6.4 de Sitter Space-Time Revisited
143
To verify that u = constant, with u(t, x, y, z) given by (6.94), are indeed null
hyperplanes in the de Sitter space-time with metric tensor g ab given via the line
element (6.93) we note that with X α = x α − w α (u) and ρ = ˙
w α X α it follows from
(6.94) that
u ,a =
e −2
√
2 k t
√
2 k ρ
,
X α
ρ
,
(6.108)
and hence
g
ab u ,a u ,b = 0 ,
(6.109)
as a consequence of (6.94). Making use of the formulas
X
α
,β = δ
α
β −
˙
w α X β
ρ
,
(6.110)
ρ ,α = ˙
w
α
+ ( ¨
w
β X
β
− ˙
w
β
˙
w
β )
X α
ρ
,
(6.111)
and
ρ ,0 =
∂ρ
∂t
= ( ¨
w
β X
β
− ˙
w
β
˙
w
β ) u ,0 ,
(6.112)
with u ,0 (and u ,α ) given in (6.108),we find that
u ,α;β = −
1
ρ
{ ˙
w
α u ,β + ˙
w
β u ,α + ( ¨
w
γ X
γ
− ˙
w
γ
˙
w
γ ) u ,α u ,β } ,
(6.113)
u ,α;0 = −
√
2 k u ,α −
1
ρ
{ ˙
w
α
+ ( ¨
w
γ X
γ
− ˙
w
γ
˙
w
γ ) u ,α } u ,0 ,
(6.114)
and
u ,0;0 = −2
√
2 k u ,0 −
1
ρ
( ¨
w
γ X
γ
− ˙
w
γ
˙
w
γ ) (u ,0 )
2 .
(6.115)
Hence we can write
u a;b = ψ a u ,b + ψ b u ,a ,
(6.116)
with
ψ a =
−
√
2 k −
1
2 ρ
( ¨
w
γ X
γ
− ˙
w
γ
˙
w
γ ) u ,0 , −
˙
w α
ρ
−
1
2 ρ
( ¨
w
γ X
γ
− ˙
w
γ
˙
w
γ ) u ,α ,
.
(6.117)