Computational Applications in Nuclear Astrophysics using JAVA

Transcription

1 Computational Applications in Nuclear Astrophysics using JAVA Lecture: Friday 10:15-11:45 Room NB 7/67 Jim Ritman and Elisabetta Prencipe Computer Lab: Friday 12:15-13:45 NB 7/67 Michael Kunkel 53

2 1) JR 2) EP 3) EP 4) JR 5) JR 6) EP 7) JR 8) EP 9) JR 10) EP 11) EP 12) JR 13) EP 14) JR Lecture Dates 54

3 Contents Introduction 1 Basics of nuclear physics 2 Big Bang 1 BB-Neutron/Proton Ratio 1 BB-Nucleosynthesis light nuclei 1 Stellar atmospheres 3 H-burning cycles 1 He-burning Supernova 1 s, r, r p, a p processes 1 Calc. of abundances for Z>Fe/Ni 2 55

4 Review of Previous Lecture Cosmic distance ladder Parallax, Star stream parallax, photometric (HR) Standard Candles: d Cepheids, SN Ia, Tully-Fischer Hubble constant Universe is neither static nor infinite! CMB T=2.73 K Friedmann equation Radiation ßà matter dominated universe: r r µ a -4 r m µ a -3 WMAP power spectrum à W, h Problems of SM (horizon, flatness) à inflation 56

5 Particle Creation after the Big Bang 57

6 Mean Free Path in the Cosmos Today a proton can travel across the universe without interaction. Thus the mean free path is larger than cm. When the universe was 1 second old, the mean free path of a photon was about the size of an atom. The corresponding large number of interactions kept the interacting constituents in thermal equilibrium. As the universe expanded the mean free path increased and at some point the interaction rate was no longer sufficient to maintain thermal equilibrium. àthere must have been (a) phase transition(s) 58

7 Assumptions of the Big-Bang Model In the early phases of the universe it was hot enough so that all particles were in thermal equilibrium (equal rates in both directions) (mass differences, e.g. 2M g 2M e generate different phase space generates different abundances) The physical laws that we know today were valid then. Continue à 59

8 Further Assumptions The number of leptons is much smaller than the number of photons. neutrinos and antineutrinos are not degenerate (Majorana n). The number of baryons is non-zero (positive). That requires baryon number conservation to be violated in the early universe. Only particles that exist today participated in the primordial nucleosynthesis. All other particles decayed before the BBN-era. The cosmological principle is valid (homogeneous and isotropic). General Relativity is the correct description of gravitation. From these assumptions we can reconstruct the primordial nucleosynthesis 60

9 Basic Equations The time scale of the expansion (Friedmann-Equation) NB! R = a Energy conservation follows from du = -PdV (adiabatic) R(t) = a(t) r G P : Scale factor : Energy density : Gravitation constant : Pressure 61

10 Chemical Potential There are four conserved quantities, which constrain the possible reaction steps: Electric charge Baryon number Electron-type Lepton number Muon-type lepton number These imply 4 independent, additive chemical potentials for the possible reactions : µ p, µ e, µ ne, µ nµ The chemical potential can be derived from the corresponding densities: N Q : Charge density N B : Baryon density N e : Electron-type lepton-density N µ : Muon-type lepton-density In general: N i = N i (µ p, µ e, µ ne, µ nµ ) 62

11 Chemical Potential Assumption in the Big-Bang-Standard-Model N Q = 0 (charge neutrality) N B, N e, N µ << N g è N B = N e = N µ = 0 One also has: à all N i must be odd functions µ i à µ i = 0 in chemical equilibrium à Distributions depend only upon the temperature 63

12 Distributions Photons: Planck-Distribution Photon Energy density Leptons: Fermi-distribution (c=1, momentum q) 64

13 Thermal Equilibrium In thermal equilibrium one has: Energy density Pressure S i : Sum over all particles in thermal equilibrium 65

14 Temperature Scale Factor Second law of thermodynamics (in equilibrium) (1) gives: (2) Combined with energy conservation it follows that: (3)

15 Temperature Scale Factor Inserting (2) in (1) we get: And thus: Change V à R 3 and Equation (3) implies: (4) The expansion is adiabatic! 67

16 Ultrarelativistic Particles For ultrarelativistic particles E = q (c = 1), energy and pressure are connected: From (2) we have: with the solution Together with (3): It follows that: 68

17 Which Particles are in Thermal Equilibrium? Only particles with mass m < kt are in thermal equilibrium with non-negligible particle number densities Below T ~ K (» m N ) à Baryongenesis will occur at about T = 1.5x10 12 K (» m p ) this includes: µ, e, n µ, n e, g, + c.c. µ, e, n µ, n e : Fermi-distribution g : Planck-distribution Depending on the mass of the particle, these leave thermal equilibrium as the temperature drops. Afterwards their particle number density is reduced by the Boltzmann-Factor exp(-m/kt). In our case this first effects the muons with m=105 MeV/c 2. 69

18 Radiation Dominated Universe In the range 5x10 9 K < T < K the universe is dominated by ultra-relativistic particles From the Friedmann-Equations we have: Which is solved by: For 10 9 K < T < 5x10 9 K numerical solutions are needed Note: R(t) ~ Öt for a radiation dominated universe 70

19 The Early Universe The universe cooled down as it expanded: T(t) ~ 1/R Particles were in thermal equilibrium as long as the reactions were faster than the expansion. i.e. particles + antiparticles «photons For kt << Mc 2 particles and antiparticles annihilate For T >> K there was a small excess of particles over antiparticles, this violated baryon/lepton-number conservation At T = K all antinucleons annihilated The remaining nucleons (B/g = h ~ 10-9 ) are the components for the BBN 71

