Relativistic Degenerate Fermi Gas

The ideal Fermi gas at $T=0$ fills every momentum state up to the Fermi momentum $p_F$ , forming a sharp Fermi sphere. The elementary treatment uses the non-relativistic dispersion $\varepsilon=p^2/2m$ — valid only while typical momenta stay small compared with $mc$ . But push the density up, by compression in the lab or by gravity in a stellar core, and the Fermi momentum climbs past $mc$ . Then the kinematics must be treated relativistically, through $\varepsilon(p)=\sqrt{p^2c^2+m^2c^4}$ , and the physics changes character. The change shows up in the equation of state. The pressure of a degenerate gas is the momentum flux of its fastest fermions, which depends on their velocity $v=\partial\varepsilon/\partial p$ . Non-relativistically $v\propto p$ and one finds $P\propto n^{5/3}$ . Ultra-relativistically the velocity saturates at $c$ , the dispersion straightens to $\varepsilon\to pc$ , and the pressure softens to $P\propto n^{4/3}$ — a weaker response to compression. The polytropic index drops from $\Gamma=\tfrac53$ to $\Gamma=\tfrac43$ , and $\tfrac43$ happens to be the exact threshold for the hydrostatic stability of a self-gravitating sphere. A small change in kinematics becomes a life-or-death matter for a star. The full $T=0$ thermodynamics is captured by Chandrasekhar's parametric functions of a single variable $x_F=p_F/mc$ : exact closed forms for the energy density, pressure and chemical potential that connect the two limits smoothly. Although strictly a $T=0$ theory, this is precisely the physically relevant regime for white dwarfs and neutron-star matter, where the Fermi temperature is enormous compared with the actual temperature and the gas is deeply degenerate. This concept is the bridge from equilibrium statistical mechanics to stellar astrophysics — the same machinery of Fermi surface, density of states and degeneracy pressure, now with relativistic kinematics — and its softened $\Gamma=\tfrac43$ equation of state is what fixes the Chandrasekhar mass, the topic this article hands forward by name.

Key takeaways

A relativistic degenerate Fermi gas is a fully filled Fermi sphere whose dispersion $\varepsilon(p)=\sqrt{p^2c^2+m^2c^4}$ interpolates between the non-relativistic $p^2/2m$ at low density and the linear, photon-like $pc$ at ultra-high density.
In the ultra-relativistic limit the pressure scales as $P\propto n^{4/3}$ rather than the non-relativistic $n^{5/3}$ ; the softer equation of state ( $\Gamma=\tfrac43$ vs $\tfrac53$ ) is what makes self-gravitating degenerate stars marginally unstable.
At $T=0$ everything is fixed by $p_F=\hbar(3\pi^2 n)^{1/3}$ : energy density, pressure and chemical potential are exact integrals of the single dimensionless variable $x_F=p_F/mc$ .
The ultra-relativistic gas is kinematically identical to a gas of massless particles, so it obeys the radiation-like equation of state $P=u/3$ — the same $\tfrac13$ as a photon gas, from kinematics, not statistics.
Because the ultra-relativistic index sits exactly at the gravitational stability threshold $\Gamma=\tfrac43$ , the theory culminates in the Chandrasekhar mass $M_{\mathrm{Ch}}\approx1.44\,M_\odot$ (derived in the white-dwarf article).
Key idea: replace the non-relativistic kinetic energy by $\varepsilon(p)=\sqrt{p^2c^2+m^2c^4}$ in the same filled Fermi sphere. Only the dispersion changes, but it softens the equation of state from $P\propto n^{5/3}$ to $P\propto n^{4/3}$ , with profound consequences for gravitational stability.

The core insight

Replace the non-relativistic kinetic energy by $\varepsilon(p)=\sqrt{p^2c^2+m^2c^4}$ in the same fully-occupied Fermi sphere. The only thing that changes is the dispersion, but it softens the equation of state from $P\propto n^{5/3}$ to $P\propto n^{4/3}$ , with profound consequences for gravitational stability.

