lecture 20: astrophysical reaction ratesinpp.ohio.edu/~meisel/phys7501/file/lecture20...is the...
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Lecture 20: Ohio University PHYS7501, Fall 2017, Z. Meisel ([email protected])
Lecture 20: Astrophysical Reaction Ratesβ’Evidence for nuclear reactions in spaceβ’Thermonuclear reaction ratesβ’Non-resonant ratesβ’Resonant ratesβ’Reaction networks
Birth of Nuclear Astrophysicsβ’Nuclear astrophysics was spawned in 1920 by a realization of Arthur Eddington (The Observatory (1920):β’Based on fossil evidence at the time, the Earth was known to be more than several hundred million years old and the sun presumably had to be at least as old
β’A plausible explanation for the sunβs power might be gravitational energy being converted into heat from the gaseous solar sphere collapsing, taking place on the Kelvin-Helmholtz timescale
β’ πππΎπΎπΎπΎ = ππππππππππ πΎπΎπΎπΎπΎπΎπΎπΎπππΎπΎπΎπΎ πΈπΈπΎπΎπΎπΎπΈπΈπΈπΈπΈπΈπΈπΈπΎπΎπΎπΎπΈπΈπΈπΈπΈπΈ πΏπΏπππΏπΏπΏπΏ π π πππππΎπΎ
= (πππππππΎπΎπΎπΎπππΎπΎππππ πΈπΈπΎπΎπΎπΎπΈπΈπΈπΈπΈπΈ)/2πΏπΏπΏπΏπΏπΏπΎπΎπΎπΎπππΏπΏπΎπΎπππΈπΈ
= πΊπΊππ2
2π π πΏπΏβ 16ππππππ for the sun
β’Instead, maybe itβs just a lump of burning coal!β’ Typical chemical bond energies are ~eVβ’ The sun has a mass of ~1030kg and a nucleus is ~10-27kg, it has ~1057 nuclei (or atoms)β’ Since the sun releases ~1039MeV/s, chemical burning would last roughlyπ‘π‘πππΏπΏπΈπΈπΎπΎ = π΄π΄πΏπΏπππΏπΏπΎπΎππ ππππ πΉπΉπΏπΏπΎπΎππ
π π πππππΎπΎ ππππ πΉπΉπΏπΏπΎπΎππ π΅π΅πΏπΏπΈπΈπΎπΎπΎπΎπΎπΎπΈπΈ~ 1057πππππππΏπΏπΏπΏβ1πΎπΎππ/πππππππΏπΏ1039πππΎπΎππ/πΏπΏβ106πΎπΎππ/πππΎπΎππ
~1012π π ~30ππππππβ’ Third timeβs the charmβ¦letβs look at nuclear energy
β’ Aston measured a 32MeV discrepancy between 4 protons and 1 Helium nucleus (4p->Ξ± actually yields ~27MeV)β’ Multiplying our fuel amount by 106 (eV->MeV) results in ~10πΊπΊππππ burn time β¦which finally does the job
2
from the virial theorem
Evidence for recent nuclear reactions in space (selected examples)
3
Solar Ξ½ attributable to hydrogen burning sequences
Bahcall, Serenelli, & Basu, ApJL (2005)
Ξ³-rays of short-lived isotopes (e.g. 44Ti) in supernova remnants
B. Grefenstette et al. Nature (2014)
Anomalous isotopic ratios in meteoritic βpre-solarβ dust grains
N.Liu et al ApJL (2017)
Neutron star merger afterglow associated with radioactive decay
Berger, Fong, & Chornock ApJL (2013)
Radioactive elements (e.g. Tc) in stellar spectra
B. Peery PASP (1971)
Match in energetics(e.g. C-fusion in X-ray superbursts)
E.F. Brown ApJL (2004)
Reaction rateβ’The way nuclei influence and are influenced by astrophysical environments is through energy generation and element transmutation (βnucleosynthesisβ). The importance of a pair of nuclides is going to be related to the rate at which they react.
