Transcript Powerpoint

Lecture 10
Nucleosynthesis During Helium
Burning and the s-Process
A. Thermodynamic Conditions (Massive Stars)
For the most part covered previously
He ~ 106 years (+- factor of three)
n=3
M   18
1 

2
4
4
  (for pure helium)
3
 M 
6
TC  4.6 10  

 Me 
E.g., M   6
solve for 
2/ 3
c1/ 3
K
(a 20 M e main sequence star)
 = 0.83


c
8
Tc  1.7 10 

 1000 g cm -3 
1/ 3
K
need temperatures > 108
to provide significant
energy generation by 3
So, typical temperatures are 2 x 108 K (higher in shell burning
later) when densities are over 1000 gm cm-3. As the core evolves the
temperature and density go up significantly.
From Schaller et al (1992) Z = 0.02 and central helium mass
fraction of 50%.
Main sequence
Mass
12
15
20
25
40
Current
Mass
11.66
14.24
18.10
20.40
20.7
Approximate
M
3
4
6
8
12
Tc/108
1.76
1.83
1.92
1.99
2.11
At 1.9 x 108 K, the temperature sensitivity of the 3
rate is approximately T20.

c/1000
1.42
1.12
0.831
0.674
0.470
g cm-3
As discussed earlier during helium burning:
dY
  3  2 Y3 3  Y (12 C)Y   (12 C)
dt
dY (12 C)
  2 Y3 3  Y (12 C)Y   (12 C)
dt
dY (16 O)
 Y (12 C)Y   (12 C)
dt
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
In massive stars evolved by Woosley & Weaver (1993)
after helium burning in the stars center (calculations
included semiconvection).
X(12C)
X(16O)
X( 20 Ne)
12
15
0.247
0.188
0.727
0.785
1.91(-3)
3.02(-3)
19
25
35
0.153
0.149
0. 147
0.815
0.813
0.807
6.44(-3)
1.16(-2)
1.84(-2)
M/M
In the sun,
12C/16O = 0.32
Caughlan and Fowler 12C(,)16O multiplied
by 1.7.
If the star contains appreciable metals there is, as we shall see
also 22Ne and 18O..
Carbon mass fraction at the end
of helium burning. For low metallicity
stars, especially of high mass, the
helium convection zone extends out
farther.
Within uncertainties, helium burning in massive stars
(over 8 solar masses) could be the origin in nature
of 12C. It is definitely the origin of 16O
Complications:
• 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(,)16O
• 12C/16O ratio may be affected by post-helium burning
evolution
Before the above reaction the composition was
almost entirely 4 He and 14 N, hence   0 (actually
a small posive value exists because of 56 Fe and the like).
After this reaction
 = 0.0019
Z
Ze
During helium core burning, 18O is later mostly
destroyed by 18O( , )22 Ne.
survives
destroyed unless preserved
by convection
destroyed
partly destroyed
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
Woosley,
Heger, and
Hoffman
(2005, in prep)
M
TC
22Ne
25Mg
26Mg
12
2.42
15
2.54
19
2.64
13.4
12.3
12.7
11.1
11.5
9.4
9.8
6.96
7.37
4.41
0.51
1.17
0.98
1.99
1.73
2.90
2.67
4.05
3.87
5.18
0.61
1.05
0.91
1.80
1.54
2.87
2.59
4.54
4.22
6.39
25
2.75
35
2.86
The remainder of the
22Ne will burn early
during carbon burning,
but then there will be
more abundant “neutron
poisons”.
These numbers are quite
sensitive to the uncertain
reaction rate for
22Ne(,n)25Mg and may b
lower limits to the 22Ne
consumption.
In 1995, an equivalent table would have been
M
22Ne
25Mg
12
15
19
25
35
14.2
10.1
6.3
3.8
2.4
2.6
4.2
5.7
6.6
7.1
26Mg
Woosley and Weaver
(1995)
3.7
6.8
9.9
11.8
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 partly
by 20Ne(n,)21Ne and 18O(,n)21Ne
This is yet another factor that can affect whether a star is a BSG or RSG
when it dies. Observational evidence favors some primordial (primary)
nitrogen production.
C. The s-Process in Massive Stars
Late during helium burning, when the temperature rises
to about 3.0 x 108 K, 22Ne is burned chiefly by the reaction
22Ne(,n)25Mg (with some competition from 22Ne(,)26Mg).
Where do the neutrons go?
Some go on 56Fe but that fraction is only:
 56 Y56
f 
 iYi
16O
22Ne
25Mg
56Fe
0.5
Y16 
 3.1 102
16
0.005
Y22 
 2.3  104
22
0.005
Y25 
 2  104
25
0.0013
Y56 
 2.3  105
56
Y16  16 1.2 10
3
Y22  22  1.3105
Y25  25  1.3103
Y56  56  2.7 104
But, 17O(,n)20Ne
destroys the 17O and
restores the neutron
30 keV neutron capture cross sections
(mostly from Bao et al, ADNDT, 2000)
Nucleus

Nucleus

(mb)
12C
* 16O
20Ne
22Ne
24Mg
25Mg
26Mg
28Si
0.0154
0.038
0.119
0.059
3.3
6.4
0.126
2.9
54Fe
56Fe
57Fe
58Fe
58Ni
64Zn
65Zn
66Zn
88Sr
27.6
11.7
40.0
12.1
41.0
59
162
35
6.2 (closed shell)
The large cross section of 25Mg is particularly significant
since it is made by 22Ne(,n) 25Mg..
* Igashira et al, ApJL, 441, L89, (1995); factor of 200 upwards revision
(mb
Y56  56
So a fraction
~10  20% capture on iron. How many
Y25  25
neutrons is this?
Yn / YFe ~ Y22 / Y56  42
where we have assumed a mass fraction of 0.02 for 22 Ne
and 0.0013 for 56 Fe.
This is about 4 - 8 neutrons per iron and obviously not
nearly enough to change e.g., Fe into Pb, but the
neutron capture cross sections of the isotopes generally
increase above the iron group and the 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 compared
with solar abundances (Rauscher et al 2001)
At the end ....
25 solar mass supernova model
post-explosion
Results depend most
sensitively upon the
reaction rate for
22
Ne ( ,n) 25 Mg.
If 22Ne does not burn
until later (i.e., carbon
burning there are much
more abundant neutron
poisons
60Ni
59Co
56Fe
57Fe
58Fe
61Ni
63Cu
65Cu
62Ni
64Ni
Here "s" stands for "slow" neutron capture
1
 1/2
  =  n
 

 ln 2

 n  Yn n
dYi
 Yi Yn  n (i)  ...
dt
 1 dYi 
1 n  

 Yi dt 

1
This means that the neutron densities are relatively small
E.g. for a 22 Ne neutron source
dYn
 0  Y Y (22 Ne) n (22 Ne)
dt
 YnY (25 Mg)n (25 Mg)
  1000, X  0.5, X(22 Ne)  0.005, X(25 Mg)  0.005
Y Y (22 Ne) n (22 Ne) Y  n (22 Ne)
Yn 

25
25
Y ( Mg)n ( Mg)
n (25 Mg)
T8
 n (22 Ne)
n (25 Mg)
nn   N AYn
n (56 Fe)
2.0
2.5
2.4(13)
1.9(11)
[1.4(6)]
[1.4(6)]
1.3(7)
1.0(9)
1.9(6)
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 uncertain
rate for (,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 lifetime of 56 Fe is about

1
dY ( Fe) 
 ( Fe)   56

dt
 Y ( Fe)

56
56
 1 year
1

Y   (
n n
56
Fe)

1
Reaction Rates (n,):
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 heavy
nuclei within kT ~ 30 keV of Qng there are very many resonances.
Occasionally, for light nuclei or near closed shells,
direct capture is important: e.g., 12C, 20,22Ne, 16O, 48Ca
More levels to make
transitions to at higher
Q and also, more
phase space for the
outgoing photon.
Q2
Q1
E3 for electric dipole
T (Q 2 )  T (Q1 )
and as a result
 n 
Tn T
Tn + T
is larger if Q is larger
 T
Rate Equations: Their Solutions
and Implications
Assume constant density, temperature, cross section, and neutron density
and ignore branching (would never assume any of these
in a modern calculation). Then
dY ( A Z ) dYA

  YA Yn  n (A) YA1 Yn  n (A  1)
dt
dt
and since nn   N A Yn and n  N A  n v  N A A vthermal
defining  
 N Y
A n
vthermal dt 
n
n
vthermal dt, one has
dnA
  nA A  nA1 A1
d
Note that  has units of inverse cross section (inverse area).
If there were locations where steady state is achieved
then
dnA
 0  nA A  nA1 A1
d
i.e.,
 n is locally a constant, and n 
1

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 because
any flux that would produce, e.g., lead in steady state
would 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 ~ 10 8 cm -3  Yn ~ nn / N A ~ 5 10 17
n experimentally at helium burning temperatures is 10 5 10 8

 n  Yn n

1
 d lnYA 

 dt 
1
~ 10 10 4 years
This can be greatly lengthened in a massive star by convection.
As a results nuclei with large cross sections will be in steady state
while those with small ones are not. This is especially so in He
shell flashes in AGB stars where the time scale for a flash may be
only a few decades.
Implicit solution:
Assuming no flow downwards from A+1 and greater
to A and below.
Ynew (A) 
Yold (A) / d  Ynew ( A  1)  n (A  1)
1 / d   n (A)
This works because in the do loop, Ynew(A-1) is updated
to its new value before evaluating Ynew(A). Matrix inversion
reduces to a recursion relation.
 (even A) = 1.
 (odd A) = 2.
 (30)=  (60) 0.1
A-56
Sample output from toy model code micros2.f
who ordered that?
e.g., 117Sn, 118Sn, 119Sm, and 120Sn are s,r isotopes. Sn is not a good
place to look for  n = const though because it is a closed shell.
(mb)
Based upon the abundances of s-only isotopes and the known neutron
capture cross sections one can subtract the s-portion of s,r isotopes
to obtain the r- and s-process yields separately.
A distribution of exposure strengths
is necessary in order to get the solar
abundances.
Massive stars do not naturally give
this.
Termination of the s-Process
Formation of an AGB star ...
H
5 Me
He
CO
Part of the ashes of each
helium shell flash are
incorporated into the fuel
for the next flash.
This naturally gives an
exponential, or power-law
distribution of exposures
H-shell burning extends helium layer
in preparation for the next flash
Convective
dredge-up
He flash He burning
ashes
CO
During each flash have a mixture
of He, C, 13C or 22Ne, new seed
nuclei and s-process from
prior flashes
Important nuclear physics modification:
s-process giants derived from AGB stars in the solar
neighborhood do not show the large 26Mg excesses one would
expect if the neutron source were 22Ne(a,n)25Mg [as it
surely is in massive stars]. Moreover these stars are too low
in mass for 22Ne(a,n)25Mg to function efficiently. A
different way of making neutrons is required. Probably
4
He(2 ,  )12 C( p,  )13 N(e )13 C
C( ,n)16 O
13
with the protons coming from mixing between the
helium burning shell and the hydrogen envelope.
Each p mixed in becomes an n.
McWilliam and Lambert, MNRAS, 230, 573 (1988) and
Malaney and Boothroyd, ApJ, 320, 866 (1987)
Hollowell and Iben, ApJL, 333, L25 (1988); ApJ, 340, 966, (1989)
many more since then
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 metallicity
until quite low values of Z.
At very low Z things can become complicated because
of the effect of neutron poisons, 12C and 16O, and primary
nitrogen 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)