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I ask you to look both ways. For the road to a knowledge of the stars leads through the atom; and important knowledge of the atom has been reached through the stars.
Arthur Eddington
Betsy Beise, University of Maryland; Philippe Chomaz, GANIL; Alexandra Gade, Michigan State University; Bob McKeown, California Institute of Technology;
Frank Timmes, University of Arizona; Bill Zajc, Columbia University; John Hardy, Texas A&M University;Dave Morrissey, Michigan State University; Witek Nazarewicz, University of Tennessee; Derek Teaney,Stony Brook University; Michael Wiescher, University of Notre Dame; Sherry Yennello, Texas A&M University
National Science Foundation; US Department of Energy's Institute for Nuclear Theory;Michigan State University; National Superconducting Cyclotron Laboratory (NSCL)
June 28-July 10, 2009National Superconducting Cyclotron Laboratory (NSCL)Michigan State University | East Lansing, Michigan
http://meetings.nscl.msu.edu/NNPSS09
topics:invited lecturers:
sponsors:
Lectures, activities, and networking opportunities fornuclear physicists 1-2 years on either side of the PhD
2009 NATIONALNUCLEAR PHYSICSSUMMER SCHOOL
Hadron Physics, Nuclear Reactions, Nuclear Structure, QCD,Neutrinos, Nuclear Astrophysics, Fundamental Symmetries
organizers: [email protected] Schatz (Chair), Wolfgang Bauer, Shari Conroy, Michael Thoennessen
Frank Timmes
Nuclear Astrophysics: Reaction Networks
Michigan State University 30Jun2009 - 02Jul2009
cococubed.asu.edu [email protected]
Outline for 01Jul2009
1. Three Vignettes
a. Energy Generation
b. Linear Algebra
c. Thermodynamic trajectories
2. Big Bang Nucleosynthesis
Stuff of the day
Synthesis of the elements in stars: forty years of progress cococubed.asu.edu/papers/wallerstein97.pdf
An answer to yesterday’s task
10 12 14 16 18 20 22
10-7
10-5
10-3
10-1
Mass F
raction
Time (s)
Cold CNO cycles, hydrostatic burn, T=15x106 K ρ=150 g cm
-3
H H H
H
H
H
α
α
α
αα α α α
C12
C12 C12 C12 C12C12C12 C12
C13
C13 C13 C13 C13C13C13 C13
N14
N14 N14 N14 N14N14N14 N14
N15
N15 N15 N15 N15N15N15 N15O16
O16
O16 O16O16O16 O16
O17
O17O17O17O17 O17
O18O18O18O18 O18
F19F19F19F19 F19
10-4
10-2
100
102
εnuc (e
rg g
-1 s-1)
ε
ε ε
ε
ε
ε
12C
13N
13C 14N
15O
15N
Cycle 1
(p, )
(p, )
(p, )(,e+ )
(,e+ )
(p, )
17O
17F
16O
18F
18O 19F
Cycle 2 Cycle 3
Cycle 4
(p, )
(p, )
(p, )
(p, )
(,e+ )
(,e+ )
(p, )
(p, )(p, )
14O 18Ne
(p, )
(p, )
(,e+ )(,e+ )
(p, )
( ,p)
19Ne( )
rp process
CNO: T9 < 0.2 Hot CNO: 0.2 < T9 < 0.5 rp process: T9 > 0.5
(,e+ )
An answer to yesterday’s task
10-11
10-7
10-3
101
105
109
1013
10-15
10-11
10-7
10-3
101
Mass F
raction
Time (s)
* hydrostatic: btemp= 3.000E+09 bden= 1.000E+09
he4 he4he4
he4he4
he4
he4
he4 he4
he4he4
he4
he4he4
he4 he4he4
he4he4 he4
c12
c12c12c12
c12
c12c12
c12
c12
c12
c12c12 c12 c12c12
c12
c12
c12
o16
o16o16o16
o16
o16o16 o16
o16
o16 o16o16o16
o16
o16
o16
o16
o16
o16
o16
ne20
ne20ne20ne20
ne20
ne20ne20
