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 Rates β€’ Evidence for nuclear reactions in space β€’ Thermonuclear reaction rates β€’ Non-resonant rates β€’ Resonant rates β€’ Reaction networks

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Page 1: Lecture 20: Astrophysical Reaction Ratesinpp.ohio.edu/~meisel/PHYS7501/file/Lecture20...is the Kronecker delta β€’Determining the reaction rate between a pair of particles 𝜎𝜎is

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

Page 2: Lecture 20: Astrophysical Reaction Ratesinpp.ohio.edu/~meisel/PHYS7501/file/Lecture20...is the Kronecker delta β€’Determining the reaction rate between a pair of particles 𝜎𝜎is

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 sun’s power might be gravitational energy being converted into heat from the gaseous solar sphere collapsing, taking place on the Kelvin-Helmholtz timescale

β€’ 𝜏𝜏𝐾𝐾𝐾𝐾 = 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝑇𝑇𝐾𝐾𝐾𝐾 𝐸𝐸𝐾𝐾𝐾𝐾𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐾𝐾𝐾𝐾𝐸𝐸𝐸𝐸𝐸𝐸 𝐿𝐿𝑇𝑇𝐿𝐿𝐿𝐿 𝑅𝑅𝑇𝑇𝑇𝑇𝐾𝐾

= (𝑃𝑃𝑇𝑇𝑇𝑇𝐾𝐾𝐾𝐾𝑇𝑇𝐾𝐾𝑇𝑇𝑇𝑇 𝐸𝐸𝐾𝐾𝐾𝐾𝐸𝐸𝐸𝐸𝐸𝐸)/2𝐿𝐿𝐿𝐿𝐿𝐿𝐾𝐾𝐾𝐾𝑇𝑇𝐿𝐿𝐾𝐾𝑇𝑇𝐸𝐸

= 𝐺𝐺𝑀𝑀2

2π‘…π‘…πΏπΏβ‰ˆ 16𝑀𝑀𝑀𝑀𝑀𝑀 for the sun

β€’Instead, maybe it’s 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 roughly𝑑𝑑𝑏𝑏𝐿𝐿𝐸𝐸𝐾𝐾 = 𝐴𝐴𝐿𝐿𝑇𝑇𝐿𝐿𝐾𝐾𝑇𝑇 π‘‡π‘‡π‘œπ‘œ 𝐹𝐹𝐿𝐿𝐾𝐾𝑇𝑇

𝑅𝑅𝑇𝑇𝑇𝑇𝐾𝐾 π‘‡π‘‡π‘œπ‘œ 𝐹𝐹𝐿𝐿𝐾𝐾𝑇𝑇 𝐡𝐡𝐿𝐿𝐸𝐸𝐾𝐾𝐾𝐾𝐾𝐾𝐸𝐸~ 1057π‘‡π‘‡π‘‡π‘‡π‘‡π‘‡πΏπΏπΏπΏβˆ—1𝐾𝐾𝑒𝑒/𝑇𝑇𝑇𝑇𝑇𝑇𝐿𝐿1039𝑀𝑀𝐾𝐾𝑒𝑒/πΏπΏβˆ—106𝐾𝐾𝑒𝑒/𝑀𝑀𝐾𝐾𝑒𝑒

~1012𝑠𝑠~30π‘˜π‘˜π‘€π‘€π‘€π‘€β€’ Third time’s the charm…let’s 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 ~10𝐺𝐺𝑀𝑀𝑀𝑀 burn time …which finally does the job

2

from the virial theorem

Page 3: Lecture 20: Astrophysical Reaction Ratesinpp.ohio.edu/~meisel/PHYS7501/file/Lecture20...is the Kronecker delta β€’Determining the reaction rate between a pair of particles 𝜎𝜎is

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 β€œpre-solar” 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)

Page 4: Lecture 20: Astrophysical Reaction Ratesinpp.ohio.edu/~meisel/PHYS7501/file/Lecture20...is the Kronecker delta β€’Determining the reaction rate between a pair of particles 𝜎𝜎is

Reaction rateβ€’The way nuclei influence and are influenced by astrophysical environments is through energy generation and element transmutation (β€œnucleosynthesis”). 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, we’ll consider one species as the projectile and the other as the target(though it doesn’t matter since all of our calculations will be in the center of mass)

β€’The reaction rate between species 1 and 2 will be:𝑅𝑅𝑅𝑅𝑑𝑑𝑅𝑅 = 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 π‘œπ‘œπ‘œπ‘œ 1 βˆ— 𝑁𝑁𝐹𝐹𝑁𝑁𝑁𝑁𝑅𝑅𝑀𝑀 𝐷𝐷𝑅𝑅𝐷𝐷𝑠𝑠𝐷𝐷𝑑𝑑𝑀𝑀 π‘œπ‘œπ‘œπ‘œ 2 βˆ— (πΌπΌπ·π·π‘‘π‘‘π‘…π‘…π‘€π‘€π‘…π‘…πΌπΌπ‘‘π‘‘π·π·π‘œπ‘œπ·π· π‘ƒπ‘ƒπ‘€π‘€π‘œπ‘œπ‘π‘π‘…π‘…π‘π‘π·π·πΉπΉπ·π·π‘‘π‘‘π‘€π‘€ π‘œπ‘œπ‘œπ‘œπ‘€π‘€ 1 + 2)

