He burning overview
• Lasts about 10% of H-burning phase • Temperatures: ~0.3 GK • Densities ~ 104 g/cm3
Reactions: 4He + 4He + 4He 12C (triple α process)
12C + 4He 16O (12C(α,γ))
Main products: carbon and oxygen (main source of these elements in the universe)
1 Helium burning 1 – the 3α process
First step: α + α 8Be unbound by ~92 keV – decays back to 2 α within 2.6E-16 s !
but small equilibrium abundance is established
Second step:
8Be + α 12C* would create 12C at excitation energy of ~7.7 MeV
1954 Fred Hoyle (now Sir Fred Hoyle) realized that the fact that there is carbon in the universe requires a resonance in 12C at ~7.7 MeV excitation energy
Experimental Nuclear Astrophysics was born
2 How did they do the experiment ?
• Used a deuterium beam on a 11B target to produce 12B via a (d,p) reaction. • 12B β-decays within 20 ms into the second excited state in 12C • This state then immediately decays under alpha emission into 8Be • Which immediately decays into 2 alpha particles
So they saw after the delay of the β-decay 3 alpha particles coming from their target after a few ms of irradiation
This proved that the state can also be formed by the 3 alpha process …
removed the major roadblock for the theory that elements are made in stars Nobel Prize in Physics 1983 for Willy Fowler (alone !)
Fred Hoyle Willy Fowler
3 How 12C is produced?
Note: 8Be ground state is a 92 keV resonance for the α+α reaction
γ decay of 12C Note: 3 into its ground state Γα/Γγ > 10 so γ-decay is very rare ! 4 Calculation of the 3α rate in stellar He burning
Under stellar He-burning conditions, production and destruction reactions for 12C*(7.6 MeV) are very fast (as state mainly α-decays !) Equilibrium: α+α 8Be 8Be + α 12C*(7.7 MeV)
Equilibrium concentration: 8Be/4He ≈ 10-9!
The 12C*(7.6 MeV) abundance is therefore given by the Saha Equation: 3/ 2 9/ 2 2π2 2π2 = 3 ρ 2 2 −Q/kT Y12C(7.6 MeV) Y4He N A e m12CkT m4HekT 2 with Q / c = m12C(7.7) − 3mα using m12C ≈ 3m4He
3 2π2 = 3/ 2 3 ρ 2 2 −Q/kT Y12C(7.6 MeV) 3 Y4He N A e m4HekT The total 3α reaction rate (per s and cm3) is then the total gamma decay rate (per s and cm3) from the 7.6 MeV state. This reaction represents the leakage out of the equilibrium ! 5 3α rate
The total 3α rate r: Γ r = Y ρN γ 12C(7.6 MeV) A And with the definition
1 2 2 2 λ α = Yα ρ N < ααα > 3 6 A one obtains: note 1/6 because 3 identical particles ! 3 2 2π Γγ 2 < ααα >= ⋅ 3/ 2 ρ 2 2 −Q/kT N A 6 3 N A e m4HekT
(Nomoto et al. A&A 149 (1985) 239)
With the exception of masses, the only information needed is the gamma width of the 7.6 MeV state in 12C. This is well known experimentally by now. 6 Energy generation in triple-alpha process
Strong temperature dependence 41 At T6=100: ε(3α) ~ T !
ε~T17
ε~T41 ε~T4
3α process: • Occurs mainly in the highest temperature regions of a star • Predominantly responsible for the luminosity of red giant stars 7 How to produce 16O via 12C(α,γ)16O?