ne20
ne20
ne20
ne20
ne20
ne20 ne20ne20
ne20
ne20
ne20
mg24
mg24mg24mg24
mg24
mg24mg24
mg24
mg24
mg24
mg24
mg24
mg24mg24
mg24
mg24
mg24
mg24
mg24
mg24
si28
si28si28si28
si28
si28si28si28
si28 si28
si28
si28si28 si28 si28
si28
si28
si28
si28
si28
s32
s32s32s32
s32
s32s32 s32 s32
s32
s32s32s32
s32 s32
s32
s32
s32
s32
s32ar36
ar36ar36ar36
ar36
ar36ar36ar36
ar36
ar36ar36
ar36
ar36
ar36ar36
ar36
ar36
ar36
ar36
ar36
ca40
ca40ca40ca40
ca40
ca40ca40ca40
ca40ca40
ca40
ca40
ca40
ca40
ca40ca40
ca40
ca40
ca40
ca40
ti44
ti44
ti44ti44ti44ti44ti44 ti44 ti44
ti44
ti44
ti44
ti44 ti44
ti44ti44
ti44
ti44
ti44ti44
cr48
cr48
cr48cr48cr48cr48cr48cr48 cr48 cr48 cr48
cr48cr48
cr48
cr48 cr48cr48
cr48cr48
cr48
fe52
fe52
fe52
fe52fe52
fe52fe52fe52 fe52 fe52 fe52fe52fe52
fe52fe52 fe52fe52
fe52fe52 fe52
ni56
ni56
ni56
ni56ni56ni56 ni56 ni56 ni56 ni56ni56ni56
ni56 ni56 ni56ni56ni56 ni56 ni56
1013
1017
1021
1025
εnuc (e
rg g
-1 s-1)
ε εεεεεεε
ε
ε
ε
ε
ε
ε
εεεε
εε
4He 12C 16O
27Al 31P 35Cl 39K 43Sc 47V 51Mn 55Co
20Ne 24Mg 28Si 32S 36Ar 40Ca 44Ti 48Cr 52Fe 56Ni
(α,γ)(αα,γ) (α,γ) (α,γ) (α,γ) (α,γ) (α,γ) (α,γ) (α,γ) (α,γ) (α,γ) (α,γ)
(p,γ)(α,p) (p,γ)(α,p) (p,γ)(α,p) (p,γ)(α,p) (p,γ)(α,p) (p,γ)(α,p) (p,γ)(α,p) (p,γ)(α,p)
Alpha-chain Network
Energy Generation
εnuc = −
∑
i
NAMic2Yi − εν (erg g−1 s−1)
An important consequence of changing the composition is the release (or absorption) of energy. The energy generation rate is
where Mi c2 is the rest mass energy of species i. Using
Mi = Aimµ + Mex,i
εnuc =∑
i
NAEbind,iYi −
∑
i
NA(ZiMex,p + NiMex,n)Yi − εν
Mex,i = ZiMex,p + NiMex,n −Bi/c2
the energy generation rate is sometimes written as
The energy lost to the freely streaming neutrinos has two parts: weak reactions and neutrino thermal processes.
For weak reactions, average energy losses are calculated for each nucleus by considering the excited state distribution, Gamow-Teller distribution, etc.
εν =
∑
i
⟨Eν⟩Yi,weak
The results are tabulated; see for example Langanke & Martinez-Pinedo (2000).
3 6 9
T9
10−80
10−60
10−40
10−20
ρYe= 10
7
ρYe= 10
8
ρYe= 10
9
ρYe= 10
10
10−4
10−2
100
102
104
3 6 9
T9
10−80
10−60
10−40
10−20
10−6
10−5
10−4
10−3
10−2
10−1
56Co e
− capture
56Co β
+
decay
56Fe e
+
capture56
Fe β− decay
λ (
s−1)
Thermal neutrinos processes include
pair neutrino:
photoneutrino:
plasma neutrino:
bremsstrahlung:
recombination:
Results are typically expressed in tables or fitting formulas, for example, Itoh et al. 1996.
e+
+ e−
→ ν + ν
e+
+ γ → e−
+ ν + ν
γplasmon → ν + ν
e−
continuum→ e
−
bound+ ν + ν
e− + N(Z, A)→ e− + N(Z, A) + ν + ν
100
103
106
109
1012
10-4
10-2
100
102
104
106
108
Density (g/cm3)
Energ
y L
oss R
ate
(erg
/g/s
)
T=5e8 K X(12
C)=X(16
O)=0.5
Pair neutrino
Photoneutrino
Plasma neutrino
Bremsstrahlung
Recombination
Total
÷x = b
Within our implicit integrations we’re solving (large) systems of linear equations. As the linear algebra generally dominates the time to obtain a solution, we’ll want to use efficient solvers.
