Chapter 2Waves in Media

2.1 Introduction The propagation of waves in a medium depends on the magnetic permeability and electricpermittivity functions µ (ω) and ε (ω) for the medium. We will generally deal with systems in which µ (ω) can beapproximated as having the value 1. A general form often used to approximate the electric permittivity is the Sellmeierequation

ε (ω) = 1 +Xm

Am

ω2m − iγmω − ω2. (2.122)

This equation is based on the assumption that the system behaves as a set of resonances. This will be observed in the twosystems we will examine in this chapter. First we will consider electromagnetic waves in ionic crystals. The discussion willbe restricted to frequencies below that of the electronic interband transitions. In this frequency range the dielectric functionis determined by the ionic displacements and the polarizabilities of the ions. The second system will be electrical conductors,either metals or plasmas. In these systems the frequency dependent conductivity determines the dielectric function.

The analysis of the models will be followed by a discussion of wave packets in dispersive materials. This discussionwill use the permittivities obtained for the two models. The propagation of electromagnetic waves in a medium involves theindex of refraction of the medium. From the deduced analytic properties of the complex index of refraction we will obtainthe Kramer­Kronig relationship between the real and imaginary parts of the index of refraction.

We will close the chapter with a discussion of reectance and transmittance of electromagnetic waves at the interfacebetween media.

2.2 Dielectric constant for ionic crystals Ionic crystals such as the alkali­halides are cubic crystals. Notethat electromagnetic fields only interact with the ‘optic modes’ for the crystal. These are the modes in which the positive andnegative ions move in opposite directions. The dielectric function of cubic crystals is independent of the direction of theelectric field and can be characterized by three parameters, the static dielectric constant εs, the optical dielectric constantεopt, and a characteristic resonant frequency ωT . For reference typical resonant frequencies are in the range 0.02eV ­ 0.04eV,optical frequencies are in the range 2eV ­ 3eV, and the electronic interband transitions begin around 4 or 5eV. An empiricaldielectric function for these crystals is

ε (ω) = εopt + ω2Tεopt − εs

ω2 + iγω − ω2T(2.123)

We will find that ωT can be identified as the transverse optical phonon frequency for the crystal. A damping factor has beenincluded to indicate what a real dielectric function might look like. This is an approximation to the Sellmeier equation, givenin Eq.2.122, in a frequency range near one resonant term and far from any other resonance. Figure 2. shows a typicaldielectric function (KBr1: εs = 4.90, εopt = 2.34, and ωT = 2.26× 1013 rad/s, 14.3meV ) with no damping.

1 Ashcroft and Mermin, Solid State Physics (Saunders College, 1976, page 553)

32

ωL/ωT

-20.0

-10.0

0.0

10.0

20.0

ε(ω) Real ε(ω) for KBr

0 1 2 3 4 5

ω / ωT

.Figure 2.

2.2.1 Longitudinal optical crystal vibrations From the permittivity we can obtain the frequency for alongitudinal optical mode of crystal vibration. These modes are characterized by a longitudinal charge density wave (that is,a charge density disturbance which propagates along the direction of the wave). This wave would generate a longitudinalelectric field wave when it generates a current density. The divergence of the current density gives the time rate of change ofthe charge density. If this is self­sustaining in the absence of externally applied fields it is a normal mode of oscillation of thesystem. The electric field set up by the microscopic charge density will satisfy ∇ ·E (r, ω) = 4πρm (r,ω) . In the absenceof an ‘external charge density’ we must have that ∇ ·D (r, ω) = 0. SinceD (r, ω) = ε (ω)E (r, ω) and∇ ·E (r, ω) 6= 0 wemust find that ε (ωL) = 0 where ωL is the frequency of the longitudinal dynamic mode. The solution is given by

ωL =

rεsεopt

ωT (2.124)

(For reference2: KBr ωL = 3.14× 1013 rad/s and (ωL /ωT )2 = 1. 9) This is known as the Lyddane­Sachs­Teller relation.

In the first approximation the longitudinal optical modes of vibration of a crystal have a constant frequency independent ofthe wavelength of the charge density wave. The location of this frequency is noted in Fig. 2. .

2.2.2 Transverse electromagnetic waves in the crystal The transverse optical modes do not generate acharge density. They do involve current densities but the electromagnetic fields generated by these currents do not contributean appreciable amount to the forces involved in the crystal vibrations. However an electromagnetic wave propagatingthrough the material can couple to the transverse optical mode of vibration.

2 Woods, Brockhouse, Cowley, and Cochran, Phys. Rev. 131 1027 (1963)

33

.Figure 3.

These modes have approximately constant frequency and wave vectors, k =2π

λTO, ranging from zero to of order

1 rad/nm. That is, λTO ≥ 2π nm. Examples of the dispersion curves for longitudinal and transverse optical andacoustic phonon modes are shown in Fig.3 . Compare the range of phonon wave vectors to a typical wave vector for an

ω = 1013 rad/s electromagnetic wave: k =ω

c=

1013rad/s

3x108 · 109nm/s≈ 3× 10−5 rad/nm. In the ω versus k curves shown

in Fig. 3 the ω versus k of an electromagnetic wave would lie on the right hand vertical axis. To search for transversemodes we consider the propagation of a transverse electromagnetic wave in the crystal. In the following we assume forconvenience that µ (ω) = 1. Since we seek transverse modes we only consider electric fields satisfying∇ ·E (r, ω) = 0.

∇× (∇×E (r, ω)) = ∇(∇ ·E)−∇2E−∇2E =

−1c

∂

∂t(∇×B)

−∇2E =−1c2

∂2

∂t2D+

4π

c

∂

∂tJo

In this case the wave equation for the electric field, (with no current sources) is given by

−∇2E (r, ω) = −ε (ω) [−iω]2

c2E (r, ω) (2.125)

= ε (ω)ω2

c2E (r, ω)

or, in wave vector space,

k2E (k, ω) = ε (ω)ω2

c2E (k, ω) (2.126)

34

so that

(k2 − ε (ω)ω2

c2)E (k, ω) = 0

Thus, for a transverse electromagnetic wave to propagate in the medium the wave vector and angular frequency must satisfythe dispersion relation

k2 = ε (ω)ω2

c2(2.127)

or, using the dielectric function without damping,

ε (ω) = εopt + ω2Tεopt − εsω2 − ω2T

=εoptω

2 − ω2T εsω2 − ω2T

µkc

ωT

¶2= ε (ω)

ω2

ω2T=

εopt[ω2

ω2T]2 − εs

ω2

ω2Tω2

ω2T− 1

or, letting x = ω2

ω2T,

εoptx2 − (εs +

µkc

ωT

¶2)x+

µkc

ωT

¶2= 0

ω2

ω2T=(εs +

³kcωT

´2)

2εopt±

(εs +

³kcωT

´2)2 − 4εopt

³kcωT

´24ε2opt

1/2

(2.128)

ω2

ω2T=

(εs +³kcωT

´2)

2εopt

1±1−

4εopt

³kcωT

´2(εs +

³kcωT

´2)2

1/2

≈(εs +

³kcωT

´2)

2εopt

1± 1∓ 12

4εopt

³kcωT

´2ε2s

whenkc

ωT−→ 0

≈(εs +

³kcωT

´2)

2εopt

1± 1∓ 12

4εopt

³kcωT

´2(εs +

³kcωT

´2)2

whenkc

ωT−→∞

As kcωT−→ 0

ω2

ω2T upper branch (+)

≈(εs +

³kcωT

´2)

2εopt2 −→ εs

εopt

ω2

ω2T lower branch (­)

≈(εs +

³kcωT

´2)

2εopt

1

2

4εopt

³kcωT

´2ε2s

−→ 0

As kcωT−→∞

35

ω2

ω2T upper branch (+)

≈(εs +

³kcωT

´2)

2εopt2 −→ εs

εopt+

³kcωT

´2εopt

ω2

ω2T lower branch (­)

≈2εopt

³kcωT

´22εopt(εs +

³kcωT

´2)−→ 1

These two dispersion relations for the electromagnetic waves are shown in Fig. 4 with the minus sign giving the lowerbranch and the plus sign the upper. We note that an electromagnetic wave in the frequency range ωT < ω < ωL ( or1 <

ω

ωT<q

εsεopt

= 1.39) can not propagate in the crystal, where ωL =q

εsεopt

ωT . The transverse optical modes of the

crystal will lie in this range of frequencies.

ω = kc/ εs1/2

0 1 2 3 4 5

kc / ωΤ

Dispersion Relation - KBr (ωT = 2.2 x10 13 rad/s)

0.00

1.00

2.00

3.00

ω /ωT

.Figure 4.

The lower branch does not appear to be photon­like (maybe phonon­like?). In contrast the dispersion relation for theupper branch approaches ω = kc/

√εopt for large k, exhibiting the standard photon­like behavior in the range of optical

frequencies. This behavior of the dispersion relation for the electromagnetic wave implies that ωT is a frequency for atransverse vibrational mode for the crystal. A wave entering the crystal with a frequency near the resonant frequency of thetransverse optical mode of the crystal would be strongly coupled to the crystal mode. The result would be a propagatingwave with a mixture of the characteristics of the mechanical and electromagnetic waves. The gap in the range of frequenciesof propagating waves is due to this coupling. The transverse optical vibration at ωT depresses the frequency of theelectromagnetic wave. The long wavelength electromagnetic wave oscillating at the frequency of the longitudinal optical

36

vibration would provide the electric field that is generated in the longitudinal mode. The ωL therefore will provide the otherend of the frequency band for the ‘stongly coupled’ mechanical­electromagnetic mode.

Finally, we note that the frequency difference between the longitudinal and transverse optical modes, ωL − ωT , can beattributed to the electric field which is generated by the longitudinal mode. Even in the long wavelength region this providesan extra restoring force for the displaced ions thereby making the longitudinal mode frequency greater than that of thetransverse mode.

2.2.3 Index of refraction for an ionic crystal Generally, when dealing with the propagation of anelectromagnetic wave in a medium, the frequency of the wave is determined by the boundary conditions and sources.

0 0.5 1 1.5 20

4

8

12

real indeximaginary indexn = 1

Index of Refraction

Normalized frequency

.Figure 5.

