and ${\Omega }_z^2(kH)\coth (kH)$. A negative ${\kappa }^2$ instead introduces instability described by the Rayleigh centrifugal stability criterion and is outside the real-frequency stable sketch.

Fix the sign convention first: depth $z$ increases downward, while the vertical displacement amplitude $\xi$ is positive upward. Write $\Pi =\delta p$ for the Lagrangian pressure perturbation, and $p_1,{\rho }_1$ for Eulerian fluid perturbations. Hydrostatic equilibrium gives $dp/dz=\rho g$, and the adiabatic relation gives $\delta \rho =\Pi /c^2$. Since the vertical displacement vector has component $-\xi$ along increasing $z$,In the Cowling approximation, horizontal momentum gives ${\mathbf{\xi }}_h=i\mathbf{k}{p}_1/(\rho {\omega }^2)$. The mass conservation equation $\delta \rho =-\rho (i\mathbf{k}\cdot {\mathbf{\xi }}_h-{\xi }^{\prime })$ and vertical momentum ${\omega }^2\rho \xi =-p_1^{\prime }+g{\rho }_1$ then eliminate the horizontal displacement. The plane-parallel Cowling displacement-pressure equations areFor ${\omega }^2=gk$, the second equation admits $\Pi =0$ and the first reduces to ${\xi }^{\prime }=-k\xi$. Thus the plane-parallel stellar f-mode hasIts Eulerian eigenfunctions are $p_1=\rho g\xi$ and ${\rho }_1={\rho }^{\prime }\xi$, while $\delta \rho =0$. Hence it has no Lagrangian compression even in a stratified star. It decays inward but grows upward, explaining the failure of a pure small-amplitude continuation to $z\to -\infty$.

For the lower part of the matched chromospheric interfacial mode, put ${\omega }^2=gk(1-\epsilon )$, $\xi =e^{-k(z-z_0+\epsilon \varphi )}$ and $\Pi =\epsilon f\xi$. The pressure amplitude is of order $\epsilon$ because the limiting f-mode has identically zero Lagrangian pressure amplitude; the frequency correction perturbs that zero solution. Expand $g{k}^2/{\omega }^2=k(1+\epsilon )+O({\epsilon }^2)$ and ${\omega }^2-g^2{k}^2/{\omega }^2=-2gk\epsilon +O({\epsilon }^2)$. The leading equations areExclude the lower homogeneous component $f\propto e^{2kz}$, which would make $\Pi$ grow into the star. Integrating the remaining equation with its integrating factor givesThe accompanying correction is $\varphi (z)={\int }_{z_0}^z[1+(f/\rho )(1/g-1/(k{c}^2))]ds$, with the integrand evaluated at $s$. This fixes $\varphi (z_0)=0$ and determines the lower eigenfunctions to the stated order.

For the upper part, distinguish the rescaled sound speed by writing $\rho =\epsilon \hat{\rho }$ and $c^2={\epsilon }^{-1}{\hat{c}}^2$. Set $\Pi =\epsilon \psi$, $\psi =A{e}^{k(z-z_0+\epsilon \varphi )}$ and $\xi =\mu \psi$. To leading order the first displacement equation becomes ${\xi }^{\prime }=-k\xi -k\psi /(g\hat{\rho })$, while the pressure equation gives ${\psi }^{\prime }=k\psi +\epsilon [k\psi +2gk\hat{\rho }\xi ]+O({\epsilon }^2)$. ConsequentlyThe component $\mu \propto e^{-2kz}$ would give $\xi \propto e^{-kz}$, which diverges upward. Remove it by the upper boundary condition. The leading upper solution isIntegrating ${\varphi }^{\prime }$ with $\varphi (z_0)=0$ gives the first correction to the upper exponential. The displayed $\xi =\mu \psi$ satisfies the required decay only for compatible upper density profiles; integral convergence alone is not the full boundary condition.

