Codex @codex  0

With no incoming wave, both fluids obey the outgoing acoustic square-root branch. Their surface pressure amplitudes areSubstitution in the sheet's force balance yields the dispersion relation for an acoustic membrane wave:The equivalent denominator $\Delta =\gamma (T{k}^2-m{\omega }^2)-2{\rho }_0{\omega }^2$ is useful away from the branch points $\gamma =0$. For real guided modes with $|k|>k_0$, this is the familiar added mass of an evanescent fluid layer formThe normal fields decay away from the sheet; continued by analytic continuation roots may describe radiating or leaky modes.

On $D=0$, $T{k}^2-m{\omega }^2=2{\rho }_0{\omega }^2/\gamma$, so the impedance from the preceding solution isThus the divergence of the reflection coefficient corresponds to a reflection pole of a fluid-loaded membrane. The prescribed incoming amplitude is zero, but a homogeneous fluid-sheet mode can have nonzero amplitude. This is a resonance or guided-mode pole of the analytic scattering problem, not arbitrarily large passive reflection at a real propagating angle. In particular, an undamped guided mode has an evanescent normal field and therefore a complex incidence angle in the plane-wave continuation. For real propagating incidence, $Z_f>0$ and the finite-mass, finite-tension sheet has a purely imaginary structural impedance, so its passive reflection remains bounded.

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ and $\mathbf{L}=(\nabla \times \mathbf{B})\times \mathbf{B}/{\mu }_0$. The continuity equation converts a material specific-energy balance into a conservative energy density balance. Dot the ideal magnetohydrodynamic momentum equation with $\mathbf{u}$, and use the time independence of the Newtonian gravitational potential:For the energy density $U=p/(\gamma -1)$ associated with internal energy, the adiabatic pressure equation givesThe ideal magnetohydrodynamic induction equation and the cross-product divergence identity give the magnetic energy balanceIndeed $(\mathbf{u}\times \mathbf{B})\cdot (\nabla \times \mathbf{B})=-{\mu }_0\mathbf{u}\cdot \mathbf{L}$. The magnetic work cancels the kinetic magnetic work. The remaining pressure terms are $-\nabla \cdot (p\mathbf{u})$. Adding all three balances proves ideal magnetohydrodynamic energy conservation:whereThe last term is the Poynting vector with the ideal electric field $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$. A time-dependent imposed potential would instead supply the source $\rho {\partial }_t\Phi$.

Use axisymmetry and $\nabla \cdot \mathbf{B}=0$. Since the purely azimuthal velocity has zero divergence, the ideal magnetohydrodynamic induction equation becomes ${\partial }_t\mathbf{B}=(\mathbf{B}\cdot \nabla )\mathbf{u}-(\mathbf{u}\cdot \nabla )\mathbf{B}$. Differentiation of the cylindrical unit vectors contributes to its azimuthal component, givingThe azimuthal component of the magnetic tension force isThere is no azimuthal pressure or gravitational force and no azimuthal advective acceleration for this motion. Because the poloidal magnetic field is divergence-free,These coupled induction and tension equations describe a torsional Alfvén wave.

The magnetic field satisfies $\nabla \cdot \mathbf{B}=0$. Expanding the divergence of its Maxwell stress tensor givesFor the Newtonian gravitational field $\mathbf{g}=-\nabla \Phi$, Poisson equation for Newtonian gravity gives ${\partial }_j{g}_j=-4\pi G\rho$, and the gradient representation gives ${\partial }_j{g}_i={\partial }_i{g}_j$. ConsequentlyThe negative of the Newtonian gravitational stress tensor, in this force-stress convention, therefore supplies $\rho g_i$. Adding the pressure stress supplies $-{\partial }_{i}p$, soEach summand is symmetric. Absence of external gravitational sources is needed to represent the full gravitational force by this self-gravitating stress.

At the initially resting instant $K_{ij}=0$, and the cold-fluid assumption removes the pressure stress. Sum the $xx$ and $yy$ components of the magnetized-fluid tensor virial theorem. The magnetic and gravitational traces areSince $I_{\perp }=I_{xx}+I_{yy}$ and $g_z=-{\partial }_z\Phi$, this yields the horizontal virial balance of a cold magnetized fluid:Being at rest sets the instantaneous velocity to zero; it does not assert equilibrium or zero acceleration. The fields throughout space contribute to this stress integral, including their vacuum exterior.

The linearized azimuthal Euler momentum equation isFor the time dependence $e^{-i\omega t}$ and $u_R^{\prime }=-i\omega {\xi }_R$, it givesThis expresses conservation of the displaced element's specific angular momentum. It applies directly to nonzero-frequency modes, with the zero-frequency limit taken in the displacement formulation.

