140
6 de Sitter Cosmology
To exhibit a solution of the field equation (6.77) we note that if W = (p/q) 2 then,
in general,
p
2 ∂ 2 W
∂ζ ∂ ¯
ζ
+
3
W = 3
p
q
4
κ .
(6.88)
Hence when κ = 0 we have a solution of (6.77) given by
H = c 0
p 2
q 2 ,
(6.89)
and, for simplicity, we will take c 0 = constant (although we can have c 0 = c 0 (u)).
Now (6.87) reads
− ds
2
= 2 p
−2 dζ d ¯
ζ + 2 p
−2
|1 + n ζ |
4 du {dr + c 0 p
3
|1 + n ζ |
−6 du} , (6.90)
with p given by (6.85). This is the solution of Ozsváth [10]. To put it in a more
familiar form make the coordinate transformation:
n ζ =
2
√
2 n + e −
√
2 n X + i
√
2 n Y
2
√
2 n − e −
√
2 n X − i
√
2 n Y
.
(6.91)
Absorbing constants into u and c 0 we arrive at
− ds
2
= dX
2
+ e
2
√
2 n X (dY
2
+ 2 du dr) + f 0 e
−
√
2 n X du
2 ,
(6.92)
with f 0 a constant. This is the form of the solution given in [10]. Its significance is
that the maximal group of isometries of this space-time has five parameters and, in
this sense, it appears to be the closest one can get to plane gravitational waves when
a cosmological constant is present [11].
6.4
de Sitter Space-Time Revisited
When introducing parametrisations, as in (6.40), exceptional cases can arise. For
example if w 0 = 0 in (6.40) then the parametrisation given is impossible if > 0.
An example of this occurs when the de Sitter line element is given in the quite
familiar form of
ds
2
= dt
2
− e
2
√
2 k t (dx
2
+ dy
2
+ dz
2 ) with k =
.
(6.93)
6 de Sitter Cosmology
To exhibit a solution of the field equation (6.77) we note that if W = (p/q) 2 then,
in general,
p
2 ∂ 2 W
∂ζ ∂ ¯
ζ
+
3
W = 3
p
q
4
κ .
(6.88)
Hence when κ = 0 we have a solution of (6.77) given by
H = c 0
p 2
q 2 ,
(6.89)
and, for simplicity, we will take c 0 = constant (although we can have c 0 = c 0 (u)).
Now (6.87) reads
− ds
2
= 2 p
−2 dζ d ¯
ζ + 2 p
−2
|1 + n ζ |
4 du {dr + c 0 p
3
|1 + n ζ |
−6 du} , (6.90)
with p given by (6.85). This is the solution of Ozsváth [10]. To put it in a more
familiar form make the coordinate transformation:
n ζ =
2
√
2 n + e −
√
2 n X + i
√
2 n Y
2
√
2 n − e −
√
2 n X − i
√
2 n Y
.
(6.91)
Absorbing constants into u and c 0 we arrive at
− ds
2
= dX
2
+ e
2
√
2 n X (dY
2
+ 2 du dr) + f 0 e
−
√
2 n X du
2 ,
(6.92)
with f 0 a constant. This is the form of the solution given in [10]. Its significance is
that the maximal group of isometries of this space-time has five parameters and, in
this sense, it appears to be the closest one can get to plane gravitational waves when
a cosmological constant is present [11].
6.4
de Sitter Space-Time Revisited
When introducing parametrisations, as in (6.40), exceptional cases can arise. For
example if w 0 = 0 in (6.40) then the parametrisation given is impossible if > 0.
An example of this occurs when the de Sitter line element is given in the quite
familiar form of
ds
2
= dt
2
− e
2
√
2 k t (dx
2
+ dy
2
+ dz
2 ) with k =
.
(6.93)