Presentation is loading. Please wait.

Presentation is loading. Please wait.

Nucleosynthesis 12 C(

Similar presentations


Presentation on theme: "Nucleosynthesis 12 C("β€” Presentation transcript:

1 Nucleosynthesis 12 C(𝛼,𝛾) 16 O at MAGIX/MESA
Stefan Lunkenheimer MAGIX Collaboration Meeting 2017

2 Topics S-Factor Simulation Outlook

3 S-Factor

4 Stages of stellar nucleosynthesis
Hydrogen Burning (PPI-III & CNO Chain) Fuel: proton π‘‡β‰ˆ2β‹… K Main product: 4 He Helium Burning Fuel: 4 He π‘‡β‰ˆ2β‹… K Main product: 12 C , 16 O

5 Helium Burning in red giants
Main reactions: 3𝛼→ 12 C+𝛾 12 C 𝛼,𝛾 16 O 12 C/ 16 O abundance ratio Further burning states Nucleosynthesis in massive stars Cp. Hammache: 12 C 𝛼,𝛾 16 O in massive star stellar evolution

6 Gamow-Peak Fusion reaction below Coulomb barrier π‘˜π‘‡βˆΌ15 𝑇=2β‹… 10 8 K Transmission probability governed by tunnel efffect Gamow-Peak 𝐸 0 Convolution of probability distribution Maxwell-Boltxmann QM Coulomb barrier transmission Depends on reaction and temperature Cp. Marialuisa Aliotta: Exotic beam studies in Nuclear Astrophyiscs

7 S-Factor Nonresonant Cross section 𝜎 𝐸 = 1 𝐸 𝑒 βˆ’ 2πœ‹ 𝑍 1 𝑍 2 𝛼𝑐 𝑣 𝑆(𝐸)
𝜎 𝐸 = 1 𝐸 𝑒 βˆ’ 2πœ‹ 𝑍 1 𝑍 2 𝛼𝑐 𝑣 𝑆(𝐸) π‘’βˆ’ Factor = probability to tunnel through Coulomb barrier 𝑣= velocity between the two nuclei 𝛼= fine structure constant 𝑍 1 ,𝑍 2 = Proton number of the nuclei 𝑆 𝐸 = Deviation Factor from trivial model

8 Gamow-Peak for 12 C 𝛼,𝛾 16 O Gamow-Peak (π‘‡β‰ˆ2β‹… 10 8 K)
𝑆 𝐸 =𝐸⋅ 𝑒 𝑏/ 𝐸 𝜎(𝐸) Gamow-Peak (π‘‡β‰ˆ2β‹… 10 8 K) 𝐸 0 = π‘β‹…π‘˜β‹…π‘‡ β‰ˆ300 keV π‘˜= Bolzmann constant 𝑏=πœ‹π›Ό 𝑍 1 𝑍 2 2πœ‡ 𝑐 2 πœ‡= 𝑀 1 𝑀 2 𝑀 1 + 𝑀 reduced mass Gamow Width Ξ”=4 𝐸 0 π‘˜π‘‡/3

9 Cross section 𝜎( 𝐸 0 )~ 10 βˆ’17 barn
Precise low-energy measurements required Direct measurements never 𝐸 cm <0.9 MeV Cp. Simulation of Ugalde 2013

10 Measurement of S-Factor
Approximate 𝑆(300 keV) Buchmann (2005) 102βˆ’198 keVβ‹…b Caughlan and Fowler (1988) 120βˆ’220 keVβ‹…b Hammer (2005) 162Β±39 keVβ‹…b

11 Measurement at MAGIX@MESA
Time reverted reaction 16 O(𝛾,𝛼) 12 C Cross section gain a factor of Γ—100 Inelastic 𝑒 βˆ’ scattering on oxygen gas Measurement of coincidence ( 𝑒 βˆ’ ,𝛼) suppress background 𝛼-Particle with low energy High Luminosity

12 Inverse Kinematik Time reversed reaction: 𝜎( 𝐸 0 )~ 10 βˆ’15 barn
High Energy resolution required MAGIX 𝐸 0 Cp. Simulation of Ugalde 2013

13 Simulation

14 Introduction MXWare (see talk Caiazza) Monte Carlo Integration
Fix Beam Energy Target at Rest Simulation acceptance 4πœ‹

