Population Nucleosynthesis - University of Cambridge

Population Nucleosynthesis

Rob Izzard

Carolune Institute for Quality Astronomywww.carolune.net/ciqua

Visitor at Saint Mary’s University, Halifax

“At CIQuA we strive to study astronomy at thehighest level possible, usually around 1500m.”

Halifax 2004 – p.1/78

Collaborators

Chris Tout, Maria Lugaro, Richard Stancliffe,Ross Church (IoA, Cambridge)Amanda Karakas (Saint Mary’s, Halifax, Canada)John Lattanzio (Monash University, Melbourne,Australia)Onno Pols, Axel Bonacic (Utrecht,The Netherlands)Lynnette Dray (Leicester, UK)Carolina Ödman (CIQuA)

Halifax 2004 – p.2/78


• Detailed single stellar evolution models• Detailed binary stellar evolution models• Single vs binary parameter space• (Traditional) Population Synthesis• Why use Population Synthesis?• Population Nucleosynthesis• AGB stars, Hot Bottoms, Binary Processes• Chemical Yields, Nuclear Reaction Rates

Halifax 2004 – p.3/78

Detailed 1D Single Stellar Evolution

Solutions to equations:

• Hydrostatic equilibrium dPdm

= −Gm4πr4

• Mass conservation drdm

= 14πr2ρ

• Nuclear energy generation dLdm

= ε

• (Radiative) Transport of the energy fluxdTdm

= − 34ac

κT 3

F(4πr2)2 (or convection prescription)

• Chemistry dXi

dt= XiXj 〈σv〉ij

Halifax 2004 – p.4/78

Run time and extra time

• Code runs in ∼ minutes-hours

• Basic chemistry: 1H, 4He, 12C, 14N, 16O, 20Ne,56Fe

• Post-processing nucleosynthesis required forextra isotopes

• Code runs in ∼ days

Halifax 2004 – p.5/78

Single Star Uncertainties

• Initial mass M

• Initial abundances Z

• Mass-loss prescription• Convective mixing prescription (MLT)• Nuclear reaction rates 〈σv〉

• Code breakdown (numerical problems)• Coupling to supernova II/Ib/c models

Halifax 2004 – p.6/78

Detailed 1D Binary Stellar Evolution

There are two types of code:

Coupled 1D models. e.g. Eggleton’s TWIN.

• Assumes stars are approximately spherical• 1D stellar structure equations• Perhaps small perturbations• Simple mass transfer

Halifax 2004 – p.7/78


There are two types of code:Coupled 1D models. e.g. Eggleton’s TWIN.

• Assumes stars are approximately spherical• 1D stellar structure equations• Perhaps small perturbations• Simple mass transfer

Halifax 2004 – p.7/78


Full 3D models, e.g. Djehuty

• 3D explicit (magneto-) hydrodynamic code• Nuclear bomb simulation code!• As close to “reality” as we can get• Requires (US military) supercomputers• Most of us are not allowed to use it!

Halifax 2004 – p.8/78

Halifax 2004 – p.9/78

Binary Star Uncertainties

Guess. . .

• Initial mass M




Halifax 2004 – p.10/78

Binary Star Uncertainties

Guess. . .

• Initial mass M




Halifax 2004 – p.10/78

Binary Star Processes

• Tidal Interaction• Roche Lobe Overflow• Common Envelope Evolution• Wind collision• Wind accretion• Thermohaline Mixing• Explosions: Type Ia SNe and novae

Halifax 2004 – p.11/78

Run time and extra time

• Code runs in at least twice the time of itssingle star equivalent

• Basic chemistry: 1H, 4He, 12C, 14N, 16O, 20Ne,56Fe

• Post-processing nucleosynthesis???• Djehuty code evolves stars in approximately

real time!

Halifax 2004 – p.12/78

Single Star Parameter Space

At its simplest, this is two dimensional

• Mass M (distribution IMF)• Metallicity Z (solar scaled or LMC/SMC)

Assume Z = Z� to reduce this to 1D.All other physics is fixed.

Halifax 2004 – p.13/78

Binary Star Parameter Space

This is never simple!

