Stellar Power Sources

ReminderReminder

nPlease return Assignment 1 to theSchool Office by 13:00 Weds. 11thFebruary (tomorrow!)nPlease return Assignment 1 to theSchool Office by 13:00 Weds. 11thFebruary (tomorrow!)

–The assignment questions will be reviewedin next week’s workshop–The assignment questions will be reviewedin next week’s workshop

Stellar Power SourcesStellar Power Sources

nPossibilitiesnPossibilities

–Chemical burning–Chemical burning

nonly a few thousand yearsnonly a few thousand years

–Gravitational contraction–Gravitational contraction

na few million yearsna few million years

–Nuclear fusion–Nuclear fusion

n~1010 years for a sun-like starn~1010 years for a sun-like star

Physics of FusionPhysics of Fusion

nThe Nuclear PotentialnThe Nuclear Potential

–Repulsive at large distances due to Coulomb–Repulsive at large distances due to Coulomb

–Attractive at short range due to Strong NuclearForce–Attractive at short range due to Strong NuclearForce

n(1fermi = 10-15 m)n(1fermi = 10-15 m)

Physics of FusionPhysics of Fusion

–Classically, for nuclei to fuse, a barrier ofheight Ec has to be overcome:–Classically, for nuclei to fuse, a barrier ofheight Ec has to be overcome:

Ec in MeV, ZA, ZB nuclear charges and rN the range of thestrong force in fermis (typically 1 fermi)

Physics of FusionPhysics of Fusion

–Typical stellar models give typical coretemperatures of ~ 107 K (kT ~ 1keV)–Typical stellar models give typical coretemperatures of ~ 107 K (kT ~ 1keV)

–Fraction of nuclei with sufficient energy toovercome EC given by Boltzmann:–Fraction of nuclei with sufficient energy toovercome EC given by Boltzmann:

Classically, fusion shouldn’t happen!

Physics of FusionPhysics of Fusion

nSolution:nSolution:

–Quantum Mechanics–Quantum Mechanics

–Protons described by SchroedingerEquation–Protons described by SchroedingerEquation

–Quantum mechanical tunnelling possible–Quantum mechanical tunnelling possible

Physics of FusionPhysics of Fusion

–Probability of barrier penetration given by:–Probability of barrier penetration given by:

Where rc is the classical closest distance of approach ofthe nuclei and  is defined by:

Physics of FusionPhysics of Fusion

–The probability of barrier penetration canbe recast in terms of energy:–The probability of barrier penetration canbe recast in terms of energy:

Where EG is the Gamow energy define by:

= fine structure constant (~1/137)

mr = reduced mass of nuclei

Physics of FusionPhysics of Fusion

–For two protons, EG =493 keV–For two protons, EG =493 keV

–In a typical stellar core, kT ~ 1 keV–In a typical stellar core, kT ~ 1 keV

–Hence the probability of barrier penetrationis:–Hence the probability of barrier penetrationis:

Still slow, but there are a lot of protons, and we have a lotof time!

Fusion Cross SectionsFusion Cross Sections

nConsider a proton moving in a medium with nprotons per unit volumenConsider a proton moving in a medium with nprotons per unit volume

–Probability of fusion occuring within a distance x= nx–Probability of fusion occuring within a distance x= nx

 = reaction cross section = reaction cross section

–Mean distance between collisions = mean freepath, l = 1/n–Mean distance between collisions = mean freepath, l = 1/n

–Mean time betweencollisions, = l/<v> = 1/ n<v>–Mean time betweencollisions, = l/<v> = 1/ n<v>

note  may depend on vnote  may depend on v

Fusion Cross SectionsFusion Cross Sections

nThe fusion cross section (Units, barns = 10-28m2) as a function of energy is given by:nThe fusion cross section (Units, barns = 10-28m2) as a function of energy is given by:

S(E) is a slowly varying function determined by thenuclear physics of the reaction

