Nucleosynthesis
12C(π½, πΏ)16O
at MAGIX/MESA
Stefan Lunkenheimer MAGIX Collaboration Meeting 2017
Nucleosynthesis 12 C(, ) 16 O at MAGIX/MESA Stefan Lunkenheimer - - PowerPoint PPT Presentation
Nucleosynthesis 12 C(, ) 16 O at MAGIX/MESA Stefan Lunkenheimer MAGIX Collaboration Meeting 2017 Topics S-Factor Simulation Outlook 2 S-Factor 3 Stages of stellar nucleosynthesis Hydrogen Burning (PPI-III & CNO Chain)
Stefan Lunkenheimer MAGIX Collaboration Meeting 2017
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3π½ β 12C + πΏ
12C π½, πΏ 16O
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ππ βΌ 15 keV @ π = 2 β 108K
tunnel efffect
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π πΉ = 1 πΉ πβ2ππ1π2π½π
π€
π(πΉ)
π€ = velocity between the two nuclei π½ = fine structure constant π1, π2 = Proton number of the nuclei
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πΉ0 = 1 2 π β π β π
2 3
β 300 keV
π1π2 π1+π2 reduced mass
Ξ = 4 πΉ0ππ/3
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π πΉ = πΉ β ππ/ πΉπ(πΉ)
measurements required
done @πΉcm < 0.9 MeV
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Approximate π(300 keV)
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π(πΉ0)~10β15barn
πΉ0
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π2 = β4πΉπΉβ² sin2 π
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π =
π2βπ2βπ2 2π
with
π + pπ π 2
invariant mass of photon and oxygen
π2π πΞ©ππΉβ² = 4π½2πΉβ²2 π4 π
2 π2, π β cos2 π
2 + 2π
1 π2, π β sin2 π
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Relation beween structural functions and the transversal / longitudinal part of the virtual photon cross section ππ, ππ
π
1 = π 4π2π½ ππ
π
2 = π 4π2π½ 1 β π2 π2 β1
(ππ + ππ) with π =
π2βπ2 2π So we get
π3π πΞ©ππΉβ² = Ξ ππ + πππ
with
Ξ =
π½π 2π2 π2 β πΉβ² πΉ β 1 1βπ
π = 1 β 2
π2βπ2 π2
tan2
π 2 β1
For π2 β 0 : ππ vanish and ππ β πtot πΏβ + 16O β π
π5π πΞ©πππΉβ²πΞ©β = Ξ πππ€ πΞ©β
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Phase space examination under T-symmetry invariance
ππβπ ππβπ = (2π½3+1)(2π½4+1) (2π½1+1)(2π½2+1) β | π|π
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| π|π
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Spinstatistic: I=0 for evenβeven nuclides ( 4He, 12C, 16O) in ground state 2π½πΏ + 1 = 2 for photon. So we get π(16O πΏ, π½ 12C) = 1 2 π2 β πHe + πC
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π2 β πHe β πC
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π2 β πO
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π2 β πO
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β π(12C(π½, πΏ)16O)
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Ugalde
π βΌ 1034 ππβ2π‘β1
πβ, π½ βAcceptance needed.
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Nonresonant cross section π(16π(πΏ, π½)12π·)
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http://magix.kph.uni-mainz.de
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Waver and Woosley Phys Rep 227 (1993) 65
In the center of mass frame 16π(πΏβ, π½)12π· πΉ3 =
π2+π3
2βπ4 2
2π
πΉ4 =
π2+π4
2βπ3 2
2π
π = πΉ2 β π2 =
π2β π3+π4 2 π2β π3βπ4 2 2π
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Cross section inelastic scattering (cp. Chapter 7.2) π2π πΞ©ππΉβ² = ππ πΞ©
β
Mott π
2 π2, π + 2π 1 π2, π tan2 π
2 With structural functions π
1, π 2
And Mott crossection (in this case) ππ πΞ© Mott
β
= 4π½2πΉβ²2 π4 cos2 π 2 We get (cp. Halzen & Martin Chapter 8) π2π πΞ©ππΉβ² = 4π½2πΉβ²2 π4 π
2 π2, π β cos2 π
2 + 2π
1 π2, π β sin2 π
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Connection between count rate and cross section π =
Ξ©
π΅ Ξ© dπ πΞ© πΞ© β πππ’ + πBG With π : Luminosity π : Number of counts π΅ Ξ© βΆ Acceptance (1 full accepted, 0 not detected)
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Definition of mean value in volume π: π = 1 π
π
π π¦ πππ¦ Estimator for mean value: π β 1 π
π=1 π
π(π¦π) Monte-Carlo Integration:
π
π π¦ πππ¦ = π β π π
π=1 π
π π¦π Β± π π π2 β π 2 Strategies for numerical improvements:
π2 β π 2
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ππ πΞ©πππΉππΞ©β πΞ©πππΉππΞ©β Transform Ξ©, πΉ β π, 1/π2, π with det πΎ = π4
π 2ππΉπΉβ²
With Monte-Carlo Integration: ππ πΞ©πππΉππΞ©β πΞ©πππΉππΞ©β = V N
π
det πΎ β ππ πΞ©πππΉππΞ©β π, 1/π2, π, Ξ©β Define ππ = π β det πΎ β
ππ πΞ©πππΉππΞ©β
So we get ππ = π4 π 2ππΉπΉβ² β π€ β Ξπ β Ξπ β Ξ cos πβ β Ξπβ β Ξ β πππ€ πΞ©β
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