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A stellar equation of state supplies pressure and internal energy as functions of density, temperature and composition, together with thermodynamic derivatives needed for stability and transport. Hydrostatic equilibrium fixes the pressure gradient, but does not determine which microscopic components provide the pressure. In ordinary dense interiors local thermodynamic equilibrium is a useful starting point. A consistent mixture iswhere ions, Electrons, radiation and interaction corrections are distinguished. The classical Electron pressure and electron degeneracy pressure are two limits of the same Electron contribution, and must not be added as if they belonged to different particles. The finite-temperature electron equation of state interpolates between them.

In a fully ionized, nondegenerate, nonrelativistic gas, $P_g=\rho k_{B}T/(\mu m_u)$ and specific thermal energy is $u_g=3k_{B}T/(2\mu m_u)$. With nuclear mass fractions $X_A$, charges $Z_A$ and mass numbers $A_A$, the mean molecular weight satisfies ${\mu }^{-1}={\sum }_A{X}_A(1+Z_A)/A_A$, and ${\mu }_e^{-1}={\sum }_A{X}_A{Z}_A/A_A$ is the mean molecular weight per electron. This regime describes much of an ordinary main sequence interior. Toward cooler layers, ionization and molecular dissociation change particle numbers and consume heat. The Saha equation relates ionization to both temperature and Electron density: there is no universal horizontal ionization boundary. These regions have larger heat capacity and can have a reduced stellar adiabatic exponent. The simple fully ionized formula is then insufficient.

Equilibrium photons give radiation pressure $P_{\gamma }=a_r{T}^4/3$ and energy per volume $a_r{T}^4$, or specific energy $u_{\gamma }=a_r{T}^4/\rho$. In the nondegenerate gas regime, equality with gas pressure gives the radiation-to-gas pressure boundarya line of slope $1/3$ on a $(\log \rho ,\log T)$ plot. Higher temperatures at fixed mass density favor photon support. A monatomic gas has stellar adiabatic exponent $5/3$, while radiation alone has $4/3$; their adiabatic exponents of a monatomic gas-radiation mixture are not obtained by assuming a fixed pressure fraction during compression. Radiation support is particularly important in massive stars.

For Electrons, Pauli exclusion principle and the Fermi-Dirac distribution determine occupation numbers. The net Electron density is $n_e=\rho /({\mu }_e{m}_u)$ and the Fermi momentum is $p_F=\hslash (3{\pi }^2{n}_e{)}^{1/3}$. Define the kinetic electron Fermi temperatureFor $T\gg T_F$ the Electrons are nearly classical; for $T\ll T_F$ they are strongly degenerate and their pressure depends primarily on density. The intermediate region requires finite-temperature electron equation of state integrals, not a discontinuous switch of formulas.

The equation of state of a cold electron gas gives, in its two limits,These are respectively the $n=3/2$ and $n=3$ pressure-density powers, explaining the approximate white dwarf polytropic mass-radius relation sequence and the Chandrasekhar limit. The kinetic energy per volume is $3P_e/2$ in the nonrelativistic limit and $3P_e$ in the ultrarelativistic limit. Ions can still supply much of the heat capacity even when the Electron pressure supplies the mechanical support.

The Electron degeneracy crossover $T\sim T_F$ has slope $2/3$ at low mass density and $1/3$ at high mass density. The electron relativistic density threshold isa vertical marker where $p_F=m_{e}c$. It is different from the thermal electron relativistic threshold $k_{B}T\sim m_e{c}^2$, near $5.93\times 10^9\mathrm{K}$, a horizontal temperature scale. Hot dilute matter can have relativistic thermal Electrons without degeneracy; cold dense matter can have relativistic degenerate Electrons without reaching that temperature.

Electron degeneracy also does not automatically imply that Electrons dominate the total pressure. Comparing the cold Electron limit with radiation gives the radiation-to-degeneracy pressure boundaryIts logarithmic slopes are $5/12$ for nonrelativistic Electrons and $1/3$ for ultrarelativistic Electrons. One must compare the pressures separately from the degeneracy criterion; extrapolating the classical gas-radiation line into a degenerate region is incorrect.

The stellar equation-of-state regime diagram uses an illustrative fully ionized carbon composition, ${\mu }_e=2$, $\mu =12/7$. It displays the electron-degeneracy crossover, both radiation-pressure comparisons, and distinct thermal and density-driven relativity scales. The curves are limiting-model comparisons, not sharp phase boundaries or a calibrated complete equation of state. Partial ionization, molecular physics and interactions modify the low-temperature regions indicated on the plot.

