Solution for HW #7
Problem 1 (G 9.25)
ω⎡
fj ⎤
Nq 2
k = ⎢1 +
⎥
∑
c ⎢⎣ 2mε 0 j ω 2j − ω 2 ⎥⎦
According to (9.169),
dω ⎛ dk ⎞
If damping is neglected, the group velocity is vg =
=⎜
⎟
dk ⎝ dω ⎠
−1
fj
2ω f j ⎫⎪⎤
dk 1 ⎡
Nq 2 ⎧⎪
= ⎢1 +
+
ω
⎨∑
∑j (ω 2 − ω 2 )2 ⎬⎥
d ω c ⎢⎣ 2mε 0 ⎪⎩ j ω 2j − ω 2
j
⎭⎪⎦⎥
ω 2j + ω 2 ⎤
1⎡
Nq 2
= ⎢1 +
∑ f j (ω 2 − ω 2 )2 ⎥ ⇒
c ⎢⎣ 2mε 0 j
⎥⎦
j
vg =
c
ω 2j + ω 2
Nq
1+
∑ f j (ω 2 − ω 2 )2
2mε 0 j
j
2
≤c
Because the second term in the denominator is greater than 0
On the other hand, the phase velocity v =
> c or < c , depending on ω
Problem 2 (G 9.19)
a) Maxwell equations in a metal is
c
2
Nq
1
fj 2
1+
∑
2mε 0 j
(ω j − ω 2 ) 2
can be both
∂2E
∂E
∇ E = με 2 + μσ
∂t
∂t
∂2B
∂B
∇ 2 B = με 2 + μσ
∂t
∂t
2
These equations are satisfied by plane waves with complex wave number
E ( x , t ) = E 0 e i ( kx − ω t )
B ( x , t ) = B 0 e i ( kx − ω t )
where k = μεω + iμσω
2
2
The text book implies that
frequencies
ω≥
σ is to be treated as a frequency-independent for
1
τ
1
σ ⎞2
⎛
k = μεω 2 + iμσω = μεω ⎜1 + i
ωε ⎟⎠
⎝
Since we were told that
σ
ωε
1 , we can expand the square root using the
binomial formulas (or Taylor Expansion)
1
1 + x ≈ 1 + x for x
2
⎛ 1 σ ⎞
k ≈ μεω ⎜1 + i
⎟
⎝ 2 ωε ⎠
The imaginary part of k , ki , is give by ki =
The skin depth is d =
For pure water,
∴
1 2
=
ki σ
σ
σ μ
με =
2ε
2 ε
ε
μ
ε = 80.1ε 0 , μ ≈ μ0 , σ =
1
2.5 × 105 Ω ⋅ m
80.1 × (8.85 × 10−12 )
d = 2 × (2.5 × 10 ) ×
= 1.19 × 104 m
−7
4π × 10
5
1
This is a huge distance ~10km!
We need to understand down to what frequency this result is valid. The high ω
condition means
σ
1
=
rad / s = 5.6 × 103 rad / s
5
−12
ε 2.5 × 10 × 80.1 × (8.85 × 10 )
ω
or
ν=
ω
= 0.89 × 103 Hz , which is at the lower end of the RF range.
2π
b) For a good conductor, neglected the first term inside the square root in the
1
equation for k , therefore k ≈ ( iσμω ) 2
i
π
Writing i = e 2 , thus we can get
i
π
i = e 4 = cos
π
4
+ i sin
π
4
=
2
2
+i
2
2
1
2
( μσω ) 2 , at the same time, the real part of k , k = ki
Therefore, ki =
2
The wavelength in a metal is
Hence, d =
λ=
2π
k
λ
2π
In plugging the number, as suggested by the book, we have to be careful. The
book
suggests
using
σ ≈ 107
1
Ω⋅m
at
the
visible
range
frequency
ω = 1015 rad / s .
If do that, we find
d=
2
μσω
=
2
≈ 1.3 × 10−8 m = 13nm
15
4π × 10
→ Text book answer!
In fact, the frequency dependence of the conductivity in the visible range is by no
means negligible. Recalling the Drude formula for the conductivity
σ=
σ0
1 − iωτ
ne2τ
is the conductivity at ω = 0 . We see that in the high
m
frequency limit ( ωτ
1)
where
σ0 =
σ =i
Assuming that still
σ
ωε
σ0
ωτ
1 , we neglect the first term in the expression for k .