β€’1’s flux is the number of 1 atoms per area per time: 𝐹𝐹1 = 𝐷𝐷1𝑣𝑣where 𝐷𝐷1 is the 1 number density and 𝑣𝑣 is the relative velocity

β€’The interaction probability is provided by the cross section 𝜎𝜎12(𝑣𝑣)β€’So, 𝑀𝑀 𝑣𝑣 = 𝐹𝐹1𝐷𝐷2𝜎𝜎12 𝑣𝑣 = 𝐷𝐷1𝐷𝐷2𝜎𝜎12 𝑣𝑣 𝑣𝑣, is the reaction rate for velocity 𝑣𝑣

4

Page 5: Lecture 20: Astrophysical Reaction Ratesinpp.ohio.edu/~meisel/PHYS7501/file/Lecture20...is the Kronecker delta β€’Determining the reaction rate between a pair of particles 𝜎𝜎is

Thermonuclear reaction rateβ€’ Of course, nature isn’t so neat and orderly. Any environment will have a finite temperature and

therefore any nuclei will have a velocity distribution 𝑃𝑃(𝑣𝑣) that’s defined by the temperature.β€’ To get an average rate given some relative velocity distribution, we need to do an average

𝑀𝑀 = οΏ½0

βˆžπ‘ƒπ‘ƒ(𝑣𝑣)𝐷𝐷1𝐷𝐷2𝜎𝜎12 𝑣𝑣 𝑣𝑣 𝑑𝑑𝑣𝑣 = 𝐷𝐷1𝐷𝐷2 οΏ½

0

βˆžπ‘ƒπ‘ƒ(𝑣𝑣)𝜎𝜎12 𝑣𝑣 𝑣𝑣 𝑑𝑑𝑣𝑣 = 𝐷𝐷1𝐷𝐷2 πœŽπœŽπ‘£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:𝑀𝑀 = 𝐾𝐾1𝐾𝐾2 𝜎𝜎𝜎𝜎 12

1+𝛿𝛿12, where 𝛿𝛿12 is the Kronecker delta

β€’ Determining the reaction rate between a pair of particles πœŽπœŽπ‘£π‘£ is the nuclear physicist’s 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€’ It’s more common to deal with matter densities and get the number density by taking into

account the fraction of mass species 𝐷𝐷 contributes: 𝐷𝐷𝐾𝐾 = πœŒπœŒπ‘π‘π΄π΄π‘‹π‘‹π‘–π‘–

πΏπΏπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘š,𝑖𝑖= πœŒπœŒπ‘π‘π΄π΄π‘Œπ‘ŒπΎπΎ

where 𝑁𝑁𝐴𝐴 is Avogadro’s number, 𝑁𝑁𝐿𝐿𝑇𝑇𝑇𝑇,𝐾𝐾 is the molar mass (a.k.a. mass in amu),and 𝑋𝑋𝐾𝐾 (≀ 1) and π‘Œπ‘ŒπΎπΎ are the mass and mole fractions, respectively 5

Similarly, the reduced reaction rate NA<Οƒv> is often used

Page 6: Lecture 20: Astrophysical Reaction Ratesinpp.ohio.edu/~meisel/PHYS7501/file/Lecture20...is the Kronecker delta β€’Determining the reaction rate between a pair of particles 𝜎𝜎is

Thermonuclear rates and the Maxwell-Boltzmann distributionβ€’ Typically, the environment we’re considering is a classical gas in thermodynamic equilibrium,

so the velocity distribution is described by the Maxwell-Boltzmann distribution

𝑃𝑃𝑀𝑀𝐡𝐡 𝑣𝑣 = 4πœ‹πœ‹π‘£π‘£2 𝐿𝐿2πœ‹πœ‹π‘˜π‘˜π΅π΅π‘‡π‘‡

3/2exp βˆ’ 𝐿𝐿𝜎𝜎2

2π‘˜π‘˜π΅π΅π‘‡π‘‡, where π‘˜π‘˜π΅π΅ is the Boltzmann constant and the pre-

factor is for normalization (i.e. ∫0βˆžπ‘ƒπ‘ƒπ‘€π‘€π΅π΅ 𝑣𝑣 𝑑𝑑𝑣𝑣 = 1)

β€’ Both species interacting in an environment will have this distribution, so taking velocities for1 and 2 in the lab-frame, πœŽπœŽπ‘£π‘£ 12 = ∫0

∞∫0βˆžπ‘ƒπ‘ƒπ‘€π‘€π΅π΅ 𝑣𝑣1 𝑃𝑃𝑀𝑀𝐡𝐡 𝑣𝑣2 𝜎𝜎12 𝑣𝑣 𝑣𝑣𝑑𝑑𝑣𝑣1𝑑𝑑𝑣𝑣2

β€’ This is less-than desirable, since we would rather be working in terms of relative velocity only,so we consider the fact that 𝑣𝑣𝐾𝐾 = 𝑣𝑣𝐾𝐾𝐿𝐿 Β± 𝑣𝑣

