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Boundary Value Problems
• Scalar Expansions as Solutions of the Laplace’s Equation:
- Laplace’s Equation: ∇2 Φ = 0.
- Coefficients: A, B, C, D, E, F are coefficients, where α, β, m, l, k are the dummies.
- 2-D Box:
Φ = Σ∞
α (Aα cos αx + Bα sin αx)(Cα cosh αy + Dα sinh αy).
Boundary Conditions: Φ(x = 0) = Φ(x = a) = 0 gives,



x
C
cosh
y
+
D
sinh
y
.
Φ = Σ∞
sin
n
n
n=0
a
a
a
- 2-D Polar:
α
−α
Φ = (A0 + B0 ln ρ)(C0 + D0 φ) + Σ∞
)(Cα cos αφ + Dα sin αφ).
α (Aα ρ + Bα ρ

Boundary Conditions: Φ(φ = 0) = Φ(φ = 2π) gives,
m
−m
Φ = A0 + B0 ln ρ + Σ∞
)Cm cos mφ.
m=1 (Am ρ + Bm ρ

- 3-D Box:

Φ = Σ∞
α Σβ (Aαβ cos αx + Bαβ sin αx)(Cαβ cos βy + Dαβ sin βy)(Eαβ cosh

p
p
α2 + β 2 z + Fαβ sinh α2 + β 2 z).

Boundary Conditions: Φ(x = 0) = Φ(x = a) = Φ(y = 0) = Φ(y = b) = 0 gives,
s 
s 


2

2 
2
2

Φ = Σn=0 Σm=0 sin
x sin
y Enm cosh
+
z + Fnm sinh
+
z .
a
b
a
b
a
b
- Spherical Harmonic:
l
l
−l−1
Φ = Σ∞
)Ylm (θ, φ).
l=0 Σm=−l (Alm r + Blm r

Boundary Conditions: azimuthal symmetry, i.e. Φ(φ) = Φ(0) ∀φ, gives, Legendre Function
l
−l−1
Φ = Σ∞
)Pl (cos θ).
l=0 (Al r + Bl r

- Ordinary Bessel Function:
Z ∞
Φ = Σ∞
dk[Am (k)Jm (kρ) + Bm (k)Nm (kρ)][Cm (k) cos mφ + Dm (k) sin mφ][Em (k)ekz + Fm (k)e−kz ].
m=0
0

Boundary Conditions: Φ(ρ = 0) and Φ(z → ∞) finite, and Φ(ρ = a) = 0 gives,
xmn
xmn

ρ)[Cmn cos mφ + Dmn sin mφ]e− a z ,
Φ = Σ∞
m=0 Σn=1 Jm (
a
where Jm (xmn ) = 0.
- Modified Bessel Function:
Z ∞
Φ = Σ∞
dk[Am (k)Im (kρ) + Bm (k)Km (kρ)][Cm (k) cos mφ + Dm (k) sin mφ][Em (k) cos kz + Fm (k) sin kz].
m=0
0

Boundary Conditions: azimuthal symmetry, Φ(ρ → ∞) finite, andΦ(z = 0) = Φ(z = L) = 0 gives,





ρ
cos

sin
z
.
Φ = Σ∞
Σ
A
K
m=0 n=1 mn m
L
L

• Orthogonality:
a

Z

n0 π
x sin
xdx =
a
a

sin
0

Z

Z

a

Z

cos
0

0

Yl,m
(θ, φ)Yl,m (θ, φ)d(cos θ)dφ = δl,l0 δm,m0

π

0
−1

Z

n0 π
a
x cos
xdx = δn,n0
a
a
2

Pl (cos θ)Pl0 (cos θ)d cos θ =
Z1 ∞

2
δl,l0
2l + 1

1
Jm (kρ)Jm (kρ)ρdρ = δ(k − k 0 )
k
0
 


Z α
xmn0
a2
xmn
ρ Jm
ρ ρdρ = [Jm+1 (xmn )]2 δn,n0
Jm
a
a
2
0
• Scalar 1/r Expansions:
- 2-D Box:
1
1
=
0
|~x − ~x |
π

