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The kinetic electron Fermi temperature is $T_F=[\sqrt{m_e^2{c}^4+p_F^2{c}^2}-m_e{c}^2]/k_B$, where the Fermi momentum $p_F=\hslash (3{\pi }^2{n}_e{)}^{1/3}$ counts the net electrons. Strong degeneracy requires $T\ll T_F$. With mean molecular weight per electron ${\mu }_e$ and $n_e=\rho /({\mu }_e{m}_u)$, $T_F=(m_e{c}^2/k_B)[\sqrt{1+(\rho /{\rho }_{\ast }{)}^{2/3}}-1]$. Its nonrelativistic and ultrarelativistic limits scale as ${\rho }^{2/3}$ and ${\rho }^{1/3}$, respectively.

For noninteracting electrons at zero temperature, define $x=p_F/(m_{e}c)$ with Fermi momentum $p_F=\hslash (3{\pi }^2{n}_e{)}^{1/3}$. The isotropic momentum flux gives electron degeneracy pressureIts nonrelativistic and ultrarelativistic limits arerespectively. In a fully ionized stellar gas, $n_e=\rho /({\mu }_e{m}_u)$, using mean molecular weight per electron ${\mu }_e$. This neglects finite-temperature effects, interactions, and changes in composition.

The Fermi energy is the highest occupied one-particle energy in the ground state of a noninteracting Fermi gas at zero temperature, measured here relative to the lowest one-particle energy. The Pauli exclusion principle allows only one fermion in each complete one-particle quantum state, including its spin, so adding more particles fills progressively higher energy eigenstates. A cold gas can therefore have substantial kinetic energy even when its temperature is zero. In the thermodynamic limit, with the same energy reference, $E_F=\mu (T=0)$ where $\mu$ is the chemical potential.^[ref]

For a uniform, three-dimensional gas of noninteracting electrons, with spin one-half, equal populations of the two spin states, no applied magnetic field, and nonrelativistic particles of mass $m=m_e$,Here $n$ is the total number density, $V$ the volume, $p_F$ the Fermi momentum, $\hslash =h/(2\pi )$ the reduced Planck constant, and $k_B$ the Boltzmann constant. The electron Fermi temperature $T_F$ expresses the same energy scale in kelvin; it is not the actual gas temperature. Strong degeneracy of a degenerate electron gas means $T\ll T_F$. The momentum formula counts states and survives changing the energy–momentum relation, whereas $E_F=p_F^2/(2m)$ requires $p_F\ll mc$.^[ref]

The reference matters: the displayed $E_F$ excludes rest energy and sets the zero at $p=0$. Including rest energy shifts both the one-particle energy and the zero-temperature chemical potential by $m{c}^2$, leaving occupations unchanged. More generally, adding a constant $C$ to every one-particle energy also adds $C$ to $\mu$; the Fermi-Dirac distribution depends only on their difference. In a finite quantum system the chemical potential at zero temperature can lie between the last occupied and first empty energy eigenstates; its equality with the filling edge is a continuum statement.

At positive temperature, $E_F$ remains the zero-temperature scale associated with the specified number density. The Fermi level $\mu (T)$ must instead satisfy the particle-number conditionwhere $g(E)$ is the extensive density of states, including both spin states. Smearing the Fermi-Dirac distribution while arbitrarily holding $\mu =E_F$ generally changes $N$. This distinction matters when interpreting electron heat capacity and any experiment that infers temperature from an occupation profile.^[ref]

Choose $n=10^{28}{\mathrm{m}}^{-3}$ for a uniform noninteracting electron model with two equally populated spin states. This is an original worked example, not a measured number density assigned to an actual material. Using $m_e\simeq 9.10938\times 10^{-31}\mathrm{kg}$, $\hslash \simeq 1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$, and $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ givesThe conversion uses $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$: an electronvolt is an energy unit, while a joule is its SI counterpart. The dimensions provide a quick check: $n^{1/3}$ is inverse length, $\hslash n^{1/3}$ has units of momentum, $p_F^2/m$ has units of energy, and $E_F/k_B$ has units of temperature. The ratio $p_F/(m_{e}c)\simeq 0.002574$ verifies the nonrelativistic limit for this model, where $c$ is the speed of light.

