Drude Model and Metallic Optical Response

We take the machinery of the conductor from electromagnetism and turn it toward light. Three companion articles own the electrodynamics of a metal: aq-em-waves-in-conductors derives the complex wavevector $\tilde{k}^2 = \mu\varepsilon\omega^2 + i\mu\sigma\omega$ from Ohm's law and the loss-tangent dial $\sigma/\varepsilon\omega$ ; aq-em-skin-depth-reflection-conductors turns the skin depth $\delta = \sqrt{2/\mu\sigma\omega}$ into the surface resistance $R_s = 1/\sigma\delta$ and the reflection loss $1 - R = \sqrt{8\varepsilon_0\omega/\sigma}$ ; and aq-em-plasma-frequency-dielectric-response builds the free-electron collective mode $\omega_p = \sqrt{n_e e^2/\varepsilon_0 m_e}$ and its lossless dielectric $\varepsilon_r = 1 - \omega_p^2/\omega^2$ with total reflection $|r| = 1$ below cutoff, and records Wood's 1933 finding that sodium turns transparent near $209\ \mathrm{nm}$ ( $\hbar\omega_p = 5.9\ \mathrm{eV}$ ). Those results are recalled here by name, not re-derived. What this article owns is the optical response of a real metal across the whole spectrum. Paul Drude in 1900 treated the conduction electrons as a classical gas — each obeying Newton's equation under the optical field and a friction force for collisions — with, crucially, no restoring spring: a conduction electron has no home to return to. That single change from the bound-electron oscillator, sending the resonance frequency $\omega_0$ to zero, produces the complex permittivity $\varepsilon(\omega) = 1 - \omega_p^2/(\omega^2 + i\gamma\omega)$ . Below the plasma frequency $\varepsilon$ is large and negative, the wave is evanescent, and the metal reflects almost everything; above $\omega_p$ it turns positive and light slips through. This article follows that one function from radio to ultraviolet, cashes it out in the sodium and gold numbers, and marks the honest edge where the free-electron picture fails: the yellow of gold is interband physics, not Drude.

Key takeaways

The Drude model is the Lorentz oscillator with the spring removed ( $\omega_0 \to 0$ ): conduction electrons respond to the optical field with inertia and collisional damping but no restoring force.
Solving the driven equation of motion gives $\varepsilon(\omega) = 1 - \dfrac{\omega_p^2}{\omega^2 + i\gamma\omega}$ , with the plasma frequency $\omega_p = \sqrt{n_e e^2/\varepsilon_0 m_e}$ setting the response scale and $\gamma = 1/\tau$ the collision rate.
Below $\omega_p$ the real permittivity is negative, the wave is evanescent, and reflectance approaches one — the metallic mirror; above $\omega_p$ the metal becomes transparent, the ultraviolet transparency of the alkalis.
The same collision rate that damps the optical response gives the DC conductivity $\sigma_0 = n_e e^2\tau/m = \varepsilon_0\omega_p^2/\gamma$ as $\omega \to 0$ — optics and bench electricity share one electron response.
Sodium: $\omega_p = 8.99\times10^{15}\ \mathrm{rad/s}$ , $\lambda_p = 209.5\ \mathrm{nm}$ ( $\hbar\omega_p = 5.9\ \mathrm{eV}$ ), $\sigma_0 = 2.22\times10^{7}\ \mathrm{S/m}$ against a measured $2.1\times10^{7}$ ; gold: $\lambda_p = 137.5\ \mathrm{nm}$ , $\hbar\omega_p = 9.0\ \mathrm{eV}$ .
Key idea: a metal is a Lorentz oscillator with zero resonance frequency — its free electrons make the permittivity negative below the plasma frequency (reflective) and positive above it (transparent), so one function $\varepsilon(\omega)$ carries metallic shine, the plasma edge, and the DC conductivity all at once.

The core insight

A metal is a Lorentz oscillator with no restoring force: the free electrons give a complex permittivity $\varepsilon(\omega) = 1 - \omega_p^2/(\omega^2 + i\gamma\omega)$ that is negative below the plasma frequency $\omega_p$ — making the metal a mirror — and positive above it, making it transparent. The same model links the optical response to the DC conductivity $\sigma_0 = n_e e^2\tau/m$ ; where it fails (the colour of gold) is exactly where quantum band structure takes over.