1/E introduced to account for low energy behaviour

Fusion Reaction RatesFusion Reaction Rates

Consider the reaction rate between twonuclei, A and B, travelling with relative speed,v, with concentrations nA and nB, with crosssection Consider the reaction rate between twonuclei, A and B, travelling with relative speed,v, with concentrations nA and nB, with crosssection 

–Mean time for an A nucleus to fuse with a B is: A = 1/ nB<v>–Mean time for an A nucleus to fuse with a B is: A = 1/ nB<v>

–Hence, the total fusion rate per unit volume is:RAB = nAnB <v>–Hence, the total fusion rate per unit volume is:RAB = nAnB <v>

Fusion Reaction RatesFusion Reaction Rates

–To obtain <v>, we note that:–To obtain <v>, we note that:

Where P(vr) is the Maxwell-Boltzmann distribution givenby:

Fusion Reaction RatesFusion Reaction Rates

–Including the function for  found earlier,the total reaction rate is then:–Including the function for  found earlier,the total reaction rate is then:

Concentrating on the integral, we note that for a givenimpact energy, E, there is a competition between theBoltzmann term and the Gamow energy term

Fusion Reaction RatesFusion Reaction Rates

exp(-E/kT)

exp(-(EG/E)1/2)

exp(-E/kT-(EG/E)1/2)

–Proton-proton reactions–Proton-proton reactions

nT = 2x107 K, EG = 290kTnT = 2x107 K, EG = 290kT

Fusion Reaction RatesFusion Reaction Rates

–Note there is a range of energies in whichfusion rates peak–Note there is a range of energies in whichfusion rates peak

–impact energy for peak rate, E0,–impact energy for peak rate, E0,

Width given by:(Via a Taylor expansion)

Fusion Reaction RatesFusion Reaction Rates

–For the proton-proton reaction shownearlier,E0 = 4.2kT = 7.2 keVE = 4.8kT = 8.2 keV–For the proton-proton reaction shownearlier,E0 = 4.2kT = 7.2 keVE = 4.8kT = 8.2 keV

Fusion Reaction RatesFusion Reaction Rates

–The total reaction rate is found from theintegral shown earlier.–The total reaction rate is found from theintegral shown earlier.

This gives:

Fusion Reaction RatesFusion Reaction Rates

–Fusion reaction rates are strongly temperaturedependent–Fusion reaction rates are strongly temperaturedependent

ne.g, p-d reaction, EG = 0.657 MeV, around 2x107 Kne.g, p-d reaction, EG = 0.657 MeV, around 2x107 K

This implies the rate varies as T 4.6

Fusion in Stars IFusion in Stars I

nHydrogen fusion mechanisms in mainsequence starsnHydrogen fusion mechanisms in mainsequence stars

–Proton-Proton Chain–Proton-Proton Chain

–CNO Cycle–CNO Cycle

–Relative rates and temperaturedependencies–Relative rates and temperaturedependencies

Hydrogen BurningHydrogen Burning

nIn a main sequence star, the principalsource of power is fusion of protonsinto helium nucleinIn a main sequence star, the principalsource of power is fusion of protonsinto helium nuclei

–4p  4He + 2e+ + 2e–4p  4He + 2e+ + 2e

–Relies on weak nuclear force to mediatereaction:p  n + e+ + e–Relies on weak nuclear force to mediatereaction:p  n + e+ + e

–Total energy release (including annihlationof positrons) 26.73 MeV–Total energy release (including annihlationof positrons) 26.73 MeV

Hydrogen BurningHydrogen Burning

nFour particle reaction unlikely, hence mightexpect a three-step process:nFour particle reaction unlikely, hence mightexpect a three-step process:

–2p  2He +–2p  2He +

–2He  d + e+ + e–2He  d + e+ + e

–2d  4He +–2d  4He +

nProblem:nProblem:

–No bound state of 2He–No bound state of 2He

–Hence, it looks like hydrogen burning is slow–Hence, it looks like hydrogen burning is slow

Proton-Proton ChainProton-Proton Chain

nA possibility:nA possibility:

–Fuse protons via the weak nuclear force togive deuterium–Fuse protons via the weak nuclear force togive deuterium