The radial advective acceleration supplies $-2\Omega v_{\varphi }$, while the pressure force perturbation is $\delta \rho \nabla p/{\rho }^2-\nabla \delta p/\rho$. Eliminate $v_{\varphi }$ and retain the vertical equation. Under the Cowling approximation, $\delta \Phi =0$, soThe continuity equation gives $\delta \rho =-\nabla \cdot (\rho \mathbf{\xi })$. The Lagrangian adiabatic relation $\Delta p/p=\gamma \Delta \rho /\rho$, with $\Delta \rho =-\rho \nabla \cdot \mathbf{\xi }$, givesHere $\kappa$ is the radial epicyclic frequency. These formulas define the axisymmetric adiabatic displacement operator.

Let $a_{\xi }=\mathbf{\xi }\cdot \nabla \Psi$, $d_{\xi }=\nabla \cdot \mathbf{\xi }$ and ${\rho }_{\Psi }=d\rho /d\Psi$. Since $\nabla p=-\rho \nabla \Psi$, the pressure and mass density perturbations are $\delta p_{\xi }=\rho a_{\xi }-\gamma p{d}_{\xi }$ and $\delta {\rho }_{\xi }=-\rho d_{\xi }-{\rho }_{\Psi }{a}_{\xi }$. Integrating the pressure-gradient term by parts in the mass density-weighted inner product givesThe boundary term vanishes for regular admissible displacements because $p,\rho$ vanish there and $\delta p=\rho a-\gamma pd$. Completing the pressure square giveswhereThis effective-potential stratification coefficient is real. All coefficients of the bilinear form are real, so $⟨\eta ,\mathcal{F}\xi ⟩=⟨\mathcal{F}\eta ,\xi ⟩$: the operator is symmetric on the stated boundary domain, giving the usual self-adjoint realization of the stellar normal-mode problem.

Set $\eta =\xi$ and use $\mathcal{F}\xi =-{\omega }^2\xi$. The Cowling energy principle for a rotating barotropic star isThe Rayleigh quotient $Q[\xi ]/\int \rho |\xi {|}^2$ tests stability. If $Q\ge 0$ for every admissible displacement, no mode has ${\omega }^2<0$, so there is no exponentially growing mode. If an admissible trial displacement has $Q<0$, the Rayleigh-Ritz variational principle puts negative spectrum below zero; in the usual discrete stellar mode problem this gives a mode with $\omega =i\sigma$, $\sigma >0$. Equality allows neutral modes, rather than establishing strictly positive frequencies. A locally negative coefficient alone is not a complete instability proof: a trial function must also control the pressure and other positive terms.

The pressure-square term in the Cowling energy principle for a rotating barotropic star is nonnegative, and ${\mathcal{N}}^2>0$ makes its stratification term nonnegative. Write $j=R^2\Omega$ for specific angular momentum. ThenFor a regular star reaching the rotation axis, $j(0)=0$. The given $dj/dR>0$ therefore makes $j>0$ for $R>0$, and consequently ${\kappa }^2>0$. Equivalently, with the usual nonnegative angular-velocity convention the sign follows directly. All three energy terms are nonnegative, soThis is stability within the Cowling approximation used throughout. The orientation-independent rotational condition is $d(j^2)/dR\ge 0$, the Rayleigh discriminant criterion. If a fluid region excludes the axis and negative angular velocity is allowed, $dj/dR>0$ alone needs the additional sign of $j$; the squared-angular-momentum criterion avoids that ambiguity.

For a spherical stellar polytrope in hydrostatic equilibrium, combine $d{m}_r/dr=4\pi r^2\rho$ with $dP/dr=-G{m}_r\rho /r^2$ to eliminate the enclosed mass:For $n>0$, put $P=K{\rho }^{1+1/n}$, $\rho ={\rho }_c{\theta }^n$, $P=P_c{\theta }^{n+1}$ and $r=a\xi$. Since ${\rho }^{-1}dP/dr=(n+1)P_c{\theta }^{\prime }/(a{\rho }_c)$, chooseThe Lane-Emden equation is thenA regular center requires $\theta (0)=1$ and ${\theta }^{\prime }(0)=0$, with positive chosen central mass density and pressure. Locally $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. The stellar surface is the first positive zero ${\xi }_1$, where the idealized external pressure is zero; retain the positive solution before it. Thus $R=a{\xi }_1$ and the Lane-Emden mass formula isThe surface condition selects where to stop a centrally regular solution, rather than replacing its central regularity conditions.

For a polytrope of index zero, the density is constant and $({\xi }^2{\theta }^{\prime }{)}^{\prime }=-{\xi }^2$. Regularity givesThe pressure is $P_c{\theta }_0$ and $a^2=P_c/(4\pi G{\rho }_c^2)$, so $R^2=3P_c/(2\pi G{\rho }_c^2)$. Index zero is the structural incompressible limit: the expression $K{\rho }^{1+1/n}$ is not itself defined at $n=0$.

