Part 3 · Light in matter

3.1 Dielectrics: the Lorentz model

Source: Postorino, week 3 slides, pp. 17–22, with Fowles ch. 6 and Frova–Mataloni

Objectives

Theorem. In a neutral, non-magnetic medium, Maxwell’s equations give the vacuum wave equation plus two source terms: one from the bound charges ( $P$ ), one from the free charges ( $J$ ).
Picture. Each bound electron is a damped oscillator driven by the light’s electric field; the nuclei are too heavy to follow.
Recipe. Complex substitution with $e^{- iω t}$ : $P = \frac{N e ^{2} / m}{ω _{0}^{2} - ω ^{2} - iγω} E$ .
Vocabulary. Bound and free charges, perfect insulator and conductor, electron density, polarisability.

From here on the slides follow Fowles, Introduction to Modern Physics, ch. 6, and Frova and Mataloni’s optics notes (PDF).

A change of sign convention

WarningFrom week 3, time goes as

e^{- iω t}

Weeks 1–2 wrote the complex time dependence as $e^{+ iω t}$ (§1.3 Linear equations and the complex substitution). From week 3 the slides write it as $e^{- iω t}$ , as in the plane wave $e^{i (k x - ω t)}$ .

Each $\partial / \partial t$ now gives $- iω$ , not $+ iω$ .
So the damped oscillator’s denominator, $ω_{0}^{2} - ω^{2} + iγω$ in week 1, becomes $ω_{0}^{2} - ω^{2} - iγω$ , written $(- ω^{2} - iγω + ω_{0}^{2})$ on the slides.

Every result is the complex conjugate of the week 1 one, and the physics is the same beyond the lectures. With $e^{- iω t}$ , absorption shows up as a positive imaginary part of $N$ and $K$ .

The wave equation with source terms

Start again from Maxwell’s equations in matter, with $D = ε_{0} E + P$ and $B = μ_{0} (H + M)$ . Take an electrically neutral, non-magnetic medium: $ρ = 0$ , $M = 0$ , so $B = μ_{0} H$ . Then

Proposition 3.1.1Maxwell’s equations, neutral non-magnetic mediumslides w3, p. 17

$\nabla \cdot E = - \frac{1}{ε _{0}} \nabla \cdot P, \nabla \cdot B = 0, \nabla \land E = - \frac{\partial B}{\partial t}, \nabla \land B = μ_{0} ε_{0} \frac{\partial E}{\partial t} + μ_{0} \frac{\partial P}{\partial t} + μ_{0} J .$ The last is $μ_{0}$ times the curl-of- $H$ equation.

As for vacuum, apply $\nabla \land$ to Faraday’s law.

Theorem 3.1.2Wave equation with source termsslides w3, p. 18

$- \nabla (\nabla \cdot E) + \nabla^{2} E = μ_{0} ε_{0} \frac{\partial ^{2} E}{\partial t ^{2}} + μ_{0} \frac{\partial ^{2} P}{\partial t ^{2}} + μ_{0} \frac{\partial J}{\partial t} .$

Proof

$\nabla \land \nabla \land E = - \frac{\partial}{\partial t} [\nabla \land B] = - \frac{\partial}{\partial t} [μ_{0} ε_{0} \frac{\partial E}{\partial t} + μ_{0} \frac{\partial P}{\partial t} + μ_{0} J] .$ Differentiate in time, use $\nabla \land \nabla \land E = \nabla (\nabla \cdot E) - \nabla^{2} E$ (Proposition 2.2.2), and change sign.

The term $- \nabla (\nabla \cdot E)$ will turn out to vanish (Proposition 3.1.8). Apart from it, this is the vacuum wave equation plus two source terms:

one depending on $P$ : the field generated by the bound charges;
one depending on $J$ : the field generated by the free charges.