β’We can figure out what that reaction rate is by considering a simple example to a gas with two species, each with their own number density, inside a volume at some temperature
β’For simplicity, weβll consider one species as the projectile and the other as the target(though it doesnβt matter since all of our calculations will be in the center of mass)
β’The reaction rate between species 1 and 2 will be:π π π π π‘π‘π π = πΉπΉπΉπΉπΉπΉπΉπΉ ππππ 1 β πππΉπΉπππππ π ππ π·π·π π π·π·π π π·π·π‘π‘ππ ππππ 2 β (πΌπΌπ·π·π‘π‘π π πππ π πΌπΌπ‘π‘π·π·πππ·π· πππππππππ π πππ·π·πΉπΉπ·π·π‘π‘ππ ππππππ 1 + 2)
β’1βs flux is the number of 1 atoms per area per time: πΉπΉ1 = π·π·1π£π£where π·π·1 is the 1 number density and π£π£ is the relative velocity
β’The interaction probability is provided by the cross section ππ12(π£π£)β’So, ππ π£π£ = πΉπΉ1π·π·2ππ12 π£π£ = π·π·1π·π·2ππ12 π£π£ π£π£, is the reaction rate for velocity π£π£
4
Thermonuclear reaction rateβ’ Of course, nature isnβt so neat and orderly. Any environment will have a finite temperature and
therefore any nuclei will have a velocity distribution ππ(π£π£) thatβs defined by the temperature.β’ To get an average rate given some relative velocity distribution, we need to do an average
ππ = οΏ½0
βππ(π£π£)π·π·1π·π·2ππ12 π£π£ π£π£ πππ£π£ = π·π·1π·π·2 οΏ½
0
βππ(π£π£)ππ12 π£π£ π£π£ πππ£π£ = π·π·1π·π·2 πππ£π£ 12
(If the reaction is endothermic then the integral lower-limit is the reaction threshold)
β’ To avoid double-counting of particles, we have to make a slight modification for identical nuclei:ππ = πΎπΎ1πΎπΎ2 ππππ 12
1+πΏπΏ12, where πΏπΏ12 is the Kronecker delta
β’ Determining the reaction rate between a pair of particles πππ£π£ is the nuclear physicistβs taskβ’ The number densities are determined be calculating the effects of all reaction rates as a
function of time in a reaction network to see how many nuclei are created/destroyedβ’ Itβs more common to deal with matter densities and get the number density by taking into
account the fraction of mass species π·π· contributes: π·π·πΎπΎ = πππππ΄π΄ππππ
πΏπΏππππππππ,ππ= πππππ΄π΄πππΎπΎ
where πππ΄π΄ is Avogadroβs number, πππΏπΏππππ,πΎπΎ is the molar mass (a.k.a. mass in amu),and πππΎπΎ (β€ 1) and πππΎπΎ are the mass and mole fractions, respectively 5
Similarly, the reduced reaction rate NA<Οv> is often used
Thermonuclear rates and the Maxwell-Boltzmann distributionβ’ Typically, the environment weβre considering is a classical gas in thermodynamic equilibrium,
so the velocity distribution is described by the Maxwell-Boltzmann distribution
πππππ΅π΅ π£π£ = 4πππ£π£2 πΏπΏ2πππππ΅π΅ππ
3/2exp β πΏπΏππ2
2πππ΅π΅ππ, where πππ΅π΅ is the Boltzmann constant and the pre-
factor is for normalization (i.e. β«0βπππππ΅π΅ π£π£ πππ£π£ = 1)
