He burning overview

• Lasts about 10% of H-burning phase • : ~0.3 GK • Densities ~ 104 g/cm3

Reactions: 4He + 4He + 4He  12C (triple α process)

12C + 4He  16O (12C(α,γ))

Main products: and (main source of these elements in the universe)

1 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 (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  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  2π2   2π2    = 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  2π2  = 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 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 • Predominantly responsible for the of 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 (C-burning or O-burning ?) • (and other) core sizes (outcome of SN explosion) • (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 burning Burning conditions:

for stars > 12 Mo (solar masses) (ZAMS=zero age star) T~ 1.3-1.7 GK ρ ~ 106 g/cm3

Why would neon burn before oxygen ??? Answer: Temperatures are sufficiently high to initiate 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 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  2π2  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 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/ 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

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 ): 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 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 )

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