• Mass M1 (IMF)• Mass M2 (or ratio q = M2/M1, flat-q)• Metallicity Z (abundances solar scaled?)• Separation a (or period P related by Kepler’s

law, flat-ln)• Eccentricity e

A five-dimensional space!Assume e = 0, Z = Z� to reduce it to 3D.

Halifax 2004 – p.14/78

Chaos in Binaries

• Evolution is chaotic• Perturb initial conditions slightly→• Leads to very different evolution!• High resolution grid required to resolve these

effects• Particularly novae and SNe Ia which occur

rarely

Halifax 2004 – p.15/78

N14 from a 7 M� primary star

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1 1

6 5 4 3 2 1 0.75 0.5 0.3 0.1Secondary Mass

0.1

1

10

100

1000

10000

100000

Initi

al P

erio

d / d

ays

M M M M M M M M

MT1 MT1 MT1 MT1 MSnII MSnII MSnII M

CT1 MT1 MT1 MT1 MSnII MSnII MSnII MSnII

CT1 CT1 CT1 CT1 CT1 CSnIaELD CSnII CSnIaELDSnIbc

CT1 CT1 CT1 CT1 CT1 CCSnIaCoal CSnIaELDSnIbc CCMSnAIC

CT1 CT1 CT1 CT1 CT1 CCSnIaCoal CCMSnAIC CCM

CT1 CT1 CT1 CT1 CT1 CNC CCT2 CCSnII

CT1 CT1 CT1 CT1 CT1 CNA1CSnIaELD CCT2 CCT2

CCA1 CCA1 CCA1 CCA1 CCA1 CNA1CSnIaELD CCT2 CCT2

CCA1 CCA1 CCA1 CCA1 CCNMA1 CC CCSnIaCoal CCSnIaELD

C C C C CC CC CC CCC

T1C T1C T1C C CT2C CC CC CT2C

T1C T1C T1C T1C T1CC T1CC T1CC T1CCC

T1 T1 T1 T1C T1CC T1CC T1CCC T1CCC

T1 T1 T1 T1 T1T2 T1T2 T1T2 T1T2

T1 T1 T1 T1 T1T2 T1T2 T1T2 T1T2

Halifax 2004 – p.16/78

Time is Money (unless you’re a student)

• At 1 hour per model (very conservative!)

• With 100 grid points per dimension. . .• That is 106 grid points. . .• Or 106 hours. . .• Or 41667 days. . .• Or 114 years. . .• Or 38 PhDs. . .• That assumes you got the physics right in the

first place. Which you didn’t!

Halifax 2004 – p.17/78


• At 1 hour per model (very conservative!)• With 100 grid points per dimension. . .

• That is 106 grid points. . .• Or 106 hours. . .• Or 41667 days. . .• Or 114 years. . .• Or 38 PhDs. . .• That assumes you got the physics right in the


Halifax 2004 – p.17/78


• At 1 hour per model (very conservative!)• With 100 grid points per dimension. . .• That is 106 grid points. . .

• Or 106 hours. . .• Or 41667 days. . .• Or 114 years. . .• Or 38 PhDs. . .• That assumes you got the physics right in the


Halifax 2004 – p.17/78


• At 1 hour per model (very conservative!)• With 100 grid points per dimension. . .• That is 106 grid points. . .• Or 106 hours. . .

• Or 41667 days. . .• Or 114 years. . .• Or 38 PhDs. . .• That assumes you got the physics right in the


Halifax 2004 – p.17/78


• At 1 hour per model (very conservative!)• With 100 grid points per dimension. . .• That is 106 grid points. . .• Or 106 hours. . .• Or 41667 days. . .

• Or 114 years. . .• Or 38 PhDs. . .• That assumes you got the physics right in the


Halifax 2004 – p.17/78


• At 1 hour per model (very conservative!)• With 100 grid points per dimension. . .• That is 106 grid points. . .• Or 106 hours. . .• Or 41667 days. . .• Or 114 years. . .

• Or 38 PhDs. . .• That assumes you got the physics right in the


Halifax 2004 – p.17/78


• At 1 hour per model (very conservative!)• With 100 grid points per dimension. . .• That is 106 grid points. . .• Or 106 hours. . .• Or 41667 days. . .• Or 114 years. . .• Or 38 PhDs. . .

