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For a single-component equilibrium system with extensive internal energy $E(S,V,N)$, scaling its entropy, volume and particle number together gives $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. The Euler theorem for homogeneous functions therefore gives $E=S{\partial }_{S}E+V{\partial }_{V}E+N{\partial }_{N}E$. The first law of thermodynamics identifies these derivatives as $T,-P,\mu$, respectively, proving $E=TS-PV+\mu N$. Dividing by volume yields the entropy density formula $s=(\rho +P-\mu n)/T$. This step fixes the entropy normalization rather than merely identifying its differential. Differentiating the Euler relation and subtracting $dE=TdS-PdV+\mu dN$ gives the Gibbs-Duhem equation $SdT-VdP+Nd\mu =0$. The argument requires extensivity; it is not an assumption about arbitrary nonadditive systems.

Neglecting interactions and thermal pairs, the Fermi-Dirac distribution $f(p)=[\exp ((\epsilon (p)-{\mu }_{\mathrm{kin}})/(k_{B}T))+1{]}^{-1}$ with kinetic energy $\epsilon (p)=\sqrt{m_e^2{c}^4+p^2{c}^2}-m_e{c}^2$ determines the Electron contribution to a stellar equation of state. The kinetic chemical potential is fixed by the electron number density, not an independently chosen pressure fraction:The electron kinetic internal energy per volume is $({\pi }^2{\hslash }^3{)}^{-1}{\int }_0^{\infty }{p}^2\epsilon (p)f(p)dp$. In the nondegenerate limit $P_e=n_e{k}_{B}T$; at zero temperature the integrals become the equation of state of a cold electron gas. These are limits of one component and must not be added together as independent electron pressures. When electron-positron thermal pair abundance matters, both charge species must be included, with charge neutrality fixing their net density.

For an inviscid fluid with specific internal energy $e$, let $E=\rho (e+|\mathbf{u}{|}^2/2)$. In a prescribed potential $\Phi$, ${\partial }_{t}E+\nabla \cdot [(E+p)\mathbf{u}]=-\rho \mathbf{u}\cdot \nabla \Phi +Q$, where $Q$ is external heat supply. A static gravitational potential can instead be included in the conserved energy.

For a fixed Newtonian gravitational potential, total energy density is $\mathcal{E}=\rho (u^2/2+\Phi )+p/(\gamma -1)+B^2/(2{\mu }_0)$. Its flux combines advected kinetic/potential energy, specific enthalpy and the Poynting vector. Dotting the ideal magnetohydrodynamic momentum equation with velocity, adding the adiabatic internal energy equation and the magnetic energy equation cancels the magnetic work. The pressure terms combine into $-\nabla \cdot (p\mathbf{u})$, proving the conservation law. A time-dependent imposed potential instead contributes $\rho {\partial }_t\Phi$.

For a two-dimensional Fermi gas of $N$ spin-one-half particles with Fermi energy $E_F$, the constant density of states makes the low-temperature internal energy correction and heat capacity at constant areaThus its low-temperature heat capacity obeys a power law with exponent one.

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 $\rho {\partial }_t\Phi$.

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.

For a spherical hydrostatic star with negligible surface pressure, the stellar virial theorem isFor a monatomic nonrelativistic perfect gas, $U=\frac{3}{2}\int PdV$, so $2U+\Omega =0$ and the total energy is $E=U+\Omega =\Omega /2$. Half the released binding energy raises the internal energy; only the other half can be radiated. With fixed mass, no nuclear supply and no external work, conservation of energy givesFor the uniform-density stellar model, this becomes the contraction-luminosity evolution equationIt relates luminosity to the contraction rate; it does not by itself prescribe $L(t)$. Given $L(t)$,For constant luminosity, $R(t)=R_0/[1+t/t_0]$ with $t_0=3G{M}^2/(10R_{0}L)$. This describes quasi-static Kelvin-Helmholtz contraction, not dynamical free fall. If the gas has a different constant specific-heat ratio ${\gamma }_g$, the appropriate stellar virial theorem is $3({\gamma }_g-1)U+\Omega =0$, so the radiated fraction of binding release is $(3{\gamma }_g-4)/[3({\gamma }_g-1)]$ rather than universally one half.

Work in one fixed inertial frame, and take positive velocity in the rocket's direction. During $d\tau$, the rocket ejects the positive amount of mass $-m^{\prime }(\tau )d\tau$. Its exhaust has inertial velocity $v(\tau )-u$, by the Galilean transformation of velocities. Once expelled, that material retains its velocity because it experiences no external force. The system's momentum is therefore the rocket momentum plus the sum of all these exhaust momenta:The minus sign comes from the positive expelled mass being $-m^{\prime }(\tau )d\tau$.

