advanced35 min studyLesson 273 in Physics

Drude conductivity and the plasma edge

Treat mobile carriers as damped inertial charges and separate conductivity from screened plasma response.

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What you’ll understand

  • Derive complex Drude conductivity from carrier momentum balance.
  • Relate conductivity to permittivity without counting carriers twice.
  • Find a screened zero crossing and state why it is not an exact reflectivity edge.

A mobile electron has inertia and loses momentum through collisions, but it does not feel the bound oscillator's restoring spring in a spatially uniform ideal conductor. Removing that spring produces the Drude model. Its direct output is a current response; the same response can also be packaged as a frequency-dependent electric permittivity. This dual description explains why a material that conducts at low frequency can have a negative real permittivity over part of an optical spectrum.

Solve for current rather than displacement

Let nn be the carrier number density, mm an effective inertial mass and −e-e the carrier charge. In a relaxation-time description, scattering removes average momentum at rate γ=1/τ\gamma=1/\tau, where τ\tau is a positive time. This phenomenological term represents an ensemble average, not a statement that every collision occurs at equal intervals. Under a weak homogeneous field, the mean velocity equation is linear and the current is charge density times mean velocity.

For e−iωte^{-i\omega t}, the velocity denominator is γ−iω\gamma-i\omega. Multiplying by the negative charge again gives a positive conductivity numerator. At low frequency the current is nearly in phase with the field and its real part dissipates energy. At high frequency the carriers cannot equilibrate within one cycle; the current becomes mainly reactive. The model omits interband transitions, detailed collision energy dependence and nonlocal transport, so it should be used only over a justified measurement range.

σ(ω)=ne2m(γ−iω)=ϵ0ωp2γ−iω,ωp2=ne2ϵ0m,σdc=ne2τm\sigma(\omega)=\frac{ne^2}{m(\gamma-i\omega)}=\frac{\epsilon_0\omega_p^2}{\gamma-i\omega},\qquad \omega_p^2=\frac{ne^2}{\epsilon_0m},\qquad \sigma_{\rm dc}=\frac{ne^2\tau}{m}
Conductivity σ has SI units S/m, plasma angular frequency ωp has s⁻¹ and τ has s. This is a local single-carrier relaxation-time approximation.↔ Scroll wide equations horizontally.

Package the same current into permittivity

One can keep the carrier current explicitly on the right side of Ampère's law or absorb it into displacement current. The two descriptions are equivalent if the carrier term appears exactly once. Calling it both a free conductivity and part of a fitted complex permittivity in the same equation counts the same response twice. Let a real, nearly constant background relative permittivity describe bound processes outside the frequency band being modeled.

Equating the current to the time derivative of a carrier polarization gives a plus sign before iσ/(ϵ0ω)i\sigma/(\epsilon_0\omega) in relative permittivity. The resulting Drude term has negative real part and positive imaginary part. In particular, negative real permittivity is not a declaration of gain: passivity is controlled by absorption, which remains nonnegative for positive damping. The formal divergence of carrier polarization at zero frequency is compatible with a finite dc current and should not be interpreted as a static bound dipole.

ϵr(ω)=ϵ∞+iσϵ0ω=ϵ∞−ωp2ω(ω+iγ)\epsilon_r(\omega)=\epsilon_\infty+\frac{i\sigma}{\epsilon_0\omega}=\epsilon_\infty-\frac{\omega_p^2}{\omega(\omega+i\gamma)}
ε∞ is a dimensionless background valid over the chosen band. At positive frequency this convention gives Im εr ≥ 0 when γ > 0; the explicit conductivity is then omitted from the same Ampère equation.↔ Scroll wide equations horizontally.

Identify the screened zero crossing

Separating real and imaginary parts shows that the real permittivity crosses zero only if the squared plasma frequency exceeds the background times the squared damping rate. In the weak-damping limit the crossing lies near the plasma frequency divided by the square root of the background permittivity. This is screening by the bound-charge response. It distinguishes the unscreened carrier parameter from an observed optical feature, even before reflection at a surface is considered.