β€’ This means the double integral can be re-stated in terms of integrating over 𝑣𝑣 and 𝑣𝑣𝐾𝐾𝐿𝐿:πœŽπœŽπ‘£π‘£ 12 = ∫0

∞∫0βˆžπ‘ƒπ‘ƒπ‘€π‘€π΅π΅ 𝑣𝑣𝐾𝐾𝐿𝐿 𝑃𝑃𝑀𝑀𝐡𝐡 𝑣𝑣 𝜎𝜎12 𝑣𝑣 𝑣𝑣𝑑𝑑𝑣𝑣𝐾𝐾𝐿𝐿𝑑𝑑𝑣𝑣

β€’ Since the cross section only depends on the relative velocity and ∫0βˆžπ‘ƒπ‘ƒπ‘€π‘€π΅π΅ 𝑣𝑣𝐾𝐾𝐿𝐿 𝑑𝑑𝑣𝑣𝐾𝐾𝐿𝐿 = 1,

πœŽπœŽπ‘£π‘£ 12 = οΏ½0

βˆžπ‘ƒπ‘ƒπ‘€π‘€π΅π΅ 𝑣𝑣 𝜎𝜎12 𝑣𝑣 𝑣𝑣 𝑑𝑑𝑣𝑣

where we should keep in mind 𝑁𝑁 in 𝑃𝑃𝑀𝑀𝐡𝐡 refers to the reduced mass πœ‡πœ‡6

Page 7: Lecture 20: Astrophysical Reaction Ratesinpp.ohio.edu/~meisel/PHYS7501/file/Lecture20...is the Kronecker delta β€’Determining the reaction rate between a pair of particles 𝜎𝜎is

Thermonuclear rates and the Maxwell-Boltzmann distributionβ€’ Inserting 𝑃𝑃𝑀𝑀𝐡𝐡 𝑣𝑣 into πœŽπœŽπ‘£π‘£ 12 = ∫0

βˆžπ‘ƒπ‘ƒπ‘€π‘€π΅π΅ 𝑣𝑣 𝜎𝜎12 𝑣𝑣 𝑣𝑣 𝑑𝑑𝑣𝑣 yields

πœŽπœŽπ‘£π‘£ 12 = 4πœ‹πœ‹ πœ‡πœ‡2πœ‹πœ‹π‘˜π‘˜π΅π΅π‘‡π‘‡

3/2∫0∞𝜎𝜎12 𝑣𝑣 𝑣𝑣 𝑣𝑣2exp βˆ’ πœ‡πœ‡πœŽπœŽ2

2π‘˜π‘˜π΅π΅π‘‡π‘‡π‘‘π‘‘π‘£π‘£

β€’ Noting the center of mass energy 𝐸𝐸 = 12πœ‡πœ‡π‘£π‘£2 and so 𝑑𝑑𝐸𝐸 = πœ‡πœ‡π‘£π‘£ 𝑑𝑑𝑣𝑣 (i.e. 𝑣𝑣2 β†’ 2

πœ‡πœ‡πΈπΈ, 𝑑𝑑𝑣𝑣 β†’ 1

πœ‡πœ‡πœŽπœŽπ‘‘π‘‘πΈπΈ),

we finally arrive at a useful equation for the astrophysical reaction rate

πœŽπœŽπ‘£π‘£ 12 =8πœ‹πœ‹πœ‡πœ‡

1π‘˜π‘˜π΅π΅π‘‡π‘‡ 3/2 οΏ½

0

∞𝜎𝜎12 𝐸𝐸 𝐸𝐸exp βˆ’

πΈπΈπ‘˜π‘˜π΅π΅π‘‡π‘‡

𝑑𝑑𝐸𝐸

β€’ Note that this is the general formula for classical gases.We’ll go over special cases (for classical gases) in a bit.

β€’ As an aside, personally I find π‘˜π‘˜π΅π΅ hard to remember,but I find it easier to remember that 11.6045 βˆ— 𝐸𝐸𝑀𝑀𝐾𝐾𝑒𝑒 = 𝑇𝑇9,where 𝐸𝐸𝑀𝑀𝐾𝐾𝑒𝑒 is energy in MeV and 𝑇𝑇9 is temperature in GK

7

Page 8: Lecture 20: Astrophysical Reaction Ratesinpp.ohio.edu/~meisel/PHYS7501/file/Lecture20...is the Kronecker delta β€’Determining the reaction rate between a pair of particles 𝜎𝜎is

Reverse (a.k.a. inverse) ratesβ€’ Even when a particular reaction direction is exothermic (𝑄𝑄 > 0), for a finite temperature,

some fraction of particles will have energies with 𝐸𝐸 β‰₯ 𝑄𝑄, and that fraction will grow with 𝑇𝑇‒ 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»122πœ‹πœ‹

2), the number of sub-states that

can be populated (2𝐽𝐽 + 1), the probability of being in a particular sub-state for each of the reactants ( 1

(2𝐽𝐽1+1)1

(2𝐽𝐽2+1)), a factor of 2 for identical particles (1 + 𝛿𝛿12), and the matrix elements

for forming the compound nucleus 𝐢𝐢 through 1 + 2 and decaying via 3 + 4:

𝜎𝜎12β†’34 = πœ‹πœ‹Ξ»122πœ‹πœ‹

2 2𝐽𝐽 + 12𝐽𝐽1 + 1 2𝐽𝐽2 + 1

(1 + 𝛿𝛿12) 3 + 4 𝐻𝐻2 𝐢𝐢 𝐢𝐢 𝐻𝐻1 1 + 2 2

β€’ The reverse process would then be

𝜎𝜎34β†’12 = πœ‹πœ‹Ξ»342πœ‹πœ‹

2 2𝐽𝐽 + 12𝐽𝐽3 + 1 2𝐽𝐽4 + 1

(1 + 𝛿𝛿34) 1 + 2 𝐻𝐻1 𝐢𝐢 𝐢𝐢 𝐻𝐻2 3 + 4 2

β€’ I.e. they’re identical except for the statistical and geometric factors out front

8

Page 9: Lecture 20: Astrophysical Reaction Ratesinpp.ohio.edu/~meisel/PHYS7501/file/Lecture20...is the Kronecker delta β€’Determining the reaction rate between a pair of particles 𝜎𝜎is

Reverse (a.k.a. inverse) ratesβ€’ Making the substitution that

λ𝑖𝑖𝑖𝑖2πœ‹πœ‹

2= ℏ2

𝑝𝑝2= ℏ2

2πœ‡πœ‡π‘–π‘–π‘–π‘–πΈπΈπ‘–π‘–π‘–π‘–and noting πœ‡πœ‡πΎπΎπ‘–π‘– = 𝐿𝐿𝑖𝑖𝐿𝐿𝑖𝑖

𝐿𝐿𝑖𝑖+𝐿𝐿𝑖𝑖,

taking the ratio of the forward and reverse cross sections yields𝜎𝜎12𝜎𝜎34

=𝐴𝐴3𝐴𝐴4𝐴𝐴1𝐴𝐴2

𝐸𝐸34𝐸𝐸12

2𝐽𝐽3 + 1 2𝐽𝐽4 + 12𝐽𝐽1 + 1 2𝐽𝐽2 + 1

(1 + 𝛿𝛿12)(1 + 𝛿𝛿34)

where 𝐴𝐴𝐾𝐾 can be used instead of 𝑁𝑁𝐾𝐾 since the units cancel

β€’ For the rates, recall πœŽπœŽπ‘£π‘£ 12 = 8πœ‹πœ‹πœ‡πœ‡12

1π‘˜π‘˜π΅π΅π‘‡π‘‡ 3/2 ∫0

∞𝜎𝜎12 𝐸𝐸12 𝐸𝐸12exp βˆ’ 𝐸𝐸12π‘˜π‘˜π΅π΅π‘‡π‘‡

,

so πœŽπœŽπ‘£π‘£ 34 = 8πœ‹πœ‹πœ‡πœ‡34

1π‘˜π‘˜π΅π΅π‘‡π‘‡ 3/2 ∫0

∞𝜎𝜎34 𝐸𝐸34 𝐸𝐸34exp βˆ’ 𝐸𝐸34π‘˜π‘˜π΅π΅π‘‡π‘‡

β€’ Since 𝐸𝐸34 = 𝐸𝐸12 + 𝑄𝑄12 (if 𝑄𝑄12 > 0), when we take the ratio πœŽπœŽπ‘£π‘£ 34/ πœŽπœŽπ‘£π‘£ 12 , the integrands mostly cancel except for the energy-independent part:

πœŽπœŽπ‘£π‘£ 34

πœŽπœŽπ‘£π‘£ 12=

2𝐽𝐽1 + 1 2𝐽𝐽2 + 12𝐽𝐽3 + 1 2𝐽𝐽4 + 1

(1 + 𝛿𝛿34)(1 + 𝛿𝛿12)

πœ‡πœ‡12πœ‡πœ‡34

3/2

exp βˆ’π‘„π‘„12π‘˜π‘˜π΅π΅π‘‡π‘‡

β€’ The exponential dominates, so 𝜎𝜎𝜎𝜎 34𝜎𝜎𝜎𝜎 12

β‰ˆ exp βˆ’ 𝑄𝑄12π‘˜π‘˜π΅π΅π‘‡π‘‡

gives the correct order of magnitude9

Page 10: Lecture 20: Astrophysical Reaction Ratesinpp.ohio.edu/~meisel/PHYS7501/file/Lecture20...is the Kronecker delta β€’Determining the reaction rate between a pair of particles 𝜎𝜎is

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: 𝜎𝜎𝜎𝜎 3Ξ³

𝜎𝜎𝜎𝜎 12β‰ˆ πœ‡πœ‡12𝐾𝐾2

π‘˜π‘˜π΅π΅π‘‡π‘‡

3/2exp βˆ’ 𝑄𝑄12

π‘˜π‘˜π΅π΅π‘‡π‘‡

β€’ This extra factor is actually a huge deal because of that extra temperature dependenceβ€’ In fact, for low 𝑄𝑄-value reactions,

the photodisintegration rate is dominant

10

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Aside: S-factors, the nuclear physics nuggets of πœŽπœŽπ‘£π‘£

