Fermi Energy, Momentum and Fermi Temperature

Every degenerate Fermi gas — the conduction electrons in a copper wire, the electrons crushed inside a white dwarf, the atoms in an ultracold trap — carries a single characteristic scale that can be read in three currencies at once: an energy, a momentum, and a temperature. The energy of the highest filled level is the Fermi energy $\varepsilon_F$ ; the momentum of a particle there is the Fermi momentum $p_F=\hbar k_F$ ; and $\varepsilon_F/k_B$ is the Fermi temperature $T_F$ . These are not three separate facts but one scale in three units, and remarkably it is fixed by the number density $n=N/V$ alone — no temperature, no interactions, just how tightly you pack the fermions. The counting that sets the scale is owned by the zero-temperature ground state (aq-sm-degenerate-fermi-gas-zero-temperature): filling every state up to a sharp cutoff produces a filled ball in momentum space whose radius is $k_F=(3\pi^2 n)^{1/3}$ for spin- $\tfrac12$ particles. This article takes that $k_F$ as given and converts it into the working scales physicists actually quote. Because $k_F\propto n^{1/3}$ is pure geometry, the dispersion relation does the rest: non-relativistically $\varepsilon_F=\hbar^2 k_F^2/2m\propto n^{2/3}$ , and the $n^{2/3}$ growth is the fingerprint of a free Fermi gas. The real payoff is the Fermi temperature. $T_F$ answers the question "is this gas quantum?" in one number. When $T\ll T_F$ only a thin shell of particles within $\sim k_BT$ of the surface can move, and the gas is degenerate; when $T\gg T_F$ occupations are tiny and Maxwell–Boltzmann statistics returns. What makes metals so counter-intuitive — their electrons refuse to contribute the classical $\tfrac32 Nk_B$ heat capacity — is that copper's electronic $T_F$ is about $8\times10^4$ K, so at room temperature the electron sea is frozen solid in the statistical sense. The same arithmetic, run at stellar densities, gives an $\varepsilon_F$ approaching an electron's rest energy and underlies the stiffness of white-dwarf matter.

Key takeaways

A degenerate Fermi gas has one characteristic scale expressed three ways: Fermi energy $\varepsilon_F$ , Fermi momentum $p_F=\hbar k_F$ , and Fermi temperature $T_F=\varepsilon_F/k_B$ .
The scale is set by density alone: $k_F=(3\pi^2 n)^{1/3}$ (spin- $\tfrac12$ ), $\varepsilon_F=\dfrac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ , so $k_F\propto n^{1/3}$ and $\varepsilon_F\propto n^{2/3}$ .
The $k_F$ itself comes from counting a filled Fermi sphere (owned by aq-sm-degenerate-fermi-gas-zero-temperature); this article turns it into the energy, momentum, velocity, and temperature scales.
$T_F$ is the degeneracy thermometer: $T\ll T_F$ means quantum-degenerate, $T\gg T_F$ means classical.
The numeric ladder spans fifteen orders of magnitude in $T_F$ : ultracold atoms $\sim10^{-7}$ K, metals $\sim10^{4}$ – $10^{5}$ K, white-dwarf electrons $\sim10^{9}$ K.
Key idea: geometry fixes $k_F=(3\pi^2 n)^{1/3}$ from density alone; the dispersion $\varepsilon=\hbar^2k^2/2m$ then converts that single length scale into an energy $\varepsilon_F\propto n^{2/3}$ and a temperature $T_F=\varepsilon_F/k_B$ that decides whether the gas is quantum or classical.

The core insight

The Fermi wavevector $k_F=(3\pi^2 n)^{1/3}$ is set by density alone — it is the radius of the filled Fermi sphere and knows nothing about the dispersion. Feeding it through the free-particle relation $\varepsilon=\hbar^2 k^2/2m$ produces the Fermi energy $\varepsilon_F\propto n^{2/3}$ , and dividing by $k_B$ gives the Fermi temperature $T_F$ , the single number that decides whether quantum statistics matters.

