Presentation is loading. Please wait.

Presentation is loading. Please wait.

1B11 Foundations of Astronomy Sun (and stellar) Models Silvia Zane, Liz Puchnarewicz

Similar presentations


Presentation on theme: "1B11 Foundations of Astronomy Sun (and stellar) Models Silvia Zane, Liz Puchnarewicz"— Presentation transcript:

1 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/

2 1B11 Physical State of the stellar (Sun) interior  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.. Fundamental assumptions:

3 1B11 1) Equation of hydrostatic equilibrium.  Consider a small volume element at distance r from the centre, cross section S=2  r, thickness dr Concept 1): Stars are self-gravitating bodies in dynamical equilibrium  balance of gravity and internal pressure forces. (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 1/R S 4 : For the Sun: P c >4.5 10 13 Nm -2 =4.5 10 8 atm

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

7 1B11 Toward the E-balance equation: The virial theorem 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. For a fully ionized gas  =5/3 and 2U+  =0 Total Energy of the star: E=U+ 

8 1B11 Toward the E-balance equation Energy loss at stellar surface as measured by stellar luminosity is compensated by energy release from nuclear reactions through the stellar interior. Sources of stellar energy: since stars lose energy by radiation, stars supported by thermal pressure require an energy source to avoid collapse.  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 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). Summary:

10 1B11 The equations of Stellar structure Boundary conditions: M=0, L=0 and r=0; M=M s L=4R S 2  T eff 4 and P=2/3g/k These equations must be solved for specified M s and composition. Using mass as independent variable (better from a theoretical point of view): “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) Uniqueness of solution: the Vogt Russel “theorem”:

11 1B11 Last ingredient: Equation of State N=number density of particles;  =mean particle mass in units of m H. Define: Perfect gas: 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 A=average atomic weight of heavier elements; each metal atom contributes ~A/2 electrons Total number of particles: If the material is assumed to be fully ionized: N=(2X+3Y/4+Z/2)  /m H (1/  ) = 2X+3Y/4+Z/2) Very good approximation is “standard” conditions!

13 1B11 Deviations from a perfect gas 1) When radiation pressure is important (very massive stars): The most important situations in which a perfect gas approximation breaks down are: 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)


Download ppt "1B11 Foundations of Astronomy Sun (and stellar) Models Silvia Zane, Liz Puchnarewicz"

Similar presentations


Ads by Google