Reminder n Please return Assignment 1 to the School Office by 13:00 Weds. 11 th February (tomorrow!) –The assignment questions will be reviewed in next week’s workshop

Stellar Power Sources n Possibilities –Chemical burning n only a few thousand years –Gravitational contraction n a few million years –Nuclear fusion n ~10 10 years for a sun-like star

Physics of Fusion n The Nuclear Potential –Repulsive at large distances due to Coulomb –Attractive at short range due to Strong Nuclear Force n (1fermi = m)

Physics of Fusion –Classically, for nuclei to fuse, a barrier of height E c has to be overcome: E c in MeV, Z A, Z B nuclear charges and r N the range of the strong force in fermis (typically 1 fermi)

Physics of Fusion –Typical stellar models give typical core temperatures of ~ 10 7 K (kT ~ 1keV) –Fraction of nuclei with sufficient energy to overcome E C given by Boltzmann: Classically, fusion shouldn’t happen!

Physics of Fusion n Solution: –Quantum Mechanics –Protons described by Schroedinger Equation –Quantum mechanical tunnelling possible

Physics of Fusion –Probability of barrier penetration given by: Where r c is the classical closest distance of approach of the nuclei and  is defined by:

Physics of Fusion –The probability of barrier penetration can be recast in terms of energy: Where E G is the Gamow energy define by:  = fine structure constant (~1/137) m r = reduced mass of nuclei

Physics of Fusion –For two protons, E G =493 keV –In a typical stellar core, kT ~ 1 keV –Hence the probability of barrier penetration is: Still slow, but there are a lot of protons, and we have a lot of time!

Fusion Cross Sections n Consider a proton moving in a medium with n protons per unit volume –Probability of fusion occuring within a distance  x = n  x  = reaction cross section  = reaction cross section –Mean distance between collisions = mean free path, l = 1/n  –Mean time between  collisions,  = l/ = 1/ n –Mean time between  collisions,  = l/ = 1/ n note  may depend on v note  may depend on v

Fusion Cross Sections n The fusion cross section (Units, barns = m 2 ) as a function of energy is given by: S(E) is a slowly varying function determined by the nuclear physics of the reaction 1/E introduced to account for low energy behaviour

Fusion Reaction Rates Consider the reaction rate between two nuclei, A and B, travelling with relative speed, v, with concentrations n A and n B, with cross section  Consider the reaction rate between two nuclei, A and B, travelling with relative speed, v, with concentrations n A and n B, with cross section  –Mean time for an A nucleus to fuse with a B is:  A = 1/ n B –Mean time for an A nucleus to fuse with a B is:  A = 1/ n B –Hence, the total fusion rate per unit volume is: R AB = n A n B –Hence, the total fusion rate per unit volume is: R AB = n A n B

Fusion Reaction Rates –To obtain, we note that: Where P(v r ) is the Maxwell-Boltzmann distribution given by:

Fusion Reaction Rates –Including the function for  found earlier, the total reaction rate is then: Concentrating on the integral, we note that for a given impact energy, E, there is a competition between the Boltzmann term and the Gamow energy term

Fusion Reaction Rates exp(-E/kT) exp(-(E G /E) 1/2 ) exp(-E/kT-(E G /E) 1/2 ) –Proton-proton reactions n T = 2x10 7 K, E G = 290kT

Fusion Reaction Rates –Note there is a range of energies in which fusion rates peak –impact energy for peak rate, E 0, Width given by:(Via a Taylor expansion)

Fusion Reaction Rates –For the proton-proton reaction shown earlier, E 0 = 4.2kT = 7.2 keV  E = 4.8kT = 8.2 keV

Fusion Reaction Rates –The total reaction rate is found from the integral shown earlier. This gives:

Fusion Reaction Rates –Fusion reaction rates are strongly temperature dependent n e.g, p-d reaction, E G = MeV, around 2x10 7 K This implies the rate varies as T 4.6

Fusion in Stars I n Hydrogen fusion mechanisms in main sequence stars –Proton-Proton Chain –CNO Cycle –Relative rates and temperature dependencies

Hydrogen Burning n In a main sequence star, the principal source of power is fusion of protons into helium nuclei –4p  4 He + 2e e –Relies on weak nuclear force to mediate reaction: p  n + e + + e –Total energy release (including annihlation of positrons) MeV

Hydrogen Burning n Four particle reaction unlikely, hence might expect a three-step process: –2p  2 He +  – 2 He  d + e + + e –2d  4 He +  n Problem: –No bound state of 2 He –Hence, it looks like hydrogen burning is slow

Proton-Proton Chain n A possibility: –Fuse protons via the weak nuclear force to give deuterium –p  n + e + + e n Requires 1.8 MeV –p + n  d n Releases 2.2 MeV –Net Result: p + p  d + e + + e

Proton-Proton Chain –Recall the rate of a reaction is given by: This integral gives (see Phillips sec n 4.1) The symbols have their previous meaning, A = reduced mass in au S(E) is in keV barns

Proton-Proton Chain –The reaction p + p  d + e + + e is mediated by the weak force and is hence slow –S ~ 4x keV barns (calculated - too small to measure!) –What is the rate of proton-proton fusion in the core of the sun?

