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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.

An entropy mode is a perturbation of entropy and density with no leading pressure or velocity perturbation in a uniform ideal fluid. For an adiabatic equation of state define $s=\delta p/p_0-\gamma \delta \rho /{\rho }_0$. An entropy perturbation can have $\delta p=0$ and $\delta \rho /{\rho }_0=-s/\gamma$, so it has zero frequency in the local fluid frame. In anisotropic reduced magnetohydrodynamics, $Ds=0$ with $D={\partial }_t+{\mathbf{u}}_{\perp }\cdot {\nabla }_{\perp }$, and turbulence mixes its variance as a passive scalar.

Insert the pressure-negligible velocity law into the steady spherical mass flux. With $x=r/R_e$,An adiabatic equation of state $p{\rho }^{-\gamma }=\text{constant}$ and the ideal gas law giveFor a monatomic gas with $\gamma =5/3$,The normalization $T_e$, and hence $p_e=n_e{k}_B{T}_e$, was not specified and remains a boundary parameter. If it is large enough for the pressure gradient to affect acceleration, replace the ballistic velocity by the full adiabatic Bernoulli equation solution. The general mass-flux and thermodynamic relations still hold, but these particular radius powers then need not hold.

For the Schwarzschild criterion, displace a small gas parcel upwards. Assume its composition is unchanged, it remains in pressure balance with its surroundings, and it moves quickly enough that heat exchange is negligible. The parcel obeys the adiabatic equation of state $P{\rho }^{-\gamma }=\mathrm{constant}$. Combining this with the ideal gas equation gives $T\propto P^{(\gamma -1)/\gamma }$, so its adiabatic temperature gradient is ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma =2/5$.

Let the surrounding temperature gradient be $\nabla =d\log T/d\log P$. For an upward displacement $\delta \log P<0$, the parcel's temperature relative to its new surroundings isIf $\nabla <{\nabla }_{\mathrm{ad}}$, this is negative: the parcel is cooler and, at equal pressure and composition, denser than its surroundings. It therefore sinks back; a downward displacement likewise gives a restoring buoyancy force. If the inequality is reversed, the displaced parcel continues to rise. The stability condition is thusEquality is the neutral boundary. There is no composition-gradient term because the ideal-gas parcel and the background have the same mean molecular weight.

In the thin upper stellar atmosphere, take a plane-parallel approximation with constant $g$, optical depth increasing inward, and zero overlying pressure $P(0)=0$. Hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. The ideal gas equation converts the stipulated opacity intoDefine $u=1+3\tau /2$, so the given grey atmosphere relation is $T^4=T_e^{4}u/2$. Integrating from the surface givesThis proves the required pressure law. In particular $P^2\propto g{T}_e^{-4\beta }$ times a function only of $u$ and $\beta$.

For the grey-atmosphere convection onset with density-linear opacity, differentiate the two profiles to obtain the radiative temperature gradient:It increases monotonically from zero, since $\beta >1$, and first reaches the adiabatic temperature gradient at $u_b^{\beta -1}=(4\beta +1)/5$. ThereforeBelow this point the radiative stratification is unstable and convection must transport some of the outward energy flux.

At the outer convection boundary, $u_b$ depends only on $\beta$, so $T_b=T_e(u_b/2{)}^{1/4}\propto T_e$ and $P_b\propto g^{1/2}{T}_e^{-2\beta }$ for fixed $\mu ,{\kappa }_0,\beta$. The matching constant in $P=K{T}^{5/2}$ is $K=P_b/T_b^{5/2}$. ConsequentlyThis temperature form of the polytropic relation follows from an ideal monatomic adiabat; its $K$ is not the density-polytrope coefficient in Question 3.

For the deep radiative boundary beneath a power-law-opacity convective envelope, negligible envelope mass and absent local energy generation give $m\simeq M$ and $L_r\simeq L$. Divide the radiative temperature equation by the pressure equation to get the stellar radiative temperature gradientThis is the gradient that would be needed if radiation alone carried $L$, and it is the quantity to compare with $2/5$ even while the actual region is convective. The deeper Kramers' opacity law gives $\kappa ={\kappa }_1\mu P{T}^{-9/2}/\mathcal{R}$. Substituting $P=K{T}^{5/2}$ yieldsAs $T$ rises inward this decreases, so the transition back to a stable radiative interior is atThus $T_{\mathrm{in}}^{7/2}\propto K^{2}L/M$. Using the outer matching and the Stefan–Boltzmann law,The mass and radius cancel. Hence, at fixed opacity coefficients and composition,This uses the same adiabat through the convection zone and distinguishes its inner transition temperature from its outer onset temperature.

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).

Use $v=|u_r|>0$ for the radial speed and distinguish the central mass $M$ from the Mach number $\mathcal{M}=v/c$. In each smooth nonzero-flow region, mass conservation, the adiabatic equation of state, and the Bernoulli equation giveHere $K>0$ is constant on a smooth isentropic flow branch, $c$ is the adiabatic sound speed, and $\mathcal{E}$ is the Bernoulli function. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal{E}>0$; flows with $\mathcal{E}\le 0$ do not have these positive $r_s,x,\lambda$ and require another scaling.

Indeed, the supplied adiabatic process equations give $D(p{\rho }^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial fluid flow. The adiabatic sound speed is $c^2=(\partial p/\partial \rho {)}_s$, and the specific enthalpy is $h=c^2/(\gamma -1)=2c^2$. The radial Euler momentum equation, $vdv/dr=-{\rho }^{-1}dp/dr-GM/r^2$, integrates to $\mathcal{E}$ because $dh=dp/\rho$ on this isentropic flow branch.

Put $D=[j(3K/2{)}^2{]}^{2/5}$. Eliminating the mass density from the continuity equation givesWith $y={\mathcal{M}}^{2/5}$, the Bernoulli equation becomesTherefore the physical constants of the spherical polytropic flow with adiabatic exponent three halves are

For $t>0$, $F^{\prime }(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the algebraic curves. For $0<\lambda <1$, there are two disconnected branches at every $x>0$, one subsonic flow with $y<1$ and one supersonic flow with $y>1$. For $\lambda =1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta )=5+10{\eta }^2+O({\eta }^3)$. For $\lambda >1$, solutions exist only where $F(x)\ge 5\lambda$: if $x_{-}<1<x_{+}$ solve $F(x_{\pm })=5\lambda$, there is a forbidden interval $x_{-}<x<x_{+}$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

The zero-temperature internal energy isUsing part (i),Consequently the adiabatic equation of state isThe pressure is degeneracy pressure: it remains nonzero at zero temperature because of the Pauli exclusion principle. By comparison, a classical ultrarelativistic ideal gas at fixed temperature obeys $pV=N{k}_{B}T$, so its corresponding volume exponent is one and its pressure vanishes as $T\to 0$.