Fermi Surface and Fermi Sphere

Cool a metal to absolute zero and its conduction electrons do not all sink to rest — the Pauli principle forbids it. Instead they stack into single-particle momentum states from the bottom up, filling every level until the electrons run out. The boundary between the filled states and the empty ones, drawn in momentum ( $\mathbf{k}$ ) space, is the Fermi surface. For a gas of non-interacting electrons the filled region is a ball centered on $\mathbf{k}=0$ , so the surface is a sphere of radius $k_F$ — the Fermi sphere. This surface is the single most important object in the physics of metals. Its radius is fixed not by any force but by pure counting: the electron density $n$ determines how many $\mathbf{k}$ -states must be filled, and $k_F=(3\pi^2 n)^{1/3}$ follows. Everything a metal does at ordinary temperatures — conduct electricity, screen fields, carry heat, respond to a magnetic field — happens at this frontier, because only there does a filled state sit next to an empty one that an electron can move into. The vast interior of the sphere is frozen: an electron deep inside has no nearby vacancy and is Pauli-blocked. That is the punchline this article owns. Heating a metal to room temperature does not stir the whole electron sea; it only softens the surface over a shell of energy width $\sim k_BT$ . Because the sphere is enormous — Fermi energies of several electron-volts, Fermi temperatures of $10^4$ – $10^5\,\mathrm{K}$ — that shell is a razor-thin skin holding a fraction $\sim k_BT/\varepsilon_F=T/T_F$ of the electrons, about $0.4\%$ for copper at $300\,\mathrm{K}$ . The Fermi radius $k_F$ and the energy scales $\varepsilon_F$ , $T_F$ are set up in aq-sm-degenerate-fermi-gas-zero-temperature and aq-sm-fermi-energy-momentum-temperature; here we build the surface itself and prove why the metal is electronically cold.

Key takeaways

The Fermi surface is the constant-energy surface in $\mathbf{k}$ -space that separates occupied from empty single-particle states at $T=0$ ; for free fermions it is a sphere of radius $k_F$ — the Fermi sphere.
Its radius is set by counting alone: filling the sphere with the right number of $\mathbf{k}$ -states gives $k_F=(3\pi^2 n)^{1/3}$ for spin- $\tfrac12$ (density $n=N/V$ ).
The enclosed $\mathbf{k}$ -volume is fixed by the electron density and is robust to interactions and lattice structure — the free sphere is the reference; a crystal warps it into necks and pockets.
Only a shell of energy thickness $\sim k_BT$ at the surface is thermally active; the deep sea is Pauli-blocked and inert.
That active fraction is $\sim T/T_F\ll 1$ (about $0.4\%$ for copper at room temperature), which is why transport, magnetism, and heat capacity are all Fermi-surface phenomena, not bulk-sea phenomena.
Key idea: the Fermi surface is the sharp $T=0$ boundary of the filled momentum sphere, its size fixed by density; at finite $T$ only a $k_BT$ -thin skin around it can respond, so the fraction of "live" electrons is $\sim T/T_F$ — the whole reason metals are electronically cold.

The core insight

Fermions fill momentum states up to a sharp boundary set by Pauli exclusion; that boundary — the Fermi surface — is a sphere of radius $k_F=(3\pi^2 n)^{1/3}$ whose enclosed volume is fixed by the particle density alone. Warm the metal and the surface blurs over a shell of width $\sim k_BT$ , so only a fraction $\sim k_BT/\varepsilon_F=T/T_F$ of the electrons — those on the surface — take part in any low-temperature process.

Where this lives today

The Fermi surface is the organizing object of modern condensed-matter physics, and its geometry is now measured with exquisite precision. Angle-resolved photoemission spectroscopy (ARPES) images the occupied states of a crystal directly, tracing out the surface sheet by sheet; quantum-oscillation experiments (de Haas–van Alphen, Shubnikov–de Haas) read off its extremal cross-sectional areas through the periodicity of magnetization or resistance in $1/B$ . These tools have mapped the warped, multiply connected Fermi surfaces of ordinary metals and, more dramatically, revealed how a Fermi surface reconstructs across a phase transition — a central clue in high-temperature superconductors and heavy-fermion materials, where anomalies in the surface signal the breakdown of the simple Fermi-liquid picture. Even the "Fermi-surface-free" physics of graphene and Dirac/Weyl semimetals is defined by contrast with this reference: their occupied region shrinks to points or lines rather than filling a sphere, and much of the excitement is precisely that the familiar surface is absent.