At sufficiently high temperature, electron-positron thermal pair abundance can become important. The pair abundance depends on density and chemical potential as well as temperature; $k_{B}T=m_e{c}^2$ is not a universal onset line. In the dilute ultrarelativistic limit, both pair species together add energy density $7a_r{T}^4/4$ to the photons' $a_r{T}^4$, and have pressure one third of their energy density. While pairs are being created, thermal energy is spent on rest mass, which can reduce the stellar adiabatic exponent below $4/3$ and contribute to pair-instability supernova physics.

At high mass density and low temperature, Interactions governed by Coulomb's law invalidate the noninteracting-ion approximation. The ionic Coulomb coupling parameter $\Gamma =Z^2{e}^2/(4\pi {ϵ}_0{a}_i{k}_{B}T)$, where $a_i=(3/(4\pi n_i){)}^{1/3}$, grows as ${\rho }^{1/3}/T$. Corrections become significant when $\Gamma$ is of order one, and a sufficiently strongly coupled plasma can crystallize. At still greater mass density, electron capture alters ${\mu }_e$ and nuclear matter replaces the ideal electron-ion model; neutron star interiors require strong-interaction and relativistic equations of state. These further regimes lie beyond the simple pressure curves plotted here. A useful stellar EOS is thermodynamically consistent across the crossovers, rather than just the maximum of unrelated pressure laws.

Write the perturbation velocity as $(u,v)$, the density amplitude as ${\sigma }_1$, and ${\gamma }_d=ϵ\Omega$. Axisymmetry removes advection by the background azimuthal flow, but the radial perturbation advects the Keplerian shear: $u{\partial }_x{U}_{g,y}=-3\Omega u/2$. Combining this with the Coriolis force gives the linearized shearing sheet equationsThe razor-thin disk Poisson kernel supplies the self-gravity term. The $\Omega /2$ coefficient, rather than $2\Omega$, is essential: it includes the perturbed advection of the background velocity.

Put $A=c^2{k}^2-2\pi G{\sigma }_0|k|$ and ${\omega }^2={\Omega }^2+A$. The determinant of the three amplitude equations isExpanding it gives the dust gravitational dispersion relation with gas dragUsing a determinant avoids division by $s$ and retains the neutral/secular branch. The radial epicyclic frequency of this Keplerian shearing sheet is $\Omega$, so the three terms in ${\omega }^2$ represent rotational support, self-gravity and dust pressure.

Use a local, homogeneous razor-thin disk approximation in a frame rotating with constant $\Omega$. The unperturbed planar velocity is zero in this frame, and the large-scale gravitational and centrifugal forces balance. Neglect viscosity, magnetic fields, thickness and background gradients across a wavelength. Take small planar disturbances, with wavelength short compared with the galaxy's background scale but long enough for a fluid description. The barotropic closure of a razor-thin disk is $P=K{\Sigma }^{\gamma }$, with $K$ fixed and a positive derivativeSolid-body rotation has no shear and has radial epicyclic frequency $\kappa =2|\Omega |$. By rotational symmetry of the local model choose a wavevector along $x$, and write perturbations proportional to $e^{i(kx-\omega t)}$.

Let ${\Sigma }_1,u,v,{\Phi }_1$ be the surface density, $x$-velocity, $y$-velocity and Newtonian gravitational potential amplitudes. The linearized continuity equation and Euler equations with the Coriolis force areThe perturbing Newtonian gravitational potential solves the Poisson equation for Newtonian gravityIts decaying solution is proportional to $e^{-|k||z|}$. The jump in its derivative is $-2|k|{\Phi }_1(0)=4\pi G{\Sigma }_1$, giving the razor-thin disk Poisson kernelEliminating $v$ from the two momentum equations, and using ${\Sigma }_1=k{\Sigma }_{0}u/\omega$, gives the density-wave branchThe complete linear system also has a zero-frequency balanced mode; it is not the growing density-wave branch, and division by $\omega$ in this elimination excludes it. The result is the uniformly rotating gas-sheet dispersion relation: pressure opposes compression at large wavenumber, self-gravity promotes compression, and rotation provides epicyclic support.