Z

~

d2 k

eik·~x
1
=
2
~
π
|k|

0

Z
dαdβ

0

eiα(x−x )+iβ(y−y )
=
α2 + β 2

0

Z

0

eiα(x−x ) e−α(y−y )
α

- 2-D Polar:
ln

 m
1
1
1 ρ&lt;

=
ln
+
Σ
cos m(φ − φ0 )
m=1
|~x − ~x0 |
ρ&lt;
m ρ&gt;

- 3-D Box:
1
1
=
|~x − ~x0 |
2π 2

Z

~

eik·~x
1
d k
=
2
~

|k|
3

Z

√ 2 2
0
0
0
eiα(x−x ) eiβ(y−y ) e− α +β (z−z )
dαdβ
α2 + β 2

- Spherical Harmonic:
 l 
r&lt;
1

l
Ylm
= Σl=0 Σm=−l
(θ0 , φ0 )Ylm (θ, φ)
l+1
0
|~x − ~x |
2l + 1 r&gt;
- Legendre Function:
 l 
r&lt;
1

=
Σ
Pl (cos θ0 )Pl (cos θ)
l=0
l+1
|~x − ~x0 |
r&gt;
- Ordinary Bessel Function:

Z ∞ 
1
1
0

0
0
dk
=
J0 (kρ)J0 (kρ ) + Σm=1 Jm (kρ)Jm (kρ ) cos m(φ − φ ) e−k(z&gt; −z&lt; )
|~x − ~x0 |
2
0

Z ∞ 

0 im(φ−φ0 )
= Σm=−∞
dk Jm (kρ)Jm (kρ )e
e−k(z&gt; −z&lt; )
0

- Modified Bessel Function:


Z
1
4 ∞
1

0
=
dk
I0 (kρ&lt; )K0 (kρ&gt; ) + Σm=1 Im (kρ&lt; )Km (kρ&gt; ) cos m(φ − φ ) cos k(z − z 0 )
|~x − ~x0 |
π 0
2

Z ∞ 
4 ∞
im(φ−φ0 )
= Σm=−∞
dk Im (kρ&lt; )Km (kρ&gt; )e
cos k(z − z 0 )
π
0

Green’s Functions
• Basics:
- Electric Potential in Green’s Function (Jackson Eq. 1.42):
Z
I
1
∂G(~x, ~x0 ) 2 0
1
∂Φ(~x0 )
Φ(~x) =
G(~x, ~x0 ) − Φ(~x0 )
d a
ρ(~x0 )G(~x, ~x0 )d3 x0 +
0
4πε V
4π ∂V ∂n
∂n0
- Reciprocity Theorem:
Z
Z
0 0 3
ρ(~x)Φ (~x )d x +
V

0

0

2

Z

σ(~x)Φ (~x )d a =

∂V

Z

0

0

Z

3

ρ (~x)Φ(~x )d x +
V

ρΦ0 +

V

Z
∂V

σΦ0 =

Z
V

σ 0 (~x)Φ(~x0 )d2 a

∂V

ρ0 Φ +

Z

σ0 Φ

∂V

- Method to Find Green’s Functions: propose a geometric shape. Make the shells all grounded
conductors and the spaces all vacuum. Impose a test charge in the region of question. Divid the
regions by the shells according to the coordinate chosen, and also divid the region across the test
charge. Then solve for the potential in the region of question (in the &gt; and &lt; manner, where the ~x0
goes for the source, i.e. the test charge) as boundary value problem. Note the the only discontinuity
for the boundary conditions is on the boundary where the test charge is located. Finally, plug in
q = 4πε0 .
1
- IMPORTANT Method: imposing the source term with |~x−~
x0 | expansion and then matching the
terms of the expansion (matches the direction as well for the vector potentials) for the boundary
value problem, is often easier than making another boundary across the source terms.
- Non-orthogonality Conditions: should be applied to the boundary that decides “&gt;” and “&lt;”,
e.g. sinh and cosh are not orthogonal and so are used when cylinder is divided in z-direction, as in
ordinary Bessel, while modified Bessel divids the cylinder in ρ-direction.
• Some Green’s Functions: “&gt;” and “&lt;” are comparison of ~x and ~x0 and they symbolizes the
”jumps” in Green’s functions,
- Inside the x-y Direction of a Box:

nπ 0

mπ 0 sinh
16π

sin
x sin
x sin
y sin
y
G(~x, ~x0 ) = Σ∞
n=1 Σm=1
ab
a
a
b
b

q


nπ 2
a

+
q

nπ 2
a


mπ 2
z&lt;
b
+


mπ 2
b

sinh
sinh

q


nπ 2
a

+


mπ 2
(c
b

q


nπ 2
a

+


mπ 2
c
b

− z&gt; )

- The Intermediate Region of Concentric Sphere (a &lt; b):


 ∗ 0 0
l
r&gt;

a2l+1
1
Ylm (θ , φ )Ylm (θ, φ)
0

l
l
G(~x, ~x ) = Σl=0 Σm=−l
r&lt; − l+1
− 2l+1
.
l+1
2l + 1
b
1 − ( ab )2l+1
r&lt;
r&gt;
- Inside the Radius of a Finite Cylinder with Radius “a” and Height “L” (Ordinary Bessel):


xmn 0
0 sinh xmn (L − z ) sinh xmn z
8 Jm ( xmn
&gt;
&lt;

a
a
a ρ)Jm ( a ρ ) 2 cos m(φ − φ )

G(~x, ~x0 ) = Σ∞
Σ
.
m=0 n=1
axmn
(Jm+1 (xmn ))2
1 + δm,0
sinh xmn
a L
- Inside the Height of a Finite Cylinder with Radius “a” and Height “L” (Modified Bessel):






2 cos m(φ − φ0 )
nπ 
nπ 0 
0

∞ 8 Im ( L ρ) Km L a Im L ρ&gt; − Km L ρ&lt; Im L a
G(~x, ~x ) = Σm=0 Σn=1
sin
z sin
z .

L
Im ( L a)
1 + δm,0
L
L

.

Paralleling Electrostatics and Magnetostatics
Electrostatics
~ = −∇Φ
~
E

~ ·D
~ =ρ

~ = −∇Φ
~ M
H
Z ~ 0
µ
J(~x ) 3 0
~
A=
d x

|~x − ~x0 |
~ −M
~
~ = 1B
H
µ0
~ ×H
~ = J~

~2 −D
~ 1) · n
~2 − E
~ 1) = 0
(D
ˆ = σ, n
ˆ × (E

~ ×M
~ = J~m , M
~ ×n
~m

ˆ=K
~2 − B
~ 1) · n
~2 − H
~ 1) = K
~
(B
ˆ = 0, n
ˆ × (H

~ = εE
~ = (1 + χe )ε0 E
~
D
Z
p~ = ~x0 ρ(~x0 )d3 x0
Z
p~ = P~ (x~0 )d3 x0

~ = (1 + χm ) 1 B
~
~ = 1B
H
µ
µ0
Z
1
~ x0 )d3 x0
m
~ =
~x0 × J(~
2
Z
~ (x~0 )d3 x0
m
~ = M

n(~
p·n
ˆ − p~)
~ x) = 1 3ˆ
E(~
4πε |~x − ~x0 |3
1 (~
p · ~x)
Φ(~x) =
4πε |~x − ~x0 |3
Z
~ x)d3 x
F~ = ρ(~x)E(~

n( m
~ ·n
ˆ − m)
~
~ x) = µ 3ˆ
B(~
4π |~x − ~x0 |3
~ × ~x)
~ x ) = µ (m
A(~
4π |~x − ~x0 |3
Z
~ x) × B(~
~ x)d3 x
F~ = J(~
Z

~ x) × B(~
~ x) d3 x
~τ = ~x × J(~

~ p · E)
~
F~ = ∇(~

~ m
~
F~ = ∇(
~ · B)
~
~τ = m
~ ×B
Z
1
~ x) · A(~
~ x)d3 x
W =
J(~
2
1
W = Σi Li Ii2 + Σi Σj Mij Ii Ij
2

Φ=

1
4πε

Z

ρ(~x0 ) 3 0
d x
|~x − ~x0 |

~ = ε0 E
~ + P~
D

linear di-ic
dipole

far

force

work

W =
W =
W =

linear di-ic

Magnetostatics
~ =∇
~ ×A
~
B

W =

Z
1
ρ(~x)Φ(~x)d3 x
2
1
Σi Σj Pij Qi Qj
2
1
Σi Σj Cij Vi Vj
2Z
1
~ x) · D(~
~ x)d3 x
E(~
2

W =

1
2

Z

~ x) · B(~
~ x)d3 x
H(~

Others:
H
- Current Density and Current: J~ = ρ~v , I = J~ · d~a, where ~a is the cross section.
R ∇
H nˆ ·M
~ 0 ·M
~ (~
~ (~
x0 ) 3 0
x) 2 0
1
1
- Scalar Potential Under Permanent Magnetization: ΦM (~x) = − 4π
|~
x−~
x0 | d x + 4π ∂V |~
x−~
x0 | d a .
V
- For source terms, integrate over uniform distribution with delta functions to get the pre-factors.

• Maxwell’s Equations:
Coulomb:

~ ·D
~ =ρ

~ ·B
~ =0
NoMagMono: ∇

~
~ ×E
~ + ∂B = 0
∂t
~

~
~ ×H
~ − D = J.
Ampere: ∇
∂t

• Continuity Equation:
~ · J~ + ∂ρ = 0.

∂t
• Potentials:
~ =∇
~ ×A
~
B

~
~ = − ∂ A − ∇Φ.
~
E
∂t

• Relations:
- Propagation regions are usually sourceless. And without the closed boundary condition in any
certain direction (i.e. wave guides), the only spacial dependent of the field is the variable in the
propagation direction.
- NoMagMono/Faraday are homogeneous (source-unrelated), and are associated with gauge invariance.
- Coulomb/Ampere are inhomogeneous (source-related), and are associated with continuity equa~ and Φ by plugging Coulomb/Ampere into the continuity equation.
tion. Get the relation between A
- Wave equations of the fields are obtain by taking curl of Faraday/Ampere, and use
~ ×∇
~ ×~a = ∇(
~ ∇
~ ·~a) − ∇
~ 2~a. Notice that n is a relation between k and ω, which is derived through

fields’ wave equations.
- Wave equations of the potentials are obtained by taking divergence of Coulomb and taking curl
of Ampere.
~ and B
~ are obtained by Faraday/Ampere. The curl should be taken on which
- Relation between E
~ and B.
~
ever is given. This relation includes the perpendicularity of E
- To test the fields are transverse, first set up the propagation direction (in the exponential), then
use Coulomb/NoMagMono, gauge condition after plugging in the potentials, or the relation between
~ and B
~ to show that kˆ · E
~ = 0 or kˆ · B
~ = 0. For propagating wave in free space, B
~ and D
~ are
E
always transverse.
- The field are usually in the transverse direction, and the vector potential is in the same direction
as the electric field.
• Gauge Invariance:
~→A
~ + ∇Λ
~
A

Φ→Φ−

~
∂Λ
.
∂t

• Gauges:
Coulomb:

~ ·A
~=0

Lorentz:

~ ·A
~ + 1 ∂Φ = 0.

c2 ∂t

• Retarded Green’s Function:
G(+) (~x, t; x~0 , t0 ) =

x0 | 
δ t0 − [t − |~x−~
]
c
.
|~x − ~x0 |

[f (~x0 , t0 )]ret 3 0
d x,
|~x − ~x0 |
R
|~
x−~
x0 | 
0
0 0
0
~ J),
~ (Φ, ρ), and [f (~x0 , t0 )]
where (ψ, f ) = (A,
ret = δ t − [t − c ] f (~x , t )dt .
- Useful delta relation:
X δ(~x − ~xi )
∂F
δ(F (~x)) =
, F (~xi ) = 0.

