Lecture 10

Nucleosynthesis During HeliumBurning and the s-Process

So, typical temperatures are 2 x 108 K (higher in shell burninglater) when densities are over 1000 gm cm-3. As the core evolves the temperature and density go up significantly.

A. Thermodynamic Conditions (Massive Stars)

Mα =181−βμ2β 4

μ=43 (for pure heliuμ ) solve for β

TC =4.6 ×106 μβMα

Me

⎛⎝⎜

⎞⎠⎟2/3

rc1/3 K

E.g., Mα =6 (α 20 Me μ αin sequence stαr)

β=0.83

Tc=1.7 ×108 rc

1000g cμ -3

⎛⎝⎜

⎞⎠⎟1/3

K

For the most part covered previouslytHe ~ 106 years (+- factor of three)

need temperatures > 108

to provide significant energy generation by 3α

n=3

Main sequence Current Approximate Tc/108 rc/1000 g cm-3

Mass Mass Mα

12 11.66 3 1.76 1.42 15 14.24 4 1.83 1.12 20 18.10 6 1.92 0.831 25 20.40 8 1.99 0.674 40 20.7 12 2.11 0.470

From Schaller et al (1992) Z = 0.02 and central helium massfraction of 50%.

At 1.9 x 108 K, the temperature sensitivity of the 3α rate is approximately T20.

As discussed earlier during helium burning:

dYαdt

=−3r2Yα3 l3α −Y(12C)Yαrlαg (

12C)

dY(12C)dt

=r2Yα3 l3α −Y(12C)Yαrlαg (

12C)

dY(16O)dt

=Y(12C)Yαrlαg (12C)

Coulomb barrier and lack of favorable resonances inhibit alpha capture on 16O.

Several general features:

• 12C production favored by large density; oxygen by lower density

• 12C produced early on, 16O later

• last few alpha particles burned most critical in setting ratio 12C/16O

• Energy generation larger for smaller 12C/16O.

B1. Principal Nucleosynthesis

If the star contains appreciable metals there is, as we shall see also 22Ne and 18O..

12 16 20M/M X( C) X( O) X( Ne)

12 0.247 0.727 1.91(-3) 15 0.188 0.785 3.02(-3)

e

19 0.153 0.815 6.44(-3) 25 0.149 0.813 1.16(-2) 35 0. 147 0.807 1.84(-2)

In massive stars evolved by Woosley & Weaver (1993)after helium burning in the stars center (calculations included semiconvection).

Caughlan and Fowler 12C(α,g)16O multipliedby 1.7.

In the sun, 12C/16O = 0.32

Carbon mass fraction at the endof helium burning. For low metallicitystars, especially of high mass, the helium convection zone extends outfarther.

Within uncertainties, helium burning in massive stars(over 8 solar masses) could be the origin in natureof 12C. It is definitely the origin of 16O

• If the helium core grows just a little bit towards the end of helium burning, the extra helium convected in greatly decreases the 12C synthesis.

• Mass loss from very massive WR stars can greatly increase the synthesis of both 12C and 16O in stars over 40 solar masses

• The uncertain rate for 12C(α,g)16O

• 12C/16O ratio may be affected by post-helium burning evolution

Complications:

Before the above reaction the composition wasalmost entirely 4He and 14N, hence h ≈ 0 (αctuαlly

α sμ αll posive vαlue exists βecαuse of 56Fe αnd the like).

After this reαction

h=0.0019ZZe

During heliuμ core βurning, 18O is lαter μ ostly

destroyed βy 18O(α,g)22Ne.

survives

destroyed unless preservedby convection

destroyed

partly destroyed

M TC 22Ne 25Mg 26Mg

12 2.42 13.4 0.51 0.61 12.3 1.17 1.0515 2.54 12.7 0.98 0.91 11.1 1.99 1.8019 2.64 11.5 1.73 1.54 9.4 2.90 2.8725 2.75 9.8 2.67 2.59 6.96 4.05 4.5435 2.86 7.37 3.87 4.22 4.41 5.18 6.39

So one expects that, depending on mass, some but not all of the 22Ne will burn.

The following table gives the temperature at the center of the given model and the mass fractions of 22Ne, 25Mg, and 26Mg each multiplied by 1000, when the helium mass fraction is 1% and zero

The remainder of the 22Ne will burn early during carbon burning,but then there will be more abundant “neutronpoisons”.

These numbers are quitesensitive to the uncertainreaction rate for 22Ne(αn)25Mg and may belower limits to the 22Neconsumption.

Woosley,Heger, and Hoffman(2005, in prep)

In 1995, an equivalent table would have been

M 22Ne 25Mg 26Mg

12 14.2 2.6 3.7 15 10.1 4.2 6.8 19 6.3 5.7 9.9 25 3.8 6.6 11.8 35 2.4 7.1 12.9

Helium burning is also the origin in nature of 22Ne and (part of)25Mg and 26Mg. The nucleus 21Ne is also made here partlyby 20Ne(n,g)21Ne and 18O(α,n)21Ne

Woosley and Weaver (1995)

This is yet another factor that can affect whether a star is a BSG or RSGwhen it dies. Observational evidence favors some primordial (primary)nitrogen production.

