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```DEPARTMENT OF PHYSICS AND ASTRONOMY
LIFECYCLES OF STARS
Option 2601
M.R. Burleigh 2601/Unit 4
Stellar Physics
 Unit 1 - Observational properties of
stars
 Unit 2 - Stellar Spectra
 Unit 3 - The Sun
 Unit 4 - Stellar Structure
 Unit 5 - Stellar Evolution
 Unit 6 - Stars of particular interest
M.R. Burleigh 2601/Unit 4
DEPARTMENT OF PHYSICS AND ASTRONOMY
Unit 4
Stellar Structure
M.R. Burleigh 2601/Unit 4
Starbirth
M.R. Burleigh 2601/Unit 4
Young Stars
M.R. Burleigh 2601/Unit 4
Globular Clusters
M.R. Burleigh 2601/Unit 4
Star Death
M.R. Burleigh 2601/Unit 4
Star Death
M.R. Burleigh 2601/Unit 4
Star Death
M.R. Burleigh 2601/Unit 4
Star Death
M.R. Burleigh 2601/Unit 4
Stellar Structure







Hydrostatic equilibrium
Equations of state
Energy transport (not derived)
Energy sources
Stellar models
Mass-Luminosity relation
Eddington Limit
M.R. Burleigh 2601/Unit 4
Hydrostatic Equilibrium
r=R
(r)
r
P + dP
M.R. Burleigh 2601/Unit 4
P
Centre
(r = 0)
dr
Hydrostatic Equilibrium
Equation of hydrostatic equilibrium:
However:
M 
dP
Gmr  r 

2
dr
r
dm
2
 4r  r 
dr
r  
R
0 dm
(1)
(2)
R
4 0 
Only need to know (r) to determine mass
M.R. Burleigh 2601/Unit 4
r r dr
2
E.g. Sun’s central pressure
G=6.67x10-11 Nm2Kg-2, M=1.989x1030kg,
R=6.96x108m
 (ave)  =3M /4R3=1410kgm-3
Surface pressure = 0
Let r = dr = R and M(r)= M
Pc~G M (ave)  / R = 2.7x1014Nm-2
M.R. Burleigh 2601/Unit 4
Equations of State
Assume material is a perfect gas
Obeys perfect gas law:
Pr   nr kT r 
Number density
of particles
M.R. Burleigh 2601/Unit 4
(3)
Boltzmann’s constant
(1.381  10-23JK-1)
Equations of State
n(r) is dependant on density and composition:
mH = 1.67  10-27 kg = mass of a hydrogen atom
 = mean molecular weight
nr  
 r 
 r mH
1
1


3
1  2

2 X  4 Y  2 Z 
Mass fractions of:
H
He
Metals (all other heavier elements)
 r kT r 
 Pr  
 r mH
M.R. Burleigh 2601/Unit 4
In massive stars, radiation pressure also contributes to
the total pressure:
a T r 
3
4
a = 7.564  10-14Jm-3K-4 = radiation constant ( = ¼ ac)
M.R. Burleigh 2601/Unit 4
E.g. Sun’s central temperature
Use Pc and  estimates, assume  ~ ½
Then Tc ~ PcmH/(ave)  k ~ 1.2x 107K
Gas dissociated into ions & electrons but
overall electrically neutral… a plasma
M.R. Burleigh 2601/Unit 4
Energy Transport
T(r) depends on how energy is
transported from interior  surface
Three processes…
1.Conduction – collision of hot energetic
atoms with cooler… poor in gases
2.Convection – mass motions of fluids,
need steep temp. gradient… happens in
some regions of most stars
M.R. Burleigh 2601/Unit 4
Energy Transport
Three processes…
1.Conduction
2.Convection
3.Radiation – high energy photons flow
outward losing energy by scattering and
absorption… opacity sources at high T
are i) electron scattering and ii)
photoionization
M.R. Burleigh 2601/Unit 4
 64r T r  dT 
L r    
 

 3 r  r   dr 
2
3
(r) (opacity) depends only upon N(r), T(r) and (r)
L(r) at the surface is the star’s bolometric luminosity
M.R. Burleigh 2601/Unit 4
(4)
For the Sun




L ~ 9.5x1029/ Joules s-1
However, we do not know  very well
Ranges from 10-3 <<  << 107
Therefore…
– 1022 << L << 1032 Joules s-1
 Measured value is 3.9x1026 implies
–  ~ 2.4x103
M.R. Burleigh 2601/Unit 4
The Virial Theorem
 Considers total energy in a star
 Gravitational contraction
 Gravitational potential energy  kinetic
energy
 Kinetic energy in bulk  Heat
M.R. Burleigh 2601/Unit 4
The Virial Theorem
Take the equation of hydrostatic equilibrium:
dP
Gmr  r 

dr
r2
and
dm
 4r 2  r 
dr
dP
mG
mG
3 dP


4

r


dm
dm
r
4r 4
But:


d
dP
dr
4r 3 P  4r 3
 4r 2 .3P
dm
dm
dm


d
dr
mG
3
2

4r P  4r .3P

dm
dm
r
M.R. Burleigh 2601/Unit 4
Integrate over the whole star:
4r
3
M
P  0M
0
3P
M Gm
dm   0
dm

r
Zero at both limits Twice the thermal
(P(m) = 0 marks the
(kinetic) energy
boundary of the star)
-2U
P,  and r are
functions of m
Gravitational
binding energy

