Nuclear Astrophysics

Transcription

1 I. Hydrostatic burning and onset of collapse Karlheinz Langanke GSI & TU Darmstadt 15

2 Stellar energy source Energy comes from nuclear reactions in the core. E = mc H 4 He + neutrinos MeV The Sun converts 600 million tons of hydrogen into 596 million tons of helium every second. The difference in mass is converted into energy. The Sun will continue burning hydrogen during 5 billions years. Energy released by H-burning: erg g 1 Solar Luminosity: erg s 1

3 Coulomb barrier dominates stellar burning As a star forms, density and temperature increase in the center. Fusion of hydrogen is the first long-term energy source that can ignite as it has the smallest Coulomb barrier: For first-generation stars (Population III) the ppi chain is the only possible sequence of reactions. 3 or 4 body reactions are very unlikely. Chain has to proceed by 2-body reactions or decays.

4 The ppi chain Step 1: p + p 2 He (not possible) p + p d + e + + ν e Step 2: d + p 3 He d + d 4 He (d abundance too low) Step 3: 3 He + p 4 Li ( 4 Li is unbound) 3 He + d 4 He + n (d abundance too low) 3 He + 3 He 4 He + 2p d + d is not going, because Y d is extremely small and d + p leads to rapid destruction. 3 He + 3 He works, because Y 3He increases as nothing destroys it.

5 The ppi chain

6 4 He as catalyst 4 He can act as catalyst initializing the ppii and ppiii chains. With which nucleus will 4 He fuse? protons: the fusion of 4 He and protons lead to 5 Li which is unbound. deuterons: the fusion of deuterons with 4 He can make stable 6 Li; however, the deuteron abundance is too low for this reaction to be significant 3 He: 3 He and 4 He can fuse to 7 Be. This is indeed the break-out reaction from the ppi chain. Once 7 Be is produced, it can either decay by electron capture or fuse with a proton. Thus, the reaction sequence branches at 7 Be into the ppii and ppiii chains.

7 The solar pp chains 1 1 H + 1 1H 2 1 H + 1 1H 2 1 H + e + + ν e 3 2 He + γ 85% 15% 3 2 He + 3 2He 4 2 He H 3 2 He + 4 2He 7 4 Be + γ (PP I) 15% 0.02% 7 4 Be + e 7 3Li + ν e 7 3 Li + 1 1H 2 4 2He (PP II) 7 4 Be + 1 1H 8 5 B + γ 8 5 B 8 4 Be + e + + ν e 8 4 Be 2 4 2He (PP III)

8 The other hydrogen burning: CNO cycle requires presence of 12 C as catalyst

9 Solar neutrino fluxes and detector thresholds Solar hydrogen burning produces neutrinos Depending on material, the detectors are blind for neutrinos with energies smaller than a threshold.

10 Neutrino astronomy In the 1950s, Ray Davis (2002 Nobel Prize Laureate) decided to measure the solar neutrinos. (Every second, 10 billion neutrinos pass through every square cm on Earth). In 1967, the detector (615 tons of C 2 Cl 4 ) was installed at Homestake Gold Mine, South Dakota (1,500 m depth). In 1968, the first measurement was a factor 3 smaller than predictions. Similar results by other experiments. Super-Kamiokande, Japan (50,000 tons pure water) Sudbury Neutrino Observatory, Canada (1,000 tons heavy water)

11 Detecting solar neutrinos Homestake: first observation of solar neutrinos detection ν e + 37 Cl e + 37 Ar blind for E ν < 814 kev Kamiokande, Super-Kamiokande: proof that neutrinos are from Sun detection ν e + e ν e + e (Cerenkov) blind for E ν < 5000 kev GALLEX: observation of pp neutrinos, in agreement with luminosity detection ν e + 71 Ga e + 71 Ge blind for E ν < 233 kev Sudbury SNO: proof of solar neutrino oscillations detection ν e + d p + p + e (charged current) detection ν x + d p + n + ν x (neutral current) neutral current reaction detects all neutrino flavors blind for E ν < 2224 kev

12 Observed solar neutrino deficit

13 The SNO proof of neutrino oscillations 8 Φ SK ES Φ SNO CC 6 Φ µτ (10 6 cm -2 s -1 ) 4 2 Φ SK+SNO x Φ SSM x Φ e (10 6 cm -2 s -1 ) Observed TOTAL neutrino flux agrees with solar model predictions!

14 End of hydrogen core burning When the hydrogen fuel in the core gets exhausted, an isothermal core of about 8% of the stellar mass can develop in the center. Continous hydrogen burning adds to the core mass which eventually rises over the Schönberg-Chandrasekhar mass limit. Then the core s temperature (and density) rise. Finally the central temperature is high enough (T c 10 8 K) to ignite helium core burning. Hydrogen burning continues in a shell outside the helium core. This (hydrogen shell burning) occurs at higher temperatures than hydrogen core burning.

