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## H and He Burning

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**H and He Burning**Prof John Lattanzio Centre for Stellar and Planetary Astrophysics School of Mathematical Sciences**Reaction Rates**• Lets look at reactions involving two different kinds of particles, i and j (or and ) • Suppose there are two of each in a box There are thus four possible reaction pairs**Rate formula**• In general each i can react with all the other j particles so the number of reaction pairs is ninj • For identical particles this becomes ninj-1 except that it counts each pair twice; so the number of pairs is really ni(ni-1)/2 in this case • Now in general ni is ~ NA ~ 1023 • So ni-1 = ni to high accuracy!**rij = ni nj sij**1 + dij New rate formula • So for identical particles the number of pairs is ni2/2to very high accuracy • But for different particles its ninj • We can combine these into one formula by using the Kronecker delta dij • dij = 1 if i = j • dij = 0 otherwise**H Burning**• There are two main ways of burning H • Proton-Proton chains (pp) • CNO cycle(s) • PP chains are the simplest • We deal with those first**The Proton-Proton Reaction**One proton becomes a neutron with a positron for charge conservation and a neutrino for energy conservation • The first step is adding two protons Positron or anti-electron Neutrino (electron) 1H + 1H 2D + b+ + ne Deuterium = heavy H Z=1 N=1 Nucleus = d = deuteron Hydrogen Z=1 N=0 Nucleus = p = proton The key is really: p n + b+ + ne**Next Step: H + D**• The next step is adding H and D Gamma ray (photon) 1H + 2D 3He + g Helium-3 Z=2 N=1 Hydrogen Z=1 N=0 Nucleus = p = proton Deuterium = heavy H Z=1 N=1 Nucleus = d = deuteron No changes of particles here…2p + 1n on each side**Last Step: 3He + H?**• You may think that the next step is to ad a proton to 3He to get 4He, the most common isotope of helium • But it turns out that the nuclear structure is such that this is highly unlikley! You cannot tell this from pure thought!**Last Step: 3He + 3He**• The last step is adding two 3He together Returning 2 protons 3He + 3He 4He + 2p Helium-3 Z=2 N=1 Helium-4 Z=2 N=2 No changes of particles here…4p + 2n on each side**Overall PPI Chain**1H + 1H 2D + b+ + ne 1H + 2D 3He + g 3He + 3He 4He + 2p**rij = ni nj sij**1 + dij What are the rates? 1H + 1H 2D + b+ + ne rpp = np2spp/2 1H + 2D 3He + g rpd =npndspd 3He + 3He 4He + 2p r33 = n32s33/2**What are the DEs for the abundances?**1H + 1H 2D + b+ + ne rpp = np2spp/2 1H + 2D 3He + g rpd =npndspd 3He + 3He 4He + 2p r33 = n32s33/2 d np = production rate – destruction rate dt = 2 r33 – 2 rpp – rpd**For Deuterium**1H + 1H 2D + b+ + ne rpp = np2spp/2 1H + 2D 3He + g rpd =npndspd 3He + 3He 4He + 2p r33 = n32s33/2 d nd = production rate – destruction rate dt = rpp – rpd**For 3He**1H + 1H 2D + b+ + ne rpp = np2spp/2 1H + 2D 3He + g rpd =npndspd 3He + 3He 4He + 2p r33 = n32s33/2 d n3 = production rate – destruction rate dt = rpd – 2 r33**For 4He**1H + 1H 2D + b+ + ne rpp = np2spp/2 1H + 2D 3He + g rpd =npndspd 3He + 3He 4He + 2p r33 = n32s33/2 d n4 = production rate – destruction rate dt = r33**Summary of rates and DEs**1H + 1H 2D + b+ + ne rpp = np2spp/2 1H + 2D 3He + g rpd =npndspd 3He + 3He 4He + 2p r33 = n32s33/2 d np d nd = rpp – rpd = 2 r33 – 2 rpp – rpd dt dt d n3 d n4 = rpd – 2 r33 = r33 dt dt**Deuterium in equilibrium**• Note that the 2D equation is self-correcting • If nd is low, the first term dominates • nd increases • If nd is high the second term dominates • nd decreases d nd = rpp – rpd dt = np2spp/2 – npndspd An equilibrium exists when RHS=0**spp**( ) np ( )eqm 2spd nd Deuterium in equilibrium d nd = rpp – rpd = 0 dt np2spp/2 = npndspd =**3He in equilibrium**• Note that the 3He equation is also self-correcting • If n3 is low, the first term dominates • n3 increases • If n3 is high the second term dominates • n3 decreases d n3 = rpd – 2r33 dt = npndspd –n32s33 An equilibrium exists when RHS=0**d np**dt PPI in equilibrium • Suppose the deuterium and 3He are in eqm d np d nd = rpp – rpd=0 = 2 r33 – 2 rpp – rpd dt dt d n3 d n4 = rpd – 2 r33=0 = r33 dt dt = rpd/2 = rpp/2 = - ¼ He-4 made at one quarter the rate of H destruction!