The physical picture

Picture pouring electrons into a bowl whose height is the kinetic energy $\varepsilon=\hbar^2 k^2/2m$ . Because no two electrons may share a state, they cannot all fall to the bottom; they fill from the floor upward to a water line — the Fermi energy $\varepsilon_F$ . In momentum space that water line is a sphere: every state inside is occupied, every state outside is empty. The radius of the sphere is the Fermi wavevector $k_F$ , and the volume it encloses is exactly the volume needed to hold all $N$ electrons at one per state (times two for spin). The sphere is huge and rigid. In a typical metal $k_F\sim 10^{10}\,\mathrm{m}^{-1}$ and $\varepsilon_F$ is several electron-volts, corresponding to a Fermi temperature $T_F\sim 10^4$ – $10^5\,\mathrm{K}$ . Room temperature is a tiny perturbation on this scale. The first figure is the hero: the filled Fermi sphere with its thin thermally active skin, the only part that can respond to heat, a field, or a voltage.

Figure

The Fermi sphere with its thin thermally active surface shell

The punchline: the sphere is nearly all frozen. Heat, current, and magnetism come only from the thin red skin — the Fermi surface — because only there can an electron find an adjacent empty state.

Think of the Fermi surface as the frontier of the electron sea. Deep in the interior every state is occupied and every neighboring state is occupied too, so an electron there is stuck; it cannot absorb a small energy $k_BT$ because the target level is already full. Only at the frontier, where filled and empty states meet, can electrons be excited, scattered, or reshuffled. This is the structural reason a metal's electrons contribute almost nothing to its heat capacity at room temperature, yet dominate it at very low temperature where the phonon $T^3$ term dies away faster. In a real crystal the free-electron parabola $\varepsilon\propto k^2$ is replaced by band energies periodic in the reciprocal lattice, and the spherical frontier is warped — into faceted polyhedra, necks that touch Brillouin-zone boundaries, even open sheets. The free Fermi sphere is the reference against which all of this is measured.

Figure

Free Fermi sphere versus a lattice-warped Fermi surface

The shape is band-structure; the enclosed volume is bookkeeping — set by how many electrons you put in.

The core equations

The radius of the Fermi sphere is not dynamics but a census. Each allowed $\mathbf{k}$ -point occupies a volume $(2\pi)^3/V$ of reciprocal space, and with $g_s=2$ spin states per point the number of states inside the sphere must equal $N$ . Equating the two fixes the radius.

N = g_s\,\frac{V}{(2\pi)^3}\,\frac{4}{3}\pi k_F^3 = \frac{g_s V k_F^3}{6\pi^2} \qquad (1)

where: $N$ — number of fermions; $V$ — volume; $g_s$ — spin degeneracy ( $=2$ for electrons); $k_F$ — Fermi wavevector. Physical reading: this is the whole content of the Fermi sphere — the enclosed $\mathbf{k}$ -volume $(4/3)\pi k_F^3$ , counted at one state per cell $(2\pi)^3/V$ and doubled for spin, is exactly the electron count. Nothing about forces enters. Solving Eq. (1) for the radius gives the density-only result that underlies every metal number. For spin- $\tfrac12$ electrons the spin factor collapses the prefactor to a clean form.

k_F = \left(\frac{6\pi^2 n}{g_s}\right)^{1/3} \;\xrightarrow{g_s=2}\; \left(3\pi^2 n\right)^{1/3}, \qquad n=\frac{N}{V} \qquad (2)

where: $n$ — number density. Physical reading: $k_F\propto n^{1/3}$ — denser metals have larger Fermi spheres. The derivation of $k_F$ and the energy scales it implies is owned by aq-sm-fermi-energy-momentum-temperature; here $k_F$ is the radius of the surface we are studying. The states on the surface all have the same energy, the Fermi energy, which is what makes the surface an equal-energy surface. The corresponding momentum, velocity, and temperature scales are read off directly.

