9.2 Electromagnetic Waves in Vacuum
9.2.1 The Wave Equation for E and B
Source Free Wave Equations:
In vacuum, Maxwell’s equations reduce to:
In Vacuum:
r
r
r
r
r
r
∂B
∂E
, ∇ ⋅ B = 0 , ∇ × B = µ0ε 0
∇⋅ E = 0, ∇× E = −
∂t
∂t
r
r
r
r
r
r
∂
∂2E
1 ∂2E
2
∇ × ∇ × E = − ∇ × B = − µ0ε 0 2 Æ − ∇ E + ∇ ∇ ⋅ E = − 2 2
∂t
∂t
c ∂t
r
r
r
∂
∂2B
∇ × ∇ × B = µ0ε 0 ∇ × E = − µ0ε 0 2
∂t
∂t
(
)
(
)
(
)
(
(
)
)
⎛ 2 1 ∂2 ⎞ r
⎛ 2 1 ∂2 ⎞ r
Æ ⎜⎜ ∇ − 2 2 ⎟⎟ E = 0 and ⎜⎜ ∇ − 2 2 ⎟⎟ B = 0 (homogeneous vector wave
c ∂t ⎠
c ∂t ⎠
⎝
⎝
equations)
In linear, isotropic, and homogeneous nonconducting media, Maxwell’s equations
reduce to:
In Linear Media:
r
r
r
r
r
r
r
∂B
∂H
∂E
= −µ
∇⋅ E = 0, ∇× E = −
, ∇× H = ε
, ∇⋅H = 0
∂t
∂t
∂t
r
r
r
r
r
r
∂
∂2E
1 ∂2E
2
∇ × ∇ × E = − µ ∇ × H = − µε 2 Æ − ∇ E + ∇ ∇ ⋅ E = − 2 2
∂t
∂t
v ∂t
(
)
(
)
(
)
⎛
⎛
1 ∂2 ⎞ r
1 ∂2 ⎞ r
Æ ⎜⎜ ∇ 2 − 2 2 ⎟⎟ E = 0 and ⎜⎜ ∇ 2 − 2 2 ⎟⎟ H = 0 (homogeneous vector wave
v ∂t ⎠
v ∂t ⎠
⎝
⎝
equations)
r
r
E and B satisfy the three-dimensional wave equation,
∇2 f =
1 ∂2 f
.
v 2 ∂t 2
The Solution of The Nonhomogeneous Wave Equations:
r
⎛ 2 1 ∂2 ⎞
1 ∂2 ⎞ r
ρ ⎛
⎜⎜ ∇ − 2 2 ⎟⎟V = − , ⎜⎜ ∇ 2 − 2 2 ⎟⎟ A = − µ0 J ?
c ∂t ⎠
c ∂t ⎠
ε0 ⎝
⎝
r
The solutions are V (r , t ) =
1
4πε 0
∫
r
ρ (r ' , tr )
r r
µ0
r r dτ ' and A(r , t ) =
4π
r − r'
r r
J (r ' , tr )
∫ rr − rr' dτ ' .
9.2.2 Monochromatic Plane Waves:
The Electromagnetic Spectrum:
The Electromagnetic Spectrum
Frequency
Wavelength
23
Photon Energy
Type
9
10
10 eV
1022
1021
γ rays
10-13m
20
6
10
10 eV
1019
1018
1Å
17
10
1 nm
x rays
3
10 eV
16
10
Ultraviolet
15
10
1 µm
1014
1 eV
13
10
Infrared
12
10
1011
10
10
1 mm
EHF (mm wave)
1 cm
SHF (micro wave)
9
10
UHF (mobile phone)
8
10
1m
VHF (TV, FM)
7
10
HF
106
MF (AM)
5
10
1 km
LF (Navigation)
4
10
VLF (RF, Navigation, Sonar)
103
ULF
2
6
10
10 m
SLF
1
10
ELF
The Visible Range
Frequency (Hz)
Color
Wavelength (nm)
1.0 X 1015
Near ultraviolet
300
14
Shortest visible blue
400
14
6.5 X 10
Blue
460
5.6 X 1014
Green
540
7.5 X 10
5.1 X 1014
Yellow
590
14
Orange
610
14
Longest visible red
760
14
Near infrared
1000
4.9 X 10
3.9 X 10
3.0 X 10
I. ORIGINAL THERMAL 2.45 GHz MICROWAVE EFFECT
Frequencies ranging from 3 MHz to 30 GHz i.e. from radio-frequencies to the infrared
are being used to process food. Depending on the chosen frequency and the particular
design of the applicator, treatment by electromagnetic energy at different wavelengths
has distinct features. For example, in microwave ovens electromagnetic waves with
centimeters wavelength freely propagate and are absorbed by solid or liquid phase food
products. The principle of microwave heating is that the changing electrical field
interacts with the molecular dipoles and charged ions. The heat generated by the
molecular rotation is due to friction of this motion.
The influence of microwave energy on chemical or biochemical reactions is strictly
thermal. The microwave energy quantum is given by the usual equation W = h ν.
Within the frequency domain of microwaves and hyper-frequencies (300 MHz - 300
GHz), the corresponding energies are respectively 1.24 10-6 eV - 1.24 10-3 eV. These
energies are much lower than the usual ionisation energies of biological compounds
(13.6 eV), of covalent bond energies like OH (5 eV), hydrogen bonds (2 eV), Van der
Waals intermolecular interactions (lower than 2 eV) and even lower than the energy
associated to Brownian motion at 370C (2.7 X 10-3eV). From this scientific point of
view, direct molecular activation of microwaves should be excluded. Some kind of step
by step accumulation of the energy, giving rise to a high-activated state should be
totally excluded due to fast relaxation. Like Peterson wrote in many of his articles: "The
question and the debate of the non thermal effect of microwave give a lot of damage for
the reputation of this technology and its application in industry". Microwaves are only
absorbed by dipoles, transforming their energy into heat.
