Part 3 · Light in matter

3.2 The complex refractive index and absorption

Source: Postorino, week 3 slides, pp. 23–29, and week 4 slides, pp. 1–5

Objectives

Vocabulary. Complex refractive index $N = n + iκ$ , complex wave vector $K = k + i α$ , extinction and absorption coefficients, absorbance, oscillator strength, normal and anomalous dispersion.
Theorem. For the Lorentz medium, $K^{2} = \frac{ω ^{2}}{c ^{2}} N^{2}$ with $N^{2} = 1 + \frac{N e ^{2}}{m ε _{0}} \frac{1}{ω _{0}^{2} - ω ^{2} - iγω}$ .
Picture. The real part governs propagation (speed $c / n$ ), the imaginary part absorption ( $e^{- αx}$ ). $κ$ peaks at each resonance; $n$ rises towards it and drops through it.
Recipe. Measure the absorbance $A = ln (I_{I} / I_{T}) = 2 αx$ and get $α (ω)$ , $κ (ω)$ .

The complex refractive index

Put $P$ (Theorem 3.1.7) and $\nabla \cdot E = 0$ (Proposition 3.1.8) into the propagation equation: $\nabla^{2} E = μ_{0} ε_{0} \frac{\partial ^{2} E}{\partial t ^{2}} + μ_{0} \frac{N e ^{2} / m}{ω _{0}^{2} - ω ^{2} - iγω} \frac{\partial ^{2} E}{\partial t ^{2}} .$ With $μ_{0} ε_{0} = 1/ c^{2}$ and, for the second term, $μ_{0} = \frac{1}{c ^{2} ε _{0}}$ :

Theorem 3.2.1Wave equation in a Lorentz mediumslides w3, p. 23

$\nabla^{2} E = \frac{1}{c ^{2}} [1 + \frac{N e ^{2}}{m ε _{0}} \frac{1}{ω _{0}^{2} - ω ^{2} - iγω}] \frac{\partial ^{2} E}{\partial t ^{2}},$ with solutions $E (x, t) = E_{0} e^{i (K x - ω t)}$ , $K \in C$ (a complex wave vector), where $K^{2} = \frac{ω ^{2}}{c ^{2}} [1 + \frac{N e ^{2}}{m ε _{0}} \frac{1}{ω _{0}^{2} - ω ^{2} - iγω}] = \frac{ω ^{2}}{c ^{2}} N^{2} .$

Proof

Substitute the trial solution: $\nabla^{2}$ gives $- K^{2}$ and $\partial^{2} / \partial t^{2}$ gives $- ω^{2}$ . Cancel the common factor $- E$ .

This is the complex analogue of the dispersion relation. In vacuum $ω = k c$ , so $k^{2} = ω^{2} / c^{2}$ ; in matter $k = \frac{ω}{c} n$ . The bracket plays the role of $n^{2}$ :

Definition 3.2.2Complex refractive indexEssentialslides w3, p. 23–25

$N^{2} = 1 + \frac{N e ^{2}}{m ε _{0}} \frac{1}{ω _{0}^{2} - ω ^{2} - iγω}, N = n + i κ, K = \frac{ω}{c} N = k + i α .$ $n$ is the refractive index, $κ$ the extinction coefficient; $k$ the wave vector, $α$ the absorption coefficient. The real and imaginary parts of $K = \frac{ω}{c} N$ give $k = \frac{ω}{c} n (propagation), α = \frac{ω}{c} κ (absorption) .$

$N^{2}$ is a frequency-dependent, complex $ε_{r}$ (Definition 2.3.1) beyond the lectures.

Summary so far. In a neutral, non-magnetic medium:

Maxwell’s equations give the wave equation with source terms.
A perfect insulator has only bound charges, so the $J$ term drops.
The microscopic mechanism, in the harmonic approximation: $P = - N e r$ . A static field gives $P = \frac{N e ^{2}}{K} E$ ; an oscillating one drives a damped oscillator, $r = \hat{r} e^{- iω t}$ , giving $P (ω)$ .
The wave equation for $E$ alone has solutions $E_{0} e^{i (K x - ω t)}$ with $K^{2} = \frac{ω ^{2}}{c ^{2}} N^{2}$ .