It turns out to be
1
2
1
2
σ0 ⎞
⎛ σ ⎞
⎛
⎛ σ0 ⎞
k ≈ μεω ⎜ i
⎟ = μεω ⎜ i ⋅ i
⎟ = i με ⎜ ⎟
⎝ ωε ⎠
⎝ ωτωε ⎠
⎝ τε ⎠
1
2
1
If
μ , ε = μ0 , ε 0 ,
1 ⎛ σ ⎞2
k ≈i ⎜ 0 ⎟
c ⎝ τε ⎠
σ 0 ne2τ ne2
=
=
= ω p2 , where ω p is the plasma frequency.
Recall that
ε 0τ mε 0τ mε 0
Then ki ≈
In a metal,
ωp
c
, or d =
c
ωp
( Notice that d does not depend on the frequency!)
ω p ≈ 3 × 1015 rad / s , so
3 × 108
d≈
m ≈ 10−7 m ∼ 100nm
15
3 × 10
which is by a factor of 10 larger than the textbook answer!
Problem 2 (G 9.28)
a = 2.28cm , b = 1.01cm
According to Eqs. (9.184, 9.185), the discrete values of k x and k y are k x =
mπ
a
,
ky =
nπ
b
,
m, n = 0,1,2,3...
The z-component of the wave number is
2
2
2
2
⎡
⎛ω ⎞
⎛ω ⎞
⎛n⎞ ⎤
2
2
2 ⎛m⎞
k z = ⎜ ⎟ − k x − k y = ⎜ ⎟ − π ⎢⎜ ⎟ + ⎜ ⎟ ⎥
⎝c⎠
⎝c⎠
⎣⎢⎝ a ⎠ ⎝ b ⎠ ⎦⎥
2
Since k z must be real, the frequency of a propagating model must satisfy the
2
⎛m⎞ ⎛n⎞
condition ω > π c ⎜ ⎟ + ⎜ ⎟
⎝ a ⎠ ⎝b⎠
or connecting to
ν=
2
ω
(because the problem gives frequency in Hz)
2π
2
2
c ⎛m⎞ ⎛n⎞
Thus, ν >
⎜ ⎟ + ⎜ ⎟ = ν nm
2 ⎝ a ⎠ ⎝b⎠
Substituting some numbers, we obtain
ν 10 = 0.66 × 1010 Hz ;
ν 20 = 1.32 × 1010 Hz
ν 30 = 1.97 × 1010 Hz ;
ν 01 = 1.94 × 1010 Hz
ν 02 = 2.97 × 1010 Hz ;
ν 11 = 1.62 × 1010 Hz
Only four frequencies ----
ν 10 ,ν 20 ,ν 01 and ν 11 are below the 1.70 × 1010 Hz .
Therefore, only those four frequencies can be excited.
Problem 2 (G 9.29)
From 9.11
S =
1
Re E × B∗
2 μ0
(
)
where E = E0 exp[i ( kz − ωt )] , B = B0 exp[i ( kz − ωt )]
and B = B0 exp[ −i ( kz − ωt )]
∗
In general, Eq. (9.180) is telling us that
Ex =
⎛ ∂Ez
i
∂B ⎞
k
+ω z ⎟
2
2 ⎜
(ω / c ) − k ⎝ ∂x
∂y ⎠
Ey =
⎛ ∂Ez
i
∂B ⎞
k
−ω z ⎟
2
2 ⎜
(ω / c) − k ⎝ ∂y
∂x ⎠
