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E 2 l E sin B N sin B S
5 (sin 5 B N sin 5 B S ) +
1
10
A 1
2
3 (sin 3 B N sin 3 B S )+ 1
=
(20.154)
1
+O GK ( l E ) ,
sin B N sin B S + 2
3 E 2 (sin 3 B N sin 3 B S )+O 1 ( E 4 ) 1
1
S E A 1 ,A 2
1
2 A 1 (1
E 2 ) l 1
=
= (20.155)
E
sin B S ) 1 1
+O 2 ( E 4 ) ,
3 E 2 sin 3 B N sin 3 B S
1
2 A 1 (1
2
E 2 ) l E (sin B N
=
sin B N
sin B S
E 2 ) 2 1
l E
sin 3 B N
sin 3 B S
sin 5 B N
sin 5 B S
1
20
2
3
+ 1
5
I AGK =
×
(1
sin B N
sin B S
sin B N
sin B S
1
+O 2 ( E 4 ) +
3 E 2 sin 3 B N
sin 3 B S
sin B N sin B S
2
×
(20.156)
+O GK ( l E ) ,
I AGK ( B N = −B S )= 1
20
l E
E 2 ) 2 ×
(1
1
3 sin 2 B N (1 + E 2 )+sin 4 B N 4
15 E 2 sin 6 B N +O 2 ( E 4 ) +
2
9 E 2 + 1
2
×
(20.157)
5
+O GK ( l E ) .
End of Proof.
Proof (Soldner, l = L − L 0 ).
1 E 2 sin 2 B
1
Λ 1 =1+ cos 2 B
2
l 2 +O S ( l 4 ) 2 =1 ,
(20.158)
E 2
d S E 2 A 1 ,A 2
+ l E
B N
1) 2 = A 1 (1
E 2 )
2 S E 2 A 1 ,A 2
1) 2
1
2 S E 2 A 1 ,A 2
d B cos B ( Λ 1
I AS =
( Λ 1
d l
E 2 sin 2 B ) 2 ,
(20.159)
(1
−l E
B S
1 E 2 sin 2 B
1
1= cos 2 B
2
l 2 +O S ( l 4 ) ,
Λ 1
E 2
(20.160)
E 2 sin 2 B ) 2
(1
1) 2 = 1
4 cos 4 B (1
l 4 +O S ( l 6 ) ,
( Λ 1
E 2 ) 2
d S E 2 A 1 ,A 2 ( Λ 1
+ l E
B N
1) 2 = 1
4
A 1
l 4
=
d B cos 5 B +O S ( l 6 )
d l
(20.161)
1
E 2
−l E
B S
E 2 l E sin B N sin B S
5 (sin 5 B N sin 5 B S ) +
1
10
A 1
2
3 (sin 3 B N sin 3 B S )+ 1
=
1
+O S ( l E ) ,
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