Increasing this chosen number density eightfold doubles the Fermi momentum and quadruples the Fermi energy, giving approximately $6.7701\mathrm{eV}$. The exponent $2/3$ has a geometric origin: the number of occupied states grows as the cube of a momentum radius, but kinetic energy of nonrelativistic particles grows as its square. This makes Fermi energy a useful bridge between a count of particles and a mechanical energy scale.

Fermi momentum is the magnitude of the momentum at the boundary of filled one-particle states in an isotropic zero-temperature Fermi gas. For the uniform three-dimensional two-component electron model, a short state count givesTo see the factors, impose periodic boundary conditions on a cubic box of side $L$, so $V=L^3$ and each momentum component is $p_i=2\pi \hslash j_i/L$ for an integer $j_i$. Each allowed momentum point occupies a cell of volume $(2\pi \hslash /L{)}^3$ in momentum space. Divide the sphere volume $4\pi p_F^3/3$ by this cell volume and multiply by two for the two spin states. Replacing the discrete count with a volume is the large-box approximation; surface and shell effects are neglected.^[ref]

The factor two is not automatic for every Fermi gas. If only one internal state is populated, its number density $n_{\sigma }$ instead satisfies $p_{F,\sigma }=\hslash (6{\pi }^2{n}_{\sigma }{)}^{1/3}$. For two equally populated states, $n_{\sigma }=n/2$ recovers the displayed result. A Fermi surface in a crystal can have a more complicated shape, so a single spherical Fermi momentum is a property of this isotropic model.

The group velocity of an excitation evaluated on the Fermi surface. For an isotropic dispersion relation $ϵ(p)$, $v_F=(dϵ/dp{)}_{p_F}$, where $p_F$ is the Fermi momentum.

The Fermi wavevector is the radius of the occupied sphere in wavevector coordinates for an isotropic Fermi gas. In the uniform two-component electron model, $k_F=(3{\pi }^{2}n{)}^{1/3}$ and $E_F={\hslash }^2{k}_F^2/(2m_e)$ for nonrelativistic particles. Its units are inverse length, whereas the Fermi momentum has units of momentum.

A Landau quasiparticle is a fermionic excitation connected continuously to a Fermi gas particle when interactions preserve the phase. Near an isotropic Fermi surface, its energy relative to the chemical potential is ${\xi }_p={ϵ}_p-\mu \simeq v_F(p-p_F)$, with Fermi momentum $p_F$ and Fermi velocity $v_F$.^[ref]

The Pauli exclusion principle restricts decay. For a three-dimensional normal Fermi liquid with regular short-range interactions, a representative low-energy lifetime model isHere $E_{\ast }$ is an energy scale and $A$ is dimensionless; sharpness requires $\hslash /{\tau }_p\ll |{\xi }_p|$ at zero temperature.^[ref]

For an original example, stipulate $A=1$ and $T=0$. At $|{\xi }_p|/E_{\ast }=0.02$, the decay energy is $0.0004E_{\ast }$, just $2\%$ of the excitation energy; halving $|{\xi }_p|$ reduces that ratio to $1\%$. Approaching the Fermi surface improves the particle description even though the microscopic interaction has not weakened. This quadratic illustration is three-dimensional; low-dimensional corrections require separate analysis.

A fully ionized helium nucleus supplies two electrons for approximately $4m_p$ of mass, so $n_e=\rho /(2m_p)$. Filling all electron momentum states up to the Fermi momentum givesFor a nonrelativistic electron, the speed is $p/m_e$. The three mean squared Cartesian components of momentum are equal by rotational symmetry and sum to $p^2$, so each is $p^2/3$. This is the isotropic tensor integral. Thus the momentum flux, or electron degeneracy pressure, isSubstituting the expression for $p_0$ and simplifying the powers of two givesHere $h$ is the Planck constant. This is the pure-helium specialization of nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture.

For a cold helium white dwarf, neglect ion thermal pressure and use this electron degeneracy pressure as the support against Newtonian gravity. Combining hydrostatic equilibrium with mass conservation givesSet $\rho ={\rho }_c{\theta }^{3/2}$ and $r=\alpha \xi$, where ${\alpha }^2=5K{\rho }_c^{-1/3}/(8\pi G)$. Direct substitution gives the Lane-Emden equationIts dimensionless solution is independent of ${\rho }_c$. Let ${\xi }_1$ be its first zero and ${\omega }_1=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)={\int }_0^{{\xi }_1}{\xi }^2{\theta }^{3/2}d\xi$. Integrating mass conservation givesEliminating ${\rho }_c$ yields the polytropic mass-radius relationThus a more massive nonrelativistic white dwarf is smaller at fixed stellar composition. The assumptions require $p_0\ll m_{e}c$ and $k_{B}T\ll p_0^2/(2m_e)$ in the region providing most support; the law is not extended into the relativistic or thermally supported regimes.

To derive the nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture, take fully ionized hydrogen and helium with hydrogen mass fraction $X$, the electron number density isHere helium nuclei have mass $4m_p$ in the approximation used in the question. Filling the two spin states of each momentum mode up to the Fermi momentum $p_0$ givesThe isotropic momentum flux gives pressure $\frac{1}{3}\int pvdn$. For nonrelativistic electrons $v=p/m_e$, so the electron degeneracy pressure isSubstitute the composition-dependent $n_e$ and simplify powers of two:Comparing $5/3=1+1/n$ gives the polytropic index $n=3/2$. The result assumes complete electron degeneracy, $p_0\ll m_{e}c$, and negligible nuclear mass corrections.

Now adopt the stated approximate partially degenerate equation of state, keeping composition and hence $K,\mu$ fixed during contraction. Set $q=m/M$ and define the mass-averaged temperature ${\overline{T}}_M={\int }_0^{1}Tdq$. The stellar virial theorem from part (a) givesFor an index-$3/2$ stellar polytrope, the permitted gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. Also, using $\bar{\rho }=3M/(4\pi R^3)$,Thus, with $A=3K{\int }_0^1[3\rho /(4\pi \bar{\rho }){]}^{2/3}dq$,The normalized density $\rho /\bar{\rho }$ is a fixed function of $q$ for this homologous polytropic structure. Therefore its integral and $K$ fix $A$, which is independent of $M$ and $R$ for a common composition. The mean here is mass-weighted, rather than a volume average.

For the mass-averaged temperature maximum in a partially degenerate polytrope, put $\alpha =2\mu GM/(7\mathcal{R})$ and $\delta =\mu A{M}^{2/3}/(3\mathcal{R})$, both positive. As a function of $s=1/R$ the temperature is the concave quadraticIt reaches its maximum when $R=2\delta /\alpha =7A{M}^{-1/3}/(3G)$, withAt large radius, contraction increases the temperature as in a thermally supported star. Once electron degeneracy provides a sufficiently large fraction of the support, further contraction can lower the thermal temperature. The formal zero-temperature radius is $R_{\mathrm{deg}}=\delta /\alpha =R_{\max }/2$, with $R_{\mathrm{deg}}\propto M^{-1/3}$. The continuation to negative temperatures would be unphysical; the adopted sequence only describes its positive-temperature part.

The contraction-limited central temperature has the same mass scaling. Indeed, if the total polytropic pressure is $P=K_{\mathrm{tot}}{\rho }^{5/3}$, the approximate equation of state gives $T=(\mu /\mathcal{R})(K_{\mathrm{tot}}-K){\rho }^{2/3}$. The fixed density profile therefore fixes $T_c/{\overline{T}}_M$. A sufficiently small mass never reaches the central temperature required for sustained hydrogen burning, producing a hydrogen-burning minimum mass. Such an object becomes a cooling, degeneracy-supported brown dwarf instead of settling onto the main sequence. More massive contracting objects can ignite hydrogen before reaching this degeneracy-limited maximum. This conclusion uses the nonrelativistic, fixed-composition model; it does not require extrapolation into the relativistic regime.

For an isotropic degenerate electron gas, pressure is one third of the momentum flux. In the ultrarelativistic limit each occupied mode has speed $c$, so the supplied momentum-space state count givesA fully ionized helium nucleus provides two electrons for approximately $4m_p$ of mass, so $n_e=\rho /(2m_p)$. Eliminating the Fermi momentum $p_0=(3h^3\rho /(16\pi m_p){)}^{1/3}$ therefore givesThe two-spin-state degeneracy is already included in the supplied state count; no further factor 2 is needed.

If electron degeneracy pressure supplies the star's pressure, this is an index-three stellar polytrope. Its central pressure and density obeyThe radius cancels, leaving $M^{2/3}=K(3a_3/(4\pi ){)}^{4/3}/(c_{3}G)$. Thus the Chandrasekhar mass from central polytropic constants isThe dependence on microscopic constants reflects the balance of fermionic pressure and gravity. The absence of radius dependence is special to the limiting $4/3$ pressure exponent. Ordinary partially relativistic white dwarfs can have a range of masses below this Chandrasekhar limit; the formula does not force every white dwarf to have this mass. As a degenerate stellar core grows toward the limiting mass, compression cannot provide an arbitrarily larger supporting pressure relative to gravity. Above it the ideal cold, nonrotating electron-supported configuration cannot remain in equilibrium; nuclear ignition, collapse or other changes in the support mechanism must intervene. Which outcome occurs depends on the core's composition and evolutionary state.

As $T\downarrow 0$, the Fermi-Dirac distribution tends to a step function: $\bar{n}(p)=1$ for $ϵ(p)<{\mu }_0$, and $0$ for $ϵ(p)>{\mu }_0$. The value at equality is immaterial to the integrals. The Fermi momentum is defined by $ϵ(p_F)={\mu }_0$: the Pauli exclusion principle fills the lowest available states and forbids multiple occupation of a state with a fixed spin. Thus the sketch is a horizontal line at one up to $p_F$, followed by a jump to zero.

In the ultrarelativistic, $ϵ(p)\simeq cp$, givingEliminate $p_F$ to obtainThe supplied expression for $ϵ$ includes rest energy; in this limit its leading term is also the leading kinetic energy.

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.

Let $f(\mathbf{p})d^{3}p$ be particle number per unit volume in a momentum element, including the spin-state count in $f$. This is the local momentum part of a phase-space distribution function, normalized by the number density $n=\int f(\mathbf{p})d^{3}p$. For an isotropic distribution with particle energy $E(p)=\sqrt{m^2{c}^4+p^2{c}^2}$, the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor $p\ll mc$, $E-m{c}^2=p^2/(2m)$ and $\mathbf{v}=\mathbf{p}/m$, giving $P=2u_{\mathrm{kin}}/3$. For $p\gg mc$, $E\simeq pc$ and $v\simeq c$, giving $P\simeq u_{\mathrm{kin}}/3$. ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.

For a classical Maxwell-Boltzmann distribution, $f=n(2\pi m{k}_{B}T{)}^{-3/2}\exp [-p^2/(2m{k}_{B}T)]$. Three Gaussian component integrals give $⟨p^2⟩=3m{k}_{B}T$, so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$, where the mean molecular weight counts ions and free electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has $P=n{k}_{B}T$ but $u_{\mathrm{kin}}\simeq 3n{k}_{B}T$; its momentum distribution is proportional to $e^{-pc/(k_{B}T)}$ instead of the nonrelativistic Gaussian.

For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse $x=pc/(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation constant and photon equation of state are thereforeLikewise $n_{\gamma }=2\zeta (3)(k_{B}T/\hslash c{)}^3/{\pi }^2$. Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.