β’ Both species interacting in an environment will have this distribution, so taking velocities for1 and 2 in the lab-frame, πππ£π£ 12 = β«0
ββ«0βπππππ΅π΅ π£π£1 πππππ΅π΅ π£π£2 ππ12 π£π£ π£π£πππ£π£1πππ£π£2
β’ This is less-than desirable, since we would rather be working in terms of relative velocity only,so we consider the fact that π£π£πΎπΎ = π£π£πΎπΎπΏπΏ Β± π£π£
β’ This means the double integral can be re-stated in terms of integrating over π£π£ and π£π£πΎπΎπΏπΏ:πππ£π£ 12 = β«0
ββ«0βπππππ΅π΅ π£π£πΎπΎπΏπΏ πππππ΅π΅ π£π£ ππ12 π£π£ π£π£πππ£π£πΎπΎπΏπΏπππ£π£
β’ Since the cross section only depends on the relative velocity and β«0βπππππ΅π΅ π£π£πΎπΎπΏπΏ πππ£π£πΎπΎπΏπΏ = 1,
πππ£π£ 12 = οΏ½0
βπππππ΅π΅ π£π£ ππ12 π£π£ π£π£ πππ£π£
where we should keep in mind ππ in πππππ΅π΅ refers to the reduced mass ππ6
Thermonuclear rates and the Maxwell-Boltzmann distributionβ’ Inserting πππππ΅π΅ π£π£ into πππ£π£ 12 = β«0
βπππππ΅π΅ π£π£ ππ12 π£π£ π£π£ πππ£π£ yields
πππ£π£ 12 = 4ππ ππ2πππππ΅π΅ππ
3/2β«0βππ12 π£π£ π£π£ π£π£2exp β ππππ2
2πππ΅π΅πππππ£π£
β’ Noting the center of mass energy πΈπΈ = 12πππ£π£2 and so πππΈπΈ = πππ£π£ πππ£π£ (i.e. π£π£2 β 2
πππΈπΈ, πππ£π£ β 1
πππππππΈπΈ),
we finally arrive at a useful equation for the astrophysical reaction rate
πππ£π£ 12 =8ππππ
1πππ΅π΅ππ 3/2 οΏ½
0
βππ12 πΈπΈ πΈπΈexp β
πΈπΈπππ΅π΅ππ
πππΈπΈ
β’ Note that this is the general formula for classical gases.Weβll go over special cases (for classical gases) in a bit.
β’ As an aside, personally I find πππ΅π΅ hard to remember,but I find it easier to remember that 11.6045 β πΈπΈπππΎπΎππ = ππ9,where πΈπΈπππΎπΎππ is energy in MeV and ππ9 is temperature in GK
7
Reverse (a.k.a. inverse) ratesβ’ Even when a particular reaction direction is exothermic (ππ > 0), for a finite temperature,
some fraction of particles will have energies with πΈπΈ β₯ ππ, and that fraction will grow with ππβ’ Recall that the cross section for an entrance channel through a compound nuclear state is the
product of the effective geometric area of the projectile ( Ξ»122ππ
2), the number of sub-states that
can be populated (2π½π½ + 1), the probability of being in a particular sub-state for each of the reactants ( 1
(2π½π½1+1)1
(2π½π½2+1)), a factor of 2 for identical particles (1 + πΏπΏ12), and the matrix elements
for forming the compound nucleus πΆπΆ through 1 + 2 and decaying via 3 + 4:
ππ12β34 = ππΞ»122ππ
2 2π½π½ + 12π½π½1 + 1 2π½π½2 + 1
(1 + πΏπΏ12) 3 + 4 π»π»2 πΆπΆ πΆπΆ π»π»1 1 + 2 2
β’ The reverse process would then be
ππ34β12 = ππΞ»342ππ
2 2π½π½ + 12π½π½3 + 1 2π½π½4 + 1
(1 + πΏπΏ34) 1 + 2 π»π»1 πΆπΆ πΆπΆ π»π»2 3 + 4 2
β’ I.e. theyβre identical except for the statistical and geometric factors out front
8
Reverse (a.k.a. inverse) ratesβ’ Making the substitution that
Ξ»ππππ2ππ
2= β2
ππ2= β2
2πππππππΈπΈππππand noting πππΎπΎππ = πΏπΏπππΏπΏππ
πΏπΏππ+πΏπΏππ,
taking the ratio of the forward and reverse cross sections yieldsππ12ππ34
=π΄π΄3π΄π΄4π΄π΄1π΄π΄2
πΈπΈ34πΈπΈ12
2π½π½3 + 1 2π½π½4 + 12π½π½1 + 1 2π½π½2 + 1
(1 + πΏπΏ12)(1 + πΏπΏ34)
where π΄π΄πΎπΎ can be used instead of πππΎπΎ since the units cancel
β’ For the rates, recall πππ£π£ 12 = 8ππππ12
1πππ΅π΅ππ 3/2 β«0
βππ12 πΈπΈ12 πΈπΈ12exp β πΈπΈ12πππ΅π΅ππ
,
so πππ£π£ 34 = 8ππππ34
1πππ΅π΅ππ 3/2 β«0
βππ34 πΈπΈ34 πΈπΈ34exp β πΈπΈ34πππ΅π΅ππ
β’ Since πΈπΈ34 = πΈπΈ12 + ππ12 (if ππ12 > 0), when we take the ratio πππ£π£ 34/ πππ£π£ 12 , the integrands mostly cancel except for the energy-independent part:
πππ£π£ 34
πππ£π£ 12=
2π½π½1 + 1 2π½π½2 + 12π½π½3 + 1 2π½π½4 + 1
(1 + πΏπΏ34)(1 + πΏπΏ12)
ππ12ππ34
3/2
exp βππ12πππ΅π΅ππ
β’ The exponential dominates, so ππππ 34ππππ 12
β exp β ππ12πππ΅π΅ππ
gives the correct order of magnitude9
Reverse (a.k.a. inverse) rates: photodisintegrationβ’ For photodisintegration, we have to take into account the fact that the photons follow a
Planck distribution and not a Maxwell-Boltzmann distribution
β’ So the reverse rate becomes: ππππ 3Ξ³
ππππ 12β ππ12πΎπΎ2
πππ΅π΅ππ
3/2exp β ππ12
πππ΅π΅ππ
β’ This extra factor is actually a huge deal because of that extra temperature dependenceβ’ In fact, for low ππ-value reactions,
the photodisintegration rate is dominant
10
Aside: S-factors, the nuclear physics nuggets of πππ£π£
β’ Often itβs useful to remove the trivial energy dependence from the cross section, in particular for charged-particle reaction rates
β’ The idea is that ππ(πΈπΈ) contains all of the interesting physics
β’ Since the energy dependence is different for different types of reactions,e.g. direct capture of a neutron as compared to direct capture of a charged particle, the factorization that is done to get ππ(πΈπΈ) depends on the reaction type
11
Rolfs & Rodney, Cauldrons in the Cosmos (1988)
Aside-er: Energy release in a reactionβ’ How much energy does a nuclear reaction rate release into the environment?β’ This is as simple as it sounds, where the rate of energy release is just the product of the energy
release per reaction times the reaction rate: ππ12 = ππ12ππ12β’ Given our pre-agreed upon units, ππ = MeV
cm3s,
β¦though some people prefer energy released per gram: Μππ12 = ππ12πΈπΈ12ππ
(units MeV/(g.s))
β’ To be on the safe side, one needs to keep in mind the reverse rate,since we saw it can be significantat high temperatures: ππ12,πΎπΎπΎπΎππ = ππ12 + ππ34 = ππ12(ππ12βππ34)
12
Wiescher & Schatz, Prog.Theory.Phys.Suppl. (2000)
Thermonuclear rate: Direct neutron-captureβ’ Recall that at low energies, those of interest for nuclear astrophysics, the neutron-capture
cross section is described by the 1/π£π£ law: πππΎπΎ πΎπΎππππ β1ππππ
β’ As such, itβs clear that πππ£π£ πΎπΎ πΎπΎππππ β πΌπΌπππ·π·π π π‘π‘π π π·π·π‘π‘β’ The cross section (and rate) can be characterized by the S-factor at thermal energies and any