Bx =
⎛ ∂Bz ω ∂Ez ⎞
i
−
⎜k
⎟
(ω / c) 2 − k 2 ⎝ ∂x c 2 ∂y ⎠
By =
⎛ ∂Bz ω ∂Ez ⎞
i
k
+
⎟
2
2 ⎜
(ω / c) − k ⎝ ∂y c 2 ∂x ⎠
In a TE model, E z = 0 while [Eq. (9.186)]
⎛ mπ x ⎞ ⎛ nπ y ⎞
Bz = B0 cos ⎜
⎟ cos ⎜
⎟
⎝ a ⎠ ⎝ b ⎠
Substituting Bz in to equations for E x , E y Bx B y , we get
Bx = −
ik
⎛ mπ
2
2 ⎜
(ω / c) − k ⎝ a
⎞
⎛ mπ x ⎞
⎛ nπ y ⎞
⎟ B0 sin ⎜
⎟ ⋅ cos ⎜
⎟
⎠
⎝ a ⎠
⎝ b ⎠
By = −
ik
⎛ nπ
2
2 ⎜
(ω / c) − k ⎝ b
⎞
⎛ mπ x ⎞
⎛ nπ y ⎞
⎟ B0 cos ⎜
⎟ ⋅ sin ⎜
⎟
⎠
⎝ a ⎠
⎝ b ⎠
Ex = −
iω
⎛ nπ
⎜
(ω / c) 2 − k 2 ⎝ b
⎞
⎛ mπ x ⎞
⎛ nπ y ⎞
E
cos
sin
⋅
⎟ 0
⎜
⎟
⎜
⎟
⎠
⎝ a ⎠
⎝ b ⎠
iω
⎛ mπ
2
2 ⎜
(ω / c) − k ⎝ a
⎞
⎛ mπ x ⎞
⎛ nπ y ⎞
⎟ B0 sin ⎜
⎟ ⋅ cos ⎜
⎟
⎠
⎝ a ⎠
⎝ b ⎠
Ey =
Then,
S =
{(
)}
1
Re E y Bz∗ − Ez By∗ xˆ + Ez By∗ − E y Bz∗ yˆ + Ex By∗ − E y Bx∗ zˆ
2 μ0
) (
) (
1
iπω B02 ⎛ m ⎞ ⎛ mπ x ⎞
⎛ mπ x ⎞ 2 ⎛ nπ y ⎞
Re{
=
⎟ sin ⎜
⎟ cos ⎜
⎟ cos ⎜
⎟ xˆ
2
2 ⎜
2μ0
(ω / c) − k ⎝ a ⎠ ⎝ a ⎠
⎝ a ⎠
⎝ b ⎠
iπω B02 ⎛ n ⎞ 2 ⎛ mπ x ⎞ ⎛ nπ y ⎞
⎛ nπ y ⎞ ˆ
+
⎟ cos ⎜
⎟ sin ⎜
⎟ cos ⎜
⎟y
2
2 ⎜
(ω / c) − k ⎝ b ⎠
⎝ a ⎠ ⎝ b ⎠
⎝ b ⎠
2
⎡⎛ n ⎞ 2
⎤
⎛m⎞
2 ⎛ mπ x ⎞
2 ⎛ nπ y ⎞
2 ⎛ mπ x ⎞
2 ⎛ nπ y ⎞
cos
sin
sin
cos
+
+
⎢⎜ ⎟
⎜
⎟
⎜
⎟ ⎥ zˆ}
⎜
⎟
⎜
⎟ ⎜ ⎟
2
2 2
b
a
b
a
a
b
⎝
⎠
⎝
⎠
⎝
⎠ ⎦⎥
⎝
⎠
⎝
⎠
⎝
⎠
⎡⎣(ω / c) − k ⎤⎦ ⎣⎢
iπω B02
2
1
ω kπ 2 B02
⎛n⎞
⎛ mπ x ⎞ 2 ⎛ nπ y ⎞
{⎜ ⎟ cos 2 ⎜
=
⎟ sin ⎜
⎟
2
2 μ0 ⎡(ω / c) 2 − k 2 ⎤ ⎝ b ⎠
⎝ a ⎠
⎝ b ⎠
⎣
⎦
2
⎛m⎞
⎛ mπ x ⎞ 2 ⎛ nπ y ⎞
+ ⎜ ⎟ sin 2 ⎜
⎟ cos ⎜
⎟}
⎝a⎠
⎝ a ⎠
⎝ b ⎠
Averaging S
over the area of the wave guide with the help of the following
formulas
a
2 ⎛ mπ x ⎞
dx
sin
=
⎜
⎟
∫0
⎝ a ⎠ 2
a
and
a
2 ⎛ nπ y ⎞
dy
cos
=
⎜
⎟
∫0
⎝ a ⎠ 2
b
gives
∫
⎡⎛ n ⎞ 2 ⎛ m ⎞ 2 ⎤
1
ω kπ 2 B02
S ⋅ da =
ab ⎢⎜ ⎟ + ⎜ ⎟ ⎥
2μ0 ⎡(ω / c)2 − k 2 ⎤ 2 ⎢⎣⎝ b ⎠ ⎝ a ⎠ ⎥⎦
⎣
⎦
Also from (9.11)