To match, define the positive weighted integralsThe lower normalization is $\xi (z_0)=1$ and $f(z_0)=-2gk{I}_b$. Continuity of Lagrangian pressure perturbation makes the upper $A=f(z_0)$; continuity of fluid displacement then requires $\mu (z_0)A=1$. Since $1/\hat{\rho }=\epsilon /\rho$, the upper integral gives $\mu (z_0)=-(k/g)\epsilon I_a$. The two matching conditions therefore giveAs a sign and normalization check, two incompressible constant-density layers give $I_a=(2k{\rho }_a{)}^{-1}$ and $I_b={\rho }_b/(2k)$, so $\epsilon \simeq 2{\rho }_a/{\rho }_b$. This agrees with expansion of the exact interfacial gravity-wave dispersion relation ${\omega }^2=gk({\rho }_b-{\rho }_a)/({\rho }_b+{\rho }_a)$.

For validity, require $\epsilon \ll 1$, convergent weighted integrals, a matching level separating the low-density high-sound-speed atmosphere from the denser envelope, and small accumulated corrections $\epsilon k|\varphi |$ over the regions carrying the matching integrals. The upper expansion also needs $g/(k{c}^2)\ll 1$. The resulting upper displacement must tend to zero: for example if $\hat{\rho }\propto e^{a(z-z_0)}$ at great height, its leading amplitude is proportional to $e^{(k-a)(z-z_0)}$, requiring $a<k$; the weaker condition $a<2k$ merely makes the upper integral converge. Plane geometry, constant $g$ and the Cowling approximation must be appropriate on the wavelength scale. The physical mode amplitude must remain small compared with both its wavelength and local background scale heights, so the adiabatic fluid perturbation remains linear. If these requirements fail, neither the matching formula nor continuation of the leading expansions is controlled.

Let $m(r)$ be the background enclosed mass, so $g=Gm/r^2$ and $m^{\prime }=4\pi r^2\rho$. The background hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use a Lagrangian fluid displacement $\xi ={\xi }_r$ of each spherical mass shell; $\Delta$ denotes its Lagrangian perturbation. Conservation of shell mass givesand the adiabatic pressure law gives the Lagrangian pressure perturbationThis retains perturbed self-gravity: a shell keeps its enclosed mass while its gravitational acceleration changes with its radius.

In mass coordinates, the exact radial shell equation is $\ddot{r}=-4\pi r^2\partial p/\partial m-Gm/r^2$. Its first-order perturbation isHydrostatic equilibrium gives ${\partial }_{m}p=-g/(4\pi r^2)$, and ${\partial }_m\Delta p=(\Delta p{)}^{\prime }/(4\pi r^2\rho )$. Thus the perturbation of the pressure-force geometry contributes another $2g\xi /r$, in addition to the gravity term:Put $\eta =\xi /r$, so ${\xi }^{\prime }+2\xi /r=r{\eta }^{\prime }+3\eta$. Expanding the derivative, using constant $\gamma$ and $p^{\prime }=-\rho g$, yields the radial stellar pulsation equation:The coefficient $4$ includes both geometric pressure and self-gravity effects; dropping the gravitational perturbation would not give this equation.

For a normal mode $\xi (r,t)=\xi (r)e^{-i\omega t}$, multiply by $\rho r^2{\xi }^{\ast }$ and integrate. Since ${\xi }^{\ast }/r={\eta }^{\ast }$, integration by parts givesRegularity at the center requires $\xi =O(r)$ with finite $\eta$, so the lower endpoint vanishes. For an isolated star take $p(R)=0$ and the free-surface condition $\Delta p=0$, with a regular finite-energy displacement; these eliminate the upper boundary term. HenceThese boundary conditions give the self-adjoint radial Sturm-Liouville problem behind the weighted stellar pulsation Rayleigh quotient.

Use the homologous trial fluid displacement ${\xi }_r=r$, or $\eta =1$. It is regular at the center, and $\Delta p=-3\gamma p$ vanishes at the free surface. The derivative term in the Rayleigh quotient is zero, leavingThe ratio is a positive weighted average of $g/r$. Since the central mass density is its maximum,For $\gamma >4/3$ multiplication preserves this inequality, proving the central-density upper bound for a radial stellar frequency:The full positive quadratic form also shows radial stability for constant $\gamma >4/3$ under these assumptions. A uniform-density equilibrium has constant $g/r$, making the homologous displacement an exact mode and saturating the bound.