β€’ Often it’s useful to remove the trivial energy dependence from the cross section, in particular for charged-particle reaction rates

β€’ The idea is that 𝑆𝑆(𝐸𝐸) 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 𝑆𝑆(𝐸𝐸) depends on the reaction type

11

Rolfs & Rodney, Cauldrons in the Cosmos (1988)

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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: πœ–πœ–12 = 𝑄𝑄12𝑀𝑀12β€’ Given our pre-agreed upon units, πœ–πœ– = MeV

cm3s,

…though some people prefer energy released per gram: Μƒπœ–πœ–12 = 𝑄𝑄12𝐸𝐸12𝜌𝜌

(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: πœ–πœ–12,𝐾𝐾𝐾𝐾𝑇𝑇 = πœ–πœ–12 + πœ–πœ–34 = 𝑄𝑄12(𝑀𝑀12βˆ’π‘€π‘€34)

12

Wiescher & Schatz, Prog.Theory.Phys.Suppl. (2000)

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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/𝑣𝑣 law: 𝜎𝜎𝐾𝐾 𝐾𝐾𝑇𝑇𝑝𝑝 ∝1πœŽπœŽπ‘›π‘›

β€’ As such, it’s clear that πœŽπœŽπ‘£π‘£ 𝐾𝐾 𝐾𝐾𝑇𝑇𝑝𝑝 β‰ˆ πΌπΌπ‘œπ‘œπ·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/𝑣𝑣 behavior is accounted for by the local derivative(s) of the S-factor:

𝜎𝜎 𝐸𝐸 =πœ‡πœ‡2𝐸𝐸

𝑆𝑆 0 + �̇�𝑆 0 𝐸𝐸½ + 12οΏ½ΜˆοΏ½π‘† 0 𝐸𝐸 + β‹―

β€’ 𝑆𝑆(0) is generally the S-factor at thermal energy (𝑣𝑣𝑇𝑇𝑑 = 2.2 Γ— 105 𝐾𝐾𝐿𝐿𝐿𝐿

= 𝐸𝐸𝐿𝐿 = 2.53 Γ— 10βˆ’8𝑀𝑀𝑅𝑅𝑀𝑀):𝑆𝑆 0 = πœŽπœŽπ‘‡π‘‡π‘‘π‘£π‘£π‘‡π‘‡π‘‘ = 2.2 Γ— 10βˆ’19πœŽπœŽπ‘‡π‘‡π‘‘π‘π‘π‘šπ‘š

3

𝑠𝑠 , where πœŽπœŽπ‘‡π‘‡π‘‘ is the thermal cross section in barns

β€’ �̇�𝑆 0 and οΏ½ΜˆοΏ½π‘† 0 are fit to cross section data near thermal energies

β€’ Employing this 𝜎𝜎 𝐸𝐸 in the general equation πœŽπœŽπ‘£π‘£ = 8πœ‹πœ‹πœ‡πœ‡

1π‘˜π‘˜π΅π΅π‘‡π‘‡ 3/2 ∫0

∞𝜎𝜎 𝐸𝐸 𝐸𝐸exp βˆ’ πΈπΈπ‘˜π‘˜π΅π΅π‘‡π‘‡

𝑑𝑑𝐸𝐸,results in

πœŽπœŽπ‘£π‘£ 𝐾𝐾 𝐾𝐾𝑇𝑇𝑝𝑝 = 𝑆𝑆 0 + 4πœ‹πœ‹οΏ½Μ‡οΏ½π‘† 0 π‘˜π‘˜π΅π΅π‘‡π‘‡ 1/2 + 3

4οΏ½ΜˆοΏ½π‘† 0 π‘˜π‘˜π΅π΅π‘‡π‘‡ + β‹―13

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Thermonuclear rate: Direct neutron-capture

14

Examples (from the ):

3He(n,p) 32Si(n,Ξ³) 208Pb(n,Ξ³)

Clear that the main feature is πœŽπœŽπ‘£π‘£ 𝐾𝐾 𝐾𝐾𝑇𝑇𝑝𝑝 β‰ˆ πΌπΌπ‘œπ‘œπ·π·π‘ π‘ π‘‘π‘‘π‘…π‘…π·π·π‘‘π‘‘ over 3 orders of magnitude in T

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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 λ𝑑𝑑𝐾𝐾𝐡𝐡𝐸𝐸𝑇𝑇𝐸𝐸𝑇𝑇𝐾𝐾𝐾𝐾, and the probability of the charged projectile tunneling through the Coulomb barrier of the target

β€’ When discussing Ξ± decay, we showed 𝑃𝑃𝑇𝑇𝐿𝐿𝐾𝐾𝐾𝐾𝐾𝐾𝑇𝑇 𝐸𝐸 = exp(βˆ’2πœ‹πœ‹πœ‚πœ‚),

where the factor with the Sommerfeld parameter is 2πœ‹πœ‹πœ‚πœ‚ = 2πœ‹πœ‹ 𝐾𝐾2

ℏ𝐾𝐾𝑍𝑍1𝑍𝑍2

πœ‡πœ‡πΎπΎ2

2𝐸𝐸

β€’ For 𝐸𝐸 in MeV and 𝐴𝐴 in atomic mass units (1u = 931.5MeV/c2):