Ref: http://www.mdpi.net/ecsoc-5/e0017/e0017.htm
~
~
~
~
E( z , t ) = E0ei (kz −ωt ) , B( z , t ) = B 0ei (kz −ωt )
( )
( )
( )
~
~
~
~
E0 = E0 x xˆ + E0 y yˆ + E0 z zˆ
r
1. ∇ ⋅ E = 0
∂
∂
∂
Ex + E y + Ez = 0
∂x
∂y
∂z
include direction, amplitude, phase;
vector phasor
[( )
]
[( )
]
]= 0
~
Æ E0 z = 0
[( )
]
∂ ~
∂ ~
∂ ~
E0 x ei (kz −ωt ) +
E0 y ei (kz −ωt ) +
E0 z ei (kz −ωt ) = 0
∂x
∂y
∂z
(E~ ) ∂∂z [e (
i kz − ωt )
0 z
( )
( )
~
The same as B 0
z
=0
r
r
∂B
2. ∇ × E = −
∂t
xˆ
yˆ
zˆ
∂
∂
∂
~
~
= − xˆik E0 y ei (kz −ωt ) + yˆ ik E0 x ei (kz −ωt )
∂x
∂y
∂z
~
~
i ( kz − ωt )
i ( kz −ωt )
E0 x e
E0 y e
0
~
~
= iω xˆ B 0 x ei (kz −ωt ) + yˆ B 0 y ei (kz −ωt )
( )
( )
( )
[( )
( )
( )
~
k E0
y
( )
~
= −ω B 0
( )
Æ
x
]
( )
~
and k E 0
( )
( )
x
( )
~
= ω B0
y
( )
( )
(( )
( ) )
k ~
k ~
k
~
~
~
~
~
B 0 = B 0 x xˆ + B 0 y yˆ = − E0 y xˆ + E0 x yˆ = zˆ × E0 x xˆ + E0 y yˆ
=
k
ω
ω
ω
ω
~
zˆ × E0
r
r
E and B are in phase and mutually perpendicular.
r
1 r
B0 = E0
c
r
r
If E points in the x direction, then B points in the y direction:
1~
~
~
~
E( z , t ) = E0ei (kz −ωt ) xˆ , B( z , t ) = E0ei (kz −ωt ) yˆ
c
1
real part: E( z , t ) = E0 cos(kz − ωt + δ )xˆ , B( z , t ) = E0 cos(kz − ωt + δ ) yˆ
c
r
let n̂ be the polarization vector and k be the propagation vector
r
transverse wave Æ nˆ ⋅ k = 0
1~ rr
1
~
~ rr
~
~
E(r, t ) = E0ei (k ⋅r −ωt )nˆ , B(r, t ) = E0ei (k ⋅r −ωt ) kˆ × nˆ = kˆ × E
c
c
r r
r r
1
real part: E(r, t ) = E0 cos k ⋅ r − ωt + δ nˆ , B(r, t ) = E0 cos k ⋅ r − ωt + δ kˆ × nˆ
c
( )
(
)
(
)(
)
9.2.3 Energy and Momentum in EM Waves
1⎛
B2 ⎞
B2
E
⎟⎟ ( ε 0 E 2 =
Æ B = ε 0 µ0 E = )
u = ⎜⎜ ε 0 E 2 +
2⎝
µ0 ⎠
µ0
c
As the wave travels, it carries the energy u = ε 0 E 2 = ε 0 E02 cos 2 (kz − ωt + δ ) along
with it. The energy flux transported by the fields is given by the Poynting vector.
Poynting vector: the energy flux density (energy per unit area, per unit time)
(
r 1 r r
S=
E×B
µ0
)
Taking electromagnetic fields as E( z, t ) = E0 cos(kz − ωt + δ )xˆ and
1
B(z , t ) = E0 cos(kz − ωt + δ ) yˆ , the energy flux density is
c
r
1 2
S=
E0 cos 2 (kx − ωt + δ )zˆ = cε 0 E02 cos 2 (kx − ωt + δ )zˆ = cuzˆ .
cµ0
r
r S
The momentum density stored in the field is P = 2 .
c
For monochromatic plane waves, we have the momentum density of
r u
P = zˆ .
c
Any macroscopic measurements will encompass many cycles of the wave oscillations,
so we want to know the average value.
Average over a period:
1
u = ε 0 E02 cos 2 (kz − ωt + δ ) = ε 0 E02
2
S =
⎛ ε E2 ⎞
E
cε
1 1
1 1
E0 B0 =
E0 0 = 0 E02 = c × ⎜⎜ 0 0 ⎟⎟
c
2 µ0
2 µ0
2
⎝ 2 ⎠
P =
1
ε 0 E02
2c
⎛ ε E2 ⎞
I ≡ S = c × ⎜⎜ 0 ⎟⎟ velocity * energy density per unit volume
⎝ 2 ⎠
When light falls on a perfect absorber, it delivers its momentum to the surface. In a
r r
time ∆t the momentum transfer is ∆p = PAc∆t , so the radiation pressure is
P=
r
r 1
1 ∆p
I
= cP = ε 0 E02 = .
2
c
A ∆t
What about the light pressure on a perfect reflector?
Exercise: 9.10