Proton-Proton Chain n Assume a typical model of the sun’s core: –T = 15x10 6 K –  = 10 5 kgm -3 –Proton fraction = 50% –hence proton density n p =3x10 31 m -3 –Also S ~ 4x keV barns and E G = 494 keV n These numbers give a rate of: R pp = 5x10 13 m -3 s -1

Proton-Proton Chain n Mean lifetime of a proton in the sun’s core: –Rate of any two protons fusing f = R pp / (1/2 n p )~3.3x s -1 –Hence, mean time for a proton pair to fuse is:  = 1/f = 3x10 17 s ~ 9x10 9 years –Slow proton-proton rate sets timescale for stellar lifetimes

Proton-Proton Chain n Following p-p fusion, further reactions to produce 4 He are rapid n The proton-proton chain can follow three branches (pathways):

Proton-Proton Chain p + p  d + e + + e p + d  3 He +  2 3 He  4 He + 2p Q eff = 26.2 MeV 85% 3 He + 4 He  7 Be +  e Be  7 Li + e p + 7 Li  4 He + 4 He Q eff = 25.7 MeV 15% p + 7 Be  8 B +  8 B  8 Be + e + + e 8 Be  4 He + 4 He Q eff = 19.1 MeV 0.02%

Proton-Proton Chain n Average energy release per p-p fusion: –Take into account: –two p-p fusions per branch –weightings of each branch –15 MeV per p-p fusion –Given number of fusions per m -3 calculated earlier, energy production rate ~ 120 Wm -3

The CNO Cycle n The proton-proton chain has a temperature dependence of ~ T 4 n Internal temperatures of more massive stars are only moderatly higher n Luminosities much greater than can be explained by the T 4 dependence

The CNO Cycle n Implications: –Another mechanism must be at work –This mechanism must have a higher order temperature dependence –Implies a higher Coulomb barrier Recall power of dependence  E G 1/3 and E G  (Z A Z b ) 2 Recall power of dependence  E G 1/3 and E G  (Z A Z b ) 2 –Such a mechanism is the CNO cycle

The CNO Cycle –Reaction Pathway: p + 12 C  13 N +  13 N  13 C + e + + e p + 13 C  14 N +  p + 14 N  15 O +  p + 15 N  12 C + 4 He Q eff = 23.8 MeV 15 O  15 N + e + + e S = 1.5 keV barns E G = 32.8 MeV S = 5.5 keV barns E G = 33 MeV S = 3.3 keV barns E G = 45.2 MeV S = 78 keV barns E G = 45.4 MeV

The CNO Cycle n Notice that: – 12 C acts as a catalyst –Rate governed by slowest step n in p-p, the first p-p fusion step n In CNO, if one considers all parameters, the p- 14 N step is slowest –Abundance of 14 N~0.6%

The CNO Cycle –Rate of p- 14 N fusion in the sun –Abundance of 14 N~0.6% n gives 2.6x N m -3 –S = 3.3 keV barns, E G = 45.2 MeV –Other parameters same as for p-p fusion: –R pN = 1.6x10 12 s -1 m -3 –CNO cycle contributes at most a few % to the power of a sun-like star –Mean lifetime of a 14 N nucleus in the sun ~ 5x10 8 yrs

The CNO Cycle n The CNO cycle is strongly temperature dependent –Using ideas from previous lecture, at the temperature of a sun-like star, and considering the p- 14 N step, R pN  T 20 –We can compare the rate of p-p vs. CNO as a function of temperature

CNO vs. p-p  T  T 2.96 sun

CNO vs. p-p n We can see that: –CNO contributes a few % of the sun’s output –In a moderately hotter stellar core (~1.8x10 7 K) CNO ~ p-p –In hot (>2x10 7 ) cores, CNO > p-p

A requirement for CNO n The CNO cycle requires the heavy elements C, N and O. –These elements have negligible abundance from the Big Bang –They are not formed in the p-p chain –What is the origin of heavy elements? –See Next Lecture - Fusion in Stars II

Next Week n Heavy element production n Star formation