String Theory Demystified

272 String Theory Demystifi ed
2 2
case. Using H + nH H + [ n( n− 1)/ 6] H = 0,
we can eliminate H from the
a a b b
d
equations and write an equation for H alone:
a
H ⎛
a 3+ = ⎜
Ha
⎝
2
3n + 6n⎞
⎟
n −1
⎠
(16.14)
Now the accelerating nature of the expansion is apparent since H a > 0. Integration
gives
Now make the defi nition
⎛
2
1 1 3+ 3n + 6n⎞
− + = ⎜
⎟ t
H () t H ( 0)
⎝ n −1
a a
⎠
t
n
n n H −1
1
=
2
3+ 3 + 6 ( 0)
a
Then it can be shown that the Hubble constant is given by
Further integration gives the scale factor:
H ( 0)
a
H () t = a
1−
tt /
a
at () =
( 1−
t t)
where a is a constant of integration and we have defi ned
n n
p =
n
− + +
2
3 3 6
3( + 3)
The acceleration of the three expanding dimensions of the universe is then
a
a
= ⎛
⎝
⎜
p
t
⎞ p + 1 1
⎠
⎟
t ( 1−
tt)
p
2
> 0
Using the relation for the ratio H b /H a it can be shown that
bt () = b( 1−tt)
q
(16.15)
(16.16)
(16.17)
(16.18)