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Drude Theory of Metals

Drude made a theory.

Physics

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Metals

Metals occupy a special place in solid state physics. They share striking properties:

  • High electrical and thermal conductivity
  • Ductility, malleability, and metallic luster

Although most solids are nonmetallic, metals have played a central role in theory from the late 19th century onward. The metallic state is considered a fundamental state of matter — over two-thirds of elements are metals.

To understand nonmetals, it’s often necessary to understand metals first (e.g. copper vs. salt).

Over the past 100+ years, physicists have built models to describe metals both qualitatively and quantitatively.

This chapter introduces the Drude model (ca. 1900):

  • Simple, intuitive picture of metallic conduction
  • Useful for rough estimates, still in use today
  • Fails to explain some key experiments and deeper phenomena

Basic Assumptions of the Drude Model

  • After the discovery of the electron (1897), Drude applied kinetic theory of gases to metals, modeling them as a gas of electrons.
  • In simplest kinetic theory:
    • Electrons are like gas molecules → rigid spheres moving in straight lines until they collide.
    • Collisions are instantaneous and no forces act between them except during collisions.

Electrons in a Metal

  • A metal contains both:
    • Light, mobile valence electrons
    • Heavy, immobile positive ions
  • When atoms condense into a metal:
    • Valence electrons become delocalized → form an electron gas
    • Core electrons + nucleus = immobile ion background

Electron Density

Drude treated conduction electrons as a gas of particles of mass mm. Their density:

n=0.6022×1024ZρmA (in cm3)n = 0.6022 \times 10^{24} \cdot \frac{Z \rho_m}{A} \text{ (in cm}^{-3}\text{)}

Where:

  • ZZ: conduction electrons per atom
  • ρm\rho_m: mass density in g/cm³
  • AA: atomic mass

Also used: Wigner-Seitz radius rsr_s — radius of a sphere per conduction electron:

VN=1n=4πrs33rs=(34πn)1/3\frac{V}{N} = \frac{1}{n} = \frac{4\pi r_s^3}{3} \Rightarrow r_s = \left( \frac{3}{4\pi n} \right)^{1/3}

Key Assumptions of Drude Theory

  1. Independent electron approximation:

    • Ignore electron-electron and electron-ion interactions during collisions.
    • Each electron moves freely between collisions.
  2. Collisions are instantaneous:

    • Modeled as abrupt velocity changes (like bouncing off hard walls).
    • No gradual scattering — just sudden redirection.
  3. Collision rate:

    • Probability of collision per time = 1/τ1/\tau
    • After each collision, an electron emerges with a random velocity, unrelated to its prior motion.
  4. Local thermal equilibrium:

    • Electrons regain thermal distribution after each collision.
    • Hotter regions → faster electrons emerge.

Despite its simplicity, the Drude model gives a surprisingly good first approximation to electron behavior in metals, especially for:

  • Estimating conductivity
  • Understanding basic scattering processes
  • Building intuition for metallic transport

Table: Free Electron Densities

ElementZZn(1022/mathrmcm3)n \\ (10^{22}/\\mathrm{cm}^3)rsr_s (Å)rs/a0r_s / a_0
Li14.701.723.25
Na12.652.083.93
Cu18.471.412.67
Be224.70.991.87
Al318.11.102.07

DC Electrical Conductivity of a Metal

Ohm’s Law:

V=IRV = IR

Drude theory aims to estimate RR from microscopic principles. To eliminate geometry dependence, we define resistivity ρ\rho via:

E=ρj\mathbf{E} = \rho \mathbf{j}

Current Density

Let current II flow through a wire of length LL, cross-sectional area AA:

j=IAj = \frac{I}{A}

Using V=ρjLR=ρLAV = \rho j L \Rightarrow R = \rho \frac{L}{A}

Drude model: electrons of charge e-e, number density nn, average velocity mathbfv\\mathbf{v}:

j=nev\mathbf{j} = -ne \mathbf{v}

Step-by-Step Conductivity Derivation

Starting from the current density equation:

mathbfj=nemathbfv\\mathbf{j} = -ne \\mathbf{v}

The average drift velocity in an electric field is:

vavg=eτmE\mathbf{v}_{\text{avg}} = -\frac{e \tau}{m} \mathbf{E}

Substituting the drift velocity into the current equation:

j=ne(eτmE)\mathbf{j} = -ne \left(-\frac{e \tau}{m} \mathbf{E}\right)

Simplifying the expression:

j=ne2τmE\mathbf{j} = \frac{ne^2 \tau}{m} \mathbf{E}

This gives us the famous Drude conductivity:

σ=ne2τm\sigma = \frac{ne^2 \tau}{m}

So from j=nevavg\mathbf{j} = -ne \mathbf{v}_{\text{avg}}:

j=(ne2τm)E=σE\mathbf{j} = \left( \frac{ne^2 \tau}{m} \right) \mathbf{E} = \sigma \mathbf{E}

with:

σ=ne2τm,ρ=1σ\sigma = \frac{ne^2 \tau}{m}, \rho = \frac{1}{\sigma}

Estimating Relaxation Time

Rearranging:

τ=mρne2\tau = \frac{m}{\rho ne^2}

This allows τ\tau to be estimated from experimental resistivity.

Resistivity vs Temperature

Table 1.2: Selected values of ρ\rho (μΩ⋅cm)

Elementρ\rho(77 K)ρ\rho(273 K)ρ\rho(373 K)ρ/T\rho/T ratio
Cu0.21.562.21.05
Ag0.31.592.31.00
Au0.62.23.21.03
Al0.42.74.31.06
Fe2.210140.96
  • Resistivity rises roughly linearly with TT at room temp.
  • Drops steeply as T0T \to 0
τ=(mρne2)(rsa0)2×1014sec\tau = \left( \frac{m}{\rho n e^2} \right) \sim \left( \frac{r_s}{a_0} \right)^2 \times 10^{-14} \, \text{sec}
  • At room temperature, τ1014sec\tau \sim 10^{-14} \, \text{sec}
  • A more intuitive idea: estimate the mean free path, =vˉτ\ell = \bar{v} \tau, the average distance an electron travels between collisions

Drude Relaxation Times (Approximate, 101410^{-14} sec)

Element77 K273 K373 K
Cu672.41.8
Ag202.01.5
Au122.11.6
Al6.40.800.55
Fe0.920.460.32
  • Relaxation time decreases with increasing temperature
  • At low TT, scattering is dominated by impurities/defects
  • At high TT, phonon scattering dominates

Time-Dependent View: Momentum and Collisions

  • Average electron velocity: v=p(t)/m\mathbf{v} = \mathbf{p}(t)/m
  • So current density is:
j=nep(t)m\mathbf{j} = -\frac{ne \mathbf{p}(t)}{m}

To model p(t)\mathbf{p}(t), assume:

  • Electrons collide randomly with a probability dt/τdt/\tau
  • Between collisions, motion evolves under external fields (electric, magnetic)

Evolution of Electron Momentum

  • Total momentum per electron satisfies:
p(t+dt)=(1dtτ)[p(t)+f(t)dt]\mathbf{p}(t+dt) = \left(1 - \frac{dt}{\tau} \right) [ \mathbf{p}(t) + \mathbf{f}(t)dt ]

where f(t)\mathbf{f}(t) is the force per electron due to external fields.

Taking the limit as dt0dt \to 0:

dp(t)dt=p(t)τ+f(t)\frac{d\mathbf{p}(t)}{dt} = -\frac{\mathbf{p}(t)}{\tau} + \mathbf{f}(t)

This is the Drude momentum equation — Newton’s second law with a damping term due to collisions.


Hall Effect and Magnetoresistance

In 1879, E. H. Hall tested whether magnetic forces act on:

  • The whole wire (current), or
  • Just the moving electrons

He reasoned: if a magnetic field deflects moving charges, they should shift sideways in the conductor → causing a transverse voltage. Hall was able to measure this.

Lorentz Force and Setup

Electric field Ex\mathbf{E}_x drives current density jxj_x in the xx-direction.
Magnetic field H\mathbf{H} points in the zz-direction.

Lorentz force on electrons:

F=ecv×H\mathbf{F} = -\frac{e}{c} \, \mathbf{v} \times \mathbf{H}
  • Electrons deflect in y-y direction
  • Accumulate on one side → build up transverse electric field Ey\mathbf{E}_y
  • This transverse field cancels Lorentz force in steady state

Measured Quantities

Two main observables:

  1. Magnetoresistance (longitudinal):

    ρ(H)=Exjx\rho(H) = \frac{E_x}{j_x}
  2. Hall Coefficient:

    RH=EyjxHR_H = \frac{E_y}{j_x H}
  • Negative RHR_H implies negative charge carriers (electrons)
  • Surprisingly, some metals show positive RHR_H

Hall Effect Analysis

Starting with the Lorentz force on moving electrons:

F=ecv×H\mathbf{F} = -\frac{e}{c} \mathbf{v} \times \mathbf{H}

Newton’s second law with all forces (electric field, magnetic field, and collisions):

dpdt=eEecv×Hpτ\frac{d\mathbf{p}}{dt} = -e\mathbf{E} - \frac{e}{c} \mathbf{v} \times \mathbf{H} - \frac{\mathbf{p}}{\tau}

In steady state, breaking into x and y components:

0=eExωcpypxτ0 = -eE_x - \omega_c p_y - \frac{p_x}{\tau}

0=eEy+ωcpxpyτ0 = -eE_y + \omega_c p_x - \frac{p_y}{\tau}

This leads to the Hall coefficient:

RH=EyjxH=1necR_H = \frac{E_y}{j_x H} = -\frac{1}{nec}

Total force per electron:

dpdt=e(E+pmc×H)pτ\frac{d\mathbf{p}}{dt} = -e \left( \mathbf{E} + \frac{\mathbf{p}}{mc} \times \mathbf{H} \right) - \frac{\mathbf{p}}{\tau}

Steady-state: dpdt=0\frac{d\mathbf{p}}{dt} = 0

Break into components:

0=eExωcpypxτ0=eEy+ωcpxpyτ\begin{aligned} 0 &= -eE_x - \omega_c p_y - \frac{p_x}{\tau} \\ 0 &= -eE_y + \omega_c p_x - \frac{p_y}{\tau} \end{aligned}

Where cyclotron frequency:

ωc=eHmc\omega_c = \frac{eH}{mc}

Current Components

From j=nev=nemp\mathbf{j} = -ne\mathbf{v} = -\frac{ne}{m}\mathbf{p}, we get:

σ0Ex=ωcτjy+jxσ0Ey=ωcτjx+jy\begin{aligned} \sigma_0 E_x &= \omega_c \tau \, j_y + j_x \\ \sigma_0 E_y &= -\omega_c \tau \, j_x + j_y \end{aligned}
  • σ0\sigma_0 is the Drude conductivity with no magnetic field
  • The angle ϕ\phi between E\mathbf{E} and j\mathbf{j} is the Hall angle: tanϕ=ωcτ\tan \phi = \omega_c \tau

Table: Hall Coefficients at High Fields

MetalValence1/RHnec-1/R_H n e c
Li10.8
Na11.1
K11.2
Rb11.1
Cs10.9
Cu11.5
Ag11.3
Au11.5
Be20.2-0.2
Mg20.4-0.4
In30.3-0.3
Al30.3-0.3
  • For alkali metals, Drude’s prediction works well: one electron per atom.

  • For multivalent metals (Be, Mg, In, Al), observed values deviate from simple theory — sometimes even suggesting positive charge carriers.

  • RHR_H depends only on carrier type and density

  • Surprisingly, in real materials:

    • RHR_H often deviates from the Drude prediction
    • Depends on magnetic field strength and sample preparation
  • Quantum theory is required to explain deviations

Drude model predicts:

RH=1necR_H = -\frac{1}{nec}

Cyclotron Frequency and Magnetic Effects

The cyclotron frequency:

ωc=eHmc=2πνc\omega_c = \frac{eH}{mc} = 2\pi \nu_c

With:

νc (GHz)=2.80×H (kG)\nu_c \ (\text{GHz}) = 2.80 \times H \ (\text{kG})
  • ωcτ\omega_c \tau is a dimensionless measure of magnetic field strength
  • If ωcτ1\omega_c \tau \ll 1: field barely perturbs orbits
  • If ωcτ1\omega_c \tau \gtrsim 1: strong deflection → major magnetic effects

AC Electrical Conductivity of a Metal

Let the electric field oscillate with frequency ω\omega:

E(t)=Re[E(ω)eiωt]\mathbf{E}(t) = \text{Re}\left[ \mathbf{E}(\omega) e^{-i\omega t} \right]

Momentum equation:

dpdt=pτeE\frac{d\mathbf{p}}{dt} = -\frac{\mathbf{p}}{\tau} - e\mathbf{E}

Assume steady-state solution:

p(t)=Re[p(ω)eiωt]\mathbf{p}(t) = \text{Re}\left[ \mathbf{p}(\omega) e^{-i\omega t} \right]

So that,

iωp(ω)=p(ω)τeE(ω)-i\omega \mathbf{p}(\omega) = -\frac{\mathbf{p}(\omega)}{\tau} - e\mathbf{E}(\omega)