deviation from 1/π£π£ behavior is accounted for by the local derivative(s) of the S-factor:
ππ πΈπΈ =ππ2πΈπΈ
ππ 0 + οΏ½ΜοΏ½π 0 πΈπΈΒ½ + 12οΏ½ΜοΏ½π 0 πΈπΈ + β―
β’ ππ(0) is generally the S-factor at thermal energy (π£π£πππ‘ = 2.2 Γ 105 πΎπΎπΏπΏπΏπΏ
= πΈπΈπΏπΏ = 2.53 Γ 10β8πππ π ππ):ππ 0 = πππππ‘π£π£πππ‘ = 2.2 Γ 10β19πππππ‘ππππ
3
π π , where πππππ‘ is the thermal cross section in barns
β’ οΏ½ΜοΏ½π 0 and οΏ½ΜοΏ½π 0 are fit to cross section data near thermal energies
β’ Employing this ππ πΈπΈ in the general equation πππ£π£ = 8ππππ
1πππ΅π΅ππ 3/2 β«0
βππ πΈπΈ πΈπΈexp β πΈπΈπππ΅π΅ππ
πππΈπΈ,results in
πππ£π£ πΎπΎ πΎπΎππππ = ππ 0 + 4πποΏ½ΜοΏ½π 0 πππ΅π΅ππ 1/2 + 3
4οΏ½ΜοΏ½π 0 πππ΅π΅ππ + β―13
Thermonuclear rate: Direct neutron-capture
14
Examples (from the ):
3He(n,p) 32Si(n,Ξ³) 208Pb(n,Ξ³)
Clear that the main feature is πππ£π£ πΎπΎ πΎπΎππππ β πΌπΌπππ·π·π π π‘π‘π π π·π·π‘π‘ over 3 orders of magnitude in T
Thermonuclear rate: Direct charged particle-captureβ’ Recall that the cross section for charged-particle capture depends on the effective geometric
area of the projectile, described by Ξ»πππΎπΎπ΅π΅πΈπΈπππΈπΈπππΎπΎπΎπΎ, and the probability of the charged projectile tunneling through the Coulomb barrier of the target
β’ When discussing Ξ± decay, we showed πππππΏπΏπΎπΎπΎπΎπΎπΎππ πΈπΈ = exp(β2ππππ),
where the factor with the Sommerfeld parameter is 2ππππ = 2ππ πΎπΎ2
βπΎπΎππ1ππ2
πππΎπΎ2
2πΈπΈ
β’ For πΈπΈ in MeV and π΄π΄ in atomic mass units (1u = 931.5MeV/c2):
2ππππ = 0.989ππ1ππ21πΈπΈ
π΄π΄1π΄π΄2(π΄π΄1+π΄π΄2)
β’ Removing the trivial energy dependency:πππΎπΎπ‘.πΎπΎππππ(πΈπΈ) = 1
πΈπΈexp β2ππππ ππ(πΈπΈ)
15Rolfs & Rodney, Cauldrons in the Cosmos (1988)
Thermonuclear rate: Direct charged particle-captureβ’ Employing Ο(πΈπΈ) = 1
πΈπΈexp β2ππππ ππ(πΈπΈ) in πππ£π£ = 8
ππππ1
πππ΅π΅ππ 3/2 β«0βππ πΈπΈ πΈπΈexp β πΈπΈ
πππ΅π΅πππππΈπΈ gives:
= 8ππππ
1πππ΅π΅ππ 3/2 β«0
β ππ πΈπΈ exp β πΈπΈπππ΅π΅ππ
β 2ππππ πππΈπΈ
β’ The integrand is a product of the probability of a charged-particle pair having energy πΈπΈ(from the Maxwell-Boltzmann distribution) and the tunneling probability for that energy
β’ The integrand maximum (found by solving for the derivative being equal to zero) is:
πΈπΈπΊπΊ = 0.122 ππ12ππ22π΄π΄1π΄π΄2
(π΄π΄1+π΄π΄2)ππ92
1/3MeV
β’ Approximating the integrand as a Gaussianresults in a distribution with the 1/π π widthβπΊπΊ= 4 πΈπΈπΊπΊπππ΅π΅ππ
3MeV
β’ πΈπΈπΊπΊ is the Gamow peak and βπΊπΊ is the widthof the Gamow window, which is roughlythe energy range for which we care abouta charged particle reaction rate for some ππ
16
Rolfs & Rodney, Cauldrons in the Cosmos (1988)
Note: Donβt be too naΓ―ve when using the Gamow window estimate.Itβs based on a roughly constant S(E), so the true window of interest could be different (T.Rauscher PRC 2010).