20 t = s T = K Soup of: Quark-Gluon- Plasma, Neutrinos, Electrons, Photons Early Universe 72

21 t = 3x10-5 s T = 2x10 12 K Thermal energy about equal to the rest mass of protons and neutrons If the temperature sinks, then no free quarks can be created out of the radiation field Quark-antiquark pairs annihilate, except for ~ 10-9 fraction of remaining quarks Where are the antiquarks?! The remaining quarks bind to form protons and neutrons (baryogenesis) Early Universe 73

22 Where is the Antimatter? Why do we exist at all? After the BB, there were equal numbers of particles and antiparticles. During the expansion and cooling down, the quarks and antiquarks should (nearly) all annihilate. This is quantified by h = net baryon / photon ratio One might expect h~10-20 by random chance Observation (WMAP) h~10-9 Why do we have 11 orders of magnitude too much matter (and correspondingly too little antimatter)?? 74

23 Where is the Antimatter? On Earth antimatter can only be produced at high energy accelerators, (or very briefly after b + decay) Cosmic rays: antiproton / proton ~10 4 ; are mostly secondary products from cosmic ray collisions with ISM. Proof for baryon-antibaryon asymmetry on galactic scale. Spatial separation?? If matter and antimatter galaxies existed in the same galaxy cluster, expect strong g-ray radiation from annihilation. The absence of that flux indicates that clusters of galaxies ( M ) are either all baryons or all antibaryons. 75

24 Sakharov Conditions The most reasonable assumption for the baryon/antibaryon asymmetry is that the very early Universe T > 38 MeV already possessed an asymmetry between matter and antimatter, which prevented the annihilation catastrophe. There are three basic ingredients necessary to generate a non-zero baryon number from an initially baryon symmetric state (Sakharov,1967): 1. baryon (and lepton) number violation 2. C and CP violation (charge conjugation C, parity P) 3. non-equilibrium conditions By RIA Novosti archive, image #25981 / Vladimir Fedorenko / CC-BY-SA 3.0, CC BY-SA 3.0, index.php?curid=

25 Sakharov Conditions 1 Baryon number conservation must be violated. Otherwise only the initial conditions could cause the observed asymmetry. GUT predict this violation, but minimal GUT models predict proton lifetimes (~10 31 a) that are experimentally ruled out. p + e + + p 0 The best limits come from Super-Kamiokande analysis gives halflife > 1.67 x years SUSY predictions years Hyper-Kamiokande will be factor 5-10 more sensitive 77

26 Sakharov Conditions 2 Even with B/L number violation, must have C and CP violation in order to produce an asymmetry. - P is maximally violated in weak interactions - CPV is observed in many neutral K and B meson decays, but that is not enough by O(10 9 ) to explain the measured h Thus, the search for further CPV processes is a major activity in current particle physics. Given CPT, CPV equivalent to time reversal viol. Only by observing CPV processes can we be sure that distant clusters of galaxies are really matter and not antimatter 78

27 Sakharov Conditions 3 CPT invariance demands particle and antiparticle masses to be equal. Also entropy is maximal when chemical potentials of non-conserved numbers = zero Thus, in thermal equilibrium the phase space density of baryons and antibaryons are necessarily identical. Non-equilibrium conditions given by the expansion. As the density and temperature drop, at some point the expansion rate will exceed the particle interaction rate 79

28 Thermal Equilibrium for Interacting Particle Species 80

29 Boltzmann Equation for Annihilation Boltzmann eq. generalized Friedmann s 2 nd eq. to describe the abundance varies with time. For one species in equilib., without annihilation and creation n is the number density (abundance) of a particle species. Boltzmann eq. for 4 species connected via 1+2 ßà 3+4 the rates n 1, n 2, n 3 and n 4 are connected via: 81

30 Boltzmann Equation Production rate of 1 is proportional to n 3 x n 4 Loss rate of 1 is proportional to n 1 x n 2 Production is higher for Bosons than Fermions (+/- sign for Bose enhancement/ Fermi blocking The amplitude/matrix Element is given by the physical process (makes right side = 0 for non interacting particles, see previous page) 82

31 Simplifications of the Boltzmann Eq. Scattering leads to very fast thermal equilibrium in the corresponding BE or FD distributions. Only a constant µ remains, and that is a function of time. In chem. equil. The chemical potentials are related by µ 1 + µ 2 = µ 3 + µ 4. Out of chem. equil. there is a differential equation for µ For cosmological applications addressed here, T is lower than E-µ. As a result the +/-1 drops out of the BE&FD distributions and we have another simplification 83

32 Simplifications of the Boltzmann Eq. The low temperature limit also implies exp[-(e-µ)/t] f<<1 thus 1+/- f 1, and the last line of the Boltzmann equation reduces to: For this we have also used energy conservation E 1 +E 2 = E 3 +E 4 Now we can solve for the number densities n i with the degeneracies of the species g i 84

33 Equilibrium Number Densities We can define the equilibrium number density n i (0) and thus Now the final line of the Boltzmann eq. becomes: 85

34 Saha Equation If we define the thermally averaged cross section: Now the Boltzmann equation reduces to: The left side is proportional to Hn 1 the right to n 1 n 2 <sv> If the reaction rate is much larger than the expansion rate, then n 2 <sv> >> H and the part in brackets must be 0 This gives us the Saha Equation 86

35 End of the Fifth Lecture 87