Where this lives today

The Drude permittivity is the foundation of plasmonics. A metal with $\varepsilon < 0$ beside a dielectric supports surface plasmon polaritons and nanoparticle resonances that confine light below the diffraction limit; these drive surface-enhanced Raman spectroscopy, plasmonic biosensors, and near-field optics, all reading the Drude $\varepsilon(\omega)$ of gold and silver. The plasma edge is also tunable: because $\omega_p^2 \propto n_e$ , doping a transparent conducting oxide such as indium tin oxide places the plasma frequency in the infrared, giving epsilon-near-zero materials and switchable infrared modulators and smart windows. At the metrology bench, Drude's own tool endures: spectroscopic ellipsometry fits a Drude term (plus interband oscillators) to measured optical constants and is the semiconductor industry's standard film-thickness and carrier-density probe. And metamaterial wire lattices synthesise an artificial $\omega_p$ at gigahertz frequencies by diluting the effective electron density — Drude's classical gas, engineered to order.

The physical picture

Picture the conduction electrons as a slab of charged jelly floating on the fixed lattice of ions. Push the slab with a field and it slides freely; the only resistance is friction from collisions with impurities, phonons, and defects. Because no spring pulls it back, its low-frequency response is enormous, which drives the permittivity deeply negative. The plasma frequency is the natural sloshing rate of that whole slab against the ion background: displace it uniformly and the exposed surface charge builds a restoring field, so the gas rings at $\omega_p$ . An optical wave slower than that rate cannot make the electrons follow in a propagating way — it is expelled and reflected; a wave faster than $\omega_p$ outruns the collective response and passes through. Damping turns a perfect mirror into a slightly lossy one. Each collision randomises an electron's velocity, converting coherent oscillation into heat, and in the DC limit this same friction is ordinary Ohmic resistance — which is why the optical $\gamma$ and the conductivity $\sigma_0$ are two faces of one coin. The whole shape of metallic reflectivity from radio to ultraviolet is encoded in the single function $\varepsilon(\omega)$ .

Figure

The real Drude permittivity ε₁(ω) crossing zero at the plasma frequency: negative below ωₚ where the wave is evanescent and the metal reflects, positive above it where light propagates and the metal is transparent

The mirror lives on the red side of that crossing, the transparent window on the blue. For a good metal the plasma frequency sits in the ultraviolet, so the whole visible spectrum falls on the negative- $\varepsilon$ side and the metal looks like a mirror. Sodium is the clean punchline: it mirrors the entire visible band, then its reflectivity collapses in the ultraviolet.

Figure

Reflectance of sodium versus wavelength: R near unity across the whole visible band, then a cliff at the plasma edge λₚ = 209.5 nm into a deep-ultraviolet transparent window

The core equations

Everything follows from one equation of motion. A conduction electron carries the inertial and frictional terms of the Lorentz oscillator but not its spring, because there is no equilibrium position to return to. Setting the restoring constant to zero is the whole content of the model.

m\ddot{\mathbf r} + m\gamma\dot{\mathbf r} = -e\mathbf E(t) \qquad (1)

where: $m$ — electron (effective) mass; $\mathbf r$ — displacement; $\gamma = 1/\tau$ — collision damping rate; $\tau$ — mean free time; $-e$ — electron charge; $\mathbf E(t) \propto e^{-i\omega t}$ — optical field. The physical reading: this is the Lorentz equation with $\omega_0 = 0$ , so every result of dielectric dispersion carries over with the resonance deleted. Solving in steady state, forming the polarization, and reading off $\mathbf D = \varepsilon_0\varepsilon\mathbf E$ (Derivation A) gives the central result. Two material numbers govern it: the plasma frequency, fixed by the electron density, and the damping rate, fixed by the collision time.

