---
title: Conduction and the Free-Electron Gas
module: Free-Electron Fermi Gas
moduleNumber: 5
lessonNumber: 1
order: 501
summary: >
  Drude's classical free-electron model gets Ohm's law right but the resistivity,
  its temperature dependence, and the heat capacity wrong. Replacing the
  Maxwell-Boltzmann distribution with the Fermi-Dirac distribution and treating
  electron-lattice collisions as wave scattering repairs all three: the Fermi
  energy, Fermi speed, and a mean free path set by thermal lattice vibrations.
topics: [Free-Electron Fermi Gas]
draft: false
sources:
  - book: Tipler & Llewellyn
    ref: "Ch. 10 — Solid State Physics; §10-2 Classical Theory of Conduction, §10-3 Free-Electron Gas in Metals, §10-4 Quantum Theory of Conduction"
  - book: Ashcroft & Mermin
    ref: "Ch. 1 — The Drude Theory of Metals"
---

Metals conduct because they contain electrons free to move through a lattice of
fixed positive ions. Paul Drude proposed this picture around 1900, three years
after the electron was discovered. It predicts Ohm's law and links electrical to
thermal conduction, but it gets three quantities wrong, and each failure is
repaired by the same two quantum corrections: the electrons obey the
[Fermi-Dirac distribution](/statistical-mechanics/quantum-statistics/quantum-statistics-bose-einstein-and-fermi-dirac)
rather than the classical one, and they scatter as waves rather than as
particles.

## The Drude model

In the classical model the free electrons move at high thermal speed even with no
applied field. At $T = 300\ \text{K}$,

$$
\langle v \rangle = \sqrt{\frac{8kT}{\pi m_e}} = 1.08\times10^5\ \text{m/s}.
$$

An applied field $\mathcal{E}$ superimposes a small **drift velocity** $v_d$
opposite to the field. If there are $n$ electrons per unit volume, the current
through area $A$ is

$$
I = neAv_d, \qquad j = \frac{I}{A} = nev_d.
$$

$$
% caption: The Drude picture: an electron moves at high thermal speed and
% scatters off lattice ions, with a small drift velocity v_d superimposed by the
% applied field; the charge in the shaded length passes area A per unit time.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % ions on a grid
  \foreach \i in {0,1,2,3,4,5}{
    \foreach \j in {0,1,2}{
      \fill[black] (\i*1.1,\j*1.0) circle (2.2pt);
    }
  }
  % zig-zag electron path drifting right
  \draw[acc, very thick] (0.2,0.6) -- (0.9,1.5) -- (1.7,0.4) -- (2.6,1.6)
    -- (3.4,0.5) -- (4.3,1.7) -- (5.1,0.6);
  \node[acc, anchor=south] at (2.7,1.75) {electron path};
  % drift direction (opposite the applied field)
  \draw[black, ->, thick] (0.3,-0.7) -- (2.0,-0.7) node[right] {drift $v_d$};
\end{tikzpicture}
$$

For copper carrying $1\ \text{A}$ in a wire of radius $0.815\ \text{mm}$, the
free-electron density (one per atom) is $n = 8.47\times10^{28}\ \text{m}^{-3}$
and the drift velocity is $v_d \approx 3.5\times10^{-5}\ \text{m/s}$ — about $0.01\
\text{mm/s}$, ten orders of magnitude below the thermal speed. Charge moves
slowly even though the signal travels near light speed.

The resistivity follows from the **mean free path** $\lambda = \langle v \rangle
\tau$, the average distance between collisions:

$$
\rho = \frac{m_e \langle v \rangle}{n e^2 \lambda}, \qquad \sigma = \frac{1}{\rho} = \frac{n e^2 \lambda}{m_e \langle v \rangle}.
$$

> **Definition (Mean free path).** The average distance $\lambda$ an electron
> travels between scattering events. Classically it is set by the geometric cross
> section $\pi r^2$ of the ions, $\lambda = 1/(n_a \pi r^2)$; for copper this
> gives $\lambda \approx 0.38\ \text{nm}$, roughly one lattice spacing.

The model fails in three ways:

- **Magnitude.** Using $\langle v \rangle$ from the Maxwell-Boltzmann
  distribution, the predicted resistivity of copper is about **7 times** the
  measured value.
- **Temperature dependence.** Experiment gives $\rho \propto T$, but the model
  gives $\rho \propto \langle v \rangle \propto T^{1/2}$.
- **Heat capacity.** A classical electron gas should add $\tfrac32 R$ to the
  molar heat capacity, so metals would have $C_v = 3R + \tfrac32 R = \tfrac92 R$.
  The measured value is very nearly $3R$; the electron contribution is only about
  $0.02R$.