Current density:

j(ω)=nep(ω)m=ne2τmE(ω)1iωτ\mathbf{j}(\omega) = -\frac{ne\mathbf{p}(\omega)}{m} = \frac{ne^2 \tau}{m} \cdot \frac{\mathbf{E}(\omega)}{1 - i\omega \tau}

Define frequency-dependent (AC) conductivity:

j(ω)=σ(ω)E(ω),σ(ω)=σ01iωτ,σ0=ne2τm\mathbf{j}(\omega) = \sigma(\omega)\mathbf{E}(\omega), \sigma(\omega) = \frac{\sigma_0}{1 - i\omega\tau}, \sigma_0 = \frac{ne^2\tau}{m}

Electromagnetic Wave Propagation

We derive wave equations from Maxwell’s laws:

E=0,H=0,×E=1cHt×H=4πcj+1cEt\nabla \cdot \mathbf{E} = 0, \nabla \cdot \mathbf{H} = 0, \nabla \times \mathbf{E} = -\frac{1}{c} \frac{\partial \mathbf{H}}{\partial t} \\ \nabla \times \mathbf{H} = \frac{4\pi}{c} \mathbf{j} + \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}

Assume eiωte^{-i\omega t} time dependence and substitute j=σ(ω)E\mathbf{j} = \sigma(\omega) \mathbf{E}:

×(×E)=2E=ω2c2(1+4πiσω)E\nabla \times (\nabla \times \mathbf{E}) = -\nabla^2 \mathbf{E} = \frac{\omega^2}{c^2} \left( 1 + \frac{4\pi i \sigma}{\omega} \right) \mathbf{E}

Wave equation with complex dielectric:

2E=ω2c2ϵ(ω)E,ϵ(ω)=1+4πiσω\nabla^2 \mathbf{E} = \frac{\omega^2}{c^2} \epsilon(\omega) \mathbf{E}, \epsilon(\omega) = 1 + \frac{4\pi i \sigma}{\omega}

High-Frequency Limit and Plasma Frequency

If ωτ1\omega \tau \gg 1, then:

σ(ω)σ0iωτ,ϵ(ω)1ωp2ω2\sigma(\omega) \approx \frac{\sigma_0}{-i\omega\tau}, \epsilon(\omega) \approx 1 - \frac{\omega_p^2}{\omega^2}

Define plasma frequency:

ωp2=4πne2m\omega_p^2 = \frac{4\pi ne^2}{m}

No propagation if ω<ωp\omega < \omega_p (since ϵ<0\epsilon < 0).

Estimate:

ωpτ=1.6×102(rsa0)3/2(1ρμ)\omega_p \tau = 1.6 \times 10^2 \left( \frac{r_s}{a_0} \right)^{3/2} \left( \frac{1}{\rho_\mu} \right)

Transparency Threshold

Frequency and wavelength at which metals become transparent:

νp=ωp2π=11.4(rsa0)3/2×1015 Hz\nu_p = \frac{\omega_p}{2\pi} = 11.4 \left( \frac{r_s}{a_0} \right)^{-3/2} \times 10^{15} \ \text{Hz} λp=cνp=0.26(rsa0)3/2×103 A˚\lambda_p = \frac{c}{\nu_p} = 0.26 \left( \frac{r_s}{a_0} \right)^{3/2} \times 10^3 \ \text{Å}

Table 1.5: Transparency Threshold (approximate)

ElementTheoretical λ\lambda (10³ Å)Observed λ\lambda (10³ Å)
Li1.52.0
Na2.02.1
K2.83.1
Rb3.13.6
Cs3.54.4

Plasma Oscillations (Plasmons)

Charge oscillations with time dependence eiωte^{-i\omega t}.

From continuity and Gauss’s law:

j=ρtj(ω)=iωρ(ω)E(ω)=4πρ(ω)\nabla \cdot \mathbf{j} = -\frac{\partial \rho}{\partial t} \Rightarrow \nabla \cdot \mathbf{j}(\omega) = i\omega \rho(\omega) \\ \nabla \cdot \mathbf{E}(\omega) = 4\pi \rho(\omega)

Using j(ω)=σ(ω)E(ω)\mathbf{j}(\omega) = \sigma(\omega) \mathbf{E}(\omega):

iωρ(ω)=4πσ(ω)ρ(ω)1+4πiσ(ω)ω=0i\omega \rho(\omega) = 4\pi \sigma(\omega) \rho(\omega) \Rightarrow 1 + \frac{4\pi i \sigma(\omega)}{\omega} = 0

Same condition as for electromagnetic wave propagation: ϵ(ω)=0\epsilon(\omega) = 0.