\varepsilon(\omega) = 1 - \frac{\omega_p^2}{\omega^2 + i\gamma\omega}, \qquad \omega_p = \sqrt{\frac{n_e e^2}{\varepsilon_0 m}} \qquad (2)

where: $\varepsilon(\omega)$ — complex relative permittivity (background $\varepsilon_\infty = 1$ for a simple metal); $\omega_p$ — plasma frequency; $n_e$ — conduction-electron density; $\varepsilon_0$ — vacuum permittivity. The physical reading: the real part $\varepsilon' = 1 - \omega_p^2/(\omega^2 + \gamma^2)$ decides reflection, and the imaginary part $\varepsilon'' = \omega_p^2\gamma/[\omega(\omega^2 + \gamma^2)]$ carries the Ohmic loss. Taken to zero frequency, the same response must reproduce Ohm's law. The current density $\mathbf J = -n_e e\dot{\mathbf r}$ compared with $\mathbf J = \sigma\mathbf E$ (Derivation C) gives the AC conductivity, whose DC limit is the classical Drude result.

\sigma_0 = \frac{n_e e^2\tau}{m} = \varepsilon_0\,\frac{\omega_p^2}{\gamma} \qquad (3)

where: $\sigma_0$ — static conductivity; $\tau = 1/\gamma$ — relaxation time. The physical reading: a single collision time $\tau$ fixes both the optical damping and the electrical resistance — the bridge between metal optics and bench conductivity. The optical observables come through the complex index. Once $\varepsilon(\omega)$ is known, the refractive index and normal-incidence reflectance follow from the Fresnel coefficient, unchanged except that $\tilde{n}$ is now complex.

\tilde{n} = n + i\kappa = \sqrt{\varepsilon(\omega)}, \qquad R = \left|\frac{\tilde{n} - 1}{\tilde{n} + 1}\right|^2 \qquad (4)

where: $n$ — refractive index; $\kappa$ — extinction coefficient; $R$ — power reflectance at normal incidence. The physical reading: where $\varepsilon$ is negative, $\tilde{n}$ is nearly pure imaginary, and this Fresnel formula returns $R$ close to one — a metal mirror. Strip the damping ( $\gamma \to 0$ ) and the plasma edge stands bare — the lossless dielectric owned by aq-em-plasma-frequency-dielectric-response, met here as the boundary between a reflecting and a transparent metal.

\varepsilon(\omega) = 1 - \frac{\omega_p^2}{\omega^2} \;\Rightarrow\; \begin{cases} \varepsilon < 0,\ R = 1 & \omega < \omega_p \\[2pt] \varepsilon > 0,\ n < 1 & \omega > \omega_p \end{cases} \qquad (5)

where: the crossing $\varepsilon(\omega_p) = 0$ marks the reflectance cliff and permits a longitudinal charge-density (bulk plasmon) oscillation. The physical reading: below $\omega_p$ the metal hands back all the light; above it, light propagates with a phase index below one.

Derivation AFrom free electrons to the complex permittivity

The goal is equation (2): the full damped optical permittivity of a metal, obtained by solving the driven equation of motion for one electron, multiplying by the density to get the polarization, and identifying $\varepsilon$ through $\mathbf D = \varepsilon_0\varepsilon\mathbf E$ . The strategy is exactly the bound-electron dispersion calculation with the resonance frequency set to zero, so the plasma frequency emerges as the coefficient combining density, charge, and mass.

Step 1. Insert a harmonic trial solution $\mathbf r = \mathbf r_0 e^{-i\omega t}$ into equation (1), turning $\dot{\mathbf r} \to -i\omega\mathbf r$ and $\ddot{\mathbf r} \to -\omega^2\mathbf r$ : $(-m\omega^2 - im\gamma\omega)\mathbf r = -e\mathbf E$ , so $\mathbf r = \dfrac{e\mathbf E}{m(\omega^2 + i\gamma\omega)}$ . Why: at optical frequencies the electron reaches a steady oscillation in step with the field, and the harmonic ansatz converts the differential equation into algebra.

Step 2. Build the polarization from the dipole moments per unit volume, $\mathbf P = -n_e e\,\mathbf r = -\dfrac{n_e e^2}{m}\dfrac{\mathbf E}{\omega^2 + i\gamma\omega}$ . Why: each displaced electron is a dipole $-e\mathbf r$ , and the free-electron polarization is their density-weighted sum — the entire optical response lives here.

