What consequences does this model predict for the optical properties of metals? Let us take up the model of the previous section: Free movement of electrons means that there is no restoring force, in other words there is no resonance frequency ω0.
We can also relate the damping constant β to a fundamental material property: Drude assumed that the presence of an electric field accelerates electrons, but they lose their energy by occasionally bumping into the ionized cores of the crysal lattice. On average, the electrons are accelerated for a certain amount of time called relaxation time τ before they undergo the next collision. During this time they pick up an average drift velocity v given by product of the acceleration a and τ. The accelerating force is, of course, due the electric field F = eE. Thus:
We will now reverse the logic; we argue that on average a collisions takes place after the relaxation time and that it is very unlikely to find electrons that are faster than the drift velocity because they should undergo their next collision. Effectively, this is like a damping force that becomes more important at higher velocity. For the sake of simplicity we assume it to be proportional to the electron velocity. From above we find F = m/τ . v. Putting that into the equation of motion with zero resonance frequency we find:
Going through the same steps with average amplitude, dipole moment and polarization as in the previous chapter, we find for the dielectric function:
We introduced the plasma frequency ωp, and the quantity ε takes care of the fact that the high frequency limit of the dielectric constant often differs from one.
A representation of the two functions for a plasma frequency corresonding to an energy of 2.5 eV is given below:
From the diagram we observe that κ should become small for frequencies higher than the plasma frequency, i.e. the electron gas behaves like a dielectic and becomes transparent. In reality this is not completely true, the figure below shows the dielectric functions for two real metals, silver (black) and gold (red).
In the real materials the exciction coefficient does not approach zero because of interband transition that absorb energy. In gold this is the case for blue light which explains the yellow colour of gold. For silver this absorption occurs at higher energy in the near UV which makes silver a very neutral reflector for visible light.
In order to calculate the reflection of the free electron gas we must use the complex Beer formula:
The figure below compares the reflection reflection curves for the examples given above. For low incident photon energies the free electron plasma in metals reflects almost ideally, around the plasma frequency metals absorb, and at higher energies they are (at least) partially transparent. This finding explains the yellow appearance of gold, because its plasma frequency is in the visible and it absorbs most of the blue and green light, whereas its reflectiton is excellent for yellow and red. This effect is more pronounced for copper, it absorbs also yellow light and reflects only red light.