The relation between the wave vector and the angular frequency is given by k = kr + iki with

kr (ω) = n (ω)ω

c= Re k (2.129)

ki (ω) = κ (ω)ω

c= Im k (2.130)

From Eq. 2.127 we find

k2 = [Re ε (ω) + i Im ε (ω)]ω2

c2(2.131)

= [n (ω) + iκ (ω)]2ω2

c2

= [n2 (ω)− κ2 (ω) + 2in (ω)κ (ω)]ω2

c2

n2 (ω)− κ2 (ω) = Re ε (ω) (2.132)

2n (ω)κ (ω) = Im ε (ω)

37

and also

n2 (ω) + κ2 (ω) = |ε (ω)|

n2 (ω) =1

2[Re (ε (ω)) + |ε (ω)|] , n (ω) > 0 (2.133)

κ2 (ω) =1

2[−Re (ε (ω)) + |ε (ω)|] , sgn [κ (ω)] = sgn [Im ε (ω)] (2.134)

For an ionic crystal

ε (ω) = εopt + ω2Tεopt − εs

ω2 + iγω − ω2T· ω

2 − iγω − ω2Tω2 − iγω − ω2T

= εopt + ω2Tεopt − εs

[ω2 − ω2T ]2 + γ2ω2

[ω2 − ω2T − iγω]

Re ε (ω) = εopt + ω2Tεopt − εs

[ω2 − ω2T ]2 + γ2ω2

[ω2 − ω2T ] (2.135)

= εopt + ω2T

εopt − εoptω2Lω2T

[ω2 − ω2T ]2 + γ2ω2

[ω2 − ω2T ]

= εopt[ω2 − ω2T ]

2 + γ2ω2 + (ω2T − ω2L)[ω2 − ω2T ]

[ω2 − ω2T ]2 + γ2ω2

= εopt[ω2 − ω2T ][ω

2 − ω2T + ω2T − ω2L] + γ2ω2

[ω2 − ω2T ]2 + γ2ω2

= εopt[ω2 − ω2T ][ω

2 − ω2L] + γ2ω2

[ω2 − ω2T ]2 + γ2ω2

Im ε (ω) = −ω4T(εopt − εs) γω

(ω2 − ω2T )2+ γ2ω2

= −ω2Tεopt

¡ω2T − ω2L

¢γω

(ω2 − ω2T )2+ γ2ω2

(2.136)

|ε (ω)| = εopt

¡ω2 − ω2L

¢2+ γ2ω2

(ω2 − ω2T )2+ γ2ω2

(2.137)

The real and imaginary parts of the index of refraction for KBr are shown in Fig. 5 .The electric field for a plane electromagnetic wave travelling in the +z direction is given by

E (z, t) =

Z ∞−∞

A (ω) exp [ikz − ωt] dω

=

Z ∞−∞

A (ω) exphiω

c(n (ω) + iκ (ω))z − ωt

idω

E (z, t) =

Z ∞−∞

A (ω) exphiω

c(n (ω) z − ct)

iexp

h−ωcκ (ω) z

idω, A3 (ω) = 0 (2.138)

In general εs > εopt and, therefore, ωκ (ω) ≥ 0 for all ω. It follows that Eq. 2.138 describes a damped plane wave. Theproperties of this wave will be analyzed in a later section.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

38

Section 2.3 Electronic plasmas

Assignment 6a: Jackson Problem 7.19 ­An approximately monochromatic plane wave packet in one dimension has the instantaneous form, u (x, 0) =

f (x) exp(ik0x) with f (x) the modulation envelope.

(1) f (x) = N1 exp

µ−α |x|

2

¶(2) f (x) = N2 exp

µ−α

2x2

4

¶(3) f (x) = N3 (1− α |x|) Θ ¡1− α2x2

¢(4) f (x) = N4Θ

¡1− α2x2

¢(a) Calculate the wave number spectrum |A (k)|2 for each of the above forms for f (x).(b) Explicitly evaluate the rms deviations from the mean,∆x and∆k.(c) Using graphs compare the functions |f (x)|2 and the functions |A (k)|. Include two graphs one comparing the functions f (x) and the other comparing the functions |A (k)|2 .

Assignment 6b: Jackson Problem 7.20 ­ A homogeneous, isotropic, nonpermeable dielectric is characterized by acomplex index of refraction n (ω) . (a) Show that the general solution for plane waves in one dimension can be written

u (x, t) =1√2π

Z ∞−∞

e−iωthA (ω) eiω n(ω)x/c +B (ω) e−iω n(ω)x/c

idω

where u (x, t) is a component of E or B. (b) If u (x, t) is real show that n (−ω) = n (ω)∗ . (c) Show that, if u (0, t) and[∂u (x, t) ∂x]x=0 are tha boundary values of u (x, t) at x = 0, the coefficients A (ω) and B (ω) areµ

A (ω)

B (ω)

¶=1

2

1√2π

Z ∞−∞

eiωt·u (0, t)∓ ic

ωn (ω)

µ∂u (x, t)

∂x

¶x=0

¸dt

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.3 Electronic plasmas A plasma is an electrically charged gas consisting of ions and electrons. It is similar to aliquid conductor, but is generally distinguished from a liquid conductor by density. A liquid conductor has a number densityof the order of 1022cm−3whereas a plasma has a number density of the order of 1018cm−3. A plasma can be thought of asa gaseous conductor. [ See S. Gartenhaus, Elements of Plasma Physics, Holt, Rinehart and Winston, New York 1964 for anintroduction to plasmas.]

We consider here a system composed of ‘free electrons’ and a uniform background of positive charge to provide chargeneutrality for the system. In most cases the radiation emitted (lost) by the accelerated ions in the plasma is negligible, asare quantum effects (which are relevant only at high densities and low temperatures.) In a gas of density 1018cm−3 theaverage separation distance is ≈ [10−18]1/3 =10−6cm = 100 Å. The ”macroscopic” treatment of electric and magneticfields in a plasma assumes that properties are averaged over distances ≥ 100 Å. The electrons in the gas can also undergoclose collisions with the ions and with other electrons. The collisions between electrons thermalize the electron ‘gas’ whilethe collisions with ions will reduce the momentum of the electron gas.

Our analysis will use the linearized Boltzmann equation for an electron density in phase space, ρ (r,v, t) (with unitscm−3(cm/ sec)−3). The equilibrium electron density will be velocity dependent but not spatially dependent, i.e., ρ0 (v)with

Rρ0 (v) d

3v = ρ0 andRvρ0 (v) d

3v = 0. (The ρ0 has units cm−3.) The linearized Boltzmann equation assumesthat the total time derivative of ρ (r,v,t) taken along a particle trajectory is proportional to ρ (r,v,t)− ρ0(v), the deviationof the electron density from equilibrium. The proportionality constant is −1/τ and is due to ‘collisions’ not included in thedynamical equations for the system. In the model considered here the excluded interactions are close collisions betweenelectrons and collisions with the ions.

39

Section 2.3 Electronic plasmas

As an example of a collision term we consider the electron­electron collisions3. The collision cross­section is a functionof the relative speed of the particles and the scattering angle. We note that the center of mass motion of the particles(conservation of momentum) and their relative speed (conservation of energy) are not changed by the collision. The collisionprocess transfers particles between the velocity pairs v, v0 and u, u0 and can be classified by v0 and, taking v− v0 asthe z axis, the angle (θ, φ) for u− u0.·

dρ (r,v,t)

dt

¸collision

=

ZZσ (|v− v0| ; θ, φ) |v− v0| ρ (r,u,t) ρ (r,u0, t) d3v0dΩ

−ZZ

σ (|v− v0| ; θ, φ) |v − v0| ρ (r,v,t) ρ (r,v0, t) d3v0dΩ (2.139)

If ρ(r,v,t) is the Boltzmann distribution ρ0 (r,v,t) , the collision term is zero. In the case that

δρ (r,v,t) = ρ (r,v,t)− ρ0(r,v,t)

is ‘small’ then the collision term can be approximated by a linear function of δρ (r,v,t) (essentially using the principle ofdetailed balance) to obtain·

dρ (r,v,t)

dt

¸collision

=

ZZσ (|v− v0| ; θ, φ) |v − v0| [δρ (r,u,t) ρ0 (r,u0, t) + ρ0 (r,u,t) δρ (r,u

0, t)] d3v0dΩ

−ZZ

σ (|v− v0| ; θ, φ) |v − v0| [δρ (r,v,t) ρ0 (r,v0, t) + ρ0 (r,v,t) δρ (r,v0, t)] d3v0dΩ (2.140)

∼ −δρ (r,v,t)τ

(2.141)

This will provide a reasonable, semi­empirical expression for the relaxation of the distribution to equilibrium. The Boltzmannequation itself is a hybrid combination of dynamics and probability.

The total time derivative along a trajectory will involve the time dependence of r and v on the trajectory.

d

dtρ (r,v,t) =

∂

∂tρ (r,v,t) + v ·∇rρ (r,v,t) + a ·∇vρ (r,v,t) = −ρ (r,v,t)− ρ0 (v)

τ. (2.142)

The acceleration of the electrons is due to the electric field generated by the spatial variation in the electron density and byany externally generated electric and magnetic fields. In particular the latter will be the transverse electric and magneticfields of an electromagnetic wave. The acceleration of an electron located at r and moving with velocity v at time t is

a = − e

m

hE (r, t)+

v

c×B (r, t)

i(2.143)

The fields satisfy Maxwell’s equations:

∇ ·E (r, t) = ε−1i (−4πe)Z[ρ (r,v,t)− ρ0 (v)] d

3v, (2.144)

∇×E (r, t) = −1c

∂

∂tB (r, t) , (2.145)

∇×B (r, t) =εic

∂

∂tE (r, t)− 4πe

c

Zvρ (r,v,t) d3v, (2.146)

∇ ·B (r, t) = 0. (2.147)

3 Kerson Huang, Statistic Mechanics (Problem of Kinetic Theory, Chapt. 3, Wiley, 1963) See also Section 5.4The quantum mechanical system is covered by Pines, The Many Body Problem (Benjamin, 1961) An interesting approach is provided by an article by

Ehrenreich and Cohen (Phys, Rev., 115, 786­790 (1959)) which is reprinted on pp 255­259.

40

Section 2.3 Electronic plasmas

We have assumed that the background charges have a constant permittivity εi.This set of equations is non­linear in the electron density function. We will restrict ourselves to systems in which the

electron charge density only deviates by a small amount from its equilibrium value. This permits us to linearize the equationsby defining

δρ (r,v,t) = ρ (r,v,t)− ρ0 (v) (2.148)

and keeping only those terms involving the δρ to first order or lower. In this approximation the transport equation becomes

∂

∂tδρ (r,v,t) + v ·∇rδρ (r,v,t) + a ·∇vρ0 (v) = −

δρ (r,v,t)

τ(2.149)

The modified Maxwell’s equations are

∇ ·E (r, t) = 4πε−1i (−e)Z

δρ (r,v,t) d3v, (2.150)

∇×B (r, t) =εic

∂

∂tE (r, t)− 4πe

c

Zv δρ (r,v,t) d3v. (2.151)

We will assume that an electric field exists in the plasma and that this electric field is the real part of

E (r, t) = E0 exp [i (k · r− ωt)] (2.152)

Neglecting the magnetic force in our transport equation we find the δρ must satisfy

∂

∂tδρ (r,v,t) + v ·∇rδρ (r,v,t)− e

mE0 exp [i (k · r− ωt)] ·∇vρ0 (v) = −

δρ (r,v,t)

τ(2.153)

We look for a electron density variation of the form

δρ (r,v,t) = ∆ (v) exp [i (k · r− ωt)] (2.154)

Using Eq. 2.153 we find that

∆ (v) =eτ

m[−iωτ + iv · kτ + 1]−1 E0 ·∇vρ0 (v) (2.155)

The spatial electron density is obtained by integrating δρ (r,v,t) over the velocity

n (r, t) = exp (ik · r− iωt)

ZZ Zeτ

m

E0 ·∇vρ0 (v)

−iωτ + iv · kτ + 1d3v (2.156)

In many cases ρ0 (v) is only a function of the square of the electron’s velocity. For simplicity we will assume this propertyfor ρ0 (v) . Then

∇vρ0 (v) =v

v

∂

∂vρ0 (v) (2.157)

and∂

∂vρ0 (v) has no angular dependence. We can also let the z for the v integration be along k, so that k · v = kv cos θ and

all the angular dependence in the integrand arises from v and k · v. We can also write

n (r, t) = i exp (ik · r− iωt)

ZZZeτ

mv

[(E0 · k)k+(E0 · x)x+ (E0 · y)y] · vωτ − v · kτ + i

dρ0 (v)

dvd3v (2.158)

= i exp (ik · r− iωt)

ZZZeτ

mv

[(E0 · k)k · vωτ − v · kτ + i

dρ0 (v)

dvd3v

+i exp (ik · r− iωt) ·ZZZeτ

mv

[(E0 · x) sin θ cosφ+ (E0 · y) sin θ sinφ]ωτ − vk cos θτ + i

dρ0 (v)

dvv2dvd(− cos θ)dφ

41

Section 2.3 Electronic plasmas

The last term on the right is zero since both integrations over φ vanish. We are then left with

n (r, t) = iE0 · kk2

exp (ik · r− iωt)

ZZZeτ

mv

k · vωτ − v · kτ + i

dρ0 (v)

dvd3v (2.159)

It suffices for our analysis to carry out an expansion of the integrand in powers of¡v · k / £ω + iτ−1

¤¢

n (r, t) = iE0 · kk2

ei(k·r−ωt)ZZZ

e

mv

k · vω + iτ−1

·1 +

k · vω + iτ−1

+

¸dρ0 (v)

dvd3v (2.160)

= iE0 · kk2

ei(k·r−ωt)Σn=1,∞ZZZ

e

mv

·k · v

ω + iτ−1

¸ndρ0 (v)

dvd3v

= iE0 · kk2

ei(k·r−ωt)Σn=1,∞Z ∞0

2πe

m

·kv

ω + iτ−1

¸ndρ0 (v)

dvvdv

Z 1

−1cosn θd(cos θ)

= iE0 · kk2

ei(k·r−ωt)Σj=2,4,6,...Z ∞0

2πe

m

µkv

ω + iτ−1

¶j2

j + 1

dρ0 (v)

dvvdv

= iE0 · kk2

ei(k·r−ωt)Z ∞0

4πe

3m

µkv

ω + iτ−1

¶2 "1 +

3

5

µkv

ω + iτ−1

¶2+ ...

