Electromagnetic Waves in Materials
Outline
Review of the Lorentz Oscillator Model
Complex index of refraction – what does it mean?
TART
Microscopic model for plasmas and metals
1
True / False
1. In the Lorentz oscillator model of an atom, the electron is
bound to the nucleus by a spring whose spring constant is the
same for any atom.
2. The following is the differential equation described by the
Lorentz oscillator model:
3. For dielectrics, we can approximate the index of refraction as
the square root of the dielectric constant.
=
2
με
ε
≈
μ0ε 0
ε0
Microscopic Description of Dielectric Constant
Nucleus
Bx
Electron
“spring”
Damping
Electron mass
field
E
Restoring force
force
(binding electron & nucleus)
3
Solution using complex variables
Lets plug-in the expressions for
and y
into the differential equation from slide 3:
Natural resonance
y
d2
d
q
2
y(t) + γ y(t) + ωo y(t) = Ey (t)
2
dt
m
dt
Ey (t) = Re{Ey ejωt }
+ +
+ +
+ +
2
ω y + jωγy +
p
+q
−q
ωo2 y
y(t) = Re{yejωt }
q
= Ey
m
q
1
y=
Ey
2
2
m (ωo − ω ) + jωγ
4
Oscillator Resonance
1
q
y=
Ey
2
2
m (ω0 − ω ) + jωγ
Ey (t) = Re{Ey ejωt }
y(t) = Re{yejωt }
Driven harmonic oscillator: Amplitude and Phase depend on frequency
Low frequency
medium
amplitude
Displacement, y
in phase with Ey
At resonance
large amplitude
Displacement, y
90º out of phase with Ey
5
High frequency
vanishing
amplitude
Displacement y and Ey
in antiphase
Polarization
Since charge displacement, y, is directly related to polarization, P, of our material
we can then rewrite the differential equation:
= o E
+ P
D
Py = N qy
For linear polarization in ŷ direction
2
d
d
N
q
( 2 + γ + ωo2 )Py (t) =
Ey (t) = o ωp2 Ey (t)
dt
m
dt
2
N
q
ωp2 =
o m
Py (t) = Re{Py ejωt }
ωp2
Py = 2
o Ey
2
(ωo − ω ) + jγω
6
Microscopic Lorentz Oscillator Model
ωo2
ωo
ωp
7
kspring
=
m
ωo2
kspring
=
m
amplitude
180°
90°
Frequency
Phase Lag
Amplitude
Low
frequency
limit
phase
(displacement vs. driving field)
Behavior of a driven (and damped) harmonic oscillator can be summarized as follows
Hi
frequency
limit
This type of response of bound charges is typical for many materials
8
Complex Refractive Index
˜ = r − ji
1
vp = √
μ
√
√
μ
c
≈√
n≡
=√
μo o
o
vp
˜
=n
˜2
o
= (n − jκ)2
= n2 − κ2 − 2jnκ
real
= n2 − κ2
o
imag
= 2nκ
o
9
(real)
(imaginary)
Absorption Coefficient
1
n= √
2
r + 2r + 2i
1
κ= √
2
−r + 2r + 2i
˜0 e−αz/2 ej(ωt−k0 nz) }
E (t, z) = Re{E
Absorption
Refractive
index
2π
α = 2ko κ = 2 κ [cm-1]
λo
10
Absorption and Reflection
˜0 e−αz/2 ej(ωt−k0 nz) }
E(t, z) = Re{E
Absorption
coefficient
I(z) = Io e−αz
Refractive
index
Beer-Lambert Law or Beer’s Law
11
TART
ω0 − γ 2 ω0 + γ 2
Different resonant frequencies
ωp
ω0
Photograph by Hey Paul on Flickr.
•
•
•
•
Transmissive Absorp. Reflective Transmissive
12
Transmissive
Absorptive
Reflective
Transmissive
ω < ω0 − γ 2
ω 0 − γ 2 <ω < ω 0
ω 0 + γ 2 <ω < ω p
ω >ωp
Plasma in Ionosphere
Plasma is an ionized gas consisting of positively charged molecules (ions) and
negatively charged electrons that are free to move.
