Chemical Evolution of the Universe

Transcription

1 Chemical Evolution of the Universe Part 5 Jochen Liske Fachbereich Physik Hamburger Sternwarte jochen.liske@uni-hamburg.de

2 Astronomical news of the week

3 Astronomical news of the week

4 Astronomical news of the week

5 Astronomical news of the week

6 Astronomical news of the week

7 Contents 1. Cosmological background 2. Primordial nucleosynthesis 3. Stellar structure, nucleosynthesis and evolution 4. Galactic chemical evolution 5. Chemical evolution in the intergalactic medium

8 Contents 1. Cosmological background 2. Primordial nucleosynthesis 3. Stellar structure, nucleosynthesis and evolution 3.1 Hertzsprung-Russell diagram 3.2 Timescales 3.3 Stellar structure 3.5 Stellar evolution 4. Galactic chemical evolution 5. Chemical evolution in the intergalactic medium

9 3.1 Hertzsprung-Russell diagram The HRD relates two fundamental properties of stars: total luminosity and flux through surface Stars occupy only certain regions in this parameter space These correspond to different evolutionary phases of stars The HRD is central to our understanding of stars and their evolution

10 3.1 Hertzsprung-Russell diagram Developed around 1910 by: Ejnar Hertzsprung ( ) Danish astronomer and chemist 1905: definition of absolute magnitude : Göttingen : Leiden Henry Norris Russell ( ) American astronomer : Princeton 1923: Russell-Saunders (LS) coupling (atomic physics) Ejnar Hertzsprung Henry Norris Russell

11 3.1 Hertzsprung-Russell diagram Different versions of the HRD: Theoretical version: L vs T eff Observational versions: Photometry: M V vs colour index (colour magnitude diagram, CMD) Spectroscopy: M V vs spectral class All versions are closely related to one another, but not identical

12 3.1 Hertzsprung-Russell diagram Different regions in the HRD correspond to different evolutionary phases: Main sequence: normal state of stars, H burning, stars spend most of their lives here Giants: later evolutionary phases White dwarfs: degenerate final state of stars of a particular mass range (the other possible final states are not represented in the HRD: neutron stars and black holes)

13 3.1 Hertzsprung-Russell diagram Different regions in the HRD correspond to different evolutionary phases: Main sequence: normal state of stars, H burning, stars spend most of their lives here Giants: later evolutionary phases White dwarfs: degenerate final state of stars of a particular mass range (the other possible final states are not represented in the HRD: neutron stars and black holes) The population density of different regions provides clues to the duration of the associated evolutionary phases

14 3.1 Hertzsprung-Russell diagram Different regions in the HRD correspond to different evolutionary phases: Main sequence: normal state of stars, H burning, stars spend most of their lives here Giants: later evolutionary phases White dwarfs: degenerate final state of stars of a particular mass range (the other possible final states are not represented in the HRD: neutron stars and black holes) The population density of different regions provides clues to the duration of the associated evolutionary phases In general: HRD extremely important to test our understanding of stellar evolution

15 3.1 Hertzsprung-Russell diagram Different regions in the HRD correspond to different evolutionary phases: Main sequence: normal state of stars, H burning, stars spend most of their lives here Giants: later evolutionary phases White dwarfs: degenerate final state of stars of a particular mass range (the other possible final states are not represented in the HRD: neutron stars and black holes) The population density of different regions provides clues to the duration of the associated evolutionary phases In general: HRD extremely important to test our understanding of stellar evolution Vice versa: assuming we understand stellar evolution, we can use properties of the HRD to infer characteristics of stellar population (age, metallicity) and/or its distance

16 3.1 Hertzsprung-Russell diagram The Main Sequence (MS) H burning in core Normal state Longest phase of a star s life More or less one-dimensional Hints at a simple relation between M, R, L, T eff Need only 1 parameter for its description?

17 3.1 Hertzsprung-Russell diagram Empirically we find the following relations between M and L, R for MS stars: L M 3.5 R M 0.8 Exponents slightly mass dependent Relations can be derived theoretically using some simplifying assumptions From L = 4 R 2 σt eff 4 log T eff = 0.5 log M + const The MS is a 1-parameter sequence Mass is the central characteristic of MS stars

18 3.1 Hertzsprung-Russell diagram Mass is the only free parameter on the MS

19 Contents 1. Cosmological background 2. Primordial nucleosynthesis 3. Stellar structure, nucleosynthesis and evolution 3.1 Hertzsprung-Russell diagram 3.2 Timescales 3.3 Stellar structure 3.5 Stellar evolution 4. Galactic chemical evolution 5. Chemical evolution in the intergalactic medium

20 3.2 Timescales Stars are spheres of gas held in quasi-static equilibrium by the balance between their own gravity and their inner pressure Energy balance of star: Time until energy reservoir is depleted: Then: decrease of L or tap into new reservoir Reason for stellar evolution

