The frequency dependent conductivity is
σ(ω) =
j(ω)
E(ω)=
nev(ω)
E(ω)= neμ
1 −iωτ
1 +ω^2 τ^2(116)
If we think of low frequencies the conductivity isσ(ω) = neμwhich is the same result we got for
diffusive transport. For high frequencies the conductivity isσ = −ineμωτ, hence the current again is
out of phase with respect to the electric field and the metal acts like an insulator.
The polarization is the number of charges multiplied with the displacement, for harmonic solutions
the displacement isx =
∫
v(ω,t)dt = v(iωω)P(ω) = −nex(ω) =
−nev(ω)
iω(117)
The electrical susceptibility is given by the polarization divided by the electric field
χ(ω) =P(ω)
0 E(ω)=
−nev(ω)
i 0 ωE(ω)=
ne
i 0 ω(
iωm
e+
1
μ)−^1 (118)
The dielectric constant can be calculated withε(ω) = 1 +χ(ω)and by using the definition of the
plasma frequencyω^2 P = ne
2
ε 0 m and the mobilityμ =eτ
m:ε(ω) = 1−
ω^2 P
ω^2 +iω(^2) Pε 0
σ(ω)ω
(119)
For low damping (largeσ) the dielectric constant can be approximated by the following equation
ε(ω) ≈ 1 −ωP^2
ω^2, (120)
this is what’s shown in fig. 71(a). At the plasma frequency the dielectric constant is zero and for higher
frequenciesεgets positive, thus the metal acts like an insulator, but what happens if the dielectric
constant is negative?
For a negative dielectric constant there is no propagating wave solution to the wave-equation only
damped solutions, but because the wave and it’s energy has to go somewhere the wave gets reflected
out of the material. In the following fig. 71(b) the reflectivity for Aluminium is shown. For photon
energies 0 − 15 eV aluminium is shiny, in the ultraviolet the reflectivity drops and the light goes
through the aluminium. To remind you: the plasma frequency is depending on the electron density,
this behaviour is used to make transparent contacts for solar cells. There a semiconductor with a low
electron density is used, so visible light can pass through it. However for DC the semiconductor is
still conducting.