Part 3 · Light in matter

3.3 Metals: the Drude model

Source: Postorino, week 4 slides, pp. 6–10

Objectives

Picture. A metal is a gas of free electrons among fixed ions; collisions erase their memory every $τ$ on average and act as viscous friction.
Theorem. $σ = N e^{2} τ / m$ ; in an oscillating field $J = \frac{σ}{1 - iω τ} E$ and $N^{2} = 1 - \frac{ω _{P}^{2}}{ω ^{2} + iω / τ}$ , with $ω_{P}^{2} = \frac{N e ^{2}}{m ε _{0}}$ .
Key. Below the plasma frequency a metal reflects; above it, it is transparent.
Recipe. Skin depth $δ = 2/ ωσ μ_{0}$ .

The Drude model

Only free charges: $P = 0$ , $J \neq = 0$ , still with $M = 0$ and $ρ = 0$ . Maxwell’s equations become $\nabla \cdot E = 0, \nabla \cdot B = 0, \nabla \land E = - \frac{\partial B}{\partial t}, \nabla \land B = μ_{0} ε_{0} \frac{\partial E}{\partial t} + μ_{0} J,$ and the propagation equation (Theorem 3.1.2) is $\nabla^{2} E = \frac{1}{c ^{2}} \frac{\partial ^{2} E}{\partial t ^{2}} + μ_{0} \frac{\partial J}{\partial t} .$

Definition 3.3.1Drude model of a conductorslides w4, p. 6

The electrons (negative) move freely and chaotically; the ions (positive) are fixed.
The electrons have no long-range (Coulomb-like) interactions with each other, only collisions with the nuclei.
After a collision the electrons lose memory: they come out in a random direction.
As in an ideal gas, their mean kinetic energy is proportional to the temperature: $⟨ \frac{1}{2} m v^{2} ⟩ = \frac{3}{2} K_{B} T$ .
$τ$ is the mean time between one collision and the next.

Conductivity

The current density is $J = - N e ⟨ v ⟩$ , with $v_{d} = ⟨ v ⟩$ the drift velocity and $N$ the number of electrons per unit volume. The random thermal motion averages to zero; only the drift carries current. The electrical conductivity $σ$ is the factor in Ohm’s law, $J = σ E$ .

Theorem 3.3.2Static conductivityslides w4, p. 7

$σ = \frac{N e ^{2} τ}{m} .$ More carriers, and longer free flights between collisions, conduct better.

Proof

The electrons feel $F = - e E$ , so $a = - e E / m$ and $v (t) = v_{0} - \frac{e E}{m} t$ , with $v_{0}$ the velocity just after a collision and $t$ the time since it. Average over the electrons: $v_{d} = ⟨ v (t)⟩ = ⟨ v_{0} ⟩ - \frac{e E}{m} ⟨ t ⟩ = - \frac{e E}{m} τ .$ $⟨ v_{0} ⟩ = 0$ because collisions erase memory and directions after one are random; $⟨ t ⟩ = τ$ , the mean time since the last collision. Then $J = - N e v_{d} = \frac{N e ^{2} τ}{m} E$ .

Oscillating field. Account for the overall effect of collisions, with their loss of memory, as a viscous friction $- β v$ : $m \frac{\partial v}{\partial t} + β v = - e E .$ Divide by $m$ and multiply by $- N e$ to get an equation for $J = - N e v$ : $\frac{\partial J}{\partial t} + \frac{β}{m} J = \frac{N e ^{2}}{m} E = \frac{σ}{τ} E .$

Proposition 3.3.3Collisions as frictionslides w4, p. 8

$β / m = 1/ τ$ , so $\frac{\partial J}{\partial t} + \frac{1}{τ} J = \frac{σ}{τ} E .$

Proof

Take a static field, $\partial J / \partial t = 0$ , and require the static conductivity: $\frac{β}{m} J = \frac{σ}{τ} E$ gives $J = \frac{m}{β} \frac{σ}{τ} E$ , which is $σ E$ only if $\frac{m}{β τ} = 1$ .

Theorem 3.3.4AC conductivityslides w4, p. 8

$J = \frac{σ}{1 - iω τ} E .$

Proof

Complex substitution, $J = \hat{J}_{0} e^{- iω t}$ , $E = \hat{E}_{0} e^{- iω t}$ : $- iω \hat{J}_{0} + \frac{1}{τ} \hat{J}_{0} = \frac{σ}{τ} \hat{E}_{0}$ . Multiply by $τ$ : $(1 - iω τ) \hat{J}_{0} = σ \hat{E}_{0}$ .

It is the Lorentz oscillator with no spring, $ω_{0} = 0$ : free electrons are not bound beyond the lectures. For $ω τ ≪ 1$ the static Ohm’s law holds; for $ω τ ≫ 1$ the electrons can’t keep up and the current lags the field by $90°$ .