Where this lives today

The relativistic degenerate Fermi gas is the baseline equation of state for every compact star. In a white dwarf, as the mass increases the central electrons cross $x_F\gtrsim1$ , the equation of state softens to $P\propto\rho^{4/3}$ , and at $\Gamma=\tfrac43$ the star becomes only marginally stable — yielding the Chandrasekhar mass $M_{\mathrm{Ch}}\approx1.44\,M_\odot$ , derived in White Dwarf Stars and the Chandrasekhar Limit. Deep in a neutron star, degenerate neutrons (plus a small fraction of relativistic electrons) reach the relativistic regime at supranuclear densities, and this ideal gas is the non-interacting scaffold onto which strong-interaction corrections are added to model the maximum mass. At the most extreme densities — the cores of the heaviest neutron stars, or hypothetical quark stars — deconfined quarks form an ultra-relativistic degenerate Fermi gas with $P\approx u/3$ , and the same kinematic softening governs whether such matter can support a stable star. Every one of these problems begins with the single substitution $\varepsilon=p^2/2m\to\sqrt{p^2c^2+m^2c^4}$ .

Origins · 1 minEquation of state and how this was discovered

Equation of state

the pressure–density–temperature relation of matter · the relativistic softening that decides a star's fate

An equation of state relates a system's pressure to its density (and temperature). For a degenerate Fermi gas at $T=0$ it is a pure pressure–density relation, $P(n)$ , fixed entirely by the dispersion. The relativistic degenerate gas has the defining feature that its equation of state softens as density rises: from the stiff $P\propto n^{5/3}$ (polytropic index $\Gamma=\tfrac53$ ) to the marginal $P\propto n^{4/3}$ ( $\Gamma=\tfrac43$ ), the exact threshold for gravitational stability. Because a star's structure is set by hydrostatic equilibrium acting on its equation of state, this softening is not a detail but the physics that caps the masses of white dwarfs and neutron stars.

History

1926

— Enrico Fermi and Paul Dirac introduce Fermi–Dirac statistics; Fowler applies it to white dwarfs.

1930

— Subrahmanyan Chandrasekhar treats the electrons relativistically and finds the softened equation of state.

1932

— Lev Landau independently argues for a maximum mass of degenerate stars from the same relativistic softening.

1935

— Chandrasekhar publishes the full relativistic equation of state and its functions f(x) and χ(x).

1939

— Oppenheimer and Volkoff apply the relativistic degenerate gas to neutron stars via the TOV equation.

The physical picture

Picture the Fermi sphere being squeezed. Adding a fermion costs energy $\varepsilon(p_F)$ , and the wall $p_F$ grows as $n^{1/3}$ . Non-relativistically the cost $\propto p_F^2\propto n^{2/3}$ rises steeply; ultra-relativistically the cost is only $\propto p_F\propto n^{1/3}$ , so the gas pushes back more weakly. That softness is the whole story — and it is set purely by the shape of the dispersion, recalled from Degenerate Fermi Gas at Zero Temperature and applied here with relativistic kinematics. The velocity of the fastest fermions saturates at $c$ . Since pressure is essentially momentum flux $\sim n\,p_F\,v_F$ , once $v_F\to c$ the flux can no longer grow as fast with density, so $P$ climbs only as $n^{4/3}$ . The punchline figure below tracks the ratio $P/u$ (pressure to kinetic energy density) as the gas is compressed: it starts at $\tfrac23$ (non-relativistic) and falls to $\tfrac13$ (ultra-relativistic) — the softening, with the ultra-relativistic value being exactly the photon-gas number.