Where this lives today

The Fermi scales are the first numbers any condensed-matter or astrophysics calculation reaches for. In metals, $\varepsilon_F$ and $g(\varepsilon_F)$ (owned by aq-sm-density-of-states-free-particles) fix the linear electronic heat capacity, the Pauli paramagnetic susceptibility, and the electrical and thermal conductivities through the Fermi velocity $v_F$ ; angle-resolved photoemission spectroscopy now maps the Fermi surface and $\varepsilon_F$ of real materials directly. In astrophysics the same arithmetic at $n\sim10^{36}\,\mathrm{m^{-3}}$ gives white-dwarf electrons a Fermi energy comparable to $m_ec^2$ , pushing them into the relativistic regime that sets the $\approx1.4\,M_\odot$ Chandrasekhar mass, while neutron-star matter has neutron $T_F\sim10^{12}$ K so it is degenerate even at $10^9$ K. In the cold-atom laboratory, trapped $^6$ Li and $^{40}$ K gases with $T_F\sim100\,$ nK are cooled to $T/T_F<0.1$ , realizing tunable Fermi gases whose momentum-space step is imaged in time-of-flight and whose $k_F$ is set by the trap and atom number.

The physical picture

Picture filling a stadium from the front rows. Pauli exclusion says one fermion per seat, so as you add particles they stack into ever-higher-momentum states. In momentum space the occupied seats form a solid ball — the Fermi sphere (owned as a geometric object by aq-sm-fermi-surface-fermi-sphere) — and its radius $k_F$ is set entirely by how many particles you poured in per unit volume. Pack them tighter and the ball grows as $n^{1/3}$ ; the topmost particle sits at the surface with energy $\varepsilon_F$ . The first figure is the hero: the momentum-space Fermi sphere, whose radius $k_F$ is the one number everything else descends from. Notice that even at absolute zero the particles are not at rest — the surface ones carry momentum $\hbar k_F$ , and the average kinetic energy is $\tfrac35\varepsilon_F$ , a few electron-volts in a metal. That stored motion is the origin of degeneracy pressure.

Figure

The Fermi sphere in momentum space with radius set by density

The punchline: one radius $k_F$ , fixed by density, generates all three scales. Compress the gas and the whole sphere inflates as $n^{1/3}$ .

The Fermi temperature is a thermometer for "quantumness." Only fermions within a thermal shell $\sim k_BT$ of the surface have empty neighbours to jump into; everything deeper is Pauli-blocked. The fraction that can respond to heat is roughly $T/T_F$ , so a gas with a huge $T_F$ behaves as if it were near absolute zero even at room temperature. The second figure is the punchline of the whole article: a logarithmic ladder of $T_F$ across real systems, with the $300$ K room-temperature line showing at a glance which gases are frozen-degenerate.

Figure

Ladder of Fermi temperatures across systems with the room-temperature line marking degeneracy

Metals and stars sit far above the room-temperature line — their electrons are deeply degenerate — while a trapped atomic gas must be cooled below its own $T_F$ to reach the quantum regime.

The core equations

The scale begins with the Fermi wavevector, which the ground-state count fixes from density alone. The formula below is the spin- $\tfrac12$ specialization; the general degeneracy- $g$ version appears in the derivation.

k_F=\left(3\pi^2 n\right)^{1/3},\qquad p_F=\hbar k_F \qquad (1)

where: $k_F$ — Fermi wavevector, the radius of the filled Fermi sphere; $n=N/V$ — total number density; $p_F=\hbar k_F$ — Fermi momentum; the $3\pi^2$ already includes the spin degeneracy $g=2$ . Physical reading: only $n$ enters, so $k_F$ is a pure geometric property of how densely the fermions are packed — the dispersion has not yet appeared. Feeding $k_F$ through the free-particle dispersion converts a length scale into an energy. This is where the mass and the $n^{2/3}$ scaling enter.