Z

ψ(~x, t) =

∂x ~
x=~
xi

~ = |~x − ~x0 |
• Jefimenko: R
 ~ 0 0  
 ˆ
ˆ  ∂ρ(~x0 , t0 ) 
R
1 ∂ J(~
x ,t )
R
0 0
[ρ(~x , t )]ret +
− 2
d x
2
0
R
cR
∂t
∂t0
ret c R
ret




Z
ˆ
~ 0 0
ˆ
R
~ x, t) = µ0
~ x0 , t0 )] × R + ∂ J(~x , t )
B(~
d3 x0 [J(~
×
.
ret
2
0

R
∂t
ret cR

~ x, t) =
E(~

1
4πε0

• Poynting:
Z
~
Wby field = J~ · E

~ + ~v × B)
~
F~ = q(E

Z

3 0

du ~ ~ ~ ~
+ ∇ · S + J · E = 0,
dt

⇒ P~field =

1 ~ ~
~ · H)
~
(E · D + B
2
~≡E
~ ×H
~
S
~
~ ×H
~ =S
~g = µεE
c
u≡

Z
~g ,

d
⇒ (P~mech + P~field )α =
dt

I
Tαβ ηβ ,

Tαβ

E2 +
Bα Bβ
= ε(Eα Eβ +

µε
2

(energy)
(energy flow)
(momentum)
B2
µε

δαβ )

• Invariance:
~ P~ , D,
~ ∇
~
E,
~ M
~ ,H
~
B,

rank 1, odd vec

even time

rank 1, even vec

odd time

ρ
J~

rank 0, even scal

even time

rank 1, odd vec

odd time.

- Cross product does not vary the ranks, but dot product does.
- If rank, trans, time symmetry match, then such term can potentially be in the Taylor expansion.
- Continuously multiplying the scaler terms can always be absorbed into the coefficient of the Taylor
expansion.

(stress).

• Sourceless Maxwell’s Equations: describes how light passes through different media,
Coulomb:

~
~ ×E
~ + ∂B = 0
∂t
~

~ ×H
~ − D = 0.
Ampere: ∇
∂t

~ ·D
~ =0

~ ·B
~ =0
NoMagMono: ∇
~ =E
~ 0 ei~k·x−iωt , then B
~ 0 = √µεkˆ × E
~ 0.
If E
1
ω = kv, v = √ , n =
µε

r

µε

;ω=
= 2πf , k =
.
µ0 ε0
T
λ

• Time Average of the Real Part:
R

ω

0

cos2 ωtdt
1
= .
R 2π
2
ω
dt
0

1
~ ·n
~ ×H
~ ∗ ].
S
ˆ = Re[ˆ
n·E
2
• Polarization: relation between the transverse fields, suppose propagation in z direction,
- Transverse (linear):
~1 = x
ˆ, ~2 = yˆ.
- Circular:
~± = ~1 ± ~2 .
- Stoke: only need 3 s’s to completely specify the polarization.
from linear
~ 2 + |~2 · E|
~ 2
s0 = |~1 · E|

from circular
~ 2 + |~∗− · E|
~ 2
s0 = |~∗+ · E|

~ 2 − |~2 · E|
~ 2
s1 = |~1 · E|
~ ∗ (~2 · E)]
~
s2 = 2Re[(~1 · E)

~ ∗ (~∗ · E)]
~
s1 = 2Re[(~∗+ · E)