Figure

Pressure-to-energy-density ratio P/u versus x_F = p_F/mc — starts at 2/3 (non-relativistic) and softens to 1/3 (ultra-relativistic), the radiation value

The ultra-relativistic gas is kinematically identical to a gas of massless particles, so it behaves like radiation: $P=u/3$ . This 'photon-gas-like' equation of state is exactly the marginal one for gravitational binding — which is why the relativistic correction is not a small refinement but the deciding factor in stellar collapse. And whether a given star is relativistic is decided entirely by $x_F=p_F/mc$ : a race between density (which raises $p_F$ ) and rest mass (which sets the scale $mc$ ). For electrons $mc$ is small, so white-dwarf electrons go relativistic at modest densities while the far heavier ions stay non-relativistic.

Figure

The two dispersions — non-relativistic ε = p²/2m (steep parabola) versus relativistic ε = √(p²c² + m²c⁴), which straightens to the linear ε → pc at large p

The core equations

The single-particle energy and velocity are the relativistic dispersion; everything downstream follows from them. Non-relativistically the velocity grows without bound as $p/m$ ; relativistically it saturates at $c$ .

\varepsilon(p)=\sqrt{p^2c^2+m^2c^4}\,,\qquad v=\frac{\partial\varepsilon}{\partial p}=\frac{pc^2}{\varepsilon} \qquad (1)

where: $\varepsilon$ — single-particle energy including rest mass; $m$ — rest mass; $v$ — group velocity, $\to c$ as $p\to\infty$ . It is this velocity saturation, not any change in the statistics, that softens the pressure. Counting states in the filled sphere at $T=0$ fixes $p_F$ from the density, and the relativity parameter $x_F$ measures where on the dispersion the Fermi surface sits.

n=\frac{g\,p_F^3}{6\pi^2\hbar^3},\qquad p_F=\hbar\Big(\tfrac{6\pi^2 n}{g}\Big)^{1/3},\qquad x_F\equiv\frac{p_F}{mc} \qquad (2)

where: $n$ — number density; $g$ — spin degeneracy ( $g=2$ for electrons, giving $p_F=\hbar(3\pi^2 n)^{1/3}$ ); $x_F$ — relativity parameter, $x_F\ll1$ non-relativistic, $x_F\gg1$ ultra-relativistic. Read $x_F$ as the single knob that tunes the gas from the $\tfrac53$ to the $\tfrac43$ polytrope. The pressure is the momentum flux, $\tfrac13 n\langle p v\rangle$ , integrated over the sphere with the relativistic velocity. It reduces to a closed form built from Chandrasekhar's function $f(x)$ .

P=\frac{g}{6\pi^2\hbar^3}\int_0^{p_F}\frac{p^4 c^2}{\sqrt{p^2c^2+m^2c^4}}\,\mathrm{d}p=\frac{g\,m^4c^5}{24\pi^2\hbar^3}\,f(x_F),\qquad f(x)=x\sqrt{1+x^2}\,(2x^2-3)+3\,\mathrm{arcsinh}\,x \qquad (3)

where: $P$ — degeneracy pressure; $f(x)$ — Chandrasekhar pressure function. The $\tfrac13$ inside is the isotropy average; substituting $p=mc\sinh\theta$ turns the integral into the elementary $f(x)$ . Its two asymptotes, $f\to\tfrac85 x^5$ and $f\to2x^4$ , are exactly the $n^{5/3}$ and $n^{4/3}$ laws. The energy density is the companion integral, written through the Chandrasekhar energy function $\chi(x)$ , and the chemical potential at $T=0$ is simply the Fermi energy.

u=\frac{g\,m^4c^5}{8\pi^2\hbar^3}\,\chi(x_F),\quad \chi(x)=8\!\int_0^{x}\!t^2\sqrt{1+t^2}\,\mathrm{d}t,\qquad \mu=\varepsilon_F=mc^2\sqrt{1+x_F^2} \qquad (4)

where: $u$ — energy density; $\chi(x)$ — Chandrasekhar energy function, written here as its defining integral (its memorized closed form is easy to garble — the base export did exactly that); $\mu$ — chemical potential, equal to the Fermi energy at $T=0$ . In the ultra-relativistic limit $\chi\to2x^4$ and $u\propto p_F^4\propto n^{4/3}$ , matching $P=u/3$ . Finally, the ultra-relativistic equation of state — the payoff — is the radiation relation with the marginal polytropic index.