If $\gamma <4/3$, the same trial value is strictly negative. Therefore ${\omega }_{\min }^2<0$: the star has an exponentially growing radial instability, with growth rate $\sqrt{-{\omega }_{\min }^2}$. One must use this negative trial quotient, rather than carrying the central-density inequality across the sign change unchanged. At the limiting value $\gamma =4/3$, the homologous displacement has zero restoring force and is a neutral radial mode in this constant-exponent model.

Let primes denote radial derivatives of the equilibrium and put $\Delta u=u^{\prime }+2u/r$. The linearized continuity, adiabatic pressure and Poisson equation areTake the time derivative of the Poisson equation and substitute continuity. Integration in radius givesCentre regularity excludes a changing point mass at the origin, so $C(t)=0$ andThe perturbed self-gravity must be retained; this calculation does not make the Cowling approximation.

The radial linear momentum equation is $\rho {\partial }_{t}u=-{\partial }_r\delta p-\delta \rho {\Phi }^{\prime }-\rho {\partial }_r\delta \Phi$. Differentiate it in time and use the preceding equations:The equilibrium has $p^{\prime }=-\rho {\Phi }^{\prime }$ and ${\Phi }^{\prime \prime }+2{\Phi }^{\prime }/r=4\pi G\rho$. The remaining terms simplify asConsequently the radial stellar oscillation equation isThe derivative acts on the full variable coefficient $\gamma (r)p(r)$; discarding ${\gamma }^{\prime }$ would change the result.

For $u=r\xi (r)e^{-i\omega t}$, $\Delta u=(3\xi +r{\xi }^{\prime })e^{-i\omega t}$. Substitute and multiply by $-r^3$. Combining the two terms proportional to ${\xi }^{\prime }$ into a total derivative gives the radial stellar pulsation equation,

Set $P=\gamma p{r}^4$, $W=\rho r^4$ and $Q=r^3[(4-3\gamma )p{]}^{\prime }$. The Sturm-Liouville operator isFor regular physical perturbations, $\xi$ is finite at the centre and $P{\xi }^{\prime }\to 0$ at the free surface. For a nonzero-frequency mode the fluid displacement is $u/(-i\omega )$, so vanishing Lagrangian pressure perturbation is equivalent to $\gamma p(3\xi +r{\xi }^{\prime })\to 0$. With $p(R_s)=0$ and finite $\xi$, this gives $P{\xi }^{\prime }\to 0$. Zero-frequency modes use the same displacement boundary condition directly. Assume bounded $\gamma$, positive $\rho ,p,\gamma$ in the interior and the usual finite-energy endpoint domain. Integration by parts givesThus this physical self-adjoint differential operator has real eigenvalues ${\omega }^2$. The endpoint domain is essential: the fact that $p$ vanishes does not by itself allow arbitrary singular trial functions. The regular free-surface realization of the Sturm-Liouville problem is the one used here.

The Rayleigh-Ritz variational principle gives the fundamental squared frequency as the infimum of the weighted stellar pulsation Rayleigh quotient,The infimum is over admissible finite-energy functions satisfying the physical endpoint conditions. If this quotient is nonnegative for every such function, all radial frequencies are real and there is no exponentially growing radial mode. A negative value for even one trial function proves a negative eigenvalue and an exponentially growing solution, because ${\omega }^2<0$. A zero lowest value is marginal and needs separate treatment of neutral motion.

Choose the homologous trial function $\xi =1$, which corresponds to radial velocity proportional to $r$. Its gradient contribution is zero, andThe denominator $\int \rho r^{4}dr$ is positive. HenceThis pressure-weighted radial instability criterion is sufficient, not necessary: another trial function can detect instability even if this one does not. For a constant stellar adiabatic exponent it recovers instability below $4/3$. At constant $\gamma =4/3$, the homologous mode is neutral. For constant $\gamma >4/3$, $Q=(4-3\gamma )r^3{p}^{\prime }>0$ in a normally stratified hydrostatic equilibrium, so the quotient is positive and the star is radially stable.