For fully degenerate electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere ${\mu }_e$ is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to $p_F$:Its two limits areThe corresponding kinetic energy densities are $3n_e[p_F^2/(2m_e)]/5$ and $3n_e{p}_{F}c/4$, respectively. Thus in the high-density relativistic limit the electron pressure scales as ${\rho }^{4/3}$ and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.

The boundaries on a stellar equation-of-state regime diagram concern the electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate electron gas is nonrelativistic well below ${\rho }_{\ast }$ and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate electron gas instead becomes thermally relativistic near $T_{\ast }$: this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.

The exact electron Fermi temperature, subtracting electron rest energy, isFor electrons with negligible thermal pairs, strong degeneracy requires $T\ll T_F$, whereas $T\gg T_F$ gives a nondegenerate gas. In the pair-rich regime the actual electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes $2/3$ and $1/3$ explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test $n_e[2\pi {\hslash }^2/(m_e{k}_{B}T){]}^{3/2}\sim 2$ gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.

The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once electrons are strongly degenerate, compare radiation pressure with $P_e(\rho )$ instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is $T=[3P_e(\rho )/a_{\mathrm{rad}}{]}^{1/4}$, with logarithmic slopes $5/12$ and $1/3$ in the nonrelativistic and ultrarelativistic limits. Degeneracy of the electrons and dominance of their pressure are distinct criteria.

For the pair curve, distinguish baryonic net electrons from thermally created electrons and Positrons. Chemical equilibrium with photons requires opposite electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives $n_{-}-n_{+}=n_b=\rho /({\mu }_e{m}_u)$, while the nondegenerate equilibrium product is $n_{-}{n}_{+}=n_0^2$. HencePairs become important when $n_0$ is comparable with or exceeds $n_b$, approximately the electron-positron thermal pair abundance curve ${\rho }_{\mathrm{pair}}(T)={\mu }_e{m}_u{n}_0(T)$. Below its density at a fixed temperature, pairs dominate over the net charge electrons. The exponential makes the low-temperature portion steep in a log-log plot. For ${\mu }_e=2$, the full zero-potential integral gives approximate pair-marker densities $1.80\times 10^3\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $10^9\mathrm{K}$ and $7.52\times 10^5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $3\times 10^9\mathrm{K}$. At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives $n_0\to [3\zeta (3)/(2{\pi }^2)](k_{B}T/\hslash c{)}^3$, so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.

The original diagram uses fully ionized helium, ${\mu }_e=2$ and ${\mu }_{\mathrm{mol}}=4/3$, only to set numerical locations. It plots the full $T_F$ curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For $T\gg T_{\ast }$ with abundant nondegenerate pairs, the combined electron/Positron energy density is $7a_{\mathrm{rad}}{T}^4/4$, so photons plus pairs have $u=11a_{\mathrm{rad}}{T}^4/4$ and $P=u/3$. Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.

Assume zero temperature, complete ionization, constant composition, noninteracting electrons, negligible ion thermal pressure, and Newtonian stellar gravity. Ions supply almost all the mass, while electron degeneracy pressure supplies support. By the Pauli exclusion principle, the two electron spin states fill a momentum sphere up to Fermi momentum $p_F$. Counting states giveswhere ${\mu }_e$ is mean molecular weight per electron. The momentum flux of this isotropic Fermi gas isWriting $x=p_F/(m_{e}c)$ and performing the integral gives the equation of state of a cold electron gasIn the nonrelativistic limit $p_F\ll m_{e}c$, $v(p)\simeq p/m_e$, soIn the ultrarelativistic limit $p_F\gg m_{e}c$, $v(p)\simeq c$, soThese are polytropic equations of state with indices $3/2$ and $3$, respectively. The relativistic softening underlies the Chandrasekhar mass limit; Coulomb corrections, thermal effects, rotation, and general relativity are excluded from this idealized derivation.