Propagation and absorption

Theorem 3.2.3The real part propagates, the imaginary part absorbsslides w3, p. 25, 27

$E (x, t) = E_{0} e^{i [(k + i α) x - ω t]} = E_{0} e^{- α x} e^{i [k x - ω t]}, I \propto ∣ E ∣^{2} = ∣ E_{0} ∣^{2} e^{- 2 α (ω) x} .$

$i \cdot i αx = - αx$ : the imaginary part of $K$ becomes a real, decaying exponential. The wave oscillating as $k x - ω t$ moves at $ω / k = c / n$ , and its amplitude dies over a length $1/ α$ ; the intensity dies twice as fast, over $1/2 α$ .

Definition 3.2.4Absorbanceslides w3, p. 26

A beam crosses a sample along $x$ . Incident field $E_{I} = E_{0} e^{i [k x - ω t]}$ , intensity $I_{I} (ω) \propto ∣ E_{0} (ω) ∣^{2}$ ; transmitted field $E_{T} = E_{0} e^{- α (ω) x} e^{i [k x - ω t]}$ , intensity $I_{T} (ω) \propto ∣ E_{0} ∣^{2} e^{- 2 α (ω) x} = ∣ E_{0} ∣^{2} e^{- A (ω)}$ . The absorbance is $A (ω) = ln (\frac{I _{I} ( ω )}{I _{T} ( ω )}) .$

It is experimentally accessible: measure what goes in and what comes out. For thickness $x$ , $A = 2 α (ω) x$ , so measuring $A$ at each frequency gives $α (ω)$ and from it $κ (ω)$ beyond the lectures.

Transparent: $α (ω) = 0$ ; the wave crosses undamped.
Opaque: $α (ω) x^{*} ≫ 1$ ; the wave dies inside a sample of thickness $x^{*}$ .

The same material can be either at different frequencies. The slide’s example: chlorophyll a absorbs near 430 and 660 nm, chlorophyll b near 450 and 640 nm, carotenoids at 450–500 nm. They absorb blue and red and let green through: that is why leaves are green.

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Figure 3.2.1 A wave enters a slab of thickness

x^{*}

. Inside, its wavelength shrinks to

λ_{0} / n

and its amplitude decays as

e^{- α x}

(

envelope

); it leaves with its vacuum wavelength and a smaller amplitude. Set

n

and

κ

and read the absorbance.

Explicit n and κ for one resonance

Theorem 3.2.5n and κ of a Lorentz mediumslides w4, p. 1; w3, p. 28

Writing $N^{2} = (n + iκ)^{2} = n^{2} - κ^{2} + i 2 nκ$ , $n^{2} - κ^{2} = 1 + \frac{N e ^{2}}{m ε _{0}} (\frac{ω _{0}^{2} - ω ^{2}}{( ω _{0}^{2} - ω ^{2} ) ^{2} + γ ^{2} ω ^{2}}) (6.34)$ $2 nκ = \frac{N e ^{2}}{m ε _{0}} (\frac{γω}{( ω _{0}^{2} - ω ^{2} ) ^{2} + γ ^{2} ω ^{2}}) (6.35)$ (numbered as in Fowles).

Proof

Separate the real and imaginary parts of the fraction by multiplying by the conjugate of its denominator: $\frac{1}{( ω _{0}^{2} - ω ^{2} ) - iγω} \cdot \frac{( ω _{0}^{2} - ω ^{2} ) + iγω}{( ω _{0}^{2} - ω ^{2} ) + iγω} = \frac{ω _{0}^{2} - ω ^{2}}{( ω _{0}^{2} - ω ^{2} ) ^{2} + ( γω ) ^{2}} + i \frac{γω}{( ω _{0}^{2} - ω ^{2} ) ^{2} + ( γω ) ^{2}} .$ Equate real and imaginary parts with $n^{2} - κ^{2} + i 2 nκ$ .

Key observation 3.2.6One electron in one stateslides w3, p. 28

$κ = Im {N}$ is a single peak at $ω_{0}$ and zero elsewhere: absorption happens only near resonance. (6.35) has the oscillator’s Lorentzian shape (Proposition 1.4.7), width $\approx γ$ .
$n = Re {N}$ is 1 far from resonance on both sides. Approaching $ω_{0}$ from below it rises above 1 (normal dispersion: $n$ grows with $ω$ ). Through the absorption line it drops steeply to a minimum below 1 (anomalous dispersion, $d n / d ω < 0$ ), then climbs back towards 1.