#dρ0 (v)

dvvdv

The leading term in this expansion can be written as

iE0 · kei(k·r−ωt) 4πe

3m(ω + iτ−1)2

Z ∞0

dρ0 (v)

dvv3dv (2.161)

= iE0 · kei(k·r−ωt) 4πe

3m(ω + iτ−1)2

·v3ρ0 (v) |∞0 − 3

Z ∞0

ρ0 (v) v2dv

¸= −iE0 · kei(k·r−ωt) e

m(ω + iτ−1)2

ZZZρ0 (v) d

3v

Using this first term as an approximation for n (r, t), one obtains

n (r, t) = −i eρ0E0 · km (ω + iτ−1)2

ei(k·r−ωt) (2.162)

independent of the form taken for the equilibrium electron density. We should note that n (r, t) = 0 if E0 · k =0.

42

Section 2.3 Electronic plasmas

Electron current density in plasma

In a similar way we can obtain the electron current density.

je (r, t) = exp (ik · r− iωt)

Zeτ

mv

E0 ·∇vρ0 (v)

−iωτ + iv · kτ + 1d3v (2.163)

We will restrict the analysis to the case in which E0 is perpendicular to k. To further simplify we will only seek the lowestorder, non­zero term in

¡v · k / £ω + iτ−1

¤¢. With these simplifications and letting E0 be along z and

je (r, t) = i exp (ik · r− iωt)

ZZZeτ

mvv

(E0 · z)z · vωτ − v · kτ + i

dρ0 (v)

dvd3v (2.164)

= i(E0 · z) exp (ik · r− iωt)

ZZZe

mvv

v cos θ

ω + iτ−1X

n=0,∞

·k · v

ω + iτ−1

¸ndρ0 (v)

dvd3v

≈ i(E0 · z) exp (ik · r− iωt)

ZZZe

mvv

v cos θ

ω + iτ−1

"1 +

k · vω + iτ−1

+ .

·k · v

ω + iτ−1

¸2.

#dρ0 (v)

dvd3v +

= i(E0 · z) exp (ik · r− iωt) [

ZZZe

mvz(v cos θ)2

ω + iτ−1

"1 +

k · vω + iτ−1

+ .

·k · v

ω + iτ−1

¸2+ ..

#dρ0 (v)

dvd3v +

ZZZeτ

mv[vxx+ vyy]

v cos θ

ω + iτ−1

"1 +

kxvx + kyvyω + iτ−1

+ .

·k · v

ω + iτ−1

¸2+ ..

#dρ0 (v)

dvd3v]

Consider the following integrations. Note that k · z = 0 so that k · v =kxvx + kyvy:

Z 2π

0

Z π

0

cosn0θvx

ivjy sin θdθdφ = v2Z 2π

0

Z π

0

cosn θ sini+j θ cosi φ sinj φ sin θdθdφ (2.165)

= v2Z 1

−1cosn

0θ sini+j θd(cos θ)

Z 2π

0

cosi φ sinj φdφ; u = cos θ

= v2Z 1

−1un

0[1− u2]

i+ j

2 du

Z 2π

0

cosi φ sinj φdφ; (2.166)

= 0 if n0 is odd (This applies to all terms along x and y)

= v2Z cosπ

cos 0

cos2 θ sini+j θd(− cos θ)Z 2π

0

cosi φ sinj φdφ if n=2 ( z terms)(2.167)

So all the terms with along x and y vanish. The terms along z vanish if i = j = 1 because

Z 2π

0

cosφ sinφdφ = 0 (2.168)

Some of the terms for i = j = 2 do not vanish. A non zero term comes from the third term in the expansion.

43

Section 2.3 Electronic plasmas

Thus, keeping the two lowest order terms,

je (r, t) ≈ i(E0 · z)zeik·r−iωt[2πZ

e

m

v

ω + iτ−1dρ0 (v)

dvv2dv

Zcos2 θd(− cos θ) +

i(E0 · z)zeik·r−iωt[Z Z

e

m

v cos2 θ

ω + iτ−1

·kxvx + kyvyω + iτ−1

¸2dρ0 (v)

dvv2dvdφd(− cos θ)

= iE0eik·r−iωt[2π

Z ∞0

e

m

v

ω + iτ−1dρ0 (v)

dvv2dv

2

3+

iE0eik·r−iωt

Z ∞0

e

m

v3

(ω + iτ−1)3dρ0 (v)

dvdv

Z π

0

Z 2π

0

cos2 θ[kv sin θ]2dφd(− cos θ) (2.169)

= iE0eik·r−iωt e

3m

4π

ω + iτ−1[v3ρ0 (v) |∞0 − 3

Z ∞0

ρ0 (v) v2dv] +

iE0 exp (ik · r− iωt)

Z ∞0

e

m

k2v5

(ω + iτ−1)3dρ0 (v)

dvdvπ

Z 1

−1cos2 θ[1− cos2 θ]d(cos θ)

= −iE0 exp (ik · r− iωt)e

m

1

ω + iτ−1

ZZZρ0 (v) d

3v +

iE0 exp (ik · r− iωt)π

Z ∞0

e

2m

k2v5

(ω + iτ−1)3dρ0 (v)

dvdv[2

3− 25]

= −iE0 exp (ik · r− iωt)e

m

1

ω + iτ−1

ZZZρ0 (v) d

3v +

iE0 exp (ik · r− iωt)π

Z ∞0

e

2m

k2v5

(ω + iτ−1)3dρ0 (v)

dvdv4

15

= −iE0 exp (ik · r− iωt)e

m

·1

ω + iτ−1ρ0 −

1

15

ZZZk2v3

(ω + iτ−1)3dρ0 (v)

dvd3v

¸≈ −i e

mE0 exp (ik · r− iωt)

ρ0ω + iτ−1

keeping only the first term (2.170)

If τ is very large (the relaxation to equilibrium takes a long time compared to the period of the field) then the electroncurrent density will lead the electric field in time by 90. That is, using −i = e−iπ/2

je (r, t) ≈ e−iπ/2e

mE0 exp (ik · r− iωt)

ρ0ω + iτ−1

(2.171)

=e

mE0

ρ0ω + iτ−1

e−iπ/2+ik·r−iωt

=e

mE0

ρ0ω[1− i

2π

τfieldτ

+ ...]e−iπ/2+ik·r−iωt

≈ =e

mE0

ρ0ωei[k·r−ωt−π/2] =

e

mEρ0ωe−iπ/2

This means that the electric field has the same phase as the current density at a later time t0 = t+π

2ω. Physically the current

density lags the force by 90 as expected.

F = −eE =eEeiπ (2.172)

je (r, t) = −F ρ0mω

e−iπ/2

= Fρ0mω

eiπ−iπ/2

= Fρ0mω

eiπ/2

44

Section 2.3 Electronic plasmas

That is, the force has the same phase as the current density at an earlier time, t0 = t− π

2ω. Finally, in the limit that k and ω

go to zero we obtain the equation for the dc current

jc = −eje = e2ρ0τ

mE0 (2.173)

This gives the standard equation for electrical conductivity

σ0 =e2ρ0τ

m. (2.174)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Assignment 7: For a metal the electron density would be approximated by the ‘fermi sea’,

ρ0 (v) = 2

µm∗

h

¶3Θ (vf − v) , vf =

¡3π2ρ0

¢1/3 m∗

(2.175)

with m∗ the effective mass of the electron. The equilibrium electron density for a charged plasma at temperature T is givenby the Boltzmann distribution

ρ0 (v) = ρ0

µm

2πkBT

¶3/2exp

µ− mv2

2kBT

¶(2.176)

Using these equilibrium densities evaluate the correction terms for the electron density, n (r, t) , and electron currentdensity, j (r, t) .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

45

Section 2.3 Electronic plasmas

2.3.1 Longitudinal plasma oscillation The electron density given by Eq. 2.162 will create a charge densitywhich generates an electric field. We seek the condition for which the electric field generated by this charge density equalsthe electric field which creates the charge density variation. In this case the field and charge density are related by Gauss’slaw (εi is the permittivity of the background ions)

∇ ·E(r,t) = −4πρ (r, t) /εiik ·E0ei(k·r−ωt) = 4πen (r, t) /εi

= 4πie2ρ0E0 · k

mεi (ω + iτ−1)2ei(k·r−ωt) (2.177)

This equation can be satisfied for E0 · k 6= 0 only if

4πe2ρ0

mεi (ω + iτ−1)2= 1 (2.178)

A plasma frequency is defined by

ω2p =4πe2ρ0mεi

(2.179)

and the undamped plasma will oscillate at this frequency. The plasma frequency is approximately¡6× 104rad · cm3/2/s

¢ ·ρ1/20 and ranges from 109rad/s for gaseous plasmas, through 1013rad/s for doped semiconductors to 1016rad/s for metals.

For reference, the ‘observed’ relaxation time constant for currents in a metal is of order 10−14s at room temperature. Thiswould be the characteristic time between electron­phonon collisions.

2.3.2 Transverse electromagnetic waves in a plasma A transverse electromagnetic wave wouldgenerate a current density in the plasma but not a charge density. The (transverse) electric field in the plasma,E(r,t) = E0 exp (ik · r− iωt) must satisfy

∇× (∇×E) = ∇× [−∂c∂tB] (2.180)

−∇2E+∇(∇ ·E) =−∂c∂t(∇×B) (2.181)

=−∂c∂t(∂D

c∂t− 4π

cJo) (2.182)

Note that for a transverse wave∇ ·E = ik ·E = 0 and (using the first order term for Jo = eje)

−∇2E =−1c2

∂2

∂t2D+

4π

c

∂

c∂t[−ie

2

mE0 exp (ik · r− iωt)

ρ0ω + iτ−1

] (2.183)

k2E0 = εiω2

c2E0 − 4πω

c2e2

mE0

ρ0ω + iτ−1

(2.184)

Hence, the dispersion relation (see Eq.127) emerges from:

(ck)2

ω2E0 = [εi − 4π

ω

e2

m

ρ0ω + iτ−1

]E0 (2.185)

= ε(ω)E0 (2.186)

46

Section 2.3 Electronic plasmas

We can identify the permittivity for the system as

ε (ω) = εi − 4πe2ρ0mω (ω + iτ−1)

. (2.187)

Using ω2p =4πe2ρ0mεi

from Eq.2.179

ε (ω) = εi

"1− ω2p

ω (ω + iτ−1)

#= εi

"1− ω2pτ

2 (ωτ − i)

ωτ (ω2τ2 + 1)

#(2.188)

= εi − 4πσ0ω (ωτ + i)

= εi − 4πσ0 (ωτ − i)

ω (ω2τ2 + 1)(2.189)

= εi − 4πσ0τ

(ω2τ2 + 1)+ i

4πσ0ω (ω2τ2 + 1)

(2.190)

= εi

"1− ω2pτ

2

(ω2τ2 + 1)

#+ i

εiω2pτ2

ωτ (ω2τ2 + 1)(2.191)

showing the explicit relationships between the permittivity, the plasma frequency, and the conductivity. Figure 6 shows

the real and imaginary parts of the permittivity for a metal with ωpτ = 10. We note that the real part of the permittivity isnegative below the plasma frequency and monotonically approaches permittivity of the background ions above the plasmafrequency.

real ε

imaginary ε

ωτ = 10

-3.00

-2.00

1.00

-1.00

0.00

ε(ω)/ε i

5.0 10.0 15.0 20.0 25.0 30.0

ωτ

Permittivity of a metal

.Figure 6. Real and imaginary parts of the electrical permittivity for a metal with ωpτ = 10.

