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n+ 1 n′
+ 1 * * * *
−{ () i ( −i) [ π n +τn]
⎡
⎣
bn′ π n′ + an′ τ n′ + bn′ τ n′ + an′ πn′
⎤
⎦}
n+ 1 n′
+ 1 * * * *
{ 2( i) ( i) {[ bn n an n]
⎡
⎣
bn′ n′ an′ n′ ⎤
⎦} { ⎡
⎣
bn′ n′ an′ n′
⎤
⎦[ bn n an n]
}}
+ − τ + π τ + π + π + τ π + τ
(3. 134)
tal que as integrais ficam,
( 2n
+ 1)
4π 4 π nn ( + 2)
I E ⎡⎣a b ⎤⎦ E ⎡⎣b b a a ⎤⎦
∞
∞
2 * 2
* *
4
=
0
Re
2 2 n n
+
2 2 0
Re
n n+ 1+
n n+
1
µω n= 1 nn ( + 1) µω n=
1 ( n+
1)
π
∑ ∑ (3. 135)
2
2π
E0
5 2 2
∞
n=
1
{ [ ]}
I =− ∑ Re (2n+ 1) an + bn
(3. 136)
µω
π
I = dθ θ θ E E − −i − i e ×
{
* n′+ 1 n+ 1 2iρ
∫ sen cos
n n′
2 Re ( ) ( )
2( µω)
(3. 137)
6 2
0
∑
[ bn n n′ an n n′ bn n n′ bn n n′ bn n n′ an n n′ an n n′ an n n′
]}
× τπ + ππ − ππ − ττ + πτ − τπ − πτ + ττ
Combinando I1 com I4 = I7, I2 com I5 = I8
e I3 com I6 = I9, e de acordo com a
1
{ i}
* * * *
expressão de força
i
= ∫ − ε ( θ θ
+
ϕ ϕ) +µ ( θ θ
+
ϕ ϕ)
F Re [ E E E E H H H H ] n da, ficamos com:
2
( 2n+
1)
∞
∞
⎛4π 2 4 π 2 ⎞⎛
* nn ( + 2)
* *
⎞
I7 = ⎜ E
2 0
ε+µ E
2 2 0 ⎟⎜
Re ⎡anbn⎤+ Re ⎡bnbn+ 1+
anan+
1⎤⎟
k µω n= 1 n( n+ 1)
⎣ ⎦
n=
1 ( n+
1)
⎣ ⎦
⎝ ⎠⎝ ⎠
2
usando a relação k⋅ k =ω εµ :
( 2n
+ 1)
∑ ∑ (3.138)
∞
∞
8 π 2 ⎛
* nn ( + 2)
* *
⎞
I7 =
⎛
E
⎞
⎜ 2 0
ε ⎟⎜
Re ⎡anbn⎤+ Re ⎡bnbn+ 1+
anan+
1⎤⎟
⎝ k n= 1 n( n+ 1)
⎣ ⎦
n=
1 ( n+
1)
⎣ ⎦
⎠⎝ ⎠
e agora por último,
π
0
∑ ∑ (3. 139)
2
4π
E0
8 2
∞
n=
1
{ [ ]}
I =− ε ∑ Re (2n+ 1) an + bn
(3. 140)
k
⎛ πε πµ ⎞
* n′+ 1 n+ 1 2iρ
I3 + I6 = ⎜ + d sen cos E Re
2 2 2 ⎟ θ θ θ
nEn′
−( −i) ( − i)
e ×
⎝ k µω
∫ ∑
⎠
[ anτπ n n′ + bnππ n n′ −anππ n n′ −anττ n n′ + anπτ n n′ −bnτπ n n′ −bnπτ n n′ + bnττ n n′
] +
[ bn n n′ an n n′ bn n n′ bn n n′ bn n n′ an n n′ an n n′ an n n′
] }
τπ + ππ − ππ − ττ + πτ − τπ − πτ + ττ ⇒0
{
(3. 141)
Dessa forma, no final dos cálculos a força devido a pressão de radiação é:
100