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Figure 3.2.2

n

and

κ

of a Lorentz medium against frequency. Drag the frequency marker; change

ω_{0}

γ

and the strength

ω_{P}^{2} = N e^{2} / m ε_{0}

. The absorption peak sits at

ω_{0}

with width about

γ

;

n

is above 1 below it and below 1 just above it.

RemarkReading the formulasbeyond the lectures

Below resonance $ω_{0}^{2} - ω^{2} > 0$ : the electrons move in phase with the field, $P$ adds to $ε_{0} E$ , and $n > 1$ . Above resonance they move in antiphase, $P$ subtracts, and $n < 1$ . This is the in-phase/antiphase switch of the driven oscillator (Figure 3.1.1). Glass has its resonances in the UV, so visible light sits on the rising, normal side: $n$ is larger for blue than for red, and a prism sends blue further.

A complex system: electrons, nuclei, molecules, with many degrees of freedom.

Theorem 3.2.7Many resonancesslides w3, p. 28

A real material has several resonances $ω_{j}$ , each with its own damping $γ_{j}$ and weight $f_{j}$ : $N^{2} = 1 + \frac{N e ^{2}}{m ε _{0}} j \sum (\frac{f _{j}}{ω _{j}^{2} - ω ^{2} - i γ _{j} ω}) . (6.36)$

The slide’s figure: a hypothetical substance with absorption bands in the infrared ( $ω_{1}$ ), visible ( $ω_{2}$ ) and ultraviolet ( $ω_{3}$ ). $κ$ peaks at each band; $n$ steps down across each one, with an anomalous-dispersion wiggle, and tends to 1 in the X-ray region.

$f_{j}$ is the oscillator strength, the fraction of electrons responding at $ω_{j}$ , with $\sum_{j} f_{j} = 1$ beyond the lectures. Its real justification is quantum mechanical: transitions between atomic levels (§5.4 Bohr's atom).

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Figure 3.2.3 Three resonances, in the infrared, visible and ultraviolet, as in the slide’s hypothetical substance.

κ

has a peak at each;

n

steps down across each one and tends to 1 at the highest frequencies.

Two remarks on the setup

N.B. The incident beam is the one at the start of the sample. The interfaces and the reflection there, which in reality also reduces the transmitted light (Proposition 2.3.9), are ignored.
N.B. The frequency does not change from one medium to another (Theorem 2.3.10); the different speed shows up as a shorter wavelength, $λ_{m} = \frac{λ _{0}}{n}, ω = k_{m} v = \frac{2 π}{λ _{m}} \frac{c}{n} = \frac{n 2 π}{λ _{0}} \frac{c}{n} = \frac{2 π}{λ _{0}} c = k_{0} c .$ In the slide’s figure the wave crosses a slab of $n = 1.5$ between $0$ and $x^{*}$ with a shorter wavelength $λ_{n}$ , and leaves with its vacuum wavelength $λ_{0}$ (Figure 3.2.1).

Polarisation mechanisms

The Lorentz model describes the electrons. In general a field polarises matter in three ways, each responding up to its own frequency range.

mechanism	what happens	response time	frequencies
orientational (dipolar)	permanent dipoles align with the applied field	gas $1 0^{- 12}$ s; small molecules in low-viscosity liquids $1 0^{- 11}$ – $1 0^{- 10}$ s; large molecules in viscous liquids $1 0^{- 6}$ s	far IR; microwaves; radio
molecular (ionic)	in molecules with ionic character the field pulls the ions apart and changes the size of the dipole	$1 0^{- 12}$ – $1 0^{- 13}$ s	infrared
electronic	the field shifts the centre of the electron charge, creating a dipole and changing its size	$1 0^{- 15}$ s	visible–UV

Electronic polarisability occurs in all materials, solid, liquid and gas, and is very fast. It is essentially independent of temperature, because the electron density changes little. This is the Lorentz model of class.
Molecular polarisability occurs in molecular systems; it is weaker and slower than the electronic kind.
Orientational polarisability occurs in polar substances such as water, whose molecules carry a permanent dipole. Thermal agitation opposes it by keeping orientations random, so it depends strongly on temperature. Its times are much longer still.

RemarkWhy time maps to frequencybeyond the lectures

A mechanism that needs a time $τ$ to respond can follow a field only up to frequencies of about $1/ τ$ . Beyond that it is “frozen out” and stops contributing to $ε_{r}$ (Frozen out). So $ε_{r}$ steps down as frequency rises: water has $ε_{r} \approx 80$ at low frequency but $n^{2} \approx 1.78$ in the visible.