\varepsilon_F=\frac{\hbar^2 k_F^2}{2m}=\frac{\hbar^2}{2m}\left(3\pi^2 n\right)^{2/3} \qquad (2)

where: $\varepsilon_F$ — Fermi energy, equal to the chemical potential at $T=0$ ; $m$ — particle mass (the band effective mass in a solid); $\hbar$ — reduced Planck constant. Physical reading: because $\varepsilon\propto k^2$ , the Fermi energy inherits $\varepsilon_F\propto n^{2/3}$ — doubling the density raises $\varepsilon_F$ by $2^{2/3}\approx1.59$ , a steeper climb than $k_F$ 's $2^{1/3}\approx1.26$ . Dividing the Fermi energy by Boltzmann's constant gives the temperature scale that governs degeneracy. The Fermi velocity, the speed of a surface particle, is the last member of the family.

T_F=\frac{\varepsilon_F}{k_B},\qquad v_F=\frac{p_F}{m}=\frac{\hbar k_F}{m} \qquad (3)

where: $T_F$ — Fermi temperature; $v_F$ — Fermi velocity, the group speed at the surface; $k_B$ — Boltzmann constant. Physical reading: $T_F$ compares the quantum energy scale to thermal energy — when $k_BT\ll\varepsilon_F$ (i.e. $T\ll T_F$ ) the gas is degenerate; $v_F$ is typically $\sim10^6$ m/s in a metal, about $1\%$ of the speed of light. For the ultrarelativistic limit reached in dense stars, the dispersion changes and so does the scaling of the energy with density, though $k_F$ is unchanged.

\varepsilon_F=\hbar c\,k_F=\hbar c\left(3\pi^2 n\right)^{1/3}\quad(\varepsilon_F\gg mc^2) \qquad (4)

where: $c$ — speed of light. Physical reading: when the Fermi energy approaches the rest energy $mc^2$ , the linear dispersion $\varepsilon\approx pc$ takes over and $\varepsilon_F\propto n^{1/3}$ — this softening of the pressure with density is what ultimately limits the mass of a white dwarf.

Derivations

Derivation AFrom the Fermi sphere radius to the three scales

The goal is to take the Fermi wavevector $k_F$ — whose value $(3\pi^2 n)^{1/3}$ is fixed by the zero-temperature state-count owned by aq-sm-degenerate-fermi-gas-zero-temperature — and convert it into the Fermi energy, momentum, velocity, and temperature. The route is purely algebraic: the geometry has already delivered $k_F(n)$ , so we only apply the free-particle dispersion $\varepsilon=\hbar^2k^2/2m$ and the definitions $p_F=\hbar k_F$ , $T_F=\varepsilon_F/k_B$ . The one thing to watch is the spin-degeneracy bookkeeping: the general count gives $k_F=(6\pi^2 n/g)^{1/3}$ , which becomes $(3\pi^2 n)^{1/3}$ only after setting $g=2$ .

Step 1. Recall the count that fixes the radius. Filling every state inside a sphere of radius $k_F$ with the $\mathbf{k}$ -space density $g/(2\pi)^3$ per unit real volume gives $N=\dfrac{gV}{(2\pi)^3}\cdot\dfrac{4}{3}\pi k_F^3$ , hence $k_F=(6\pi^2 n/g)^{1/3}$ . Why: this step is the geometry, owned by the ground-state article; we import its result rather than re-derive it, and it is dispersion-independent, so $k_F\propto n^{1/3}$ for any free gas.

Step 2. Specialize to spin- $\tfrac12$ . With $g=2$ , $6\pi^2 n/g=3\pi^2 n$ , so $k_F=(3\pi^2 n)^{1/3}$ and $p_F=\hbar k_F$ . Why: electrons and other spin- $\tfrac12$ fermions carry a two-fold spin degeneracy per orbital state, which halves the required $k_F^3$ ; the familiar $3\pi^2$ is $6\pi^2/2$ .