–p  n + e+ + e–p  n + e+ + e

nRequires 1.8 MeVnRequires 1.8 MeV

–p + n  d–p + n  d

nReleases 2.2 MeVnReleases 2.2 MeV

–Net Result: p + p  d + e+ + e–Net Result: p + p  d + e+ + e

Proton-Proton ChainProton-Proton Chain

–Recall the rate of a reaction is given by:–Recall the rate of a reaction is given by:

This integral gives (see Phillips secn 4.1)

The symbols have their previous meaning, A = reducedmass in auS(E) is in keV barns

Proton-Proton ChainProton-Proton Chain

–The reaction p + p  d + e+ + eis mediated by the weak force and is henceslow–The reaction p + p  d + e+ + eis mediated by the weak force and is henceslow

–S ~ 4x10-22 keV barns (calculated - toosmall to measure!)–S ~ 4x10-22 keV barns (calculated - toosmall to measure!)

–What is the rate of proton-proton fusion inthe core of the sun?–What is the rate of proton-proton fusion inthe core of the sun?

Proton-Proton ChainProton-Proton Chain

nAssume a typical model of the sun’s core:nAssume a typical model of the sun’s core:

–T = 15x106 K–T = 15x106 K

–  = 105 kgm-3–  = 105 kgm-3

–Proton fraction = 50%–Proton fraction = 50%

–hence proton density np =3x1031 m-3–hence proton density np =3x1031 m-3

–Also S ~ 4x10-22 keV barns and EG = 494 keV–Also S ~ 4x10-22 keV barns and EG = 494 keV

nThese numbers give a rate of:Rpp = 5x1013 m-3s-1nThese numbers give a rate of:Rpp = 5x1013 m-3s-1

Proton-Proton ChainProton-Proton Chain

nMean lifetime of a proton in the sun’score:nMean lifetime of a proton in the sun’score:

–Rate of any two protons fusingf = Rpp / (1/2 np)~3.3x10-18 s-1–Rate of any two protons fusingf = Rpp / (1/2 np)~3.3x10-18 s-1

–Hence, mean time for a proton pair to fuseis: = 1/f = 3x1017 s ~ 9x109 years–Hence, mean time for a proton pair to fuseis: = 1/f = 3x1017 s ~ 9x109 years

–Slow proton-proton rate sets timescale forstellar lifetimes–Slow proton-proton rate sets timescale forstellar lifetimes

Proton-Proton ChainProton-Proton Chain

nFollowing p-p fusion, further reactionsto produce 4He are rapidnFollowing p-p fusion, further reactionsto produce 4He are rapid

nThe proton-proton chain can followthree branches (pathways):nThe proton-proton chain can followthree branches (pathways):

Proton-Proton ChainProton-Proton Chain

p + p  d + e+ + e

p + d  3He +

2 3He  4He + 2p

Qeff = 26.2 MeV

85%

3He + 4He  7Be + 

e- + 7Be  7Li + e

p + 7Li  4He + 4He

Qeff = 25.7 MeV

15%

p + 7Be  8B + 

8B  8Be + e+ + e

8Be  4He + 4He

Qeff = 19.1 MeV

0.02%

Proton-Proton ChainProton-Proton Chain

nAverage energy release per p-p fusion:nAverage energy release per p-p fusion:

–Take into account:–Take into account:

–two p-p fusions per branch–two p-p fusions per branch

–weightings of each branch–weightings of each branch

–15 MeV per p-p fusion–15 MeV per p-p fusion

–Given number of fusions per m-3 calculatedearlier, energy production rate ~ 120 Wm-3–Given number of fusions per m-3 calculatedearlier, energy production rate ~ 120 Wm-3

The CNO CycleThe CNO Cycle

nThe proton-proton chain has atemperature dependence of ~ T 4nThe proton-proton chain has atemperature dependence of ~ T 4

nInternal temperatures of more massivestars are only moderatly highernInternal temperatures of more massivestars are only moderatly higher

nLuminosities much greater than can beexplained by the T 4 dependencenLuminosities much greater than can beexplained by the T 4 dependence