All three come from the same mistake: electrons are not a classical gas.

## The Fermi gas

Because electrons obey the [exclusion
principle](/atomic-physics/many-electron-atoms/periodic-table-atomic-spectra), no more
than two (opposite spins) share a level. At $T = 0$ they fill the lowest levels
up to the **Fermi energy** $E_F$. For a three-dimensional box of volume $V$ with
$N$ electrons, filling states up to $E_F$ gives

$$
E_F = \frac{h^2}{2m}\left(\frac{3N}{8\pi V}\right)^{2/3},
$$

and the average electron energy is

$$
\langle E \rangle = \frac{3}{5}E_F,
$$

not the classical $\tfrac32 kT$. For copper, $N/V = 8.47\times10^{28}\
\text{m}^{-3}$ gives $E_F = 7.06\ \text{eV}$. Defining the **Fermi temperature**
$T_F = E_F/k$, copper has $T_F \approx 82{,}000\ \text{K}$ — far above any
temperature at which it is solid.

> **Definition (Fermi energy and temperature).** The **Fermi energy** $E_F$ is
> the energy of the highest occupied state at $T = 0$; the **Fermi temperature**
> $T_F = E_F/k$ is the temperature at which $kT$ would equal $E_F$. Because $T \ll
> T_F$ for all metals, the electron gas is strongly **degenerate**: its
> distribution barely differs from the $T = 0$ step.

$$
% caption: The number of electrons per energy, n(E) ∝ √E, filled to the Fermi
% energy at T = 0 (solid edge); raising T smears only a slice of width ~kT near
% E_F (dashed), so most electrons cannot absorb thermal energy.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (7.2,0) node[below] {$E$};
  \draw[->, black] (0,0) -- (0,3.6) node[left] {$n(E)$};
  % sqrt curve
  \draw[acc, thick, domain=0:5.0, samples=80, variable=\x]
    plot ({\x},{1.35*sqrt(\x)});
  % filled region shaded up to E_F at x=5
  \fill[acc!14] (0,0) -- plot[domain=0:5.0, samples=80, variable=\x] ({\x},{1.35*sqrt(\x)}) -- (5.0,0) -- cycle;
  % vertical drop at E_F
  \draw[acc, thick] (5.0,0) -- (5.0,3.02);
  \draw[black, dashed] (5.0,0) -- (5.0,-0.05) node[below] {$E_F$};
  % smeared tail
  \draw[black, dashed, domain=4.3:6.2, samples=40, variable=\x]
    plot ({\x},{3.02/(1+exp((\x-5.0)*6))});
  \node[black, anchor=west, font=\scriptsize] at (5.15,1.4) {width $\sim kT$};
\end{tikzpicture}
$$

Only electrons within about $kT$ of $E_F$ have empty states to move into, so only
they can absorb thermal energy. At $T = 300\ \text{K}$, $kT = 0.026\ \text{eV}
\ll E_F$, so only a tiny fraction $\sim kT/E_F$ participate. This immediately
fixes the heat capacity: a full calculation (Sommerfeld) gives

$$
C_v(\text{electrons}) = \frac{\pi^2}{2}\frac{T}{T_F}R.
$$

For copper at $300\ \text{K}$ this is $\approx 0.018R$, matching experiment,
because the huge $T_F$ suppresses it. The electron gas is nearly frozen out of
the heat capacity.

$$
% caption: The classical Dulong-Petit prediction adds a constant (3/2)R electron
% term at all temperatures; the quantum electron contribution is instead linear
% in T and tiny, so metals show only the (3R) lattice value near room temperature.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.6,0) node[below] {$T$};
  \draw[->, black] (0,0) -- (0,3.8) node[left] {$C_v$};
  % classical constant electron term (3/2 R) as dashed high line
  \draw[black, dashed] (0,3.0) -- (6.2,3.0);
  \node[black, anchor=west, font=\scriptsize] at (3.2,3.45) {classical: $\tfrac92 R$};
  % lattice 3R line
  \draw[black, thick] (0.4,2.0) -- (6.2,2.0);
  \node[black, anchor=west, font=\scriptsize] at (4.6,1.8) {lattice $3R$};
  % quantum electron term: linear small
  \draw[acc, very thick, domain=0:6.0, samples=40, variable=\x]
    plot ({\x},{0.09*\x});
  \node[acc, anchor=west, font=\scriptsize] at (4.4,0.65) {quantum electrons $\propto T$};
\end{tikzpicture}
$$

