Chapter 2Baryons, Cosmology, Dark Matter and Energy2.1 Hubble expansionWe are all aware that at the present time the universe is expanding. However, what willbe its ultimate fate? Will it continue to expand forever, or will the expansion slow andfinally reverse? In order to see what role the constituent matter and energy – baryons,photons, neutrinos, and other stuff not yet identified – of our universe may play in answeringthis question, we explore their effects in an expanding homogeneous and isotropic universe.Consider a small test mass m which sits on the surface of a spherical chunk of this universehaving radius R. If the mean energy density of the universe is ρ, then the mass containedinside the spherical volume isM(R) =43πR3ρThe potential energy of the test mass, as seen by an observer at the center of the sphere, isU = −GM(R)mRwhile its kinetic energy isT =12mv2=12m dRdt!2By Hubble’s Law the expansion velocity is given byv = HRwhere H =1RdRdtis the Hubble constant. Although the While the size of H has been debatedin the past, recent determinations give a rather precise value of 71 ± 4 km/s/Mpc. (Oneparsec = 3.262 light years.) The total energy of the test particle is thenEtot= T + U =12mR2(H2−83πρG)and the fate of the universe depends on the sign of this number, or equivalently with therelation of the density to a critical valueρcrit=3H28πG∼ 1.88 × 10−29h2g/cm3where h ∼ 0.71 ± 0.04 is (today’s) Hubble constant in units of 100 km/s/Mpc. This meansρ∼<ρcrit⇒ continued expansionρ∼>ρcrit⇒ ultimate contraction2.2 Photon, baryon, and neutrino contributions to mass/energy densitySo how does the measured mass/energy density of the universe match up to ρcrit? We can1certainly do one immediate calculation, for photons. You are probably aware that pho-tons remained in thermal equilibrium with the matter as long as there were free protonsand electrons. But just as we calculated the n + p ↔ d + γ equilibrium, we can evaluatethe p + e−↔ H + γ equilibrium, where H denotes the hydrogen atom. Given the ioniza-tion potential of H of 13.6 eV, one can calculate when the photons cool to the point thatphotocapture can no longer efficiently break up newly formed atoms. One can show thiscorresponds to a temperature of about 1 eV and to a time about 380,000 years after theBig Bang. After this point, the photons decouple from the matter as they no longer see freecharges to scatter off. This decoupled background of photons is now redshifted to microwaveenergies.For the photon number densitynγ= 2Zd3q(2π)31exp(q/Tγ) − 1= 2ζ(3)T3γ/π2∼ 408/cm3where ζ(3) ∼ 1.20206 is the Riemann zeta function and Tγthe today’s cosmic microwavebackground temperature, measured (with great accuracy) to be about 2.73 K. Similarly forthe energy density in photonsργ= 2Zd3q(2π)3qexp(q/Tγ) − 1= π2T4γ/15 ∼ 4.6 × 10−34g/cm3It follows that photons contribute only 0.0000485 of the closure density.Now what we did in BBN allows us to estimate the baryonic (or nucleonic) contributionto the ρ as well. The baryon to photon number density is η, which either BBN or cosmicmicrowave background studies finds to beηBBN= (5.9 ± 0.8) × 10−10ηCM B= (6.14 ± 0.25) × 1010So these values are in great agreement. Using the CMB value, we then findnnucleons= ηCM Bnγ= 2.51 × 10−7/cm3and thus multiplying by the average nucleon mass (a detail – but we know the n/p ratio is1/7 for doing this average)ρb= 4.19 × 10−31g/cm3∼ 0.0442ρcritThat is, baryons provide only 4.4% of the closure mass. Clearly the electron contributionto ρ, ρe∼ (6me/7mN)ρb, is then neglible, about 2 ×10−5of ρcrit, comparable to the photoncontribution.2One can count the “visible” nucleons, by integrating over all of the luminous matter in starsand gas clouds. Interestingly this yieldsρvis∼ 0.02ρcritThus one concludes roughly half of the baryons are not visible. Presumably these nucleonsare some place – perhaps nonluminous gas clouds – because we believe BBN, and becausethe BBN prediction for η is now confirmed by CMB results. This problem is sometimescalled the dark baryons problem – though there are even more intriguing “dark” problems.A second dark problem has to do with large-scale gravitational interactions of galaxies,galaxy clusters, etc. For some time it has been clear that the total ρ is much larger thanthat coming from photons and baryons (and electrons). For example, Doppler studies ofthe rotation rates of spiral galaxies indicate that these systems are much more massive thantheir luminosities seem to suggestρrot∼ 20ρvisThis is too large a discrepancy to attribute jus t to the dark baryons. The origin of the “darkmatter” responsible for this discrepancy is a matter of current study: there are several pos-sibilities. But regardless of the origin of the dark matter, it appears that the matter/energydensity of our universe is a lot closer to ρcritthan one would guess from our calculations ofρband ργ.Just as we have a CMB, there will be a relic neutrino spectrum left over from the big bang.These neutrinos would have decoupled when temperatures were s lightly above 1 MeV. Sincethat first second of the big bang, no further interactions have occ urred. If we had somemeans to detect these neutrinos, they would tell us about conditions at that very early time,e.g., their tempe rature fluctuations (probably exceeding tiny!) over the sky would tell usabout the structure of the universe at 1 sec.We do the calculation of the ne utrino contribution to ρ making two assumptions. First isthe assumption that we have three flavors (thus 6 neutrinos in all), as the standard modeltells us, all of which are light. We will see that this is know from both cosmology, and froma combination of tritium β decay and recent discoveries of neutrino oscillations. The upperbound on the masses of the light neutrinos is about 1 eV.With this assumption neutrinos are relativistic when they decoupled. It follows that eachneutrino flavor (e.g., νeand ¯νe) contributes:nν= 2Zd3q(2π)31exp(q/Tν) + 1= 3ζ(3)T3ν/(2π2)It thus followsnν=34(TνTγ)3nγ3What about Tν? In the very early universe electrons, positrons, neutrinos, and photonswould all be relativistic and in equilibrium, characterized by a single temperature. Thenthere is an epoch around 1 MeV when the neutrinos have decoupled, but the electrons arerelativistic and in equilibrium with the photons. Let the