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H and He Burning

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  1. H and He Burning Prof John Lattanzio Centre for Stellar and Planetary Astrophysics School of Mathematical Sciences

  2. 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

  3. 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!

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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!

  9. 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

  10. Overall PPI Chain 1H + 1H  2D + b+ + ne 1H + 2D  3He + g 3He + 3He  4He + 2p

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. spp ( ) np ( )eqm 2spd nd Deuterium in equilibrium d nd = rpp – rpd = 0 dt np2spp/2 = npndspd =

  19. 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

  20. 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!

  21. 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

  22. More on PPI: 2D DE for deuterium

  23. 2D equation Now definet

  24. 2D equation Can almost always assume the 2D is in equilibrium tp(D) ~ secs or mins! Can remove 2D from the equations

  25. H burns into He4 in ~1011 years D2 destroyed in 10-5y Or 300 secs Or 5 mins 2D equilibrium: T = 10 million K

  26. H burns into He4 in ~ 1010 years 2D equilibrium: T = 20 million K D2 destroyed in 10-6.5y Or 10 secs

  27. (3He/H)e ~ 10-2 to 10-10 low T high T 3He equation Assume 2D in equilibrium: rpp = rpd

  28. 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

  29. 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

  30. Exercise

  31. PP Chains at various T • As T rises: • H burns more quickly • D2 eqm happens earlier • He3 eqm happens earlier • He3 eqm value decreases

  32. 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

  33. Neutrinos? PP Chains Fate of 3He determines PPI or PPII and III Fate of7Be determines PPII and III

  34. DEs for the PP chains

  35. 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

  36. H burns into He4 in ~1011 years 7Li and 7Be: T = 10 million K Li and Be reach eqm values in ~106 years

  37. H burns into He4 in ~1010 years 7Li and 7Be: T = 20 million K Li and Be reach eqm values in ~104 years

  38. PP Chains PPI PPIII PPII

  39. PP Chains

  40. Chart of the Nuclides: The “big boys/girls” periodic table Z = number of protons N = number of neutrons

  41. PP Chains PPI PPII PPIII

  42. 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

  43. 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

  44. 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

  45. 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

  46. 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

  47. 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

  48. The CN cycle But sometimes we get: 15N(p,g)16O There is a "branching"

  49. ON cycle • Once there is some 16O present… (or 16O may be present initially in any case…)

  50. The CNO bi-cycle