2πœ‹πœ‹πœ‚πœ‚ = 0.989𝑍𝑍1𝑍𝑍21𝐸𝐸

𝐴𝐴1𝐴𝐴2(𝐴𝐴1+𝐴𝐴2)

β€’ Removing the trivial energy dependency:πœŽπœŽπΎπΎπ‘‘.𝐾𝐾𝑇𝑇𝑝𝑝(𝐸𝐸) = 1

𝐸𝐸exp βˆ’2πœ‹πœ‹πœ‚πœ‚ 𝑆𝑆(𝐸𝐸)

15Rolfs & Rodney, Cauldrons in the Cosmos (1988)

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Thermonuclear rate: Direct charged particle-captureβ€’ Employing Οƒ(𝐸𝐸) = 1

𝐸𝐸exp βˆ’2πœ‹πœ‹πœ‚πœ‚ 𝑆𝑆(𝐸𝐸) in πœŽπœŽπ‘£π‘£ = 8

πœ‹πœ‹πœ‡πœ‡1

π‘˜π‘˜π΅π΅π‘‡π‘‡ 3/2 ∫0∞𝜎𝜎 𝐸𝐸 𝐸𝐸exp βˆ’ 𝐸𝐸

π‘˜π‘˜π΅π΅π‘‡π‘‡π‘‘π‘‘πΈπΈ gives:

= 8πœ‹πœ‹πœ‡πœ‡

1π‘˜π‘˜π΅π΅π‘‡π‘‡ 3/2 ∫0

∞ 𝑆𝑆 𝐸𝐸 exp βˆ’ πΈπΈπ‘˜π‘˜π΅π΅π‘‡π‘‡

βˆ’ 2πœ‹πœ‹πœ‚πœ‚ 𝑑𝑑𝐸𝐸

β€’ The integrand is a product of the probability of a charged-particle pair having energy 𝐸𝐸(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:

𝐸𝐸𝐺𝐺 = 0.122 𝑍𝑍12𝑍𝑍22𝐴𝐴1𝐴𝐴2

(𝐴𝐴1+𝐴𝐴2)𝑇𝑇92

1/3MeV

β€’ Approximating the integrand as a Gaussianresults in a distribution with the 1/𝑅𝑅 widthβˆ†πΊπΊ= 4 πΈπΈπΊπΊπ‘˜π‘˜π΅π΅π‘‡π‘‡

3MeV

β€’ 𝐸𝐸𝐺𝐺 is the Gamow peak and βˆ†πΊπΊ is the widthof the Gamow window, which is roughlythe energy range for which we care abouta charged particle reaction rate for some 𝑇𝑇

16

Rolfs & Rodney, Cauldrons in the Cosmos (1988)

Note: Don’t be too naΓ―ve when using the Gamow window estimate.It’s based on a roughly constant S(E), so the true window of interest could be different (T.Rauscher PRC 2010).

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Thermonuclear rate: Direct charged particle-captureβ€’ The S factor is again represented with a Taylor expansion 𝑆𝑆 𝐸𝐸 = 𝑆𝑆 0 + �̇�𝑆 0 𝐸𝐸 + 1

2οΏ½ΜˆοΏ½π‘† 𝐸𝐸 𝐸𝐸2 + 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 𝑆𝑆(𝐸𝐸) 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

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Thermonuclear rate: Narrow resonance(s)β€’Recall the Breit-Wigner form we found for the resonant reaction cross section𝜎𝜎𝐡𝐡𝐡𝐡,𝑋𝑋 𝑇𝑇,𝑏𝑏 π‘Œπ‘Œ(𝐸𝐸) = πœ‹πœ‹ Ξ»

πœ‹πœ‹

2 2𝐽𝐽+1(2π½π½π‘Žπ‘Ž+1)(2𝐽𝐽𝑋𝑋+1)

Ξ“π‘Žπ‘Žπ‘‹π‘‹(𝐸𝐸)Γ𝑏𝑏𝑏𝑏(𝐸𝐸)(πΈπΈβˆ’πΈπΈπ‘…π‘…)2+(Ξ“(𝐸𝐸))2/4

β€’Employing this in the general form πœŽπœŽπ‘£π‘£ = 8πœ‹πœ‹πœ‡πœ‡

1π‘˜π‘˜π΅π΅π‘‡π‘‡ 3/2 ∫0

∞𝜎𝜎 𝐸𝐸 𝐸𝐸exp βˆ’ πΈπΈπ‘˜π‘˜π΅π΅π‘‡π‘‡

𝑑𝑑𝐸𝐸,we realize that the contributions of the integrand are pretty negligible outside of 𝐸𝐸𝑅𝑅