**PPI Energy Generation**• One can show (but not quickly!) that the energy generation rate for PPI is • e ~e0 r X2 T64 • where T6 = T / 106K This is PPI only... There is also PPII and PPIII**More on PPI: 2D**DE for deuterium**2D equation**Now definet**2D equation**Can almost always assume the 2D is in equilibrium tp(D) ~ secs or mins! Can remove 2D from the equations**H burns into He4**in ~1011 years D2 destroyed in 10-5y Or 300 secs Or 5 mins 2D equilibrium: T = 10 million K**H burns into He4**in ~ 1010 years 2D equilibrium: T = 20 million K D2 destroyed in 10-6.5y Or 10 secs**(3He/H)e ~ 10-2 to 10-10**low T high T 3He equation Assume 2D in equilibrium: rpp = rpd**H burns into He4**in ~1011 years 3He equilibrium: T = 10 million K He3 increases due to D2 destruction Reaches eqm later 3Heeqm ~ 10-3.5**H burns into He4**in ~1010 years 3He equilibrium: T = 20 million K He3 increases due to D2 destruction Reaches eqm later 3Heeqm ~ 10-5.2**PP Chains at various T**• As T rises: • H burns more quickly • D2 eqm happens earlier • He3 eqm happens earlier • He3 eqm value decreases**PP Chains (I, II and III)**• PPI Chain • But once there is some 4He… • And 7Be can • Capture a proton • Or capture an electron • Makes PPII and PPIII chains**Neutrinos?**PP Chains Fate of 3He determines PPI or PPII and III Fate of7Be determines PPII and III**Some simplifications!**2D in equilibrium (seconds) 7Be and 7Li in equilibrium in a few years – few thousand years Further details This enables some simplifications At low T we get PPI At higher T we get PPII and PPIII Equations for PP Chains**H burns into He4**in ~1011 years 7Li and 7Be: T = 10 million K Li and Be reach eqm values in ~106 years**H burns into He4**in ~1010 years 7Li and 7Be: T = 20 million K Li and Be reach eqm values in ~104 years**PP Chains**PPI PPIII PPII**Chart of the Nuclides:**The “big boys/girls” periodic table Z = number of protons N = number of neutrons**PP Chains**PPI PPII PPIII**the heavy species**X(a,b)Y X(a,b)Y lighter particles Notation • Time to become real nuclear physicists! • We have been writing a+X Y+b • From now on we will write X(a,b)Y**Examples**• 2D +p 3He + g becomes 2D(p,g)3He • 3He + 3He 4He + 2 1H becomes 3He(3He,2p)4He • 1H + 1H 2D + g becomes 1H(p,g)2D • 3He + 4He 7Be + g becomes 3He(a,g)7Be • 12C +p 13N + g becomes 12C(p,g)13N • 12C +4He 16O + g becomes 12C(a,g)16O**Decays**• Decays are different • 8B 8Be + b+ + n becomes 8B(b+n)8Be • NB No comma! • 13N 13C + b+ + n becomes 13N(b+n)13C**CN Cycle**• At slightly higher temperatures the dominant H burning reactions are the CN cycle 12C + p 13N + g 13N 13C + b+ + ne 13C + p 14N + g 14N + p 15O + g 15O 15N + b+ + ne 15N + p 12C + a**CN Cycle**• Add up both sides: 12C + p 13N + g 13N 13C + b+ + ne 13C + p 14N + g 14N + p 15O + g 15O 15N + b+ + ne 4H into He-4 15N + p 12C + a 4p a + energy**CN Cycle**• Temperature sensitivity? • It starts at about 15-20 million K • PP starts at 5-10 million K • Approximate rate formula • e ~e0 r XH XCNT620 • e ~e0 r X2 T64 for PP**The CN cycle**But sometimes we get: 15N(p,g)16O There is a "branching"**ON cycle**• Once there is some 16O present… (or 16O may be present initially in any case…)