Equations 2.133 and 2.134 can be used to evaluate the real and imaginary parts of the index of refraction. The real andimaginary parts of the index of refraction are shown plotted versus ωτ in Fig. 7 . In this plot the product, ωpτ , of the plasmafrequency and the relaxation time, τ , has been taken to be 10 and the permittivity of the background ions has been taken tobe 1

47

Section 2.4 Propagation of plane waves in a medium

0 5 10 15 200

2

4

6

8

real indeximaginary indexn = 1

Index of Refraction, metallic

.Figure 7

We note that in the region near zero frequency the imaginary part of the permittivity is very large (approaching infinityat zero frequency). In this region the real and imaginary parts of the index of refraction are nearly equal. Indeed the realpart of the index of refraction lies below the imaginary part for all frequencies below the plasma frequency. Note that theenergy dissipated is proportional to the imaginary part of the permittivity (see Eq. 111 in Chapter 1). It will therefore notbe meaningful to speak of a wave propagating in the medium at frequencies below the plasma frequency.

2.4 Propagation of plane waves in a medium We now have two models for the index of refraction. Ournext task is to investigate the propagation of a plane electromagnetic wave in an ionic crystal and in a metal. Reviewing thepermittivities obtained for the models we note that the real part of the permittivity is an even function of ω and the imaginarypart is an odd function of ω. The properties of the real and imaginary parts of the indices of refraction are obtained fromthose of the permittivities and Eqs. 2.133 and 2.134. Like the permittivity we find that the real part of the index of refractionis an even function of ω and the imaginary part is an odd function of ω. Using these properties a plane polarized wavetravelling in the +z direction can be described by the expression

E (z, t) =

Z ∞0

E0 (ω) exp(i[Re k(ω) + i Im k(ω)]z − ωt) dω + c.c.

=

Z ∞0

E0 (ω) exp(iω

c(n (ω) z − ct) exp(−ω

cκ (ω) z) dω + c.c. (2.192)

We will restrict our analysis to the case in which |E0 (ω)| has a single, ‘narrow’ peak of width ∆ω at ω0. The peak will beconsidered narrow if

∆ω

n (ω0)

dn (ω0)

dω0¿ 1.

2.4.1 Group velocity For a wave composed of a narrow range of frequencies the wave vector k (ω) can beexpanded about the central frequency to obtain an approximate expression for the propagation of the wave. In our example

48

Section 2.4 Propagation of plane waves in a medium

the central frequency is ω0.

ω

cn (ω) =

ω0cn (ω0) +

·n (ω0) + ω0

dn (ω0)

dω0

¸ω − ω0

c+1

2

·2dn (ω0)

dω0+ ω0

d2n (ω0)

dω20

¸(ω − ω0)

2

c+ ... (2.193)

A similar expression holds for ωκ (ω) /c. Working to first order in ω − ω0 we find

E (z, t) = exp(iω0c(n (ω0) z − ct) exp(−ω0

cκ (ω0) z)

×Z ∞0

E0 (ω) exp

µiω − ω0

c

½·n (ω0) + ω0

dn (ω0)

dω0

¸z − ct

¾¶dω + c.c. (2.194)

where we have neglected the variation in the imaginary part of the index of refraction. In this approximation the wave travelswith the group velocity

vg (ω0) = c

·n (ω0) + ω0

dn (ω0)

dω0

¸−1(2.195)

i.e.

E (z, t) =

Z ∞0

E0 (ω) exp

µiω − ω0vg (ω0)

z − vg (ω0) t¶dω

· exp(iω0c(n (ω0) z − ct) exp(−ω0

cκ (ω0) z) + c.c.

= F(z − vg (ω0) t) exp(iω0c(n (ω0) z − ct) exp(−ω0

cκ (ω0) z) + c.c. (2.196)

This should describe a ‘broad’ wave packet with the envelope function travelling with speed vg (ω0) and the underlyingwave travelling with speed c/n (ω0) . Figure 8 shows the group velocity for the index of refraction of an ionic crystal shownin Fig. 5 . The normalized frequency is ω/ωT with the approximation (for the peaked E0 (ω)) that ω ≈ ω0. The groupvelocity approaches 0 at ω/ωT = 1 where n (ωT ) diverges.

0 0.25 0.5 0.75 10

0.25

0.5Group velocity below resonance

normalized frequency

v/c

.Figure 8. Group velocity for an ionic crystal below ωT .

49

Section 2.4 Propagation of plane waves in a medium

Fig. 9 shows the group velocity, above the resonance, in a region where n (ω ≈ ω0) is small and the phase velocity,c/n (ω0) is greater than the speed of light. The normalized frequency is ω/ωT .

1.4 1.55 1.7 1.85 20

0.25

0.5Group 'velocity' above resonance

Normalized frequency

v/c

.Figure 9. Group velocity of an ionic crystal above ωT .

The plasma presents a different aspect of the wave propagation problem. The group velocity can exceed the speed oflight in the region where the index of refraction (see Fig. 7) is small and slowly varying. However, in this region there isno wave propagation as illustrated by Fig.10 . This shows the phase change for the wave as the intensity falls by 1/e. Thenormalized frequency is ωτ , where ωpτ = 10.

0 2 4 6 8 10 120

1

2

3

4Phase change in one damping length

normalized frequency

Phas

e ch

ange

, rad

ians

.Figure 10. Phase change in plasma

50

Section 2.4 Propagation of plane waves in a medium

For reference the decay length, in units of cτ , for the transverse wave in a plasma is shown in Fig. 11 . For metallicplasmas ωp ≈ 1016rad/s and the figure would correspond to a system with a damping constant of τ ≈ 10−15s.In this casecτ ≈ 100nm.

0 5 10 15 200

2

4

6

8

Normalized frequency

Nor

mal

ized

dam

ping

leng

th

.Figure 11. Plasma decay length / cτ .

2.4.2 Causality Our concept of causality sequences events using the speed of light as the limiting factor. Let the timeand location of two events be (r, t) and (r0, t0) . The event at (r0, t0) is in the past of the event at (r, t) if |r− r0| < c (t− t0) .This means that t is so much larger than t0that the signal (emitted at (r0, t0)) has long since reached the observation point r.If t is not quite large enough (but t > t0) the event is in the present of (r, t) and |r− r0| > c |t− t0|. That is, the signal hasnot yet reached the observation point, r. The event is in the future of (r, t) if t0 > t. and |r− r0| < c (t0 − t). Thismeans that the signal has not yet been generated.. Causality requires that a given event can only be inuenced by eventsoccurring in its past. Therefore no signal may propagate faster than the speed of light. In particular, a wave in a dispersivemedium will propagate with a speed which does not exceed the vacuum speed of light. This places constraints on the indexof refraction and the electric permittivity. Let u (x, t) be a wave propagating in a medium with a permittivity ε (ω) and letn (ω) = ε (ω)

1/2. For example

u (x, t) =

Ze−iωt

·A (ω) exp

µiωn (ω)x

c

¶+B (ω) exp

µ−iωn (ω)x

c

¶¸dω (2.197)

Let u (0, t) and [∂u (x, t) /∂x]x=0 be given. Then

u (0, t) =

Z ∞−∞

e−iωt[A (ω) +B (ω)]dω and (2.198)Z ∞−∞

eiω0tu (0, t) dt =

Z ∞−∞

Z ∞−∞

e−i(ω−ω0)t[A (ω) +B (ω)]dωdt

= 2π

Z ∞−∞[A (ω) +B (ω)]δ(ω − ω0)dω

1

2π

Z ∞−∞

eiω0tu (0, t) dt = [A (ω0) +B (ω0)]

51

Section 2.4 Propagation of plane waves in a medium

and similarly

∂

∂x0u (x, t) |x0=o =

Z ∞−∞

e−iωt[iωn (ω)

cA (ω)− i

ωn (ω)

cB (ω)]dω (2.199)

Z ∞−∞

eiω0t ∂

∂x0u (x, t) |x0=odt =

Z ∞−∞

Z ∞−∞

iωn (ω)

ce−i(ω−ω

0)t[A (ω)−B (ω)]dωdt

= 2π

Z ∞−∞[A (ω)−B (ω)]i

ωn (ω)

cδ(ω − ω0)dω and

−i c

ω0n (ω0) 2π

Z ∞−∞

eiω0t ∂

∂x0u (x, t) |x0=odt = [A (ω0)−B (ω0)]

Finally,

A (ω0) =1

2 · 2πZ ∞−∞

eiω0tu (0, t)− i

c

ω0n (ω0)∂

∂x0u (x, t) |x0=odt (2.200)

B (ω0) =1

2 · 2πZ ∞−∞

eiω0tu (0, t) + i

c

ω0n (ω0)∂

∂x0u (x, t) |x0=odt (2.201)

In the integrals for A (ω) and B (ω) let ω0 be replaced by ω. And let t be replaced by t0 to distinguish it from the value oftime on the left hand side in the following expression. :

u (x, t) =1

4π

Z ∞−∞

Z ∞−∞

e−iω(t−t0) exp

µiωn (ω)x

c

¶½u (0, t0)− ic

ωn (ω)[∂u (x0, t0) /∂x0]x0=0

¾dωdt0

+1

4π

Z ∞−∞

Z ∞−∞

e−iω(t−t0) exp

µ−iωn (ω)x

c

¶½u (0, t0) +

ic

ωn (ω)[∂u (x0, t0) /∂x0]x0=0

¾dωdt0(2.202)

Next, the integration over ω can be converted into a contour integration in the complex ω plane.

u (x, t) =1

4π

Z ZC

e−iω(t−t0) exp

µiωn (ω)x

c

¶½u (0, t0)− ic

ωn (ω)[∂u (x0, t0) /∂x0]x0=0

¾dωdt0

+1

4π

Z ZC

e−iω(t−t0) exp

µ−iωn (ω)x

c

¶½u (0, t0) +

ic

ωn (ω)[∂u (x0, t0) /∂x0]x0=0

¾dωdt0 (2.203)

with C denoting the contour which passes above any poles or cuts on the real ω axis. To observe the constraints imposed bycausality it suffices to consider the following expressions for x > 0,

g (x, t− t0) = − 1

4π

ZC

e−iω(t−t0) exp

µiωn (ω)x

c

¶ic

ωn (ω)dω (2.204)

and

∂

∂xg (x, t− t0) = f (x, t− t0) =

1

4π

ZC

e−iω(t−t0) exp

µiωn (ω)x

c

¶dω (2.205)

A general property of ε (ω) and n (ω) is that for large |ω| they approach one. To obtain well behaved integrands forlarge |ω| we will subtract off this asymptotic dependence. But first we consider the functions with n (ω) = 1,call themg0 (x, t− t0) and f0 (x, t− t0) . The following are evaluated using a contour in the lower half plane and the Cauchy residue

52

Section 2.4 Propagation of plane waves in a medium

theorem. The residue is that of a simple pole at ω = 0.

g0 (x, t− t0) = − 1

4π

ZC

e−iω(t−t0) exp

³iωx

c

´ ic

ωdω = −2πi

4π[−icΘ

³t− t0 − x

c

´] (2.206)

= − c

2Θ³t− t0 − x

c

´Note: countour direction C is negative (2.207)

f0 (x, t− t0) =∂

∂xg (x, t− t0) =

∂

∂x

c

2Θ³t− t0 − x

c

´=1

2δ³t− t0 − x

c

´(2.208)

=1

4π

ZC

e−iω(t−t0) exp

³iωx

c

´dω =

1

2δ³t− t0 − x

c

´(2.209)

These expressions are consistent with the relationship between g (x, t) and f (x, t) . As expected g0 (x, t− t0) andf0 (x, t− t0) are ‘causal’ in that a disturbance at the origin at time t0 does not effect the wave at x until the time t satisfyingc (t− t0) = x. We now consider g (x, t− t0)− g0 (x, t− t0)

g (x, t− t0)− g0 (x, t− t0) = − ic

4π

ZC

e−iω(t−t0−x/c)

ω

½exp (iω [n (ω)− 1]x/c)

n (ω)− 1¾dω (2.210)

This is the fourier transform of a function which vanishes as ω → ±∞. This transform is zero for x > c (t− t0) if

γ (ω) =1

ω

½exp (iω [n (ω)− 1]x/c)

n (ω)− 1¾

(2.211)

has no singularities or cuts in the upper half complex ω plane.