C. The s-Process in Massive Stars

Late during helium burning, when the temperature risesto about 3.0 x 108 K, 22Ne is burned chiefly by the reaction22Ne(α,n)25Mg (with some competition from 22Ne(α,g)26Mg).

Where do the neutrons go? Some go on 56Fe but that fraction is only:

56 56

i i

Yf

Ys

s=∑

2 316 16 16

4 522 22 22

4 325 25 25

5 456 56 56

0.5 3.1 10 1.2 10160.005 2.3 10 1.3 10

220.005 2 10 1.3 10

250.0013 2.3 10 2.7 10

56

Y Y

Y Y

Y Y

Y Y

s

s

s

s

− −

− −

− −

− −

≈ = × ≈ ×

≈ = × ≈ ×

≈ = × ≈ ×

≈ = × ≈ ×

But, 17O(α,n)20Nedestroys the 17O and restores the neutron

16O

22Ne

25Mg

56Fe

30 keV neutron capture cross sections (mostly from Bao et al, ADNDT, 2000)

Nucleus s Nucleus smb) (mb)

12C 0.0154 54Fe 27.616O 0.038 56Fe 11.720Ne 0.119 57Fe 40.022Ne 0.059 58Fe 12.124Mg 3.3 58Ni 41.025Mg 6.4 64Zn 5926Mg 0.126 65Zn 16228Si 2.9 66Zn 35 88Sr 6.2 (closed shell)

The large cross section of 25Mg is particularly significantsince it is made by 22Ne(α,n) 25Mg..

*

* Igashira et al, ApJL, 441, L89, (1995); factor of 200 upwards revision

56 56

25 25

n 22 56

22

56

YSo a fraction ~10 20% capture on iron. How many

neutrons is this? Y / ~ / 42

where we have assumed a mass fraction of 0.02 for Ne

and 0.0013 for Fe.

Fe

Y

Y Y Y

ss

−

=

This is about and obviously notnearly enough to change e.g., Fe into Pb, but the neutron capture cross sections of the isotopes generally increase above th

4 - 8

e ir

on group and t e

h

neutrons per iron

solar abundances decrease. A significant s-process occurs that produces significant quantities of the isotopes with A < 90.

Composition of a 25 solar mass star at the end of helium burning comparedwith solar abundances (Rauscher et al 2001)

25 solar mass supernova model post-explosion

At the end ....

22 25

Results depend most sensitively upon the reaction rate for

Ne ( ,n) Mg.α

If 22Ne does not burnuntil later (i.e., carbonburning there are muchmore abundant neutronpoisons

56Fe 57Fe 58Fe

59Co

60Ni 61Ni 62Ni 64Ni

63Cu 65Cu

dYi

dt=−YiYn rlng (i) + ...

t ng =1Yi

dYidt

⎛⎝⎜

⎞⎠⎟−1

Here "s" stands for "slow" neutron capture

tβ =tng tβ =1lβ

=t1/2ln 2

tng = rYnlng( )−1

This μ eαns thαt the neutron densities αre relαtively sμ αll

E.g. for α 22Ne neutron source

dYndt

≈0 ≈ rYαY(22Ne)lαn(

22Ne)

− rYnY(25Mg)lng (

25Mg)

r ≈1000, Xα ≈0.5, X(22Ne)≈0.005, X(25Mg)≈0.005

Yn ≈YαY(

22Ne)lαn(22Ne)

Y(25Mg)lng (25Mg)

≈Yαlαn(

22Ne)lng (

25Mg)

T8 lαn(22Ne) lng (

25Mg) nn=rNAYn lng (56Fe)

2.0 2.4(−13) [1.4(6)] 1.3(7) 1.9(6)2.5 1.9(−11) [1.4(6)] 1.0(9) 1.9(6)3.0 5.6(−10) 1.4(6) 3.0(10) 1.9(6)

Most of the s-process takes place around T8 = 2.5 - 3, so the neutron density is about 109 - 1010 cm-3 (depends on uncertainrate for (α,n) on 22Ne and on how much 22Ne has burned).

At these neutron densities the time between capture,even for heavy elements with bigger cross sections than iron,is days. For 56Fe itself it is a few years

Eg. at T8 =3 (nn : 1010 ), the lifetiμ e of 56Fe is αβout

t(56Fe) =1

Y(56Fe)dY(56Fe)

dt⎛⎝⎜

⎞⎠⎟−1

= rYnlng (56Fe)( )

−1

≈1 yeαr

Reaction Rates (n,g):

Either measured (Bao et al, ADNDT, 76, 70, 2000) or calculated using Hauser-Feshbach theory (Woosley et al., ADNDT, 22, 371, (1976) Holmes et al., ADNDT, 18, 305, (1976); Rauscher et al. ADNDT, 75, 1, (2000))

The calculations are usually good to a factor of two. For heavynuclei within kT ~ 30 keV of Qnthere are very many resonances.