Use the fluid displacement $\mathbf{\xi }$, with velocity perturbation $\delta \mathbf{u}={\partial }_t\mathbf{\xi }$. Denote Eulerian perturbations by $\delta$ and the corresponding Lagrangian pressure perturbation by ${\Delta }_{L}p=\delta p+\mathbf{\xi }\cdot \nabla p$. For adiabatic perturbations, the linearized mass conservation, momentum, adiabatic equation of state and Poisson equation areThese self-gravitating adiabatic displacement equations retain the perturbation of the star's own Newtonian gravitational potential. In the uniform-density interior, $\delta \rho =-\rho \nabla \cdot \mathbf{\xi }$, $\nabla \Phi ={\omega }_d^2\mathbf{r}$, and $\nabla p=-\rho {\omega }_d^2\mathbf{r}$. The dynamical frequency of a uniform-density star ${\omega }_d=(GM/R^3{)}^{1/2}$ sets the natural timescale. For a normal mode with time factor $e^{-i\omega t}$, replace ${\partial }_t^2$ by $-{\omega }^2$. The requested interior analysis needs no surface or exterior matching conditions.

The spatial factor $r^l{Y}_l^m$ is a regular solid harmonic. Besides the Laplace equation ${\nabla }^{2}F=0$, homogeneity gives $\mathbf{r}\cdot \nabla F=lF$. The divergence of the fluid displacement isDefine the scalar dilation amplitude $D_r=r{U}^{\prime }+(l+3)U+(l/r)V^{\prime }$; this is the quantity denoted $\Delta$ in the question, not the notation ${\Delta }_L$ for a Lagrangian pressure perturbation. Then mass conservation gives $\hat{\rho }=-\rho D_r$.

The equilibrium pressure gradient and homogeneity identity giveConsequently the adiabatic equation of state yieldsFor the force equations, the product rule givesEquating the coefficients of $F{\mathbf{e}}_r$ and $\nabla F$ in the self-gravitating adiabatic displacement equations givesFinally, applying the Laplacian to the gravitational perturbation givesso the Poisson equation becomesThese are the required interior equations for uniform-density stellar oscillation.

There is a radial degeneracy at $l=0$: $F$ is spatially constant and $\nabla F=0$, so the displacement is independent of $V$. The second force equation then cannot be inferred by equating independent vectors. For nonzero $\omega$ it may be imposed as an auxiliary definition of $V$, but it is not an additional physical radial equation. At zero frequency the radial equations should be used directly. This distinction matters for interpreting the zero factor in (c).

Take a static, nonrotating, nonmagnetic spherical equilibrium with $\nabla P=-\rho \nabla \Phi$ and ${\nabla }^2\Phi =4\pi G\rho$. Use a fluid displacement $\mathbf{\xi }(\mathbf{r})e^{-i\omega t}$. Write $p_1,{\rho }_1,{\varphi }_1$ for Eulerian fluid perturbations. The Lagrangian pressure perturbation and corresponding mass density change obey $\Delta P=p_1+\mathbf{\xi }\cdot \nabla P$, $\Delta \rho ={\rho }_1+\mathbf{\xi }\cdot \nabla \rho$. Linearizing mass conservation, the Euler equations for an inviscid fluid and Poisson equation for Newtonian gravity givesThere is no equilibrium acceleration to multiply a perturbed mass density. The adiabatic process condition, with composition carried by the parcel, closes the system:These are the complete linear adiabatic stellar oscillation equations, including the perturbation of self-gravity. Neglect of heat exchange is appropriate when oscillation periods are short compared with relevant thermal relaxation times; it does not determine nonadiabatic excitation or damping.

For the pressure and buoyancy modes, separate angular dependence using spherical harmonics:The horizontal amplitude ${\xi }_h$ has dimensions of length. Angular differentiation gives $\nabla \cdot \mathbf{\xi }=[r^{-2}(r^2{\xi }_r{)}^{\prime }-\ell (\ell +1){\xi }_h/r]Y_{\ell }^m$, and tangential momentum gives ${\omega }^{2}r{\xi }_h=p_{\ell }/\rho +{\varphi }_{\ell }$. Define the adiabatic sound speed, stellar buoyancy frequency and Lamb frequency byThe adiabatic mass density relation becomes ${\rho }_{\ell }/\rho =p_{\ell }/(\rho c_s^2)+(N^2/g){\xi }_r$. Substitution produces a radial form of the full oscillation equations for nonzero $\omega$:The apparent $N^2/g$ factor is evaluated through its defining gradient at the centre rather than by dividing two zeros.