Nukleosynthese in der Nuklearen Astrophysik

Transcription

1 Nukleosynthese in der Nuklearen Astrophysik Freitag 11 Uhr c.t. - 13:00 Raum NB 2/170 Tobias Stockmanns und Marius Mertens t.stockmanns@fz-juelich.de m.mertens@fz-juelich.de

2 TS MM MM Brückentag TS TS TS TS TS PANDA-Meeting MM TS + Seminarvorträge TS MM TS + Seminarvortrag? Termine 2

3 Inhalt Einführung Grundlagen der Kernphysik 2 Urknall 1 Urknall-Nukleosynthese 1 Stellaratmosphäre 2 H-Verbrennung 1 He-Verbrennung 1 Supernova 1 s,r,rp, ap Prozesse 1 Solarneutrinos 1 Neutrinomasse/oszillationen 1 3

4 Themenauswahl Vermessung und Bedeutung der Hintergrundstrahlung für die Kosmologie Cobe, WMAP und Planck "Dunkle Materie und Dunkle Energie "Der Lebenszyklus von Sternen "Bestimmung von Elementhäufigkeiten in Sternen und Meteoriten "Aktuelle Messungen von astrophysikalischen Parametern in kernphysikalischen Experimenten Das LUNA-Experiment Messungen mit dem R3B-Experiment Messungen zum p-prozess 4

5 Reaction Rates

6 Other pp Chains

7 Lebensdauer 12 C + 1 H 13 N + γ + 1,95 MeV 1, Jahre 13 N 13 C + e + + νe + 1,37 MeV 7 Minuten 13 C + 1 H 14 N + γ + 7,54 MeV 2, Jahre 14 N + 1 H 15 O + γ + 7,35 MeV 3, Jahre 15 O 15 N + e + + ν e + 1,86 MeV 82 Sekunden 15 N + 1 H 12 C + 4 He + 4,96 MeV 1, Jahre

8 The Hot CNO Cycle

9 Comparing pp chain and CNO Cycle 99% of the Sun s energy is produced in the pp chain, only about 1% via the CNO cycle. Would the Sun be only (10-20)% more massive, the CNO cycle would be dominant

10 VORLESUNG 8 10

11 The end of hydrogen burning 11

12 What happens when the H runs out? Mean molecular weight m increases, reducing the pressure Temperature increases to maintain hydrodynamic stability Y e decreases due to 4p 4 He The opacity k decreases due to both effects (T,Y) The luminosity increases: main sequence brightening The conditions do not allow He to burn, thus L(r)=0 in the core Thus dt/dr=0 and the core is isothermal For hydrostatic stability the density must increase in the center

13 The Schönberg-Chandrasekhar mass limit There is a maximum pressure (from the rest of the star at larger radii) that a isothermal core can support. Using the virial theorem this scales as: As the mass of the isothermal core increases the maximal pressure that it can withstand decreases. At some point it can not support the layers above it. The mass limit depends upon the mass fractions (mean molecular weight) in the core and envelope For the sun ( X = 0.708,Y = 0.272, Z = 0.02) and assuming complete ionization, the envelope has μ env = In the isothermal core assume X=0. Then μ iso = If more than 8% of the suns mass is in the isothermal core it will collapse (unless there is another source of pressure:e - degeneracy!)

14 Red Giants During shell burning the temperature is higher than for core burning. More energy is released there than can be transported outward. As a result the system expands and cools down. At the lower temperature some of the gas recombines to H, leading to a higher opacity. This is positive feedback leading to a runaway expansion. Radius increases by a factor ~400. The much larger surface area means the temperature must drop. (Red color) Energy transport in the high opacity region is mainly convective

15 Scales 15

16 Helium Flash Initially the conditions are not sufficient for helium burning in stars of certain mass so the core keeps on contracting. The density becomes so high that the Pauli principle becomes important for the electrons. i.e. The Fermi energy is higher than the temperature. (For stars between 0.5 and 2.5 solar masses) In this case the density and pressure do not depend upon the temperature. (degenerate core). As the temperature finally reaches the level to helium burning this proceeds as an explosion because the increased energy does not reduce the density as in the normal burning phase, instead the temperature and thus the energy production rate rises dramatically. The energy production rate can rise to time the solar rate. This is absorbed in the outer layers. Finally the temperature is higher than the Fermi energy so the star can expand and cool down into a stable burning process Helium cores reach temperatures of ~ 1 x 10 9 K and densities of g/cm 3

17 Ignition of Heavier Nuclei Heavier nuclei have higher Coulomb barriers, thus the temperature must be higher for hydrostatic burning. If this temperature is above the maximum achievable for the isothermal core then the species will not burn. Minimum core mass is called the ignition mass : For M > 2.25M the star is massive enough to proceed directly to a stage of core helium burning and shell hydrogen burning. If the star s mass is too small to ignite helium burning directly, (M < 2.25M ), hydrogen burning begins in the unburned fuel surrounding the depleted core (shell burning).

18 Mystery of C abundance 20

19 Triple alpha process 21

20 The triple a process 22

21 Formation of 8 Be 23

22 Equilibrium between 8 Be and a particles 24

23 Hoyle state 25

24 Hoyle State 26

25 The consequence of the triple a process 27

26

27

28 Impact of conservaton laws 30

29 Selection rules for spin-less projectile and target 31

30 Abundances 32

31 He burning in red giants 33

32 The 12 C(a, ) 16 O reaction 34

33 The 12 C(a, ) 16 O reaction 35

34 The 12 C(a, ) 16 O reaction 36

35 Interference and the 12 C(a, ) 16 O S-factor 37

36 Interference and the 12 C(a, ) 16 O S-factor 38

37 Interference and the 12 C(a, ) 16 O S-factor 39

38 The rate of the 12 C(a, ) 16 O reaction 40

39 The rate of the 12 C(a, ) 16 O reaction 41

40 Blocking of quiescent He burning 42

41 The structure 20 Ne 43

42 The structure of 20 Ne 44

43 The rate of the 16 O(a, ) 20 Ne reaction 45

44 Production of 12 C and 16 O 46

45 Why is there so much C and O (the basic building blocks of life)? This is a fast reaction since a resonance is in the Gamov peak This is a slow reaction. If these resonances were higher it would be faster and there would be no carbon left, if they were lower it would be so slow that there would be no oxygen. He Be dn He Be dn Be dt dt 1 1 dn N dt N He N Be He Be Be He Be He Q 2 He J p = 2 - p=(-1) L Since this level is 2 -, it is not possible for an alpha to be absorbed into this level L p 2 - Otherwise there would be no oxygen left. Only in very heavy stars (very high T) can this proceed (via the 3 - state)

46 Reactions on the ashes of the CNO cycle 48

47 14 N and neutron production 49

48 The Messier 3 (M3) globular cluster 50

49 The Messier 3 (M3) globular cluster 51

50 52