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Nuclear reactions and stellar Luminosity

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We have shown that to provide the observed luminosity of the sun for billions of ... See http://burro.astr.cwru.edu/Academics/Astr221/StarPhys/coulomb.html ... – PowerPoint PPT presentation

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Title: Nuclear reactions and stellar Luminosity


1
Nuclear reactions and stellar Luminosity
  • We have shown that to provide the observed
    luminosity of the sun for billions of years
    requires nuclear fusion reactions (burning).
    However, it is improper to say the nuclear
    reaction rate determines the luminosity. The
    luminosity is determined by the temperature
    structure within the star
  • which we know fairly well from the virial theorem
  • and the rate of flow of energy
  • heat energy flows from hotter region to cooler
    region by
  • conduction
  • convection
  • radiation
  • So the nuclear reactions are important in
    determining
  • The position on the evolutionary track for the
    main sequence phase
  • the lifetime on the main sequence
  • But not the luminosity of the sun at its current
    radius

2
Stellar evolution to main sequence
  • Another way of thinking about this is to imagine
    the evolutionary track of the sun on the H-R
    diagram
  • because of this, we present only brief comments
    on the details of the nuclear reactions. As long
    as there are more tightly bound nuclei, the star
    can contract, thus raising the temperature, until
    the nuclear reaction rate equals the luminosity
    from the star
  • This party continues till we hit Fe (iron)
  • Most tightly bound nucleus
  • End of the line for contraction-induced
    luminosity
  • Leads to supernovae

3
Nuclear reactions
  • Speaking practically, all we need is the nuclear
    reaction energy generation rate e(r,T). This
    will depend on the composition, T, and density of
    the gas.
  • For stars, the abundance of an element is
    described by Xatom, the fraction of the total
    mass from that element.
  • X, Y, and Z are the mass fractions of Hydrogen,
    Helium, and everything else (sometimes referred
    to as metals)
  • The initial sun had X 0.7, Y 0.28, and Z
    0.02
  • Xoxygen 0.2
  • For a human being, Xoxygen 60 and X 10
  • Because of the enormous coulomb barrier, by far
    the easiest to burn (i.e. burns at the lowest
    temperatures) of abundant atoms is H.
  • substantial H burning requires 3 million K
  • sets limit of lowest mass star

4
Reaction rate, energy generation
  • The rate of nuclear fusion energy generation is
    given by the energy released times the reaction
    rate
  • Reaction rates are a generic problem in
    astropysics. They depend on
  • the density of each reactant n1, n2
  • the relative speed of the reactants, v
  • the cross-section s (probability of a fusion
    reaction)
  • consider the case where reactant-2 is stationary
    and reactant-1 moves with velocity v
  • this would be the situation for electrons
    colliding with atoms, why?
  • The volume swept out in time dt by one particle
    of reactant-1 is
  • dV svdt
  • The number of reactions is
  • dN n2 dV
  • the reaction rate per unit volume per unit time
    is
  • r n1 dN/dt n1 n2sv

5
Energy generation rate
  • If the energy liberated per reaction is Q, the
    energy generated per unit volume per unit time is
    just rQ.
  • So e rQ/r
  • recall definition of e
  • For T.E., the velocities will be Maxwellian, and
    we have to take the average over the velocity
    distribution
  • where normalized M-Bmann distribution
  • and E (1/2)mv2
  • Using the measured nuclear cross sections, and
    mass deficits (accounting for neutrinos, which
    simply leave the star without heating), the
    energy generation is calculated as a function of
    T. For the purposes of constructing a stellar
    model, the T dependence is fit with a power law
    near the temperature of interest.
  • e e 0raTb
  • a 1 for 2-body reaction (net Luminosity r2)
  • b 4 to 10 (4 for p-p chain, 10 for CNO bi-cycle)

6
Coulomb barrier, tunneling, Gamow peak
  • Since the interior is ionized, the nuclei are
    positively charged, and the repulsive potential
    is enormous slide 7
  • Electric potential energy for 2 protons separated
    by 1 fermi is about U 1.4 MeV (see MathCAD and
    slide 7).
  • kT 1 eV for 10,000 K, so for 10 million K it
    equals 1 keV
  • classical probability e -(U/kT) essentially
    zero
  • reaction only occurs through Q.M. tunneling.
  • Tunneling probability exponential in ratio of
    barrier height U and energy of particle E
  • U e-2/r
  • r de Broglie wavelength h/p
  • p mv
  • E (1/2)mv2
  • s exp(-2p2U/E) exp(-const.E -1/2)
  • Product of Maxwellian distribution and tunneling
    gives Gamow Peak slide 8 and MathCAD

7
Coulomb barrier
  • For discussion of coulomb barrier
  • See http//hyperphysics.phy-astr.gsu.edu/hbase/nuc
    ene/coubar.html

8
tunneling, Gamow peak
  • For discussion of Gamow tunneling peak
  • See http//burro.astr.cwru.edu/Academics/Astr221/S
    tarPhys/coulomb.html
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