Step 3. Assemble $\mathbf D = \varepsilon_0\mathbf E + \mathbf P = \varepsilon_0\varepsilon(\omega)\mathbf E$ and read off $\varepsilon(\omega) = 1 - \dfrac{n_e e^2}{\varepsilon_0 m}\dfrac{1}{\omega^2 + i\gamma\omega}$ . Why: the permittivity measures how much the polarization adds to the vacuum displacement, and dividing out $\varepsilon_0\mathbf E$ isolates it.

Step 4. Identify the plasma frequency $\omega_p^2 = n_e e^2/\varepsilon_0 m$ , giving $\varepsilon(\omega) = 1 - \omega_p^2/(\omega^2 + i\gamma\omega)$ . Setting $\gamma \to 0$ recovers the lossless $1 - \omega_p^2/\omega^2$ ; restoring the spring $\omega_0 \neq 0$ recovers the Lorentz oscillator. Why: this is the owned result — the single function that, evaluated in each frequency band below, produces mirror, plasmon, and transparent window in turn. *Verified: $\varepsilon(\omega)$ , its real and imaginary parts, and the $\omega_0 \to 0$ reduction of the Lorentz susceptibility in verify_aq-optics-drude-model-metallic-response.py.*

Go deeper · Derivation B · 1 minThe plasma edge and the reflection verdict

Derivation BThe plasma edge and the reflection verdict

The goal is equation (5): to show that a lossless Drude metal reflects perfectly below $\omega_p$ and turns transparent above it, and to pin the sodium edge at $\lambda_p = 209.5\ \mathrm{nm}$ . The strategy is to feed the collisionless permittivity into the complex index and the normal-incidence reflectance and read each regime honestly; the subtle point, recalled from aq-em-plasma-frequency-dielectric-response, is that "no propagation" does not mean "no field."

Step 1. Set $\gamma = 0$ : $\varepsilon(\omega) = 1 - \omega_p^2/\omega^2$ , real, negative for $\omega < \omega_p$ and positive above. Why: for a good metal the damping is a small correction over most of the spectrum, so the sign of the real permittivity alone decides whether light reflects or passes.

Step 2. Below the plasma frequency the index is purely imaginary, $\tilde{n} = i\sqrt{\omega_p^2/\omega^2 - 1} = i\kappa$ : an evanescent wave, penetrating only to the collisionless skin depth $\delta \simeq c/\omega_p$ — for sodium tens of nanometres, set by inertia, not dissipation. Why: a negative permittivity forces the wavevector imaginary, so the field decays into the metal rather than travelling through it.

Step 3. Put that imaginary index into the Fresnel coefficient: $R = |(i\kappa - 1)/(i\kappa + 1)|^2 = (1 + \kappa^2)/(1 + \kappa^2) = 1$ . Total reflection for any nonzero $\kappa$ — the perfect mirror of the lossless limit. Why: an imaginary index cannot transmit time-averaged power, so energy conservation forces all of it back; damping only shaves this slightly below one.

Step 4. Above the plasma frequency $\tilde{n} = \sqrt{1 - \omega_p^2/\omega^2}$ is real and below one: the wave propagates and the metal is transparent, the ultraviolet window of the alkalis. At exactly $\omega_p$ , $\varepsilon = 0$ permits a longitudinal bulk-plasmon oscillation. For sodium, $\lambda_p = 2\pi c/\omega_p = 209.5\ \mathrm{nm}$ . Why: this is the owned edge — the boundary wavelength that Wood measured in 1933 and that this article reproduces from the electron density. *Verified: $R = 1$ below cutoff for the lossless index, $\varepsilon(\omega_p) = 0$ , and the sodium edge $\lambda_p = 209.5\ \mathrm{nm}$ in verify_aq-optics-drude-model-metallic-response.py.*

Go deeper · Derivation C · 1 minRecovering the DC conductivity

Derivation CRecovering the DC conductivity

The goal is equation (3): to show the optical permittivity already contains Ohm's law, so the collision time that damps light is the same one that resists current. The strategy is to compute the current density from the electron velocity, compare it with $\mathbf J = \sigma\mathbf E$ to define the AC conductivity, and take the zero-frequency limit.