• That assumes you got the physics right in thefirst place. Which you didn’t!

Halifax 2004 – p.17/78


• At 1 hour per model (very conservative!)• With 100 grid points per dimension. . .• That is 106 grid points. . .• Or 106 hours. . .• Or 41667 days. . .• Or 114 years. . .• Or 38 PhDs. . .• That assumes you got the physics right in the

first place. Which you didn’t!Halifax 2004 – p.17/78

Population Synthesis

What if we can reduce the code run time by afactor of 107? Evolution of a stellar populationwould take a matter of hours. You could start amodel run, go to the pub, and it’ll be finished bythe time you wake up . . .

Halifax 2004 – p.18/78

Population Synthesis

• Fit results from detailed models to simplefunctions

• Fit timescales τi, radius R, luminosity L, coremass Mc etc. as f(M,Z, t)

• Couple to binary star model (tides, RLOF, CE,wind accretion etc.)

• Model of single or binary star, from birth todeath, takes < 0.1s

• Millions of stars per day!• Lose internal stellar structure information.

Halifax 2004 – p.19/78

Why Population Synthesis?

• It is the only way to explore the completeparameter space

• Easily experiment with new physics e.g.change mass-loss prescription, commonenvelope removal efficiency etc.

• Compare to observations e.g. number ratiosof stellar types, supernova rates etc. todetermine the value of input parameters

• Feed these results back to detailed modellers• Tells them what they should be getting!

Halifax 2004 – p.20/78

Flow diagram of (my?) life

selection effectsModel

ObservationalSurvey

χ comparison2

n constraints

Detailedmodels

Synthetic Modelsm free parameters

Feedback

Halifax 2004 – p.21/78

Limitations

• Fit accuracy ∼ 5%: good enough for mostapplications

• Limited variables for comparison withobservations (L, R, M , stellar type, event –e.g. SN or nova – rates; also M )

• But this is enough to ID many types ofbinaries e.g. X-ray binaries, symbiotic stars,double degenerate pairs, Algols etc. andconstrain (some) free parameters in thephysics

Halifax 2004 – p.22/78


• Introduce nucleosynthesis into syntheticmodel

• Comparison variable set extends to L, R, M ,M , stellar type, event rate and surfaceabundances of more than 130 isotopes.

• Provides extra constraints on the models’ freeparameters.

• Fast/accurate nucleosynthesis model(observations σ ∼ 0.1 dex)

• Synthetic AGB (Iben, Renzini, Groenewegen, Forestinietc) Halifax 2004 – p.23/78

Nucleosynthesis in Stars

Halifax 2004 – p.24/78

TPAGB star

Halifax 2004 – p.25/78

Nucleosynthesis in TPAGB Stars

• Helium burning converts 4He into 12C, 16O and20Ne during each pulse (more on this later!)

• The convective hydrogen envelope mixes intothe helium-burnt region

• Carbon, oxygen and neon are brought to thesurface during each pulse

• This is the “Third Dredge Up”• NB Difficult to model! Need high resolution,

tough numerics.

Halifax 2004 – p.26/78

Thermal Pulses

Halifax 2004 – p.27/78

Third Dredge Up

• Occurs when core mass Mc exceeds athreshold Mc,min

• Amount of material mixed: λ = ∆MDredge/∆MH

• Include in models by fitting the parameters λand Mc,min

• Fit intershell abundances to detailed models:assume f(M,Z)

Halifax 2004 – p.28/78

Carbon Stars

• Surface carbon increases: star becomes a“Carbon Star”

• Easily visible (bright, ID by photometry) andcomplete SMC/LMC surveys exist

V713 Monocerotis Halifax 2004 – p.29/78

Carbon DUP

-2.8

-2.7

-2.6

-2.5

-2.4

-2.3

-2.2

-2.1

-2

0 5 10 15 20 25

log 1

0(X

C12

)

Time / 105 yrs

Detailed M=3 Z=0.02

Halifax 2004 – p.30/78

Calibrating λ and Mc,min

• λ and Mc,min are fitted to Amanda’s detailedmodels: Are they correct?

