PHYS 555 Problem Set 3 Solutions 1. (a) The universe passed from matter domination to Λ domination when ρm = ρΛ. Thus if 0 3 0 a0 = 1 today, Ωmρcrit/a = ΩΛρcrit gives 1/3 amatt Λ = (Ωm/ΩΛ) = 0.72 (1) 0 where Ωm = Ωb + Ωdm = 0.27 and ΩΛ = 0.73. Since Tγ = Tγ /a, the temperature at matter-Λ equality is Tγ matt Λ = 2.723 K/0.72 = 3.79 K . (2) (b) The heaviest mν consistent with the given data is when the neutrinos are non-relativistic. Non-relativistic neutrinos contribute to the dark matter, so we can raise their mass only until Ων = Ωdm = 0.23 before we would change Ωdm. The neutrinos froze out when they 2 3 were relativistic, so their number density below freeze out is ρν = (ζ(3)˜g/π )Tν , where Tν is redshifted from the temperature at freeze out. Thus ζ(3) ρ0 = g˜ (T 0)3m ν π2 ν ν ν ζ(3) 3 · 6 4 3 = ( )T 0 m π2 4 11 γ ν −12 4 = 2.57 × 10 (mν/eV) eV . (3) 0 0 2 −11 4 Equating ρν with ρdm = 0.23ρcrit = (h/0.72) 0.97 × 10 eV gives 2 mν = (h/0.72) 3.76 eV (4) as the largest (common) neutrino mass possible. This is another example of how cos- mology can be used to place limits on neutrino physics - recall that the light element abundances constrain the number of neutrino species that were in equilibrium during BBN. Slightly more stringent limits on the sum of all neutrino masses are found when their effect on structure formation is taken into account. 0 By the way, the above expression for ρν gives 0 0 Ων = ρν/ρcrit 0.722 m = 0.061 ν , h eV m Ω h2 = ν ν 31.5 eV P m ≡ ν , (5) 94.6 eV where the last line allows for the neutrinos to have different masses. This expression is often found in the literature. 1 2. Now we consider a universe with keV mass neutrinos and no other forms of dark matter, but otherwise identical properties to ours. In this universe the neutrinos still freeze out at 1/3 roughly 1 MeV, the photon temperature is still boosted to (11/4) Tν below 0.511 MeV (due + − to e e → γγ), but now the neutrinos switch from ρrad to ρmatt earlier than in our universe and the subsequent evolution is altered. (a) Assuming matter-radiation equality occurs when the keV neutrinos are non-relativistic, ρm = ρb + ρν. Since the baryon to photon ratio η is the same and since nν ' nγ, the 9 3 neutrinos are of order 10 times more numerous than the baryons. Thus ρν = O(10 )ρb, and we can ignore ρb compared to ρν = nνmν. So matter-radiation equality ζ(3) 3 4 π2 · 6 · T 3m = T 4 (6) π2 4 11 γ ν 15 γ occurs at a photon temperature of 270ζ(3) T = m = 303 eV. (7) m 11π4 ν This corresponds to a neutrino temperature of 216 eV, meaning that matter-radiation equality was reached just after the neutrinos became non-relativistic. We are therefore close to violating our assumption that the neutrinos were non-relativistic at matter- radiation equality, but we will not treat this more carefully here. −3 3 (b) We continue to ignore ρb in ρm. ρm now decreases as a = (Tγ/Tm) until ρm equals 0 2 −11 4 ρΛ. Since ρΛ = 0.72ρcrit = (h/0.72) 3.0 × 10 eV , we have matter-Λ equality when π2 T 3 (303 eV)4 Λ = (h/0.72)2 3.0 × 10−11 eV4 , (8) 15 Tm which occurs at a temperature of 2/3 −7 TΛ = (h/0.72) 1.76 × 10 Tm = (h/0.72)2/3 0.62 K . (9) This is lower than the 3.79 K photon temperature for matter-Λ equality in our universe. The keV neutrino universe becomes matter dominated earlier than ours, and requires 3 a greater increase in the scale factor for (a keV/a) to reach ρΛ than does a universe in 4 which ρ goes as (a keV/a) for part of the way. (c) When H has the value H0 that we measure in our universe, the energy density in −4 radiation is of order 10 ρcrit. It is even less in the keV neutrino universe, so when H = H0 there, ρ again consists of 73% Λ and 27% matter. So ΩΛ = 0.73 and Ωm = 0.27 as they are with light neutrinos. We will work out Ωb last, but for now we note that Ωdm = Ων = 0.27 is 0.04 larger than the light neutrino case because the ρb that used to account for this is now small. Next, note that since ΩΛ and Ωm are the same, the increase in the scale factor from matter-Λ equality to when H = H0 is the same. This gives us 0 the photon temperature when H = H0 with keV neutrinos, Tγ = amatt Λ0.62 K = 2/3 (h/0.72) 0.45 K. Finally, η and the photon number density nγ give us nb, from which 2 we find Ωb: nbmb ηnγmb Ωb = = ρcrit ρcrit 2 0 3 η(ζ(3)/π ) · 2 · (T ) mb = γ ρcrit η(h/0.72) 1.42 × 10−14 eV3m = b (h/0.72)2 4.2 × 10−11 eV4 0.72 0.72 = η 3.17 × 105 = 2.1 × 10−4 (10) h h for η = 6.5 × 10−10. In summary, Ωdm = Ων = 0.27 , (11) ΩΛ = 0.73 , (12) 0.72 Ω = 2.1 × 10−4 , (13) b h h 2/3 and T 0 = 0.45 K. (14) ν 0.72 0 Ωdm has increased by the 0.04 that Ωb decreased, and the ‘current’ temperatures Tγ and 0 Tν - the temperatures at which H = H0 with keV mass neutrinos - are lower than the temperatures at which H = H0 with light neutrinos. 3.