21 3.2 Timescales Nuclear timescale t N x 0.1 x M ʘ c 2 / L ʘ yr For M = 30 M ʘ L = 10 5 L ʘ : t N 3 x 10 6 yr Kelvin-Helmholtz (thermal) timescale (energy from contraction) t KH E therm / L ʘ = 1/2 E pot / L ʘ = 3/10 G M ʘ2 / R ʘ / L ʘ 10 7 yr Hydrostatic (dynamical) timescale t hyd R / c sound = R / (P/) 1/ s

22 3.2 Timescales t N >> t KH >> t hyd Most stars, most of the time, are in hydrostatic and thermal equilibrium, with slow changes in structure and composition occurring on the (long) nuclear timescale Most stars, most of the time, are on the Main Sequence Nevertheless, evolution on timescales t KH and t hyd is also seen: t KH : pre-ms evolution t hyd : pulsations (variable stars), collapse, supernova

23 3.3 Stellar structure Stars are spheres of gas held in quasi-static equilibrium by the balance between their own gravity and their inner pressure Only the outer layers are visible (photosphere and chromosphere) Size 10 3 of the star s size Mass of the star s mass Source of energy: nuclear fusion, not directly observable Determines stellar evolution Need theoretical modeling

24 3.3 Stellar structure Goal: Determination of M(r), ρ(r), P(r), T(r), L(r) Assumptions: Non-rotating No strong magnetic field No close companion Spherically symmetric distribution of gas Hydrostatic and local thermal equilibrium Constant production of energy Time-independent, quasi-static problem Fundamentals: Differential equations (hydrodynamics) Material equations Boundary conditions (at r = 0 and r = R)

25 3.3 Stellar structure Goal: Determination of M(r), ρ(r), P(r), T(r), L(r) Assumptions: Non-rotating No strong magnetic field No close companion Spherically symmetric distribution of gas Hydrostatic and local thermal equilibrium Constant production of energy Time-independent, quasi-static problem Fundamentals: Differential equations (hydrodynamics) Material equations Boundary conditions (at r = 0 and r = R)

26 3.3 Stellar structure Mass distribution 1st differential equation:

27 3.3 Stellar structure Hydrostatic equilibrium Pressure = gravitation 2nd differential equation:

28 Contents 1. Cosmological background 2. Primordial nucleosynthesis 3. Stellar structure, nucleosynthesis and evolution 3.1 Hertzsprung-Russell diagram 3.2 Timescales 3.3 Stellar structure 3.5 Stellar evolution 4. Galactic chemical evolution 5. Chemical evolution in the intergalactic medium

29 What is the energy source of stars? By considering erosion processes in the English countryside Charles Darwin already concluded that the Earth and therefore the Sun are > 3 x 10 8 yr old (today: > 10 9 yr) Radiation power of Sun 4 x W Problem: which processes can deliver this much energy over such extended periods of time? Total energy 4 x W x 10 9 yr x 3 x 10 7 s/yr J

30 Chemical energy: Typical energy a few ev (1.6 x J) per reaction Sun consists of ~10 57 atoms Total energy of ~ J Only lasts for s 8 x yr

31 Chemical energy: Typical energy a few ev (1.6 x J) per reaction Sun consists of ~10 57 atoms Total energy of ~ J Only lasts for s 8 x yr

32 Kelvin-Helmholtz mechanism (energy from contraction): Virial theorem: E therm = 1/2 E pot = 3/10 G M 2 / R = J Only lasts for ~10 7 yr

33 Kelvin-Helmholtz mechanism (energy from contraction): Virial theorem: E therm = 1/2 E pot = 3/10 G M 2 / R = J Only lasts for ~10 7 yr

34 Binding energy per nucleon has a maximum at iron group elements (8.8 MeV/nucleon) Energy can be released by moving towards the maximum Fission of massive nuclei Fusion of light nuclei Most energy released: 4 1 H 4 He ΔE = 0.7% mc 2 = 26.7 MeV

35 Binding energy per nucleon has a maximum at iron group elements (8.8 MeV/nucleon) Energy can be released by moving towards the maximum Fission of massive nuclei Fusion of light nuclei Most energy released: 4 1 H 4 He ΔE = 0.7% mc 2 = 26.7 MeV

36 Simplest example of nuclear fusion: Fusion of H to He (4 1 H 4 He) 4 H nuclei are heavier than one 4 He nucleus Mass defect = Δm = 4.8 x kg = 0.7% of original mass (in this case) 1 kg H kg He ΔE = 0.7% mc 2 = MeV = 4.2 x J for 4 1 H 4 He Assumption: 10% of the Sun s mass are fused to He Total energy x 0.1 x M ʘ c J Lasts for yr