The matter radiates its own field, which adds to the incoming one. The problem is tackled in two limits:

a perfect insulator: only bound charges, $P \neq = 0$ , $J = 0$ (this section and §3.2 The complex refractive index and absorption);
a perfect conductor: only free charges, $P = 0$ , $J \neq = 0$ (§3.3 Metals: the Drude model). The slide says “only bound charges” here too, a slip for “free”.

The Lorentz model

Only bound charges: $P \neq = 0$ , $J = 0$ . The propagation equation becomes $- \nabla (\nabla \cdot E) + \nabla^{2} E = μ_{0} ε_{0} \frac{\partial ^{2} E}{\partial t ^{2}} + μ_{0} \frac{\partial ^{2} P}{\partial t ^{2}} .$ There are too many unknowns. A model of the polarisability writes $P$ in terms of the applied field.

Definition 3.1.3Induced polarisationslides w3, p. 19

$P = - N e r$ is the dipole moment per unit volume that the field induces by acting on the bound charges, where $N$ is the electron density (electrons per unit volume), $e$ the electron charge, and $r$ the displacement from its equilibrium position of the centre of the electron cloud.

With $E = 0$ the electron cloud is centred on the nucleus: the atom is unpolarised. With $E \neq = 0$ the cloud shifts against the field, the two centres of charge separate, and the atom becomes a small dipole.

Key observation 3.1.4Only the electrons respondslides w3, p. 20

The masses differ hugely, $M_{P} / m_{e} = 1836$ (Key observation 1.1.2), so in practice only the electrons feel the applied field. They feel the force $F_{el} = - e E$ , which deforms the electron charge around the nucleus and moves its centre a distance $r$ from the nucleus.

The electron stays bound to the atom. In the harmonic approximation the binding is an elastic restoring force $- K r$ , opposite to $F_{el}$ .

Proposition 3.1.5Static polarisationslides w3, p. 20

For a static field, at equilibrium $F_{tot} = - K r - e E = 0$ , so $r = - \frac{e E}{K}, P = - N e r = \frac{N e ^{2}}{K} E .$ $P$ is parallel and proportional to $E$ .

From $D = ε_{0} E + P = ε_{0} ε_{r} E$ , the static relative permittivity is $ε_{r} = 1 + \frac{N e ^{2}}{ε _{0} K}$ beyond the lectures.

Oscillating fields. Now the electron’s motion matters. Newton’s law: $m \frac{\partial ^{2} r}{\partial t ^{2}} = - K r - e E \Rightarrow m \frac{\partial ^{2} r}{\partial t ^{2}} + K r = - e E,$ a forced harmonic oscillator: an ideal resonant system. To make it realistic, add viscous friction $- β \partial r / \partial t$ , with $β = γm$ .

Theorem 3.1.6The Lorentz oscillatorslides w3, p. 21

$\frac{\partial ^{2} r}{\partial t ^{2}} + γ \frac{\partial r}{\partial t} + ω_{0}^{2} r = - \frac{e E}{m}, ω_{0} = K / m :$ a forced damped harmonic oscillator.

Each bound electron is a damped oscillator driven by the light’s electric field (Theorem 1.4.3). $ω_{0}$ is its resonance, and $γ$ the damping from all the ways it loses energy.

Polarisation as a function of frequency

Theorem 3.1.7Polarisation in an oscillating fieldslides w3, p. 22

$P = - N e r = \frac{N e ^{2} / m}{ω _{0}^{2} - ω ^{2} - iγω} E .$

Proof

Complex substitution with $e^{- iω t}$ : $E (r, t) = E_{0} e^{i (kr - ω t)} = \hat{E}_{0} e^{- iω t}$ , absorbing the spatial phase into $\hat{E}_{0}$ , and $r (t) = \hat{r} e^{- iω t}$ . Each time derivative brings $- iω$ : $(- ω^{2} - iγω + ω_{0}^{2}) \hat{r} = - \frac{e E ^ _{0}}{m} \Rightarrow \hat{r} = \frac{- e E ^ _{0}}{( - ω ^{2} - iγω + ω _{0}^{2} ) m} .$ Multiply by $- N e$ .

