144
6 de Sitter Cosmology
It follows immediately from (6.116) that u ,a is geodesic and shear-free, on account
of the Robinson–Trautman [3] test for these properties, and in addition u ,a is
expansion-free since
g
ab ψ a u ,b = 0 ,
(6.118)
which can be readily verified directly. Thus we have confirmed that in the de
Sitter space-time with line element (6.93) the hypersurfaces u = constant, with
u(t, x, y, z) given implicitly by (6.94) are null hyperplanes.
Following (6.107) we can write
X
1
=
1
√
2 k
e
−
√
2 k t sin ξ cos η ,
X
2
=
1
√
2 k
e
−
√
2 k t sin ξ sin η ,
(6.119)
X
3
=
1
√
2 k
e
−
√
2 k t cos ξ ,
and then
ρ =
1
√
2 k
e
−
√
2 k t A with A = ˙
w
1 sin ξ cos η + ˙
w
2 sin ξ sin η + ˙
w
3 cos ξ .
(6.120)
We shall make use of these in the form
X
α
=
ρ
A
(sin ξ cos η, sin ξ sin η, cos ξ) with
ρ
A
=
1
√
2 k
e
−
√
2 k t .
(6.121)
From this we find that
dx
α dx
α
=
d
ρ
A
2 +
ρ
A
2
(dξ
2
+ sin
2 ξ dη
2 )
+2 dρ du +
˙
w
α
˙
w
α
−
2 ρ
A
∂A
∂u
du
2 .
(6.122)
Substituting into the line element (6.93), after first noting that
d
ρ
A
= −
√
2 k
ρ
A
dt and e
√
2 k t
=
1
√
2 k
A
ρ
,
(6.123)
6 de Sitter Cosmology
It follows immediately from (6.116) that u ,a is geodesic and shear-free, on account
of the Robinson–Trautman [3] test for these properties, and in addition u ,a is
expansion-free since
g
ab ψ a u ,b = 0 ,
(6.118)
which can be readily verified directly. Thus we have confirmed that in the de
Sitter space-time with line element (6.93) the hypersurfaces u = constant, with
u(t, x, y, z) given implicitly by (6.94) are null hyperplanes.
Following (6.107) we can write
X
1
=
1
√
2 k
e
−
√
2 k t sin ξ cos η ,
X
2
=
1
√
2 k
e
−
√
2 k t sin ξ sin η ,
(6.119)
X
3
=
1
√
2 k
e
−
√
2 k t cos ξ ,
and then
ρ =
1
√
2 k
e
−
√
2 k t A with A = ˙
w
1 sin ξ cos η + ˙
w
2 sin ξ sin η + ˙
w
3 cos ξ .
(6.120)
We shall make use of these in the form
X
α
=
ρ
A
(sin ξ cos η, sin ξ sin η, cos ξ) with
ρ
A
=
1
√
2 k
e
−
√
2 k t .
(6.121)
From this we find that
dx
α dx
α
=
d
ρ
A
2 +
ρ
A
2
(dξ
2
+ sin
2 ξ dη
2 )
+2 dρ du +
˙
w
α
˙
w
α
−
2 ρ
A
∂A
∂u
du
2 .
(6.122)
Substituting into the line element (6.93), after first noting that
d
ρ
A
= −
√
2 k
ρ
A
dt and e
√
2 k t
=
1
√
2 k
A
ρ
,
(6.123)