Step 1. From Step 1 of Derivation A, the velocity is $\dot{\mathbf r} = -i\omega\mathbf r = \dfrac{-e\mathbf E}{m(\gamma - i\omega)}$ after dividing through by $-i\omega$ . Why: current is carried by moving charge, so the velocity, not the displacement, is what Ohm's law needs.

Step 2. The current density is $\mathbf J = -n_e e\dot{\mathbf r} = \dfrac{n_e e^2}{m(\gamma - i\omega)}\mathbf E \equiv \sigma(\omega)\mathbf E$ , defining the AC Drude conductivity $\sigma(\omega) = \dfrac{n_e e^2/m}{\gamma - i\omega}$ . Why: comparing the microscopic current with the phenomenological $\sigma\mathbf E$ identifies the conductivity directly.

Step 3. Take $\omega \to 0$ : $\sigma_0 = n_e e^2/(m\gamma) = n_e e^2\tau/m$ , the classical Drude conductivity, and since $\omega_p^2 = n_e e^2/\varepsilon_0 m$ this is also $\varepsilon_0\omega_p^2/\gamma$ . Why: at zero frequency the electrons reach a steady drift and the response is pure Ohmic resistance — the DC end of the same curve whose optical end is the mirror.

Step 4. Confirm consistency: $\varepsilon(\omega) = 1 + i\sigma(\omega)/(\varepsilon_0\omega)$ reproduces equation (2), so the conductivity and permittivity descriptions are one theory. Why: this closes the loop with aq-em-waves-in-conductors, whose $\sigma$ -based treatment and this article's $\varepsilon(\omega)$ are the two ends of one Drude response. *Verified: $\sigma_0 = n_e e^2\tau/m = \varepsilon_0\omega_p^2/\gamma$ and the $\varepsilon$ – $\sigma$ dictionary in verify_aq-optics-drude-model-metallic-response.py.*

Limiting cases

Case	Result	Meaning
$\omega \ll \gamma$ (DC / far IR)	$\varepsilon \approx 1 + i\sigma_0/\varepsilon_0\omega$	conductive response; recovers Ohm's law, $\sigma_0 = \varepsilon_0\omega_p^2/\gamma$
$\gamma \ll \omega < \omega_p$	$\varepsilon \approx 1 - \omega_p^2/\omega^2 < 0$	evanescent, $R \to 1$ — the visible mirror
$\omega > \omega_p$	$\varepsilon > 0$ , $n < 1$	transparent (UV transparency of Na, K); plasmon at $\varepsilon = 0$
weak damping, general	$\varepsilon'' \approx \omega_p^2\gamma/\omega^3$	free-carrier absorption strongest in the infrared
$\omega \gg \omega_p$	$\varepsilon \to 1$ , $n \to 1$	metal fades to vacuum (X-rays pass through)

Common trap“The Drude model predicts the colour of gold”

It does not, and mistaking the free-electron plasma edge for a colour is the classic error. Drude's model has no band structure, so for gold it predicts a plasma frequency at $\hbar\omega_p = 9.0\ \mathrm{eV}$ ( $\lambda_p = 137.5\ \mathrm{nm}$ , deep ultraviolet) and therefore a colourless mirror across the whole visible — just like silver, which has almost the same free-electron plasma energy. Gold's yellow comes from what the model cannot see: interband transitions that lift electrons out of the filled $d$ -band into the conduction band, switching on around $2.4\ \mathrm{eV}$ and absorbing the blue-green, so the reflected light is warm. Silver's interband edge sits higher, near $3.9\ \mathrm{eV}$ , which is why silver stays a neutral mirror while gold and copper are coloured. A realistic model adds Lorentzian oscillators for these transitions on top of the Drude background; Drude alone gives shine and the plasma edge, never the hue.

Worked examples

1 · The sodium plasma edge (confidence builder)

Sodium has one conduction electron per atom, giving $n_e = 2.541\times10^{28}\ \mathrm{m^{-3}}$ . Find the plasma frequency, the plasma wavelength, and the photon energy of the edge, and say what sodium looks like across the spectrum.

