Degeneracy Pressure of Fermions

Cool a gas of identical fermions to absolute zero and, unlike any classical gas, it refuses to fall silent. It pushes outward with a pressure of astronomical size — in a metal, of order $10^{10}$ pascals, a hundred thousand atmospheres — even though nothing is moving thermally at all. This is degeneracy pressure, and it is one of the most direct mechanical consequences of the Pauli exclusion principle anywhere in physics. The reason is that fermions cannot pile into the lowest state the way bosons do. At $T=0$ the gas settles into a filled Fermi sphere: every momentum state up to $p_F$ occupied exactly once per spin, and none above. Filling states means occupying fast-moving ones, and fast particles carry momentum across any surface — which is precisely what pressure measures. The pressure does not vanish as $T\to 0$ ; it is the dominant term, with thermal corrections appearing only at order $(T/T_F)^2$ . The size of the pressure is fixed entirely by density and by the dispersion relation. Nonrelativistic fermions, with $\varepsilon = p^2/2m$ , give $P_0 \propto n^{5/3}$ . When $p_F$ grows past $mc$ and the dispersion straightens to $\varepsilon \approx pc$ , the law softens to $P_0 \propto n^{4/3}$ . That single change of exponent is decisive: a $4/3$ law cannot resist gravity without bound, and the balance between softened degeneracy pressure and self-gravity fixes the Chandrasekhar mass of a white dwarf — the topic this article hands forward, recalling it by name. Degeneracy pressure is the macroscopic face of the degenerate Fermi gas. It rests on the Fermi energy, the Fermi surface, and the free-particle density of states, and it supplies the $T=0$ baseline on which electronic heat capacity, the chemical potential's drift, and metallic compressibility are all built.

Key takeaways

At $T=0$ the exclusion principle forces fermions to fill momentum states up to $p_F$ , so even with no thermal motion the gas exerts a finite pressure $P_0 > 0$ .
For a nonrelativistic spin- $\tfrac12$ gas, $P_0 = \tfrac{2}{5} n\,\varepsilon_F = \frac{(3\pi^2)^{2/3}}{5}\frac{\hbar^2}{m} n^{5/3}$ .
The pressure–energy relation is fixed by the dispersion: $PV = \tfrac23 E$ (nonrelativistic, $\varepsilon \propto p^2$ ) and $PV = \tfrac13 E$ (ultrarelativistic, $\varepsilon \propto p$ ).
In the ultrarelativistic limit the law softens to $P_0 \propto n^{4/3}$ , and $P/u$ drops from $\tfrac23$ to $\tfrac13$ — the same watershed that governs a photon gas.
Degeneracy pressure sets the bulk modulus of metals ( $K = \tfrac53 P_0$ ) and supports white dwarfs and neutron stars against gravity.
The $n^{4/3}$ softening produces the Chandrasekhar mass limit (about $1.4\,M_\odot$ for electrons), beyond which no degenerate equilibrium exists.
Key idea: at $T=0$ the exclusion principle forces fermions into high-momentum states; the resulting momentum flux is a finite pressure $P_0 \propto n^{5/3}$ (nonrelativistic) set entirely by density and dispersion, not by temperature — and its relativistic softening to $n^{4/3}$ is what caps the mass of a white dwarf.

The core insight

At zero temperature the Pauli exclusion principle forces fermions into high-momentum states; the resulting momentum flux is a finite degeneracy pressure $P_0 \propto n^{5/3}$ (nonrelativistic), fixed entirely by density and the dispersion relation, and its softening to $n^{4/3}$ in the relativistic limit is what sets the Chandrasekhar mass of a white dwarf.

Where this lives today

Degeneracy pressure is the load-bearing wall of compact-star physics. White dwarfs are held up entirely by electron degeneracy pressure; when a white dwarf in a binary accretes matter and approaches the Chandrasekhar mass, the softened $n^{4/3}$ law can no longer resist gravity and the star detonates as a Type Ia supernova — the "standard candle" whose uniform brightness let cosmologists discover the accelerating expansion of the universe. In neutron stars the same physics repeats with neutrons ( $m = m_n$ in place of $m_e$ ), pushed to nuclear densities, with a relativistically softened maximum mass — the Tolman–Oppenheimer–Volkoff limit — now being pinned down by gravitational-wave observations of merging neutron stars. Closer to the laboratory, the $E_0 \propto N^{5/3}V^{-2/3}$ scaling of degeneracy energy governs trapped ultracold Fermi gases, quantum dots, and the electron liquid in every metal and semiconductor. The Thomas–Fermi model of atoms — the ancestor of modern density-functional theory that underlies most computational chemistry and materials science — is at heart a statement that electron degeneracy pressure balances the electrostatic attraction of the nucleus. Whenever fermions are squeezed, the same exclusion-driven pressure computed here pushes back.