The polarisation is now expressed through the electric field, by a function of frequency, so the propagation equation can be written entirely in $E$ . Check beyond the lectures: at $ω = 0$ it is $\frac{N e ^{2}}{m ω _{0}^{2}} E = \frac{N e ^{2}}{K} E$ , the static result.

Loading the demo…

Figure 3.1.1 One atom in the light’s field

E (t)

: the

electron cloud

is pulled against the field and springs back. Set the light’s frequency against the electron’s own

ω_{0}

and press Play. Well below resonance the cloud follows the field (in phase, so the dipole adds to it); at resonance it lags by a quarter period and swings widest; above resonance it moves in antiphase.

Proposition 3.1.8The divergence term vanishesslides w3, p. 22

For a light wave, $\nabla \cdot E = 0$ in the medium too.

Proof

The polarisation is a function of $ω$ times the field, $P = f (ω) E$ , so $\nabla \cdot P = f (ω) \nabla \cdot E$ . The first Maxwell equation reads $\nabla \cdot E = - \frac{1}{ε _{0}} f (ω) \nabla \cdot E \Rightarrow (1 + \frac{f ( ω )}{ε _{0}}) \nabla \cdot E = 0.$ So either $1 + f (ω) / ε_{0} = 0$ , or $\nabla \cdot E = 0$ .

WarningSlide slip

The slide gives the alternative as “ $f (ω) = 0$ ”. From the line above it is $1 + f (ω) / ε_{0} = 0$ . That would make $ε_{r} = 0$ , which happens only at isolated frequencies (longitudinal oscillations, not light waves). Otherwise $\nabla \cdot E = 0$ , which is the case used.

Exercises

Exercise 3.1.1beyond the lectures

A static field of $1 0^{6}$ V/m acts on hydrogen-like atoms with $K = 1.6 \times 1 0^{2}$ N/m (an electron with $ω_{0} \approx 1.3 \times 1 0^{16}$ rad/s). How far does the cloud move?

Solution

$r = e E / K = 1.6 \times 1 0^{- 19} \cdot 1 0^{6} /160 = 1 0^{- 18}$ m: a hundred-millionth of the atom’s size. Even huge laboratory fields barely polarise an atom, which is why the linear (harmonic) model works.

Exercise 3.1.2beyond the lectures

Show that $ω_{0} = K / m$ with $K = 160$ N/m gives an ultraviolet resonance.

Solution

$ω_{0} = 160/9.1 \times 1 0^{- 31} = 1.33 \times 1 0^{16}$ rad/s, $λ_{0} = 2 π c / ω_{0} = 142$ nm: far ultraviolet. Visible light is below resonance, so glass-like materials have $n > 1$ and are transparent (§3.2 The complex refractive index and absorption).

Exercise 3.1.3beyond the lectures

Write the Lorentz result in week 1’s $e^{+ iω t}$ convention, and check that the two descriptions give the same real displacement.

Solution

With $e^{+ iω t}$ : $\hat{r}^{'} = \frac{- e E ^ _{0} / m}{ω _{0}^{2} - ω ^{2} + iγω}$ , the conjugate of the $e^{- iω t}$ result (for real $\hat{E}_{0}$ ). Then $Re {\hat{r}^{'} e^{iω t}} = Re {(\hat{r} e^{- iω t})^{*}} = Re {\hat{r} e^{- iω t}}$ : the real part of a conjugate is the same.

Exercise 3.1.4beyond the lectures

Why do the nuclei not contribute to the polarisation at optical frequencies, though they do in the infrared?

Solution

They are about 2000 times heavier (and much more for whole ions), so their resonances are at far lower frequencies: $ω_{0} = K / M$ . At optical frequencies they are driven far above resonance and barely move. In the infrared, near their own resonances, ionic polarisation appears (Polarisation mechanisms).