15 Kinematik Momentum transfer π‘ž 2 =βˆ’4𝐸 𝐸 β€² sin 2 πœƒ 2
Photon Energy 𝜈= π‘Š 2 βˆ’ 𝑀 2 βˆ’ π‘ž 2 2𝑀 with π‘Š 2 = p 𝛾 πœ‡ + p 𝑂 πœ‡ invariant mass of photon and oxygen 𝑀= Oxygen mass Inelastic scattering cross section 𝑑 2 𝜎 𝑑Ω𝑑𝐸′ = 4𝛼 2 𝐸 β€²2 π‘ž 4 π‘Š 2 π‘ž 2 ,𝜈 β‹… cos 2 πœƒ π‘Š 1 π‘ž 2 ,𝜈 β‹…sin 2 πœƒ 2

16 Virtual Photon flux Relation beween structural functions and the transversal / longitudinal part of the virtual photon cross section 𝜎 𝑇 , 𝜎 𝐿 π‘Š 1 = πœ… 4 πœ‹ 2 𝛼 𝜎 𝑇 π‘Š 2 = πœ… 4 πœ‹ 2 𝛼 1βˆ’ 𝜈 2 π‘ž 2 βˆ’1 ( 𝜎 𝐿 + 𝜎 𝑇 ) with πœ…= π‘Š 2 βˆ’ 𝑀 2 2𝑀 So we get 𝑑 3 𝜎 𝑑Ω𝑑 𝐸 β€² =Ξ“ 𝜎 𝑇 +πœ€ 𝜎 𝐿 with Ξ“= π›Όπœ… 2 πœ‹ 2 π‘ž 2 β‹… 𝐸 β€² 𝐸 β‹… 1 1βˆ’πœ€ πœ€= 1βˆ’2 𝜈 2 βˆ’ π‘ž 2 π‘ž 2 tan 2 πœƒ βˆ’1 For π‘ž 2 β†’0 : 𝜎 𝐿 vanish and 𝜎 𝑇 β†’ 𝜎 tot 𝛾 βˆ— O →𝑋 𝑑 5 𝜎 𝑑 Ξ© 𝑒 𝑑 𝐸 β€² 𝑑 Ξ© βˆ— =Ξ“ 𝑑 𝜎 𝑣 𝑑 Ξ© βˆ— Cp. Halzen & Martin: Quarks and Leptons

17 Time reversal Factor Direct cross section -> Measurement
Compare with inverse cross section -> extract the S-Factor Calculate time reversal factor

18 Time reversal Factor Phase space examination under T-symmetry invariance 𝜎 𝑖→𝑓 𝜎 𝑓→𝑖 = (2 𝐼 3 +1)(2 𝐼 4 +1) (2 𝐼 1 +1)(2 𝐼 2 +1) β‹… | 𝑝 | 𝑓 2 | 𝑝 | 𝑖 2 Spinstatistic: I=0 for even–even nuclides ( 4 He, 12 C, 16 O ) in ground state 2 𝐼 𝛾 +1 =2 for photon. So we get 𝜎( 16 O 𝛾,𝛼 12 C) = π‘Š 2 βˆ’ π‘š He + π‘š C π‘Š 2 βˆ’ π‘š He βˆ’ π‘š C π‘Š 2 βˆ’ π‘š O 2 π‘Š 2 βˆ’ π‘š O 2 β‹…πœŽ( 12 C (𝛼,𝛾) 16 O) Cp. Mayer-Kuckuk Kernphysik: Chapter 7.3

19 Result of first simulations
Nonresonant cross section 𝜎( 16 𝑂(𝛾,𝛼) 12 𝐢) Simulation correlate to the results of Ugalde 4πœ‹βˆ’ Simulation ∼0.1 mHz Reaction Rate by 𝐸 0 with 𝐿∼ 𝑐 π‘š βˆ’2 𝑠 βˆ’1 Worst case Luminosity (see later talks) Now simulation with 𝑒 βˆ’ , π›Όβˆ’Acceptance needed.