37 Simplest example of nuclear fusion: Fusion of H to He (4 1 H 4 He) 4 H nuclei are heavier than one 4 He nucleus Mass defect = Δm = 4.8 x kg = 0.7% of original mass (in this case) 1 kg H kg He ΔE = 0.7% mc 2 = MeV = 4.2 x J for 4 1 H 4 He Assumption: 10% of the Sun s mass are fused to He Total energy x 0.1 x M ʘ c J Lasts for yr

38 Nuclear fusion Problem: particles must overcome Coulomb barrier: Attractive nuclear potential only kicks in at r x A 1/3 m E coul (r 0 ) 1 MeV

39 Thermal energy (T central 10 7 K): E therm = 3/2 k B T 1 kev << E Coul (r 0 ) No fusion reactions possible? But: particles with E = E Coul 1 MeV are nevertheless possible: Maxwell-Boltzmann distribution: However: P reaction exp( E Coul /E therm ) But only ~10 57 particles in the sun No fusion reactions possible after all?

40 Solution: quantum-mechanical tunnelling (G. Gamov 1928): For pp reactions and E = E therm 1 kev: P QM Controlled fusion Probability increases with increasing energy More massive nuclei require higher temperatures for fusion Separate burning phases

41 Probability of fusion reaction: P reaction (E) P MB x P QM Gamov Peak (still far below E Coul (r 0 )) Total reaction rate = P reaction (E) de

42 Hydrogen burning Most stars fuse H to He (90%) Direct reaction 4 1 H 4 He is extremely unlikely Need reactions with intermediate steps reaction chains For 3 x 10 6 K < T < 1.7 x 10 7 K: pp chain D 3 He 4 He

43 ppi chain: 1 H + 1 H 2 H + e + + ν e 2 H + 1 H 3 He + γ 3 He + 3 He 4 He H Reaction rate determined by the slowest reaction. Which is it?

44 ppi chain: 1 H + 1 H 2 H + e + + ν e 2 H + 1 H 3 He + γ 3 He + 3 He 4 He H Reaction rate determined by the slowest reaction. Which is it?

45 ppi chain: 1 H + 1 H 2 H + e + + ν e 2 H + 1 H 3 He + γ 3 He + 3 He 4 He H Reaction rate determined by the slowest reaction. Which is it? First one, because it is mediated by the weak force

46 There are, in fact, multiple pp chains:

47 Alternative reaction chain, in the presence of C, N, O: CNO chain C,N,O = catalyst for 4 1 H 4 He Dominates at T > 1.7 x 10 7 K Sun: 1% of energy production

48 Energy production rates: pp chain: ε T 4 CNO chain: ε T 12 18

49 Helium burning When most H is burned Mean molecular weight, T, radiation pressure P drops Core collapses Central T and P increase At T > 10 8 K He fusion: 4 He + 4 He 8 Be + γ 8 Be + 4 He 12 C + γ Triple Alpha Process ΔE 3α = 7.3 MeV

50 8 Be production is endothermic 8 Be lifetime in ground state s ( 8 Be 2 4 He) Need high 4 He density (> 10 5 kg m -3 ) Energy production: ΔE 3α = 7.3 MeV per 12 C Per unit mass: only 10% of ΔE CNO Extreme T-dependence: ε ρt 30 Additional reactions during He burning by α capture: 12 C + 4 He 16 O + γ ΔE = 7.16 MeV 16 O + 4 He 20 Ne + γ ΔE = 4.73 MeV (rare) C and O most abundant elements after H und He ( ashes of He burning )

51 Carbon burning Once He is burned up, T and P increase in core Fusion of heavier elements becomes possible At T > 6 x 10 8 K: C fusion Complex problem with many fusion channels and imprecisely known reaction rates: 16 O He 20 Ne + 4 He 12 C + 12 C 23 Na + 1 H 23 Mg + n 24 Mg + γ

52 Oxygen burning At T > 2 x 10 9 K: O fusion Fusion channels: 28 Si + 4 He 31 P + 1 H 16 O + 16 O 31 S + n 32 S + γ 24 Mg He At T > 4 x 10 9 K: Si fusion Production of elements up to 56 Fe ( 62 Ni has a problem with photodisintegration) Beyond Fe group elements the binding energy per nucleon drops fusion can no longer produce energy Collapse of star to neutron star or black hole + supernova

53 Energy production rates at T > 10 7 K extremely T dependent Fusion proceeds quickly Quick death of star

54 Fusion reaction of heaviest elements are taking place in the core Previous reactions are taking place in shells around the core Onion-like structure

55 Energy produced is transported to the outside L(r) = local power being transported to the outside ε = energy production rate per unit mass (determined by nuclear physics) 3rd differential equation: