1. The Research School of Astronomy and Astrophysics Mount Stromlo Observatory Lecture derived largely from Michael Balogh’s course notes

2. The tight relationship between Mass and luminosity his a key to the underlying physics.

3. Stellar interiors It is nearly impossible to learn anything directly about the interior of stars –The optical depth increases very quickly with distance from the surface, so photons of all wavelengths originate near the surface –An important exception are neutrinos, which interact very weakly with matter and so can escape from the Sun unimpeded –Also, solar oscillations give us a way to probe the interior by measuring the sound speed at various depths Therefore most of what we know is obtained by constructing a theoretical model containing all known physical laws, and consistent with observable surface features. –This generally requires detailed computer programs that only became feasible to run in the mid-60s (though most of the physics was already understood).

4. Equations of Stellar Structure Hydrostatic equilibrium: Mass conservation: Equation of state: There are 4 variables (P, ,M r and T) and 3 equations. To solve the stellar structure we will need to know something about the energy production and transportation

5. The Solar luminosity is 3.8x10 26 J/s –From age-dating moon rocks, we know the solar system has been around for ~4.6 billion years. If the luminosity of the Sun has been approximately constant, the total energy released would be 4.8x10 43 J. The luminosity of a star can be related to the total amount of energy produced per unit mass,  : For a spherically symmetric star, we can calculate the luminosity of a thin radial shell: Where L r is the luminosity due to all energy generated within radius r. –This is the fourth fundamental equation of stellar structure –So we need to know something about the energy generation mechanism

6. Gravitational potential So: how much energy can we get out of gravity? Assume the Sun was originally much larger than it is today, and contracted. This releases gravitational potential energy. The initial potential energy is ~0 (since R is large). The change in energy is therefore: The Solar luminosity is 3.8x10 26 J/s, so this energy would be radiated in only 10 15 s, or ~10 7 years. This is the Kelvin-Helmholtz timescale

7. Chemical energy Chemical reactions are based on the interactions of orbital electrons in atoms. Typical energy differences between atomic orbitals are ~10 eV. e.g. assume the Sun is pure hydrogen. The total number of atoms is therefore If each atom releases 10 eV of energy due to chemical reactions, this means the total amount of chemical energy available is This is ~100 times less than the gravitational potential energy available, and would be radiated in only 100,000 years at the present solar luminosity

8. The Equation Einstein showed that mass and energy are equivalent, and related by: Thus atomic masses can be expressed as energies When equating rest masses to energies, it is customary to omit the factor c 2 and leave it implicitly assumed.

9. Binding energy There is also a binding energy associated with the nucleons themselves. Making a larger nucleus out of smaller ones is a process known as fusion. For example: The mass of 4 H atoms is 4.031280u. The mass of He is 4.002603u. The mass difference is 0.028677u, equivalent to 26.71 MeV! –The … is other stuff like neutrinos, gamma rays, positrons… –~0.7% of the H mass is converted into energy

10. Nuclear energy: fusion In contrast with chemical reactions, nuclear reactions (which change one type of nucleus into another) typically release energies in the MeV range, 1 million times larger. E.g. Assume the Sun was originally 100% hydrogen, and converted the central 10% of H into helium. This would release an energy: Assuming the Sun’s luminosity has been constant at 3.8x10 26 J/s, it would take ~10 billion years to radiate all this energy. –Nuclear energy can sustain the solar luminosity over its likely lifetime

11. Proton-proton chain (PPI) The net reaction is: each of the above reactions occurs at its own rate. The first step is the slowest because it operates by the weak force. The rate of this reaction determines the rate of Helium production

12. Proton-proton chain (PPII and PPIII) Alternatively, helium-3 can react with helium-4 directly: Yet another route is via the collision between a proton and the beryllium-7 nucleus This reaction only occurs 0.3% of the time in the Sun. In the Sun, this reaction occurs 31% of the time; PPI occurs 69% of the time.

13. Nuclear energy Since the rate equations can be very complex, it is common to approximate it as a power- law over a particular temperature range Here, X i and X x are the mass fractions of the incident and target particles, respectively The parameter  can have a wide range of values depending on the set of reactions If each reaction releases an energy  the amount of energy released per unit mass is just The sum over all reactions gives the nuclear reaction contribution to  in our fundamental equation

14. The PP chain The nuclear energy generation rate for the PP chain, including all three branches: Near T~1.5x10 7 K (i.e. the central temperature of the Sun):

15. Example At the centre of the Sun,  5 ~1.5 and T 6 ~15 so If we imagine a core containing 10% of the Sun’s mass, composed entirely of 75% hydrogen (X=0.75), the total energy produced by the PP reaction is: Recall the solar luminosity is 3.8x10 26 J/s

16. The CNO cycle There is a second, independent cycle in which carbon, nitrogen and oxygen act as catalysts. The main branch (accounting for 99.6% of CNO reactions) is: at T~1.5x10 7 K

17. The CNO cycle Compared with the PP cycle, the CNO cycle is weaker at ~10 6 K, but much more strongly temperature dependent. Assuming X CNO ~0.03X, the two cycles are approximately equally energetic when which is just a little hotter than the central solar temperature.