Occasionally, for light nuclei or near closed shells,direct capture is important: e.g., 12C, 20,22Ne, 16O, 48Ca

Q2

Q1

Tg(Q2 ) > Tg(Q1)

αnd αs α result

s ng ∝Tn TgTn + Tg

≈Tg

is lαrger if Q is lαrger

More levels to maketransitions to at higherQ and also, morephase space for the outgoing photon.

Eg3 for electric dipole

Note that t has units of inverse cross section (inverse area).

Rate Equations: Their Solutions and Implications

Assume constant density, temperature, cross section, and neutron densityand ignore branching (would never assume any of these in a modern calculation). Then

dY (A Z )dt

≡dYA

dt=−YAYn rlng (A) +YA−1Yn rlng (A−1)

αnd since nn =rNA Yn αnd lng =NA s ngv ≈NAsAvtherμαl

defining t ≡ rNAYn vtherμαldt=∫ nn vtherμαl∫ dt, one hαs

dnA

dt= −nAsA + nA−1sA−1

If there were locations where steady state is achievedthen

dnA

dt≈0 =nAsA−nA−1sA−1

i.e., s n is locαlly α constαnt, αnd n ∝1s

Attaining steady state requires a time scale longer than a few times the destruction lifetime of the species in the steady state group. One has “local” steady state becauseany flux that would produce, e.g., lead in steady statewould totally destroy all the lighter s-process species.

The flow stagnates at various “waiting points” along the s-process path, particularly at the closed shell nuclei.

Eg., nn ~ 108 cm-3 ⇒ rYn ~nn / NA~5×10−17

lng experiμ entαlly αt heliuμ βurning teμ perαtures is 105 −108

t ng = rYnlng( )−1=

dlnYA

dt⎛⎝⎜

⎞⎠⎟−1

~10 −104 yeαrs

This cαn βe greαtly lengthened in α μ αssive stαr βy convection.

As a results nuclei with large cross sections will be in steady statewhile those with small ones are not. This is especially so in Heshell flashes in AGB stars where the time scale for a flash may beonly a few decades.

Implicit solution:

Ynew (A) =  Yold (A) / dt +Ynew(A−1)sng (A−1)

1 / dt + sng (A)

Assuming no flow downwards from A+1 and greaterto A and below.

This works because in the do loop, Ynew(A-1) is updatedto its new value before evaluating Ynew(A). Matrix inversionreduces to a recursion relation.

Sample output from toy model code micros2.f

s (even A) = 1.σ (odd A) = 2.σ (30)=σ (60)=0.1

Abu

ndan

ce

A-56

e.g., 117Sn, 118Sn, 119Sm, and 120Sn are s,r isotopes. Sn is not a goodplace to look for s n = const though because it is a closed shell.

who ordered that?

(mb)

Based upon the abundances of s-only isotopes and the known neutroncapture cross sections one can subtract the s-portion of s,r isotopesto obtain the r- and s-process yields separately.

A distribution of exposure strengthsis necessary in order to get the solarabundances.

Massive stars do not naturally give this.

Termination of the s-Process

Formation of an AGB star ...

5 Me

He

CO

H

H-shell burning extends helium layerin preparation for the next flash

During each flash have a mixtureof He, C, 13C or 22Ne, new seednuclei and s-process from prior flashesCO

He flash

Part of the ashes of each helium shell flash are incorporated into the fuelfor the next flash.

This naturally gives anexponential, or power-lawdistribution of exposures

He burningashes

Convectivedredge-up

Important nuclear physics modification:

s-process giants derived from AGB stars in the solar neighborhood do not show the large 26Mg excesses one wouldexpect if the neutron source were 22Ne(a,n)25Mg [as it surely is in massive stars]. Moreover these stars are too lowin mass for 22Ne(a,n)25Mg to function efficiently. Adifferent way of making neutrons is required. Probably

McWilliam and Lambert, MNRAS, 230, 573 (1988) andMalaney and Boothroyd, ApJ, 320, 866 (1987)Hollowell and Iben, ApJL, 333, L25 (1988); ApJ, 340, 966, (1989)many more since then

4 He(2α,g)12C(p,g)13N(e+n)13C13C(α,n)16O

with the protons coming from mixing between the helium burning shell and the hydrogen envelope.Each p mixed in becomes an n.

The metallicity history of the s-process can be quite complicated.

In the simplest case in massive stars with a 22Ne neutron source it is independent of metallicityuntil quite low values of Z.

At very low Z things can become complicated becauseof the effect of neutron poisons, 12C and 16O, and primarynitrogen production in massive stars. In AB stars the mixing between H and He shells is Z dependent. Some very metal poor stars are actually very s-process rich.

Thompson et al, ApJ, 677, 566, (2008)

Carbon enhanced metal poor star ([Fe/H] = -2.62,in binary in halo)