⎫
1 ⎧
1
u = Re ⎨ε 0 E ⋅ E ∗ +
B ⋅ B∗ ⎬
4 ⎩
μ0
⎭
2
⎧⎪⎛ n ⎞2
⎫
⎛ m⎞
2 ⎛ mπ x ⎞
2 ⎛ nπ y ⎞
2 ⎛ mπ x ⎞
2 ⎛ nπ y ⎞ ⎪
=
+
cos
sin
sin
cos
⎨
⎜ ⎟
⎜
⎟
⎜
⎟ ⎜ ⎟
⎜
⎟
⎜
⎟⎬
4 ⎡(ω / c)2 − k 2 ⎤ 2 ⎪⎩⎝ b ⎠
⎝ a ⎠
⎝ a ⎠ ⎝a⎠
⎝ a ⎠
⎝ a ⎠ ⎪⎭
⎣
⎦
ε0
ω 2π 2 B02
2
1
k 2π 2 B02 ⎛ n ⎞
⎛ mπ x ⎞ 2 ⎛ nπ y ⎞
2
2 ⎛ mπ x ⎞
2 ⎛ nπ y ⎞
+
{B0 cos ⎜
[ ⎟ cos2 ⎜
⎟ cos ⎜
⎟+
⎟ sin ⎜
⎟
2
2 ⎜
4 μ0
⎝ a ⎠
⎝ a ⎠ (ω / c) − k ⎝ b ⎠
⎝ a ⎠
⎝ a ⎠
2
⎛m⎞
⎛ mπ x ⎞ 2 ⎛ nπ y ⎞
+ ⎜ ⎟ sin 2 ⎜
⎟ cos ⎜
⎟]}
⎝a⎠
⎝ a ⎠
⎝ a ⎠
Again, averaging
u over the area
∫
2
2 ⎫
⎧
⎡⎛ n ⎞ 2 ⎛ m ⎞ 2 ⎤
ab ⎪ ε 0
ω 2π 2 B02
1 2
1
k 2π 2 B02 ⎡⎛ n ⎞ ⎛ m ⎞ ⎤ ⎪
u da = ⎨
B0 +
⎢⎜ ⎟ + ⎜ ⎟ ⎥ +
⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎬
4 ⎪ 4 ⎡(ω / c)2 − k 2 ⎤ 2 ⎢⎣⎝ b ⎠ ⎝ a ⎠ ⎥⎦ 4 μ0
4 μ0 (ω / c)2 − k 2 ⎢⎣⎝ b ⎠ ⎝ a ⎠ ⎥⎦ ⎪
⎦
⎩ ⎣
⎭
This is the energy per unit length of the wave guide.
2
⎛ω ⎞
⎛ω ⎞
2
Using ⎜ ⎟ − k = ⎜ nm ⎟
⎝c⎠
⎝ c ⎠
2
2
2
,
1
⎛ n⎞ ⎛ m⎞ ⎛ω ⎞
ε 0 μ0 = 2 and ⎜ ⎟ + ⎜ ⎟ = ⎜ nm ⎟
c
⎝ b ⎠ ⎝ a ⎠ ⎝ πc ⎠
∫
ω kabc 2 2
S ⋅ da =
B0
2
8μ0ωnm
2
, it will
gives that
∫
The ration
ω 2 ab 2
u da =
B0
2
8μ0ωnm
∫ S ⋅ da = energy / time = length is equal to the group velocity
∫ u da energy / length time
∫ S ⋅ da = kc
∫ u da ω
2
Since k =
=
c
ω
2
ω 2 − ωnm
1
2
ω 2 − ωnm
, the group velocity is
c
⎛
ω
dω ⎛ dk ⎞
vg =
=⎜
⎟ = c ⎜⎜
2
2
dk ⎝ dω ⎠
⎝ ω − ωnm
−1
−1
2
⎞
ω 2 − ωnm
⎟ =c
⎟
ω
⎠
Thus
∫ S ⋅ da = v
∫ u da
g
(Q.E.D)
Bounce (G 9.28)
Inside the cavity, assuming that
E = E 0 ( x , y , z ) e − iω t , B = B0 ( x, y , z )e − iωt
⎧∇ ⋅ E0 = 0
⎪∇ ⋅ B = 0
⎪
0
⎨
⎪∇ × E0 = iω B0
⎪⎩∇ × B0 = −iω E0
The set of Maxwell Equation is supplemented by the boundary conditions
E = 0 and B⊥ = 0
to be satisfied at all surfaces.