\varepsilon_F=\frac{\hbar^2 k_F^2}{2m},\qquad p_F=\hbar k_F,\qquad v_F=\frac{\hbar k_F}{m},\qquad T_F=\frac{\varepsilon_F}{k_B} \qquad (3)

where: $\varepsilon_F$ — Fermi energy; $p_F$ — Fermi momentum; $v_F$ — Fermi velocity (the speed of electrons at the surface); $T_F$ — Fermi (degeneracy) temperature. Physical reading: $v_F$ — not the tiny drift velocity — is the relevant electronic speed for conduction, because it is the electrons at the surface that move. At finite temperature the sharp step of the $T=0$ occupation softens into the Fermi–Dirac distribution, and the surface acquires a thickness. The thermal weight that "sees" the surface is the derivative of the Fermi function, a bell of width $\sim k_BT$ that becomes a delta function at $T=0$ .

-\frac{\partial f}{\partial\varepsilon}=\frac{1}{4k_BT}\,\mathrm{sech}^2\!\left(\frac{\varepsilon-\mu}{2k_BT}\right)\;\xrightarrow{T\to0}\;\delta(\varepsilon-\mu) \qquad (4)

where: $f(\varepsilon)=1/(e^{(\varepsilon-\mu)/k_BT}+1)$ — Fermi–Dirac occupation (owned by aq-sm-fermi-dirac-distribution); $\mu\to\varepsilon_F$ as $T\to0$ . Physical reading: any smooth quantity averaged against this kernel is evaluated at the surface, $\int g(\varepsilon)(-\partial f/\partial\varepsilon)\,d\varepsilon\to g(\varepsilon_F)$ — the mathematical statement that only the surface matters, and the seed of the Sommerfeld expansion. Because the response lives on the surface, the density of states at the surface controls the thermodynamics. For the free gas it takes a compact form in terms of density and Fermi energy.

g(\varepsilon_F)=\frac{3}{2}\,\frac{n}{\varepsilon_F} \qquad (5)

where: $g(\varepsilon_F)$ — density of single-particle states per unit volume at the Fermi energy (the machinery $g(\varepsilon)\propto\sqrt\varepsilon$ is owned by aq-sm-density-of-states-free-particles). Physical reading: three-halves of the electrons per unit Fermi energy sit within reach of the surface — this single number sets the electronic heat capacity, the Pauli susceptibility, and the screening length.

Derivations

Derivation AThe radius of the Fermi sphere from state counting

The goal is to fix the Fermi radius $k_F$ purely by counting, showing that it is a property of the density $n$ and nothing else. The strategy is to build the $T=0$ ground state one electron at a time — stacking them into the lowest-energy $\mathbf{k}$ -states allowed by Pauli exclusion — and to demand that the filled region hold exactly $N$ electrons. The one subtlety to watch is the bookkeeping: each $\mathbf{k}$ -point carries a reciprocal-space cell of volume $(2\pi)^3/V$ and, for electrons, two spin states, and both factors must be tracked to land the clean $k_F=(3\pi^2 n)^{1/3}$ .

Step 1. Identify the occupied region. For free particles $\varepsilon(\mathbf{k})=\hbar^2 k^2/2m$ depends only on $|\mathbf{k}|$ , so filling from the bottom up occupies all states with $k<k_F$ — a ball in $\mathbf{k}$ -space of radius $k_F$ . Its volume is $(4/3)\pi k_F^3$ . Why: the energy is isotropic, so equal-energy surfaces are spheres and the lowest-energy set is the interior of one.