P=\frac{u}{3}\quad(x_F\gg1),\qquad \Gamma\equiv\frac{n}{P}\frac{\mathrm dP}{\mathrm dn}\to\frac{4}{3} \qquad (5)

where: $\Gamma$ — polytropic/adiabatic index; $\tfrac43$ is the gravitational stability threshold. The identity $P=u/3$ is kinematic (from $\varepsilon=pc$ , $v=c$ ), which is why it matches a photon gas exactly despite the opposite statistics.

Limiting cases

All limits follow from the single parameter $x_F=p_F/mc$ :

Case	Result	Meaning
Non-relativistic ( $x_F\ll1$ )	$P\propto n^{5/3}$ , $\Gamma=\tfrac53$ , $P=\tfrac23 u_{\mathrm{kin}}$	Recovers the ordinary degenerate Fermi gas. Pressure stiff against compression; supports a star at any mass.
Ultra-relativistic ( $x_F\gg1$ )	$P\propto n^{4/3}$ , $\Gamma=\tfrac43$ , $P=\tfrac13 u$	Dispersion linear, $\varepsilon\to pc$ . Marginal gravitational stability; leads to the Chandrasekhar mass.
Massless fermions ( $m\to0$ )	$P=u/3$ for all densities	Exactly the ultra-relativistic limit at every density; same equation of state as radiation but with Fermi statistics.
Finite $T\ll T_F$	$P(T)=P(0)[1+\mathcal O((k_BT/\varepsilon_F)^2)]$	Sommerfeld corrections tiny in stellar interiors ( $T_F\sim10^9$ – $10^{10}\,$ K); the $T=0$ analysis suffices.

Derivation AUltra-relativistic equation of state and the radiation-like P equals u over three

The goal is to prove that when $p_F\gg mc$ the degenerate gas becomes kinematically equivalent to a gas of massless fermions, obeying $P=u/3$ and $P\propto n^{4/3}$ . We set $m\to0$ so that $\varepsilon=pc$ and $v=c$ for every occupied state, then evaluate the energy-density and pressure integrals — which now become elementary — and take their ratio. The surprise is that the answer is identical to a photon gas even though we are dealing with fermions.

Step 1. Energy density. With $\varepsilon=pc$ , $u=\tfrac{g}{2\pi^2\hbar^3}\int_0^{p_F}(pc)p^2\,\mathrm dp=\tfrac{g c\,p_F^4}{8\pi^2\hbar^3}$ . Why: the linear dispersion makes the integrand $p^3$ , which integrates to $p_F^4/4$ — the energy density scales as the fourth power of the Fermi momentum.

Step 2. Pressure. With $v=c$ , $P=\tfrac{g}{6\pi^2\hbar^3}\int_0^{p_F}\tfrac{p^4c^2}{pc}\,\mathrm dp=\tfrac{g c\,p_F^4}{24\pi^2\hbar^3}$ . Why: setting $v=c$ collapses the pressure integrand to $p^3 c$ , the same integral as the energy up to the overall $\tfrac13$ from the isotropy average.

Step 3. The radiation relation. Dividing, $P/u=(1/24)/(1/8)=\tfrac13$ . Why: exactly the photon-gas relation — but produced here by relativistic kinematics, not by any assumption about statistics. Massless particles of any spin give $P=u/3$ .