 2U    2U    0
Total energy of a star:
E U 
 E U  0
M.R. Burleigh 2601/Unit 4
Gravitational Contraction
Gravitational contraction 
But, using Virial theorem:
1
U   
2
 ½  excess must be lost by radiation
1) Star gets hotter
2) Energy is radiated to space
3) Total energy of the star decreases (becomes
more –ve  more tightly bound)
M.R. Burleigh 2601/Unit 4
Stellar Thermonuclear Reactions
 Light elements “burn” to form heavier
elements
 Stellar cores have high enough T and 
for nuclear fusion
 Work (after 1938) by Hans Bethe and
Fred Hoyle
M.R. Burleigh 2601/Unit 4
Stellar Thermonuclear Reactions
 Energy release can be calculated from
E=mc2
– e.g. 4 x 11H atoms  1 x 42He atom
 4 x 1.6729x10-27kg = 6.6916x10-27kg
 1 x 6.6443x10-27kg
 E = 4.26x10-12J
M.R. Burleigh 2601/Unit 4
Stellar Thermonuclear Reactions
 In the Sun ~10% of its volume is at the
T and  required for fusion
 Total energy available is…
– Energy per reaction x mass/mass in each
reaction
 Etot = 4.26x10-12 x 2x1029/6.6916x10-27 =
1.27x1044J
 L = 3.9x1026  t ~ 3.3x1017s ~ 1010yrs
M.R. Burleigh 2601/Unit 4
Stellar Thermonuclear Reactions
Proton – proton chain (PPI, T < 2  107K)
1
1H
11 H 12 H  e   
1.44MeV
2
1H
11 H 32 He  
5.49MeV
3
3
He

2
2
CNO cycle
He 42 He 11 H 11 H
12
6 C
11 H 13
7 N 
13
7 N

13
C

e

6
13
6 C
11 N 14
7 N 
14
7 N
11 H 15
8 O 
15
8 O
15
N
M.R. Burleigh 2601/Unit 4 7

15
N

e

7
4
11 H 12
C

6
2 He
12.9MeV
The PPI Chain
M.R. Burleigh 2601/Unit 4
PPI Chain
a
c
1
1H
11 H 12 H  e   
2
1H
11 H 32 He  
3
3
He

2
2
He 42 He 11 H 11 H
M.R. Burleigh 2601/Unit 4
b
The CNO Cycle
M.R. Burleigh 2601/Unit 4
CNO Cycle
12
6
C  H  N 
1
1
13
7
13
7

N 13
C

e

6
13
6
C 11 H 14
7 N 
14
7
N 11 H 15
8 O 
15
8
15
7

O 15
N

e

7
4
N 11 H 12
C

6
2 He
M.R. Burleigh 2601/Unit 4
Triple Alpha
High level reactions ~108K
4
2
He  He  Be  
8
4
Be  24He126C  
4
2
8
4
22
24
168O,10
Ne,12
Mg upto Fe
M.R. Burleigh 2601/Unit 4
Stellar Models: Equations
Hydrostatic equilibrium:
Mass equation:
dP
Gmr  r 
(1)

2
dr
r
dm
 4r 2  r  (2)
dr
Equation of state:
Pr   nr kT r  (3)
n = number density of particles
nr  
 r 
 r mH
 = mean molecular weight
M.R. Burleigh 2601/Unit 4
a 4
Prad r   T r 
3
a = radiation constant = 7.564  10-14Jm-3K-4
 = ¼ ac
64r 2T 3 r  dT
Lr   
3 r  r  dr
 = opacity
Energy generation:
dL
2
 4r  r  r  (5)
dr
ε = rate of energy production (Js-1kg-1)
M.R. Burleigh 2601/Unit 4
(4)
Boundary Conditions
Need to apply boundary conditions to the equations to use them,
i.e. fix/know values at certain values of r (centre or surface)
e.g.
r=0
M(r) = 0
L(r) = 0
r=R
M(r) = M
L(r) = L
And (r), P(r)  0
M.R. Burleigh 2601/Unit 4
T(r) = Teff
From (1) write dP  P and dr  r
Then:
P = PS – PC = 0 - PC
Surface
and
Centre
r = R
M
 PC 
R
For a perfect gas P  T
M.R. Burleigh 2601/Unit 4
M
 TC 
R
M
 TC 
R
From (4)
3
4


T
T
RT


L  R 2  C  C   C
   R  
Also:
M
 3
R
Substitute
M
TC 
R
4 M

R 4TC4
L 
M
4
R  
3
M
R

L

M

Observed relationship is L  M3.3
( is dependant on T and )
M.R. Burleigh 2601/Unit 4
Eddington Limit
Hydrostatic equilibrium assumes no net outward motion of
material from the star, but the outward flow of radiation
imparts a force on the material
L
4r 2c
 Force =
T L
2
4r c
T = cross-section of electron-photon scattering = 6.7  10-29m2
This is opposed by gravitational force =
M.R. Burleigh 2601/Unit 4
mH MG
2
r
The forces are equal at the Eddington limit
mH MG  T L
4cmH GM

L
2
2
T
r
4r c
M
 LE  1.3 10
M Sun
38
erg s-1
So if L > LE material is expelled
M max  20
M.R. Burleigh 2601/Unit 4
M Sun

2
Stellar Structure







Hydrostatic equilibrium
Equations of state
Energy transport (not derived)
Energy sources
Stellar models
Mass-Luminosity relation
Eddington Limit
M.R. Burleigh 2601/Unit 4
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