We’ll briefly look at dense, direct sparse, and iterative sparse solvers.
Matrix à is reduced to upper triangular form in tandem with a right-hand side b by Gaussian elimination, and backsubstitution on the upper triangular matrix yields the solution to ÷x = b.
This is the method you probably first learned.
If the arithmetic is exact, then the answer computed in this manner will be exact, if no zeros appear on the diagonal.
2 −1 1
−2 2 −3
2 −4 3
x1
x2
x3
=
3
−7
3
1 −1/2 1/2
−2 2 −3
2 −4 3
x1
x2
x3
=
3/2
−7
3
1 −1/2 1/2
0 1 −2
0 −3 2
x1
x2
x3
=
3/2
−4
0
1 −1/2 1/2
0 1 −2
0 0 −4
x1
x2
x3
=
3/2
−4
−12
1 −1/2 1/2
0 1 −2
0 0 1
x1
x2
x3
=
3/2
−4
3
x3 = 3 x2 = 2 x1 = 1
But computer arithmetic is not exact, so there will always be some truncation and rounding error in the answer.
If a “small” number appears on the diagonal, then its use as the pivot may lead to computing differences between big numbers and little numbers with a subsequent loss of precision.
A way around this problem is to ensure small pivots are not used by swapping rows (partial pivoting) or rows and columns (full pivoting), so as to use a particularly desirable pivot element.
What is a desirable pivot? It is not completely known theoretically. It is known, both theoretically and in practice, that simply picking the largest available element as the pivot is a very good choice.
Fact: Gauss elimination with no pivoting is numerically unstable in the presence of any roundoff error, even when a small pivot is not encountered. Never do Gauss elimination without pivoting!
LEQS is a dense matrix, Gaussian elimination routine.
The maximum element in each row serves as the pivot element, but no row or column interchanges are performed, so LEQS may be unstable on matrices that are not diagonally dominant.
A small amount of effort is expended to minimize calculations with matrix elements that are zero.
All Gaussian elimination routines have the disadvantage that for a sequence of right-hand sides, the entire matrix must be decomposed for each right-hand side.
The origin of LEQS is somewhat obscure, but circa 1962. It may be the most common linear algebra solver used for evolving nuclear reaction networks.
Ford-Seattle
1962
A · x =
(
L · U
)
· x = L ·
(
U · x
)
= b
A frequently used form of Gauss Elimination is LU decomposition.
The basic idea is to find two matrices L and U such that LU = Ã, where L is a lower triangular matrix (zero above the diagonal) and U is an upper triangular matrix (zero below the diagonal).
Once we have computed L and U we the solve L·y=b then U·x=y,a process that takes O(n2) operations.
While the factorization stage still requires O(n3) operations, it need be done only once. We can solve with as many right-hand sides as we care to, one at a time. This is a superior advantage.
A linear system is called sparse if only a relatively small number of its matrix elements aij are nonzero.
It is wasteful to use general methods on such problems, because most of storage the O(n3) operations involve zero operands.
Direct methods for sparse matrices are not that different from dense LU decomposition methods; they are just cleverly applied with due attention to the bookkeeping of zero elements.
The basic approach that all solvers use are 1) Symbolic decomposition 2) Numerical decomposition 3) Backsubstitution 4) Iterative polishing
MA28 is the Coke classic of sparse matrix solvers. hsl.rl.ac.uk/archive/hslarchive.html
UMFPACK is a modern, direct sparse matrix solver. www.cise.ufl.edu/research/sparse/umfpack
Iterative methods for sparse systems only reference the matrix à only through multiplication of a vector. “Matrix free” methods.
Iterative methods can be slow to converge and the number of iterations to reach a given level of accuracy is not known a priori.