The physical picture

Picture momentum space as a stack of cells, each holding at most $g$ fermions ( $g=2$ for spin- $\tfrac12$ ). At $T=0$ you must fill outward from the origin, because the central cells are already full. You are forced to occupy fast-moving states — not because the particles are hot, but because exclusion leaves nowhere slower to go. Those fast particles stream momentum across any surface, and that streaming is pressure. Bosons at $T=0$ do the opposite: they all collapse into the single lowest state and press with essentially nothing. The same $\hbar$ that lets bosons condense forces fermions apart. Now squeeze the box. Shrinking the volume raises every allowed momentum like $1/L$ , so each occupied state climbs in energy and the total energy rises. Pressure is the gas fighting that climb: $P = -\partial E/\partial V > 0$ . The gas resists compression not through any force between particles but because a smaller box has a higher floor of allowed energies. That is why the pressure is called degeneracy pressure — it is exclusion, not repulsion. The first figure is the whole density law at a glance: pressure versus density on log–log axes. On the left, in the ordinary-metal and low-density-star regime, the line climbs with slope $5/3$ . Where $p_F$ reaches $mc$ the line bends and continues at the shallower slope $4/3$ — the relativistic softening. That kink is the geometry behind the Chandrasekhar limit: a pressure that grows only as $n^{4/3}$ eventually loses to gravity.

Figure

Degeneracy pressure vs density — slope 5/3 (nonrelativistic) bending to slope 4/3 (ultrarelativistic) at the relativistic kink

One landmark makes the two regimes concrete. In an ordinary metal the electrons are firmly nonrelativistic ( $\varepsilon_F \sim$ eV $\ll m_ec^2 = 511$ keV), so the $5/3$ law rules and the pressure is enormous but stable — the bulk modulus $K = \tfrac53 P_0$ is what you feel when you try to compress a solid. In a white-dwarf core the density climbs so high that $p_F$ approaches $mc$ , the line bends to $4/3$ , and the star's fate hangs on that shallower slope.

Figure

Filled Fermi sphere at T = 0 — every momentum state inside |p| ≤ p_F occupied, none outside; the fast states at the edge carry the pressure

The core equations

Pressure at fixed particle number and entropy is minus the volume derivative of the energy. At $T=0$ the entropy is already zero, so this is simply how the ground-state energy responds to squeezing the box.

P = -\left(\frac{\partial E}{\partial V}\right)_{N,S} \qquad (1)

where: $P$ — pressure; $E$ — total internal energy; $V$ — volume; $N$ — particle number (fixed); $S$ — entropy ( $=0$ at $T=0$ ). Read it as: any process that raises the energy when you compress produces a positive, outward pressure. Filling the Fermi sphere fixes its radius. Counting one $\mathbf{k}$ -cell per $(2\pi/L)^3$ of reciprocal space, with $g$ fermions each, and demanding the total be $N$ , gives the Fermi momentum in terms of the density alone.

p_F = \hbar\left(\frac{6\pi^2 n}{g}\right)^{1/3}, \qquad n = \frac{N}{V} \qquad (2)

where: $p_F$ — Fermi momentum (edge of the filled sphere); $n$ — number density; $g$ — spin degeneracy ( $g=2$ for electrons); $\hbar$ — reduced Planck constant. This is owned by Fermi Energy, Momentum and Fermi Temperature and Density of States for Free Particles; we recall it and build on it. Note $p_F \propto n^{1/3}$ — compress the gas and the fastest particle speeds up. The ground-state energy is the dispersion integrated over the filled sphere. Its scaling with volume is all we need to get the pressure, and it depends only on the exponent in $\varepsilon(p)$ .

E_0 = g\,\frac{V}{(2\pi\hbar)^3}\int_{|\mathbf{p}|\le p_F}\varepsilon(p)\,\mathrm{d}^3p \qquad (3)

where: $E_0$ — total energy at $T=0$ ; $\varepsilon(p)$ — single-particle dispersion; $(2\pi\hbar)^3$ — phase-space volume per quantum state. Because $p_F \propto V^{-1/3}$ at fixed $N$ , the whole integral scales as a pure power of $V$ , and that power sets the pressure. The dispersion fixes the pressure–energy relation directly. A $p^2$ law gives one factor; a linear $pc$ law gives another. Both are exact at $T=0$ , and they are the crux of the relativistic watershed.