A mode is exponentially unstable when ${\omega }^2<0$. With $\Omega =0$ and $c>0$, this occurs atThe uniform $k=0$ perturbation is marginal rather than growing in this local calculation. With $c=0$, rotation fails to stabilize sufficiently short waves:These limits show why both pressure and rotation are needed for stability at all wavelengths.

For nonzero pressure and rotation, put $q=|k|$. Complete the square:Its minimum lies at $q_{\ast }=\pi G{\Sigma }_0/c^2$. Thus a growing mode exists precisely whenThe printed inequality has the opposite physical meaning: it is the condition for no exponentially growing density wave. Equality is marginal. In the usual gas Toomre stability criterion, $Q=\kappa c/(\pi G{\Sigma }_0)=2|\Omega |c/(\pi G{\Sigma }_0)$, so instability is $Q<1$ and stability is $Q\ge 1$. The unstable wavenumber band of a rotating gas sheet iswhen $Q<1$.

If ${\Sigma }_0$ and $\Omega \ne 0$ remain fixed while $c^2$ decreases slowly, first reach marginality atThe first wavelength to become unstable just below this threshold is the marginal fragmentation wavelength of a rotating sheet:Density maxima are separated by approximately this wavelength. The expected fragment size is of this order; an overdense half-wave has width about ${\ell }_{\mathrm{crit}}/2$. Linear theory fixes a preferred wavelength, not an exact nonlinear clump radius or shape. Further cooling shifts the fastest-growing wavelength to $2c^2/(G{\Sigma }_0)$. A rough fragment mass is consequently of order ${\Sigma }_0{\ell }_{\mathrm{crit}}^2$, with a geometrical factor depending on the nonlinear fragmentation pattern.

A convenient normalization of the Papkovich–Neuber representation uses a harmonic vector $\mathbf{h}$ and harmonic scalar $\varphi$:Indeed, $\Delta (\mathbf{x}\cdot \mathbf{h})=2\nabla \cdot \mathbf{h}$, so $\nabla \cdot \mathbf{u}=0$ and $\mu \Delta \mathbf{u}=\nabla p$. This is a rescaling of the usual harmonic-potential representation of Stokes flow.

Measure $\mathbf{x}$ from the sphere center, set $r=|\mathbf{x}|$ and $\mathbf{n}=\mathbf{x}/r$. A translational vector harmonic $\mathbf{U}/r$ provides the decaying $1/r$ force field, while the scalar dipole $(\mathbf{U}\cdot \mathbf{x})/r^3$ adjusts the surface velocity without changing that leading far field. For rotation, $(\mathbf{\Omega }\times \mathbf{x})/r^3$ is harmonic, divergence-free and perpendicular to $\mathbf{x}$. Thus tryThe translational velocity is $\frac{A}{2r}(I+\mathbf{nn})\mathbf{U}-\frac{C}{2r^3}(I-3\mathbf{nn})\mathbf{U}$. Matching its independent tangential and radial components at $r=a$ gives $A=3a/2$, $C=-a^3/2$; matching rotation gives $B=a^3$. ConsequentlyThis superposes the translating sphere in Stokes flow and the rotating sphere in Stokes flow. It is exactly $\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}$ at the sphere and tends to zero at infinity; the additive ambient pressure is set to zero.

Take $z$ positive upward, and write $b=\tilde{\rho }g$. For a straight steady viscous buoyant conduit, the inner and outer axial equations areRegularity gives $w_i^{\prime }(0)=0$, the wall has $w_o(R)=0$, and velocity and tangential traction are continuous at $r=a_0$: $w_i=w_o$, $\lambda \mu w_i^{\prime }=\mu w_o^{\prime }$. With no interfacial tension specified, normal traction is continuous; for this straight steady flow that sets $P_i=P_o$ at the interface. Thus the modified gradients are common in this case.

When their value is zero, integration givesConsequentlyThe large center-to-interface speed ratio explains why the inner parabolic component dominates at small viscosity ratio.

For a slowly varying radius, set $B=b-P_{i,z}$. The local inner solution has $w_i=w_a+B(a^2-r^2)/(4\lambda \mu )$, soThe shear-driven outer interface speed has scale $a^{2}B\ln (R/a)/\mu$. The pressure-driven return flow needed for zero total flux adds a velocity of order $q/R^2$ and a shear-driven contribution of order $a^{2}B/\mu$. Relative to the inner parabolic velocity these terms are $O(\lambda \ln (R/a))$ and $O((a/R{)}^2)$, which are small under the stated assumptions. Axial viscous derivatives are also smaller because the plume is slender. ThereforeFor clarity, the outer pressure-gradient estimate must include both return flux and transmitted interface shear. Their flux scales are $R^4{P}_{o,z}/\mu$, $a^2{R}^{2}B/\mu$ and $q$. Zero total flux hence impliesThis justifies the negligible outer pressure gradient. We subsequently choose its negligible pressure reference as zero, as in the reduced model.

