Transcript star model

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=2r, 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=2r, thickness dr
M r  dr  M r   dM r / dr dr  4r 2 dr
dM r
2
 4r
dr
(2)
1B11 First consequence: upper limit on central P
 From (1) and (2):


dPr / dM r  GM r / 4r 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 / 4r dM r 
2
 GM / 4R dM
MS
r
22
S
r
 GM S2 / 8RS4
0
PC  PS  GM S2 / 8RS4  GM S2 / 8RS4
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 
4r 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 4r 2 dr    GM r / r 4r 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 1U    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  4r 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
2
  4r   r  T

gravitational energy term
dr
dt


1B11 The equations of Stellar structure
Summary:
dP
GM r 

dr
r2
dM r
 4r 2 
dr
dL
dS 

 4r 2    r  T

dr
dt 

dT
 3kL

dr 16acr 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 2Teff4 and
P=2/3g/k
•These equations must be solved
for specified Ms and composition.
dP
dM
dr
dM
dL
dM
dT
dM

GM
4r 4
 1 /( 4r 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)