Too narrow! tail of the 1- resonance
tails of subthreshold Relevant resonances temperature:
T9~0.1-0.2 E0=0.3 MeV
12Cg.s.: 0+ 4He g.s.:0+ s-wave 0+ no states ! p-wave 1- d-wave 2+ Rolfs&Rodney 8 Production of 16O
some tails of resonances just make the reaction strong enough …
resonance (high lying)
resonance (sub threshold) resonance (sub threshold)
E2 DC Experimental complications: E1 E1 E2 • very low cross section makes direct measurement impossible • subthreshold resonances cannot be measured at resonance energy • Interference between the E1 and the E2 components 9 Uncertainty in the 12C(α,γ) rate
The single most important nuclear physics uncertainty in astrophysics
Affects: • C/O ratio~0.6 further stellar evolution (C-burning or O-burning ?) • iron (and other) core sizes (outcome of SN explosion) • Nucleosynthesis (see next slide)
Some current results for S(300 keV):
SE2=53+13-18 keV b (Tischhauser et al. PRL88(2002)2501
SE1=79+21-21 keV b (Azuma et al. PRC50 (1994) 1194)
Ongoing experimental studies, e.g. via triple-alpha decays studies at IGISOL, Jyväskylä
10 Massive star nucleosynthesis model as a function of 12C(α,γ) rate Weaver and Woosley Phys Rep 227 (1993) 65
• This demonstrates the sensitivity • One could deduce a preference for a total S(300) of ~120-220 (But of course we cannot be sure that the astrophysical model is right) 11 Blocking of helium burning: 16O(α,γ)20Ne
Unnatural parity!
At higher T, resonances at higher energies (3-,1-,0+) start to contribute and enhance the rate
too far from the threshold
Only direct capture possible at relevant stellar temperatures low cross section 12 Ratio of α capture rates on 12C and 16O
Rolfs&Rodney
Relevant region
13 Helium burning: summary
Resonance in Gamow window - C is made !
No resonance in Gamow window – C survives !
14 But some C is converted into O … Carbon burning
Burning conditions:
for stars > 8 Mo (solar masses) (ZAMS) T~ 0.6-0.7 GK ρ ~ 105-106 g/cm3 Major reaction sequences:
dominates by far
of course p’s, n’s, and α’s are recaptured … 23Mg can β-decay into 23Na
Composition at the end of burning:
mainly 20Ne, 24Mg, with some 21,22Ne, 23Na, 24,25,26Mg, 26,27Al 16 20 O is still present in quantities comparable with Ne (not burning … yet) 15 Carbon burning
High level density
cross section and S-factor vary smoothly Rolfs&Rodney
16 Carbon burning
Iliadis 17 Carbon burning
Iliadis 18 Neon burning Burning conditions:
for stars > 12 Mo (solar masses) (ZAMS=zero age main sequence star) T~ 1.3-1.7 GK ρ ~ 106 g/cm3
Why would neon burn before oxygen ??? Answer: Temperatures are sufficiently high to initiate photodisintegration of 20Ne
20Ne+γ 16O + α equilibrium is established 16O+α 20Ne + γ
this is followed by (using the liberated helium)
20Ne+α 24Mg + γ
The net effect: 2 20Ne 16O + 24Mg + 4.59 MeV
19 Neon burning
Iliadis
20 Photodisintegration
(Rolfs, Fig. 8.5.) 21 Calculations of inverse reaction rates
A reaction rate for a process like: 20Ne + γ 16O+α can be calculated from the inverse reaction rate: 16O+α 20Ne + γ
A simple relationship between the rates of a reaction rate and its inverse process (if all particles are thermalized)
Derivation of “detailed balance principle”:
Consider the reaction A+B C with Q-value Q in thermal equilibrium. Then the abundance ratios are given by the Saha equation:
3/ 2 3/ 2 n n g g m m kT A B = A B A B −Q / kT 2 e nC gC mC 2π
22 A+B C with Q-value Q in thermal equilibrium
In equilibrium the abundances are constant per definition. Therefore in addition dn C = n n < συ > −λ n = 0 dt A B C C
λC nAnB or = < συ > nC
If <σv> is the A+B C reaction rate, and λC is the C A+B decay rate Therefore the rate ratio is defined by the Saha equation as well ! Using both results one finds
3/ 2 3/ 2 λ g g m m kT C = A B A B −Q / kT 2 e < σv > gC mC 2π
µ or using mC~ mA+mB and introducing the reduced mass 23 Detailed balance and partition functions
Detailed balance:
3/ 2 λ g g µkT C = A B −Q / kT 2 e < σv > gC 2π
Need to know: - partition functions g - mass m of all participating particles can calculate the rate for the inverse process for every reaction Partition functions g:
For a particle in a given state i this is just gi = 2Ji +1 However, in an astrophysical environment some fraction of the particles can be in thermally excited states with different spins. The partition function is then given by:
−Ei / kT g = ∑ gie i 24 Photodisintegration
( , ) X(α,γ)Y exp ( , ) Y(γ,α)X 𝜆𝜆 𝛾𝛾 𝑎𝑎 𝑄𝑄 ∝ − 𝜆𝜆 𝑎𝑎 𝛾𝛾 𝑘𝑘𝑘𝑘 • Strong Q-value dependence • Q-value = mass difference = energy needed for the photodisintegration • Photodisintegration increases the relative abundance of even-even nuclei over that of even-odd nuclei • Reduces nuclei to their most stable form
12 11 12 13 13 12 C(γ,n) C Sn( C)= 18721.66 keV >> Sn( C)= 4946.31 keV C(γ,n) C 12 11 12 13 13 12 C(γ,p) B Sp( C)= 17532.86 keV >> Sp( N)= 1943.49 keV N(γ,p) C
More energy required Less energy required from γ to remove a n or p N=Z=6 from γ to remove a n or p Photodisintegration even-even Photodisintegration more probable less probable more stable
25 Oxygen burning Burning conditions:
T~ 2 x 109 K ρ ~ 107 g/cm3
Major reaction sequences:
(5%)
(56%)
(5%)
(34%)
plus recapture of n,p,d,α
Main products:
28Si,32S (90%) and some 33,34S,35,37Cl,36.38Ar, 39,41K, 40,42Ca 26 Oxygen burning
Iliadis
27 Silicon burning Burning conditions:
T~ 3-4 GK ρ ~ 109 g/cm3
Reaction sequences:
• fundamentally different to all other burning stages • direct burning 28Si+28Si would require too high temperatures (Coulomb barrier) • complex network of fast (γ,n), (γ,p), (γ,α), (n,γ), (p,γ), and (α,γ) reactions • The net effect : 2 28Si --> 56Ni 4 • photodisintegration plays a dominant role (Nγ � T )
need new concept to describe burning:
Nuclear Statistical Equilibrium (NSE)
Quasi Statistical Equilibrium (QSE)
28 The endpoint of the Silicon burning
Rolfs&Rodney
29 Silicon burning
Iliadis
30 Silicon burning
Iliadis 31 Reaction chains in Si burning
From Iliadis
λ(α,γ)
λ(γ,α)
32 Nuclear Statistical Equilibrium
In NSE, each nucleus is in equilibrium with protons and neutrons
Means: the reaction Z * p + N * n <−−−> (Z,N) is in equilibrium
Or more precisely: Z ⋅ µ p + N ⋅ µn = µ(Z,N) for all nuclei (Z,N)
NSE is established when both, photodisintegration rates of the type (Z,N) + γ −−> (Z-1,N) + p (Z,N) + γ −−> (Z,N-1) + n (Z,N) + γ −−> (N-2,N-2) + α and capture reactions of the types
(Z,N) + p −−> (Z+1,N) (Z,N) + n −−> (Z,N+1) are fast (Z,N) + α −−> (Z+2,N+2)
33 NSE: timescale
NSE is established on the timescale of the slowest reaction
18 10 16 10 102 g/cm3 107 g/cm3 approximation by Khokhlov MNRAS 239 (1989) 808 14 10 12 10 NSE if the timescale is 10 10 shorter than the timescale
8 10 for the temperature and 6 10 density being sufficiently 4 3 hours 10 high. 2 10 0 10 -2 10 -4 timeNSE to achieve (s) 10 -6 10 max Si burning -8 10 temperature -10 10 0.0e+00 2.0e+09 4.0e+09 6.0e+09 8.0e+09 1.0e+10 temperature (GK) for temperatures above ~5 GK even explosive events achieve full NSE 34 Nuclear Abundances in NSE
The ratio of the nuclear abundances in NSE to the abundance of free protons and neutrons is entirely determined by
Z ⋅ µ p + N ⋅ µn = µ(Z,N) which only depends on the chemical potentials
3 / 2 n h 2 µ = mc 2 + kT ln π g 2 mkT
So all one needs are density, temperature, and for each nucleus mass and partition function (one does not need reaction rates !! - except for determining whether equilibrium is indeed established) 35 Abundance ratio
Solving the two equations on the previous page yields for the abundance of nucleus (Z,N):
3 ( A−1) A3/ 2 2π2 2 = Z N ρ A−1 B(Z,N)/kT Y (Z, N) Yp Yn G(Z, N)( N A ) A e 2 mu kT with the nuclear binding energy B(Z,N)
Some features of this equation:
• in NSE there is a mix of free nucleons and nuclei • higher density favors (heavier) nuclei • higher temperature favors free nucleons (or lighter nuclei) • nuclei with high binding energy are strongly favored
36 Constraints
To solve for Y(Z,N) two additional constraints need to be taken into account:
Mass conservation ∑ AiYi =1 i
Proton/Neutron Ratio ∑ ZiYi = Ye i
In general, weak interactions are much slower than strong interactions.
