electromagnetism - How Does $epsilon$ Relate to the Dampened Harmonic Motion of Electrons?

I realize that the permittivity $\epsilon$ of a substance is easily calculated based on diffraction angles, but I am not satisfied with merely measuring it experimentally. I wish to understand its origin within the laws and equations of physics. I realize that it is a vector as demonstrated by the anisotropy of birefringence and $\epsilon$ also depends on wavelength. I did some reading in Griffiths' Intro to Electrodynamics p. 400-404:

The dampened harmonic motion of electrons can account or the frequency dependance of the index of refraction, and it expains why $n$ is ordinarily a slowly increasing function of $\omega$, with occasional anomalous regions where it precipitously drops.

Where $n$ is the index of refraction and $n=\frac{c k}{\omega}$.

I found some vague references to this phenomena as it relates to negative refraction.

Answer

Lubos hit the key point: you need to calculate the polarization of an atom/molecule so that is the starting point. As you may already know, when a dielectric is subjected to the impinging E-field of an EM wave, there are dipoles generated that contribute to the total internal field. The resultant field for most materials is given by $(\epsilon−\epsilon_0)E=P$. However, classically, the polarization will depend on the relative displacement between the electron cloud and the nucleus and this displacement can be calculated by thinking of the electron as a harmonic oscillator. That is, the electron cloud will oscillate about the nucleus. There are three terms that must come into play to describe the displacement of the electron.

1. The electron cloud bound to the nucleus must have some sort of restoring force: $−mω_0^2x$ where $ω_0$ is the resonate frequency and m is the mass of the electron.


2. The impinging EM wave will exert a time varying force, say $cosωt$, on the electrons: $eE(t) = eEcosωt$ where $ω$ is the driving frequency.


3. For a gas, atoms are far enough apart that we can “ignore” interactions between them. However, for atoms and molecules in close proximity, one cannot ignore these interactions which behave as “frictional” type forces. That is, the electron oscillators will dissipation some of their energy as heat. Therefore, there must be some type of velocity term: $mβ\frac{dx}{dt}$ where $β$ is a damping constant.

If we now stuff all of this into Newton’s Second law, we have an equation for the electron displacement:

$$m\frac{d^2 x}{dt^2} = −mω_0^2x - mβ\frac{dx}{dt} + eE_0cosωt$$

Physically, we expect that the electron will oscillate at the same frequency as the impinging EM wave, so that the equation above has the solution $x(t) = Acosωt$. Substituting this assumed solution and solving for the amplitude, we get

$$x(t) = \frac{eE(t)}{m(ω_0^2-ω^2+iβ ω)}$$

The electric polarization is the density of dipole moments: $P(t) = ex(t)N $ where $N =$ number of dipoles. Solving for the electric permittivity using $(ϵ−ϵ_0)E=P$,

$$ϵ=ϵ_0 + \frac{P(t)}{E(t)} = ϵ_0 + \frac{Ne^2}{m(ω_0^2-ω^2+iβω)}$$

The way the electric permittivity is related to the index of refraction is as follows: most materials are nonmagnetic at optical frequencies (the relative permeability is very close to one). So to a good approximation, the index of refraction depends only on the relative permittivity $ϵ_r: n^2(ω) = ϵ_r = \frac{ϵ}{ϵ_0}$. Therefore, the dispersion relationship from a damped harmonic oscillator view point looks like

$$n^2(ω) = ϵ_r = \frac{ϵ}{ϵ_0} = 1 + \frac{Ne^2}{m ϵ_0} \frac{1}{ω_0^2-ω^2+iβω}$$

You can see that the index of refraction is frequency dependent. Note that this is valid for only a single resonate frequency; a given substance has several such resonate frequencies and this equation will need to be modified but the quantum mechanical solution looks very similar to the one above. I will not explain this since I think that I have answered your question.