15 Which reaction can start helium burning? Consider a supply of protons and 4 He. We first note again that 5 Li is unbound. Although this nucleus is continuously formed by p+ 4 He reactions, the scattering is elastic and the formed 5 Li nuclei decay within s. As a consequence 4 He survives in the core until sufficiently large temperatures are achieved to overcome the larger Coulomb barrier between 4 He nuclei. Unfortunately the 8 Be ground state, formed by elastic 4 He+ 4 He scattering, is a resonance too and decays within s back to two 4 He nuclei.

16 The Salpeter-Hoyle suggestion In 1952 Salpeter pointed out that the 8 Be lifetime might be sufficiently large that there is a chance that it captures another 4 He nucleus. This triple-alpha reaction 3 4 He 12 C + γ can then form 12 C and supply energy. However, the simultaneous collision of 3 4 He (α-particles) is too rare to give the burning rate necessary in stellar models. So Hoyle predicted a resonance in 12 C to speed up the collision. And indeed, this Hoyle state was experimentally observed shortly after its prediction. 12 C can then react with another 4 He nucleus forming 16 O via 12 C + α 16 O + γ These two reactions make up helium burning.

17 Helium burning reactions Critical Reactions in He-burning Oxygen-16 Energy source in stellar He burning Energy release determined by associated reaction rates

18 At the end of helium burning Nucleosynthesis yields from stars may be divided into production by stars above or below 9M. stars with M 9M the stars are expected to shed their envelopes during helium burning and become white dwarfs. Most of the matter returned to the ISM is unprocessed. stars with M > 9M these stars will ignite carbon burning under non-degenerate conditions. The subsequent evolution proceeds in most cases to core collapse. These stars make the bulk of newly processed matter that is returned to the ISM.

19 Carbon Burning Burning conditions: for stars > 8 M o (solar masses) (ZAMS) T~ Mio U ~ g/cm 3 Major reaction sequences: dominates by far of course p s, n s, and a s are recaptured 23 Mg can b-decay into 23 Na Composition at the end of burning: mainly 20 Ne, 24 Mg, with some 21,22 Ne, 23 Na, 24,25,26 Mg, 26,27 Al of course 16 O is still present in quantities comparable with 20 Ne (not burning yet) 21

20 Neon Burning Neon burning is very similar to carbon burning. Burning conditions: for stars > 12 M o (solar masses) (ZAMS) T~ Bio K U ~ 10 6 g/cm 3 Why would neon burn before oxygen??? Answer: Temperatures are sufficiently high to initiate photodisintegration of 20 Ne 20 Ne+J Æ 16 O + D 16 O+D Æ 20 Ne + J equilibrium is established this is followed by (using the liberated helium) 20 Ne+D Æ 24 Mg + J so net effect: 22

21 Oxygen Burning Burning conditions: T~ 2 Bio U ~ 10 7 g/cm 3 Major reaction sequences: (5%) (56%) (5%) (34%) plus recapture of n,p,d,d Main products: 28 Si, 32 S (90%) and some 33,34 S, 35,37 Cl, Ar, 39,41 K, 40,42 Ca 27

22 Silicon Burning Silicon burning is very similar to oxygen burning. Burning conditions: T~ 3-4 Bio U ~ 10 9 g/cm 3 Reaction sequences: Silicon burning is fundamentally different to all other burning stages. Complex network of fast (J,n), (Jp), (J,a), (n,j), (p,j), and (a,j) reactions The net effect of Si burning is: 2 28 Si --> 56 Ni, need new concept to describe burning: Nuclear Statistical Equilibrium (NSE) Quasi Statistical Equilibrium (QSE) 28

23 Evolution of massive star Nuclear burning stages (e.g., 20 solar mass star) Fuel Main Product Secondary Product T (10 9 K) Time (yr) Main Reaction H He 14 N CNO 4 H Æ 4 He He O, C 18 O, 22 Ne s-process He 4 Æ 12 C 12 CDJ) 16 O C Ne, Mg Na C + 12 C Ne O, Mg Al, P NeJD) 16 O 20 NeDJ) 24 Mg O Si Si, S Fe Cl, Ar, K, Ca Ti, V, Cr, Mn, Co, Ni O + 16 O 28 SiJD)

24 Kippenhahn diagram for a 22 M star (A. Heger and S. Woosley)

25 Presupernova star Star has an onion-like structure. Iron is the final product of the different burning processes. As the mass of the iron core grows it becomes unstable and collapses once it grows above around 1.4 solar masses. E b /A Ca 56 Fe 86 Kr 107 Ag Mg 127 I Yb O 208 Pb 238U 12 C 7 4 He Li He H 1 H A

26 The early iron core The core is made of heavy nuclei (iron-mass range A 45 65) and electrons. There are Y e electrons per nucleon. The mass of the core M c is determined by the nucleons. There is no nuclear energy source which adds to the pressure. Thus, the pressure is mainly due to the degenerate electrons, with a small correction from the electrostatic interaction between electrons and nuclei. As long as M c < M ch = 1.44(2Y e ) 2 M (plus slight corrections for finite temperature), the core can be stabilized by the degeneracy pressure of the electrons. M ch is the famous Chandrasekhar mass which sets the limit mass of White Dwarfs - dead stars which are stabilized by the degeneracy pressure of electrons.