Changes in Ye can therefore be calculated from beta decays and electron captures on the NSE abundances for the current, given Ye
In many cases weak interactions are so slow that Ye iis roughly fixed.
37 Entropy has a maximum at equilibrium
There are two ways for a system of nuclei to increase entropy:
1. Generate energy (more Photon states) by creating heavier, more bound nuclei 2. Increase number of free nucleons by destroying heavier nuclei
These are conflicting goals, one creating heavier nuclei around iron/nickel and the other one destroying them
The system settles in a compromise with a mix of nucleons and most bound nuclei
for FIXED temperature:
high entropy per baryon (low ρ, high T) more nucleons low entropy per baryon (high ρ, low T) more heavy nuclei
(entropy per baryon (if photons dominate): ~T3/ρ)
38 NSE composition (Ye=0.5)
~ entropy per baryon per entropy ~
after Meyer, Phys Rep. 227 (1993) 257 “Entropy and nucleosynthesis” 39 NSE during Silicon burning
• Nuclei heavier than 24Mg are in NSE • High density environment favors heavy nuclei over free nucleons
• Ye ~0.46 in core Si burning due to some electron captures
main product 56Fe (26/56 ~ 0.46)
formation of an iron core
(in explosive Si burning no time for weak interactions, Ye~ 0.5 and therefore final product 56Ni)
40 Summary: stellar burning
>0.8M0
>8M0
>12M0
Why do timescales get smaller ?
Note: Kelvin-Helmholtz timescale for red supergiant ~10,000 years, so for massive stars, no surface temperature - luminosity change for C-burning and beyond 41 What happens to a star at different burning stages?
Rolfs&Rodney
42 Properties of stars during hydrogen burning
Hydrogen burning is first major hydrostatic burning phase of a star: Hydrostatic equilibrium: a fluid element is “held in place” by a pressure gradient that balances gravity
Force from pressure:
Fp = PdA − (P + dP)dA = −dPdA
Force from gravity: 2 FG = −GM (r)ρ(r) dAdr / r
need: For balance: FG = FP dP GM (r)ρ(r) = − dr r 2
Clayton Fig. 2-14 43 The origin of pressure: equation of state
Under the simplest assumption of an ideal gas: need high temperature !
Keeping the star hot:
The star cools at the surface - energy loss is luminosity L To keep the temperature constant everywhere luminosity must be generated
In general, for the luminosity of a spherical shell at radius r in the star: dL(r) = 4πr 2 ρε (energy equation) dr where ε is the energy generation rate (sum of all energy sources and losses) per g and s
Luminosity is generated in the center region of the star (L(r) rises) by nuclear reactions and then transported to the surface (L(r)=const) 44 Energy transport requires a temperature gradient
Star has a temperature gradient HOT HEAT COLD (hot in the core, cooler at the surface) As pressure and temperature drop towards the surface, the density drops as well.