20 Outlook

21 Simulation Finish simulation Preliminary results electron acceptance
𝛼-Particle acceptance Preliminary results Need measurement on angles smaller than Spectrometer coverage 0 degree scattering -> New Theoretic calculations

22 𝛼-Detection Low kinetic energy Needs specialized detector
∼20 MeV Needs specialized detector Silicon-Strip-Detector Choose and Test Silicon-Strip-Detectors in the Lab

23

24 BACKUP

25 Production factor Waver and Woosley Phys Rep 227 (1993) 65

26 Two-Body Reaction In the center of mass frame 16 𝑂( 𝛾 βˆ— ,𝛼) 12 𝐢
𝐸 3 = π‘Š 2 + π‘š 3 2 βˆ’ π‘š π‘Š 𝐸 4 = π‘Š 2 + π‘š 4 2 βˆ’ π‘š π‘Š 𝑝 = 𝐸 2 βˆ’ π‘š 2 = π‘Š 2 βˆ’ π‘š 3 + π‘š π‘Š 2 βˆ’ π‘š 3 βˆ’ π‘š π‘Š

27 Electron scattering Cross section inelastic scattering (cp. Chapter 7.2) 𝑑 2 𝜎 𝑑Ω𝑑𝐸′ = π‘‘πœŽ 𝑑Ω βˆ— Mott π‘Š 2 π‘ž 2 ,𝜈 +2 π‘Š 1 π‘ž 2 ,𝜈 tan 2 πœƒ 2 With structural functions π‘Š 1 , π‘Š 2 And Mott crossection (in this case) π‘‘πœŽ 𝑑Ω Mott βˆ— = 4𝛼 2 𝐸 β€²2 π‘ž 4 cos 2 πœƒ 2 We get (cp. Halzen & Martin Chapter 8) 𝑑 2 𝜎 𝑑Ω𝑑𝐸′ = 4𝛼 2 𝐸 β€²2 π‘ž 4 π‘Š 2 π‘ž 2 ,𝜈 β‹… cos 2 πœƒ 2 +2 π‘Š 1 π‘ž 2 ,𝜈 β‹…sin 2 πœƒ 2

28 Basic of Simulation Connection between count rate and cross section 𝑁= Ξ© 𝐴 Ξ© d𝜎 𝑑Ω 𝑑Ω⋅ 𝐿𝑑𝑑+ 𝑁 BG With 𝐿 : Luminosity 𝑁 : Number of counts 𝐴 Ξ© : Acceptance (1 full accepted, 0 not detected)

29 Monte Carlo Integration
Definition of mean value in volume 𝑉: 𝑓 = 1 𝑉 𝑉 𝑓 π‘₯ 𝑑 𝑛 π‘₯ Estimator for mean value: 𝑓 β‰ˆ 1 𝑁 𝑖=1 𝑁 𝑓( π‘₯ 𝑖 ) Monte-Carlo Integration: 𝑉 𝑓 π‘₯ 𝑑 𝑛 π‘₯ = 𝑓 β‰ˆ 𝑉 𝑁 𝑖=1 𝑁 𝑓 π‘₯ 𝑖 Β± 𝑉 𝑁 𝑓 2 βˆ’ 𝑓 2 Strategies for numerical improvements: Improve convergence 1/ 𝑁 Improve variance 𝑓 2 βˆ’ 𝑓 2

30 Cross section simulation
π‘‘πœŽ 𝑑 Ξ© 𝑒 𝑑 𝐸 𝑒 𝑑 Ξ© βˆ— 𝑑 Ξ© 𝑒 𝑑 𝐸 𝑒 𝑑 Ξ© βˆ— Transform Ξ©,𝐸 β†’ π‘Š,1/ π‘ž 2 ,πœ™ with det 𝐽= π‘ž 4 π‘Š 2𝑀𝐸 𝐸 β€² With Monte-Carlo Integration: π‘‘πœŽ 𝑑 Ξ© 𝑒 𝑑 𝐸 𝑒 𝑑 Ξ© βˆ— 𝑑 Ξ© 𝑒 𝑑 𝐸 𝑒 𝑑 Ξ© βˆ— = V N 𝑖 det 𝐽⋅ π‘‘πœŽ 𝑑 Ξ© 𝑒 𝑑 𝐸 𝑒 𝑑 Ξ© βˆ— π‘Š,1/ π‘ž 2 ,πœ™, Ξ© βˆ— Define πœ” 𝑖 =𝑉⋅ det 𝐽⋅ π‘‘πœŽ 𝑑 Ξ© 𝑒 𝑑 𝐸 𝑒 𝑑 Ξ© βˆ— So we get πœ” 𝑖 = π‘ž 4 π‘Š 2𝑀𝐸 𝐸 β€² β‹…π‘£β‹…Ξ”πœ™β‹…Ξ”π‘Šβ‹…Ξ” cos πœƒ βˆ— β‹…Ξ” πœ™ βˆ— β‹…Ξ“ β‹… 𝑑 𝜎 𝑣 𝑑 Ξ© βˆ—


Download ppt "Nucleosynthesis 12 C("

Similar presentations


Ads by Google