4.5 Solutions to Problems
101
= D yξ D zξ D yζ Rαμ + D yη D zη D yζ Rαμ + D yζ D zζ D yζ αμ
= (cos ψ sin φ + cos θ cos φ sin ψ)(sin ψ sin θ)(− sin θ cos φ) Rαμ
+ (− sin ψ sin φ + cos θ cos φ cos ψ)(cos ψ sin θ)(− sin θ cos φ) Rαμ
+ (− sin θ cos φ)(cos θ)(− sin θ cos φ) αμ
= (− cos θ sin
2 θ cos 2 φ − 2 sin
2 θ cos ψ sin ψ cos φ sin φ) Rαμ
+cos θ sin
2 θ cos 2 φ αμ
= −(cos θ − cos 3 θ)
Rαμ
2
+ (cos θ − cos 3 θ)
αμ
2
,
(4.41)
α
(2)
zzz =
p
q
r
D zp D zq D zr
∂α p q
∂q 1
∂μ r
∂q 1
= D zξ D zξ D zζ Rαμ + D zη D zη D zζ Rαμ + D zζ D zζ D zζ αμ
= (sin ψ sin θ) 2 (cos θ) Rαμ + (cos ψ sin θ) 2 (cos θ) Rαμ + (cos 3 θ) αμ
= cos θ sin
2 θ Rαμ + cos 3 θ αμ
= (cos θ − cos 3 θ) Rαμ + cos 3 θ αμ.
(4.42)
In the above derivation of Eqs. (4.40), (4.41), and (4.42), the average over φ and
ψ is carried out with the assumption of uniform distribution. Using the results of
Eqs. (4.40), (4.41), and (4.42), B and C are given by
B =
χ
(2)
yyz
χ
(2)
yzy
=
α
(2)
yyz
α
(2)
yzy
=
(cos θ + cos 3 θ)
Rαμ
2
+ (cos θ − cos 3 θ)
αμ
2
−(cos θ − cos 3 θ)
Rαμ
2
+ (cos θ − cos 3 θ)
αμ
2
=
(1 + R) cos θ − (1 − R) cos 3 θ
(1 − R)(cos θ − cos 3 θ)
,
(4.14)
C =
χ
(2)
zzz
χ
(2)
yyz
=
α
(2)
zzz
α
(2)
yyz
=
(cos θ − cos 3 θ) Rαμ + cos 3 θ αμ
(cos θ + cos 3 θ)
Rαμ
2
+ (cos θ − cos 3 θ)
αμ
2
=
2{R cos θ + (1 − R) cos 3 θ }
(1 + R) cos θ − (1 − R) cos 3 θ
.
(4.15)
We notice that the final results of B and C in Eqs. (4.14) and (4.15) do not include
N , α and μ.