Step 3. Apply the dispersion to get the energy. Substituting $k=k_F$ into $\varepsilon=\hbar^2k^2/2m$ gives $\varepsilon_F=\hbar^2 k_F^2/2m=(\hbar^2/2m)(3\pi^2 n)^{2/3}$ . Why: the highest occupied state sits at $k=k_F$ , so its energy is the Fermi energy; because $\varepsilon\propto k^2$ and $k_F\propto n^{1/3}$ , the energy scales as $n^{2/3}$ .

Step 4. Read off the temperature and velocity. Dividing by $k_B$ gives $T_F=\varepsilon_F/k_B$ ; differentiating the dispersion, the surface group speed is $v_F=\partial\varepsilon/\partial p|_{p_F}=p_F/m=\hbar k_F/m$ . Why: $T_F$ recasts $\varepsilon_F$ as the temperature at which thermal energy would match the Fermi scale, and $v_F$ is the speed of the particles that actually carry current and heat — the only ones near an empty state. Verified: the count $k_F^3=6\pi^2 n/g$ , its $g=2$ reduction to $(3\pi^2 n)^{1/3}$ , the $\varepsilon_F=(\hbar^2/2m)(3\pi^2 n)^{2/3}$ form, and the $n^{2/3}$ / $n^{1/3}$ scalings are checked symbolically in verify_fermi-energy-momentum-temperature.py.

Go deeper · Derivation B · 1 minWhy the average energy is three-fifths of the Fermi energy

Derivation BWhy the average energy is three-fifths of the Fermi energy

The goal is to prove that a degenerate Fermi gas at $T=0$ carries a mean energy per particle of exactly $\tfrac35\varepsilon_F$ — not zero, despite being the ground state — and to connect this to the zero-point degeneracy pressure. The strategy is to average the single-particle energy $\varepsilon=\hbar^2k^2/2m$ over the uniformly filled Fermi sphere, which reduces to a ratio of two radial integrals. This is the owned structural result of this article: it turns the abstract "highest filled level" into a thermodynamic energy and shows Pauli exclusion stores macroscopic kinetic energy at absolute zero. The full pressure treatment is forwarded to aq-sm-degeneracy-pressure-fermions.

Step 1. Write the sphere-averaged energy as a ratio of integrals. Every occupied state contributes energy $\propto k^2$ with a shell weight $\propto k^2\,dk$ , so $\langle\varepsilon\rangle=\dfrac{\hbar^2}{2m}\dfrac{\int_0^{k_F}k^2\cdot k^2\,dk}{\int_0^{k_F}k^2\,dk}$ . Why: the density of states in $k$ -space is uniform, so averaging over the filled ball is just averaging $k^2$ with the spherical-shell measure $4\pi k^2\,dk$ (the $4\pi$ cancels top and bottom).

Step 2. Evaluate the radial integrals. The numerator is $\int_0^{k_F}k^4\,dk=k_F^5/5$ and the denominator is $\int_0^{k_F}k^2\,dk=k_F^3/3$ , so their ratio is $\tfrac{3}{5}k_F^2$ . Why: these are elementary power integrals; the ratio $\tfrac{k_F^5/5}{k_F^3/3}=\tfrac35 k_F^2$ is the mean-square wavevector over a filled ball.

Step 3. Convert to the Fermi energy. Multiplying by $\hbar^2/2m$ gives $\langle\varepsilon\rangle=\tfrac35\cdot\tfrac{\hbar^2 k_F^2}{2m}=\tfrac35\varepsilon_F$ , so the total energy is $E_0=\tfrac35 N\varepsilon_F$ . Why: the mean single-particle energy is three-fifths of the maximum, because the filled sphere has more states at large $k$ (the $k^2$ shell weight) but is capped at $k_F$ .

