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Most Basic Observations Of Sun Chapter 10 The Interiors of Stars Sun has maintained (i) its size, (ii) energy it emits every second, (iii) its surface temperature all this while. • Sun is a self-gravitating objects − each part of the sun exerts force on each other part of the sun. • Gravity always attracts There must be other kinds of forces acting against Gravity to prevent the sun from collapsing onto itself. These inward and outward forces must be in perfect balance! Hydrostatic Equilibrium • Sun is radiating the same amount of energy every second − 3 x 1026 Joule It must be generating the right amount of energy every second. Energy Generation Rate It must be making sure that each of its layer transports just the right amount of energy to maintain this energy output. Energy Transport Rate • Sun is maintaining the temperature of its surface, and is also maintaining some fixed temperature at each of its radial distance from the Sun. Since the transfer of fixed (energy) radiation from the center to the surface in each layer would depend crucially on how temperature in each layer varies, it must maintain the required temperature gradient in all layers. Temperature Gradient • Mass conservation Stellar Structure Equations Hydrostatic Equilibrium Mass Conservation Energy Transport Hydrostatic Equilibrium Ex. 10.1.1 Make a crude estimate of the pressure at the center of the Sun using the Hydrostatic Equilibrium. Rigorous calculation shows the pressure is ( ) Pressure Equation of State Completely neutral gas, Completely ionized gas, In terms of mass fractions, for a neutral gas In terms of mass fractions, for a completely ionized gas Go through P. 293 Ex. 10.1.2 Make an estimate of the temperature at the center of the Sun using the Gas pressure term alone and using the value for pressure we derived in the previous example. Stellar Energy Sources Gravitational Energy Kelvin-Helmholtz Timescale Assuming a spherical star of Radius R and M, the gravitational potential energy of the star, Kelvin-Helmholtz time scale ~ 107 years Too short! From dating of meteorites we know that our solar system is about 4.5 billion year old. Nuclear Time Scale Total Mass of four (4) Hydrogen atoms Mass of one Helium atom Ex. 10.3.2 Assuming Sun was all Hydrogen, and that inner 10% of mass is hot enough to convert Hydrogen into Helium, how long Sun can produce its current luminosity? ~ 1010 year Reasonable! A proton experiences repulsive Coulomb force and attractive nuclear force due to another proton. Temperatures required for the proton to have sufficient energy to cross the Coulomb barrier Considering Quantum tunneling, temperatures required for the proton to be in the range of attractive nuclear force Gamow Peak – Reaction Rate of Nuclear Processes Reaction rate in power-law (two particle interaction) r0 is a constant, Xi and Xx are mass fractions of incident and target nuclei, α´ and β are determined from reaction rates. Energy liberated per kilogram of material per second, ε0 is the energy liberated per reaction. In the form of power-law, α = α’-1. By summing εix for all reactions occurring inside stars, we find the total energy generation rate. It has units of W kg-1. Stellar Nucleosynthesis Conservation Laws for Nuclear Reactions In any nuclear reaction, total charge, total leptons, total nucleons must be conserved. by A ZX Nuclei will be represented , where Z is the number of protons, A is the number of nucleons, and X is the chemical symbol of the element. The Proton-Proton Chains 0.42 MeV PP I 5.49 MeV 12.89 MeV Totally 26.22 MeV! PP II e- e+ annihilation : 1.02 MeV PP III where fpp is the pp chain screening factor, ψpp is a correction factor that accounts for three chains of pp, and Cpp ≈ 1 involves higher order correction terms, T6 = T/106 K. Written in power-law form near T = 1.5 X 107 K The CNO cycle Figure Copyright: http://csep10.phys.utk.edu/astr162/lect/energy/pp-cno.gif Triple Alpha Process - Helium Burning Written as a power-law centered around T = 108 K Figure Source: http://en.wikibooks.org/wiki/Introduction_to_Astrophysics/Main_Sequence _Stars Carbon and Oxygen Burning Hydrogen fuses into Helium, Helium into Carbon, Carbon into Oxygen, Oxygen into Neon…. Nuclear reactions have strong dependence on temperatures. Heavier the element (1) Higher the temperatures required for its fusion (2) Stronger becomes the dependence (of energy production) on temperature. Can we go on making heavier elements out of lighter ones and generate energy in the process? OR Is there a limit to this chain? Binding Energy Per Nucleon Radiative Temperature Gradient Condition for Convection Stellar Structure Equations Hydrostatic Equilibrium Mass Conservation Energy Transport Constitutive Relations Boundary Conditions Stellar Mass-Luminosity Relation Comparison of Numerical Model and Dimensional Analysis (DA) Figure: 3.2 A. Choudhuri (2010) Stellar Mass-Luminosity Relation Based on Observational Data of Binary stars A linear fit gives a slope of ~ 3.7 (not too different from 3)! Stellar Luminosity-Effective Temperature Relation Comparison of Numerical Model and Dimensional Analysis (DA) Solid: Model Dashed : DA Stellar Luminosity-Effective Temperature Relation Based on Observational Data of Nearby stars Figure: 3.6 A. Choudhuri (2010) A linear fit gives a slope of ~ 5.6 (not too different from 6)! Limits on Stellar Mass Lower Limit on a Stellar Mass If it is not hot enough, it is not good enough to be termed a star! ~ 0.072 Mʘ below which it cannot fuse