## Quantum conduction

The other two failures are fixed by the wave nature of the electron. In an
electric field the whole Fermi distribution shifts slightly in velocity, and the
net current comes from the electrons near the Fermi surface. So the classical
resistivity formula still holds if $\langle v \rangle$ is replaced by the
**Fermi speed**

$$
u_F = \sqrt{\frac{2E_F}{m_e}},
$$

about $1.6\times10^6\ \text{m/s}$ for copper — some 15 times the classical
thermal speed. That would make the predicted resistivity worse, not better,
unless the mean free path is also reconsidered.

$$
% caption: In an electric field the entire Fermi distribution in velocity shifts
% by a small amount, so the net current is carried by the electrons near the
% Fermi speed u_F rather than by all of them; the shift is exaggerated here.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-3.2,0) -- (3.2,0) node[below] {$v_x$};
  \draw[->, black] (0,0) -- (0,2.4) node[left] {$f_{FD}$};
  % unshifted: symmetric box
  \draw[black, thick] (-2.0,0) -- (-2.0,1.6) -- (2.0,1.6) -- (2.0,0);
  \node[black, anchor=south, font=\scriptsize] at (-1.3,1.62) {unshifted};
  \node[anchor=north, font=\scriptsize] at (2.0,0) {$u_F$};
  % shifted box (dashed)
  \draw[acc, thick, dashed] (-1.6,0) -- (-1.6,1.3) -- (2.4,1.3) -- (2.4,0);
  \node[acc, anchor=south, font=\scriptsize] at (2.4,1.32) {shifted};
\end{tikzpicture}
$$

The key correction: a perfectly ordered lattice does **not** scatter an electron
wave at all — the mean free path would be infinite. Scattering arises only from
**deviations** from perfect order, chiefly the thermal vibrations of the ions.
Treating the ions as points vibrating with mean-square displacement $\langle r^2
\rangle$, the scattering cross section is $\pi \langle r^2 \rangle$, and the
[equipartition theorem](/statistical-mechanics/foundations/classical-statistics-and-equipartition)
gives $\tfrac12 M\omega^2 \langle r^2 \rangle = kT$, so $\langle r^2 \rangle
\propto T$. The mean free path becomes

$$
\lambda = \frac{1}{n\,\pi\langle r^2 \rangle} \propto \frac{1}{T}.
$$

$$
% caption: Quantum scattering: the relevant cross section is not the ion's
% geometric area but the area π⟨r²⟩ swept by its thermal vibration, which grows
% with temperature and shrinks the mean free path as 1/T.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % classical: big solid ion
  \begin{scope}
    \draw[black, thick, fill=black!8] (0,0) circle (0.7);
    \fill[black] (0,0) circle (2pt);
    \node[anchor=north] at (0,-0.85) {classical: ion radius $r$};
  \end{scope}
  % quantum: point vibrating, area from amplitude
  \begin{scope}[xshift=5.0cm]
    \draw[acc, thick, dashed] (0,0) circle (0.55);
    \fill[black] (0.12,0.08) circle (1.6pt);
    \fill[black] (-0.14,-0.1) circle (1.6pt);
    \fill[black] (0.05,-0.15) circle (1.6pt);
    \node[anchor=north] at (0,-0.85) {quantum: vibration area};
  \end{scope}
\end{tikzpicture}
$$

Two results fall out. Since $\rho \propto 1/\lambda \propto T$, the observed
**linear** temperature dependence is recovered. And the numerical value comes out
about 100 times larger for $\lambda$ than the ion-size estimate — for copper
$\lambda \approx 39\ \text{nm}$ — which cancels the extra factor from $u_F$ and
brings the resistivity into agreement with experiment.

Impurities break the lattice order too, and their effect is temperature
independent. The resistivity splits into two additive parts (Matthiessen's rule),

$$
\rho = \rho_t + \rho_I,
$$

with $\rho_t$ from thermal vibrations and $\rho_I$ from impurities. As $T \to 0$
the thermal part vanishes and the resistivity approaches the constant residual
$\rho_I$ — a purer sample has a lower floor.

The free-electron gas explains conduction, heat capacity, and thermal
conductivity, but it treats every metal alike and says nothing about why some
solids are insulators or semiconductors. The
[Sommerfeld model](/condensed-matter/free-electron-fermi-gas/sommerfeld-model-and-heat-capacity)
of the next lesson makes the counting quantitative, and the eventual distinction
between metals, insulators, and semiconductors requires the effect of the
periodic lattice on the electron energies themselves — the
[band theory](/condensed-matter/band-theory/blochs-theorem-and-energy-bands) that
closes this half of the course.