lepton era BBN high T neutrinos in equilibrium reaction proceeds if GF=Fermi constant LHS falls with T more rapidly than RHS!
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3 lepton era BBN high T neutrinos in equilibrium GF=Fermi constant reaction proceeds if LHS falls with T more rapidly than RHS!
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6 PAIR ANNIHILATION
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8 t=10 s t=0.2 s neutrinos decouple
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10 We cannot measure the CNB directly, but due to its effect on the expansion rate of the early Universe the theoretical expectations for small-scale power spectrum of the CMB depend on Nν. According to the latest results from the Planck satellite (03/2013) Nν = 3.30±0.27 at 68% c.l., in very good agreement with the Standard Model expectation.
11 CDM WDM HDM
12 Samples from the Planck TT+lowP posterior in the mν H0 plane, colour-coded by σ8. Higher mν damps the matter fluctuation amplitude σ8, but also decreases H0. Solid black contours show the constraint from Planck TT+lowP+lensing (which mildly prefers larger masses), and filled contours show the constraints from Planck TT+lowP+lensing+BAO.
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14 8.10 A Big Bang Pillar: Light Element Nucleosynthesis The CMB tells us a great deal about the state of the Universe at the time of last scattering (tls = 0.37 Myr). However, the opacity of the early Universe prevents us from directly seeing what the Universe was like at t < tls. Looking at the last scattering surface is like looking at the surface of the Sun, and we would like to find out what conditions are like in the opaque regions so tantalizing hidden from our direct view. Light Element Nucleosynthesis provides such a probe. Theoretically, many properties of the early Universe should be quite simple. For example, when radiation is strongly dominant over matter, i.e. at scale factors the expansion of the Universe has the simple power-law form a(t)~t 1/2. The temperature of the blackbody photons, which decreases as T~a -1 (before and after e + e - annihilation of course) as the Universe expands, is given by the relation or The mean energy per photon was
15 The Large Hadron Collider in Geneva was designed to accelerate protons to an energy of 7 TeV=7,000,000 MeV. Therefore 20 femtoseconds (t 20 x sec) after the Big Bang, the entire Universe was like the LHC! An atomic nucleus contains Z protons and N neutrons, and the total number of nucleons is the atomic mass number, A=Z+N. The binding energy B (B/A=binding energy per nucleon) of a nucleus is the energy required to pull it apart into its component protons and neutrons. Equivalently, is the energy released when a nucleus is fused together from individual protons and neutrons. For instance Note that 4 He, with a total binding energy of B=28.30 MeV, and a B/A=7.07 MeV, is relatively tightly bound compared to other light nuclei. The most tightly bound nuclei are 56 Fe and 62 Ni, with B/A~ 8.8 MeV. Thus nuclei more massive than iron or nickel can release energy by fission -- splitting into lighter nuclei. Nuclei less massive than iron or nickel can release energy by fusion -- merging into heavier nuclei. =D
16 One thing we can say about BBN, after taking a look at the present-day Universe, it that it must have been very inefficient. From an energy viewpoint, the preferred Universe would be one in which the baryonic matter consistent of an Iron-Nickel alloy. Instead, 3/4 of the baryonic density (by mass) is still in the form of H, and 1/4 is in the form of 4 He. The primordial Helium fraction in the Universe (that is, the helium fraction before nucleosynthesis begins in stars) is usually expressed as the dimensionless number The Sun s atmosphere has Y = However, the Sun is made of recycled interstellar gas, which was contaminated by helium formed in earlier generations of stars. When we look at astronomical objects of different sorts, we find a minimum value of Y = NEUTRONS AND PROTONS The basic building blocks for nucleosynthesis are protons and neutrons. The rest-mass energy of a neutron is greater than that of a proton by the factor A free neutron is unstable, decaying via the reaction β-decay The decay time of a free neutron is τn=890 s. That is, if you start with a population of free neutrons, after a time t, only a fraction f = exp(-t/τn) will remain.