• Construct the luminosity function of carbonstars by modelling a population

• Define λmin such that λ = max (λmin, λfit)

• Define ∆Mc,min so Mc,min = M fitc,min + ∆Mc,min

• Plot dN/dmag for different λmin and ∆Mc,min

• Use χ2 test: best fit to LMC (Z = 0.008) andSMC (Z = 0.004).

Halifax 2004 – p.31/78

CSLFs

0

0.2

0.4

0.6

0.8

1

-7-6-5-4-3-2

30 20 10 5 2.5 1 0.5

Nor

mal

ized

Pro

babi

lity

Bolometric Magnitude

Luminosity / 103 L �

LMCSMC

Halifax 2004 – p.32/78

Best Fit for the SMC

0

50

100

150

200

250

300

-7-6-5-4-3-2

30 20 10 5 2.5 1 0.5

Star

s pe

r bin

Bolometric Magnitude

Luminosity / 103 L �

∆Mcmin=-0.07 λmin=0.65 SIN δtf=0.1 VW500Observations

Halifax 2004 – p.33/78

Best Fit for the SMC

• λmin = 0.65 and ∆Mc,min = −0.07 M�.• So third dredge-up occurs earlier and is more

efficient than the detailed models predict.• New (detailed) models by Richard Stancliffe

can fit the LMC models nicely.• But they still fail for the SMC.• The models are still wrong!• This is why synthetic models are good.

Halifax 2004 – p.34/78

Hot Bottoms?

Halifax 2004 – p.35/78

Hot Bottoms

• In stars with M & 4 M� the base of theconvective envelope is hot enough that HotBottom Burning occurs

• Hydrogen burning occurs in the convectiveregion!

• CNO, NeNa and MgAl cycles may operate,depending on the temperature

• Higher mass→ higher temperature• CNO cycling prevents a star from evolving to

a carbon starHalifax 2004 – p.36/78

Not a Carbon Star

-3.6

-3.5

-3.4

-3.3

-3.2

-3.1

-3

-2.9

-2.8

-2.7

-2.6

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

log 1

0(X

C12

)

Time / 105 yrs

Detailed M=6 Z=0.02

Halifax 2004 – p.37/78

Quick Analytic Burning

X0

τcτc

X0 X

1 1XXβ

0

Mix Mix

Time

Mix

X1 10

X X

BURN BURN β

Halifax 2004 – p.38/78


X0

X0

τcτc

1X

1X

Time

Mix

BURN β

Mix

........

Halifax 2004 – p.39/78


• burn→ mix→ burn→ mix. . . replaced by• A single burn→ mix step• Calibrate amount of burning and burn time to

Amanda’s detailed models• Require quick, but accurate, solution to

nuclear burning because differential equationsolving takes too long

• Iterative method is slow, consider analyticsolution

Halifax 2004 – p.40/78

The CNO cycle12C +1H→13 N + γ13N→13 C + e+ + ν13C +1H→14 N + γ14N +1H→15 O + γ15N +1H→12 C + α15N +1H→16 O + γ15O→15 N + e+ + ν16O +1H→17 F + γ17F→17 O + e+ + ν

17O +1H→14 N +4He17O +1H→18 F + γ18F→18 O + e+ + ν

18O +1H→15 N +4He

Halifax 2004 – p.41/78

The CN cycle

• 15N +1H→16 O + γ is slow at low temperature(or short burn times)

• CNO splits to CN and ON cycles• Express CN cycle as eigenvalue problem

d

dt

12C13C14N

=

−1/τ12 0 1/τ14

1/τ12 −1/τ13 0

0 1/τ13 −1/τ14

12C13C14N

• Solve for 12,13C, 14N as (simple) f(t).Halifax 2004 – p.42/78

The ON Cycle

• Long burn times: CN cycle→ eq. ∼ 98% 14N

• ON cycle activates

d

dt

14N16O17O

=

−1/τ12 0 1/τ14

1/τ12 −1/τ13 0

0 1/τ13 −1/τ14

14N16O17O

• Oxygen (slowly!) destroyed.• Again 14N is the result.

Halifax 2004 – p.43/78

CNO Equilibrium

1e-04

0.001

0.01

0.1

1

7.6 7.7 7.8 7.9 8 8.1 8.2 0

5

10

15

20

25

Equ

ilibr

ium

abu

ndan

ce

NC

/ N

O

log10 (T / K)

14N

10 × 17O

10 × 13C

12C

16O

10 × 15N

NC / NO

Halifax 2004 – p.44/78

Synthetic vs Detailed

-2.5

-2.45

-2.4

-2.35

-2.3

-2.25

-2.2

-2.15

-2.1

-2.05

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

log 1

0(X

N14

)

Time / 105 yrs

Detailed M=6 Z=0.02Syn M=6 Z=0.02

Halifax 2004 – p.45/78

NeNa cycle

21Ne(p,γ)−−→ 22Na

(β+) ↑ ↓ (β+)21Na 22Ne

(p, γ) ↑ ↓ (p, γ)

19F(p,γ)−−→ 20Ne

(p,α)←−− 23Na

(p,γ)−−→ MgAl

Halifax 2004 – p.46/78

Simpler NeNa cycle

21Ne(p,γ)−−−→ 22Ne

(p, γ) ↑ ↓ (p, γ)

20Ne(p,α)←−−− 23Na

Halifax 2004 – p.47/78

Another eigenvalue problem

ddt

20Ne21Ne22Ne23Na

=

− 1τ20

0 0 1τ23

1τ20

1−τ21

0 0

0 1τ21

− 1τ22

0

0 0 1τ22

− 1τ23

20Ne21Ne22Ne23Na

→analytic solution. . .

Halifax 2004 – p.48/78

Example: 21Ne

-8

-7.5

-7

-6.5

-6

-5.5

-5

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

log 1

0(X

Ne2

1)

Time / 105 yrs


Halifax 2004 – p.49/78

MgAl cycle. . .

25Mg(p,γ)−−→ 26Al

(p,γ)−−→ 27Si

(β+) ↑25Al

(p, γ) ↑

↓ (β+)26Mg↓ (p, γ)

��

��

(β+)

23Na(p,γ)−−→ 24Mg

(p,α)←−− 27Al

(p,γ)−−→ 28Si

Halifax 2004 – p.50/78

Becomes the MgAl Chain

•d24Mg

dt = −24Mgτ24

,

•d25Mg

dt = −25Mgτ25

+24Mgτ24

,

• d26Aldt =

25Mgτ25−

26Alτβ26−

26Alτ26′

=25Mgτ25−

26Alτ ′26

,

•d26Mg

dt =26Alτβ26−

26Mgτ26

• d27Aldt =

26Mgτ26

+26Alτ26′

> 0

Halifax 2004 – p.51/78

Example: 24Mg

-3.246

-3.244

-3.242

-3.24

-3.238

-3.236

-3.234

-3.232

-3.23

-3.228

-3.226

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

log 1

0(X

Mg2

4)

Time / 105 yrs


Halifax 2004 – p.52/78

Chemical Yields

• Galactic Chemical Evolution models requirecalculations of yields

• This is the amount of mass ejected as eachisotope from a population of stars

• Synthetic models can be used to calculatethe yields from single and binary stars

• Uncertainties due to variable initialdistributions / physics

Halifax 2004 – p.53/78

Chemical Yields

One definition (my definition)

yield of i =MASS OUT AS ISOTOPE i

MASS INTO STARS

• Mass into stars is different for single andbinary populations, this definition takes thatinto account.

• Binary interaction reduces number of GB andAGB stars, so reduces the yield of isotopesproduced in GB and AGB stars.

Halifax 2004 – p.54/78

Nitrogen Yield (Integrated)

Integrated nitrogen yield (mass out as 14N / massinput to stars)

• Single Stars 1.294× 10−3

• Binary Stars 9.878× 10−4

• Difference due to binaries :−24%

Halifax 2004 – p.55/78

Reaction Rates in the Intershell

• Intershell composition→ envelope pollution

• Reaction rates determine amount of isotopicproduction

• Many are quite uncertain (e.g. 25Mg (α, n)28Si fac. 105!)

• Can we quantify the uncertainty in the yields?