A popular method, generalized minimum residuals, seeks a minimization of the function
One way to generate a good “guess” is to solve some portion of Ã, call it matrix Z, that is easy to solve. Z is called the preconditioner.
⇤f(x) = AT
·
�
A · x � b
⇥
f(x) =1
2r · r =
1
2
�
�
�A · x− b
�
�
�
2
BiCG is described by Barret et al in “Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods”. netlib2.cs.utk.edu/linalg/html_templates/Templates.html
SPARSKIT is a modern, iterative sparse matrix solver. www-users.cs.umn.edu/~saad/software/SPARSKIT/sparskit.html
We’ve been holding the temperature and density constant;
Such conditions are called “hydrostatic” burning, since the local energy release doesn’t change the temperature or density, as during the hydrostatic phases of a star’s evolution.
dT
dt= 0
dρ
dt= 0
We will now relax these assumptions, andconsider any thermodynamic profile.
Density
Temperature
t=0
t=1
t=2
t=3
t=4
We could evolve temperature and density ODEs separately and then evolve the reaction network, or we can evolve the temperature and density ODEs simultaneously with the network.
The first is “operator splitting” and assumes a loose coupling between physical processes. Operator splitting is easy to implement and, by far, the most common choice.
An example when the assumption of loose coupling breaks down.
Time Step
Tem
pera
ture
n n+1 n+2
TooCool
TooHot
JustRight
• Photodisintegration dominates• Negative energy generation rate• Material cools
• Electron capture dominates• Positive energy generation rate• Material heats
What thetemperatureshould do
A Hydrodynamic-Burning Instability
The second is “unsplit”. It avoids the coupling issues, but is more difficult to implement, particularly for implicit integrations.
Our networks uses the unsplit method. Energy, temperature and density burning ODEs are appended to the reaction network.
T
T
ρ
ρ
These entries from the
temperature and density
derivatives of the reaction
rates
enuc
enuc
These entries from the
abundance derivatives
As an example, during the Big Bang the photon temperature of the expanding universe obeys the ODE:
dT
dt=
√
8πGaf(x)
3c2T 3
[
xdg/dT
3g− 1
]
−1
f(x) = 1 +45
2π4
∫
∞
0
[
√
x2 + y2 +y2
3√
x2 + y2
]
exp[
√
x2 + y2 + 1]
−1
y2dy
g(x) = 1 + Nν
7
8
[
4
11f(x)
]4/3
+
∫
∞
0
√
x2 + y2
[
exp(
√
x2 + y2
)
+ 1]
−1
y2dy
x =
mec2
kT
where
We can know the temperature at any given time because the dominant constituents are either massless or relativistic.
What we don’t know a priori is the density ρb of ordinary matter in the expanding universe. How shall we parameterize our ignorance?
A common way to express the baryon density is in terms of a baryon-to-photon ratio; the number of photons for every particle.
Mass density and number density
Number density of photons
Baryon density in terms of the photon temperature and the free parameter nb/nγ
ρb =
nb
NA
nγ =30ζ(3)
π4
aT3
γ
k
ρb =nb
nγ
=30ζ(3)
π4kNA
T3
γ
How our universe cools and gets less dense, for a given nb/nγ ratio.
Expansion factor R/R0
10-12 10-10 10-8 10-6 10-4 10-2
1
102
104
106
108
1010
1012T
em
pera
ture
(K
)
1010
1
10-10
10-20
10-30
10-40
All C
onst
itue
nts
in E
quilib
rium
PairsAnnihilate
RadiationUniverse
MatterUniverse
1 104 108 1012 1016
Time (sec)
Dens
ity
(g/c
m3)
Normalized to ρ0 = 2x10-29 g/cm3 and T
0 = 3 K
T=Tν
Tγ
Tν
ρν
ρb
ργ
ρe
ρν=ρe
Now
A typical Big Bang reaction network.