The problem with a cut can be illustrated by the integral

C (t) =

Z ∞−iδ−∞−iδ

ω−1/2 exp (−iωt) dω (2.212)

.

where the path of integration is taken to be just below the real ω axis. To use Cauchy’s theorem to evaluate the integral fort < 0 we would attempt to close the path in the upper half plane. The difficulty is that ω−1/2 ‘cuts the plane’. Let ω = |ω| eiφwith −π < φ ≤ π then as ω approaches the negative real axis from below ω1/2 → −i |ω|1/2 while as ω approaches the

53

Section 2.4 Propagation of plane waves in a medium

negative real axis form above ω1/2 → +i |ω|1/2 . This discontinuity occurs because the function z2 maps the complex planewith −π < φ ≤ π into the ‘two sheets’ −2π < φ0 ≤ 2π.

To avoid the ‘second sheet’ when the path is closed, the path must not cross the cut. When the path is closed in the upperhalf plane the segments of the path are, in the limit R → ∞, (1) −R − iδ → +R − iδ, (2) z = R · eiφ from +R − iδ to−R+ iδ, (3) −R+ iδ → iδ, (4) z = δ · eiφ from +iδ to −iδ, and (5) −iδ → −R− iδ. This path is shown in Fig. . We willnow evaluate C (t) . For t > 0 C (t) = 0 and for t < 0 we obtain the integral of the discontinuity if the integrand across thecut.

C (t) = −Z 0

−∞

exp (−iωt)(−ω)1/2

(−2i) dω = 2i√−tZ ∞0

u−1/2 exp (−iu) du = (1 + i)

r2π

− t, t < 0 (2.213)

We note that a cut in the upper half plane would lead to a non­zero value for the integral in the case that the path was to beclosed in the upper half plane.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Omit: Assignment 8: The claim is that the cut in the plane defining ω1/2 can be taken along any line going from zero

to infinity in the upper half plane. (Note the physics of a problem determines whether the path of integration is taken aboveor below the cut. There’s no physics in this problem.) Check whether this claim is correct by taking ω = |ω| exp (iφ) with−3π/2 < φ ≤ π/2.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The requirement that γ (ω) have no cuts or poles in the upper half complex ω plane is satisfied if n (ω) = ε (ω)1/2 has no

poles or cuts in the upper half complex plane. In this case ε (ω) can neither have a pole nor a zero in the upper half plane.This condition is satisfied by the model permittivities for the ionic crystals and for the charged plasmas.

A similar argument applies to f (x, t− t0)− f0 (x, t− t0) . This function is the fourier transform

f (x, t− t0)− f0 (x, t− t0) =1

4π

ZC

e−iω(t−t0−x/c)

·exp

µiω [n (ω)− 1]x/c

c

¶− 1¸dω (2.214)

Again the function being transformed vanishes as ω → ±∞ and the integral will be zero if t > 0 and ε (ω) has no poles orzeros in the upper half plane.

54

Section 2.4 Propagation of plane waves in a medium

2.4.3 Wave dispersion in a medium In order to examine the dispersion character for a wave in a medium wetake u (0, t) = u0δ (t) and [∂u (x, t) /∂t]x=0 = 0. These boundary conditions will provide identical waves travelling in the+x and −x directions. Since the Fourier transform of a delta function is a constant this acts as a ‘white light’ source. FromEq. 200,

u (x, t) =1

4π

Z ZC

e−iω(t−t0)[exp

µiωn (ω)x

c

¶+ exp

µ−iωn (ω)x

c

¶]u (0, t0) dωdt0 (2.215)

u (x, t) =1

2π

Z ZC

e−iω(t−t0) cos

µωn (ω)x

c

¶u0δ (t

0) dωdt0, t > 0

=u02π

ZC

e−iωt cosµωn (ω)x

c

¶dω, t > 0 (2.216)

Since u (x, t) = u (−x, t) it suffices to consider the wave travelling in either the +x or −x direction.

u (x, t) =u04π

ZC

exp

µ−iω

·t− n (ω)x

c

¸¶dω +

u04π

ZC

exp

µ−iω

·t+

n (ω)x

c

¸¶dω t > 0 (2.217)

= u+ (x, t) + u− (x, t)

We will chose the wave travelling in the +x direction for which the transform is

u+ (x, t) =u04π

ZC

exp

µ−iω

·t− n (ω)x

c

¸¶dω, t > 0 (2.218)

=u04π

Z ∞−∞

exp

µ−iω

·t− n (ω)x

c

¸¶dω

=u04π

Z ∞−∞[cos

µ−ω

·t− n (ω)x

c

¸¶+ i sin

µ−ω

·t− n (ω)x

c

¸¶]dω

Since n (ω) = n (−ω) only the real part of the last expression survives to give

u+ (x, t) =u02πRe

Z ∞0

exp

µ−iω

·t− n (ω)x

c

¸¶dω, t > 0 (2.219)

In the following we will assume, incorrectly, that the imaginary part of the permittivity vanishes. That is,

k = [Re ε(ω) + i0]1/2ω

c= [n (ω) + ikap(ω)]

ω

c(2.220)

and kap(ω) = 0. Thus, using Eq. 2.189,

n (ω) = ε1/2i

"1− ω2pτ

2

(ω2τ2 + 1)

#1/2

= ε1/2i

"1− ω2p

ω2

#1/2since τ −→∞ (2.221)

so that

ωn (ω) = ε1/2i

£|ω2 − ω2p|¤1/2

(2.222)

55

Section 2.4 Propagation of plane waves in a medium

This will allow us to use a special case of the ‘method of steepest descent’4 which is less tedious. The method is known asthe ‘stationary phase approximation’. The approximation will be illustrated using the electric permittivity of an undampedplasma, τ →∞.

For an undamped plasma with a plasma frequency ωp we have that ωn (ω) =qω2 − ω2p for ω > ωp. For−ωp ≤ ω ≤ ωp

we obtain ωn (ω) = iqω2p − ω2. It is convenient to define a time t0 = x/c which is the time it takes for a light signal

travelling in a vacuum to go from the origin to the observation point x. We also let ω = ξωp and t = υt0. We havedetermined that this system satisfies causality so that υ ≥ 1. With these definitions the function can be written as

u+ (x, t) =u0ωp2π

Re

Z ∞1

exp

µ−iωpt0

·υξ −

qξ2 − 1

¸¶dξ

+u0ωp2π

Re

Z 1

0

exp

µ−ωpt0

·iυξ +

q1− ξ2

¸¶dξ (2.223)

Our problem is to obtain an estimate for the value of this function if ωpt0 À 1. The second integral is exponentially decayingwith distance and so will be assumed to be negligible.

The phase, φ(ω), occurring in the first integral is ωpt0f (υ, ξ) with

f (υ, ξ) = υξ −qξ2 − 1 (2.224)

Fig. 13 gives f (υ, ξ) = φ(ω)/ωpt0 for three observation times, 1.1t0, 1.2t0, and 1.3t0.

1 3 5 7 9

ω/ω p

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

φ/(ω pt 0)

t = 1.3 t 0

t = 1.2 t 0

t = 1.1 t 0

.Fig. 13

Regions in which the phase varies ‘rapidly’ will contribute little to the integral. For large ωpt0 the major contribution to

4 This is a technique for obtaining an approximation to an integral of the form

Cexp [αu (t)] dt

for large α. The saddle points of u (z) , z complex, are located and the path of integration is distorted so as to pass through these saddle points. In goingthrough a saddle point the path is along the line for which the curvature is negative. The integral is approximated by using only the neighborhoods of thesaddle points and approximating u (z) by a quadratic in the distance from the saddle point. The classic reference for this is Courant & Hilbert, Methods ofMathematical Physics, Vol. 1 (Interscience Publishers, pp526­532) Another reference is Approximation methods in quantum mechanics by A. B. Migdaland V. Krainov (pp 6­8, W. A. Benjamin, 1969)

56

Section 2.4 Propagation of plane waves in a medium

the integral will come from the region where the magnitude of the phase is minimum or has zero slope. This occurs at

∂f

∂ξ= 0 = υ − ξp

ξ2 − 1at ξ = ξo (2.225)

ξo ≡ω

ωp=

υ√υ2 − 1 . =

1r1− [ to

t]2

(2.226)

where we note that

ξ2o − 1 =1

υ2 − 1 (2.227)

The frequency for which the slope vanishes is shown in Fig. 14

The location of the zeroof the slope of the phase

1.0 1.1 1.2 1.3 1.4 1.5

t/t 0

1

10

100

ω/ω p

.Fig. 14

For t → t0 the frequency of the minimum of the magnitude of the phase approaches infinity. On the other extreme ast→∞ the frequency of the minimum approaches the plasma frequency. We deduce then that the frequency observed at theobservation point starts high and decreases with time. The range of frequencies which contribute to the observed signal alsodecreases with time. This can be obtained by considering the reciprocal of the curvature of the phase as a function of ω.

∂2φ (ω;x, t)

∂ω2=

t0ωp(ξ2 − 1)−1.5

=t0ωp

¡ν2 − 1¢3/2 at ξ = ξo

∂2φ

∂ξ2= ωpt0

¡ν2 − 1¢3/2 at ξ = ξo (2.228)

57

Section 2.4 Propagation of plane waves in a medium

Note also that

φ(ξo) = ωpt0f (υ, ξ) |ξo = ωpt0[υξo −qξ2o − 1]

= ωpt0[υυ√

υ2 − 1 −1√

υ2 − 1 ]

= ωpt0pυ2 − 1 (2.229)

1.00 1.20 1.40 1.60 1.80 2.00

t/t 0

0

1

2

3

4

5

(d2φ /dω2)(ω p/t 0)

Reciprocal phase curvature at minimum

.Fig. 15

This is shown in Fig. 15 . The approximation to u+ (x, t) is obtained by expanding the phase about the minimum,keeping only the quadratic terms5

φ(ξ) = φ(ξo) + φ0(ξo)(ξ − ξo) +1

2φ00(ξo)(ξ − ξo)

2 + ...