Hydrogen. Upper Limit on a Stellar Mass If it is too hot, it must shed some mass! ~ 90 Mʘ above which it can hardly stay bound. Eddington Luminosity Chapter 11 The Sun Changes on Main Sequence Internal Structure of the Sun Effect of Nucleosynthesis H, He, and 23He with radial distance Temperature, Pressure with radial distance Energy Produced with radial distance Mass and Density with radial distance Critical Temp. Gradient (Convec.) with radial distance Is (Are) there any direct evidence(s) of our stellar structure theory? • Solar Neutrinos • Solar Oscillations The Solar Neutrino Problem In 1970, Raymond Davis first designed an experiment to measure the solar neutrino flux − basically a HUGE underground tank (~ one and a half km below the ground) containing LARGE amount (~377,000 liters) of cleaning fluid (C2Cl4) . Homestake Gold Mine (Ray Davis) Result of 24 years of Homestake Experiment (1970-1994) It turned out that the average neutrino flux detected at the Homestake (~ 2.5 SNU) was smaller by a factor of three than expected from the Sun. Super-Kamiokande Experiment This underground Japanese experiment contained 50,000 tons of pure water . The blue Cerenkov light emitted by electrons would be detected by the 11,200 inwardly-directed photomultiplier tubes. The number of neutrinos detected were still less by a factor of two. The other experiments like Soviet American Gallium Experiment (SAGE) and the Gran Sasso Underground Laboraroty in Italy (GALLEX) which measured the lowenergy neutrinos produced in majority in p-p chain confirmed the discrepancy in the neutron flux. Sun in different wavelengths Image Credit: NASA’s Solar Dynamics Observatory Photosphere • What are the typical temperatures here ? 4400 K (outermost layer) – 9400 K (innermost layer) • What are the typical densities here ? ~ 10-3 kg/m3 (innermost layer) • What is happening here ? Zooming-in on the photosphere Just under the surface, we see “granules”, ~1500 km in size, lasting for 8-10 min before dissolving. A few thousand km below the surface, we see “supergranules”, ~30,000 km in size, lasting for about a day before dissolving. Henry Benard Convection Cells Top-View (Real) Image Source: Wiki Side-View (Simulation) Image Source: Wiki Solar Convection Wiggles on the „blue‟ side of the line is coming from the bright regions on convection tiles, implying a radial velocity of 0.4 km s-1, those on the „red‟ side are coming from the dark regions. The “life-times” of this granules tell us about the time a convective cell take to cover a distance one mixing length (back and forth). The Chromosphere • How much is the “thickness” of this region ? ~ 1600 km • What are the typical temperatures here ? 4400 K (innermost layer) – 10000 K (outermost layer) • What are the typical densities here ? ~ 10-7 kg/m3 • What is happening here ? Chromosphere in Hα Filter Image Credit: Dave Tyler http://www.daystarfilters.com/filtergram.shtml The Transition Region • How much is the “thickness” of this region ? ~ 100 km • What are the typical temperatures here ? 10000 K (innermost layer) – 106 K • What are the typical densities here ? ~ 10-16 kg/m3 • What is happening here ? The Corona • How much is the “thickness” of this region ? ~ Does not have a well-defined outer boundary • What are the typical temperatures here ? ~106 K • What are the typical densities here ? ~ 10-17 kg/m3 • What is happening here ? K Corona – White light produced due to scattering of photospheric radiation by free electrons. In the region 1 Rʘ to 2.3 Rʘ F Corona – Scattering of photospheric radiation due to dust grains beyond 2.3 Rʘ. E Corona – Between the K Corona, and the F Corona. Source of emission lines produced due to highly ionized atoms through out the Corona. Forbidden lines. Radio emission, X-ray emission from Corona Coronal Holes Sun in X-ray Image Credit: http://osoncje.tripod.com/planeti/Sun/x-ray_sun.jpg Solar Magnetic Field Solar Wind Stars emit not only radiation, emit matter too! Van Allen Radiation Belts Image Source: Wiki Alaska Aurora Borealis Image Source: United States Air Force photo by Senior Airman Joshua Strang The Parker Wind Model Solar wind means Sun is not able to hold on to some of its mass --- charged particles escape Sun’s gravity and fly away from Sun. Considering again a balance between pressure and gravity, if the pressure is too high (since temperatures are too high), and as the gravity falls with inverse square of the distance, particles would escape the Sun. some Eugene Parker developed a simple model of Solar wind considering isothermal wind (1958). For typical temperatures and number densities of inner corona, T = 1.5 x 106 K, n0 = 3 x 1013 m-3 at about 1.4 Rʘ. Using our expression from Parker Wind Model, we get, This doesn‟t match with interstellar pressures and number densities, in general. The assumption of Hydrostatic Equilibrium is incorrect! What heats the upper atmosphere to such high temperatures? Still an unsolved problem! We only have guesses, and are no where close to the answer! Two candidates: • Acoustic heating • Magnetic heating Acoustic Heating Temperature in the Chromosphere changes only by a factor of two, but the density changes by 4 orders of magnitude. Hence, eventually acoustic waves propagate faster than local sound speed, and a shock is developed, which heats the gas. Sunspots Image Credit: Charles Chandler New Jersey Institute of Technology's New Solar Telescope New Dalton Minimum. CLIMATE CHANGE AND ITS CAUSES – by Nicola Scafetta (2010) Assignments: 10.4, 10.7, 10.9, 10.13, 10.14, 10.21, 10.22, 10.23