Step 4. Density scaling. Since $p_F\propto n^{1/3}$ , $P\propto p_F^4\propto n^{4/3}$ , hence $\Gamma=\tfrac43$ . Why: the fourth power of $p_F$ combined with $p_F\propto n^{1/3}$ gives the soft $\tfrac43$ index — the marginal case for gravitational stability, and the reason relativity caps stellar masses. *Verified: $u=gcp_F^4/8\pi^2\hbar^3$ , $P=gcp_F^4/24\pi^2\hbar^3$ , the ratio $P/u=1/3$ exactly, and the $n^{4/3}$ exponent ( $\mathrm d\ln P/\mathrm d\ln n=4/3$ ) are all checked symbolically in verify_relativistic-degenerate-fermi-gas.py.*

Go deeper · Derivation B · 1 minFull parametric equation of state from the Chandrasekhar functions

Derivation BFull parametric equation of state from the Chandrasekhar functions

The goal is the exact $T=0$ equation of state that interpolates between $n^{5/3}$ and $n^{4/3}$ , expressed through Chandrasekhar's functions $f(x_F)$ for the pressure and $\chi(x_F)$ for the energy density. The strategy is the substitution $p=mc\sinh\theta$ , which rationalizes the square root in the dispersion and turns both integrals into elementary hyperbolic ones; we then read off the closed form for $f$ and verify its two asymptotic limits.

Step 1. Pressure integral. Substituting $p=mc\sinh\theta$ , $\mathrm dp=mc\cosh\theta\,\mathrm d\theta$ , and $\varepsilon=mc^2\cosh\theta$ turns the pressure integral into $P=\tfrac{g m^4c^5}{6\pi^2\hbar^3}\int_0^{\theta_F}\sinh^4\theta\,\mathrm d\theta$ . Why: the square root becomes $\cosh\theta$ , cancelling cleanly, and $p^4\,\mathrm dp$ becomes $\sinh^4\theta\cosh\theta\,\mathrm d\theta$ ; the $\cosh$ in the denominator leaves a pure $\sinh^4$ .

Step 2. Closed form. The standard reduction of $\int\sinh^4\theta\,\mathrm d\theta$ gives Chandrasekhar's $f(x)=x\sqrt{1+x^2}(2x^2-3)+3\,\mathrm{arcsinh}\,x$ , with $x=\sinh\theta_F=x_F$ . Why: this is the exact pressure function; unlike a memorized numeric it is an algebraic identity you can differentiate to recover the integrand.

Step 3. Energy function. The energy density integral, $u=\tfrac{g m^4c^5}{8\pi^2\hbar^3}\chi(x_F)$ with $\chi(x)=8\int_0^x t^2\sqrt{1+t^2}\,\mathrm dt$ , follows the same substitution with $\varepsilon\,p^2\,\mathrm dp$ . Why: keeping $\varepsilon=mc^2\cosh\theta$ in the numerator gives $\sinh^2\theta\cosh^2\theta$ ; we present $\chi$ as its defining integral to avoid the garbled closed form the AllQFT base article carried.

Step 4. Asymptotic limits. Expanding, $f\to\tfrac85 x^5$ as $x\to0$ (giving $P\propto n^{5/3}$ ) and $f\to2x^4$ as $x\to\infty$ (giving $P\propto n^{4/3}$ ). Why: these two limits are exactly the non-relativistic and ultra-relativistic polytropes, so the single function $f$ contains both regimes and the smooth crossover between them. *Verified: $f(x)=8\int t^4/\sqrt{1+t^2}\,\mathrm dt$ equals the closed form; the asymptotes $f\to\tfrac85 x^5$ and $f\to2x^4$ ; and $\chi'(x)=8x^2\sqrt{1+x^2}$ with $\chi\to2x^4$ at large $x$ are all checked symbolically in verify_relativistic-degenerate-fermi-gas.py.*

Worked examples

Example 1 — The density at which electrons become relativistic.

Step 1. Relativity sets in at $x_F=1$ , i.e. $p_F=m_e c$ . From $p_F=\hbar(3\pi^2 n)^{1/3}$ ( $g=2$ ), invert: $n=(m_e c/\hbar)^3/(3\pi^2)$ . With $m_e c/\hbar=(9.109\times10^{-31})(2.998\times10^{8})/(1.055\times10^{-34})=2.59\times10^{12}\ \mathrm{m^{-1}}$ , cube it: $(2.59\times10^{12})^3=1.74\times10^{37}$ .