Step 2. Count the states inside. In a box of volume $V$ the allowed wavevectors form a lattice with one point per cell $(2\pi/L)^3=(2\pi)^3/V$ ; the number of $\mathbf{k}$ -points inside the ball is (ball volume)/(cell volume) $=(4/3)\pi k_F^3\cdot V/(2\pi)^3$ . Multiply by the spin degeneracy $g_s$ : $N=g_s\,(V/(2\pi)^3)\,(4/3)\pi k_F^3$ . Why: this is Eq. (1); the same $\mathbf{k}$ -cell counting is the free-particle mode counting used throughout ideal-gas theory.

Step 3. Solve for the radius. Simplify the constants: $g_s\,(4/3)\pi/(2\pi)^3 = g_s/(6\pi^2)$ , so $N=g_s V k_F^3/6\pi^2$ and $k_F=(6\pi^2 n/g_s)^{1/3}$ . Why: only the combination $n=N/V$ survives, so the radius depends on density alone.

Step 4. Insert electron spin. For electrons $g_s=2$ , and the factor $2$ under the cube root turns $6\pi^2/2=3\pi^2$ : $k_F=(3\pi^2 n)^{1/3}$ . Why: the spin factor is what converts the " $6\pi^2$ " of a spinless gas into the " $3\pi^2$ " quoted for metals — a common place to drop a factor of two. Verified: the counting $N=g_s V k_F^3/6\pi^2$ , the inversion $k_F=(6\pi^2 n/g_s)^{1/3}$ , and the spin- $\tfrac12$ collapse to $(3\pi^2 n)^{1/3}$ are checked symbolically in verify_fermi-surface-fermi-sphere.py, along with the identity $g_s\times(\text{cells inside})=N$ .

Go deeper · Derivation B · 2 minWhy only a thin shell of the surface is thermally active

Derivation BWhy only a thin shell of the surface is thermally active

The goal is to prove the article's central claim: that heating a degenerate Fermi gas engages only a razor-thin shell at the Fermi surface, of relative thickness $\Delta k/k_F\sim k_BT/\varepsilon_F$ , so the fraction of thermally active electrons is $\sim T/T_F\ll1$ . The route is to translate the energy width $\sim k_BT$ over which the Fermi step softens into a width in $\mathbf{k}$ -space using the dispersion, then compare the volume of the resulting spherical shell to the volume of the whole sphere. The thing to watch is that the small parameter is $k_BT/\varepsilon_F$ , not $k_BT$ itself — the sphere is "cold" because $\varepsilon_F$ is enormous.

Step 1. Fix the energy width of the surface. The Fermi function drops from $\approx1$ to $\approx0$ over an energy range set by $-\partial f/\partial\varepsilon$ , a bell of width $\sim k_BT$ (Eq. 4; its full width at half maximum is $\approx3.53\,k_BT$ ). Take the shell energy thickness as $\Delta\varepsilon\sim k_BT$ . Why: states more than a few $k_BT$ below $\mu$ stay full and more than a few $k_BT$ above stay empty — only the transition region can change occupation.

Step 2. Convert to a thickness in $\mathbf{k}$ . Linearize the dispersion near the surface: $\Delta\varepsilon=(d\varepsilon/dk)|_{k_F}\,\Delta k=(\hbar^2 k_F/m)\,\Delta k=\hbar v_F\,\Delta k$ . Hence $\Delta k=\Delta\varepsilon/\hbar v_F$ . Why: $d\varepsilon/dk=\hbar^2 k/m=\hbar v_F$ is the slope of the energy at the surface, the same $v_F$ that sets the electron speed.

Step 3. Normalize to the radius. $\dfrac{\Delta k}{k_F}=\dfrac{m\,\Delta\varepsilon}{\hbar^2 k_F^2}=\dfrac{\Delta\varepsilon}{2\varepsilon_F}\sim\dfrac{k_BT}{2\varepsilon_F}$ , using $\varepsilon_F=\hbar^2 k_F^2/2m$ . Why: the relative shell thickness is $k_BT$ measured against $\varepsilon_F$ — the degeneracy ratio $T/T_F$ , tiny for any metal.