Step 1. Plasma frequency: $\omega_p = \sqrt{n_e e^2/\varepsilon_0 m} = \sqrt{(2.541\times10^{28})(1.602\times10^{-19})^2 / [(8.854\times10^{-12})(9.109\times10^{-31})]} = \mathbf{8.99\times10^{15}\ rad/s}$ .

Step 2. Plasma wavelength: $\lambda_p = 2\pi c/\omega_p = 2\pi(2.998\times10^{8})/(8.99\times10^{15}) = \mathbf{209.5\ nm}$ , in the deep ultraviolet. Photon energy of the edge: $\hbar\omega_p = (1.055\times10^{-34})(8.99\times10^{15})/(1.602\times10^{-19}) = \mathbf{5.9\ eV}$ .

Step 3. Read the spectrum. Every visible wavelength ( $400$ – $700\ \mathrm{nm}$ ) is far longer than $209.5\ \mathrm{nm}$ , so $\omega < \omega_p$ : the permittivity is negative, the wave is evanescent, and $R \approx 1$ — sodium is a mirror for all visible colours. Only at wavelengths shorter than $209.5\ \mathrm{nm}$ does $\omega$ climb past $\omega_p$ and the metal turn transparent.

Step 4. This is Wood's 1933 result, and it was one of the model's first quantitative triumphs: the free-electron density alone predicts the ultraviolet transparency edge, with no adjustable parameter. $\omega_p = 8.99\times10^{15}\ \mathrm{rad/s}$ , $\lambda_p = 209.5\ \mathrm{nm}$ , $\hbar\omega_p = 5.9\ \mathrm{eV}$ .

2 · Sodium's conductivity and damping (exam level)

Take sodium's relaxation time as $\tau = 3.1\times10^{-14}\ \mathrm{s}$ . Find the predicted DC conductivity, compare it with the measured $2.1\times10^{7}\ \mathrm{S/m}$ , and check that sodium is a weakly damped (good) metal at optical frequencies.

Step 1. DC conductivity from equation (3): $\sigma_0 = n_e e^2\tau/m = (2.541\times10^{28})(1.602\times10^{-19})^2(3.1\times10^{-14})/(9.109\times10^{-31}) = \mathbf{2.22\times10^{7}\ S/m}$ .

Step 2. Compare with experiment: the measured value is $2.1\times10^{7}\ \mathrm{S/m}$ , so the classical model is $6\%$ high. The residual gap is where classical physics runs out — Fermi–Dirac statistics and the band effective mass, which Sommerfeld's 1927 revision supplied without changing $\varepsilon(\omega)$ itself.

Step 3. Damping rate: $\gamma = 1/\tau = 3.2\times10^{13}\ \mathrm{s^{-1}}$ . Compare with the plasma frequency: $\gamma/\omega_p = (3.2\times10^{13})/(8.99\times10^{15}) = 3.6\times10^{-3}$ , four orders below one. Sodium is deep in the weakly damped regime, so the visible reflectance is only marginally below the lossless $R = 1$ .

Step 4. Cross-check through equation (3): $\sigma_0 = \varepsilon_0\omega_p^2/\gamma = (8.854\times10^{-12})(8.99\times10^{15})^2/(3.2\times10^{13}) = 2.22\times10^{7}\ \mathrm{S/m}$ , matching Step 1 — the optical and electrical routes agree, as one theory demands. $\sigma_0 = 2.22\times10^{7}\ \mathrm{S/m}$ (vs $2.1\times10^{7}$ measured), $\gamma/\omega_p = 3.6\times10^{-3}$ .

Physics of the portrait

The model belongs to the man who first made light and electricity report to one set of books.