PV = \frac{2}{3}E \;\;(\text{nonrel.,}\ \varepsilon\propto p^2), \qquad PV = \frac{1}{3}E \;\;(\text{ultrarel.,}\ \varepsilon\propto p) \qquad (4)

where: $PV$ — pressure times volume; $E$ — internal energy. The factor is $s/3$ where $\varepsilon \propto p^s$ : quadratic dispersion ( $s=2$ ) gives $\tfrac23$ , linear ( $s=1$ ) gives $\tfrac13$ . Equivalently $P = (s/3)\,u$ with $u = E/V$ the energy density. For the nonrelativistic gas the energy per particle is a clean fraction of the Fermi energy, so the pressure collapses to a compact form set by $n$ and $\varepsilon_F$ alone.

E_0 = \frac{3}{5}N\varepsilon_F, \qquad \varepsilon_F = \frac{p_F^2}{2m}, \qquad P_0 = \frac{2}{5}n\,\varepsilon_F = \frac{(3\pi^2)^{2/3}}{5}\frac{\hbar^2}{m}\,n^{5/3} \qquad (5)

where: $\varepsilon_F$ — Fermi energy; $m$ — fermion mass; the average energy per particle is $\tfrac35\varepsilon_F$ , not $\tfrac32 k_BT$ . The steep $n^{5/3}$ law means the pressure stiffens fast under compression — this is why fermionic matter is nearly incompressible. When $p_F \gg mc$ the dispersion straightens and the pressure law softens by one third of a power. The prefactor now carries $\hbar c$ instead of $\hbar^2/m$ , and the mass drops out entirely.

P_0 = \frac{(3\pi^2)^{1/3}}{4}\,\hbar c\,n^{4/3} \qquad (6)

where: $c$ — speed of light; the $n^{4/3}$ scaling is the softer, ultrarelativistic law. Because the exponent is now $4/3$ , the pressure grows more slowly than the $n^{5/3}$ metal — and this modest change is what a self-gravitating star cannot survive. The stiffness of the gas — its resistance to compression — is the bulk modulus, and for the nonrelativistic law it is a fixed multiple of the pressure itself.

K = -V\frac{\partial P}{\partial V} = \frac{5}{3}P_0 = \frac{2}{3}n\,\varepsilon_F \qquad (7)

where: $K$ — bulk modulus (inverse compressibility). Since $P_0 \propto V^{-5/3}$ at fixed $N$ , differentiating pulls down a factor $\tfrac53$ . In a metal $K$ from the electron gas is comparable to the measured bulk modulus — degeneracy pressure is much of what makes solids hard.

Derivation ANonrelativistic degeneracy pressure from the filled Fermi sphere

The goal is to derive the boxed law $P_0 = \tfrac25 n\varepsilon_F = \frac{(3\pi^2)^{2/3}}{5}\frac{\hbar^2}{m}n^{5/3}$ from scratch: count the particles in the sphere to fix $p_F$ , integrate the kinetic energy over the sphere to get $E_0$ , and then convert energy to pressure using the $PV=\tfrac23 E$ relation. The strategy is deliberately elementary — three integrals over a solid sphere in momentum space at $T=0$ , where the Fermi–Dirac occupation is a sharp step and the integrals become ordinary. Watch how the $p^4$ weight in the energy integral (versus $p^2$ in the number integral) is exactly what produces the $\tfrac35$ and, through it, the $n^{5/3}$ scaling.

Step 1. Count the particles. With $g$ states per cell, $N = \frac{gV}{(2\pi\hbar)^3}\int_0^{p_F} 4\pi p^2\,\mathrm{d}p = \frac{gV}{6\pi^2\hbar^3}p_F^3$ , which inverts to $p_F = \hbar(6\pi^2 n/g)^{1/3}$ . Why: the number is the spin-weighted phase-space volume of the sphere; the $4\pi p^2$ is the shell area, and integrating $p^2$ gives $p_F^3/3$ .

Step 2. Integrate the kinetic energy. With $\varepsilon = p^2/2m$ , $E_0 = \frac{gV}{(2\pi\hbar)^3}\int_0^{p_F}\frac{p^2}{2m}4\pi p^2\,\mathrm{d}p = \frac{gV}{2m}\frac{p_F^5}{10\pi^2\hbar^3}$ . Why: the integrand is now $p^4$ , so it integrates to $p_F^5/5$ ; the extra two powers of $p$ push weight to the fast edge of the sphere.

Step 3. Form the energy per particle. Dividing the two integrals, $\frac{E_0}{N} = \frac{3}{5}\frac{p_F^2}{2m} = \frac{3}{5}\varepsilon_F$ . Why: the ratio of $p_F^5/5$ to $p_F^3/3$ is $\tfrac35 p_F^2$ ; the average particle carries $\tfrac35$ of the Fermi energy, a purely geometric fraction.