~ ∗ (~∗ · E)]
~
s2 = 2Im[(~∗ · E)

~ ∗ (~2 · E)]
~
s3 = 2Im[(~1 · E)

+

s3 =

|~∗+

~ 2 − |~∗− · E|
~ 2.
· E|

• Boundary Conditions: E for transmitted, E’ for transmitted, E” for reflected,
~ +D
~ 00 − D
~ 0) · n
(D
ˆ=0
~ +B
~ 00 − B
~ 0) · n
(B
ˆ=0
~ +E
~ 00 − E
~ 0) × n
(E
ˆ=0
~ +H
~ 00 − H
~ 0) × n
(H
ˆ=0

~ + εE
~ 00 − ε0 E
~ 0) · n
(linear di-ic) ⇒ (εE
ˆ=0
~ + ~k × E
~ 00 − k~0 × E
~ 0) · n
(~k × D
ˆ=0
~ +E
~ 00 − E
~ 0) × n
(E
ˆ=0
1
~ + 1 ~k × E
~ 00 − 1 ~k ×0 E
~ 0) × n
ˆ = 0.
( ~k × D
µ
µ
µ0

• Snell: i is the incident angle, and r is the refracted angle.
sin i
k0
n0
=
= , k 00 = −k.
sin r
k
n

• Rate: T + R = 1.
- Transmission Rate: T ≡

1 ~0
~ 0∗
2 |E ×H |
1 ~
~∗
2 |E×H |

~0|
|S
~
|S|

µ0 ε0
µ0

E~ 0

µε ~ .
E
µ
1

=
=

E~ 00
~ 00 |
.
- Reflection Rate: R ≡ |S|S|
~
~ = E
- Fresnel Equations: i is the incident angle, and r is the refracted angle.
~ ⊥ plane of incidence:
E

~ k plane of incidence:
E

E0 + E000 − E00 = 0
E0 cos i − E000 cos i − E00 cos r = 0
s
s
r
r
r
r
ε
ε 00
ε0 0
ε
ε 00
ε0 0
E0 cos r = 0
E =0
E0 cos i −
E0 cos i −
E0 +
E0 −
0
µ
µ
µ
µ
µ
µ0 0

E00
2n cos i

=
E0
n cos i + µµ0 n02 − n2 sin i

n cos i − µµ0 n02 − n2 sin i
E000

=
E0
n cos i + µµ0 n02 − n2 sin i

E00
=
E0

2nn0 cos i

µ 02
02
2
µ0 n cos i + n n − n sin i

µ 02
02
2
E000
µ0 n cos i − n n − n sin i

= µ
02
02
2
E0
µ0 n cos i + n n − n sin i

• Brewster’s Angle: R = 0.
ε = ε0 ,

only possible in E ⊥ plane of incidence

0

only possible in E k plane of incidence.

µ=µ,
• Total Internal Reflection:

n &gt; n0 , i &gt; i0 = sin−1 (

n0
).
n

• Frequency Dispersion (Simple Model): ω is the driven frequency, µ is not ω dependent.
ε(ω) = ε0 +

Z
N e2 X 2
(ωj − ω 2 − iωγj )−1 ,
me j

where N is # of molecules and Z is # of electron in one molecule. The γj are damping constants.
In a wave function, attenuation constant α indicates an exponential decay:
α
α2
ω2
ε
ω2
ε
k = β + i , where β 2 −
= 2 Re[ ], βα = 2 Im[ ].
2
4
c
ε0
c
ε0
~ x, ω) = σ(ω)E(~
~ x, ω).
- Conductivity σ: εtotal (ω) = ε(bounded) (ω) + i ωσ . Ohm’s law: J(~
2
σ
σ
Poor conductor: 2ε
√kµε ; Good conductor: 2ε
√kµε , γ(free) ≈ 0 (Start from ωk 2 =
- Low frequency limit (ω  max(ωj )): ε(ω) ≈ ε(bounded) + i

NZ

(free) e

(free) −iω

me ω γ

1
µε ).