18. The triple-alpha process The burning of helium occurs via the triple alpha process: The intermediate product 8-beryllium is very unstable, and will decay if not immediately struck by another Helium. Thus, this is almost a 3-body interaction Note the very strong temperature dependence. A 10% increase in T increases the energy generation by a factor 50.

19. Nucleosynthesis At the temperatures conducive to helium burning, other reactions can take place by the capturing of  -particles (He atoms). Further atom building in solar-type stars is prohibited by the increasingly strong Coulomb barrier (Not enough T and density to get an alpha into a nucleus) –However, in more massive stars, the central temperature may be significantly higher, allowing further reactions to occur. –Iron is the most stable nucleus. Thus, iron will be the ultimate product of nucleosynthesis, if temperatures are high enough to overcome the Coulomb barrier and fuse together.

20. Energy transport To solve for T, P, M and L as a function of radius r, we need one more equation. We need a differential equation relating temperature to radius, which is related to the transport of energy from the core to the surface. –Radiation: the photons carry the energy as they move through the star, and are absorbed at a rate that depends on the opacity. Slow –Convection: buoyant, hot mass will rise. Very efficient where it works. –Conduction: collisions between particles transfer kinetic energy of particles. This is usually not important because gas densities are too low. Slow. Slow.

21. The Solar model In this way we can build up a model of the interior structure of the Sun In general the differential equations are solved numerically Boundary conditions: –in the simplest case, , P and T =0 at r=R –M,L=0 at r=0

22. Convection in the Sun For the solar model we can plot dlnP/dlnT as a function of radius. When this is <2.5, convection will occur. This is the region where a boyant blob will continue to remain boyant as it rises inside the star.

23. The Solar interior The interior can be divided into three regions: 1.Core: In the very centre, the temperature is high enough to sustain nuclear reactions, via the PP chain 2.The radiative zone: energy is transported via radiation 3.The convective zone: the temperature here is lower, and the opacity is higher due to recombination. Thus convection takes over as the main mechanism for energy transport.

24. The solar model: evolution In the core, the mass fraction of H has decreased from X=0.75 to X=0.34 after 4.6Gyr –Accordingly, the mass fraction of He has increased from 0.24 to 0.64 At the surface, the fraction of H has also increased by ~0.03, due to the diffusive settling of more massive elements. The change in core H fraction has a direct effect on the luminosity and temperature, by changing the nuclear reaction rates and the amount of pressure support

25. Energy production Although nuclear reaction rates are higher where the temperature is higher, most of the energy is not produced at the centre of the Sun, because: –The amount of mass in a shell at radius r is i.e. there is more mass per unit volume at large radius (assuming constant density) –The mass fraction of hydrogen (X) at the centre has been depleted due to fusion, and the rate equations depend on X 2.

26. The Evolutionary Cycle of a Star like the Sun The Main Sequence The Sub-Giant Branch The Red Giant Branch The Horizontal Branch The Asymptotic Giant Branch. Planetary Nebulae and White Dwarf stars

27. 8.1 Evolutionary Tracks on the H-R Diagram Plot how the observations of a star change during its lifetime.

28. 8.2 Main Sequence Evolution Stars start their lives on the Main Sequence with approximately 75% H and 25% He. Core burning H to He H-He envelope In the center: as H burns to He: the fraction of He (and mean molecular weight ) goes up to maintain the pressure support the core contracts the temperature & Pressure increase the rate of energy generation increases due to the sensitivity of energy generation to Temperature.

29. At the surface: the increased energy generation means that the luminosity goes up the envelope is heated and expands so the radius of the star increases Duration: about 10 billion years for a 1 solar mass star

30. H-He envelope H-burning shell He core, gradually contracting and growing from “ash” from the shell the core contracts and heats up a new shell-source of burning hydrogen is ignited just outside the core He “ash” from the shell source increases the mass of the contracting core the energy generation rate slowly increases, heating the envelope 8.3 Sub-Giant Branch When the core runs out of hydrogen it loses its source of pressure support. In the center:

31. At the surface: the luminosity increases slightly radius of the star continues to increase the effective temperature decreases. Duration: about 1 billion years for a 1 solar mass star

32. He core supported by electron degeneracy pressure, still growing from He ash, contracting and heating up Increasingly rapidly-burning H- shell Rapidly expanding H-He envelope 8.4 Red Giant Branch : In the center: the core becomes dense enough to be supported by electron degeneracy pressure it continues to grow and contract as it gains He-ash from the shell source the rate of He-burning increases as the shell is heated by sinking into the center Not all the energy from the shell source can be carried by radiation - the envelope becomes convective.