(n,γ)
(d,γ)
(n,γ)
(n,γ)
(p,γ)(p,γ)
(p,γ)
(ν,e-)(e+,ν)
free decay
(d,p)
(n,p)
(d,n)
To Be
7
From
Be7
From
Be8
From
Li7
To Li7
Exoergic Direction
(d,p)(t,pn)
(He3,2p)
(He4 ,γ)
(d,n)
(He3,pn)
(t,2n)
(n,H
e4 )
(γ,H
e4 )
(p,H
e4 )
(He4 ,γ)
p d
He3 He4
t
n
For times < 15 s the temperature > 3 billion K, and our universe is still a soup of protons, neutrons, electrons and more exotic matter. Anything more complex is blasted apart by high energy photons.
101
102
103
104
105
10-15
10-13
10-11
10-9
10-7
10-5
10-3
10-1
101
Mass F
raction
Time (s)
* big bang: eta= 6.140E-10 Nnu= 3.000E+00 H0= 7.100E+01 Tcmb= 2.730E+00
nnnnnnnnnnnn nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
n
n
n
n
nn
n
n
n
pppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppp p p p p p p p
dddddddd
dddd
dddd
ddddd
ddddddddd
dddddddddddddddddddddddddddd
dd
dd d d d d d
t
ttttttttt
tt
t tt t
ttttt
ttt t
ttttttttttttttttttttttttttttttttt
tt
tt
t t t t t
he3
he3he3he3he3he3he3he3he3he3
he3he3
he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3 he3 he3 he3 he3 he3
he4he4he4
he4he4he4he4he4he4he4
he4he4
he4he4
he4he4he4he4he4
he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4
he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4 he4 he4 he4 he4 he4
li6li6li6li6li6
li6li6li6li6li6li6li6li6li6li6li6li6li6li6li6
li6li6
li6li6
li6 li6 li6 li6 li6
li7
li7li7
li7li7li7li7li7
li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7
li7li7
li7li7 li7 li7 li7 li7 li7
be7be7be7be7be7be7be7be7be7be7be7be7be7be7be7
be7
be7
be7be7 be7 be7 be7 be7 be7
b11b11b11b11b11b11b11b11b11b11b11b11b11
b11b11
b11
By 3 min deuterium survives after it is fused and is quickly turned into helium. Without deuterium all the neutrons would decay and our universe would be pure hydrogen.
101
102
103
104
105
10-15
10-13
10-11
10-9
10-7
10-5
10-3
10-1
101
Mass F
raction
Time (s)
* big bang: eta= 6.140E-10 Nnu= 3.000E+00 H0= 7.100E+01 Tcmb= 2.730E+00
nnnnnnnnnnnn nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
n
n
n
n
nn
n
n
n
pppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppp p p p p p p p
dddddddd
dddd
dddd
ddddd
ddddddddd
dddddddddddddddddddddddddddd
dd
dd d d d d d
t
ttttttttt
tt
t tt t
ttttt
ttt t
ttttttttttttttttttttttttttttttttt
tt
tt
t t t t t
he3
he3he3he3he3he3he3he3he3he3
he3he3
he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3 he3 he3 he3 he3 he3
he4he4he4
he4he4he4he4he4he4he4
he4he4
he4he4
he4he4he4he4he4
he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4
he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4 he4 he4 he4 he4 he4
li6li6li6li6li6
li6li6li6li6li6li6li6li6li6li6li6li6li6li6li6
li6li6
li6li6
li6 li6 li6 li6 li6
li7
li7li7
li7li7li7li7li7
li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7
li7li7
li7li7 li7 li7 li7 li7 li7
be7be7be7be7be7be7be7be7be7be7be7be7be7be7be7
be7
be7
be7be7 be7 be7 be7 be7 be7
b11b11b11b11b11b11b11b11b11b11b11b11b11
b11b11
b11
Carbon and oxygen are not produced since: (1) there are no stable isotopes with 5 or 8 nucleons, (2) the Coulomb barrier starts to be significant, (3) the low density suppresses the fusion of helium to carbon.
By 35 min nucleosynthesis is essentially complete.