Thermonuclear rate: Direct charged particle-captureβ’ The S factor is again represented with a Taylor expansion ππ πΈπΈ = ππ 0 + οΏ½ΜοΏ½π 0 πΈπΈ + 1
2οΏ½ΜοΏ½π πΈπΈ πΈπΈ2 + β―
β’ As with neutron-capture, it is determined experimentally. Here, by dividing the trivial energy dependence from the measured cross section as a function of energy[Or doing something equivalent with a fancy R-matrix fit, e.g. A.Kontos et al PRC2013)]
β’ The energy dependent terms of ππ(πΈπΈ) winds up making the reaction rate integral have several temperature-dependent factors, which make for a rather complicated form:(See C. Iliadis, Nuclear Physics of Stars or Fowler, Caughlan, & Zimmerman, Annu.Rev.Astron.&Astro. (1967))
β’ This is the inspiration for the functional formof parameterized reaction rates used in databases,e.g. JINA REACLIB (R. Cyburt et al. ApJS (2010))
17
Thermonuclear rate: Narrow resonance(s)β’Recall the Breit-Wigner form we found for the resonant reaction cross sectionπππ΅π΅π΅π΅,ππ ππ,ππ ππ(πΈπΈ) = ππ Ξ»
ππ
2 2π½π½+1(2π½π½ππ+1)(2π½π½ππ+1)
Ξππππ(πΈπΈ)Ξππππ(πΈπΈ)(πΈπΈβπΈπΈπ π )2+(Ξ(πΈπΈ))2/4
β’Employing this in the general form πππ£π£ = 8ππππ
1πππ΅π΅ππ 3/2 β«0
βππ πΈπΈ πΈπΈexp β πΈπΈπππ΅π΅ππ
πππΈπΈ,we realize that the contributions of the integrand are pretty negligible outside of πΈπΈπ π
β’So, we make the approximation πππ£π£ πΈπΈπΎπΎπΏπΏ = 8ππππ
πΈπΈπ π πππ΅π΅ππ 3/2 exp β πΈπΈπ π
πππ΅π΅ππβ«0βπππ΅π΅π΅π΅ πΈπΈ πππΈπΈ
β’Noting that Ξ» changes little over the resonance Ξ» β Ξ»π π , writing the statistical factor as ππ, and noting the widths Ξ are essentially constant over the resonance we findβ«0βπππ΅π΅π΅π΅ πΈπΈ πππΈπΈ β ππ Ξ»π π
ππ
2ΞππππΞππππ β«0
β 1πΈπΈβπΈπΈπ π 2+(Ξ)2/4
πππΈπΈ = 2Ξ»π π 2ππΞππππΞππππ
Ξ
β’For obfuscation purposes, we substitute in πΎπΎ for ΞππππΞππππΞ
and call ΟπΎπΎ the resonance strengthβ’If we know the cross section at the peak of the resonance, we can make the approximation that the integral is half the width times the height: β«0
βπππ΅π΅π΅π΅ πΈπΈ πππΈπΈ β ππΞππ(πΈπΈπ π )
β’The resonant rate becomes: πππ£π£ πΈπΈπΎπΎπΏπΏ = 2πππππππ΅π΅ππ
3/2β2 πππΎπΎ π π exp β πΈπΈπ π
πππ΅π΅ππ 18
Thermonuclear rate: Narrow resonance(s)β’For a reaction with several resonances, the total rate is just the sum.Using units of MeV for πΈπΈπ π ,πΎπΎ and πππΎπΎ π π ,πΎπΎ and amu for π΄π΄πΎπΎ,
πππ΄π΄ πππ£π£ ππ πΈπΈπΎπΎπΏπΏ. =1.54 Γ 1011
π΄π΄1π΄π΄2π΄π΄1 + π΄π΄2
ππ93/2οΏ½
πΎπΎ=1
ππ
πππΎπΎ π π ,πΎπΎexp β11.6045πΈπΈπ π ,πΎπΎ
ππ9cm3
mol s
β’But how do I know which resonances matter?β’You might guess the ones in theGamow window β¦but it depends onthe relative widths for theexit and entrance channels.