Step 4. Convert to pressure. Since $\varepsilon \propto p^2$ gives $PV = \tfrac23 E$ exactly, $P_0 = \tfrac23\frac{E_0}{V} = \tfrac23 \cdot \tfrac35 n\varepsilon_F = \tfrac25 n\varepsilon_F$ . Why: the $\tfrac23$ is the $s/3$ factor with $s=2$ (proved in the next card); multiplying by $\tfrac35\varepsilon_F$ per particle gives $\tfrac25 n\varepsilon_F$ .

Step 5. Insert $\varepsilon_F$ . With $g=2$ , $\varepsilon_F = \frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ , so $P_0 = \frac{2}{5}n\cdot\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3} = \frac{(3\pi^2)^{2/3}}{5}\frac{\hbar^2}{m}n^{5/3}$ . Why: the extra $n^{2/3}$ from $\varepsilon_F$ multiplies the $n$ out front, producing the $n^{5/3}$ density law — the logarithmic slope $\mathrm{d}\ln P_0/\mathrm{d}\ln n = 5/3$ . *Verified: $p_F$ , $E_0/N=\tfrac35\varepsilon_F$ , the prefactor $(3\pi^2)^{2/3}/5$ , and the exponent $5/3$ are all reproduced symbolically in verify_degeneracy-pressure-fermions.py, matching Pathria & Beale §8.1 and Landau–Lifshitz §57.*

Go deeper · Derivation B · 1 minThe pressure–energy relation PV equals two-thirds E from the grand potential

Derivation BThe pressure–energy relation PV equals two-thirds E from the grand potential

Here the goal is to prove that $PV = \tfrac23 E$ for any nonrelativistic ideal quantum gas in three dimensions — bosons or fermions, any temperature — and to show that replacing the dispersion by $\varepsilon \propto p$ turns it into $PV = \tfrac13 E$ . The route is the grand potential $\Omega = -PV$ , written as an energy integral against the free-particle density of states $g(\varepsilon) \propto \varepsilon^{1/2}$ , with a single integration by parts doing all the work. The payoff is conceptual: the factor $\tfrac23$ versus $\tfrac13$ is not a fermion fact at all — it is fixed by the power of $\varepsilon$ in the density of states, which in turn is fixed by the dispersion. This is the mechanism behind the relativistic watershed.

Step 1. Write the grand potential. For fermions, $PV = -\Omega = k_BT\int_0^\infty g(\varepsilon)\ln\!\big(1+e^{(\mu-\varepsilon)/k_BT}\big)\mathrm{d}\varepsilon$ , with $g(\varepsilon) = A V\varepsilon^{1/2}$ and $A = \frac{g}{4\pi^2}(2m/\hbar^2)^{3/2}$ . Why: $\Omega = -k_BT\sum \ln(1+e^{(\mu-\varepsilon)/k_BT})$ for a Fermi gas; converting the sum to an integral introduces the $\varepsilon^{1/2}$ density of states of a 3D free particle.

Step 2. Integrate by parts. Take $u = \ln(1+e^{(\mu-\varepsilon)/k_BT})$ and $\mathrm{d}v = \varepsilon^{1/2}\mathrm{d}\varepsilon$ , so $v = \tfrac23\varepsilon^{3/2}$ . The boundary term vanishes at both ends, leaving $PV = \tfrac23\int_0^\infty \varepsilon\,g(\varepsilon)\,\frac{1}{e^{(\varepsilon-\mu)/k_BT}+1}\mathrm{d}\varepsilon$ . Why: the derivative of the logarithm is exactly the Fermi function, and the antiderivative of $\varepsilon^{1/2}$ carries the pure number $\tfrac23$ — the source of the factor.

Step 3. Recognize the energy. The remaining integral is $E = \int_0^\infty \varepsilon\,g(\varepsilon)\,f(\varepsilon)\,\mathrm{d}\varepsilon$ , the total internal energy. Hence $PV = \tfrac23 E$ . Why: energy is the density of states weighted by occupation and by $\varepsilon$ ; that is precisely what integration by parts produced.

Step 4. Repeat for linear dispersion. For $\varepsilon \approx pc$ the density of states is $g(\varepsilon) \propto \varepsilon^2$ , so $v = \tfrac13\varepsilon^3$ and the same manipulation gives $PV = \tfrac13 E$ . Why: the antiderivative of $\varepsilon^2$ carries $\tfrac13$ instead of $\tfrac23$ ; the ratio $P/u$ halves, from $\tfrac23$ to $\tfrac13$ , as the gas turns relativistic — this is the watershed. *Verified: the generic $P = (s/3)(E/V)$ scaling with $s=2$ (nonrelativistic) and $s=1$ (ultrarelativistic), and the identity $P/u = \tfrac23$ being exactly double $\tfrac13$ , are checked symbolically in verify_degeneracy-pressure-fermions.py; matches Kardar SPP Ch. 7 and Reif §9.16.*