Plasma exists naturally in what we call ionosphere (80 km ~ 120 km above the
surface of the Earth). Here the UV light from the Sun ionizes air molecules.
Aurora Australis
Image is in the public domain
13
Plasmas (which we will assume to be lossless, γ = 0 )
… have no restoring force for electrons, ωo = 0
2
ωp
= o 1 − 2
ω
Reflecting
1.0
What happens when
the dielectric constant
is negative?
<0
ω > ωp
>0
0.5
/o
ω < ωp
Transparent
0
-0.5
1
If ε < 0 then n is imaginary
1.5
ω/ωp
14
2
Optical Response of Plasmas
= n − jκ
o
Transparent
Reflecting
1
κ
n
n, κ
0.5
n=
ωp2
1− 2
ω
for
ω > ωp
ωp2
−1
2
ω
for
ω < ωp
κ=
0
-0.2
0.5
1
1.5
ω/ωp
15
2
Plasma Frequency
N e2
ωp =
o m
(1012 M −3 )(1.602x10−19 C)2
=
(8.854 × 10−12 F/M )(9.109 × 10−31 Kg)
= 5.64 × 107 rad/sec
= 2π × (8.98M Hz)
AM radio is in the range 520-1610 kHz
FM radio in in the range 87.5 to 108MHz
16
Reflected
Transmitted
AM radio
transmitter
The Ionosphere and Radio Wave Propagation
The ionosphere is important for radio wave (AM only) propagation....
Ionosphere is composed of the D, E, and F layers.
The D layer is good at absorbing AM radio waves.
D layer disappears at night...the E and F layers bounce the waves back to the Earth.
This explains why radio stations adjust their power output at sunset and sunrise.
17
Why do metals reflect light?
© Kyle Hounsell. All rights reserved. This content is excluded from our Creative
Commons license. For more information, see http://ocw.mit.edu/fairuse.
18
Why is there a discontinuity here?
o or γ must change for this to be true
Metals are Lossy
Metals have loss
but have no restoring force for electrons
0.5
0.4
= o
1−
ωp2
ω2
r = o
1−
i = o
= r + ii
− jωγ
ωp2
r
i
0.2
0.1
ωp
-5
ω2 + γ 2
γωp2
ω(ω 2 + γ 2 )
0.3
Drude Model for metals
r , i
γ=0
ωo = 0
-10
-20
-25
-30
19
r
ω
Behavior of Metals
γ/2
100
6
3
T
n
Reflection(%)
n, κ
R
4
2
80
κ
5
ωp
40
T
R
20
n
1
60
ω
ωp
20
ω
Image by Kate Hopkins http://www.flickr.com/photos/accidentalhedonist/5200667428/ on flickr
T
Dielectric
Metal
Plasma
A
ω0 − γ 2
R
ω0
ω0 = 0
ω0 + γ 2
ωp
ω0 + γ 2
ωp
ω0 = 0
ωp
γ =0
21
T
Key Takeaways
Nucleus
Electron
Lorentz Oscillator Model
d2 y
dy
2
m 2 = −mωo y + qEy − mγ
dt
dt
“spring”
ωp2
P (ω) = 2
o E(ω)
2
(ωo − ω ) + jωγ
ñ = n − jκ
α = 2ko κ
= r − ji
E(t, z) = Re{Ẽ0 e−αz/2 ej(ωt−k0 nz) }
Decay
Absorption
coefficient
I(z) =
Refractive
index
• Transmissive
• Absorptive
Io e−αz Beers Law
• Reflective
• Transmissive
22
ω < ω0 − γ 2
ω 0 − γ 2 <ω < ω 0
ω 0 + γ 2 <ω < ω p
ω >ωp
MIT OpenCourseWare
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6.007 Electromagnetic Energy: From Motors to Lasers
Spring 2011
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