Step 4. Extract the pressure scale. For any non-relativistic ideal gas $PV=\tfrac23 E$ , so $P_0=\tfrac23 E_0/V=\tfrac25 n\varepsilon_F\propto n^{5/3}$ . Why: differentiating $E_0\propto V\,(N/V)^{5/3}\propto V^{-2/3}$ at fixed $N$ gives $P_0=-\partial E_0/\partial V=\tfrac23 E_0/V$ ; the steep $n^{5/3}$ dependence is what makes degenerate matter so stiff, a point developed in aq-sm-degeneracy-pressure-fermions. Verified: the mean $\langle k^2\rangle=\tfrac35 k_F^2$ , the resulting $E_0=\tfrac35 N\varepsilon_F$ , and $P_0=(2/5)n\varepsilon_F$ with $PV=\tfrac23 E$ are checked symbolically in verify_fermi-energy-momentum-temperature.py.

Limiting cases

The density fixes $k_F$ for any dispersion, but how it becomes an energy depends on the regime.

Case	Result	Meaning
Non-relativistic ( $\varepsilon_F\ll mc^2$ )	$\varepsilon_F=\dfrac{\hbar^2 k_F^2}{2m}\propto n^{2/3}$	Metals: $T_F\sim10^4$ – $10^5$ K, so degenerate at all ordinary temperatures
Ultrarelativistic ( $\varepsilon_F\gg mc^2$ )	$\varepsilon_F=\hbar c\,k_F\propto n^{1/3}$	Dense stellar matter: softer pressure $P\propto n^{4/3}$ , the seed of the Chandrasekhar limit
Classical, $T\gg T_F$	$n\lambda_T^3\ll1$ , Fermi–Dirac $\to$ Maxwell–Boltzmann	Fermi sphere no longer sharp; ordinary gas behaviour
Degenerate, $T\ll T_F$	$\mu(T)\approx\varepsilon_F\big[1-\tfrac{\pi^2}{12}(T/T_F)^2\big]$	Sharp Fermi surface, Sommerfeld corrections in $T/T_F$

Worked examples

Example 1 — The Fermi scales of copper from scratch.

Step 1. Setup. Copper is monovalent, contributing one conduction electron per atom, giving $n=8.49\times10^{28}\,\mathrm{m^{-3}}$ . We compute $k_F$ , then $\varepsilon_F$ , $T_F$ , and $v_F$ using $\hbar=1.0546\times10^{-34}$ J·s, $m_e=9.109\times10^{-31}$ kg, $k_B=1.381\times10^{-23}$ J/K.

Step 2. Fermi wavevector. $3\pi^2 n=3\pi^2(8.49\times10^{28})=2.514\times10^{30}\,\mathrm{m^{-3}}$ , so $k_F=(2.514\times10^{30})^{1/3}=\mathbf{1.36\times10^{10}\ m^{-1}}$ .

Step 3. Fermi energy. $\varepsilon_F=\dfrac{(1.0546\times10^{-34})^2(1.36\times10^{10})^2}{2(9.109\times10^{-31})}=1.13\times10^{-18}\,\mathrm{J}=\mathbf{7.0\ eV}$ , matching the tabulated copper value.

Step 4. Temperature and velocity. $T_F=\varepsilon_F/k_B=(1.13\times10^{-18})/(1.381\times10^{-23})=\mathbf{8.2\times10^{4}\ K}$ , and $v_F=\hbar k_F/m_e=(1.0546\times10^{-34})(1.36\times10^{10})/(9.109\times10^{-31})=\mathbf{1.57\times10^{6}\ m/s}$ . Since $300\,\mathrm{K}/T_F\approx3.7\times10^{-3}$ , well under a percent of the electrons are thermally active — copper's electron sea is deeply degenerate.

Example 2 — How the scales respond to compression.

Step 1. Setup. Sodium has $n=2.65\times10^{28}\,\mathrm{m^{-3}}$ , roughly a third of copper's density. Predict $\varepsilon_F$ and confirm the $n^{2/3}$ scaling against a direct computation.

Step 2. Scaling prediction. Since $\varepsilon_F\propto n^{2/3}$ , sodium's Fermi energy should be $\varepsilon_F^{\mathrm{Na}}=7.0\,\mathrm{eV}\times(2.65/8.49)^{2/3}=7.0\times(0.3121)^{2/3}=7.0\times0.461=\mathbf{3.2\ eV}$ .