33. At the surface: The luminosity increases dramatically the effective temperature decreases slightly the stellar radius increases dramatically. The star loses some mass via stellar winds Duration: about 1 billion years for a 1 solar mass star “Forbidden region”

34. 8.5 Horizontal Branch Once the temperature and density of the core are high enough Helium-burning via the triple-alpha process is ignited. This ignition process is known as the helium flash. He-burning core H-burning shell Steady envelope In the center: the core first expands and cools then the star burns helium steadily in the core the energy generation rate is lower than on the RGB

35. At the surface: the luminosity is steady and much less than on the RGB the radius is smaller than the RGB Duration: about 0.1 billion years. (the “Helium- burning Main Sequence”) Helium flash

36. 8.6 Asymptotic Giant Branch When all the Helium in the core is burnt: In the center: the core contracts until supported by electron degeneracy pressure a He-burning shell is ignited (Like an RGB star but with He-burning shell). Rapidly expanding envelope H-burning shell He-burning shell Carbon core

37. At the surface: The luminosity increases the effective temperature stays (approximately) the same the stellar radius increases dramatically. The star loses mass via pulsations

38. 8.7 Planetary Nebulae In low-mass stars the core never gets hot enough to ignite Carbon-burning. However, the rate of He-burning in the shell source is unstable, so the star pulsates and the envelope of the star is ejected in a planetary nebula. The core of the star remains behind as a white dwarf star that gradually cools. The gas in planetary nebulae is heated by UV-photons from hot, young white dwarf stars. They are typically expanding at speeds of 10-30km/s. They will glow, giving off optical photons for 10,000-50,000 years until they are too diffuse and too far from the White Dwarf star to be heated.

44. Direct observations of the core: neutrinos Because of their low cross section, neutrinos can escape from the Sun’s core to the Earth without further interaction One type of neutrino detector on Earth uses an isotope of chlorine, which will (rarely) interact with a neutrino to produce a radioactive isotope of argon. This reaction requires the neutrino to have an energy of 0.814 MeV or more, and can only detect neutrinos from the “side- reactions” in the PP chain: PPI I PPI II The Homestake detector contains ~400,000 L of cleaning fluid 2x10 30 atoms of Cl isotope Detect one Argon atom every 2-3 days.

45. Direct observations of the core: neutrinos More recently, the GALLEX (also SAGE) experiments uses 30 tons of natural gallium in a 100 ton aqueous gallium chloride solution to detect neutrinos via: This is sensitive to lower neutrino energies (0.233 MeV) and can detect neutrinos from the main branch of the PP chain

46. The Solar neutrino problem Both the Homestake and GALLEX experiments detected fewer neutrinos (by a factor 2-3) than were expected from the PP-chain reactions. This problem existed for about 30 years.

47. The solution to the problem was suggested by results from the Super- Kamiokande detector in Japan –Uses 40,000 tons of normal water. Neutrinos scatter electrons in the water to velocities exceeding the speed of light in water, producing a flash of light. –Results showed that electron neutrinos produced in the upper atmosphere can change into tau- or muon- neutrinos –This means neutrinos must have some mass and can therefore oscillate between flavours.

48. The Solar neutrino problem solved The Sudbury Neutrino Observatory uses heavy water, and was able to directly detect the flux of all types of neutrinos from the Sun. The flux of all neutrino types is in agreement with the predicted electron neutrino production from the PP reactions The results are completely consistent with the standard solar model.

49. Solar Oscillations also probe Core Stars ring like bells with different modes, and these are related to the structure of the star, and its composition. Successful comparison between predicted mode frequencies gives sound speed, and density as and measured normal functions of depth.

50. Until recently, the model of the sun seemed to reproduce the solar oscillations. –Can ‘see’ bottom of Convective Zone. –Accuracy has been sufficient to correct errors in opacity codes. Recent Revision of solar abundance now causes some problems…

51. Useful Stars- Cepheids Massive Stars (3-20 M sun ) (Young Stars) Pulsate from 3-100 days (Pulsation rate depends on Mass) Located in Instability Strip LMC Cepheids

52. RR-Lyrae Pulsate 4-14 hours – in an area Known as the instability strip. Stars which have low metal content – old stars Basically all the same brightness Use as distance measurers Main Sequence These Stars Are Dead Subgiants Red Giants

53. Stars Larger than the Sun