Step 2. Divide by $3\pi^2=29.6$ : $n=1.74\times10^{37}/29.6=\mathbf{5.9\times10^{35}\ m^{-3}}$ . This is the electron density at which the degenerate gas crosses from the $\tfrac53$ to the $\tfrac43$ polytrope.

Step 3. Converting to mass density with $\mu_e=2$ : $\rho=2\times(1.673\times10^{-27})\times(5.9\times10^{35})\approx2.0\times10^{9}\ \mathrm{kg\,m^{-3}}=2\times10^6\ \mathrm{g\,cm^{-3}}$ — a typical white-dwarf core density, confirming that white-dwarf electrons are precisely on the boundary between the two regimes.

Example 2 — Confirming P = u/3 numerically across the crossover.

Step 1. Compute $P/u_{\mathrm{kin}}$ from the exact integrals at a deeply non-relativistic point, $x_F=0.01$ : numerically integrating $P=\tfrac13\int p^4 c^2/\varepsilon\,\mathrm dp$ and $u_{\mathrm{kin}}=\int(\varepsilon-mc^2)p^2\,\mathrm dp$ up to $p_F=0.01\,m_ec$ gives $P/u_{\mathrm{kin}}=\mathbf{0.667}\approx\tfrac23$ .

Step 2. Repeat deep in the ultra-relativistic regime, $x_F=100$ : the same integrals give $P/u_{\mathrm{kin}}=\mathbf{0.338}\approx\tfrac13$ .

Step 3. So the ratio softens from $\tfrac23$ to $\tfrac13$ as the gas is compressed — exactly the curve of the punchline figure. The ultra-relativistic value equals the photon-gas $\tfrac13$ , the kinematic fingerprint of the linear dispersion, and the source of the $\Gamma=\tfrac43$ instability that fixes the Chandrasekhar mass.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., §8.4 (relativistic Fermi gas, white dwarfs) · Chandrasekhar, An Introduction to the Study of Stellar Structure, Ch. X–XI · Kardar, Statistical Physics of Particles, Ch. 7 · Landau & Lifshitz, Statistical Physics Part 1, §61, §106 · Shapiro & Teukolsky, Black Holes, White Dwarfs and Neutron Stars, Ch. 2–3.

Provenance. Dispersion, $p_F$ , $u$ , $P$ and the Chandrasekhar functions $f(x)$ , $\chi(x)$ follow Pathria & Beale §8.4 and Chandrasekhar Ch. X–XI. This article owns the relativistic equation of state (the softening $n^{5/3}\to n^{4/3}$ , the radiation-like $P=u/3$ , and the Chandrasekhar functions), recalls the $T=0$ Fermi sphere from Degenerate Fermi Gas at Zero Temperature and Degeneracy Pressure of Fermions by name, and forward-teases the Chandrasekhar mass $M_{\mathrm{Ch}}\approx1.44\,M_\odot$ to White Dwarf Stars and the Chandrasekhar Limit — which owns its full derivation. Presentation note: the AllQFT base carried a garbled fragment for the energy function $\chi(x)$ ; here $\chi$ is stated via its provably-correct defining integral, $\chi(x)=8\int_0^x t^2\sqrt{1+t^2}\,\mathrm dt$ , machine-verified. The asymptotics $f\to\tfrac85 x^5$ , $f\to2x^4$ , and $P=u/3$ are all recomputed in verify_relativistic-degenerate-fermi-gas.py.

Builds on: Degenerate Fermi Gas at Zero Temperature · Fermi Energy, Momentum and Fermi Temperature · Density of States for Free Particles · Degeneracy Pressure of Fermions · Quantum Statistics of Ideal Gases: Occupation Numbers Leads to: White Dwarf Stars and the Chandrasekhar Limit · Equation of State of Dense Matter · Sommerfeld Expansion

Why is P = u/3 for an ultra-relativistic gas the same as for photons, even though electrons are fermions?

