Titelseite Standard Solar Model 08.12.2005 1
Abstract Cross section, S factor and lifetime ppi chain ppii and ppiii chains CNO circle Expected solar neutrino spectrum 2
Solar Model Establish a model for the energy generation in the sun: Observables Photon luminosity: 3.810 26 W= 3.810 33 ergs -1 Age of the sun: about 5 billion years Solar radius of r s =6.96110 10 cm Solar mass m s =1.989 10 33 g Assumptions Spherical symmetry Hydrostatic and thermal equilibrium Equation of state of an ideal gas Present surface abundances of elements similar to the primordial composition but modified by previous stellar generations 3
Reaction rate The rate of nuclear reaction between two particles a and b with number density N a, N b and relative velocity v is: rab = NaN bσ(v)v The velocity v i of each nucleus (i) is Maxwell-Boltzmann distributed: 2 m mv i N(v i ) = N0 4πvi exp 2πkT 2kT Thus for two particles a and b: σ v = N (v )N (v ) σ(v)vdv dv 0 0 In the centre of mass system and after integration over the centre of mass velocity: 1/ 2 3/ 2 2 a a b b a b 8 1 E σ v = σ(e)e exp de 3/ 2 (kt) 0 πµ kt 4
Coulomb barrier For the p+p reaction the Coulomb barrier is about 550keV, but the typical energy in the sun is only 1.35keV. The reaction proceeds by tunneling. Penetration through the Coulomb barrier gives the penetration probability P: P = exp{ 2πη} With the Sommerfeld factor: η = Z Z 1 2 2 e v 5
Cross section The energy dependence of the cross section is dominated by the penetration through the Coulomb barrier. We find the astrophysical S-factor S(E), where this dependence is factorized: S(E) σ (E) : = exp( 2 πη) E η = Z Z 1 2 2 e v The S factor varies only slowly with the energy in non resonant reactions. 6
Typical examples for the S-factor and the cross section 7
Gamow window Using the definition for the S-factor: σ v 8 = S(E) exp( 2 E / kt)de 3 πη πµ (kt) 0 The shaded peak is proportional to the energy integral over the product of the Maxell-Boltzmann distribution and the tunneling probability. Its maximum is the most effective energy for stellar burning and one has to know the cross section for this E 0 : E =1.22[keV](Z Z µ T ) 2 2 2 1/ 3 0 1 2 6 8
The ppi chain In a star, fusion of hydrogen is the first long-term energy source that can ignite itself: For first-generation stars, protons are the source for the possible reactions. 3 or 4 body reactions are very unlikely. The chain has to be proceed by 2-body reactions or decays. 9
The ppi chain d+d is not going, because the deuteron abuncance is extremely small and d + p leads to rapid destruction. 3 He+ 3 He works, because the abuncance of 3 He increases as nothing destroys it. 10
The pp fusion reaction The only possible bound state for two nucleons is the deuteron. This requires a change of charge in the p(p,e + )D reaction. Strong and electromagnetic interaction require conservation of nucleonic charge, thus this reaction occurs via weak interaction. This makes the reaction rather slow at solar temperatures and the cross section is not observed in experiments but estimated theoretically in perturbation theory: λ J ;T T H J ;TT f f Zf W i i Zi 2 11
The quantum numbers of p+p and D The quantum numbers of the initial and final nuclear states are: The low-energy fusion should occur from p+p s-waves (L = 0) For p+p s-waves, the Pauli principle requires S = 0. (S, T = 1 are symmetric, S, T = 0 antisymmetric under exchange of particles.) The tensor interaction implies the presence of an L = 2 component in the deuteron ground state. Fermi transitions are forbidden byj = 0 selection rule. Gamow-Teller transition is allowed and dominates the cross section. 12
Gamow-Teller matrix element The nuclear wavefunctions factorize in space and spin-isospin parts and the matrix element factorizes as well: A ; k k f f Zf σ ± i; i Zi k = 1 J T T t J TT 2 = Ψ ( ) Ψ ( ) = 1, = 0 = 0, = 1 2 A k k L = 0 r f i = { 0,2} r S f Tf σ t L Si Ti k = 1 2 3 = M M = M 2 2 2 2 space spin space * 2 = Λ = space 0 L = 0( ) = 0( ) f Li M u r u r r dr 13
Overlap of the p+p and D wave function 14
S factor for pp For the pp reaction the S factor cannot be measured experimentally due to the low cross section at solar temperatures. One finds theoretically: S 11 (0) = (4.00 ± 0.05) 10 22 That is equal to kevb σ v = cm s 43 3 11 1.2 10 / at T6 = 15 The lifetime for protons in the sun follows as: 10 τ = 10 a 15
