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Theoretical Astrophysics and Cosmology Week 3

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Theoretical Astrophysics and Cosmology Week 3 Theoretical Astrophysics and Spring Semester 2017 Cosmology Prof. L. Mayer, Prof. A. Refregier Week 3 Fulvio Scaccabarozzi Issued: 15 March 2017 Andrina Nicola Due: 22 March 2017 Exercise 1 Neutrinos are coupled to the thermal bath through weak interaction processes like − + e + e νe +ν ¯e: (1) The cross section for these processes is given by k T 2 σ ' 5 × 10−43 B cm2: (2) weak 1MeV As the temperature T of the universe drops, the reaction rate drops much more rapidly than the expansion rate and the neutrinos decouple, i.e. stop interacting. A given particle will start to decouple as soon as its reaction rate falls below the expansion rate of the universe. We can get an estimate of the moment of decoupling of a given particle by determining when the characteristic expansion time equals the time it takes for a reaction to occur i.e. 1 1 = ; (3) Γ H where Γ(T ) = nν(T )hσ(T )vi is the reaction rate and H is the expansion rate of the universe. We make the assumption that neutrinos are massless, i.e. v = c and we also assume that the neutrinos are in thermal equilibrium at temperature T before decoupling. This means that the number density of neutrinos before decoupling is given by 3 3 ζ(3) kBT nν(T ) = gν ; (4) 4 π2 ~c since the neutrinos are fermionic particles. The number of degrees of freedom for neutrinos is given by gν = 2Nν, where the number of neutrino species in Nν = 3. This is due to the fact that each neutrino species has only one degree of freedom, because neutrinos are always left-handed, whereas antineutrinos are always right-handed. We can estimate the expansion rate of the universe at neutrino decoupling by assuming that the universe is radiation dominated and that the relativistic particles present at these times are the photons, the electrons/positrons and the three neutrino species, i.e. 8πG H2(T ) = ρ (T ) ; (5) 3c2 r 1 2 4 π (kBT ) 7 where ρr(T ) = g∗ 3 and g∗ = 2 + (4 + 6) = 10:75. 30 (~c) 8 Setting H(T ) = Γ(T ) (6) then gives us the decoupling temperature of the neutrinos as Tdec = 0:9 MeV: (7) We can estimate the redshift of neutrino decoupling by noting that the neutrino and photon temperature was equal at the time of decoupling. Since we know that the photon temperature 1 scales as a we have T 1 + z = : (8) T0 1 + z0 Setting T0 = 2:73 at z0 = 0 gives us an estimate for the neutrino decoupling redshift of 10 zdec ' 10 : (9) The scale factor during radiation domination is given by 1 4 a 32πGρr;0 1 = t 2 : (10) a0 3 −34 −3 Inserting ρr;0 ≈ 7:8 × 10 g cm gives us the time of neutrino decoupling as tdec ' 1 s: (11) 1 By noting that the temperature of the neutrinos after decoupling scales as a , we can estimate the temperature of the neutrino background at redshift z = 1000 as 1 + z1000 T1000 = Tdec ' 1000 K: (12) 1 + zdec Exercise 2 The following equation follows from the conservation of the stress-energy tensor Tµν: dρ ρ + P=c2 + 3 = 0: (13) da a If multiplied by c2, it can be written in terms of energy density = ρc2: d + P + 3 = 0: (14) da a We multiply it by the volume V / a3 d a3 + 3a2 + 3P a2 = 0 da (15) d dV ( V ) = −p ; da tot da to get the first law of thermodynamics where the entropy is conserved: dU + pdV = 0 ! dS = 0. 2 Exercise 3 In order to find the temperature Ts at which baryons and antibaryons stop annihilating sig- nificantly we need to find the freeze-out temperature of the annihilation reaction ¯ b + b γ + γ: (16) We thus need to find the temperature at which the reaction rate equals the expansion rate i.e. 1 1 = ; (17) Γ H where Γ(T ) = nb(T )hσ(T )v(T )i is the reaction rate and H(T ) is the expansion rate of the universe. In order to find the reaction rate we note that a given species starts to become non relativis- tic when the temperature of the Universe falls below its rest mass. The mass of the heaviest baryons is approximately mb ≈ 1 GeV and therefore the dominant (by mass) baryonic/anti- baryonic species slowly become non relativistic when T ≤ 1 GeV. Their number density can thus be expressed as: 3=2 2 mbkBT mbc nb = gb 2 exp − : (18) 2π~ kBT From the Friedmann equation we know that the Hubble parameter is given by 8πGρ(T ) H2 = : (19) 3c2 Since the radiation density dominates at high T , we may assume this is the only source of energy in the Universe (radiation-dominated epoch), i.e. we can set 2 4 π (kBT ) ρ(T ) = ρr(T ) = gr ; (20) 30 (~c)3 where gr denotes the effective degrees of freedom of relativistic species which is given by X 7 X g = g + g ; (21) ? B 8 F B F where F denotes Fermionic species and B denotes Bosonic species. From Eq. 17 we therefore get 3=2 2 3 4 1=2 mbkBT mbc 8π Ggr (kBT ) gb 2 exp − hσ(T )v(T )i = 3 5 : (22) 2π~ kBT 90 ~ c From the exercise sheet we know that ~2c hσ(T )v(T )i = 2 ; (23) mπ where mπ denotes the mass of the pion. By solving this equation numerically for gr = gγ + 7 43 7 8 (3gν + 3gν¯ + ge + ge¯) = 4 and gb = 8 (gp + gp¯ + gn + gn¯) = 7, one obtains Ts ≈ 11 MeV. The ratio between the number density of baryons or antibaryons to the number density of photons at Ts is: 3=2 2 mbkB Ts mbc gb 2 exp − nb 2π~ kB Ts −35 = 3 = 3:5 · 10 ; nγ 2:404 kB Ts π2 ~c (24) n + n b ab ≈ 7 · 10−35: nγ 3 This ratio will remain constant regardless of the expansion of the Universe, since both number densities drop with a−3. The relic number density of baryons and anti-baryons would therefore be 3 · 10−32 cm−3. If the Universe had no baryon/anti-baryon symmetry, the baryonic matter density would be many orders of magnitude smaller and thus the total matter density would be significantly reduced. The dark matter will be the only matter component relevant for the dynamics of the Universe as a whole. This effect will not modify the succession of the phases of the Universe. Nevertheless it will modify the length of the periods, when a certain component dominates. With the matter component reduced by about 20%, the radiation era will last longer and the change from the matter dominated to Dark Energy dominated era will take place earlier. 4.
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