• Difficult problem: many rates/stars, takes too long

• Perhaps with a synthetic model?

• Synthetic Helium Burning!

• Convective intershell: much like HBB.Halifax 2004 – p.56/78

Helium Flash: T

0

5e+07

1e+08

1.5e+08

2e+08

2.5e+08Pulse 1 (Flash Only) : Temperature

320 330 340 350 360 370 380 390Time/years

0.855

0.856

0.857

0.858

0.859

0.86

0.861

0.862

0.863

0.864

0.865

Mas

s/M

sun

Halifax 2004 – p.57/78

Helium Flash: ρ

0

2000

4000

6000

8000

10000

12000

14000

16000

18000Pulse 1 (Flash Only) : Density

320 330 340 350 360 370 380 390Time/years

0.855

0.856

0.857

0.858

0.859

0.86

0.861

0.862

0.863

0.864

0.865

Mas

s/M

sun

Halifax 2004 – p.58/78

Helium Flash: convection

0

0.2

0.4

0.6

0.8

1Pulse 1 (Flash Only) : Convective Zones

320 330 340 350 360 370 380 390Time/years

0.855

0.856

0.857

0.858

0.859

0.86

0.861

0.862

0.863

0.864

0.865

Mas

s/M

sun

Halifax 2004 – p.59/78

Helium Burning Reactions

Material is processed by complete hydrogenburning, so is 0% 1H, ∼ 98%4He, CNO→ 14N,other trace metals (perhaps NeNa/MgAl cycled).Two reaction sets operate• α-capture e.g. 4He(αα, γ)12C, 12C(α, γ)16O,

16O(α, γ)20Ne etc.• n-capture e.g. 20Ne(n, γ)21Ne, 24Mg(n, γ)25Mg,

28Si(n, γ)29Si

Detailed models give final abundances 4He = 0.7,12C = 0.26, 16O = 0.004.

Halifax 2004 – p.60/78

Helium Flash: 4He

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Pulse 1 (Flash Only) : 4He

320 330 340 350 360 370 380 390Time/years

0.855

0.856

0.857

0.858

0.859

0.86

0.861

0.862

0.863

0.864

0.865

Mas

s/M

sun

Halifax 2004 – p.61/78

Helium Flash: 12C

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8Pulse 1 (Flash Only) : 12C

320 330 340 350 360 370 380 390Time/years

0.855

0.856

0.857

0.858

0.859

0.86

0.861

0.862

0.863

0.864

0.865

Mas

s/M

sun

Halifax 2004 – p.62/78

α captures

d4He

dt= −3 〈σv〉

3α(4He)3−〈σv〉

α12

4He12C−〈σv〉α16

4He16O−...,

d12C

dt= 〈σv〉3α (4He)3 − 〈σv〉α12

4He12C ,

d16O

dt= 〈σv〉α12

4He12C− 〈σv〉α164He16O

d20Ne

dt= 〈σv〉α16

4He16O− 〈σv〉α204He20Ne

. . .Halifax 2004 – p.63/78

First approximation. . .

d4He

dt= −3 〈σv〉

3α(4He)3−〈σv〉

α12

4He12C− 〈σv〉α16

4He16O,

d12C

dt= 〈σv〉3α (4He)3−〈σv〉α12

4He12C ,

d16O

dt= 〈σv〉α12

4He12C−〈σv〉α164He16O

d20Ne

dt= 〈σv〉α16

4He16O−〈σv〉α204He20Ne

Halifax 2004 – p.64/78

But it’s not really like HBB!

• Some of the material ingested into theconvective pocket is not from hydrogenburning

• In fact it is radiatively He-burnt material!

