Constraints on Nuclear Fusion Reactions

• Binary reactions dominate• Energy Conservation (Energy release)• Charge conservation (d=2H nucleus)

– pp d+e++Energy +??

• Angular momentum conservation– Proton, electron spin = (1/2)[h/(2π)]– Deuteron spin = h/(2π)

• Neutrino predicted by E.Fermi – pp d+e++ +Energy– Detected in 1950s by Reines and Cowen next to a nuclear

reactors

• Weak Interactions (lepton number conservation)– n p + e+ anti-. Half Life 12,33 year– Weak interactions are Weak Reaction Rates are slow

Light Nuclei, http://www.nndc.bnl.gov/chart/

QuickTime™ and a decompressor

are needed to see this picture.

No stable A=5 nuclei

pp chains in Sun

€

pp→D+ e+ +ν e

pD→3He+γ

€

3He+3He→4He+H +H

€

3He+4He→7Be+γ

€

7Be+ e−→7Li+ν e7Li+ p→7Be→4He+4He

€

7Be+ p→8B+γ8B→8Be+ e+ +ν e8Be→4He+4He

69%

(pp I)

(pp II) (pp III)

31%

99.7%0.3%

pp chainsH+H+H+H4He+2e+2e++2

First Step• pp D + e+ + + MeV

– Weak interaction process• D=(pn) = nucleus of 2H.• Neutron half life = 10.23 min

– Suppressed by pp Coulomb repulsion• Quantum Tunneling required• Protons are waves

– If Kinetic energy = E-U > 0= h/p

– If Kinetic energy = E-U < 0– Wave is exponentially damped exp(-r)

= [U-E]1/2.

Billiard Ball model of pp collisions

• Proton density , Temperature T• Velocity distribution (Maxwell-Boltzman)

– P(v;T) = exp(-mv2/(2KT)) / [2πkT]3/2.

– <mv2/2>= <mvx2/2> +<mvy

2/2> +<mvz2/2> =3kT/2

– vrms ≈ [kT/m]1/2.

• Collision rate per unit volume: R. ≈ (m)2.– Rate R ≈ 2v|≈ 2 [kT/m]1/2.Volume = v t

Ballistic Burn Time of Sun

• Number density of sun = /MH.– Average = 8 1029 /m3.– Central = 1032 /m3.

• Temperature, rms Velocity vrms ≈ [kT/mc2]1/2

– Surface kT = 0.5 eV. vrms ≈ 7000 m/s– Central kT = 1KeV. vrms ≈ 3 105 m/s

• Proton Mass = 0.9 GeV• Time for all protons to collide / R = /[ |v|]

– Surface 1/[|v|] ≈ 10-4 s– Center ≈ 30 ns

• Crucial role of Coulomb Repulsion and Weak interactions to regulate nuclear fusion rate in sun

pp fusion slowed by coulomb repulsion

• Kinetic energy required to bring two protons to within 1 fm– e2/(4π0 r) = 1.4 MeV– Center of sun kT ≈ 1 KeV– Maxwell Boltzmann distribution

• P(v) = exp{-[mv2/(2kT)]}• P[-(1.4MeV)/(1 KeV)] ≈ exp(-1000)

• Classical pp Collision rate in sun ≈ 0– No pp fusion, even worse for heavier elements

• Quantum barrier penetration