Applying second curl to the Faraday Law (and skipping the “0” subscripts),
ω
⎛ ω⎞
∇ × (∇ × E ) = ∇(∇ ⋅ E ) − ∇ E = iω∇ × B = iω ⎜ −i 2 ⎟ E = 2 E
c
⎝ c ⎠
2
2
∴
∇ E=−
2
ω2
c2
E
or, in components,
⎧ 2
ω2
⎪∇ Ex = − c 2 Ex
⎪
ω2
⎪ 2
⎨∇ E y = − 2 E y
c
⎪
⎪ 2
ω2
∇
=
−
E
Ez
⎪
z
2
c
⎩
Seek for a solution in following form
E x = X ( x )Y ( y ) Z ( z )
Thus, X '' YZ + XY '' Z + XYZ '' = −
ω2
c2
XYZ
X '' Y '' Z ''
ω2
+
+
=− 2
X
Y
Z
c
X ''
Y ''
Z ''
= −k y2 ,
= −k z2
= −k x2 ,
X
Y
Z
where k + k + k =
2
x
2
y
2
z
ω2
c2
According to the boundary condition E = 0 ,
E x ( x,0, z ) = E x ( x, b, z ) = E x ( x, y ,0) = E x ( x, y, d )
Therefore,
⎛ nπ y ⎞ ⎛ lπ z ⎞
Ex ( x, y, z ) = [ A sin(k x x) + B cos( k x x) ] sin ⎜
⎟ sin ⎜
⎟
⎝ b ⎠ ⎝ d ⎠
The same goes for Ey and Ez,
⎛ mπ x ⎞ ⎛ lπ z ⎞
E y ( x, y, z ) = ⎡⎣C sin(k y y ) + D cos(k y y ) ⎤⎦ sin ⎜
⎟ sin ⎜
⎟
⎝ a ⎠ ⎝ d ⎠
⎛ mπ x ⎞ ⎛ nπ y ⎞
Ez ( x, y, z ) = [ E sin(k z z ) + F cos(k z z )] sin ⎜
⎟ sin ⎜
⎟
⎝ a ⎠ ⎝ b ⎠
We still need to satisfy the Gauss’s Law at any point inside the cavity.
∇⋅E = 0=
∂E x ∂E y ∂E z
+
+
=0
∂x
∂y
∂z
⇒
k x [ A cos(k x x) − B sin(k x x)] sin(k y y )sin( k z z ) + k y ⎡⎣C cos(k y y ) − D sin(k y y ) ⎤⎦
× sin(k x x)sin(k z z ) + k z [ E cos( k z z ) − F sin( k z z ) ]sin( k x x)sin( k y y ) = 0
Since this has to be satisfied at any point, it also has to be satisfied at x = 0 , where the
last two term in ∇ ⋅ E vanish and the remainder is
k x A sin(k y y )sin(k z z ) = 0
∴A=0
Likewise, putting in y = 0, z = 0 , thus,
E = B cos(k x x )sin(k y y )sin(k z z ) xˆ + D sin(k x x)cos(k y y )sin(k z z ) yˆ + F sin(k x x)
× sin(k y y )cos(k z z ) zˆ
∇ ⋅ E = 0 is how satisfied provided that k x B + k y D + k z F = 0
The frequency satisfies
ω2
2
2
⎛ mπ ⎞ ⎛ nπ ⎞ ⎛ lπ ⎞
k
k
k
=
+
+
=
⎜
⎟ +⎜
⎟ +⎜ ⎟
c2
a
b
⎝
⎠ ⎝
⎠ ⎝d ⎠
2
x
2
y
2
z
2
ωmnl
2
⎛ mπ ⎞ ⎛ nπ ⎞ ⎛ lπ ⎞
=c ⎜
⎟ +⎜
⎟ +⎜ ⎟
⎝ a ⎠ ⎝ b ⎠ ⎝d ⎠
The magnetic field is obtained from E using
i
ω
c
B = ∇× E
I will not write down the component of B
2
2