Because the relation is purely kinematic. Pressure is $\tfrac13 n\langle p v\rangle$ and energy density is $n\langle\varepsilon\rangle$ ; when the dispersion is linear ( $\varepsilon=pc$ , $v=c$ ) the two integrals differ only by the isotropy factor $\tfrac13$ , giving $P=u/3$ regardless of whether the occupation numbers follow Bose or Fermi statistics. Statistics change how the states are filled — a step function for degenerate fermions, a Planck distribution for photons — but not the $P=u/3$ ratio, which depends only on $\varepsilon\propto p$ . Any gas of effectively massless particles obeys it.

Does 'ultra-relativistic degenerate gas' mean the particles are moving near the speed of light thermally?

No — it means the particles at the top of the Fermi sea have momenta $p_F\gg mc$ , so their velocities approach $c$ , entirely because of degeneracy (Pauli filling), not temperature. The gas is at $T=0$ ; there is no thermal motion at all. The fastest particles are fast because exclusion forced them into high-momentum states, and 'ultra-relativistic' simply says $p_F c$ dominates $mc^2$ for those states. It is degeneracy-driven, not heat-driven, relativity.

How can the same particle be non-relativistic in one star and relativistic in another?

It depends entirely on the density through $x_F=p_F/mc=(\hbar/mc)(3\pi^2 n)^{1/3}$ . In a low-mass white dwarf the electron density is modest, $x_F\ll1$ , and the electrons are non-relativistic. In a massive white dwarf approaching the Chandrasekhar limit the central density is far higher, $x_F\gtrsim1$ , and the same electrons are ultra-relativistic. The rest mass $m$ is fixed; what changes is how tightly the gas is packed, which sets where on the dispersion curve the Fermi surface sits.

Why present χ(x) as an integral instead of its closed form?

Because the closed form is easy to state incorrectly — and the AllQFT source we upgraded from did exactly that, carrying a garbled fragment. The defining integral $\chi(x)=8\int_0^x t^2\sqrt{1+t^2}\,\mathrm dt$ is unambiguous: you can differentiate it to recover the integrand $\chi'(x)=8x^2\sqrt{1+x^2}$ and check its limits directly. For the pressure we do give the closed form $f(x)$ , because it is short and we verified it symbolically. When a memorized closed form is long and error-prone, quoting the integral is the honest, checkable choice.

Is PV = ⅔E still true for a relativistic degenerate gas?

Only in the two clean limits. Non-relativistically ( $\varepsilon\propto p^2$ , density of states $\propto\varepsilon^{1/2}$ ) you get $PV=\tfrac23 E_{\mathrm{kin}}$ exactly; ultra-relativistically ( $\varepsilon\propto p$ , density of states $\propto\varepsilon^2$ ) you get $PV=\tfrac13 E$ . In between, at general $x_F$ , neither simple fraction holds — the exact relation is carried by the Chandrasekhar functions, and the thermodynamic identity is $u_{\mathrm{tot}}+P=n\mu$ with $\mu=mc^2\sqrt{1+x_F^2}$ . The $\tfrac23$ and $\tfrac13$ are the endpoints of a smooth crossover, not a universal law.

Does this gas ever produce electron–positron pairs at high enough density?

Not within this treatment, and for a good reason: pair production is controlled by the chemical potential relative to $mc^2$ and by temperature, and at $T=0$ with a fixed number of electrons there is no thermal energy to create pairs. The chemical potential $\mu=mc^2\sqrt{1+x_F^2}$ can exceed $mc^2$ , but the gas is a cold, number-conserving degenerate sea. In real hot, dense environments (supernova cores) pair production does matter, but that is a finite-temperature effect outside the scope of the ideal $T=0$ relativistic degenerate gas.