Why water is a polar molecule (slides w4, p. 4, from sciencenotes.org): oxygen and hydrogen have different electronegativities; oxygen’s two lone pairs repel each other and the bonding electrons, which bends the molecule to 104.5°; so the oxygen side carries a partial negative charge ( $δ^{-}$ ) and the hydrogen side a partial positive one ( $δ^{+}$ ). A microwave field makes these dipoles rotate back and forth: orientational polarisation, whose damping heats the water (Skin depth and the microwave oven).

The solar spectrum

The Sun’s irradiance against wavelength, from 250 nm into the infrared:

above the atmosphere it closely follows a 5778 K blackbody curve, peaking around 500 nm (§5.1 Line spectra and blackbody radiation). The Sun’s surface is at about 5777 K on average; its core at about $1.57 \times 1 0^{7}$ K.
at sea level deep notches are cut out by the atmosphere’s absorption bands: O₃ in the UV, O₂ near 760 nm, H₂O near 940, 1130 and 1400 nm and beyond, CO₂ near 2000 nm. Each is a resonance of those molecules: an $ω_{j}$ in (6.36) where $κ$ peaks.

Key observation 3.2.8Fraunhofer linesslides w4, p. 5

Absorption lines of “cold” gases lying between the source and the detector.
The Sun’s surface gases are cooler than the layers below, so they absorb, at their own resonance frequencies, light coming from the hotter interior.

The small jagged dips near the peak are Fraunhofer lines. The slide also shows the Sun’s structure from the inside out: core, radiative zone, tachocline, convective zone, photosphere (the visible surface), and the atmosphere (chromosphere, transition zone, corona).

Exercises

Exercise 3.2.1beyond the lectures

A Lorentz medium has $ω_{0} = 1 0^{16}$ rad/s, $γ = 1 0^{14}$ s⁻¹ and $N e^{2} / m ε_{0} = 1 0^{32}$ s⁻². Find $n$ at $ω = 3 \times 1 0^{15}$ rad/s and decide whether it absorbs there.

Solution

Far from resonance $γω ≪ ω_{0}^{2} - ω^{2}$ , so $κ \approx 0$ : no absorption. $n^{2} \approx 1 + \frac{1 0 ^{32}}{1 0 ^{32} - 9 \times 1 0 ^{30}} = 2.099$ , so $n \approx 1.45$ , like glass.

Exercise 3.2.2beyond the lectures

Show that (6.35) makes $κ > 0$ for every $ω > 0$ in the $e^{- iω t}$ convention, and say what a negative $κ$ would mean.

Solution

The right side is positive and $n > 0$ , so $κ > 0$ and the field decays as $e^{- αx}$ , $α = ωκ / c > 0$ . A negative $κ$ would mean the wave grows: gain, as in a laser medium, which a passive damped oscillator can’t give.

Exercise 3.2.3beyond the lectures

A solution transmits 10 % of the light through a 1 cm cell. Find the absorbance and $α$ . What fraction gets through 2 cm?

Solution

$A = ln 10 = 2.30$ . $A = 2 αx$ gives $α = 1.15$ cm⁻¹. In 2 cm, $A = 4.6$ and $I_{T} / I_{I} = e^{- 4.6} = 1%$ .

Exercise 3.2.4beyond the lectures

At $ω = ω_{0}$ , find $N^{2}$ and, for weak absorption, estimate $κ$ .

Solution

$N^{2} = 1 + \frac{ω _{P}^{2}}{- iγ ω _{0}} = 1 + i \frac{ω _{P}^{2}}{γ ω _{0}}$ , with $ω_{P}^{2} = N e^{2} / m ε_{0}$ . If $ω_{P}^{2} / γ ω_{0} = ϵ ≪ 1$ , $N \approx 1 + i ϵ /2$ : $n \approx 1$ , $κ \approx ω_{P}^{2} / (2 γ ω_{0})$ , the peak of the Lorentzian.

Exercise 3.2.5beyond the lectures

Light of 600 nm in vacuum enters water ( $n = 1.33$ ). Find its wavelength inside, and check $ω = k_{m} v$ .

Solution

$λ_{m} = 600/1.33 = 451$ nm. $k_{m} = 2 π / λ_{m}$ , $v = c /1.33$ : $k_{m} v = \frac{2 π \cdot 1.33}{600 nm} \cdot \frac{c}{1.33} = \frac{2 π c}{600 nm} = ω$ .