Origins · 1 minPaul Drude and how this was discovered

Paul Drude

1863–1906 · Leipzig · Giessen · Berlin

Paul Drude trained as an optical theorist under Woldemar Voigt at Göttingen, mastering reflection and absorption in metals — his ellipsometric technique for measuring optical constants is the direct ancestor of the semiconductor industry's workhorse metrology. Then, within three years of J. J. Thomson's 1897 electron, Drude built the first microscopic theory of conduction: his 1900 papers modelled the metal's electrons as a gas of colliding particles, yielding Ohm's law with $\sigma = n_e e^2\tau/m$ , the frequency-dependent optical response this article develops, and — his era's sensation — the Wiedemann–Franz ratio of thermal to electrical conductivity, correct to the observed order. That triumph rested on a lucky cancellation of two errors that Sommerfeld's 1927 Fermi–Dirac revision later sorted out. His Lehrbuch der Optik (1900) became a standard text, and as editor of Annalen der Physik he handled the 1905 submissions of an unknown patent clerk in Bern. Weeks after his election to the Prussian Academy in 1906, at the height of his powers and only forty-two, Drude took his own life in Berlin — leaving physics the electron gas that quantum mechanics would spend the next generation vindicating and repairing.

History

1897

— J. J. Thomson identifies the electron, the free particle Drude's metal will be built from

1900

— Drude models the conduction electrons as a classical gas, deriving Ohm's law and the optical response σ(ω) and ε(ω)

1905

— Lorentz recasts the theory with a fuller statistical treatment, the Drude–Lorentz model

1927

— Sommerfeld rebuilds the electron gas on Fermi–Dirac statistics, curing its specific-heat failure while keeping ε(ω)

1933

— Wood discovers that sodium and the other alkali metals turn transparent in the ultraviolet

1933

— Zener explains Wood's transparency as light outrunning the electrons above the plasma frequency

Going further

Texts. Ashcroft & Mermin, Solid State Physics, Ch. 1 (the Drude model) · Fox, Optical Properties of Solids, 2nd ed., Ch. 7 (free electrons, plasma edge, interband) · Jackson, Classical Electrodynamics, 3rd ed., §7.5 · Saleh & Teich, Fundamentals of Photonics, 3rd ed., §5.5 · Maier, Plasmonics: Fundamentals and Applications, Ch. 1.

Provenance. The equation of motion, susceptibility, and $\varepsilon(\omega)$ follow the Lorentz–Drude treatment (Ashcroft & Mermin Ch. 1; Jackson §7.5). Numerical values are recomputed in verify_aq-optics-drude-model-metallic-response.py: sodium $n_e = 2.541\times10^{28}\ \mathrm{m^{-3}} \Rightarrow \omega_p = 8.99\times10^{15}\ \mathrm{rad/s}$ , $\lambda_p = 209.5\ \mathrm{nm}$ , $\hbar\omega_p = 5.9\ \mathrm{eV}$ ; $\tau = 3.1\times10^{-14}\ \mathrm{s} \Rightarrow \sigma_0 = 2.22\times10^{7}\ \mathrm{S/m}$ against the measured $2.1\times10^{7}$ ; gold $n_e = 5.90\times10^{28} \Rightarrow \lambda_p = 137.5\ \mathrm{nm}$ , $\hbar\omega_p = 9.0\ \mathrm{eV}$ . The conductor electrodynamics ( $\tilde{k}$ , skin depth, $R_s$ , Hagen–Rubens, the lossless plasma edge) is recalled by name from aq-em-waves-in-conductors, aq-em-skin-depth-reflection-conductors, and aq-em-plasma-frequency-dielectric-response; the interband physics of gold's colour is optical, from Fox Ch. 7. No fabricated constants.

Prerequisites. The Lorentz Oscillator Model · Dielectric Susceptibility and Permittivity · Electric Polarization and the Displacement Field · Complex Refractive Index and the Extinction Coefficient. Leads to. Absorption Coefficient and the Beer–Lambert Law · Dispersion: Normal and Anomalous · The Refractive Index · Plane Waves in Homogeneous Isotropic Media.

Why is a metal just a Lorentz oscillator with zero resonance frequency?

Because a conduction electron has no equilibrium position to be pulled toward. In The Lorentz Oscillator Model, a bound electron feels inertia, friction, and a restoring spring $m\omega_0^2\mathbf r$ , and its susceptibility peaks at the resonance $\omega_0$ . A conduction electron keeps the inertia and friction but loses the spring, being free to drift through the lattice, so $\omega_0 = 0$ . Deleting that one term turns the resonant Lorentzian into the Drude form $\varepsilon = 1 - \omega_p^2/(\omega^2 + i\gamma\omega)$ : the resonance slides to zero frequency, and what was a bump in the dispersion becomes the huge low-frequency polarizability that drives $\varepsilon$ negative. Everything else carries over unchanged.