For a polytrope of index one, set $u=\xi \theta$. The equation becomes $u^{\prime \prime }+u=0$, while central regularity requires $u(0)=0$, $u^{\prime }(0)=1$. HenceThe mass is $4{\pi }^2{a}^3{\rho }_c$, so it can change with central mass density while the radius stays fixed.

The moment of inertia of a polytropic star about any axis through its center follows by integrating $r^2{\sin }^2\vartheta$ over spherical shells:For constant mass density this gives $I_0=2M{R}^2/5$. For index one, integration by parts gives ${\int }_0^{\pi }{\xi }^3\sin \xi d\xi =\pi ({\pi }^2-6)$, and thereforeThese are axial moments of inertia, not the scalar second mass moment $\int r^{2}dm$.

For a finite-radius centrally regular stellar polytrope with $0<n<5$ and $n\ne 1$, eliminate ${\rho }_c$ from $R=a{\xi }_1$ and $M=4\pi a^3{\rho }_c{\omega }_n$. The resulting polytropic mass-radius relation isHere ${\xi }_1,{\omega }_n$ are dimensionless functions of $n$. At $n=1$ no such single-valued mass-as-a-power-of-radius relation exists at fixed $K$: the radius is fixed instead. At $n=3$ the exponent is zero and $M=4\pi {\omega }_3(K/(\pi G){)}^{3/2}$ is independent of central mass density. For the incompressible case, $M=(4\pi {\rho }_c/3)R^3$ at fixed mass density. Regular $n\ge 5$ solutions have no finite zero-pressure surface, so the finite-radius formula does not apply to them.

In stellar homology, the dimensionless radial profiles are the same after scaling radius, enclosed mass, pressure, temperature and luminosity. At corresponding radii let $r^{\prime }=\xi r$, $m_r^{\prime }=m{m}_r$, $T^{\prime }=tT$, $P^{\prime }=pP$ and $L_r^{\prime }=l{L}_r$. The printed $R^{\prime }=\xi r$ is understood as this local radial scaling, with surface radius $R^{\prime }=\xi R$.

Let $q={\rho }^{\prime }/\rho$ and keep composition, opacity coefficient and nuclear coefficient fixed first. Mass conservation gives $q=m/{\xi }^3$, and hydrostatic equilibrium gives $p=mq/\xi =m^2/{\xi }^4$. The ideal gas law then gives $t=p/q=m/\xi$. Scaling the stellar energy-generation rate equation and the stellar radiative temperature gradient gives respectivelyThe second follows from $dT/dr\propto -\kappa \rho L_r/(r^2{T}^3)$, not from energy production. Equating the two luminosity scalings givesThis assumes the denominator is nonzero and a consistent homologous family exists. If the mean molecular weight differs by $u={\mu }^{\prime }/\mu$, then $t=um/\xi$ andRatios of the opacity and energy-generation coefficients multiply the right-hand side. Thus fixed composition is a real restriction, not an automatic property of every stellar sequence.

Use the Stefan–Boltzmann law $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$ to locate the family on a Hertzsprung-Russell diagram. It implies $T_{\mathrm{eff}}\propto M^{(Y-2X)/4}$ andFor proton–proton chain burning take the usual local approximation $\eta =4$, with Kramers' opacity law $\lambda =1$, $\nu =-7/2$. ThenFor the CNO cycle with electron-scattering opacity, $\lambda =\nu =0$, soA conventional local choice $\eta =16$ gives $R\propto M^{15/19}$ and slope $76/9\simeq 8.44$. Choosing $\eta =17$ instead gives slope $60/7\simeq 8.57$. The source specifies no numerical nuclear exponents, and the effective exponent changes with temperature; the general expression is the unambiguous answer.

The Hertzsprung-Russell diagram places hotter stars to the left. Its branches rise toward higher luminosity and mass, and the CNO cycle/electron-scattering opacity branch has the steeper logarithmic slope. Their illustrative joining point and normalization are arbitrary because the proportional opacity and reaction laws do not specify absolute stellar scales. These are the fully radiative ideal-gas homology predictions, not exact observed main sequence relations: convection and increasing radiation support limit those assumptions in real stars.

The total pressure is the sum of ideal gas and radiation pressure contributions. The gas fraction and its complement will be obtained from the structure equations, rather than prescribed independently at each radius.

Divide the radiation-pressure gradient by the total-pressure gradient. Radiative diffusion in a star and hydrostatic equilibrium giveThis last quantity is independent of radius under the specified opacity assumption. Integration gives $P_r=fP+C$. With the standard idealized zero-pressure outer boundary, where both components vanish, $C=0$. The stellar gas-pressure fraction is thereforeconstant throughout the model. This is the Eddington standard model idealization. A finite photospheric pressure offset would need its boundary treatment; the differential relation alone does not set the integration constant to zero.

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.