• Abundances 4He ∼ 0.55, 12C ∼ 0.4,16O ∼ 0.015 are a function of mass

• Contribution due to this “dredge up” is smallbut contributes to the 12C, 16O, 20Ne

• Not easy to model with a simple algorithm :(

Halifax 2004 – p.65/78

16O(α, γ)20Ne Burn Rate

0

1e+09

2e+09

3e+09

4e+09

5e+09

6e+09

7e+09

8e+09

9e+09Pulse 1 (Flash Only) : 16O(α,γ)20Ne rate

320 330 340 350 360 370 380 390Time/years

0.855

0.856

0.857

0.858

0.859

0.86

0.861

0.862

0.863

0.864

0.865

Mas

s/M

sun

Halifax 2004 – p.66/78

Prototype Model

• Assume constant abundance radiativedredge-up material

• Fit T and ρ (max) of convective pulse vs t

• Burn a fraction of the intershell at T , ρ for thepulse duration (∼ 25 years)

• Fit a constant mixing rate (burnt + unburnt) toAmanda’s model results

Halifax 2004 – p.67/78

Iterative Burning

• Traditional technique, required because of4He3 term and significant 12C-burning

• At each timestep use iterative relaxationmethod to calculate abundances

• n-capture equilibrium• Code in Perl, easy to experiment or change

(or break!)• slower than C or the evil F so must be efficient• Iterative solution agrees with Runge-Kutta

solutionHalifax 2004 – p.68/78

M = 5 M�, Z = 0.02, TP19, mix rate 1.15× 10−7

Isotope Amanda Rob4He 0.7007 0.706412C 0.2666 0.265616O 0.004212 0.00260820Ne 0.001585 0.00156621Ne 2.869× 10−5 2.328× 10−5

22Ne 0.01794 0.0192124Mg 0.0001174 0.0000983825Mg 0.002673 0.00150426Mg 0.003016 0.001613

Halifax 2004 – p.69/78

Why simple HBB model worked

• T and ρ are ∼constant over an interpulseperiod

• Burning shell is thin compared to Mconv

• Total amount burned < Mconv

• Hydrogen abundance ∼ constant over aninterpulse

Halifax 2004 – p.70/78

Why simple He-burning model fails

• T and ρ are definitely not constant in mass ortime

• Burning shell is not thin compared to Mconv?• Total amount burned�Mconv

• Helium abundance is not constant• Mixing rate is not constant?• Not because neutrons are out of eq.• Burning by iterative network is much slower

(but quicker than RK!)

Halifax 2004 – p.71/78

Conclusions

• Need to model radiative burning properly:need mass grid

• Convective burning tricky without T , ρ, Y andmixing as f(t)

• We may as well use detailed models. . .• It’s just too complicated for synthetic models.• Easier to fit results of detailed model runs.

Halifax 2004 – p.72/78

Thermal Pulse 19 T

0

5e+07

1e+08

1.5e+08

2e+08

2.5e+08

3e+08

3.5e+08Pulse 3 (Flash Only) : Temperature

340 345 350 355 360 365 370 375Time/years

0.865

0.866

0.867

0.868

0.869

0.87

0.871

0.872

0.873

0.874

Mas

s/M

sun

Halifax 2004 – p.73/78

Thermal Pulse 19 ρ

0

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000Pulse 3 (Flash Only) : Density

340 345 350 355 360 365 370 375Time/years

0.865

0.866

0.867

0.868

0.869

0.87

0.871

0.872

0.873

0.874

Mas

s/M

sun

Halifax 2004 – p.74/78

Thermal Pulse 19 Convection

0

0.2

0.4

0.6

0.8

1Pulse 3 (Flash Only) : Convective Zones

340 345 350 355 360 365 370 375Time/years

0.865

0.866

0.867

0.868

0.869

0.87

0.871

0.872

0.873

0.874

Mas

s/M

sun

Halifax 2004 – p.75/78

Thermal Pulse 19 4He

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Pulse 3 (Flash Only) : 4He

340 345 350 355 360 365 370 375Time/years

0.865

0.866

0.867

0.868

0.869

0.87

0.871

0.872

0.873

0.874

Mas

s/M

sun

Halifax 2004 – p.76/78

Thermal Pulse 19 29Si

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0.00014

0.00016

0.00018Pulse 3 (Flash Only) : 29Si

340 345 350 355 360 365 370 375Time/years

0.865

0.866

0.867

0.868

0.869

0.87

0.871

0.872

0.873

0.874

Mas

s/M

sun

Halifax 2004 – p.77/78

The end

Halifax 2004 – p.78/78

Documents

Population Nucleosynthesis - University of Cambridge