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1B11 Foundations of Astronomy Sun (and stellar) Models

1B11 Foundations of Astronomy Sun (and stellar) Models. Silvia Zane, Liz Puchnarewicz emp@mssl.ucl.ac.uk www.ucl.ac.uk/webct www.mssl.ucl.ac.uk/. 1B11 Physical State of the stellar (Sun) interior. Fundamental assumptions:.

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1B11 Foundations of Astronomy Sun (and stellar) Models

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  1. 1B11 Foundations of AstronomySun (and stellar) Models Silvia Zane, Liz Puchnarewicz emp@mssl.ucl.ac.uk www.ucl.ac.uk/webct www.mssl.ucl.ac.uk/

  2. 1B11 Physical State of the stellar (Sun) interior Fundamental assumptions: • Although stars evolve, their properties change so slowly that at any time it is a good approximation to neglect the rate of change of these properties. • Stars are spherical and symmetrical about their centre; all physical quantities depend just on r, the distance from the centre..

  3. 1B11 1) Equation of hydrostatic equilibrium. Concept 1): Stars are self-gravitating bodies in dynamical equilibrium balance of gravity and internal pressure forces. • Consider a small volume element at distance r from the centre, cross section S=2r, thickness dr (1)

  4. 1B11 2) Equation of distribution of mass. • Consider the same small volume element at distance r from the centre, cross section S=2r, thickness dr (2)

  5. 1B11 First consequence: upper limit on central P • From (1) and (2): At all points within the star r<Rs; hence 1/r4>1/RS4: For the Sun: Pc>4.5 1013 Nm-2=4.5 108 atm

  6. 1B11 Toward the E-balance equation: The virial theorem • Thermal energy/unit volume u=nfkT/2=(/mH)fkT/2 • Ratio of specific heats =cP/cV=(f+2)/f (f=3:=5/3) • U= total thermal Energy; = total gravitational energy

  7. 1B11 Toward the E-balance equation: The virial theorem For a fully ionized gas =5/3 and 2U+=0 Total Energy of the star: E=U+  • E is negative and equal to /2 or –U • A decrease in E leads to a decrease in  but an increase in U and hence T. • A star, with no hidden energy sources, c.omposed of a perfect gas, contracts and heat up as it radiates energy Stars have a negative “heat capacity” = they heat up when their total energy decreases.

  8. 1B11 Toward the E-balance equation Sources of stellar energy: since stars lose energy by radiation, stars supported by thermal pressure require an energy source to avoid collapse. • Energy loss at stellar surface as measured by stellar luminosity is compensated by energy release from nuclear reactions through the stellar interior. r=nuclear energy released per unit mass per s. Depends on T,  and chemical composition • During rapid evolutionary phases (contraction/expansion): TdS/dt accounts for the gravitational energy term

  9. 1B11 The equations of Stellar structure Summary: • P,k,r are functions of ,T, chemical composition (basic physics provides these expressions) • In total: 4, coupled, non-linear partial differential equations (+ 3 constitutive relations) for 7 unknowns: P, ,T, M, L, k, r as a function of r. • These completely determine the structure of a star of given composition, subject to suitable boundary conditions. • in general, only numerical solutions can be obtained (=computer).

  10. 1B11 The equations of Stellar structure Using mass as independent variable (better from a theoretical point of view): • Boundary conditions: M=0, L=0 and r=0; M=Ms L=4RS 2Teff4 and P=2/3g/k • These equations must be solved for specified Ms and composition. Uniqueness of solution: the Vogt Russel “theorem”: “For a given chemical composition, only a single equilibrium configuration exists for each mass; thus the internal structure is fixed”. This “theorem” has not been proven and is not rigorously true; there are unknown exceptions (for very special cases)

  11. 1B11 Last ingredient: Equation of State Perfect gas: • N=number density of particles; =mean particle mass in units of mH. • Define: • X= mass fraction of H (Sun=0.70) • Y= mass fraction of He (Sun =0.28) • Z= mass fraction of heavy elements (metals) (Sun=0.02) • X+Y+Z=1

  12. 1B11 Last ingredient: Equation of State If the material is assumed to be fully ionized: • A=average atomic weight of heavier elements; each metal atom contributes ~A/2 electrons • Total number of particles: • N=(2X+3Y/4+Z/2) /mH • (1/) = 2X+3Y/4+Z/2) • Very good approximation is “standard” conditions!

  13. 1B11 Deviations from a perfect gas The most important situations in which a perfect gas approximation breaks down are: 1)When radiation pressure is important (very massive stars): 2)In stellar interiors where electrons becomes degenerate (very compact stars, with extremely high density): here the number density of electrons is limited by the Pauli exclusion principle)

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