A real reflectivity spectrum also depends on damping, background dispersion, the external medium, sample thickness and surface quality. Therefore the algebraic zero of real permittivity is a useful material diagnostic rather than an exact universal boundary between perfect reflection and perfect transmission. In the collisionless local model with positive permeability, a negative real permittivity gives an evanescent bulk wave instead of ordinary propagating light. Finite damping makes propagation and attenuation complex rather than dividing frequencies into two perfectly sharp categories.

Re⁡ϵr=ϵ∞−ωp2ω2+γ2,Im⁡ϵr=ωp2γω(ω2+γ2),ωzero2=ωp2ϵ∞−γ2\operatorname{Re}\epsilon_r=\epsilon_\infty-\frac{\omega_p^2}{\omega^2+\gamma^2},\quad \operatorname{Im}\epsilon_r=\frac{\omega_p^2\gamma}{\omega(\omega^2+\gamma^2)},\quad \omega_{\rm zero}^2=\frac{\omega_p^2}{\epsilon_\infty}-\gamma^2
The zero formula assumes constant positive ε∞ and a positive right side. It locates Re εr = 0, not an exact reflection minimum or a lossless propagation threshold in a dissipative slab.↔ Scroll wide equations horizontally.

A broadened screened plasma crossing

Use a reference angular frequency Ω\Omega and parameters ϵ∞=4\epsilon_\infty=4, ωp=5Ω\omega_p=5\Omega and γ=1.5Ω\gamma=1.5\Omega. Find the positive zero of the real permittivity and the imaginary permittivity there.

  1. The screened squared scale is ωp2/ϵ∞=25Ω2/4=6.25Ω2\omega_p^2/\epsilon_\infty=25\Omega^2/4=6.25\Omega^2. Subtract γ2=2.25Ω2\gamma^2=2.25\Omega^2 to obtain ωzero2=4Ω2\omega_{\rm zero}^2=4\Omega^2.

  2. The positive crossing is 2Ω2\Omega. At that frequency the real part is 4−25/(4+2.25)=04-25/(4+2.25)=0, confirming the substitution directly.

  3. Evaluate the imaginary part: 25(1.5)/[2(4+2.25)]=325(1.5)/[2(4+2.25)]=3. The factors of reference frequency cancel, as they must for a dimensionless relative permittivity.

The result

The real part crosses zero at 2Ω, where εr = 3i. Significant absorption remains at the crossing.

Explore the model

Interactive lab

Interactive lab · waves

In-phase and quadrature amplitudes

Reset and set BOTH Phase difference = 0° and Amplitude ratio = 1; predict reinforcement. For a second fresh comparison, Reset and set Phase difference = 90° and Amplitude ratio = 1; predict a shifted sum. The two waves have equal fixed frequency and are mathematical amplitudes only. This lab can illustrate in-phase and quadrature components, but it does not calculate carrier conductivity, a plasma edge or frequency dispersion.

In-phase and quadrature amplitudesTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forPhase-sensitive components require a convention; the Drude equation supplies their frequency dependence.

Check your understanding

If ε∞ = 9 and damping is negligible, where is the Drude zero of real relative permittivity?

Try it yourself

A Drude conductor has σdc=100 S/m\sigma_{\rm dc}=100\,\mathrm{S/m} and is driven at ωτ=2\omega\tau=2. Find the real and imaginary conductivities and the mean Joule heating for a peak field amplitude E=3 V/mE=3\,\mathrm{V/m}.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Divide the dc conductivity by 1 − 2i and use one half times Re σ times the squared peak field.

Compare with the solution

Rationalizing gives σ=100(1+2i)/5=(20+40i) S/m\sigma=100(1+2i)/5=(20+40i)\,\mathrm{S/m}. Only the real component contributes to cycle-averaged dissipation. The heating is 12(20)(32)=90 W/m3\tfrac12(20)(3^2)=90\,\mathrm{W/m^3}. The imaginary component stores and returns energy within a cycle; using the magnitude of conductivity would incorrectly include that reversible response as heat. This calculation assumes a uniform local field and stationary sinusoidal driving, with the quoted field being the amplitude inside the conductor.

Take it with you

  • Drude carriers have damping and inertia but no uniform restoring spring.
  • Conductivity and carrier permittivity are two representations of the same current.
  • A screened zero crossing depends on background response and damping.

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