27 Initial collapse conditions If we approximate the pressure by the one of a relativistic degenerate electron gas (P ρ 4/3 ) one has P/ρ 1 4 Y eµ e Hence P/ρ is given in MeV, with µ e 1.1(ρ 7 Y e ) 1/3 MeV 1 MeV = erg/g Note that the electron chemical potential µ e is nearly 1 MeV at ρ 7 = 1 and hence reaches the nuclear energy scale. Hence it might be energetically favorable to capture high-energy electrons by nuclei.

28 Onset of collapse However, there are two processes which make the situation unstable. 1 Silicon burning is continuing in a shell around the iron core. This adds mass to the iron core, thus M c grows. 2 Electrons can be captured by nuclei. e + (Z, A) (Z 1, A) + ν e This reduces the pressure and cools the core, as the neutrinos leave. In other words, Y e and hence the Chandrasekhar mass M ch is reduced. The core finally collapses.

29 Electron capture The cross section for electron capture on free protons at rest is σ p = E 2 ν cm 2 where E ν is the energy of the emitted neutrino in MeV. The rate of electron capture on free protons then is r = σ p N A Y p = 0.016ρ 7 E 2 ν Y p [s 1 ] Although the capture cross section for those nuclei present in the core is usually smaller than the one of free protons (due to the larger energy threshold between parent and daughter nucleus), the abundance of free protons is quite low, so that the total electron capture rate is dominated by nuclei. This is an interesting nuclear structure problem which was first tackled within the Independent Particle Model (IPM) and then within the interacting shell model.

30 Presupernova evolution T = MeV, ρ = g cm 3. Composition of iron group nuclei (A = 45 65) Important processes: electron capture: e + (N, Z ) (N + 1, Z 1) + ν e β decay: (N, Z ) (N 1, Z + 1) + e + ν e Dominated by allowed transitions (Fermi and Gamow-Teller) Calculated within large-scale shell model, constrained by data from charge-exchange (d, 2 He), (n,p) experiments

31 Laboratory vs stellar electron capture capture of K-shell electrons to tail of GT strength distribution; parent nucleus in ground state capture of electrons from high-energy tail of FD distribution; capture of strong GT transitions possible; thermal ensemble of initial states

32 Shell model and (d, 2 He) GT strengths C. Bäumer et al. PRC 68, (2003) B(GT + ) V(d, 2 He) 51 Ti λ (s 1 ) LMP (d, 2 He) T (10 9 K) Old (n, p) data V(n,p) Alford et al. (1993) 0.4 B(GT + ) 0.3 large shell model calculation E x [MeV] B(GT+)

33 Consequences of electron capture Electron captures reduce the number of electrons and hence the pressure force against the gravitational collapse. Neutrinos carry energy out of the core. The core is kept cool and at a low entropy. Low entropy implies high order. Nuclei survive during the collapse phase. Electron captures also reduce the number of protons, but keeps the number of nucleons conserved. Nuclei become more neutron rich. Large neutron excess favors heavy nuclei. The nuclear composition is driven to more neutron-rich and heavy nuclei. Why do the neutron-rich nuclei not decay? β-decays are strongly suppressed by Pauli blocking as the electron chemical is high and effectively closes the phase space.

34 Pauli blocking of Gamow-Teller transition g 9/2 N=40 Blocked GT f 5/2 p 1/2 Unblocked Correlations Finite T g 9/2 f 5/2 p GT 1/2 p 3/2 p 3/2 f 7/2 neutrons protons Core f 7/2 neutrons protons Core Unblocking mechanism: correlations and finite temperature calculation of rate in SMMC + RPA model

35 Overcoming the shell gap by correlations MeV (1 +,2-) 0.5 ± 0.3 MeV (1 +,2-) 1.03 MeV (1 + ) 1.22 MeV (2 -) 1.63 MeV (1 +) 1.86 MeV (1 +) 2.22 MeV (1 +) 2.4 MeV (2 +) H 12 B (g.s.) λec (s 1 ) (d, 2 He) Shell Model Diagonalization SMMC+RPA E x [MeV] 76 Se: Gamow-Teller strength does not vanish data from KVI Groningen (Frekers et al.) ρ = g cm 3 Ye = Temperature (GK) Rates on basis of experiment and theory agree reasonably well for relevant temperatures.

36 How do shell-model rates compare to previous rates? λ ec (s 1 ) Fe Ni T 9 LMP FFN ρ 7 = Co 59 Co 54 Mn ρ 7 = ρ 7 =10.7 ρ 7 = Co T 9 ρ 7 =4.32 ρ 7 =33 A < 65: SM rates smaller Rec (s 1 ) protons nuclei ρ (g cm 3 ) A > 65: SM rates larger Thus at low densities electron capture is slower than previously believed, but at high densities previous simulations had switched off the main cooling mechanism (capture on nuclei).