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1B11 Foundations of Astronomy Sun (and stellar) Models Silvia Zane, Liz Puchnarewicz [email protected] www.ucl.ac.uk/webct www.mssl.ucl.ac.uk/ 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.. 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=2r, thickness dr Pr dr Pr S GM r Sdr / r dP GM r 2 dr r 2 0 (1) 1B11 2) Equation of distribution of mass. Consider the same small volume element at distance r from the centre, cross section S=2r, thickness dr M r dr M r dM r / dr dr 4r 2 dr dM r 2 4r dr (2) 1B11 First consequence: upper limit on central P From (1) and (2): dPr / dM r GM r / 4r 2 2 dM r dP / dM / dM P P GM / 4 r r C S r r r MS MS 0 0 At all points within the star r<Rs; hence 1/r4>1/RS4: GM MS 0 r / 4r dM r 2 GM / 4R dM MS r 22 S r GM S2 / 8RS4 0 PC PS GM S2 / 8RS4 GM S2 / 8RS4 For the Sun: Pc>4.5 1013 Nm-2=4.5 108 atm 1B11 Toward the E-balance equation: The virial theorem dP G 3 dP GM r 4r dP GM dr r 4 r Pr 3 r RS , P PS r 0, P PC RS r 2 RS dr r2 RS 3 Pr 4r 2 dr GM r / r 4r 2 dr 0 RS 0 2 2 3 P 4 r dr GM / r 4 r dr r r 0 0 • Thermal energy/unit volume u=nfkT/2=(/mH)fkT/2 • Ratio of specific heats =cP/cV=(f+2)/f (f=3:=5/3) u 1/( 1)kT / mH P / 1 3 1U 0 • U= total thermal Energy; = total gravitational energy 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 U / 2 • 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. 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. RS L r 4r 2 dr 0 r=nuclear energy released per unit mass per s. Depends on T, and chemical composition • During rapid evolutionary phases (contraction/expansion): dL dS TdS/dt accounts for the 4r 2 r T gravitational energy term dr dt 1B11 The equations of Stellar structure Summary: dP GM r dr r2 dM r 4r 2 dr dL dS 4r 2 r T dr dt dT 3kL dr 16acr 2T 3 • 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). 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 2Teff4 and P=2/3g/k •These equations must be solved for specified Ms and composition. dP dM dr dM dL dM dT dM GM 4r 4 1 /( 4r 2 ) r T dS dt 3kL /( 64 2 acr 4T 3 ) 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) 1B11 Last ingredient: Equation of State Perfect gas: P NkT m H kT • 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 1B11 Last ingredient: Equation of State If the material is assumed to be fully ionized: ELEMENT Hydrogen NO OF ATOMS X/mH NO OF ELECTRONS X/mH Helium Metals Y/(4mH) Z/(AmH) 2Y/(4mH) (1/2)AZ/(AmH) • 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! 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): 4 P kT /( mH ) aT / 3 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)