Go deeper · Derivation C · 2 minThe ultrarelativistic softening and why gravity wins above a critical mass

Derivation CThe ultrarelativistic softening and why gravity wins above a critical mass

The goal here is to derive $P_0 = \frac{(3\pi^2)^{1/3}}{4}\hbar c\,n^{4/3}$ in the limit $p_F \gg mc$ , and then to use pure scaling to show why this softened law admits only a single equilibrium mass for a self-gravitating star. The plan is to redo the sphere integral with $\varepsilon = pc$ instead of $p^2/2m$ , then compare the density dependence of the degeneracy pressure with that of gravity. The crucial observation is dimensional, not detailed: both the ultrarelativistic pressure and the gravitational pressure scale as $R^{-4}$ , so the stellar radius cancels and the balance can hold for exactly one mass — the Chandrasekhar mass, which this article recalls by name and hands to its own home article rather than deriving in full.

Step 1. Energy with linear dispersion. $E_0 = \frac{gV}{(2\pi\hbar)^3}\int_0^{p_F}(pc)\,4\pi p^2\,\mathrm{d}p = \frac{gVc}{8\pi^2\hbar^3}p_F^4$ , so $\frac{E_0}{N} = \frac34 p_F c = \frac34\varepsilon_F$ with $\varepsilon_F = p_F c$ . Why: the integrand is now $p^3$ , integrating to $p_F^4/4$ ; dividing by the $p_F^3/3$ number integral gives $\tfrac34 p_F c$ .

Step 2. Pressure. Using $PV = \tfrac13 E$ , $P_0 = \tfrac13\frac{E_0}{V} = \tfrac14 n\varepsilon_F = \frac{(3\pi^2)^{1/3}}{4}\hbar c\,n^{4/3}$ with $p_F = \hbar(3\pi^2 n)^{1/3}$ for $g=2$ . Why: substituting $\varepsilon_F = p_F c$ and $p_F \propto n^{1/3}$ yields $n\cdot n^{1/3} = n^{4/3}$ — the logarithmic slope is now $4/3$ , checked in the script.

Step 3. Compare with gravity. Model the star's gravitational pressure as $P_{\rm grav} \sim GM^2/R^4$ . With $n \sim M/R^3$ , the ultrarelativistic pressure is $P_0 \sim \hbar c\,(M/R^3)^{4/3} = \hbar c\,M^{4/3}/R^4$ . Why: both pressures scale as $R^{-4}$ , so when you set them equal the radius cancels entirely, leaving an equation in $M$ alone.

Step 4. Read off the consequence. Equating $M^{4/3}/R^4$ against $M^2/R^4$ gives one critical mass, independent of $R$ : below it degeneracy pressure can always be made to match gravity at some radius, above it gravity always wins and collapse proceeds. In the nonrelativistic case $P_0 \sim M^{5/3}/R^5$ scales as $R^{-5} \ne R^{-4}$ , so a stable $R$ -dependent equilibrium always exists — no mass limit. Why: the mismatch of $R$ -powers ( $-5$ vs $-4$ ) is precisely why nonrelativistic white dwarfs are stable while ultrarelativistic ones face a hard ceiling. *Verified: the ultrarelativistic prefactor $(3\pi^2)^{1/3}/4$ , the exponent $4/3$ , and the $R^{-4}$ scaling of both $P_{\rm grav}$ and $P_0^{\rm (ur)}$ (versus $R^{-5}$ nonrelativistic) are checked symbolically; the standard Chandrasekhar mass $M_{\rm Ch}\approx 1.4\,M_\odot$ ( $5.83/\mu_e^2\,M_\odot$ , $\mu_e=2$ ) is checked numerically in verify_degeneracy-pressure-fermions.py; matches Chandrasekhar and Pathria §8.4.*

Limiting cases

The degeneracy pressure is set by the dispersion relation and, at finite temperature, by the ratio $T/T_F$ .

Case	Result	Meaning
Nonrelativistic, $T=0$	$P_0 = \frac{(3\pi^2)^{2/3}}{5}\frac{\hbar^2}{m}n^{5/3} = \tfrac25 n\varepsilon_F$	Ground-state pressure of a spin- $\tfrac12$ free Fermi gas; the dominant term for electrons in metals and low-density white dwarfs.
Ultrarelativistic, $T=0$	$P_0 = \frac{(3\pi^2)^{1/3}}{4}\hbar c\,n^{4/3} = \tfrac14 n\varepsilon_F$	Holds when $p_F \gg mc$ ; the softer $n^{4/3}$ law is responsible for the Chandrasekhar limit.
Low- $T$ correction (Sommerfeld)	$P(T) = P_0\left[1 + \frac{5\pi^2}{12}\left(\frac{k_BT}{\varepsilon_F}\right)^2 + \cdots\right]$	Thermal corrections enter only at order $(T/T_F)^2$ , and with a positive sign — pressure rises slightly with $T$ ; degeneracy pressure dominates at low $T$ .
Classical limit $T \gg T_F$	$P \to n k_B T$	For $n\lambda_T^3 \ll 1$ the gas becomes Maxwell–Boltzmann; the exclusion-driven pressure smoothly merges into the ideal-gas law.