≈ φ(ξo) +1

2φ00(ξo)(ξ − ξo)

2

= ωpt0pυ2 − 1 + 1

2ωpt0[υ

2 − 1]3/2(ξ − ξo)2 (2.230)

Putting the above expression into the integrand we have,

u+ (x, t) ≈ u0ωp2π

Re

·Z ∞1

exp

·−i[φ(ξo) +

1

2φ00(ξo)(ξ − ξo)

2]

¸dξ

¸(2.231)

5 Integrals of the form

C (z) =z

0cos

π

2t2 dt

S (z) =z

0sin

π

2t2 dt

are Fresnel integrals. (Abramowitz and Stegun, Hanbook of Mathematical Functions, p.300, Dover) Both integrals go to 0.5 as z →∞.

58

Section 2.4 Propagation of plane waves in a medium

u+ (x, t) ≈ u0ωp2π

Re [ exp (−iφ(ξo))Z ∞1

exp

·−i12φ00(ξo) (ξ − ξ0)

2

¸dξ ]

=u0ωp2π

Re [ exp (−iφ(ξo)) [1

πφ00(ξo)]

−1/2Z ∞W (ξ=1)

exph−iπ2W 2

idW ]

=u0ωp2π

Re [ exp (−iφ(ξo)) [1

πφ00(ξo)]

−1/2Z ∞W (ξ=1)

cos(π2W 2)− i sin(

π

2W 2)dW ] (2.232)

where we used

W 2 =1

πφ00(ξo) (ξ − ξ0)

2 and (2.233)

W (ξ = 1) = [1

πωpt0]

1/2[υ2 − 1]3/4 (1− ξ0)2

= [1

πωpt0]

1/2[t2

t2o− 1]3/4 (1− ξ0)

2

≈ 0 for larget

towhere ξ0 ≈ 1 (2.234)

One can do the integrals (see the footnote) to obtain

u+ (x, t) ≈ u0ωp2π

Re [ exp (−iφ(ξo)) [1

πφ00(ξo)]

−1/2(1− i)/2] (2.235)

u+ (x, t) ≈ u04√πt0

Re

"exp

¡−iωpt0√υ2 − 1¢(υ2 − 1)3/4

(1− i)

#

=

√2u0

4√πt0

Re

"exp

¡−iωpt0√υ2 − 1¢(υ2 − 1)3/4

exp(iπ/4)

#

=u0

2√2π (x/c)1/4

cos

µωp

qt2 − (x/c)2 + π/4

¶³t2 − (x/c)2

´3/4 (2.236)

This approximation is good for ωpt0 À 1. Considering the range of plasma frequencies, from 109rad/s for gaseous plasmasthrough 1016rad/s for metallic plasmas, the distance of the observation point from the source should be much greater than10 cm for the gaseous plasmas and 1µm for the metals. The plot in Fig. 16 shows u+ (x, t) for a metallic plasma withωp = 10

16rad/s and ωpto = 1000. at three times, t. Note that the frequency decreases as time increases.

59

Section 2.5 Analyticity and the Kramers­Kronig relationship

0 5 10 15 20 25 306

4

2

0

2

4

6

x in microns

4.834

5.761−

u x to,( )

u x 2 to⋅,( )

u x 4 to⋅,( )

300 x

10 6−

Figure 16(2.237)

2.5 Analyticity and the Kramers­Kronig relationship We have determined that causality requires ε (ω)be analytic and have no zeros in the upper half complex ω plane. In the case that the local approximation holds, this can beseen by considering the causal expression

D (r, t) = E (r, t) +

Z ∞0

G (τ)E (r, t− τ) dτ (2.238)

In this expression G (τ) is a property of the system and is generally expected to vanish as τ → ∞. (The exception to thisbehavior is furnished by the electrical conductors for which G (τ) → 4πσ0, the DC conductivity6.) The Fourier transform,in time, of this expression along with the relationship between the frequency components of D and E yieldsZ ∞

−∞D (r, t) eiωtdt =

Z ∞−∞

E (r, t) eiωtdt+

Z ∞−∞

Z ∞0

G (τ)E (r, t− τ) eiωtdtdτ (2.239)

D (r, ω) = E (r, ω) +

Z ∞−∞

Z ∞0

G (τ) eiωτE (r, t− τ) eiω(t−τ)d(t− τ)dτ

ε (ω)E (r, ω) = E (r, ω) +

Z ∞0

G (τ) eiωτE (r, ω) dτ

ε (ω)− 1 =Z ∞0

G (τ) exp (iωτ) dτ (2.240)

6 In the case of a charged plasma

G (τ) = − 1

2π C

4πσ0

ω (ωτ + i)e−i ω t dω

= 4πσ0 1− e−t/τ

For τ → 0 G (τ) → 0 which can be interpreted as demonstrating the finite time it takes for a system to repond to a stimulus. While for τ → ∞G (τ)→ 4πσ0.

60

Section 2.5 Analyticity and the Kramers­Kronig relationship

Since D and E are real G (τ) will also be real. It follows, as previously noted, that

ε (ω) = ε (−ω)∗ . (2.241)

It is also possible to obtain a relationship between G and its derivatives at τ = 0 and the high frequency behavior ofε (ω) . To arrive at the relationship we note thatZ ∞

0

G (τ) exp (iωτ) dτ

=1

iω

µG (τ) exp (iωτ) |∞0+ −

Z ∞0

dG (τ)

dτexp (iωτ) dτ

¶(2.242)

and Z ∞0

dnG (τ)

dτnexp (iωτ) dτ

=1

iω

µdnG (τ)

dτnexp (iωτ) |∞0+ −

Z ∞0

dn+1G (τ)

dτn+1exp (iωτ) dτ

¶(2.243)

For all materials which are not DC conductors (and for which limn→∞dn+1G(τ)dτn+1 = 0)

ε (ω)− 1 =∞Xn=1

(−1)n(iω)n

dn−1G (τ 0)dτ 0 n−1

|τ 0=0+ (2.244)

The asymptotic behavior of ε (ω) for large ω can be deduced from this expression. It is reasonable to expect that theresponse of the system takes a finite time, i.e., the system does not instantaneously change. In the case that this is true wehave that G (0) = 0 and the first two terms in the expansion are

ε (ω)− 1 ≈ −ω−2G0 (0)− iω−3G00 (0) (2.245)

Re(ε (ω)− 1) ≈ −1ω2

G0 (0) (2.246)

Im(ε (ω)− 1) ≈ −1ω3

G00 (0) (2.247)

That is, the real part of ε (ω)− 1 vanishes as ω−2 and the imaginary part of ε (ω) vanishes as ω−3.Finally, for complex ω in the upper half (Imω > 0)

ε (ω)− 1 =Z ∞0

G (τ) eiωτdτ ≤Z ∞0

G (τ) |eiτ Reω|e−τ Imωdτ ≤Z ∞0

G (τ) dτ (2.248)

IfR∞0

G (τ) dτ is finite then ε (ω) − 1 will be finite and an analytic function of complex ω in the upper half (Imω > 0)complex plane. This has some important consequences. The ε (ω)− 1 for ω real exists and is the limit of a function whichis analytic in the upper half plane.

2.5.1 The Kramers­Kronig relations Since ε (ω)− 1 is an analytic function in the upper half complex planethen, by Cauchy’s residue theorem,

ε (z) = 1 +1

2πi

IC

ε (z0)− 1z0 − z

dz0 (2.249)

with the imaginary part of all points on the curve C restricted to be greater than or equal to zero. Let the path lie along thereal axis and let it be closed by the semi­circle of infinite radius. (ε (ω) − 1 will vanish on this semi­circle.) In addition letz = ω + iδ,with 1À δ > 0 then

ε (ω) = 1 + limδ→0

1

2πi

Z ∞−∞

ε (ω0)− 1ω0 − ω − iδ

dω0 (2.250)

61

Section 2.5 Analyticity and the Kramers­Kronig relationship

This can be rewritten using the relationships between path integrals, principal value integrals, and delta functions. Theprincipal value of the integral, where f (z0) is analytic in upper half plane, is:

P

Zf (z0)z0 − z

dz0 =IC

f (z0)z0 − z

dz0 −ZCδ

f (z0)z0 − z

dz0 (2.251)

where Cδ is a circular curve of radius δ0 which ”surrounds” the point, z. If z lies on the real axis then Cδ is a semi­circle ofradius δ0 taken below the point z and

P

Zf (z0)z0 − z

dz0 =

IC

f (z0)z0 − (z + iδ)

dz0 −ZCδ

f (z0)z0 − z

dz0 (2.252)

= 2πif(z + iδ)−ZCδ

f (z0)z0 − z

d(z0 − z) (2.253)

= 2πif(z + iδ)−Z 2π

π

f¡z + δ0eiφ

¢δ0eiφ

d(δ0eiφ) with z0 − z = δ0eiφ (2.254)

= 2πif(z + iδ)−Z 2π

π

f(z) + f 0 (z)δ0eiφ

2+ ...

1idφ (2.255)

= 2πif(z + iδ)− if(z)π − f 0 (z)δ0

2[ei2π − eiπ] + ... (2.256)

Now let .δ0 go to 0. Finally, let δ −→ 0. Then,

P

Zf (z0)z0 − z

dz0 = if(z)π (2.257)

Applying this to our problem, the integral taken below the pole equals the principal part of the integral plus πi times theintegral with a delta function

ε (ω) = 1 +1

2πi[P

Zε (ω0)− 1ω0 − ω

dω0 + iπ(ε (ω)− 1)]

= 1 +1

2πi[P

Zε (ω0)− 1ω0 − ω

dω0 +Z

ε (ω0)− 1ω0 − ω

πiδ(ω0 − ω)dω0]

= 1 +1

2πiP

Zε (ω0)− 1ω0 − ω

dω0 +1

2[ε (ω)− 1]

or

ε (ω) = 1 +1

πiP

Zε (ω0)− 1ω0 − ω

dω0 (2.258)

Using the fact that the integrand vanishes on the upper half circle at ω0 =∞, this yields relationships between the real andimaginary parts of ε (ω)

Re ε (ω) = 1 +1

πP

Z ∞−∞

Im ε (ω0)ω0 − ω

dω0 (2.259)

Im ε (ω) = − 1πP

Z ∞−∞

Re [ε (ω0)− 1]ω0 − ω

dω0 (2.260)

62

Section 2.5 Analyticity and the Kramers­Kronig relationship

which, using Re ε (ω) = Re ε (−ω) and Im ε (ω) = − Im ε (−ω) ,can be written as

Re ε (ω) = 1 +1

πP [

Z 0

−∞

Im ε (ω0)ω0 − ω

dω0 +Z ∞0

Im ε (ω0)ω0 − ω

dω0] (2.261)

= 1 +1

πP [

Z ∞0

Im ε (−ω00)−ω00 − ω

dω00 +Z ∞0

Im ε (ω0)ω0 − ω

dω0]

= 1 +1

πP [

Z ∞0

Im ε (ω00)ω00 + ω

dω00 +Z ∞0

Im ε (ω0)ω0 − ω

dω0]

= 1 +2

πP

Z ∞0

ω0 Im ε (ω0)ω0 2 − ω2

dω0

Im ε (ω) = − 1πP [

Z 0

−∞

Re ε (ω0)ω0 − ω

dω0 +Z ∞0

Re ε (ω0)ω0 − ω

dω0] (2.262)

= − 1πP [

Z ∞0

−Re ε (ω00)ω00 + ω

dω00 +Z ∞0

Re ε (ω0)ω0 − ω

dω0]

= −2ωπ

P

Z ∞0

Re [ε (ω0)− 1]ω0 2 − ω2

dω0

These are the classic Kramers­Kronig relations between the real and imaginary parts of the electric permittivity. Similarrelations hold for the frequency response function for any causal system. Similar relationships are very useful in quantummechanics (particularly in scattering theory), circuit analysis in electrical engineering, etc.

