7.6 Solutions to Problems
189
7.6.2 Bulk Term χ B
[Problem 7.2] Derive χ B
G,yyz in Eq. (7.54) from χ B0
G in Eqs. (7.52) and (7.53).
(Hint) During the derivation, use the following two relations,
cos θ α
I (ω 2 )
sin θ α
I (ω 2 )
=
q α (ω 2 )
k x (ω 2 )
,
f
β
x
f
β
z
= ε
β ,
and L I (ω f ) in Eq. (5.30).
The expression of χ
(2)
eff,G,SSP in Eq. (7.49) including χ B0 can be expanded using
Eq. (7.48) as follows,
χ
(2)
eff,G,SSP ((, ω 1 , ω 2 ) =
L G,y (()L I,y (ω 1 ) L I,z (ω 2 ) sin θ
α
I (ω 2 ) χ
(2)
q0 G,yyz ((, ω 1 , ω 2 )
+ L G,y (() L I,y (ω 1 ) L I,x (ω 2 ) cos θ
α
I (ω 2 ) χ
(2)
q0 G,yyx ((, ω 1 , ω 2 )
(7.49)
= L G,y (()L I,y (ω 1 ) L I,z (ω 2 ) sin θ
α
I (ω 2 )
·
χ
ID
yyz ((, ω 1 , ω 2 )+χ
IQ
yyz ((, ω 1 , ω 2 )+χ
IQB
yyz ((, ω 1 , ω 2 )+χ
B0
yyz ((, ω 1 , ω 2 )
+ L G,y (() L I,y (ω 1 ) L I,x (ω 2 ) cos θ
α
I (ω 2 ) χ
B0
G,yyx ((, ω 1 , ω 2 ),
(7.124)
since the yyx element of χ
(2)
q0 G involves only the χ B0
G term. Equation (7.124)
should be equivalent to Eqs. (7.51) and (7.50). Apparent differences are found in
the terms related to χ B0 and χ B , by comparing the two expressions. To make the
two expressions equivalent, the following relation is required,
L G,y (()L I,y (ω 1 ) L I,z (ω 2 ) sin θ
α
I (ω 2 ) χ
B
G,yyz ((, ω 1 , ω 2 )
= L G,y (()L I,y (ω 1 ) L I,z (ω 2 ) sin θ
α
I (ω 2 ) χ
B0
G,yyz ((, ω 1 , ω 2 )
+ L G,y (() L I,y (ω 1 ) L I,x (ω 2 ) cos θ
α
I (ω 2 ) χ
B0
G,yyx ((, ω 1 , ω 2 ).
Therefore,
χ
B
G,yyz ((, ω 1 , ω 2 )=χ
B0
G,yyz ((, ω 1 , ω 2 )+
L I,x (ω 2 )
L I,z (ω 2 )
cos θ α
I (ω 2 )
sin θ α
I (ω 2 )
χ
B0
G,yyx ((, ω 1 , ω 2 ).
(7.125)