The D(p,) 3 He cross section The LUNA collaboration has measured the cross section for the D-p reaction at the solar Gamow energies: S 12 (0)=2.5 10-4 kevb. 16
The 3 He( 3 He,2p) 4 He cross section The LUNA results for the 3 He( 3 He,2p) 4 He S factor: S 33 (0)=5.15 10 3 kevb 17
Relative Deuteron abundance The Deuteron is produced via the fusion of two protons and is burnt by fusion with a proton. D denotes the deuteron abundance, H the hydrogen abundance: dd dt 2 H = r r = σ v HD σ v 2 11 12 11 12 In equilibrium: dd/dt=0 D H e = 11 = 2 σ v σ v 12 5.6 10 18 for T 6 = 5 10 K 18
Lifetimes In a reaction of two particles a and b, the lifetime of nucleus a against destruction by b is given by: τ (a) = b N b 1 σv If we assume density of =100g/cm 3 and mass fractions X(H)=X(He)=0.5 τ = 10 0.9 10 a p p (p) The time evolution of D, assuming constant H abundance: Equilibrium is reached in about 1.6s. ab τ (d) = 1.6s 2 2 H t / τ (d) H = σ + 11 0 σ 11 2 2 p D(t) v e D v 19
Relative 3 He abundance 3 He is made by the proton-deuteron reaction and destroyed by the reaction with a proton: 3 He( 3 He,2p) 4 He 3 d He dt ( 3 ) = HD σv He σv 12 33 As D exist in equilibrium: 3 2 d He H ( ) 2 3 = σv He σv dt 2 11 33 Assume equilibrium for 3 He: 1/ 2 3 He σv 11 = H 2 σv e 33 2 20
The D and 3 He equilibrium abundances 21
Equilibrium for 3 He 22
The ppi chain 23
4 He as catalyst Once 4 He has been synthesized it 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. 24
The three pp chains 25
The 3 He( 4 He,) 7 Be cross section Measurement of the 3 He 4 He S factor results in: S 34 =0.54keVb and lifetime t=10 6 a 26
Lifetime of 7 Be(e -, e ) 7 Li For 7 Be the Q value for the + decay is Q=862keV, thus smaller than the required energy to create m(e - ) and m(e + ). The fully ionized 7 Be is stable. As an atom, 7 Be has the possibility to capture an electron from the K- shell, with an lifetime of t=76.9days in the laboratory. In the sun not all 7 Be are fully ionized. the probability to find a K-shell electron depends on density and temperature and is about 30%. the electrons are distributed over the whole space, thus the probability to find one near a nucleus is much reduced. The resulting lifetime for 7 Be in the sun is higher than on earth: t=118days. 27
7 Be: proton capture 7 Be(p,) 8 B is a well investigated reaction, because the 8 B ground state decays fast via + to 8 Be *. Due to the high Q-value of this reaction, the e is the highest energetic solar neutrino with energies up to E =14MeV, and therefore easiest to be detected in the solar neutrino experiments. The measured S 17 =0.022keVb is quite low and only one in 30.000 solar circles produces the 8 B neutrinos. 28
The men behind the pp chain 29
The CNO cycle The CNO cycle requires presence of 12 C as a catalyst. In the core of the sun the mass fraction X of the most important elements is: 1 X( H) = 0.341 X( He) = 7.74 10 3 6 = 4 X( He) 0.639 X( Be) = 1.65 10 7 11 X( C) = 2.61 10 12 5 X( N) = 6.34 10 14 3 X( O) = 8.48 10 16 3 30
Hydrogen burning: pp vs CNO The slowest reaction in a cycle determines the efficiency of the cycle. In the pp chain the slowest reaction is the generation of deuterium: p(p,e + )D because it is mediated via weak interaction. All the other reactions are at least 10 4 times faster. With increasing number of protons in the nucleus the Coulomb barrier gains influence. Thus the slowest reaction in the CNO cycle is the proton capture on 14 N mediated via electromagnetic interaction: 14 N(p,) 15 O The temperature dependence of these reactions around the solar T is: σ v T (3 E / kt 2)/3 0 with E = 1.2204( Z Z AT ) kev 2 2 2 1/ 3 0 1 2 6 3.9 20 σ v( pp) T and σ v( CNO) T 31
Energy geration: pp vs CNO With a core temperature of T=15.6 10 6 K the energy geration in the sun is dominated by the pp chain. Not even 2% are due to the CNO cycle: 32
Solar neutrinos Solar neutrinos are generated in the following reactions: p p d e + + + + Be + e Li + ν 7 7 Be + e Li + ν 7 7 * 8 8 * + + + B Be e 13 13 + + + N C e 15 15 + + + O N e ν ν ν ν e e e e e e E E E E E E ν ν ν ν ν ν 0.420MeV = 0.861MeV = 0.383MeV < 15MeV 1.199MeV 1.732MeV 33
Solar neutrino spectrum 34
Theory vs Experiment 2005 35