The significance of these relations is that it provides a way to measure or calculate one characteristic of a system and thendetermine a completely different characteristic of the system. For example the existence of an absorption peak can be seen tolead to anomalous dispersion, i.e., dn (ω) /dω < 0. This is easily demonstrated. Assume that the system has an absorptionpeak at ω = ω0. For simplicity we take

Im ε (ω) =π

2αω0δ (ω − ω0) + ... (2.263)

where the additional terms vary slowly in the region of ω0 and α is a dimensionless constant. Equation 2.261 gives the realpart of ε (ω) in the region of ω0 (but not at ω0),

Re ε (ω) = 1 +2

πP

Z ∞0

π2αω0δ (ω

0 − ω0)ω0

ω0 2 − ω2dω0 +∆ε (2.264)

= 1 +αω20

ω20 − ω2+∆ε (2.265)

with ∆ε (ω) slowly varying in the neighborhood of ω0. The contribution of the absorption peak to the real part of ε (ω) isseen as a rapid variation in Re ε (ω) from positive below ω0 to negative above ω0.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Assignment 9a: Jackson­ 7.22, (2nd edition 7.14)Use the Kramers­Kronig relations to calculate the real part of ε (ω)given the imaginary part of ε (ω) for positive ω as

(a) Im ε (ω) = λ [θ (ω − ω1)− θ (ω − ω2)] , ω2 > ω1 > 0

(b) Im ε (ω) =λγω

(ω20 − ω2)2+ γ2ω2

In each case sketch the behavior of Im ε (ω) and Re ε (ω) as functions of ω.Comment on the similarities or differences of theresults as compared with curves in Fig. 7.8 of Jackson.

Assignment 9b:Jackson, 7.23, (2nd edition, 7.15) Let ε(ω) be the permittivity of a system containing free electrons. In

this case, as seen in the lectures covering the permittivity of a charged plasma, there is a singularity at ω = 0. The singularity

63

Section 2.5 Analyticity and the Kramers­Kronig relationship

is removed in the function

f (ω) ≡ ε (ω)− 4πσ0iω

with σ0 the DC conu\ductivity of the electrons. Since the function f (ω) is analytic for Imω ≥ 0 it satisfies Kramer­Kronigtype relations. Obtain the modified Kramer­Kronig relations for ε (ω) .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

It is also possible to use known values of ε (ω) to aid with the calculation of the unknown values. For example supposewe have measured or calculated the imaginary part of the index of refraction (i.e., the absorption for the system) and the realpart of ε (ω1) is known. Then, from Eq. 2.261, ε (ω1) satisfies

Re ε (ω1) = 1 +2

πP

Z ∞0

ω0 Im ε (ω0)ω0 2 − ω21

dω0 (2.266)

It follows that

Re ε (ω)−Re ε (ω1) =2

πP

Z ∞0

ω0 Im ε (ω0) [1

ω0 2 − ω2− 1

ω0 2 − ω21]dω0

=2

π

¡ω2 − ω21

¢P

Z ∞0

ω0 Im ε (ω0)(ω0 2 − ω2) (ω0 2 − ω21)

dω0 (2.267)

This form for ε (ω) reduces the integral’s dependence on the values of Im ε (ω) at large ω. This is known as a ‘subtracteddispersion relation’. This process of subtraction can be repeated for each known value of ε (ω) . A similar relationship canbe generated, starting with Eq. 2.262, for the imaginary part ε (ω) .

2.5.2 Sum rules The Kramer­Kronig relations can be used to obtain some general properties of the integrals of thereal and imaginary parts of the permittivity. We have noted that, at high frequency, the permittivity varies as ω−2+O

¡ω−3

¢.

Physically at high frequencies the binding forces for the electrons becomes negligible and the system responds as a plasma.With this interpretation we define the plasma frequency for the system

ω2p ≡ limω→∞

£ω2 (1− ε (ω))

¤(2.268)

If Im ε (ω) varies as ω−3 +O¡ω−4

¢for high frequencies then Eq. 2.261 gives

ω2p = limω→∞

£ω2 (1−Re ε (ω))− ω2i Im ε (ω)

¤(2.269)

= limω→∞

£ω2 (1−Re ε (ω))− ω2i(ω−3 +O

¡ω−4

¢)¤

= limω→∞

·−ω2 2

πP

Z ∞0

ω0 Im ε (ω0)ω0 2 − ω2

dω0 − i(ω−1 +O¡ω−2

¢)

¸

= limω→∞

− 2πP

Z ∞0

ω0 Im ε (ω0)

[ω0 2

ω2− 1]

dω0 − i(ω−1 +O¡ω−2

¢)

ω2p =2

π

Z ∞0

ω Im ε (ω) dω (2.270)

This is the sum rule for oscillator strengths.

It is interesting to consider the relationship between the coefficients in Sellmeier’s equation, Eq. 2.122 , for the

64

Section 2.5 Analyticity and the Kramers­Kronig relationship

permittivity and the plasma frequency defined in Eq. 2.270.The sum rule requires that

ω2p =2

π

Xm

Z ∞0

ω ImAm

ω2m − iγmω − ω2dω (2.271)

=2

π

Xm

Z ∞0

ω Im[Am[ω

2m − ω2 + iγmω]

[ω2m − ω2 − iγmω][ω2m − ω2 + iγmω]

dω

=2

π

Xm

Z ∞0

ω[Amγmω

(ω2m − ω2)2 + (γmω)2dω

=1

π

Xm

Z ∞−∞

Amγmω2

(ω2m − ω2)2+ (γmω)

2 dω (2.272)

This integrand has simple poles at ω± and ω∗± with

ω± = iγ

2±rω2m −

γ2

4

In this case we rewrite the integral as

ω2p =1

π

Xm

Z ∞−∞

Amγmω2

(ω − ω+) (ω − ω−)¡ω − ω∗+

¢ ¡ω − ω∗−

¢dω (2.273)

and evaluate it using Cauchy’s residue theorem. Closing the path in the upper half complex ω plane we obtain the residues ofthe poles in the upper half plane (at ω+ and ω−)

ω2p = 2iXm

Amγmω2+

(ω+ − ω−)¡ω+ − ω∗+

¢ ¡ω+ − ω∗−

¢+2i

Xm

Amγmω2−

(ω− − ω+)¡ω− − ω∗+

¢ ¡ω− − ω∗−

¢ (2.274)

= 2iXm

[Amγmω

2+

(ω+ − ω−) 2iγm2

¡ω+ − ω∗−

¢ + Amγmω2−

(ω− − ω+)¡ω− − ω∗+

¢2iγm2

= 2Xm

"Amω

2+

(ω+ − ω−)¡ω+ − ω∗−

¢ + Amω2−

(ω− − ω+)¡ω− − ω∗+

¢#

=Xm

·Amω+

(ω+ − ω−)− Amω−(ω+ − ω−)

¸ since ω+ = −ω∗− and ω∗+ = −ω−

=Xm

Am (2.275)

The definition of the plasma frequency given in Eq. 2.268 is quite general and would reect the contributions of theresponses of all charged particles in the system. The coefficients in Sellmeier’s equation, which approximates the permittivityas a sum of resonances, gives the ‘strength’ of each resonance. In principle the sources of the resonances include opticallyactive phonons, plasmons, electronic excitations, electronic ionizations, etc.

A second sum rule is obtained if Re ε (ω) − 1 ∼ −ω2p/ω2 + O¡ω−4

¢and Im ε (ω) ∼ O

¡ω−3

¢for large ω. The

asymptotic relationship between the real and imaginary parts of ε (ω) can be obtained using Equation 2.262. We write thisexpression as

Im ε (ω) =2

πω

"Z ωm

0

Re [ε (ω0)− 1]1− (ω0/ω)2 dω0 + P

Z ∞ωm

Re [ε (ω0)− 1]1− (ω0/ω)2 dω0

#(2.276)

65

Section 2.5 Analyticity and the Kramers­Kronig relationship

with ω À ωm and ωm sufficiently large that the asymptotic form for the real part of ε (ω) provides a valid approximation.The approximated equation is

Im ε (ω) ≈ 2

πω

Z ωm

0

Re [ε (ω0)− 1]h1 + (ω0/ω)2

idω0

+2

πωP

Z ∞ωm

1

1− (ω0/ω)2£−ω2p/ω02 +O

¡ω0−4

¢¤dω0

Im ε (ω) ≈ 2

πω

·Z ωm

0

Re [ε (ω0)− 1] dω0 − [ω2p/ωm +O¡ω−2

¢¸(2.277)

Note that for an integral over finite limits with f (x) = f (−x)

P

Z a

−a

f (x)

xdx = lim

δ→0

"Z −δ−a

f (x)

xdx+

Z a

δ

f (x)

xdx

#

= limδ→0

·−Z a

δ

f (−x)x

dx+

Z a

δ

f (x)

xdx

¸= 0 (2.278)

Since Im ε (ω) ∼ O¡ω−3

¢for large ω it follows that

Z ωm

0

Re [ε (ω0)− 1] dω0 − ω2p/ωm = 0 (2.279)

or

ω−1m

Z ωm

0

Re ε (ω0) dω0 = 1 + ω2p/ω2m (2.280)

This is sometimes called a superconvergence relation. This latter relation, Eq. 2.280, does not hold for conductors while thesum rule for oscillator strengths, Eq. 2.270, will hold for conductors.

2.5.3 Some useful relations between path integrals The Kramers­Kronig relations involve principal valueintegrals. For these integrals the integrand will have simple poles on the real axis and the integral is taken along the real axisexcluding the location of the pole.

66

Section 2.5 Analyticity and the Kramers­Kronig relationship

.Fig. 16

For example let f (x) have no singularities on the real axis then

P

Z ∞−∞

f (x)

xdx = lim

δ→0

"Z −δ−∞

f (x)

xdx+

Z ∞δ

f (x)

xdx

#(2.281)

Often it is easier to evaluate the integral using Cauchy’s residue theorem. This requires a closed path and the principal valueintegration leaves a gap in the path. This problem is surmounted using the identity illustrated in Fig. 16 . The integrand isassumed to have a simple pole located at the ‘x’ on the real axis The identity holds because one can show that the integralsalong the half circles in the two contours cancel. In this case

P

Z ∞−∞

f (x)

xdx =

1

2

"Z ∞−∞, C+

f (x)

xdx+

Z ∞−∞, C−

f (x)

xdx

#(2.282)

The principal value integral has been replaced by two integrals but Cauchy’s residue theorem can often be used to evaluateeach integral.

67

Section 2.6 Vector characteristic of electromagnetic wave

.Fig.17

Another relationship between the path integrals is often useful. This identity is illustrated in Fig. 17 where again ‘x’marks the location of the pole. The direction of integration along the circular path around the pole has been selected so as tocancel the half circle for C+ and add the half circle for C−.

If both identities are combined the principal value integral is seen to equal (1) the integral along path C+ plus one halfthe integral around the pole or (2) the integral along the path C− minus one half the integral around the pole.

2.6 Vector characteristic of electromagnetic wave The final property of electromagnetic waves weshall consider is their vector characteristic. In an isotropic medium the electric field is perpendicular to the direction ofpropagation of the wave. A plane wave with the electric field given by E(r,t) = E1 (r, t) ε1is said to be linearly polarizedwith polarization vector ε1. Similarly E(r,t) = E2 (r, t) ε2 (with ε1 · ε2 = 0) describes a linearly polarized wave withpolarization vector ε2. A general plane electromagnetic wave with wave vectork (k · εi = 0, k×ε1 = kε2) is given by

E = (E1ε1 +E2ε2) exp (i [k · r− ωt]) (2.283)

B =c

ωk×E (2.284)

In general the amplitudes E1 and E2 are complex numbers.If the phases of E1and E2 differ by a nπ the wave is linearly polarized along the line at an angle of θ =

tan−1 [(−1)nE2/E1] with ε1. In this case the amplitude of the wave ispE21 +E22 .

