6.4 de Sitter Space-Time Revisited
145
we arrive at
− ds
2
=
1
2 k 2 (dξ
2
+ sin
2 ξ dη
2 ) +
1
k 2
A
ρ
2
dρ du
+
1
2 k 2
A
ρ
2
˙
w
α
˙
w
α
−
2 ρ
A
˙
A
du
2 ,
(6.124)
with the dot as always denoting differentiation with respect to u. Next make the
coordinate transformation
ρ = −
1
2 k 2 r
,
(6.125)
and (6.124) becomes
−ds
2
=
1
2 k 2 (dξ
2
+ sin
2 ξ dη
2 ) + 2 A
2 du
dr +
k
2
˙
w
α
˙
w
α r
2
+ A
−1 ˙
A r
du
.
(6.126)
Finally put
k ζ = e
iη tan
ξ
2
,
(6.127)
which results in
A = (1 + k
2 ζ ¯
ζ )
−1
k ( ˙
w
1
− i ˙
w
2 ) ζ + k ( ˙
w
1
+ i ˙
w
2 ) ¯
ζ + ˙
w
3 (1 − k
2 ζ ¯
ζ )
= p
−1 q ,
(6.128)
with
p = 1 + k
2 ζ ¯
ζ ,
(6.129)
q = ¯
β ζ + β ¯
ζ + α (1 − k
2 ζ ¯
ζ ) ,
(6.130)
and
β = k ( ˙
w
1
+ i ˙
w
2 ) , α = ˙
w
3 .
(6.131)
Now (6.126) takes the Ozsvàth–Robinson–Rózga [4] form
− ds
2
= 2 p
−2 dζ d ¯
ζ + 2 p
−2 q
2 du
dr +
1
2
κ r
2
+ q
−1
˙
q r
du
,
(6.132)