101
102
103
104
105
10-15
10-13
10-11
10-9
10-7
10-5
10-3
10-1
101
Mass F
raction
Time (s)
* big bang: eta= 6.140E-10 Nnu= 3.000E+00 H0= 7.100E+01 Tcmb= 2.730E+00
nnnnnnnnnnnn nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
n
n
n
n
nn
n
n
n
pppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppp p p p p p p p
dddddddd
dddd
dddd
ddddd
ddddddddd
dddddddddddddddddddddddddddd
dd
dd d d d d d
t
ttttttttt
tt
t tt t
ttttt
ttt t
ttttttttttttttttttttttttttttttttt
tt
tt
t t t t t
he3
he3he3he3he3he3he3he3he3he3
he3he3
he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3he3 he3 he3 he3 he3 he3
he4he4he4
he4he4he4he4he4he4he4
he4he4
he4he4
he4he4he4he4he4
he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4
he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4he4 he4 he4 he4 he4 he4
li6li6li6li6li6
li6li6li6li6li6li6li6li6li6li6li6li6li6li6li6
li6li6
li6li6
li6 li6 li6 li6 li6
li7
li7li7
li7li7li7li7li7
li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7li7
li7li7
li7li7 li7 li7 li7 li7 li7
be7be7be7be7be7be7be7be7be7be7be7be7be7be7be7
be7
be7
be7be7 be7 be7 be7 be7 be7
b11b11b11b11b11b11b11b11b11b11b11b11b11
b11b11
b11
A key unknown in big bang nucleosynthesis had been the density of ordinary matter ρb. The cosmic microwave background provides by far the most precise determination.
WMAP
10-33
10-32
10-31
10-30
10-29
10-28
10-11
10-9
10-7
10-5
10-3
10-1
Baryon mass density (g/cm3)
Mass fra
ction
H0=70.5 km/s/Mpc T
cmb=2.725 K N
ν=3.0 τ
neutron=885.7 s
d d d dd
dd
dd
d
d
d
d
he3he3he3he3he3he3
he3he3he3he3he3he3he3he3he3he3he3he3he3
αα
αα α α α α α α α α α α α α α α α
li6li6 li6 li6
li7
li7
li7li7 li7
li7
li7li7
li7li7
li7 li7 li7 li7 li7 li7 li7 li7 li7
c11c11
c11c11
c11c11
c11c11
c11
ρcrit
The observed light element abundances is consistent with a WMAP determined baryon density, with the possible exception of 7Li.
Several independent measurements lead to a consistent constraint.
WMAP.22
.23
.24
.25
.26
H0=70.5 km/s/Mpc T
cmb=2.725 K N
ν=3.0 τ
neutron=885.7 s
~ ~
4He
10-5
10-4
Mass F
raction
~ ~
2H
3He
2×10-31
4×10-31
6×10-31
10-9
10-8
Baryon mass density
7Li
Download, compile, and run the Big Bang nucleosynthesis code from cococubed.asu.edu/code_pages/burn.shtml Plot the evolution of the abundances.
Download, compile, and run the Big Bang thermodynamics code from cococubed.asu.edu/code_pages/burn.shtml Plot the evolution of the photon and neutrino temperatures.
Tasks for the day
Questions and Discussion
Betsy Beise, University of Maryland; Philippe Chomaz, GANIL; Alexandra Gade, Michigan State University; Bob McKeown, California Institute of Technology;
Frank Timmes, University of Arizona; Bill Zajc, Columbia University; John Hardy, Texas A&M University;Dave Morrissey, Michigan State University; Witek Nazarewicz, University of Tennessee; Derek Teaney,Stony Brook University; Michael Wiescher, University of Notre Dame; Sherry Yennello, Texas A&M University
National Science Foundation; US Department of Energy's Institute for Nuclear Theory;Michigan State University; National Superconducting Cyclotron Laboratory (NSCL)
June 28-July 10, 2009National Superconducting Cyclotron Laboratory (NSCL)Michigan State University | East Lansing, Michigan
http://meetings.nscl.msu.edu/NNPSS09
topics:invited lecturers:
sponsors:
Lectures, activities, and networking opportunities fornuclear physicists 1-2 years on either side of the PhD
2009 NATIONALNUCLEAR PHYSICSSUMMER SCHOOL
Hadron Physics, Nuclear Reactions, Nuclear Structure, QCD,Neutrinos, Nuclear Astrophysics, Fundamental Symmetries
organizers: [email protected] Schatz (Chair), Wolfgang Bauer, Shari Conroy, Michael Thoennessen