β’For high temperatures, particle emissionmakes the Gamow Window conceptespecially dubious
β’Nonetheless, the GW is good first guess asto which resonances are likely important
19
Iliadis, Nuclear Physics of Stars (2007)
Thermonuclear rate: Broad resonanceβ’ In the past weβve specified any resonance
with Ξ/πΈπΈπ π β³ 10% as βbroadββ’ For such cases, we obviously need to take
into account the energy dependence ofthe widths when performing the integralover energy.See the resonant reactions lecture for adiscussion on these energy dependencies.
β’ Broad resonances are a pain becausethey make extrapolation of thereaction rate at very low energies(via the S-factor) a sketchy enterprise
β’ One approach to representthese rates analytically is as a non-resonantcontribution (for πΈπΈ βͺ πΈπΈπ π ) plus ananalytic contribution, resulting in: πππ£π£ πππΈπΈππππππ πΈπΈπΎπΎπΏπΏ = πππ£π£ πππΎπΎπΈπΈπΎπΎπΎπΎππ + π π 1ππππexp β ππ2
πππ΅π΅ππ,
where π π πΎπΎ and ππ are fit to the rate calculated at several ππ by integrating over the cross section 20
Rolfs & Rodney, Cauldrons in the Cosmos (1988)
Thermonuclear rate: Subthreshold resonance
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Rolfs & Rodney, Cauldrons in the Cosmos (1988)
β’Broad resonances lurking below the reaction threshold can contribute to the rate as well
β’This can cause a huge boost over the non-resonant rateContributions from a hypothetical sub-threshold resonance with various strengths for 9Be(p,Ξ±):
E.F. Brown, ApJ (1998)
Aside: Stellar Enhancement Factors (SEFs)β’Itβs important to note that the rate calculated using the laboratory cross section is only for the ground-state of the target
β’However, in a hot environment, excited states in the target will be thermally populatedβ’The stellar enhancement factor (SEF) is the ratio of the stellar rate (which is what we want) to the rate determined from the laboratory measurement.
β’The SEF is the ratio of theπππΈπΈπΉπΉ =
πππ‘π‘π π πΉπΉπΉπΉπ π ππ π π π π π‘π‘π π πΏπΏπ π πππππππ π π‘π‘ππππππ π π π π π‘π‘π π =
βπππππΈπΈπΈπΈπΎπΎππ πΏπΏπππππππΎπΎπΏπΏ πΉπΉπππ π πΌπΌπ‘π‘π·π·πππ·π· ππππ πππ π πππππ π π‘π‘ πππππ π πΌπΌπ·π·π π π π π·π·π·π· π π π·π· πΈπΈπΉπΉπΌπΌπ·π·π‘π‘π π ππ πππ‘π‘π π π‘π‘π π (π π π π π‘π‘π π ππππππ π‘π‘π‘π π π‘π‘ πππ‘π‘π π π‘π‘π π )πΏπΏπ π πππππππ π π‘π‘ππππππ π π π π π‘π‘π π
β’The excited state populations are determined bythe partition function, and theyβre typicallyestimated using the statistical model(See e.g. A. Sallaska et al. ApJS 2013)
β’A commonly used set of partition functions usedto calculate SEFs comes fromRauscher & Thielemann ADNDT 2000
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Charged particle reactions with Aβ€40
A. Sallaska et al. ApJS (2013)
Total thermonuclear reaction rateβ’Any and all of the aforementioned rate types can contribute to the overall reaction rate
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Iliadis, Nuclear Physics of Stars (2007)
Where can I get thermonuclear reaction rates?
