ReminderReminder

Please return Assignment 1 to the Please return Assignment 1 to the School Office by 13:00 Weds. 11School Office by 13:00 Weds. 11thth February (tomorrow!)February (tomorrow!)– The assignment questions will be The assignment questions will be

reviewed in next week’s workshopreviewed in next week’s workshop

Stellar Power SourcesStellar Power Sources

PossibilitiesPossibilities– Chemical burningChemical burning

only a few thousand yearsonly a few thousand years

– Gravitational contractionGravitational contraction a few million yearsa few million years

– Nuclear fusionNuclear fusion ~10~101010 years for a sun-like star years for a sun-like star

Physics of FusionPhysics of Fusion

The Nuclear PotentialThe Nuclear Potential– Repulsive at large distances due to CoulombRepulsive at large distances due to Coulomb– Attractive at short range due to Strong Attractive at short range due to Strong

Nuclear ForceNuclear Force (1fermi = 10(1fermi = 10-15-15 m) m)

0 1 2 3 4

r (fermis)

Physics of FusionPhysics of Fusion

– Classically, for nuclei to fuse, a barrier Classically, for nuclei to fuse, a barrier of height of height EEcc has to be overcome:has to be overcome:

EZ Zrc

A B

N

144.

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

Physics of FusionPhysics of Fusion

– Typical stellar models give typical core Typical stellar models give typical core temperatures of ~ 10temperatures of ~ 1077 K (kT ~ 1keV) K (kT ~ 1keV)

– Fraction of nuclei with sufficient energy Fraction of nuclei with sufficient energy to overcome to overcome EECC given by Boltzmann:given by Boltzmann:

f EC

exp .1441 10 103

625

Classically, fusion shouldn’t happen!

Physics of FusionPhysics of Fusion

Solution:Solution:– Quantum MechanicsQuantum Mechanics– Protons described by Schroedinger Protons described by Schroedinger

EquationEquation– Quantum mechanical tunnelling Quantum mechanical tunnelling

possiblepossible

Physics of FusionPhysics of Fusion

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

P r rc N exp ( )2

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

2 2

2 2

2

r r

Em

Er

c

and

Physics of FusionPhysics of Fusion

– The probability of barrier penetration The probability of barrier penetration can be recast in terms of energy:can be recast in terms of energy:

PE

EG

exp12

Where EG is the Gamow energy define by:

E Z Z m cG A B r 2 22

= fine structure constant (~1/137)mr = reduced mass of nuclei

Physics of FusionPhysics of Fusion

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

– In a typical stellar core, In a typical stellar core, kTkT ~ 1 keV ~ 1 keV– Hence the probability of barrier Hence the probability of barrier

penetration is:penetration is:

P

exp exp .493

1 22 28 1012 10

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

Fusion Cross SectionsFusion Cross Sections Consider a proton moving in a medium with Consider a proton moving in a medium with nn protons per unit volume protons per unit volume

– Probability of fusion occuring within a distance Probability of fusion occuring within a distance xx = = nnxx = reaction cross section= reaction cross section

– Mean distance between collisions = mean free path, Mean distance between collisions = mean free path, ll = 1/ = 1/nn– Mean time betweenMean time betweencollisions, collisions, = = ll/</<v>v> = 1/ = 1/ nn<<vv> > note note may depend on may depend on vv

Fusion Cross SectionsFusion Cross Sections

The fusion cross section (Units, barns = 10The fusion cross section (Units, barns = 10--

2828 m m22) as a function of energy is given by:) as a function of energy is given by:

E

S EE

EE

G

exp12

S(E) is a slowly varying function determined by the nuclear physics of the reaction1/E introduced to account for low energy behaviour

Fusion Reaction RatesFusion Reaction Rates Consider the reaction rate between two nuclei, A and Consider the reaction rate between two nuclei, A and

B, travelling with relative speed, B, travelling with relative speed, vv, with , with concentrations concentrations nnAA and and nnBB, with cross section , with cross section – Mean time for an A nucleus to fuse with a B is:Mean time for an A nucleus to fuse with a B is:

AA = 1/ = 1/ nnBB<<vv> > – Hence, the total fusion rate per unit volume is:Hence, the total fusion rate per unit volume is:

RRABAB = = nnAAnnB B <<vv> >

Fusion Reaction RatesFusion Reaction Rates

– To obtain <To obtain <vv>, we note that: >, we note that:

v v P v dvr r r

0Where P(vr) is the Maxwell-Boltzmann distribution given by:

P v dv T EkT dEr r 3

2 exp

Fusion Reaction RatesFusion Reaction Rates

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

R n nm kT

S EEkT

EE

dEAB A Br

G

8 1

12 3

212

0( )exp

Concentrating on the integral, we note that for a given impact energy, E, there is a competition between the Boltzmann term and the Gamow energy term

Fusion Reaction RatesFusion Reaction Rates

1 2 3 4 5 6 7 8 9 10

Impact Energy (units of kT)

exp(-E/kT)

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

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

– Proton-proton reactions Proton-proton reactions TT = 2x10 = 2x1077 K, K, EEGG = 290k= 290kTT