17 Since the energy Qn released by the decay of a neutron into a proton is larger than mec 2 =0.5 MeV, the remainder of the energy is carried by the kinetic energy of the electron and that of the anti-neutrino. With a decay time of 15 min, the existence of a neutron is as fleeting as fame: neutrons are still around today because they ve been tied up in deuterium, helium and other atoms in the early Universe. Let s consider the state of the Universe when its age was t=0.1 s. At that time, the temperature was T 3 x K, and the mean energy per photon was <hν>~10 MeV. This energy is much greater than the rest-mass energy of electrons and positrons, so there were electron-positron pairs present at t=0.1 s, created by pair production At t=0.1 s, neutrons and protons were in equilibrium with each other, via the interactions As long as n and p are kept in equilibrium by the reactions above, their number density is given by the Maxwell-Boltzmann equation Since the statistical weights of protons and neutrons are the same, gn=gp=2, the neutron to proton ratio is
18 t=1 sec 2ct=6x10 10 cm~r! n freeze-out If neutrons and protons remained in equilibrium, by the time the Universe was 6 min old, there would be one neutron for every million protons. However, neutrons and protons do not remain in equilibrium that long. The interactions that mediate between neutrons and protons in the early Universe involve neutrinos and the weak nuclear force with cross-section In the early radiation-dominated Universe
19 The number density of neutrinos falls as nν~a -3 ~t -3/2, and hence the rate Γ with which neutrons and protons interact with neutrinos via the weak force falls rapidly while the Hubble parameter H~t -1. When Γ=H, the neutrinos decouple from the neutrons and protons, and the ratio of neutrons to protons gets frozen (at least until the neutrons start to decay, at time t~τn). An exact calculation of the temperature Tfreeze at which Γ=H shows that and that tfreeze~ 1 s. Once the temperature drops below Tfreeze, the neutron-to-proton ratio is frozen at the value At times tfreeze < t << τn there was one neutron for every five protons in the Universe! The scarcity of neutrons relative to protons explains why BBN was so incomplete, leaving 3/4 of the baryons in the form of unfused protons. A neutron will fuse with a proton much more readily than a proton with another proton. Compare, e.g., the proton-proton fusion reaction to Deuterium The involvement of a neutrino tells us that this reaction has got a minuscule weakinteraction cross-section, of order σw. By contrast, the neutron-proton fusion
20 involves no neutrinos and is an interaction involving the strong nuclear force (much larger cross-section than σw). Also, since protons are all positively charged, they must surmount the Coulomb barrier in order to fuse. Note that proton-proton fusion is occurring in the Sun even as you read this sentence. Fusion in the Sun, however, it is a very slow process. If you pick out any particular proton in the Sun s core, it has only one chance in 10 billion of undergoing fusion during the next year! Only exceptionally fast protons have a (tiny) probability of quantum tunnelling through the Coulomb barrier of another proton an fusing with it. The core of the Sun is, however, in hydrostatic equilibrium and its temperature and density change only slowly with time. BBN, by contrast, is a race against time. After less than an hour, the temperature and density have dropped too low for fusion to occur!
21 What is the maximum value of Y, the fraction of the baryon mass in 4 He? Suppose that every neutron present after neutron-proton freeze-out is incorporated into a 4 He nucleus. Given a neutron-to-proton ratio of nn/np=1/5, we can consider a representative group of 2 neutrons and 10 protons. The 2 neutrons can fuse with 2 protons to form a single 4 He nucleus. The remaining 8 protons will remain unfused. The (maximum) mass fraction of 4 He will then be More generally, if f nn/np, then the maximum possible value of Y is N N P P P P P P P P P P Let s move to the next stage of BBN, just after proton-neutron freezeout is complete and neutrinos have decoupled from the rest of the Universe. The time is t~2 s. The essential first step in BBN is the fusion of a proton and a neutron to form a Deuterium nucleus: The energy released (and carried away by a gamma-ray) is the binding energy of Deuterium, B=2.22 MeV. Conversely, a photon with energy > B can photodissociate a deuterium nucleus into its component proton and neutron.
22 Around the time of Deuterium synthesis, the relative numbers of free protons, neutrons, and Deuterium nuclei are given by the analogous of Saha equation which tells us that Deuterium is favored in the limit kt 0, and free neutrons and protons are favored in the limit kt. Let us define the temperature TD at which Deuterium synthesis takes place as the temperature at which nd/nn=1; that is, the temperature at which half the free neutrons have fused into Deuterium nuclei. We can now write the D-to-n ratio as a function of T and the baryon-to-photon ratio η=5.5x (note that D-synthesis happens after the e + -e - annihilation era). Before the start of Deuterium synthesis, 5 out of 6 baryons (or ~83%) are in the form of unbound protons. Thus Substituting, one gets Setting the LHS=1 we derive the temperature of D-synthesis as ktd=0.066 MeV or TD=7.6 x 10 8 K. The temperature drops to this value when the age of the Universe is td=200 s. Note that the time delay until the start of nucleosynthesis is not negligible compared to the decay time of a free neutron, τn=890 s...
23 By the time D-synthesis gets under way, neutron decay has slightly decreased the neutron-to-proton ratio from nn/np=1/5 to This in turn lowers the maximum possible 4 He mass fraction from Ymax=0.33 to Ymax=0.27! BEYOND DEUTERIUM Once a significant amount of D forms, many possible nuclear reactions become available... D, 3 H, 3 He all efficiently converted into 4 He
24 Because there are no stable nuclei with A=5, the orderly march to heavier and heavie nuclei reaches a roadblock. Fusion of 4 He with a proton or a neutron will not work since 5 He and 5 Li are not stable nuclei. Small amounts of 6 Li, 7 Li, and 7 Be, two stable isotopes of Lithium and one of Berillium, are made by reactions such as 3 The synthesis of nuclei with A>7 is hindered by the absence of stable nuclei with A=8. For example, if 8 Be is made by the reaction then the 8 Be nucleus falls back apart into a pair of 4 He nuclei with a decay time of only 3 x s.
25 H=protons N=neutrons Yp= 4 He
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