β€’So, we make the approximation πœŽπœŽπ‘£π‘£ 𝐸𝐸𝐾𝐾𝐿𝐿 = 8πœ‹πœ‹πœ‡πœ‡

πΈπΈπ‘…π‘…π‘˜π‘˜π΅π΅π‘‡π‘‡ 3/2 exp βˆ’ 𝐸𝐸𝑅𝑅

π‘˜π‘˜π΅π΅π‘‡π‘‡βˆ«0∞𝜎𝜎𝐡𝐡𝐡𝐡 𝐸𝐸 𝑑𝑑𝐸𝐸

β€’Noting that Ξ» changes little over the resonance Ξ» β†’ λ𝑅𝑅, writing the statistical factor as πœ”πœ”, and noting the widths Ξ“ are essentially constant over the resonance we find∫0∞𝜎𝜎𝐡𝐡𝐡𝐡 𝐸𝐸 𝑑𝑑𝐸𝐸 β‰ˆ πœ‹πœ‹ λ𝑅𝑅

πœ‹πœ‹

2Ξ“π‘‡π‘‡π‘‹π‘‹Ξ“π‘π‘π‘Œπ‘Œ ∫0

∞ 1πΈπΈβˆ’πΈπΈπ‘…π‘… 2+(Ξ“)2/4

𝑑𝑑𝐸𝐸 = 2λ𝑅𝑅2πœ”πœ”Ξ“π‘Žπ‘Žπ‘‹π‘‹Ξ“π‘π‘π‘π‘

Ξ“

β€’For obfuscation purposes, we substitute in 𝛾𝛾 for Ξ“π‘Žπ‘Žπ‘‹π‘‹Ξ“π‘π‘π‘π‘Ξ“

and call ω𝛾𝛾 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: ∫0

∞𝜎𝜎𝐡𝐡𝐡𝐡 𝐸𝐸 𝑑𝑑𝐸𝐸 β‰ˆ πœ‹πœ‹Ξ“πœŽπœŽ(𝐸𝐸𝑅𝑅)

β€’The resonant rate becomes: πœŽπœŽπ‘£π‘£ 𝐸𝐸𝐾𝐾𝐿𝐿 = 2πœ‹πœ‹πœ‡πœ‡π‘˜π‘˜π΅π΅π‘‡π‘‡

3/2ℏ2 πœ”πœ”π›Ύπ›Ύ 𝑅𝑅exp βˆ’ 𝐸𝐸𝑅𝑅

π‘˜π‘˜π΅π΅π‘‡π‘‡ 18

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Thermonuclear rate: Narrow resonance(s)β€’For a reaction with several resonances, the total rate is just the sum.Using units of MeV for 𝐸𝐸𝑅𝑅,𝐾𝐾 and πœ”πœ”π›Ύπ›Ύ 𝑅𝑅,𝐾𝐾 and amu for 𝐴𝐴𝐾𝐾,

𝑁𝑁𝐴𝐴 πœŽπœŽπ‘£π‘£ 𝑁𝑁 𝐸𝐸𝐾𝐾𝐿𝐿. =1.54 Γ— 1011

𝐴𝐴1𝐴𝐴2𝐴𝐴1 + 𝐴𝐴2

𝑇𝑇93/2οΏ½

𝐾𝐾=1

𝑁𝑁

πœ”πœ”π›Ύπ›Ύ 𝑅𝑅,𝐾𝐾exp βˆ’11.6045𝐸𝐸𝑅𝑅,𝐾𝐾

𝑇𝑇9cm3

mol s

β€’But how do I know which resonances matter?β€’You might guess the ones in theGamow window …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)

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Thermonuclear rate: Broad resonanceβ€’ In the past we’ve specified any resonance

with Ξ“/𝐸𝐸𝑅𝑅 ≳ 10% as β€œbroad”‒ 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 𝐸𝐸 β‰ͺ 𝐸𝐸𝑅𝑅) plus ananalytic contribution, resulting in: πœŽπœŽπ‘£π‘£ 𝑏𝑏𝐸𝐸𝑇𝑇𝑇𝑇𝑑𝑑 𝐸𝐸𝐾𝐾𝐿𝐿 = πœŽπœŽπ‘£π‘£ 𝑑𝑑𝐾𝐾𝐸𝐸𝐾𝐾𝐾𝐾𝑇𝑇 + 𝑅𝑅1𝑇𝑇𝑏𝑏exp βˆ’ 𝑇𝑇2

π‘˜π‘˜π΅π΅π‘‡π‘‡,

where 𝑅𝑅𝐾𝐾 and 𝑁𝑁 are fit to the rate calculated at several 𝑇𝑇 by integrating over the cross section 20

Rolfs & Rodney, Cauldrons in the Cosmos (1988)

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Thermonuclear rate: Subthreshold resonance

21

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)

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Aside: Stellar Enhancement Factors (SEFs)β€’It’s 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 the𝑆𝑆𝐸𝐸𝐹𝐹 =

𝑆𝑆𝑑𝑑𝑅𝑅𝐹𝐹𝐹𝐹𝑅𝑅𝑀𝑀 π‘…π‘…π‘…π‘…π‘‘π‘‘π‘…π‘…πΏπΏπ‘…π‘…π‘π‘π‘œπ‘œπ‘€π‘€π‘…π‘…π‘‘π‘‘π‘œπ‘œπ‘€π‘€π‘€π‘€ 𝑅𝑅𝑅𝑅𝑑𝑑𝑅𝑅 =

