Proc. Neutrino Astrophysics - MPP Theory Group

152
We can therefore calculate their energy density as for the photons, and convert this to the
equivalent cosmic density parameter. The result is
Ων,0 = uν,0
c 2 ρcr
= 2.8 × 10 −6 h −2
per neutrino species. Likewise, their number density is given by the Fermi distribution with
temperature Tν,0,
nν,0 = 3ζ(3)
2π2 � �3 kTν,0
≈ 113cm
¯hc
−3
(18)
per neutrino species. Suppose now there is one neutrino species which has rest mass mν > 0,
while the other species are massless. For the dark matter to be dominated by neutrinos, mν
needs to satisfy
mν nν,0 ! = Ω0 ρcr , (19)
hence,
(17)
mν c 2 ≃ 95eV (Ω0h 2 ) . (20)
Adding the energy density of the two massless neutrino species to that of the photons, the
total equivalent mass density in relativistic particles is
The Horizon
ΩR,0 = 3.0 × 10 −5 h −2 . (21)
The size of causally connected regions of the Universe is called the horizon size. It is given
by the distance a photon can travel in the time since the Big Bang. Since the appropriate
time scale is provided by the inverse Hubble parameter, the horizon size is d ′ H = cH(a)−1 ,
and the comoving horizon size is
dH(a) =
c c
= Ω
aH(a) H0
−1/2 a n/2−1 , (22)
where we have inserted the Einstein-de Sitter limit (10) of Friedmann’s equation. cH −1
0 =
3h −1 Gpc is called the Hubble radius.
The previous calculations show that the matter density today is completely dominated
by ordinary rather than relativistic matter. But since the relativistic matter density grows
faster with decreasing scale factor a than the ordinary matter density, there had to be a time
aeq ≪ 1 before which relativistic matter dominated. The condition
yields
ΩR,0 a −4
eq
!
= Ω0 a −3
eq
(23)
aeq = 3.0 × 10 −5 (Ω0h 2 ) −1 . (24)
We shall see later that the horizon size at the time aeq plays a very important rôle for
structure formation. Under the simplifying assumption that matter dominated completely
for all a ≥ aeq (i.e. ignoring the contribution from radiation to the matter density), eq. (22)
yields
dH(aeq) = c
H0
Ω −1/2
0 a 1/2
eq ≈ 13(Ω0 h 2 ) −1 Mpc . (25)