2

.

q
ω2
Ze2
1
- High frequency limit (ω  max(ωj )): ε(ω) ≈ ε0 − ε0 ωp2 , where ωp2 = N
ω 2 − ωp2 .
ε0 m e ; k = c
~ x, t) under causality of ε(ω), D(~
~ x, ω) = ε(ω)E(~
~ x, ω).
- Kramers-Kronig Relations: linear response of D(~
dω ˆ
ωˆ
- Phase velocity ~vp = k; group velocity ~vg = k.
k

dk

Electrodynamics: Waveguides &amp; Resonance Cavities
• Basics: S indicates boundary, n
ˆ is the unit normal of the boundaries; ω is the driven frequency
(externally imposed), kz with ±ˆ
z being the propagating direction. Also the following of fields are
assumed (t for transverse):
 ~


~
E(x, y) ±ikz z−iωt
∂2
E
2
2
2
2
∇ − 2
e
= (∇t − kz + µεω ) ~ = 0.
~
∂t
B
B(x, y)
TM mode (Bz = 0)
Useful BCs:

Ez |S = 0

Impedance:

kz
Z=
εω

TE mode (Ez = 0)

∂Bz
=0
∂n
S

µω
Z=
kz

±ikz ~
±ikz ~
~t =
∇t Ez H
∇t Hz
2
2
µεω − kz
µεω 2 − kz2
~ t = ±1 zˆ × E
~t
~ t = ∓Z zˆ × H
~ t.
True only for e±ikz z−iωt :
H
E
Z
• Non-perfect Conductor and (Time
Averaged) Power Loss: σ is conductivity and c stands
q
Works for all following cases:

for conductor. With skin depth δ =

~t =
E

2
µc ωσ ,

ξ indicates the depth into the conductor.

~ c ≈ 1 (1 − i)ˆ
~ k e− δξ ei δξ .
~c = H
~ k e− δξ ei δξ , E
n×H
H
δσ
q
k2
- Waveguides: βλ is the attenuation constant and ωλ ≡ ω 2 − µε
. The power lost when the fields
propagates is,
1 dP dPloss
dP
P
P (z) = P0 e−2βλ z , βλ = −
,
=−
, vg = .
2P dz
dz
dz
U
TM mode
TE mode
Z
Z
1 ω2
1 ω2
2
ε
|E
|
da
U
=
µ
|Hz |2 da
Energy U =
z
2 ωλ2 S
2 ωλ2
S
r
r
r
r
Z
2
2
2 Z
ω
ωλ2
1 εω
1 µω
2
λ
1

1

Power P =
|E
|
da
P
=
|Hz |2 da
z
2 µ ωλ2
ω2 S
2 ε ωλ2
ω2 S

2


I
I
2
dPloss
1 ω2
1 ∂Ez
dPloss
1 ω2
1
ωλ2
~ t Hz |2 + ωλ |Hz |dl.
=
dl
=
1

n
×

dz
2δσ ωλ2 ∂S µ2 ωλ2 ∂n
dz
2δσ ωλ2 ∂S µεωλ2
ω2
ω2
q
2
Energy
µεω 2 − pπ
. The energy lost over time
- Resonance Cavity: Q ≡ ω0 Stored
L
Power Loss and γλ ≡
is,
t
1
1 dU
dU
U (t) = U0 e−ω0 Q ,
=−
, Ploss = −
.
Q
ω0 U dt
dt
TM mode
TE mode


2  Z


2  Z
L

L

Energy U =
1+
ε |Ez |2 da
U=
1+
µ |Hz |2 da
4
γλ L
4
γ
L
λ
S
S
I Z L

Z
1
~ 2 dzdl+ 2 |ˆ
~ 2 da .
Ploss =

n × H|
n × H|
sides
ends
2δσ ∂S 0
S

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