12
M. V. John
m ≡
ρ m
ρ c
=
2
3(1 + w)
,
, d.e. ≡
ρ d.e.
ρ c
=
1 + 3w
3(1 + w)
.
(14)
Since this universe is flat, the total density parameter = ρ/ρ c = 1. When matter
in the present universe is considered to be nonrelativistic with w = 0, the above
equations predict m = 2/3 and d.e. = 1/3. The ratio between these densities is
a constant of the order of unity throughout the expansion history (for a a 0 ), and
this avoids the coincidence problem. Being a coasting evolution, it naturally has no
synchronicity problem. It solves all other cosmological problems, as demonstrated
in the previous works.
With the aid of the solution (7), one can see that the complex-extended Milne
model has Euclidean (++++) signature at t = 0, but it changes to the usual Lorentzian
(+ − − −) one for a a 0 . Hence, the complex extension of Milne model leads
naturally to a signature change in the early universe, a possibility discussed extensively in the literature [35, 36]. The famous Hartle–Hawking ‘no boundary’ boundary
condition [37] in quantum cosmology envisages a change of signature in the early
universe. In this regime, there is no time and the space-time is purely spatial. Even in
the classical Einstein equations, it is argued that the metric is Lorentzian not because
it is demanded by the field equations; instead, it is a condition imposed on the metric
before one looks for solutions [35]. In our case, the complex Milne model undergoes
a signature change in a very natural way.
3.1 Wheeler–DeWitt Equation
In this section, we write down the Wheeler–DeWitt equation for the complexextended Milne model. For this, we first note that the classical Milne model follows
from the Lagrangian
L = −
3π
4G
˙ ˆ
a
2
ˆ
a
N
+ N ˆ
a
.
(15)
Here N is called the lapse function, for which one can fix some convenient gauge.
Writing down the Euler–Lagrange equations with respect to the variables N and ˆ
a
and fixing the gauge N = 1 leads to Eqs. (1) and (2), respectively. The canonically
conjugate momentum is
ˆ
π a =
∂ L
∂ ˙ ˆ
a
= −
3π
2G
˙ ˆ
a ˆ
a
N
(16)
The canonical Hamiltonian can now be constructed as
H c = ˆ
π a ˆ
a − L = N
−
G
3π
ˆ
π
2
a
ˆ
a
+
3π
4G
ˆ
a
≡ N H.
(17)
M. V. John
m ≡
ρ m
ρ c
=
2
3(1 + w)
,
, d.e. ≡
ρ d.e.
ρ c
=
1 + 3w
3(1 + w)
.
(14)
Since this universe is flat, the total density parameter = ρ/ρ c = 1. When matter
in the present universe is considered to be nonrelativistic with w = 0, the above
equations predict m = 2/3 and d.e. = 1/3. The ratio between these densities is
a constant of the order of unity throughout the expansion history (for a a 0 ), and
this avoids the coincidence problem. Being a coasting evolution, it naturally has no
synchronicity problem. It solves all other cosmological problems, as demonstrated
in the previous works.
With the aid of the solution (7), one can see that the complex-extended Milne
model has Euclidean (++++) signature at t = 0, but it changes to the usual Lorentzian
(+ − − −) one for a a 0 . Hence, the complex extension of Milne model leads
naturally to a signature change in the early universe, a possibility discussed extensively in the literature [35, 36]. The famous Hartle–Hawking ‘no boundary’ boundary
condition [37] in quantum cosmology envisages a change of signature in the early
universe. In this regime, there is no time and the space-time is purely spatial. Even in
the classical Einstein equations, it is argued that the metric is Lorentzian not because
it is demanded by the field equations; instead, it is a condition imposed on the metric
before one looks for solutions [35]. In our case, the complex Milne model undergoes
a signature change in a very natural way.
3.1 Wheeler–DeWitt Equation
In this section, we write down the Wheeler–DeWitt equation for the complexextended Milne model. For this, we first note that the classical Milne model follows
from the Lagrangian
L = −
3π
4G
˙ ˆ
a
2
ˆ
a
N
+ N ˆ
a
.
(15)
Here N is called the lapse function, for which one can fix some convenient gauge.
Writing down the Euler–Lagrange equations with respect to the variables N and ˆ
a
and fixing the gauge N = 1 leads to Eqs. (1) and (2), respectively. The canonically
conjugate momentum is
ˆ
π a =
∂ L
∂ ˙ ˆ
a
= −
3π
2G
˙ ˆ
a ˆ
a
N
(16)
The canonical Hamiltonian can now be constructed as
H c = ˆ
π a ˆ
a − L = N
−
G
3π
ˆ
π
2
a
ˆ
a
+
3π
4G
ˆ
a
≡ N H.
(17)