Common trapDegeneracy pressure is a thermal or repulsive effect, so it should scale like nkT

Both instincts are wrong. Degeneracy pressure is neither thermal nor caused by any force between particles — it is a kinematic consequence of antisymmetry, present at $T=0$ where $nk_BT$ would vanish. Setting $P = nk_BT$ for metallic electrons underestimates the true pressure by the factor $T/T_F \sim 10^{-2}$ , because the relevant energy is $\varepsilon_F$ ( $\sim$ eV), not $k_BT$ ( $\sim 0.025$ eV). A second trap is using the nonrelativistic $n^{5/3}$ law in a dense stellar core where the electrons are relativistic and the correct law is $n^{4/3}$ — the difference is exactly what allows a mass limit to exist.

Worked examples

1 · Degeneracy pressure of conduction electrons in copper (confidence builder)

We compute the electron degeneracy pressure in a real metal and compare it against the classical estimate to see the scale of the effect.

Step 1. Copper's conduction-electron density is $n = 8.5\times10^{28}\ \mathrm{m^{-3}}$ . The Fermi energy is $\varepsilon_F = \frac{\hbar^2}{2m_e}(3\pi^2 n)^{2/3}$ . With $3\pi^2 n = 2.517\times10^{30}\ \mathrm{m^{-3}}$ , $(3\pi^2 n)^{2/3} = 1.845\times10^{20}\ \mathrm{m^{-2}}$ , and $\frac{\hbar^2}{2m_e} = \frac{(1.0546\times10^{-34})^2}{2(9.109\times10^{-31})} = 6.104\times10^{-39}\ \mathrm{J\,m^2}$ , giving $\varepsilon_F = 1.126\times10^{-18}\ \mathrm{J} = \mathbf{7.03\ eV}$ .

Step 2. The degeneracy pressure is $P_0 = \tfrac25 n\varepsilon_F = 0.4\times(8.5\times10^{28})\times(1.126\times10^{-18}) = \mathbf{3.83\times10^{10}\ Pa}$ .

Step 3. In atmospheres that is $3.83\times10^{10}/1.013\times10^{5} = \mathbf{3.8\times10^{5}\ atm}$ — hundreds of thousands of atmospheres, pushing outward at absolute zero.

Step 4. Compare the classical estimate $nk_BT$ at $300$ K: $8.5\times10^{28}\times1.381\times10^{-23}\times300 = 3.5\times10^{8}\ \mathrm{Pa}$ , about one hundred times smaller. The exclusion-driven pressure dwarfs the thermal one because $\varepsilon_F/k_BT \approx 270$ . This is the mechanical fingerprint of degeneracy, and much of a metal's stiffness — its bulk modulus $K = \tfrac53 P_0 \approx 6.4\times10^{10}\ \mathrm{Pa}$ — comes straight from it.

2 · Why a white dwarf softens: the electron gas turns relativistic (exam level)

We estimate the density at which electrons in a white dwarf become relativistic, and confirm that the pressure law must switch from $n^{5/3}$ to $n^{4/3}$ there.

Step 1. Electrons turn relativistic when $p_F c \approx m_e c^2$ , i.e. $p_F \approx m_e c = 2.73\times10^{-22}\ \mathrm{kg\,m/s}$ . From $p_F = \hbar(3\pi^2 n)^{1/3}$ , this means $3\pi^2 n = (p_F/\hbar)^3 = (2.59\times10^{12})^3 = 1.74\times10^{37}\ \mathrm{m^{-3}}$ , so $n \approx \mathbf{5.9\times10^{35}\ m^{-3}}$ .

Step 2. Convert to mass density. With about one nucleon per two electrons (helium/carbon core, $\mu_e = 2$ ), the mass density is $\rho \approx \mu_e m_H n = 2\times1.67\times10^{-27}\times5.9\times10^{35} = \mathbf{2.0\times10^{9}\ kg/m^3}$ — roughly two tonnes per cubic centimetre, a typical white-dwarf core density.

Step 3. Below this density the electrons are nonrelativistic and $P_0 \propto n^{5/3}$ ; above it they are relativistic and the law softens to $P_0 \propto n^{4/3}$ . At the crossover both give the same pressure, and the log–log curve of the hero figure bends from slope $\tfrac53$ to slope $\tfrac43$ .

