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A 1 sin Δ i Λ
(1 − E 2 cos 2 Δ i ) 1 / 2 = f ( Δ i )
i
∈{
1 , 2
}
,
(19.44)
f ( Δ i )= c 2
1 / 2
A 1 (1
E 2 )
cos Δ i
2 E ln 1+ E cos Δ i
1
E 2 cos 2 Δ i +
n
1
1
E cos Δ i
A 1 sin 2 Δ i
n 2 (1
A 1 (1
E 2 )
cos Δ i
2 E ln 1+ E cos Δ i
1
E 2 cos 2 Δ i ) = c 2
E 2 cos 2 Δ i +
(19.45)
n
1
1
E cos Δ i
sin 2 Δ i
n (1 − E 2 cos 2 Δ i ) +(1
E 2 )
1 / 2
c = A 1
cos Δ i
1 − E 2 cos 2 Δ i +
2 E ln 1+ E cos Δ i
1
n
,
1 − E cos Δ i
c ( Δ 1 )= c ( Δ 2 ) .
Let us substitute the two functions h ( Δ i )and g ( Δ i ):
sin 2 Δ i
h ( Δ i )= h i :=
E 2 cos 2 Δ i ,
(19.46)
1
g ( Δ i )= g i := (1 − E 2 )
.
cos Δ i
2 E ln 1+ E cos Δ i
1
E 2 cos 2 Δ i +
1
1
E cos Δ i
For c , we then arrive at
h i
n + g i 1 / 2
c = A 1
n
.
(19.47)
For n, we then arrive at
h 1
n + g 1 1 / 2
h 2
n + g 2 1 / 2
A 1
n
= A 1
n
n = h 1
h 2
(19.48)
g 2
g 1
sin 2 Δ 1
sin 2 Δ 2
1 −E 2 cos 2 Δ 2
1 −E 2 cos 2 Δ 1
E 2 )
+ 2 E ln 1+ E cos Δ 2
.
n =
(19.49)
cos Δ 2
cos Δ 1
1 −E 2 cos 2 Δ 2
1 E cos Δ 1
1+ E cos Δ 1
(1
1 −E 2 cos 2 Δ 2
1 −E cos Δ 2
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