Plasma frequency

Theorem 3.3.5Refractive index of a metalslides w4, p. 9

Putting $J$ into the propagation equation, $\nabla^{2} E = \frac{1}{c ^{2}} \frac{\partial ^{2} E}{\partial t ^{2}} + \frac{μ _{0} σ}{1 - iω τ} \frac{\partial E}{\partial t},$ with solutions $E_{0} e^{i (K x - ω t)}$ and $K^{2} = \frac{ω ^{2}}{c ^{2}} N^{2}$ , where $N^{2} = 1 - \frac{ω _{P}^{2}}{ω ^{2} + iω / τ}, ω_{P} = \frac{μ _{0} σ c ^{2}}{τ} = \frac{N e ^{2}}{m ε _{0}} .$

Proof

Substituting gives $- K^{2} = - \frac{ω ^{2}}{c ^{2}} - iω \frac{μ _{0} σ}{1 - iω τ}$ , so $K^{2} = \frac{ω ^{2}}{c ^{2}} + i \frac{ω μ _{0} σ}{1 - iω τ}$ . The algebra beyond the lectures: factor out $ω^{2} / c^{2}$ , $N^{2} = 1 + \frac{i c ^{2} μ _{0} σ}{ω ( 1 - iω τ )} .$ Divide top and bottom by $τ$ and use $\frac{i}{ω ( 1/ τ - iω )} = \frac{i}{- i ( ω ^{2} + iω / τ )} = - \frac{1}{ω ^{2} + iω / τ}$ . This gives the boxed form with $ω_{P}^{2} = c^{2} μ_{0} σ / τ$ ; then $σ = N e^{2} τ / m$ and $μ_{0} c^{2} = 1/ ε_{0}$ give $ω_{P}^{2} = N e^{2} / m ε_{0}$ .

Key observation 3.3.6The plasma frequencyslides w4, p. 9

The plasma frequency is the natural oscillation frequency of the electron gas. Displace all the electrons a little with respect to the ions, and the charge imbalance pulls them back: they oscillate collectively at $ω_{P}$ .

The same $\frac{N e ^{2}}{m ε _{0}}$ appears in front of the Lorentz formula: $N_{Lorentz}^{2} = 1 + \frac{ω _{P}^{2}}{ω _{0}^{2} - ω ^{2} - iγω}$ . Drude is the case $ω_{0} = 0$ , $γ = 1/ τ$ beyond the lectures.

n and κ for a metal

Proposition 3.3.7n and κ of a Drude metalslides w4, p. 10

$n^{2} - κ^{2} = 1 - \frac{ω _{P}^{2}}{ω ^{2} + τ ^{- 2}}, 2 nκ = \frac{ω _{P}^{2}}{ω ^{2} + τ ^{- 2}} (\frac{1}{ω τ}) .$

Proof

Multiply by the conjugate: $\frac{1}{ω ^{2} + iω / τ} = \frac{ω ^{2} - iω / τ}{ω ^{4} + ω ^{2} / τ ^{2}} = \frac{ω - i / τ}{ω ( ω ^{2} + τ ^{- 2} )},$ and equate real and imaginary parts of $N^{2}$ with $n^{2} - κ^{2}$ and $2 nκ$ .

Key observation 3.3.8Opaque below, transparent aboveslides w4, p. 10

Below $ω_{P}$ : opaque, high-reflection region. $n^{2} - κ^{2} < 0$ (for $ω τ ≫ 1$ ), so $κ > n$ : the wave decays within a short distance and almost all the light is reflected. That is why metals shine.
Above $ω_{P}$ : transparent region. $n$ rises towards 1 and $κ$ falls to about 0: the metal lets light through.

Loading the demo…

Figure 3.3.1

n

κ

and the reflectance

R

(scaled up 3×) of a Drude metal. Drag the frequency across

ω_{P}

: below it

κ > n

and

R

is close to 100 %; above it the metal becomes transparent. Shorter collision times (larger

1/ τ

) blur the edge.

Numbers beyond the lectures: copper has $N \approx 8.5 \times 1 0^{28}$ m⁻³, so $ω_{P} \approx 1.6 \times 1 0^{16}$ rad/s, $λ_{P} = 2 π c / ω_{P} \approx 115$ nm, in the UV. Metals reflect visible light; alkali metals become transparent in the ultraviolet. The normal-incidence reflectivity (Proposition 2.3.9) generalises to $R = \frac{( n - 1 ) ^{2} + κ ^{2}}{( n + 1 ) ^{2} + κ ^{2}},$ which tends to 1 when $κ ≫ n$ .

Skin depth and the microwave oven

Proposition 3.3.9Skin depthslides w4, p. 10

$δ = \frac{2}{ωσ μ _{0}} .$ For Cu the slide gives $δ = 1 0^{- 4}$ mm $= 0.1$ µm.