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REACLIB database
https://groups.nscl.msu.edu/jina/reaclib/db/
β’The Joint Institute for Nuclear Astrophysics hosts REACLIB, which contains thermonuclear reaction rates in parameterized and tabular form
β’Rates are based on published experimentally-constrained and otherwise purely theoretical rates
β’REACLIB is a standard in the field,e.g. it is employed as the default in the MESA stellar modeling software
Nuclear reaction networks
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R.V. Wagoner ApJS (1969)
β’Astrophysical environments typically containmany nuclei, each of which could in principleinteract with the other(in practice usually only the photons and light projectiles matter)
β’To evaluate what happens, we need tosolve a reaction network
β’The basic idea is to see how the abundance changesfor each isotope at each step in time based on theproduction and destruction via all mechanismsand often to include the energy generationfrom said reactions
β’For each species π·π·,πππππΎπΎπππ‘π‘
= οΏ½ππ,ππ
πππππππππππππ΄π΄ πππ£π£ ππππβπΎπΎ + οΏ½ππ
Ξ»ππβπΎπΎππππ β οΏ½πΏπΏ
πππΎπΎπππΏπΏπππππ΄π΄ πππ£π£ πΎπΎπΏπΏβπππΎπΎπΈπΈ + οΏ½πΎπΎ
Ξ»πΎπΎβπΎπΎππππ
Production Destruction
Nuclear reaction networks
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β’You might consider solving the problem explicitly,stepping through time updating each πππΎπΎ
β’The issue is that the ordinary differential equationsππππππππππ
= ππ(ππ)weβre trying to solve are very βstiffβ,i.e. theyβre very sensitive to changes(because reaction rates have steep temperature dependencies)so we would need super tiny time-steps
β’Instead, an implicit approach is neededβ’We note that the change in abundance βπππΎπΎ is equalto the change in abundance at the next time-step times the time-step ππ(πππΎπΎ(π‘π‘ + βπ‘π‘))βπ‘π‘
β’The change in abundance at the next timestep is equal to the change at the present time plus the sum of the change in the change due to other speciesβ abundances changingππ πππΎπΎ π‘π‘ + βπ‘π‘ = ππ πππΎπΎ π‘π‘ + βππ
ππππ(ππππ ππ )ππππ
βππππ β¦where ππππ(ππππ ππ )ππππ
are the elements of the Jacobian matrix
β’Solving for the change in all abundances results in βππ = ππ(π‘π‘)οΏ½1βππβ π½π½
β1
β’Inverting the matrix (the ~ bits) is where most of the computational cost comes in
F.X. Timmes ApJS (1999)
Further Readingβ’ Chapter 12: Modern Nuclear Chemistry (Loveland, Morrissey, & Seaborg)β’ Chapter 3: Nuclear Physics of Stars (C. Iliadis)β’ Chapters 3 & 4: Cauldrons in the Cosmos (C. Rolfs & W. Rodney)β’ Lecture Materials on Nuclear Astrophysics (H. Schatz)β’ Chapters 10 & 11: Stellar Astrophysics (E.F. Brown)β’ βThermonuclear Reaction Ratesβ, Fowler, Caughlan, & Zimmerman, Annu.Rev.Astron.&Astro. (1967)β’ R.V. Wagoner, Astrophys. J. Suppl. Ser. (1969)
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