An alternate representation of the wave can be given in terms of the complex basis vectors7

ε± = 2−1/2 (ε1 ± iε2) (2.285)

7

ε± · ε∗± = 1

ε± · ε∗∓ = 0

ε∗± = ε∓

68

Section 2.6 Vector characteristic of electromagnetic wave

Using these the electric field for a general plane wave is

E = 2−1/2 [(E1 − iE2) ε+ + (E1 + iE2) ε−] exp [i (k · r− ωt)] (2.286)

A wave described by E(r, t) = E0ε± exp (i [k · r− ωt]) is a circularly polarized wave. The components of this wave are

E1 (r,t) = 2−1/2 |E0| cos (k · r− ωt+ φ0) (2.287)

E2 (r,t) = ∓2−1/2 |E0| sin (k · r− ωt+ φ0) (2.288)

With the upper sign of ε± the wave is said to be left hand circularly polarized and with the lower sign it is right handcircularly polarized. This is determined by the direction of rotation of the electric field in the advancing wave (at a fixed timenot at a point). An observer (when facing the oncoming wave) sees the E rotating counterclockwise ( clockwise) at a fixedspace point for the ε+ (ε−) polarization In modern terminology the left hand circularly polarized wave has positive helicitywhile the right hand circularly polarized wave has negative helicity. The sign is the sign of the projection of the angularmomentum of the wave on the direction of wave propagation.

The general polarized wave has elliptical polarization. It is most easily seen using the circularly polarized waves. Thesemimajor axis is reached when the counter­circulating electric fields are aligned. Its length is therefore

a =√2 (|E+|+ |E−|)

. The semiminor axis is reached when the fields are anti­aligned resulting in the length

b =√2 ||E+|− |E−||

. If the waves are in phase the major axis is along ε1 and the minor axis is along ε2. Let φ+be the phase of the positivehelicity wave and φ− the phase of the negative helicity wave. If θ± are the angles between the electric fields and ε1then

θ+ = k · r− ωt+ φ+ and

θ− = −k · r+ ωt− φ−.

The alignment of these two vectors resulting in the semimajor axis occurs for θ+ − θ− = 0, 2Nπ. This has solutions

k · r− ωt = −0.5 ¡φ+ + φ−¢, −0.5 ¡φ+ + φ−

¢+Nπ

. The resultant angles with ε1 are

θmajor = 0.5¡φ+ − φ−

¢, 0.5

¡φ+ − φ−

¢+Nπ.

2.6.1 Stoke’s parameters The polarization of an electromagnetic wave can be determined by intensitymeasurements. These measurements yield the magnitude of the amplitudes of ε± and ε1,2 and the relative phases of theseamplitudes. In terms of the field expression the amplitudes are

a±eiδ± = ε∗± ·E (r, t) (2.289)

ajeiδj = εj ·E (r, t) (2.290)

The measurements are the following:

1. The total intensity

s0 = a21 + a22 = a2+ + a2− = (2.291)

2. The difference in linear polarization intensities along ε1 and ε2

s1 = a21 − a22 = 2a+a− cos (δ− − δ+) (2.292)

69

Section 2.7 Reection and refraction at a plane interface between dielectrics

3. The difference in linear polarization intensities at 45 to ε1 and ε2

s2 = 2a1a2 cos (δ2 − δ1) = 2a+a− sin (δ− − δ+) (2.293)

4. The difference in positive and negative helicity intensities

s3 = a2+ − a2− = 2a1a2 sin (δ2 − δ1) (2.294)

These parameters are the Stoke’s parameters. They are not all independent since they only depend on the relative phaseof the amplitudes. Ideally they satisfy

s20 = s21 + s22 + s23 (2.295)

however there is no such thing as a monochromatic wave. Any spread in the frequencies will lead to a variation in theamplitudes, both the magnitudes and phases. Time average measurements yield that

s20 ≥ s21 + s22 + s23 (2.296)

Natural light, even if a nearly monochromatic segment is analyzed, satisfies

s1 = s2 = s3 = 0 (2.297)

The Stokes parameters are quadratic in the field strength and can be determined through intensity measurements only, inconjuction with a linear polarizer and a quarter­wave plate or the equivalent. Their measurement determines completely thestate of polarization of the wave.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Assignment 10: Jackson­ 7.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.7 Reection and refraction at a plane interface between dielectrics Consider a planeelectromagnetic wave (wave vector k and angular frequency ω) travelling in a medium with a permittivity εi (ω) andpermeability µi (ω) . We assume that this medium fills the space z < 0 and that the region z > 0 is filled with a mediumwith a permittivity εt (ω) and permeability µt (ω) . From our investigations of Maxwell’s equations we have learned thatthe tangential components of E and H and the normal components of D and B are continuous at the interface between thematerials. Without any loss of generality we can assume that the electric field part of the wave is described by the real partof

Ei (r, t) = Ei exp [i (k1x+ k3z − ωt)] , (2.298)

i.e., the plane of incidence is perpendicular to the y axis. The continuity conditions at the interface require that the transmittedand reected electric fields have the same spatial variation as the incident field along the planar interface,

Et (r, t) = Et exp [i (k1x+ k03z − ωt)] (2.299)

Er (r, t) = Er exp [i (k1x− k3z − ωt)] (2.300)

with

k21 + k23 = εi (ω)µi (ω)ω2

c2(2.301)

k21 + k0 23 = εt (ω)µt (ω)ω2

c2(2.302)

This requirement is seen to yield Snell’s law of refraction along with the ‘law of reection’ at interfaces between media.

70

Section 2.7 Reection and refraction at a plane interface between dielectrics

To relate the various fields to the electric fields we consider the spatial and temporal fourier transform of the source freeMaxwell’s equations along with the constituent equations for an infinite medium.

B0 =c

ωk×E0 (2.303)

H0 = µ (ω)−1 B0 (2.304)

D0 = ε (ω) E0 (2.305)

It is convenient to consider two cases, the first with the incident electric field perpendicular to the plane of incidence

Ei = Eie2 (2.306)

and the second with the incident electric field parallel to the plane of incidence

Ei = Eik3e1 − k1e3

k(2.307)

In each case the electric fields of the transmitted and reected waves will lie in the same planes as the incident wave.

2.7.1 Electric field perpendicular to plane of incidence With the incident electric field perpendicular tothe plane of incidence the electric fields of the transmitted and reected waves will have the amplitudes

Et = Ete2 (2.308)

Er = Ere2 (2.309)

The continuity requirement for the component of the electric field which is tangential to the interface gives

Ei +Er = Et (2.310)

The tangential components of the magnetic field vector, H, for the incident, transmitted8, and reected waves are

[Hi]1 = −cµi (ω)−1 k3Ei/ω (2.311)

[Ht]1 = −cµt (ω)−1 k03Et/ω (2.312)

[Hr]1 = cµi (ω)−1

k3Ei/ω (2.313)

The continuity requirement for the component of the magnetic field which is tangential to the interface gives

µi (ω)−1

k3 [Ei −Er] = µt (ω)−1

k03Et (2.314)

The solution of Equations 2.310 and 2.314 is

Er =µt (ω) k3 − µi (ω) k

03

µt (ω) k3 + µi (ω) k03

Ei (2.315)

Et =2µt (ω) k3

µt (ω) k3 + µi (ω) k03

Ei (2.316)

The average component of the incident Poynting vector which is perpendicular to the interface is

[Si]avg = Re (k3/µi (ω)) |Ei|2 (2.317)

The corresponding average component of the reected Poynting vector is

[Sr]avg = −Re (k3/µi (ω)) |Ei|2¯µt (ω) k3 − µi (ω) k

03

µt (ω) k3 + µi (ω) k03

¯2(2.318)

8 It is assumed here that k03 is real. If it is complex then the imaginary part is positive, i.e., the wave is damped out going into the second medium.

71

Section 2.7 Reection and refraction at a plane interface between dielectrics

Generally the ratio of this reected component to the incident component gives the fraction of the incident intensity which isreected

R⊥ =¯µt (ω) k3 − µi (ω) k

03

µt (ω) k3 + µi (ω) k03

¯2(2.319)

The component of the average transmitted Poynting vector which is perpendicular to the interface is

[St]avg = Re

µk03

µt (ω)

¶ ¯2µt (ω) k3

µt (ω) k3 + µi (ω) k03

¯2|Ei|2 (2.320)

and the fraction of the incident intensity which is transmitted is

T⊥ =k03/µt (ω) + c.c.

k3/µi (ω) + c.c.

¯2µt (ω) k3

µt (ω) k3 + µi (ω) k03

¯2. (2.321)

In the case that k3 and the permeabilities are real we have that

1−R⊥ =2µi (ω)µτ (ω) k3 (k

03 + c.c.)

|µt (ω) k3 + µi (ω) k03|2

= T⊥ (2.322)

If k3 is complex (as is generally true) an interference between incident and reected waves occurs.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Assignment 11: The material of a thin film, in air, has a permittivity given by ε = 10 + 0.5i. (Assume the permeabilitiesare one.) Light with a wave number k0 (real) is directed at the film at normal incidence. (a) By explicit calculation of eachpart evaluate the fraction of the intensity which is (1) reected, (2) transmitted, (3) absorbed for film thickness d satisfying0 < k0d < 2π. (b) Repeat part (a) with ε = 10− 0.5i. (−absorption is creation) (c) Compare and discuss the structures seenin plots of your results.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.7.2 Electric field parallel to the plane of incidence With the electric field parallel to the plane ofincidence the magnetic fields will be perpendicular to this plane. The magnetic field of the incident wave has an amplitude

Hi = Hie2 (2.323)

and the magnetic fields of the reected and transmitted waves have amplitudes

Hr = Hre2 (2.324)

Ht = Hte2. (2.325)

The fourier transform of the source free Ampere­Maxwell law relates the electric and magnetic fields

k×H = −ε (ω) ωcE (2.326)

This gives the components of the electric fields which are parallel to e1,

(Ei)1 =ck3

ωεi (ω)Hi (2.327)

(Er)1 = − ck3ωεi (ω)

Hr (2.328)

(Et)1 =ck03

ωεt (ω)Ht (2.329)

72

Section 2.7 Reection and refraction at a plane interface between dielectrics

Continuity of the tangential electric fields yields

(Ei)1 + (Er)1 = (Et)1 (2.330)

while continuity of the tangential magnetic fields gives

εi (ω)

k3[(Ei)1 − (Er)1] =

εt (ω)

k03(Et)1 (2.331)

The solution of these equations is

(Er)1 =εi (ω) k

03 − εt (ω) k3

εi (ω) k03 + εt (ω) k3(Ei)1 (2.332)

(Et)1 =2εi (ω) k

03

εi (ω) k03 + εt (ω) k3(Ei)1 (2.333)

At normal incidence Equations 2.315 and 2.316 and Equations 2.332 and 2.333 give identical results

Er =[µt (ω) εi (ω)]

1/2 − [µi (ω) εt (ω)]1/2[µt (ω) εi (ω)]

1/2+ [µi (ω) εt (ω)]

1/2Ei (2.334)

Et =2 [µt (ω) εi (ω)]

1/2

[µt (ω) εi (ω)]1/2 + [µi (ω) εt (ω)]

1/2Ei (2.335)

2.7.3 Systems with multiple surfaces In the case that the electromagnetic wave passes through multiplesurfaces there will generally be Both incident and reected waves on each side of the interfaces. The most common exampleof these systems involve the thin film coating of optical elements. The problem is to design and manufacture a coating whichwill provide a given reectance or transmittance between elements. The importance of the problem is reected in the largenumber of formulations of the problem and the multitude of computer codes designed to solve the problem.

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