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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.

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Permittivity is surely not a "vector". It is a scalar constant or, at most - in anisotropic materials - a tensor. Microscopically, permittivity boils down to the related constant "polarizability" which quantifies how much a molecule is polarized when exposed to an external electric field. In this way, one creates some density of dipoles $\vec P\sim \vec E$ and they modify the original electric field. Maxwell's equations in materials are usually written directly in terms of $\vec E,\vec D$ where $\vec D = \vec E+\vec P$, schematically. – Luboš Motl Jun 23 '13 at 8:18
duplicate of – Ben Crowell Jun 23 '13 at 15:48

3 Answers 3

up vote 5 down vote accepted

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.

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nice answer Carlos! – Dale Jul 13 '13 at 0:22
@Carlos: You have a typo in your answer. The polarization and E-field relation in the first paragraph is reversed. Really nice answer, more physical than the book's explanation. – Juan Sep 22 '13 at 18:41
@Juan: thanks for correcting me. I've made the changes. – Carlos Sep 23 '13 at 11:35

You may wish to look at the Lorentz-Lorenz equation (please see ) and the book by Born and Wolf, which is referenced there and provides a derivation of the equation. The equation has limited applicability (as it does not take into account quantum theory), but it is a good first step.

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As well as Born&Wolf, you might like to look at Chapter 11 of volume 2 of the "Feynman lectures on physics", called "inside dielectrics". This is an excellent beginner's classical description of the mechanisms behind a material's polarisability.

Actually it's not the $damped$ motion of electrons that gives rise to the usual, real valued permittivity. Damping (i.e. energy loss) begets the imaginary part of permittivity and leads to the attenuation of light as it propagates through the material. The permittivity's real part simply arises from light's being "slowed down" by the material as follows. A photon is propagating through free space and is then absorbed by one of the material's constituent (atom or molecule), fleetingly raising the latter into a higher energy state. A very short time later, this excited constituent emits a photon in the same direction as the first. The second one propagates on, is absorbed by another matter constituent, and a third is emitted a short time later, again with exactly the same momentum as the second. This repeated absorption, delay, emission can be modelled by saying the original photon is just running slower than it would in free space. Call me pedantic, but I like to say that it's not light that is propagating through the material, but a quantum superposition of photons and excited matter states.

This process is somewhat like fluorescence, but with two crucial differences:

  1. It is $extremely$ fast: the delay between absorption and re-emission is femtoseconds at the very most.

  2. The absorber returns $exactly$ to its initial state and gains no energy nor momentum, so there is no wavelength shift and no direction change between the absorbed and re-emitted photons. Mostly there is no angular momentum transfer either, and this means that the light's polarisation stays the same - see the chapter "Angular Momentum" in volume 3 of the Feynman lectures to understand the relationship between polarisation and angular momentum. The exception to the last point is, of course, in a birefringent material.

Damping arises when the excited state is weakly coupled to a band of other, usually vibrational states in the matter. The weak coupling means that mostly photons are absorbed and re-emitted loss mostly as described above, but every so often the excited state couples to the vibration states and the photon concerned is thus permanently lost.

The lossless (real-valued) permittivity as a function of wavelength is described by the Sellmeier equations. With loss, these become the Ketteler–Helmholtz model. I have a couple of papers on this subject:

Vance RWC, Ladouceur F. One photon electrodynamics in optical fiber with fluorophore systems. II. One-polariton propagation in matter and fibers from the one-photon correspondence principle. J Opt Soc Am 2007; 24(4): 942-958.

Vance RWC, Ladouceur F. One photon electrodynamics in optical fiber with fluorophore systems. III. One-polariton propagation in fluorophore-driven fibers. J Opt Soc Am 2007; 24(6): 1369-1382.

A further paper aside from my own on the full Ketteler–Helmholtz model is:

G. Dubost and A. Bellossi, “Modèle physique de la membrane cellulaire dans les spectres infrarouge, visible etultraviolet,” Rev. Sci. Tech. Déf. 64, 113–127 (2004).

I'm really sorry: I haven't come across an english language description - but I notice Carlos has just given a good classical description in his answer. Also, as advised by Ben Cromwell, see

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