Differentiate, applying the product rule to the rocket momentum and the fundamental theorem of calculus to the integral:The momentum conservation law thus gives the nonrelativistic rocket equation

The same bookkeeping for kinetic energy givesIts derivative iswhere the rocket equation cancels the first two terms. Thus kinetic energy increases during mass ejection, with $\boxed{K^{\prime }=-u^2{m}^{\prime }/2>0}$ when $u>0$ and $m^{\prime }<0$. This does not violate conservation of energy: the increase comes from stored internal energy.

Now $m$ denotes the instantaneous rest mass of the rocket, including its fuel and stored internal energy, and $v$ is measured in a fixed inertial frame. Assume the exhaust is expelled backwards with speed $0<u<c$ in the rocket's instantaneous rest frame. By the relativistic velocity-addition formula, material emitted at time $\tau$ has inertial velocityIt coasts at this velocity after emission.

The crucial change is that the expelled material's rest mass is not simply $-dm$: some of the rocket's rest energy becomes exhaust motion. Instead, use conservation of relativistic energy. Put $q(t)=m(t)\gamma (t)$, so the rocket's energy is $c^{2}q(t)$. The exhaust emitted during $d\tau$ carries energy $-c^2{q}^{\prime }(\tau )d\tau$. For material moving at velocity $w$, the relativistic energy-momentum relation gives momentum $wE/c^2$, since $E={\gamma }_{\mathrm{ex}}\mu c^2$ and $p={\gamma }_{\mathrm{ex}}\mu w$. Hence the modified total momentum isMeanwhile the total relativistic energy is $c^2[q(t)-{\int }_0^t{q}^{\prime }(\tau )d\tau ]=c^{2}q(0)$. These expressions consistently account for both energy and momentum of the expelled material.

Differentiate the momentum formula. The momentum conservation law givesTo simplify it, useMultiplying by ${\gamma }^2(1-uv/c^2)$, the equation becomesThe two terms containing $uv/c^2$ cancel. Dividing by $\gamma >0$ proves the relativistic rocket equation

For fixed finite velocities, the nonrelativistic limit $c\to \infty$ gives ${\gamma }^2=(1-v^2/c^2{)}^{-1}\to 1$. Hence the relativistic rocket equation reduces towhich is the nonrelativistic rocket equation.

The conserved quantities called energy in the two calculations include different contributions. The nonrelativistic calculation considered only kinetic energy, which increases as fuel releases internal energy; total energy including that internal reservoir is still conserved. Relativistic energy includes rest energy, and therefore includes the fuel's stored internal energy through the rocket's rest mass. That rest mass decreases as energy is transferred into the exhaust. Thus conservation of relativistic energy and the nonconservation of kinetic energy express the same physical conversion.

Write $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative and $q=\mathbf{u}\cdot \mathbf{B}$ for the cross-helicity density, so $H_c={\int }_{V}qdV$. The ideal magnetohydrodynamic induction equation of ideal magnetohydrodynamics, with $\nabla \cdot \mathbf{B}=0$, givesThe dot product of the Euler momentum equation with the magnetic field has no Lorentz force contribution, because $\mathbf{B}\cdot [(\nabla \times \mathbf{B})\times \mathbf{B}]=0$. Consequently,Here $h=e+p/\rho$ is the specific enthalpy, $T$ the temperature, and $s$ the specific entropy; $e$ is the internal energy per unit mass. The first law of thermodynamics in this convention is $de=Tds-pd(1/\rho )$, so $dh=Tds+dp/\rho$. Substitution, followed by $\nabla \cdot \mathbf{B}=0$, yields the cross-helicity conservation lawThe vector triple product identity $\mathbf{u}\times (\mathbf{u}\times \mathbf{B})=\mathbf{u}q-u^2\mathbf{B}$ puts this in the equivalent form

For the fixed volume, the divergence theorem gives the exact cross-helicity balanceThus cross-helicity is constant when the net source equals the outward flux. Simple sufficient hypotheses are $\mathbf{B}\cdot \nabla s=0$ throughout the volume and $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ on its boundary. The first condition includes a homentropic flow; the second prevents both relevant boundary fluxes. Periodic boundary conditions, or sufficiently rapid decay at infinity, provide alternatives. These statements concern smooth ideal magnetohydrodynamics: material specific entropy conservation alone, $Ds/Dt=0$, does not eliminate spatial specific entropy gradients along the magnetic field lines.

For a reversible change, the first law of thermodynamics is $TdS=dU+pdV$. Using the monatomic internal energy,Along the constant-volume leg, $Q_{AB}=\frac{3}{2}{V}_0(2p_0-p_0)=\frac{3}{2}{p}_0{V}_0>0$, so heat is absorbed. Along the constant-pressure return leg, $Q_{CA}=\frac{5}{2}{p}_0(V_0-2V_0)=-\frac{5}{2}{p}_0{V}_0$, so heat is emitted. The work signs are zero on $AB$ and negative, or work on the gas, on $CA$.