1B11

Foundations of AstronomySun (and stellar) Models

Silvia Zane, Liz Puchnarewiczemp@mssl.ucl.ac.ukwww.ucl.ac.uk/webctwww.mssl.ucl.ac.uk/

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:

1B11 1) Equation of hydrostatic equilibrium.

Consider a small volume element at distance r from the centre, cross section S=2r, thickness dr

Concept 1): Stars are self-gravitating bodies in dynamical equilibrium balance of gravity and internal pressure forces.

0/ 2 rSdrGMSPP rrdrr

2r

GM

dr

dP r(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

drrdrdrdMMM rrdrr24/

24 rdr

dM r (2)

1B11 First consequence: upper limit on central P

From (1) and (2): 24// rGMdMdP rrr

r

M

rSCr

M

rr dMrGMPPdMdMdPSS

0

2

0

4///

At all points within the star r<Rs; hence 1/r4>1/RS4:

42

0

22

0

2 8/4/4/ SSr

M

Srr

M

r RGMdMRGMdMrGMSS

4242 8/8/ SSSSSC RGMRGMPP

For the Sun: Pc>4.5 1013 Nm-2=4.5 108 atm

1B11 Toward the E-balance equation: The virial theorem

drrrGMdrrPPrSS

SS

C

R

r

R

r

PPRr

PPrr2

00

2,

,03 4/434

• Thermal energy/unit volume u=nfkT/2=(/mH)fkT/2• Ratio of specific heats =cP/cV=(f+2)/f (f=3:=5/3)

013 U

2r

GM

dr

dP r 2

34r

G

dr

dPGMdPr r

drrrGMdrrPSS R

r

R

r2

00

2 4/43

1//)1/(1 PmkTu H

• U= total thermal Energy; = total gravitational energy

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

2/ UE

Stars have a negative “heat capacity” = they heat up when their total energy decreases.

For a fully ionized gas =5/3 and 2U+=0Total Energy of the star: E=U+

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.

drrLSR

r2

0

4

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):

dt

dSTr

dr

dLr 24 TdS/dt accounts for the

gravitational energy term

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:

32

2

2

2

16

3

4

4

Tacr

kL

dr

dT

dt

dSTr

dr

dL

rdr

dMr

GM

dr

dP

r

r

r

1B11 The equations of Stellar structure

• 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.

Using mass as independent variable (better from a theoretical point of view):

)64/(3

)4/(1

4

342

2

4

TacrkLdM

dTdt

dST

dM

dL

rdM

drr

GM

dM

dP

r

“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”:

1B11 Last ingredient: Equation of State

• N=number density of particles; =mean particle mass in units of mH.

• Define:

Perfect gas:kT

mNkTP

H

• 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

• 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) /mH

• (1/) = 2X+3Y/4+Z/2)

• Very good approximation is “standard” conditions!

ELEMENT NO OF ATOMS NO OF ELECTRONS Hydrogen X/mH X/mH

Helium Y/(4mH) 2Y/(4mH) Metals Z/(AmH) (1/2)AZ/(AmH)

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)

3/)/( 4aTmkTP H