1) Radiative transport (“photon diffusion” - mean free path ~1cm in sun): to carry a luminosity of L, a temperature gradient dT/dr is needed:
3 a: radiation density constant 2 4acT dT L(r) = −4πr =7.56591e-15 erg/cm3/K4 3κρ dr κ is the “opacity”, for example luminosity L in a layer r gets attenuated by photon absorption with a cross section σ: −σ nr −κ ρ r L = L0e = L0e 2) Convective energy transport takes over when necessary temperature gradient too steep (hot gas moves up, cool gas moves down, within convective zone) not discussed here, but needs a temperature gradient as well 45 Radiative vs convective
Stars with M<1.2 M0 have radiative core and convective outer layer (like the sun):
convective
radiative
pp chain dominates no steep temperature gradient in the core radiative core Outer layer cooler hydrogen is neutral opaque to UV photons convection
Stars with M>1.2 M0 have convective core and radiative outer layer:
(convective core about 50% of mass
for 15M0 star)
High mass CNO cycle dominates in the core steeper energy vs T dependence
temperature gradient steep enough to make the core convective 46 The structure of a star in hydrostatic equilibrium
Equations so far: dP GM (r)ρ = − (hydrostatic equilibrium) dr r 2 dL(r) = 4πr 2 ρε (energy) dr 4acT 3 dT L(r) = −4πr 2 (radiative energy transfer) 3κρ dr dM (r) In addition of course: = 4πρr 2 (mass) dr and an equation of state
BUT: solution not trivial, especially as ε,κ in general depend strongly on composition, temperature, and density 47 Example: The sun
But - thanks to helioseismology one does not have to rely on theoretical calculations, but can directly measure the internal structure of the sun
oscillations with periods of 1-20 minutes
max 0.1 m/s
48 Temperature gradient in the sun
49 Density profile of the sun
50 Hydrogen profile of the sun
Convective zone (const abundances)
More hydrogen on the surface, Less in the core
Christensen-Dalsgaard, Space Science Review 85 (1998) 19 51 Hertzsprung-Russell diagram
TEMPERATURE
Not uniformly distributed At least clusters: 95% of all • Main sequence stars are • White dwarfs in the main • Red giants sequence • Supergiants
Application: Stellar distances
Main-sequence star: O 30,000 - 60,000 K Blue stars If the spectral class B 10,000 - 30,000 K Blue-white stars known: A 7,500 - 10,000 K White stars H-R gives the L F 6,000 - 7,500 K Yellow-white stars compare to the observed G 5,000 - 6,000 K Yellow stars (like the Sun) brightness K 3,500 - 5,000K Yellow-orange stars distance d M < 3,500 K Red stars 52 Stefan’s law
Surface temperature T, radius R and luminosity L of a star are related to each other:
= 4 constant σ = 5.67E-5 ergs K-4s-1cm-2 2 4 In terms𝐿𝐿 of the𝜋𝜋 𝑅𝑅sun’s𝜎𝜎 𝑇𝑇properties𝑠𝑠 : = 2 4 𝐿𝐿 𝑅𝑅 𝑇𝑇 𝐿𝐿𝑆𝑆𝑆𝑆𝑆𝑆 𝑅𝑅𝑆𝑆𝑆𝑆𝑆𝑆 𝑇𝑇𝑆𝑆𝑆𝑆𝑆𝑆 Main sequence stars: . . 5 5 3 5 𝐿𝐿 ∝ 𝑇𝑇𝑠𝑠 𝐿𝐿 ∝ 𝑀𝑀 • Most massive stars are the most luminous
• L vs Ts dependence can be theoretically derived for stars whose internal structure similar to the sun Rolfs&Rodney
53 Hertzsprung-Russell diagram
Perryman et al. A&A 304 (1995) 69 HIPPARCOS distance measurements
Magnitude m = Measure of stars brightness
Definition: difference in magnitudes m from ratio of brightnesses b:
b1 m2 − m1 = 2.5 log b2
(star that is x100 brighter has by 5 lower magnitude)
absolute scale historically defined: Sirius: -1.5, Main Sequence Sun: -26.2 ~90% of stars in naked eye easy: <0, limit: <4 H-burning phase Absolute magnitude is a measure effective surface temperature of luminosity = magnitude that star would have at 10 pc distance Sun: +4.77 54 Temperature, Luminosity, Mass relation during H-burning
• surface temperature (can be measured from contineous spectrum) • luminosity (can be measured if distance is known) (recall Stefan’s Law L~R2 T4, so this rather a R-T relation)
HR Diagram HR Diagram
M-L relation: Stefan’s Law L~M4 (very rough approximation cutoff at exponent rather 3-4) ~0.08 Mo
55 (from Chaisson McMillan) Main Sequence evolution
Main sequence lifetime:
H Fuel reservoir F~M F −3 4 lifetime τ = ∝ M Luminosity L~M MS L
Recall from Homework: H-burning lifetime of sun ~ 1010 years
−3 M note: very approximate τ ≈ 10 MS 10 years exponent is really M ⊕ between 2 and 3
so a 10 solar mass star lives only for 10-100 Mio years a 100 solar mass star only for 10-100 thousand years !
56