Convection Convection Matt Penrice Astronomy 501 University of Victoria.
Nucleosynthesis University of Victoria Astr 501 Matt Penrice.
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Transcript of Nucleosynthesis University of Victoria Astr 501 Matt Penrice.
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)
€
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)
€
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
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