Fusion Reaction RatesFusion Reaction Rates

– Note there is a range of energies in Note there is a range of energies in which fusion rates peakwhich fusion rates peak

– impact energy for peak rate, impact energy for peak rate, EE0,0,

EE kTG

0

213

4

( )

Width given by:(Via a Taylor expansion)

E E kTG

4

3 212

13

16

56( )

Fusion Reaction RatesFusion Reaction Rates

– For the proton-proton reaction shown For the proton-proton reaction shown earlier,earlier,EE00 = 4.2k = 4.2kTT = 7.2 keV = 7.2 keVEE = 4.8k = 4.8kTT = 8.2 keV = 8.2 keV

Fusion Reaction RatesFusion Reaction Rates

– The total reaction rate is found from The total reaction rate is found from the integral shown earlier.the integral shown earlier.

R n nm kT

S EEkT

EE

dEAB A Br

G

8 1

12 3

212

0( )exp

This gives:

R n n S EEkTAB A B

G

( )exp 34

13

Fusion Reaction RatesFusion Reaction Rates

– Fusion reaction rates are strongly Fusion reaction rates are strongly temperature dependenttemperature dependent e.g, p-d reaction, Ee.g, p-d reaction, EGG = 0.657 MeV, around 2x10 = 0.657 MeV, around 2x1077 K K

dR

dTEkT

R

T

R

Tpd G pd pd

446

13

.

This implies the rate varies as T 4.6

Fusion in Stars IFusion in Stars I

Hydrogen fusion mechanisms in Hydrogen fusion mechanisms in main sequence starsmain sequence stars– Proton-Proton ChainProton-Proton Chain– CNO CycleCNO Cycle– Relative rates and temperature Relative rates and temperature

dependenciesdependencies

Hydrogen BurningHydrogen Burning

In a main sequence star, the In a main sequence star, the principal source of power is fusion principal source of power is fusion of protons into helium nucleiof protons into helium nuclei– 4p 4p 44He + 2eHe + 2e++ + 2 + 2ee

– Relies on weak nuclear force to Relies on weak nuclear force to mediate reaction:mediate reaction:p p n + e n + e++ + + ee

– Total energy release (including Total energy release (including annihlation of positrons) 26.73 MeVannihlation of positrons) 26.73 MeV

Hydrogen BurningHydrogen Burning

Four particle reaction unlikely, hence Four particle reaction unlikely, hence might expect a three-step process:might expect a three-step process:– 2p 2p 22He +He +– 22He He d + e d + e++ + + ee

– 2d 2d 44He +He + Problem:Problem:

– No bound state ofNo bound state of 22HeHe

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

Proton-Proton ChainProton-Proton Chain

A possibility:A possibility:– Fuse protons via the weak nuclear Fuse protons via the weak nuclear

force to give deuteriumforce to give deuterium– p p n + e n + e++ + + ee

Requires 1.8 MeVRequires 1.8 MeV

– p + n p + n d d Releases 2.2 MeVReleases 2.2 MeV

– Net Result:Net Result: p + p p + p d + e d + e++ + + ee

Proton-Proton ChainProton-Proton Chain

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

R n nm kT

S EEkT

EE

dEAB A Br

G

8 1

12 3

212

0( )exp

This integral gives (see Phillips secn 4.1)

Rn n

AZ ZS E

EkT

EkTAB

A B

A B

G G

648 104

34

240

23

13

1. ( ) exp m s-3

The symbols have their previous meaning, A = reduced mass in au S(E) is in keV barns

Proton-Proton ChainProton-Proton Chain

– The reactionThe reaction p + p p + p d + e d + e++ + + ee

is mediated by the weak force and is is mediated by the weak force and is hence slowhence slow

– SS ~ 4x10 ~ 4x10-22-22 keV barns (calculated - keV barns (calculated - too small to measure!)too small to measure!)

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

Proton-Proton ChainProton-Proton Chain

Assume a typical model of the sun’s Assume a typical model of the sun’s core:core:– TT = 15x10 = 15x1066 K K– = 10= 1055 kgm kgm-3-3

– Proton fraction = 50%Proton fraction = 50%– hence proton density hence proton density nnpp =3x10=3x103131 m m-3-3

– Also S ~ 4x10Also S ~ 4x10-22-22 keV barns and keV barns and EEGG = 494 keV = 494 keV These numbers give a rate of:These numbers give a rate of:

RRpppp == 5x105x1013 13 mm-3-3ss-1-1

Proton-Proton ChainProton-Proton Chain

Mean lifetime of a proton in the Mean lifetime of a proton in the sun’s core:sun’s core:– Rate of any two protons fusingRate of any two protons fusing

ff = = RRpppp / (1/2 / (1/2 nnpp)~3.3x10)~3.3x10-18-18 s s-1-1

– Hence, mean time for a proton pair to Hence, mean time for a proton pair to fuse is:fuse is: = 1/ = 1/ff = 3x10 = 3x101717 s ~ 9x10 s ~ 9x1099 years years

– Slow proton-proton rate sets Slow proton-proton rate sets timescale for stellar lifetimestimescale for stellar lifetimes