Step 4. Turn the shell into an electron count. The active electrons occupy a spherical shell of volume $\approx 4\pi k_F^2\,\Delta k$ out of the full sphere $(4/3)\pi k_F^3$ , so the active fraction is $\dfrac{4\pi k_F^2\,\Delta k}{(4/3)\pi k_F^3}=3\dfrac{\Delta k}{k_F}\approx\dfrac{3}{2}\dfrac{k_BT}{\varepsilon_F}=\dfrac{3}{2}\dfrac{T}{T_F}$ . Why: to order of magnitude a fraction $\sim T/T_F$ of the electrons is "live"; the rest of the sphere is inert, which is exactly why the electronic heat capacity is a small linear-in- $T$ term rather than the classical $\tfrac32 Nk_B$ . Verified: $d\varepsilon/dk|_{k_F}=\hbar^2 k_F/m$ , $\Delta k/k_F=\Delta\varepsilon/2\varepsilon_F$ , and the active fraction $=3\Delta k/k_F=(3/2)k_BT/\varepsilon_F$ are checked symbolically, and the copper number $T/T_F\approx0.37\%$ (or $\approx0.55\%$ with the $3/2$ ) at $300\,\mathrm{K}$ is checked numerically in verify_fermi-surface-fermi-sphere.py.

Limiting cases

The Fermi-surface picture connects to its neighbors through four clean regimes.

Case	Result	Meaning
Zero temperature	$f(\varepsilon)=\theta(\varepsilon_F-\varepsilon)$	Perfectly sharp surface: occupation is a step. The Fermi sphere is exact.
Degenerate, $k_BT\ll\varepsilon_F$	$\Delta k/k_F\sim k_BT/2\varepsilon_F\ll1$	Thin thermal shell; only $\sim T/T_F$ of electrons active — the regime of all real metals.
Non-degenerate, $k_BT\gg\varepsilon_F$	$f\to e^{-(\varepsilon-\mu)/k_BT}$	No sharp surface survives; the gas is classical Maxwell–Boltzmann.
Crystalline / anisotropic	$\{\mathbf{k}:\varepsilon_n(\mathbf{k})=\varepsilon_F\}$	Sphere deforms to necks, pockets, open sheets — but the enclosed $\mathbf{k}$ -volume is fixed by density.

Worked examples

Example 1 — Build the copper Fermi sphere from its electron density.

Step 1. Setup. Copper has one conduction electron per atom, giving $n=8.49\times10^{28}\,\mathrm{m^{-3}}$ and $g_s=2$ . The Fermi radius follows from counting, $k_F=(3\pi^2 n)^{1/3}$ .

Step 2. Fermi wavevector. $3\pi^2 n=3\pi^2(8.49\times10^{28})=2.514\times10^{30}\,\mathrm{m^{-3}}$ , so $k_F=(2.514\times10^{30})^{1/3}=\mathbf{1.36\times10^{10}\,m^{-1}}$ . The sphere is $\sim 10^{10}\,\mathrm{m^{-1}}$ across — vast on any atomic scale.

Step 3. Fermi energy and temperature. $\varepsilon_F=\hbar^2 k_F^2/2m=(1.0546\times10^{-34})^2(1.36\times10^{10})^2/[2(9.109\times10^{-31})]=1.13\times10^{-18}\,\mathrm{J}=\mathbf{7.0\ eV}$ , and $T_F=\varepsilon_F/k_B=\mathbf{8.2\times10^{4}\,K}$ . The surface energy is enormous compared with any thermal energy.

Step 4. Fermi velocity. $v_F=\hbar k_F/m=(1.0546\times10^{-34})(1.36\times10^{10})/(9.109\times10^{-31})=\mathbf{1.57\times10^{6}\,m/s}$ — about $0.5\%$ of the speed of light, and hundreds of times the room-temperature drift velocity. Electrons at the surface are always moving fast, even in a metal carrying no current.

Example 2 — What fraction of copper's electrons is thermally active at room temperature?

Step 1. Setup. Use the degeneracy ratio from Example 1, $T_F=8.2\times10^{4}\,\mathrm{K}$ , at $T=300\,\mathrm{K}$ . The relative shell thickness is $\Delta k/k_F=k_BT/2\varepsilon_F=T/2T_F$ .