Where it comes frombeyond the lectures

At low frequency, $ω τ ≪ 1$ and $ω ≪ σ / ε_{0}$ , the second term in $K^{2}$ dominates: $K^{2} \approx iω μ_{0} σ \Rightarrow K = ω μ_{0} σ e^{iπ /4} = \frac{ω μ _{0} σ}{2} (1 + i) .$ So $α = ω μ_{0} σ /2 = 1/ δ$ , and the amplitude goes as $e^{- x / δ}$ .

WarningSlide slip

The slide calls $δ$ “the distance at which the intensity falls by $1/ e$ ”. With this formula it is where the amplitude falls by $1/ e$ ; the intensity falls by $1/ e$ at $δ /2$ .

For copper, $σ = 6 \times 1 0^{7}$ S/m: $δ$ is 8.4 mm at 60 Hz, 66 µm at 1 MHz and 2 µm at 1 GHz; it reaches 0.1 µm only near $4 \times 1 0^{11}$ Hz, so the slide’s value is for very high frequencies beyond the lectures. High-frequency currents flow only in a thin skin of a wire.

Example 3.3.10The microwave ovenslides w4, p. 10

It runs at 2450 MHz, $λ_{0} = c / ν \approx 12$ cm. Its metal walls reflect the waves, which form standing waves inside (Proposition 2.4.4). The energy maxima are 6 cm apart, $λ_{0} /2$ : that is why food has hot and cold spots and the plate turns. The water in the food absorbs through orientational polarisation (Polarisation mechanisms).

The general case

Key observation 3.3.11Real materials have both behavioursslides w4, p. 10

For a material with non-ideal electronic properties, free and bound charges both contribute: $N^{2} = 1 - \frac{ω _{P}^{2}}{ω ^{2} + iω / τ} + \frac{N e ^{2}}{m ε _{0}} j \sum (\frac{f _{j}}{ω _{j}^{2} - ω ^{2} - i γ _{j} ω}) .$

Exercises

Exercise 3.3.1beyond the lectures

Aluminium has $N = 1.8 \times 1 0^{29}$ m⁻³. Find its plasma frequency and the wavelength below which it becomes transparent.

Solution

$ω_{P} = N e^{2} / m ε_{0} = 1.8 \times 1 0^{29} \cdot 2.56 \times 1 0^{- 38} / (9.11 \times 1 0^{- 31} \cdot 8.85 \times 1 0^{- 12}) = 2.4 \times 1 0^{16}$ rad/s. $λ_{P} = 2 π c / ω_{P} \approx 79$ nm: transparent only in the far UV.

Exercise 3.3.2beyond the lectures

Copper has $σ = 6 \times 1 0^{7}$ S/m. Compute the skin depth at 50 Hz and at 2.45 GHz. Why is a thin metal coating enough to shield microwaves?

Solution

$δ = 2/ (ωσ μ_{0})$ . At 50 Hz, $ω = 314$ s⁻¹ and $δ = 9.2$ mm. At 2.45 GHz, $ω = 1.54 \times 1 0^{10}$ s⁻¹ and $δ = 1.3$ µm. A coating a few µm thick already attenuates the amplitude by $e^{- 3}$ or more.

Exercise 3.3.3beyond the lectures

From $σ = 6 \times 1 0^{7}$ S/m and $N = 8.5 \times 1 0^{28}$ m⁻³, find copper’s collision time $τ$ and the drift velocity in a field of 1 V/m.

Solution

$τ = mσ / (N e^{2}) = 9.11 \times 1 0^{- 31} \cdot 6 \times 1 0^{7} / (8.5 \times 1 0^{28} \cdot 2.56 \times 1 0^{- 38}) = 2.5 \times 1 0^{- 14}$ s. $v_{d} = e E τ / m = 1.6 \times 1 0^{- 19} \cdot 2.5 \times 1 0^{- 14} /9.11 \times 1 0^{- 31} = 4.4 \times 1 0^{- 3}$ m/s: millimetres per second.

Exercise 3.3.4beyond the lectures

Show that for $ω τ ≫ 1$ and $ω < ω_{P}$ , $N$ is almost purely imaginary, and find the decay length of the field.

Solution

$N^{2} \approx 1 - ω_{P}^{2} / ω^{2} < 0$ , so $N \approx i ω_{P}^{2} / ω^{2} - 1$ : $n \approx 0$ , $κ = ω_{P}^{2} / ω^{2} - 1$ . Then $α = ωκ / c = ω_{P}^{2} - ω^{2} / c$ , and for $ω ≪ ω_{P}$ the decay length is $c / ω_{P}$ , about 20 nm for copper. With $n \approx 0$ , $R \approx (1 + κ^{2}) / (1 + κ^{2}) = 1$ : total reflection.