Step 3. Direct check. Computing from scratch, $3\pi^2 n=7.847\times10^{29}$ , $k_F=9.22\times10^{9}\,\mathrm{m^{-1}}$ , and $\varepsilon_F=\hbar^2k_F^2/2m_e=5.19\times10^{-19}\,\mathrm{J}=\mathbf{3.24\ eV}$ — consistent with the scaling estimate.

Step 4. Interpret the compression. If you squeezed sodium to copper's density (a factor $8.49/2.65=3.2$ in $n$ ), $\varepsilon_F$ would rise by $3.2^{2/3}=2.17$ and the degeneracy pressure by $3.2^{5/3}=6.9$ . This $n^{5/3}$ stiffness — far steeper than a classical gas's $n^{1}$ — is exactly why compressing degenerate matter costs so much energy and why it can hold up a star.

Origins · 1 minfermi wavevector and how this was discovered

fermi wavevector

central concept, 1926–1928 · The Fermi wavevector k_F is the momentum-space radius of the filled Fermi sphere, the single length scale from which the energy, momentum, and temperature scales all descend. It follows from the statistics Enrico Fermi introduced in early 1926 for an ideal gas obeying the exclusion principle. Arnold Sommerfeld applied Fermi–Dirac statistics to the electrons in a metal in 1927, replacing the failed classical free-electron model of Drude and turning k_F into the working scale of solid-state physics.

History

1926

— Fermi formulates the statistics of an ideal gas of exclusion-obeying particles (Rendiconti Lincei)

1926

— Dirac independently derives the same statistics from antisymmetric wavefunctions (Proc. Roy. Soc.)

1927

— Sommerfeld applies Fermi–Dirac statistics to metallic electrons, defining the Fermi energy and Fermi surface

1928

— Bloch extends the free-electron gas to periodic lattices, launching band theory and the modern Fermi surface

1930

— Chandrasekhar finds that relativistic degeneracy softens the pressure, limiting the mass of white dwarfs

Going further

Texts: Pathria & Beale, Statistical Mechanics (3rd ed.), §8.1; Ashcroft & Mermin, Solid State Physics, Ch. 2 (Table 2.1 for metallic $\varepsilon_F$ , $T_F$ , $k_F$ , $v_F$ ); Kardar, Statistical Physics of Particles, Ch. 7; Reif, Fundamentals of Statistical and Thermal Physics, §9.16; Landau & Lifshitz, Statistical Physics Part 1, §57.
Builds on: Degenerate Fermi Gas at Zero Temperature (owns the ground-state count that fixes $k_F$ ) · Density of States for Free Particles (owns $g(\varepsilon)$ ) · Fermi Surface and Fermi Sphere (owns the sphere geometry) · Quantum Statistics: Occupation Numbers.
Leads to: Sommerfeld Expansion · Electronic Heat Capacity of Metals · Degeneracy Pressure of Fermions · Chemical Potential of a Fermi Gas vs Temperature.
Provenance: AllQFT `sm` export (Pathria & Beale, §8.1), aq-sm-fermi-energy-momentum-temperature, ref STAT.IQGAS.FERME.1219. All scale relations, scalings, and the copper/sodium/white-dwarf/cold-atom numbers python-verified (22 checks, all PASS): Cu $\varepsilon_F=7.0$ eV, $T_F=8.2\times10^4$ K, $v_F=1.57\times10^6$ m/s; Na $\varepsilon_F=3.24$ eV reproduce the Ashcroft–Mermin values.

Are the Fermi energy, momentum, and temperature really the same thing?

