The Sun and the Stars
The Sun and the Stars
Dr Matt Burleigh
The Sun and the Stars
The Hertzsprung-Russell Diagram (E. Hertzsprung and H.N. Russell)
Plot of surface temperature versus luminosity, or colour (e.g. B-V) versus absolute magnitude M,
and various other combinations
bright
What are the physical
mechanisms responsible
for the L,T sequence?
dim
Hot
Cool
Dr Matt Burleigh
The Sun and the Stars
The Harvard Classification scheme is an excitation-ionization scheme
The physical properties of stellar
atmospheres are determined by :
i) Temperature, T
ii) Pressure, P
iii) chemical composition, 
Dr Matt Burleigh
Hydrostatic Equilibrium
A star is in hydrostatic equilibrium  no net force acting on gas (i.e. gravity balanced by pressure)
r=R
(r)
r
P + dP
P
Centre
(r = 0)
dr
Dr Matt Burleigh
The Sun and the Stars
In hydrostatic equilibrium  no net force acting on gas (i.e. gravity balanced by pressure)
Consider a slab of gas area A, thickness r
Force due to gravity Fg, is given by
 GM 
Fg  ma   (r ) Ar  2 
 r 
This must be balanced by the upward pressure (remember P=F/A), that is
 GM 
AP    (r ) Ar  2 
 r 
rearrange,
P
 GM 
   (r ) 2 
r
 r 
as r  0, then
dP
 GM
   (r ) 2
dr
 R

   g ( r )

where g is the surface gravity,
(ie. gravitational acceleration at r=R)
If we express the pressure, P in terms of the optical depth, ,
then
d   dr
æ GM öæ -dt ö
dP = -r ç 2 ÷ç
÷
è r øè kr ø
dP =
g
k
dt
Gas pressure depends on opacity ,
and surface gravity g
Dr Matt Burleigh
E.g. Sun’s central pressure
What is the pressure at the centre of the Sun?
First calculate average density:
æ M ö
r avg = ç
÷
è 4 / 3p R3 ø
M=1.989x1030kg, R=6.96x108m
 (avg)  = 3M /4R3=1410 kgm-3
Then calculate pressure using Hydrostatic Equilibrium:
æ GM ö
dP
= -r (r) ç 2 ÷
è R ø
dr
Integrate from Pc to P R and from R=R0=0 to R= R,
If surface pressure = 0, then
Pc~ G M (ave)  / R = 2.7x1014 Nm-2
Dr Matt Burleigh
The Sun and the Stars
We assume that particle densities are high enough that the gas is in thermodynamic equilibrium
(i.e. all processes and their inverses occur at the same rate).
P  nkT
The gas obeys the ideal gas law,
Where the particle number density n is related to the mass density , and chemical composition
 (the mean molecular weight), by
1

For pure hydrogen, =1
ionised hydrogen, =1/2

mH n

k is Boltzmann’s constant (1.381  10-23JK-1)
mH is proton mass (mass of H atom) 1.67x10-27kg
(for ionized H there are roughly equal nos. of e- and p+, and me << mp)
For stellar interiors, the gas is mostly ionised, and therefore in general
1
3
1
 2 X  Y  Z  1.6

4
2
Where, X is the mass fraction of H
Y
“”
He
Z
“”
metals (all other elements)
(mass fraction is the % by mass of
one species relative to the total.
Ionized H gives two particles per mH)
Dr Matt Burleigh
The Sun and the Stars
What is the temperature at the centre of the Sun?
P  nkT
Combine ideal gas law,
And mean molecular weight
Re-arranged for n
Gives
Hence
n(r) =
P (r) =
\T ( r ) =
1


mH n

r (r)
m (r)mH
r ( r ) kT ( r )
m ( r ) mH
P(r)m ( r ) mH
r (r) k
Use Pc and avg estimates, assume  ~ ½ (ie pure ionized hydrogen)
Then Tc ~ PcmH/(avg)  k ~ 1.2x 107K
Gas dissociated into ions & electrons but overall electrically neutral… a plasma
Dr Matt Burleigh
The Sun and the Stars
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
Energy release can be calculated from E=mc2
– e.g. in H fusion 4 x 11H atoms  1 x 42He atom
Mass of 4 x 11H atoms = 4 x 1.6729x10-27kg = 6.6916x10-27kg
Mass of 1 x 42He atom = 1 x 6.646477x10-27kg
E = (4 x 11H atoms - 42He atom)c2 = (6.6916x10-27kg - 6.646477x10-27kg)c2 =
4x10-12J
Dr Matt Burleigh
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 (total mass available for fusion / mass in each reaction)
= Energy per reaction x (0.1 x mass of Sun / mass of 4 x 11H atoms)
(where mass of Sun is 2x1030kg)
•
•
•
•
Hence, Etot = 4.26x10-12 x (2x1029/6.6916x10-27) = 1.27x1044J
Lifetime of Sun is then the time taken to fuse all the available mass, which is given
by the total energy available Etot divided by the solar luminosity L
So, lifetime tsun= Etot/L = 1.27x1044J / 3.9x1026 Js-1
 tsun ~ 3.3x1017s ~ 1010yrs
Dr Matt Burleigh
Stellar Thermonuclear Reactions
Proton – proton chain (PPI, T < 2  107K)
3
2
2
1
1
1
H +11 H ®12 H + e+ + n
1.44MeV
2
1
H +11 H ®32 He + g
5.49MeV
He +32 He ®24 He +11 H +11 H
12.9MeV
is deuterium,
PPI occurs 91% of time in Sun, although there are other PP chains
H
Dr Matt Burleigh
b
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
Dr Matt Burleigh
Stellar Thermonuclear Reactions
CNO cycle
12
6 C
11 H 13
7 N 
13
7 N
13
6 C
11 N 14
7 N 
14
7 N
11 H 15
8 O 
15
8 O
15
7 N
•
•
•
•
•
•

13
C

e

6

15
N

e

7
4
11 H 12
C

6
2 He
The Carbon cycle (CNO cycle) also converts H to He but requires a C nucleus as a catalyst
Requires temperatures >1.6x107K
Occurs in Sun but minor compared to PPI
More important fusion process for stellar masses >1.1M sun
Since requires a C nucleus, only occurs in Pop I stars
Second and fifth steps occur because 13N and 15O are unstable isotopes with half lives of only a few
minutes
Dr Matt Burleigh
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
Dr Matt Burleigh
Stellar Thermonuclear Reactions
At very high temperatures, ~108K, other
processes fuse Helium (alpha particles) into
heavier elements
Triple alpha process:
4
2
He  He  Be  
8
4
Be  24He126C  
•
•
4
2
8
4
The triple alpha process is first stage of helium burning in an evolved star that has left the main sequence
In most massive stars, then burn C, Ne, O and Mg up to Fe
Dr Matt Burleigh