Step 4. The consequence is the mass limit. A more massive white dwarf is denser, so more of its electrons are relativistic, so more of the star lives on the soft $4/3$ branch — and a $4/3$ pressure cannot outrun gravity. The exact balance gives the Chandrasekhar mass $M_{\rm Ch} \approx 1.4\,M_\odot$ ( $5.83/\mu_e^2\,M_\odot$ with $\mu_e=2$ ), beyond which no degenerate equilibrium exists. The full derivation is owned by White Dwarf Stars and the Chandrasekhar Limit; here we have only shown why the softening makes such a limit inevitable.

Origins · 1 minDegeneracy pressure and how this was discovered

Degeneracy pressure

the T = 0 pressure of a Fermi gas · exclusion, not heat

Degeneracy pressure is the pressure a Fermi gas exerts at absolute zero — the direct thermodynamic face of the Pauli exclusion principle. Because no two fermions may share a state, filling $N$ particles forces occupation of momentum states up to the Fermi momentum, and those high-momentum fermions carry kinetic energy and hammer the walls even with zero thermal input, giving $P_0 = \tfrac25 n\varepsilon_F \propto n^{5/3}$ . The concept reached its most dramatic application in 1930, when the nineteen-year-old Subrahmanyan Chandrasekhar, sailing from India to England, combined this pressure with special relativity and found that in a massive white dwarf the electrons turn relativistic, the law softens from $n^{5/3}$ to $n^{4/3}$ , and above about $1.4$ solar masses no equilibrium survives — the star must collapse. Arthur Eddington publicly dismissed the result at a 1935 Royal Astronomical Society meeting, but Chandrasekhar was entirely right; degeneracy pressure and its relativistic softening are now the cornerstone of stellar death, the trigger for Type Ia supernovae, and the boundary between white dwarfs and neutron stars (the full mass limit is derived in aq-sm-white-dwarf-chandrasekhar-limit).

History

1926

— Fowler applies the new Fermi–Dirac statistics to white dwarfs, identifying them as degenerate electron gases.

1926

— Fermi and, independently, Dirac formulate the quantum statistics of identical particles obeying exclusion.

1930

— Chandrasekhar, en route to Cambridge, derives the relativistic softening and the mass limit for white dwarfs.

1931

— Chandrasekhar publishes the maximum-mass result, about 1.4 solar masses.

1935

— Eddington publicly disputes the limit at a Royal Astronomical Society meeting; Chandrasekhar is later vindicated.

1983

— Chandrasekhar shares the Nobel Prize in Physics for the theory of stellar structure and evolution.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., Ch. 8 (Ideal Fermi Systems) · Kardar, Statistical Physics of Particles, Ch. 7 · Reif, Fundamentals of Statistical and Thermal Physics, §9.16 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 7 · Landau & Lifshitz, Statistical Physics Part 1, §57–61 · Chandrasekhar, An Introduction to the Study of Stellar Structure, Ch. 11.

Provenance. All formulas are standard textbook results. A verifier should: (1) reproduce $p_F = \hbar(6\pi^2 n/g)^{1/3}$ and $\varepsilon_F = \frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ for $g=2$ from Fermi-sphere counting; (2) confirm $E_0 = \tfrac35 N\varepsilon_F$ and $P_0 = \tfrac25 n\varepsilon_F = (3\pi^2)^{2/3}\hbar^2 n^{5/3}/(5m)$ (Pathria & Beale §8.1, Landau–Lifshitz §57); (3) verify $PV = \tfrac23 E$ via integration by parts of the grand potential with $g(\varepsilon) \propto \varepsilon^{1/2}$ (Kardar SPP Ch. 7, Reif §9.16); (4) verify the ultrarelativistic $P_0 = (3\pi^2)^{1/3}\hbar c\,n^{4/3}/4$ and $PV = \tfrac13 E$ (Chandrasekhar; Pathria §8.4); (5) check the Sommerfeld pressure coefficient $5\pi^2/12$ and the $R^{-4}$ scaling that yields the Chandrasekhar mass $\approx 1.4\,M_\odot$ (verify_degeneracy-pressure-fermions.py, 19 checks).

Builds on: Degenerate Fermi Gas at Zero Temperature · Fermi Energy, Momentum and Fermi Temperature · Density of States for Free Particles · Quantum Statistics of Ideal Gases: Occupation Numbers Leads to: Sommerfeld Expansion · Electronic Heat Capacity of Metals · Chemical Potential of a Fermi Gas vs Temperature · Fermi Surface and Fermi Sphere

If the pressure is finite at T = 0, doesn't that violate the third law of thermodynamics?