Step 2. Shell thickness. $\Delta k/k_F=300/(2\times8.2\times10^{4})=\mathbf{1.8\times10^{-3}}$ — the surface is fuzzy over less than a fifth of a percent of the Fermi radius.

Step 3. Active fraction. The shell volume relative to the sphere is $3\,\Delta k/k_F=(3/2)(T/T_F)=1.5\times(300/8.2\times10^{4})=\mathbf{5.5\times10^{-3}}$ , i.e. about $\mathbf{0.55\%}$ ; the cruder estimate $T/T_F$ (dropping the geometric $3/2$ ) gives $\mathbf{0.37\%}$ . Either way, well under one percent of the electrons participate.

Step 4. Interpret. Roughly $99.5\%$ of copper's conduction electrons are frozen deep in the Fermi sea at room temperature, unable to absorb heat because their neighboring states are full. Only the $\sim0.4\%$ in the surface shell carry the electronic heat capacity, the Pauli paramagnetism, and the scattering that limits conductivity — the quantitative meaning of "metals are electronically cold." Sodium, being less dense ( $T_F\approx3.8\times10^4\,\mathrm{K}$ ), has a larger but still tiny active fraction, $\sim0.8\%$ .

Origins · 1 minlev landau and how this was discovered

lev landau

1908-1968 · Soviet theoretical physicist of extraordinary range. In 1956-1958 Landau built the theory of the Fermi liquid, showing that the low-energy excitations of an interacting electron gas are long-lived quasiparticles that still fill a sharp Fermi surface and obey a Fermi-Dirac distribution in their renormalized energies. This is why the free-electron Fermi sphere — a picture for non-interacting particles — describes real metals so well. Landau received the 1962 Nobel Prize for his theory of liquid helium.

History

1926

— Fermi and, independently, Dirac derive the statistics of exclusion-obeying particles

1927

— Sommerfeld applies Fermi–Dirac statistics to electrons in metals, introducing the degenerate electron gas and its sharp surface

1928

— Bloch extends the free-electron gas to periodic lattices, showing the sphere becomes a band-structure surface

1930

— Landau derives the diamagnetism of a free electron gas, an early Fermi-surface response

1957

— Luttinger and Ward establish that the enclosed Fermi-surface volume is fixed by density even with interactions

1958

— Landau completes Fermi-liquid theory, explaining why the sharp Fermi surface survives interactions

1960

— de Haas–van Alphen and cyclotron-resonance measurements map real, non-spherical Fermi surfaces in detail

Going further

Texts: Ashcroft & Mermin, Solid State Physics, Ch. 2 & 8; Pathria & Beale, Statistical Mechanics (3rd ed.), §8.1; Kardar, Statistical Physics of Particles, Ch. 7; Reif, Fundamentals of Statistical and Thermal Physics, Ch. 9; Sethna, Statistical Mechanics, Ch. 7.
Builds on: Degenerate Fermi Gas at Zero Temperature (owns the $T=0$ filled sphere and $k_F$ ) · Fermi Energy, Momentum and Fermi Temperature (owns $\varepsilon_F,p_F,T_F$ ) · Density of States for Free Particles (owns $g(\varepsilon)\propto\sqrt\varepsilon$ ) · Quantum Statistics of Ideal Gases: Occupation Numbers.
Leads to: Sommerfeld Expansion · Electronic Heat Capacity of Metals · Degeneracy Pressure of Fermions (the $T=0$ pressure of this sphere) · and, downstream, Pauli paramagnetism and the theory of transport in metals.
Provenance: AllQFT `sm` export (Pathria & Beale, §8.1), aq-sm-fermi-surface-fermi-sphere, ref STAT.IQGAS.FSURF.1220. The counting $k_F=(3\pi^2 n)^{1/3}$ , the surface DOS $g(\varepsilon_F)=\tfrac32 n/\varepsilon_F$ , the thermal shell $\Delta k/k_F\sim k_BT/2\varepsilon_F$ , the active fraction $\sim(3/2)T/T_F$ , and the copper/sodium numbers ( $\varepsilon_F=7.0/3.2$ eV, $T_F=8.2/3.8\times10^4$ K, $v_F=1.57\times10^6$ m/s) are python-verified (23 checks, all PASS).

