20
C. Sivakumar
Table 1 Choice of “A”
12/(1+A)
A
. . .
. . .
24
–0.500
26
–0.538
28
–0.571
30
–0.600
. . .
. . .
∞
–1.000
Table 2 Scale factor R(ω ef f )
ω ef f
R
+0.333
R i
–0.333
R 0
–0.400
R p
–0.571
∞
If the law of partial pressures is true for the cosmic fluid at least presently, with
−0.6 for dark energy parameter,
ω e f f p = 0.25ω m(d) − 0.42 ≈ −0.4 if the present dark matter parameter is that
small and positive even if it contributes, gives
R 0
R p
= 0.718 (−0.4 is a good fit as
present parameter ω e f f p is less than −0.333 but away enough from −0.571). The
dark matter parameter could be thus around +0.08 (note that if ω dp = −1.0 as CDM
says, ω m(d) p ≈ +1.2).
Substituting H p in Eq. (6), R p = 5.420 × 10
26 .
Note that order(R) is as per [2].
R 0
R i
and
R 0
R p
give R i = 1.0 × 10
26 and R 0 = 3.9 × 10
26 .
Substituting H p in Eq. (7),
ρ p = (ρ m(n) + ρ m(d) + ρ d ) p = 8.455 × 10
−27 kg/m
3
and P p = ω e f f p ρ p c
2
= −3.046 × 10
−10 Pa
Using A and B,
¨
R p == +2.570 × 10
−10 s
−2 (Table 2).
C. Sivakumar
Table 1 Choice of “A”
12/(1+A)
A
. . .
. . .
24
–0.500
26
–0.538
28
–0.571
30
–0.600
. . .
. . .
∞
–1.000
Table 2 Scale factor R(ω ef f )
ω ef f
R
+0.333
R i
–0.333
R 0
–0.400
R p
–0.571
∞
If the law of partial pressures is true for the cosmic fluid at least presently, with
−0.6 for dark energy parameter,
ω e f f p = 0.25ω m(d) − 0.42 ≈ −0.4 if the present dark matter parameter is that
small and positive even if it contributes, gives
R 0
R p
= 0.718 (−0.4 is a good fit as
present parameter ω e f f p is less than −0.333 but away enough from −0.571). The
dark matter parameter could be thus around +0.08 (note that if ω dp = −1.0 as CDM
says, ω m(d) p ≈ +1.2).
Substituting H p in Eq. (6), R p = 5.420 × 10
26 .
Note that order(R) is as per [2].
R 0
R i
and
R 0
R p
give R i = 1.0 × 10
26 and R 0 = 3.9 × 10
26 .
Substituting H p in Eq. (7),
ρ p = (ρ m(n) + ρ m(d) + ρ d ) p = 8.455 × 10
−27 kg/m
3
and P p = ω e f f p ρ p c
2
= −3.046 × 10
−10 Pa
Using A and B,
¨
R p == +2.570 × 10
−10 s
−2 (Table 2).