Proton-Proton ChainProton-Proton Chain

Following p-p fusion, further Following p-p fusion, further reactions to produce reactions to produce 44He are rapidHe are rapid

The proton-proton chain can follow The proton-proton chain can follow three branches (pathways):three branches (pathways):

Proton-Proton ChainProton-Proton Chainp + p d + e+ + e

p + d 3He +

2 3He 4He + 2p

Qeff = 26.2 MeV85%

3He + 4He 7Be +

e- + 7Be 7Li + e

p + 7Li 4He + 4He

Qeff = 25.7 MeV15%

p + 7Be 8B +

8B 8Be + e+ + e

8Be 4He + 4He

Qeff = 19.1 MeV0.02%

Proton-Proton ChainProton-Proton Chain

Average energy release per p-p Average energy release per p-p fusion:fusion:– Take into account:Take into account:– two p-p fusions per branchtwo p-p fusions per branch– weightings of each branchweightings of each branch– 15 MeV per p-p fusion15 MeV per p-p fusion– Given number of fusions per mGiven number of fusions per m-3-3

calculated earlier, energy production calculated earlier, energy production rate ~ 120 Wmrate ~ 120 Wm-3-3

The CNO CycleThe CNO Cycle

The proton-proton chain has a The proton-proton chain has a temperature dependence of ~ temperature dependence of ~ T T 44

Internal temperatures of more Internal temperatures of more massive stars are only moderatly massive stars are only moderatly higherhigher

Luminosities much greater than Luminosities much greater than can be explained by the can be explained by the T T 44 dependencedependence

The CNO CycleThe CNO Cycle

Implications:Implications:– Another mechanism must be at workAnother mechanism must be at work– This mechanism must have a higher This mechanism must have a higher

order temperature dependenceorder temperature dependence– Implies a higher Coulomb barrierImplies a higher Coulomb barrier

Recall power of dependence Recall power of dependence EEGG1/31/3

and and EEGG ( (ZZAAZZbb))22

– Such a mechanism is the CNO cycleSuch 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 MeVS = 5.5 keV barnsEG = 33 MeVS = 3.3 keV barnsEG = 45.2 MeVS = 78 keV barns EG = 45.4 MeV

The CNO CycleThe CNO Cycle

Notice that:Notice that:– 1212C acts as a catalystC acts as a catalyst– Rate governed by slowest stepRate governed by slowest step

in p-p, the first p-p fusion stepin p-p, the first p-p fusion step In CNO, if one considers all parameters, In CNO, if one considers all parameters,

the the p-p-1414N step is slowestN step is slowest

– Abundance of Abundance of 1414N~0.6%N~0.6%

The CNO CycleThe CNO Cycle

– Rate of p-Rate of p-1414N fusion in the sunN fusion in the sun– Abundance of Abundance of 1414N~0.6%N~0.6%

gives 2.6x10gives 2.6x102828 1414N mN m-3-3

– S = 3.3 keV barns, S = 3.3 keV barns, EEGG = 45.2 MeV = 45.2 MeV– Other parameters same as for p-p fusion:Other parameters same as for p-p fusion:– RRpNpN = 1.6x10 = 1.6x101212 s s-1-1mm-3-3

– CNO cycle contributes at most a few % to CNO cycle contributes at most a few % to the power of a sun-like starthe power of a sun-like star

– Mean lifetime of a Mean lifetime of a 1414N nucleus in the sunN nucleus in the sun ~ 5x10 ~ 5x1088 yrs yrs

The CNO CycleThe CNO Cycle

The CNO cycle is strongly The CNO cycle is strongly temperature dependenttemperature dependent– Using ideas from previous lecture, at Using ideas from previous lecture, at

the temperature of a sun-like star, the temperature of a sun-like star, and considering the p-and considering the p-1414N step, N step, RRpN pN T T 2020

– We can compare the rate of p-p vs. We can compare the rate of p-p vs. CNO as a function of temperatureCNO as a function of temperature

CNO vs. p-p CNO vs. p-p

1.E+08

1.E+10

1.E+12

1.E+14

1.E+16

1.E+18

1.E+20

1.E+22

1.E+24

1.E+26

1.E+07 1.E+08Temperature (K)

Fusi

on rate

(s-1

m-3)

p-pCNO

T 15.63

T 15.63

T 2.96T 2.96

sun

CNO vs. p-pCNO vs. p-p

We can see that:We can see that:– CNO contributes a few % of the sun’s CNO contributes a few % of the sun’s

outputoutput– In a moderately hotter stellar core In a moderately hotter stellar core

(~1.8x10(~1.8x1077 K) CNO ~ p-p K) CNO ~ p-p– In hot (>2x10In hot (>2x1077) cores, CNO > p-p) cores, CNO > p-p

A requirement for CNOA requirement for CNO

The CNO cycle requires the heavy The CNO cycle requires the heavy elements C, N and O.elements C, N and O.– These elements have negligible These elements have negligible

abundance from the Big Bangabundance from the Big Bang– They are not formed in the p-p chainThey 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 IISee Next Lecture - Fusion in Stars II

Next WeekNext Week

Heavy element productionHeavy element production Star formationStar formation