Simple Plasma Oscillation Model

Assume slab of NN electrons displaced by dd:

  • Surface charge: σ=±nde\sigma = \pm nde
  • Electric field: E=4πndeE = 4\pi nde
  • Equation of motion: Nmd¨=Ne(4πnde)=4πne2NdNm \ddot{d} = -Ne(4\pi nde) = -4\pi ne^2 N d

→ Harmonic oscillator at ωp\omega_p.

Plasmons observable via electron energy loss spectroscopy.


Thermal Conductivity of a Metal

Wiedemann–Franz Law

One of the Drude model’s key successes was explaining the Wiedemann–Franz law:

κσT\frac{\kappa}{\sigma} \propto T

This states that the ratio of thermal conductivity κ\kappa to electrical conductivity σ\sigma is proportional to TT. The constant of proportionality is the Lorenz number:

κσT=L2.22×108 WΩ/K2\frac{\kappa}{\sigma T} = L \approx 2.22 \times 10^{-8}~\text{W} \cdot \Omega/\text{K}^2

This relationship holds remarkably well across many metals.

Experimental Values

Elementκ\kappa (273 K)κ/σT\kappa/\sigma T (273 K)κ/σT\kappa/\sigma T (373 K)
Cu3.852.202.29
Ag4.182.312.38
Au3.102.322.36
Fe0.802.612.88
Tl0.502.752.75

Units:

  • κ\kappa: W/cm·K
  • κ/σT\kappa/\sigma T: W·Ω/K²

Heat Conduction in Metals

Drude assumed:

  • Heat is carried by conduction electrons, not ions
  • Electron collisions randomize direction
  • Temperature gradient leads to a net thermal current

Thermal current density (Fourier’s Law):

jq=κT\vec{j}^q = -\kappa \nabla T

Microscopic Picture

Electrons from hotter side carry more energy than those from colder side. Assume:

  • Electrons come equally from left (xvtx - vt) and right (x+vtx + vt)
  • Energy per electron: ε(T)\varepsilon(T)

Thermal current density:

jq=12nv[ε(T[xvt])ε(T[x+vt])]j^q = \frac{1}{2}nv[\varepsilon(T[x - vt]) - \varepsilon(T[x + vt])]

For small gradients:

jq=nv2τdεdT(dTdx)j^q = nv^2 \tau \frac{d\varepsilon}{dT} \left(-\frac{dT}{dx}\right)

In 3D:

jq=13v2τcv(T)(Eq. 1.50)j^q = \frac{1}{3}v^2 \tau c_v (-\nabla T) \quad \text{(Eq. 1.50)} κ=13v2τcv\kappa = \frac{1}{3}v^2 \tau c_v

Wiedemann–Franz from Drude

Using:

  • cv=32nkBc_v = \frac{3}{2}n k_B
  • v2=3kBTmv^2 = \frac{3k_B T}{m}

We get:

κσ=32(kBe)2T\frac{\kappa}{\sigma} = \frac{3}{2}\left(\frac{k_B}{e}\right)^2 T κσT=32(kBe)2=1.11×108 WΩ/K2\frac{\kappa}{\sigma T} = \frac{3}{2}\left(\frac{k_B}{e}\right)^2 = 1.11 \times 10^{-8}~\text{W} \cdot \Omega/\text{K}^2

This is half the experimental value. Drude’s estimate was off due to errors in both cvc_v and v2v^2, which accidentally canceled.

Thermoelectric Field (Seebeck Effect)

A temp gradient also produces an electric field:

E=QT\vec{E} = Q \nabla T

For no net current (vQ+vE=0v_Q + v_E = 0), use:

  • Drift from temp gradient:

    vQ=τddx(v22)=τvdTdx(1D)v_Q = -\tau \frac{d}{dx} \left( \frac{v^2}{2} \right) = -\tau v \frac{dT}{dx} \text{(1D)} vQ=τ6dv2dTT(3D)v_Q = -\frac{\tau}{6} \frac{dv^2}{dT} \nabla T \text{(3D)}
  • Drift from electric field:

    vE=eEτmv_E = -\frac{eE\tau}{m}

To cancel, require:

Q=13eddT(mv22)=cv3neQ = -\frac{1}{3e} \frac{d}{dT} \left( \frac{mv^2}{2} \right) = -\frac{c_v}{3ne} Q=kB2e0.43×104 V/KQ = -\frac{k_B}{2e} \approx -0.43 \times 10^{-4}~\text{V/K}