Near the moving interface the outer radial source flow is $u_r=a{a}_t/r$: the interface's axial advection is smaller because its axial velocity is small compared with the inner core speed. Its radial strain is ${\partial }_r{u}_r=-a{a}_t/r^2$. Continuity of normal traction at $r=a$, neglecting the much smaller inner normal viscous stress, givesWith $A=a^2/a_0^2$, this is $P_i=\mu A_t/A$ after dropping $P_o$. Conservation of plume volume is $\pi (a^2{)}_t+q_z=0$.

Let the axial and velocity scales beSince $\mu U_0/(b{\ell }^2)=1$, substitution gives the conduit equationThe buoyant flux and the viscous normal-stress pressure are both retained at this scaling.

Linearizing about unit area gives $a_T+2a_Z-a_{TZZ}=0$. ThusWave crests propagate upward. The group velocity is $2(1-k^2)/(1+k^2{)}^2$: it is downward for $|k|>1$, so the phase direction is not the direction of every wave packet.

For a travelling wave $f(\zeta )$, integrating once and using its uniform far field givesDivide by $f^2$, multiply by $f^{\prime }/f$ and integrate again. A convenient potential, defined up to an additive constant, isAt the positive crest $f=\alpha f_0$, the derivative is zero. Subtracting the far-field potential yieldsThis is the solitary-wave amplitude-speed relation for the conduit equation. A nontrivial elevation wave has $\alpha >1$ and $c>2f_0$; as $\alpha \downarrow 1$, the speed tends to the long-wave speed $2f_0$. Positive area is required throughout the derivation.

At rest, all time derivatives and advective terms vanish. The continuity equation and the stated energy equation are then identities, while the momentum equation reduces to $\boxed{\nabla (p_0+\chi )=0}$. The reference state need not have uniform mass density or entropy. Assume smooth positive reference pressure and mass density, and retain only first-order perturbations.

The linearized continuity equation, momentum balance and adiabatic energy relation areThe perturbation of $c^2$ multiplies a vanishing reference material derivative, so it does not enter at first order. Differentiate the last equation in time, then substitute the linearized momentum equation:Expanding the divergence yields the stratified acoustic pressure equationFor a perfect gas, $c_0^2{\rho }_0=\gamma p_0$. Put $a(\mathbf{x})=p_0^{1/\gamma }/{\rho }_0$. ThenThus the divergence-form stratified acoustic pressure equation isThis is homogeneous linear propagation through the static medium, including its inhomogeneity. In the acoustic analogy of part (a), $W$ has no first-order contribution when viscosity is neglected and the reference velocity is zero. Nevertheless the chosen left-hand operator lacks the gradient terms in the genuine propagation equation. Those effects must consequently appear in $Q_{tt}$. They describe propagation of an existing disturbance, rather than an independent source. Since reference-field choices also change the division between the operator and forcing, $Q_{tt}$ cannot be identified unambiguously with newly generated noise. Its presence in the right-hand side is a consequence of the chosen analogy, not a source-classification theorem.

Scale time by ${\omega }^{-1}$, length by $k^{-1}$, velocity by $\omega /k$, pressure by $\mu \omega$, and streamfunction by $\omega /k^2$. Write $A=\alpha /k$, $\xi =x-t$, and drop stars from dimensionless coordinates. The surface is $y=ϵ\sin \xi$ and its material velocity in the sheet frame is $(0,-ϵ\cos \xi )$. The no-slip boundary condition is thereforeThe stationary matrix identifies a preferred frame. In the sheet frame it translates with $\mathcal{U}{\mathbf{e}}_x$, so matrix-relative Brinkman velocity gives the consistent equationsEquivalently one may keep the printed right-hand side $A^2\mathbf{u}$ after defining $\tilde{p}=p-A^2\mathcal{U}x$; then the far-field pressure has the gradient needed to balance that term. Using the printed equation with a uniform far-field flow and no such adjustment is inconsistent. The distinction first matters at second order, since the first-order swimming speed is zero. We solve the upper half-space; the lower half-space is its reflected counterpart.