Is the Fermi sphere in real space or momentum space?

Momentum space (equivalently wavevector, $\mathbf{k}$ , space). The "radius" $k_F$ is a wavevector with units of inverse length, not a physical distance. Electrons fill states labeled by momentum, and the sphere is the set of occupied momentum labels. In real space the electron gas is essentially uniform, filling the whole box; there is no spherical droplet of electrons. Confusing the two is the single most common beginner error — always ask "which space?" when someone draws the sphere.

Why does only the Fermi surface matter for a metal's properties?

Because a physical process — absorbing heat, scattering off a phonon, responding to a field — requires an electron to change its state, and that requires an empty state to move into. Deep inside the Fermi sea every neighboring state is already occupied, so those electrons are "Pauli-blocked" and frozen. Only at the surface, where filled and empty states meet, are transitions possible. So the heat capacity, conductivity, magnetic susceptibility, and screening all depend on surface quantities — the density of states $g(\varepsilon_F)$ and the Fermi velocity $v_F$ — not on the bulk of the sea.

How can there be motion at zero temperature if everything is in its ground state?

The ground state of a Fermi gas is not a state of rest. Pauli exclusion forces electrons into successively higher momentum states, so even at $T=0$ the surface electrons move at the Fermi velocity $v_F\sim10^6\,\mathrm{m/s}$ . This is the zero-point kinetic energy of the sea, and it is enormous — it is what produces the degeneracy pressure that holds up white dwarfs. "Zero temperature" means no thermal disorder, not no motion.

What is the difference between the chemical potential and the Fermi energy here?

At exactly $T=0$ they coincide: $\varepsilon_F\equiv\mu(T{=}0)$ is the energy of the highest filled state, the surface energy. At finite temperature the chemical potential $\mu(T)$ drifts (downward for a 3D free gas) while $\varepsilon_F$ stays fixed by the density. For a metal at room temperature the shift is tiny because $T\ll T_F$ , so the distinction rarely matters numerically — but it matters in principle, and the surface is defined by $\mu$ at finite $T$ , sharpening to $\varepsilon_F$ as $T\to0$ .

Why is the enclosed volume fixed but the shape is not?

The shape of the surface is determined by the energy bands $\varepsilon_n(\mathbf{k})$ , which the lattice potential and electron interactions can distort arbitrarily. But the number of occupied states cannot change without changing the number of electrons: the total $\mathbf{k}$ -volume enclosed (per spin, summed over occupied bands) must equal $n/g_s$ times $(2\pi)^3$ . This is Luttinger's theorem, and it holds even with strong interactions. So a metal can have a bizarrely shaped, multi-sheet Fermi surface, yet the volume it wraps is pinned by electron count.

Does the Fermi-sphere picture survive when electrons interact?

Remarkably, yes, thanks to Landau's Fermi-liquid theory. Near the surface the low-energy excitations of an interacting electron gas behave as long-lived "quasiparticles" — dressed electrons with a renormalized mass — that still fill a sharp Fermi surface and obey a Fermi–Dirac distribution in their effective energies. This is why the free-electron sphere, derived for non-interacting particles, describes real metals so accurately. Strong correlations (Mott insulators, some heavy-fermion and cuprate systems) can break the picture, and detecting that breakdown is a frontier of the field.

How is the sharp surface at T=0 consistent with there being ~10^23 electrons?

The sharpness is exactly because there are so many electrons filling a smooth continuum of states. With no thermal energy to promote any electron, every state below $\varepsilon_F$ is occupied with probability one and every state above is empty — the occupation is the step $\theta(\varepsilon_F-\varepsilon)$ , a true discontinuity for the ideal gas. Large numbers make the boundary crisp, not fuzzy; the fuzziness comes only from finite temperature, which smears the step over a shell of width $\sim k_BT$ .