Chapter 5: Waves
23
3.
E
z
and
B
z
are zero everywhere: the Transversal electromagnetic mode (TEM). Than holds:
k
=
±
ω
√
ε
μ
and
v
f
=
v
g
, just as if here were no waveguide. Further
k
∈
IR
, so there exists no cutoff
frequency.
In a rectangular, 3 dimensional resonating cavity with edges
a
,
b
and
c
the possible wave numbers are given
by:
k
x
=
n
1
π
a
,
k
y
=
n
2
π
b
,
k
z
=
n
3
π
c
This results in the possible frequencies
f
=
vk/
2
π
in the cavity:
f
=
v
2
n
2
x
a
2
+
n
2
y
b
2
+
n
2
z
c
2
For a cubic cavity, with
a
=
b
=
c
, the possible number of oscillating modes
N
L
for longitudinal waves is
given by:
N
L
=
4
π
a
3
f
3
3
v
3
Because transversal waves have two possible polarizations holds for them:
N
T
= 2
N
L
.
5.6
Nonlinear wave equations
The
Van der Pol
equation is given by:
d
2
x
dt
2

εω
0
(1

β
x
2
)
dx
dt
+
ω
2
0
x
= 0
β
x
2
can be ignored for very small values of the amplitude. Substitution of
x
∼
e
i
ω
t
gives:
ω
=
1
2
ω
0
(
i
ε
±
2
1

1
2
ε
2
)
. The lowestorder instabilities grow as
1
2
εω
0
. While
x
is growing, the 2nd term becomes larger
and diminishes the growth. Oscillations on a time scale
∼
ω

1
0
can exist. If
x
is expanded as
x
=
x
(0)
+
ε
x
(1)
+
ε
2
x
(2)
+
· · ·
and this is substituted one obtains, besides periodic,
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 Spring '10
 Ye
 Energy, Trigraph, dx dt

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