132
6 de Sitter Cosmology
we have from (6.6) that
η ij ξ
i ξ
j
= 0 .
(6.30)
Differentiating this with respect to x k results in
u ,k =
ξ k
R
= k k (say) ,
(6.31)
with
R = η ij ˙
w
i ξ
j and thus η ij ˙
w
i k
j
= +1 .
(6.32)
As always a dot indicates differentiation with respect to u. It is clear from (6.30) and
(6.31) that the hypersurfaces u = constant are null. Straightforward calculations
yield
ξ
i
,j = δ
i
j − ˙
w
i k j ,
(6.33)
and
R ,i = ˙
w i + A k i ,
(6.34)
with ˙
w i = η ij ˙
w j and
A = − ˙
w i ˙
w
i
+ R ¨
w i k
i .
(6.35)
Using (6.31)–(6.35) we arrive at
k i,j =
1
R
(η ij − ˙
w i k j − ˙
w j k i − A k i k j ) = k j,i ,
(6.36)
which displays the algebraic structure [3] guaranteeing that k i is geodesic and
shear-free with expansion ϑ = k i ,i /2 = 1/R = 0. With a semicolon, as before,
indicating covariant differentiation with respect to the Riemannian connection
(6.12) associated with the metric tensor given via the line element (6.10) we find
that
k i;j = (R
−1
+ λ
−1 η
kl λ ,k k l )η ij − (R
−1
˙
w i + λ
−1 λ ,i )k j
−(R
−1
˙
w j + λ
−1 λ ,j )k i − A R
−1 k i k j .
(6.37)
6 de Sitter Cosmology
we have from (6.6) that
η ij ξ
i ξ
j
= 0 .
(6.30)
Differentiating this with respect to x k results in
u ,k =
ξ k
R
= k k (say) ,
(6.31)
with
R = η ij ˙
w
i ξ
j and thus η ij ˙
w
i k
j
= +1 .
(6.32)
As always a dot indicates differentiation with respect to u. It is clear from (6.30) and
(6.31) that the hypersurfaces u = constant are null. Straightforward calculations
yield
ξ
i
,j = δ
i
j − ˙
w
i k j ,
(6.33)
and
R ,i = ˙
w i + A k i ,
(6.34)
with ˙
w i = η ij ˙
w j and
A = − ˙
w i ˙
w
i
+ R ¨
w i k
i .
(6.35)
Using (6.31)–(6.35) we arrive at
k i,j =
1
R
(η ij − ˙
w i k j − ˙
w j k i − A k i k j ) = k j,i ,
(6.36)
which displays the algebraic structure [3] guaranteeing that k i is geodesic and
shear-free with expansion ϑ = k i ,i /2 = 1/R = 0. With a semicolon, as before,
indicating covariant differentiation with respect to the Riemannian connection
(6.12) associated with the metric tensor given via the line element (6.10) we find
that
k i;j = (R
−1
+ λ
−1 η
kl λ ,k k l )η ij − (R
−1
˙
w i + λ
−1 λ ,i )k j
−(R
−1
˙
w j + λ
−1 λ ,j )k i − A R
−1 k i k j .
(6.37)
