NucleosynthesisUniversity of Victoria

Astr 501Matt Penrice

OutlineBasics of nucleosynthesis

Nucleosynthesis in stars

Nuclesynthesis on earth

Conclusion

Energy ProductionStars are powered by nuclear fusion

Energy conservation

Endothermic vs Exothermic

€

Qn = M n,1 +M n,2 −M n,3 −M n,4( )c 2

Cross SectionClassically the cross section depends on the

geometric area of the nuclei

Quantum mechanically the cross section depends on the de Broglie wavelength

€

σ =π Rp + Rt( )2

€

σ =πλ 2

€

λ =mp + mt

mt

h

2mp E l( )2

Reaction RateNuclear reaction rates depend on the number

density of projectile and target particles as well as the velocity and cross section

Nuclei in the star will have a velocity distribution therefore

Taking the velocity distribution and identical particles into account we arrive at

€

r = N tN pvσ v( )

€

σv = φ v( )vσ v( )dv0

∞

∫

€

r = N tN p σv 1+δ tp( )−1

Mean LifetimeWe also need to know the mean lifetime of

nuclei in the star

Therefore we can define the mean lifetime as

€

dNx

dt

⎛

⎝ ⎜

⎞

⎠ ⎟= −

1

τ y X( )Nx

€

τy X( ) =1

Ny σv

Maxwell-Boltzmann Distribution

Assume the stellar gas is in thermodynamic equilibrium and nondegenerate and nonrelativistic

Reaction rate per particle revisited

After some hand waving (RM,CMV,CME)

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φ v( ) = 4πv 2 m

2πkT

⎛

⎝ ⎜

⎞

⎠ ⎟3 / 2

exp −mv 2

2kT

⎛

⎝ ⎜

⎞

⎠ ⎟

€

σv = φ(v t )0

∞

∫0

∞

∫ φ(v p )σ v( )vdv pdv t

€

σv =8

πμ

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 21

kT( )3 / 2 σ E( )E exp −

E

kT

⎛

⎝ ⎜

⎞

⎠ ⎟dE

0

∞

∫

Coulomb BarrierPositively charged nuclei repel each other which

forms an energy barrier

€

Vc =Z1Z2e

2

r

Cauldrons in the Cosmos, Rolfs, Rodney

Classical ProblemClassically for the p+p reaction to occur the

energy of the protons must exceed 550 keV which corresponds to a central stellar temperature of 6.4 GK

This temperature is much higher then what is expected for stellar interiors as well as raising the issue of the fusion causing an explosion rather than a controlled burn

We know that stars burn so how can we resolve this issue?

Q&M To The Rescue

Classically a particle with energy E< Ec cannot penetrate the Coulomb barrier. However quantum mechanically it has a finite possibility to “tunnel” through the Coulomb barrier

€

P =ψ Rn( )

2

ψ Rc( )2

Some NumbersFor a temperature of 0.01 GK if we only consider

the high energy wing of the M-B distribution the probability of overcoming the barrier is P=3*10^-375

For a temperature of 0.01 GK the tunneling probability is P=9*10^-10

S-Factor The cross section for charged particle reactions drops

sharply for energies below the Coulomb barrier

S(E) is known as the astrophysical S-factor which is slowly varying for nonresonant reactions

€

σ E( ) =1

Eexp −2πη( )S E( )

Cauldrons in the Cosmos, Rolfs, Rodney

Gamow PeakReaction rate per particle pair again

€

σv =8

πμ

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 21

kT( )3 / 2 S E( )exp −

E

kT−

b

E1/ 2

⎛

⎝ ⎜

⎞

⎠ ⎟dE

0

∞

∫

€

b = 0.989Z1Z2μ1/ 2 MeV( )

1/ 2

Cauldrons in the Cosmos, Rolfs, Rodney

Narrow ResonanceIf the particles in the entrance channel form an

excited state decay can occur

This decay can only occur for unique energies and is defined as a resonant phenomenon

Breit-Wigner formula (single-level resonance)

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σ E( ) = πλ 22J +1

(2J1 +1)(2J2 +1)(1+δ12)

ΓaΓb

(E − ER )2 + (Γ/2)2

Narrow Resonance IINarrow resonance reaction rate per particle pair

€

σv =2π

μkT

⎛

⎝ ⎜

⎞

⎠ ⎟

3 / 2

h2 ωγ( )Rexp −

ER

kT

⎛

⎝ ⎜

⎞

⎠ ⎟f

€

ωγ =ωΓaΓb

Γ

Nuclesynthesis On EarthHydrogen Bombs

National Ignition Facility

ConclusionNucleosynthesis in stars requires quantum

tunneling to overcome the Coulomb barrier

This allows for controlled nuclear burning giving rise to the long lived low mass stars we see today

Narrow resonance becomes important in lower mass stars where the temperature is lower