βˆ‘π‘‡π‘‡π‘‡π‘‡πΈπΈπΈπΈπΎπΎπ‘‡π‘‡ 𝐿𝐿𝑇𝑇𝑇𝑇𝑇𝑇𝐾𝐾𝐿𝐿 πΉπΉπ‘€π‘€π‘…π‘…πΌπΌπ‘‘π‘‘π·π·π‘œπ‘œπ·π· π‘œπ‘œπ‘œπ‘œ 𝑇𝑇𝑅𝑅𝑀𝑀𝑇𝑇𝑅𝑅𝑑𝑑 𝑆𝑆𝑆𝑆𝑅𝑅𝐼𝐼𝐷𝐷𝑅𝑅𝑠𝑠 𝐷𝐷𝐷𝐷 𝑅𝑅𝐷𝐷 𝐸𝐸𝐹𝐹𝐼𝐼𝐷𝐷𝑑𝑑𝑅𝑅𝑑𝑑 𝑆𝑆𝑑𝑑𝑅𝑅𝑑𝑑𝑅𝑅 (𝑅𝑅𝑅𝑅𝑑𝑑𝑅𝑅 π‘œπ‘œπ‘œπ‘œπ‘€π‘€ 𝑑𝑑𝑑𝑅𝑅𝑑𝑑 𝑆𝑆𝑑𝑑𝑅𝑅𝑑𝑑𝑅𝑅)πΏπΏπ‘…π‘…π‘π‘π‘œπ‘œπ‘€π‘€π‘…π‘…π‘‘π‘‘π‘œπ‘œπ‘€π‘€π‘€π‘€ 𝑅𝑅𝑅𝑅𝑑𝑑𝑅𝑅

β€’The excited state populations are determined bythe partition function, and they’re 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

22

Charged particle reactions with A≀40

A. Sallaska et al. ApJS (2013)

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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)

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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

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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 𝐷𝐷,π‘‘π‘‘π‘Œπ‘ŒπΎπΎπ‘‘π‘‘π‘‘π‘‘

= �𝑖𝑖,π‘˜π‘˜

π‘Œπ‘Œπ‘–π‘–π‘Œπ‘Œπ‘˜π‘˜πœŒπœŒπ‘π‘π΄π΄ πœŽπœŽπ‘£π‘£ π‘–π‘–π‘˜π‘˜β†’πΎπΎ + �𝑇𝑇

Ξ»π‘‡π‘‡β†’πΎπΎπ‘Œπ‘Œπ‘‡π‘‡ βˆ’ �𝐿𝐿

π‘Œπ‘ŒπΎπΎπ‘Œπ‘ŒπΏπΏπœŒπœŒπ‘π‘π΄π΄ πœŽπœŽπ‘£π‘£ 𝐾𝐾𝐿𝐿→𝑇𝑇𝐾𝐾𝐸𝐸 + �𝐾𝐾

Ξ»πΎπΎβ†’πΎπΎπ‘Œπ‘Œπ‘‡π‘‡

Production Destruction

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Nuclear reaction networks

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β€’You might consider solving the problem explicitly,stepping through time updating each π‘Œπ‘ŒπΎπΎ

β€’The issue is that the ordinary differential equationsπ‘‘π‘‘π‘Œπ‘Œπ‘–π‘–π‘‘π‘‘π‘‡π‘‡

= π‘œπ‘œ(π‘Œπ‘Œ)we’re trying to solve are very β€œstiff”,i.e. they’re 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 βˆ†π‘Œπ‘ŒπΎπΎ is equalto the change in abundance at the next time-step times the time-step π‘œπ‘œ(π‘Œπ‘ŒπΎπΎ(𝑑𝑑 + βˆ†π‘‘π‘‘))βˆ†π‘‘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 species’ abundances changingπ‘œπ‘œ π‘Œπ‘ŒπΎπΎ 𝑑𝑑 + βˆ†π‘‘π‘‘ = π‘œπ‘œ π‘Œπ‘ŒπΎπΎ 𝑑𝑑 + βˆ‘π‘–π‘–

πœ•πœ•π‘œπ‘œ(π‘Œπ‘Œπ‘–π‘– 𝑇𝑇 )π‘Œπ‘Œπ‘–π‘–

βˆ†π‘Œπ‘Œπ‘–π‘– …where πœ•πœ•π‘œπ‘œ(π‘Œπ‘Œπ‘–π‘– 𝑇𝑇 )π‘Œπ‘Œπ‘–π‘–

are the elements of the Jacobian matrix

β€’Solving for the change in all abundances results in βˆ†π‘Œπ‘Œ = π‘œπ‘œ(𝑑𝑑)οΏ½1βˆ†π‘‡π‘‡βˆ’ 𝐽𝐽

βˆ’1

β€’Inverting the matrix (the ~ bits) is where most of the computational cost comes in

F.X. Timmes ApJS (1999)

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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€’ β€œThermonuclear Reaction Rates”, Fowler, Caughlan, & Zimmerman, Annu.Rev.Astron.&Astro. (1967)β€’ R.V. Wagoner, Astrophys. J. Suppl. Ser. (1969)

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