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Stellar Structure Section 5: The Physics of Stellar Interiors Lecture 11 – Total pressure: final remarks Stellar energy sources Nuclear binding energy.

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Presentation on theme: "Stellar Structure Section 5: The Physics of Stellar Interiors Lecture 11 – Total pressure: final remarks Stellar energy sources Nuclear binding energy."— Presentation transcript:

1 Stellar Structure Section 5: The Physics of Stellar Interiors Lecture 11 – Total pressure: final remarks Stellar energy sources Nuclear binding energy Charged particle reactions Important reactions in stars: … H to He … He to C

2 Other effects on the pressure Have looked at pressure in zero-temperature degenerate gas Do we need to make any corrections to this? Relativistic effects in non-degenerate gases (see blackboard): the pressure behaves like an ideal gas at all temperatures the thermal energy depends on the kinetic energy of the particles (but is the same function of P in the NR and ER limits as for degenerate gases) Thermal effects: produce a Maxwell-Boltzmann tail at high p. Total pressure does have temperature terms (see blackboard), but the thermal corrections to the degenerate pressure formula are small

3 Total pressure when electrons are degenerate For (most) ionized gases, the electron density is larger than the ion density, so even the non-degenerate electron pressure is larger than the ion pressure: n e kT > n i kT In the degenerate case, the electron pressure is much greater than n e kT, so the ion pressure is negligible The radiation pressure is generally smaller than the ion pressure, especially at high densities Thus, to a good approximation, when electrons are degenerate, we have: (5.34)

4 Stellar energy sources: nuclear binding energy How much nuclear energy is available in practice? The binding energy of a compound nucleus is the energy equivalent of the mass difference between the mass of the nucleus and the sum of the masses of its components, Q(Z,N) (see blackboard) More useful is the binding energy per nucleon (or packing fraction), defined as Q(Z,N)/(Z+N) (or Q/A) – and shown schematically and actually in Handout 5 Energy can be released by fusion and fission Fusion of H to He releases the most energy/nucleon (see blackboard) – about 80% of the total available (H Fe)

5 Charged particle reactions Main forces between charged nucleons are repulsive electromagnetic (Coulomb) force attractive nuclear force (strong interaction) Weak interactions important if electrons or positrons involved Gravitational forces (almost) always negligible Nuclear force is short-range, Coulomb force is long-range Overall effect: deep nuclear potential well, plus Coulomb potential barrier – Handout 6, top Classically – reaction only if particle energy > barrier height (otherwise, particle pass in hyperbolic orbits)

6 Barrier penetration and resonances Quantum mechanics allows barrier penetration even for low- energy particles (Handout 6 again) Probability of penetration, and probability of reaction once inside nucleus, both depend on relative energies of colliding particles, so total cross-section is an integral over velocity, weighted by the velocity distribution (see blackboard) Thinking of particle as a wave, some of the energy is reflected For low energy particles, the reflection coefficient is high, except for energies corresponding to energy levels in the compound nucleus formed by the reaction For these resonant reactions, the interaction cross-section is much higher

7 Reaction rates Resonances are important for low energy particles with a normally low probability of interaction The reflection coefficient is high because of the high Coulomb barrier or the virtual discontinuity at the edge of the nuclear potential well For higher energy particles, these effects are weaker, and resonances are much less important (see blackboard) For non-resonant interactions, we can write down a (rather complicated) expression for N, the number of reactions kg -1 s -1 The main uncertainty is the specific nuclear factor, S The reaction rate rises with temperature, but falls again at very high temperature (see blackboard for N, and its T dependence)

8 Important reactions in stars Most of the energy produced in stars comes from two reactions: 4 H 1 He 4 3 He 4 C 12 H to He : Bethe & von Weizsäcker proposed CNO cycle – see Handout 6 C, N, O act as catalysts, with relative abundances of isotopes remaining fixed in equilibrium, while He/H increases ratios of catalysts in equilibrium not same as on Earth; e.g. C 12 /C 13 4 (~90 on Earth) allows identification of processed material in stars or gas clouds

9 H to He reactions – the pp chain (Handout 6) First step in pp chain is very improbable: one proton needs to decay into a neutron and a positron (plus a neutrino – weak interaction) while the other proton is nearby – i.e. in a timescale ~ s, much shorter than the weak interaction timescale Only ~ 1 in proton collisions result in the formation of a D nucleus! Acts as bottleneck to whole process, and makes CNO cycle competitive. Not shown to be viable until 1952 (Salpeter) Other -decays occur in side-chains; all release neutrinos Neutrinos escape freely (solar neutrino problem) pp-chain dominates at low T, CNO takes over at higher T: pp T 4 and CNO T 17 (5.38)

10 He to C reactions (Handout 6) He fusion can occur when the temperature is ~10 8 K Triple-alpha process is also highly unusual (although no weak interactions involved) Intermediate product, Be 8, is highly unstable to decay back into two alpha particles: lifetime ~ s Also – reaction resonant, but slightly endothermic: requires energy But mean time for collisions between alpha particles < Be 8 lifetime Second reaction, to form C, also resonant, forming excited state of C 12, whose energy ~ combined energy of Be and He nuclei

11 The power of prediction! Excited state in C 12 predicted by Fred Hoyle in on visit to W A Fowlers nuclear physics group in CalTech Prediction based on observed abundances of C 12 and O 16 C 12 destroyed again by adding alpha particle to form O 16 Need C 12 to be formed fast enough to avoid complete loss into oxygen: requires resonance at specific energy Persuaded members of Fowlers group to look for resonance at 7.68 MeV in C 12 – discovery* converted Fowler into nuclear astrophysicist (for which he won 1983 Nobel Prize…) Reaction so fast it is essentially 3-body: 3 2 T 40 (5.39) * Phys. Rev. 92, , 1953

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