If a lossless metal has R = 1 below the plasma frequency, why do real mirrors lose a few percent?

Because real metals have damping, and damping is what the lossless calculation throws away. Restoring $\gamma$ makes the index slightly less than pure imaginary, so a sliver of power enters the metal and is dissipated as Ohmic heat within a skin depth. The size of that sliver is owned by aq-em-skin-depth-reflection-conductors as the Hagen–Rubens law, $1 - R = \sqrt{8\varepsilon_0\omega/\sigma}$ , which grows as $\sqrt{\omega}$ : parts in $10^{4}$ at microwave frequencies, rising toward a percent in the infrared. So "shiny" and "slightly lossy" are not in tension — the metal reflects because $\varepsilon$ is negative and eats the little it fails to reflect because $\gamma$ is not quite zero.

Is the plasma frequency here the same one as in a gas plasma or the ionosphere?

Yes — it is the identical quantity, $\omega_p = \sqrt{n_e e^2/\varepsilon_0 m_e}$ , owned by aq-em-plasma-frequency-dielectric-response, doing the same job: dividing reflection from transparency at $\omega = \omega_p$ . What differs is only the electron density. The ionosphere carries about $10^{12}$ electrons per cubic metre, so its plasma frequency lands near $9\ \mathrm{MHz}$ and it mirrors shortwave radio; a metal carries around $10^{29}$ , twenty-two orders denser, so its plasma frequency lands in the ultraviolet and it mirrors visible light. A shortwave skip off the sky and the shine of a spoon are the same inequality $\omega < \omega_p$ at wildly different densities.

What is the difference between this Drude ε(ω) and the plasma article's ε = 1 − ωₚ²/ω²?

The damping term. aq-em-plasma-frequency-dielectric-response studies a collisionless free-electron gas, so it keeps $\gamma \to 0$ and works with the real, lossless $\varepsilon = 1 - \omega_p^2/\omega^2$ — perfect for the ionosphere and interstellar space, where collisions are rare. A metal is dense and collisional, so this article keeps the full $\varepsilon = 1 - \omega_p^2/(\omega^2 + i\gamma\omega)$ , whose imaginary part carries the Ohmic loss and whose zero-frequency limit is the DC conductivity. The lossless version is the skeleton — it locates the plasma edge and gives the reflection verdict; the damped version puts flesh on it, accounting for absorption, the small mirror losses, and the bridge to electrical resistance.

How can the same collision rate γ set both the optical loss and the DC resistance?

Because both are the same electrons scattering. The friction term $m\gamma\dot{\mathbf r}$ represents collisions that randomise an electron's velocity every $\tau = 1/\gamma$ on average. At optical frequencies that friction damps the oscillation and appears as the imaginary part of $\varepsilon$ , hence absorption; at zero frequency the same friction limits the steady drift velocity in a static field and appears as electrical resistance, $\sigma_0 = n_e e^2\tau/m$ . The permittivity and conductivity are not two theories but two limits of one response function $\sigma(\omega) = (n_e e^2/m)/(\gamma - i\omega)$ , related by $\varepsilon = 1 + i\sigma/\varepsilon_0\omega$ . Measure $\sigma_0$ on a bench and you have fixed the optical damping.

Why does free-carrier absorption fall off so steeply toward the visible?

Because the absorptive part scales as $\varepsilon'' \approx \omega_p^2\gamma/\omega^3$ in the weakly damped regime, dropping as the cube of the frequency. Physically, the faster the field oscillates, the less time the electrons have to be knocked off course within a cycle, so each collision steals a smaller fraction of the coherent motion — absorption is strongest in the far infrared and weak by the visible. This $\omega^{-3}$ tail is why doped semiconductors and transparent conducting oxides are clear in the visible but absorbing in the infrared, and why the free-carrier contribution to a metal's colour is negligible next to the interband transitions.