Primordial Black Holes and Cosmological Phase Transitions Report ...

PBHs and Cosmological Phase Transitions 5
Section 2.3). The conservation of entropy leads to the useful relation (e.g.
Schmid et al., 1999)
dT 3s
= − . (15)
d ln R ds/dT
In the following we consider the solutions of equation (2) when a single component
dominates the energy density. Taking into account that, according to
observation, we live in a flat Universe, we consider κ = 0. Note that even if
κ = 0 at early times (when the scale factor is smaller) we can neglect the term
κ/R 2 in equation (2) as long as w>−1/3 (e.g. Yao et al., 2006).
Inserting equation (10) into equation (2), with κ = 0 and Λ = 0, one obtains
(e.g. Yao et al., 2006)
2
R(t) ∝ t 3(1+w) . (16)
For a radiation–dominated Universe (w =1/3), equation (16) becomes
R(t) ∝ t 1/2 . (17)
The radiation and matter densities in the Universe decrease as the expansion
dilutes the number of atoms and photons. Radiation is also diminished due to
the cosmological redshift, so its density falls faster than that of matter. When
the age of the Universe was ∼ 10 6 years it became matter–dominated. Now it
is appropriate to assume an EoS corresponding to a pressureless gas (w = 0).
For this matter–dominated Universe, equation (16) becomes
R(t) ∝ t 2/3 . (18)
This might also be the case if the Universe experienced a dust–like phase during
a phase transition on the radiation–dominated epoch (e.g. Carr et. al., 1994).
If the Universe is dominated by a positive cosmological constant Λ then we
will have an EoS with w 0 leads to (e.g. Yao et al., 2006)
Λ
R(t) ∝ exp
3 ct
(19)
which corresponds to an exponential expansion of the Universe.
Assuming that light propagates in Relativistic Cosmology in the same way
as it does in General Relativity we will consider now how an observer O receives
light from a receding galaxy. Without loss of generality we will take O to be
at the origin of coordinates r = 0. Inserting the conditions for a radial null
geodesic into the line element (1) we have (e.g. d’Inverno, 1993)
dt
R(t)
dr
= ±
(1 − kr) 1/2
(20)