They are one physical scale expressed in three units. Start from the Fermi momentum $p_F=\hbar k_F$ ; square it and divide by $2m$ to get the Fermi energy $\varepsilon_F=p_F^2/2m$ ; divide that by $k_B$ to get the Fermi temperature $T_F=\varepsilon_F/k_B$ . Each makes a different piece of physics obvious: $p_F$ (or $v_F=p_F/m$ ) governs how fast surface electrons move and carry current, $\varepsilon_F$ sets the energy budget and the chemical potential at $T=0$ , and $T_F$ tells you at a glance whether thermal energy $k_BT$ can compete with the quantum scale. They are not independent numbers, just three windows on the same $k_F$ .

Why does k_F depend only on density and not on the particle mass?

Because $k_F$ comes purely from counting states, and counting is geometry. You fill a sphere in momentum space until it holds $N$ states, and the $\mathbf{k}$ -space density of states $g/(2\pi)^3$ per unit real volume has no mass in it — it is set by the box quantization alone. So $k_F=(3\pi^2 n)^{1/3}$ for electrons, neutrons, or trapped atoms of any mass, as long as they are spin- $\tfrac12$ . The mass enters only at the next step, when you convert $k_F$ to an energy through $\varepsilon=\hbar^2k^2/2m$ : a heavier particle has the same $k_F$ but a smaller $\varepsilon_F$ and $T_F$ .

If it is the ground state, why isn't the energy zero at T = 0?

A classical gas at $T=0$ would sit motionless with zero kinetic energy, but fermions cannot all pile into the lowest momentum state — Pauli exclusion allows only one per state (two with spin). To accommodate $N$ particles they must stack up to $k_F$ , so even the ground state has particles moving with momenta up to $\hbar k_F$ . Averaging over the filled sphere gives a mean energy of $\tfrac35\varepsilon_F$ per particle, a few electron-volts in a metal. This irreducible zero-point kinetic energy is the origin of degeneracy pressure and is what physically supports white dwarfs and neutron stars.

What does it mean that only a fraction T/T_F of electrons are "active"?

At temperature $T$ , a fermion can absorb thermal energy only if there is an empty state within $\sim k_BT$ above it. Deep in the Fermi sea every nearby state is already full, so those electrons are Pauli-blocked and cannot respond. Only the electrons within a shell of thickness $\sim k_BT$ around the Fermi surface — a fraction $\sim k_BT/\varepsilon_F=T/T_F$ of the total — can be excited. For copper at room temperature that is under half a percent, which is why the electronic heat capacity is a small linear-in- $T$ term rather than the classical $\tfrac32 Nk_B$ , as developed in aq-sm-electronic-heat-capacity-metals.

When does the ultrarelativistic formula take over?

The non-relativistic relation $\varepsilon_F=\hbar^2 k_F^2/2m$ is accurate as long as $\varepsilon_F\ll mc^2$ . As density rises, $k_F\propto n^{1/3}$ grows and eventually the Fermi energy approaches the rest energy — for electrons, $m_ec^2\approx0.511$ MeV, reached at $n\sim10^{36}\,\mathrm{m^{-3}}$ , the interior density of a massive white dwarf. Beyond that the dispersion becomes linear, $\varepsilon\approx pc$ , so $\varepsilon_F=\hbar c\,k_F\propto n^{1/3}$ . The pressure then scales as $n^{4/3}$ instead of $n^{5/3}$ , a softening that has no stable balance against gravity above a critical mass — the physical origin of the Chandrasekhar limit.

Is the Fermi energy the same as the chemical potential?

Only at exactly $T=0$ , where by definition $\varepsilon_F=\mu(T{=}0)$ . At finite temperature the chemical potential drifts: for a 3D free gas it drops as $\mu(T)\approx\varepsilon_F[1-\tfrac{\pi^2}{12}(T/T_F)^2]$ and can eventually go negative in the non-degenerate regime, whereas $\varepsilon_F$ is a fixed property of the density through $\varepsilon_F=(\hbar^2/2m)(3\pi^2 n)^{2/3}$ . In a metal at room temperature the correction is $\sim10^{-5}$ , so people use $\mu\approx\varepsilon_F$ freely — harmless there, but wrong in principle and badly wrong once $T$ is a sizeable fraction of $T_F$ .