The CNO CycleThe CNO Cycle

nImplications:nImplications:

–Another mechanism must be at work–Another mechanism must be at work

–This mechanism must have a higher ordertemperature dependence–This mechanism must have a higher ordertemperature dependence

–Implies a higher Coulomb barrier–Implies a higher Coulomb barrier

Recall power of dependence  EG1/3and EG  (ZAZb)2Recall power of dependence  EG1/3and EG  (ZAZb)2

–Such a mechanism is the CNO cycle–Such a mechanism is the CNO cycle

The CNO CycleThe CNO Cycle

–Reaction Pathway:–Reaction Pathway:

p + 12C  13N + 

13N  13C + e+ + e

p + 13C  14N + 

p + 14N  15O + 

p + 15N  12C + 4He

Qeff = 23.8 MeV

15O  15N + e+ + e

S = 1.5 keV barnsEG = 32.8 MeV

S = 5.5 keV barnsEG = 33 MeV

S = 3.3 keV barnsEG = 45.2 MeV

S = 78 keV barnsEG = 45.4 MeV

The CNO CycleThe CNO Cycle

nNotice that:nNotice that:

–12C acts as a catalyst–12C acts as a catalyst

–Rate governed by slowest step–Rate governed by slowest step

nin p-p, the first p-p fusion stepnin p-p, the first p-p fusion step

nIn CNO, if one considers all parameters, thep-14N step is slowestnIn CNO, if one considers all parameters, thep-14N step is slowest

–Abundance of 14N~0.6%–Abundance of 14N~0.6%

The CNO CycleThe CNO Cycle

–Rate of p-14N fusion in the sun–Rate of p-14N fusion in the sun

–Abundance of 14N~0.6%–Abundance of 14N~0.6%

ngives 2.6x1028 14N m-3ngives 2.6x1028 14N m-3

–S = 3.3 keV barns, EG = 45.2 MeV–S = 3.3 keV barns, EG = 45.2 MeV

–Other parameters same as for p-p fusion:–Other parameters same as for p-p fusion:

–RpN = 1.6x1012 s-1m-3–RpN = 1.6x1012 s-1m-3

–CNO cycle contributes at most a few % to thepower of a sun-like star–CNO cycle contributes at most a few % to thepower of a sun-like star

–Mean lifetime of a 14N nucleus in the sun ~ 5x108 yrs–Mean lifetime of a 14N nucleus in the sun ~ 5x108 yrs

The CNO CycleThe CNO Cycle

nThe CNO cycle is strongly temperaturedependentnThe CNO cycle is strongly temperaturedependent

–Using ideas from previous lecture, at thetemperature of a sun-like star, andconsidering the p-14N step,RpN T 20–Using ideas from previous lecture, at thetemperature of a sun-like star, andconsidering the p-14N step,RpN T 20

–We can compare the rate of p-p vs. CNOas a function of temperature–We can compare the rate of p-p vs. CNOas a function of temperature

CNO vs. p-pCNO vs. p-p

T 15.63

T 2.96

sun

CNO vs. p-pCNO vs. p-p

nWe can see that:nWe can see that:

–CNO contributes a few % of the sun’soutput–CNO contributes a few % of the sun’soutput

–In a moderately hotter stellar core(~1.8x107 K) CNO ~ p-p–In a moderately hotter stellar core(~1.8x107 K) CNO ~ p-p

–In hot (>2x107) cores, CNO > p-p–In hot (>2x107) cores, CNO > p-p

A requirement for CNOA requirement for CNO

nThe CNO cycle requires the heavyelements C, N and O.nThe CNO cycle requires the heavyelements C, N and O.

–These elements have negligible abundancefrom the Big Bang–These elements have negligible abundancefrom the Big Bang

–They are not formed in the p-p chain–They are not formed in the p-p chain

–What is the origin of heavy elements?–What is the origin of heavy elements?

–See Next Lecture - Fusion in Stars II–See Next Lecture - Fusion in Stars II

Next WeekNext Week

nHeavy element productionnHeavy element production

nStar formationnStar formation