Not at all — the third law constrains entropy, not pressure. It says $S \to 0$ as $T \to 0$ , and the degenerate Fermi gas obeys this perfectly: at $T=0$ the gas is in a single, unique ground state (the filled Fermi sphere), so $S=0$ . Pressure, by contrast, is $-(\partial E/\partial V)_S$ , and there is no law forbidding a system from having nonzero energy or pressure in its ground state. In fact the ground-state energy $E_0 = \tfrac35 N\varepsilon_F$ is large, and its volume dependence is exactly the finite degeneracy pressure. A classical gas has $P = nk_BT \to 0$ only because it has a continuum of accessible states down to zero energy; fermions do not.

Where does the factor 3/5 in E₀ = (3/5) N ε_F actually come from?

It is the ratio of two moments of the Fermi sphere. The particle number weights momentum states by $p^2$ (the shell area $4\pi p^2$ ), giving $\int_0^{p_F} p^2\,\mathrm{d}p = p_F^3/3$ . The energy weights them by $p^2 \cdot \varepsilon(p) \propto p^4$ , giving $\int_0^{p_F} p^4\,\mathrm{d}p = p_F^5/5$ . Dividing energy by number, the $p_F^5/5$ over $p_F^3/3$ leaves $\tfrac35 p_F^2$ , and with $\varepsilon_F = p_F^2/2m$ that is $\tfrac35\varepsilon_F$ . So the $\tfrac35$ is pure geometry — it comes from integrating $p^4$ against $p^2$ over a solid sphere, nothing more.

Why does the same 1/3 factor appear for both a photon gas and an ultrarelativistic Fermi gas?

Because the pressure–energy relation $P = (s/3)u$ depends only on the dispersion exponent $s$ , not on the statistics. A photon has $\varepsilon = pc$ (so $s=1$ ), and an ultrarelativistic electron also has $\varepsilon \approx pc$ (so $s=1$ ); both therefore give $P = u/3$ , regardless of whether they are bosons or fermions. The Bose or Fermi character changes the occupation of each mode — and hence the numerical value of $u(T)$ — but the ratio $P/u$ is fixed by kinematics alone. This is why the radiation-pressure article and this one share the identical $\tfrac13$ : they are two faces of the same ultrarelativistic equation of state.

Does the Sommerfeld correction make the pressure rise or fall with temperature?

It makes it rise: $P(T) = P_0[1 + \frac{5\pi^2}{12}(k_BT/\varepsilon_F)^2 + \cdots]$ , with a strictly positive coefficient. Warming the gas promotes a few electrons from just below the Fermi surface to just above it, and because those promoted electrons occupy higher-momentum states, they add a little extra momentum flux — hence a little extra pressure. The effect is tiny in metals (the correction is $\sim(T/T_F)^2 \sim 10^{-4}$ at room temperature), which is why degeneracy pressure is treated as temperature-independent for stellar structure. Note the sign is opposite to the chemical potential, which decreases with $T$ ; both come from the same Sommerfeld expansion but of different integrals.

Could you get the n^5/3 law just from dimensional analysis, without doing the integrals?

Almost entirely, yes. The only length scale is the interparticle spacing $n^{-1/3}$ , so $p_F \sim \hbar n^{1/3}$ by the uncertainty principle. Nonrelativistically the energy per particle is $\varepsilon \sim p_F^2/2m \sim \hbar^2 n^{2/3}/m$ , and pressure is energy density, $P \sim n\varepsilon \sim \hbar^2 n^{5/3}/m$ — the $n^{5/3}$ law falls out immediately. The full integrals are needed only to pin down the pure number $(3\pi^2)^{2/3}/5$ out front. For the ultrarelativistic case, replace $p_F^2/2m$ by $p_F c \sim \hbar c\,n^{1/3}$ , giving $P \sim \hbar c\,n^{4/3}$ — again the exponent is fixed by dimensions, and only the prefactor $(3\pi^2)^{1/3}/4$ requires the sphere integral.

Why do neutron stars need to reach nuclear density when white dwarfs don't?

Because neutron degeneracy pressure at a given density is smaller than electron degeneracy pressure by the mass ratio. The nonrelativistic law $P_0 = \frac{(3\pi^2)^{2/3}}{5}\frac{\hbar^2}{m}n^{5/3}$ scales as $1/m$ , and a neutron is about $1839$ times heavier than an electron. So to generate a pressure comparable to what electrons provide in a white dwarf, the neutron gas must be far denser — pushed up to nuclear densities of order $10^{17}\ \mathrm{kg/m^3}$ , where the neutrons are nearly touching. The same relativistic softening then produces an analogous maximum mass (the Tolman–Oppenheimer–Volkoff limit), though nuclear interactions, which we have ignored for the ideal gas, substantially modify the exact value.