Transcript stellar interiors instructor notes

10. The Interiors of Stars
Goals:
1. Develop the basic equations describing
equilibrium conditions applying in stellar
interiors.
2. Estimate the physical conditions that must exist
at the centre of a typical star, the Sun.
3. Outline potential sources of energy generation
in stars and investigate the basics of nuclear
reactions as a means of providing a selfsustained energy source.
4. Develop the equations of energy transport in
stars.
Fact or Fiction?
Consider an inhabited planet
completely enshrouded by
clouds, upon which astronomy
is not a scientific discipline
since it is not possible to see
into space from the planet’s
surface. Yet, one can construct
abstract mathematical models
of massive spheres of hot gas
in equilibrium and deduce
their properties from what is
known about the physics of
matter. Would it come as a
surprise to scientists on such a
planet if the constant canopy
of surrounding clouds one day
parted and stars came into
view?
Models of Stars
The parameters used for studying and modeling stellar interiors
include:
r
= radial distance from the centre of the star
M(r) = mass interior to r
T(r) = temperature at r
P(r) = pressure at r
L(r) = luminosity at r
ε(r) = energy generation at r
κ(r) = opacity at r
ρ(r) = density at r
In modern models mass M is used as the dependent variable
rather than radial distance r, but it is more informative to initiate
the study of stellar interiors using the geometrical variable r.
At the “natural” boundaries of the star the corresponding values
are:
At the centre:
r
= 0
At the surface:
r
= R
M(r) = 0
M(r)
= M*
T(r) = Tc
T(r)
= 0 (or Teff)
P(r) = Pc
P(r)
= 0
L(r) = Lc
L(r)
= L*
ρ(r) = ρc
ρ(r)
= 0
Rotation and magnetic fields are usually ignored in most models
(i.e. spherical symmetry is imposed), as well as any temporal
changes (i.e. radial pulsation).
Let us examine the equations of stellar structure.
Hydrostatic Equilibrium
For balance at any point in the interior of a star, the weight of a
block of matter of unit cross-sectional area and thickness dr must
be balanced by the buoyancy force of the gas pressure, i.e.:
Mass of block = density × volume = ρ dr
Weight of the block = mass × local gravity = ρg dr
But local gravity, g = GM(r)/r2
Buoyant Force = pressure difference (top – bottom) = –dP
So:
G M ( r )  dr
 dP 
r2
or
dP  G M ( r ) 

dr
r2
Example:
Obtain a crude estimate for the pressure at the centre of the Sun.
Assume 1 M = 1.9891 × 1033 gm, 1 R = 6.9598 × 1010 cm and:
M Sun 
1.98911033 gm
3
 Sun   4


1
.
4086
gm/cm
3
3
10
4

R
Sun
3
3  6.959810 cm


Solution (instructor):
Convert the equation of hydrostatic equilibrium into a difference
equation, evaluate it at r = ½R, assume that the mean density and
M = ½M* applies there as well. Then:
dP  GM ( r )   G 12 M *  2GM*M *  GM*2


 2 4 3 
2
5
2
dr
r
R* 2
R* 3 R* 
2 R*
Set dP = Ps – Pc = 0 – Pc = –Pc and dr = rs – rc = R* – 0 = R*

 Pc  GM*2

5
R*
2 R*
GM*2
and Pc 
4
2 R*
So, for the Sun:
Pc ≈ GM2/ R4
= 6.6726 × 10–8 × (1.989 × 1033)2/ (6.9598 × 1010)4
= 5.63 × 1015 dynes/cm2
The textbook method of solution gives a value of:
Pc = ~2.7 × 1015 dynes/cm2
A more rigorous solution involving integration with the standard
solar model gives a more reliable estimate of
Pc = 2.5 × 1017 dynes/cm2
two orders of magnitude larger!
Since 1 atmosphere = 1.013 × 106 dynes/cm2, the pressure at the
centre of the Sun is equivalent to 2.5 × 1011 atmospheres!
Conservation of Mass
The mass of a star must increase uniformly from the interior to
the surface, there cannot be any holes! For each element of
thickness dr the volume must increase by the mass of the shell
encompassed, i.e. by the spherical area × thickness dr. The
resulting equation is:
dM ( r )  4 r  dr
2
or
dM ( r )
2
 4 r 
dr
Ideal Gas Law
The ideal gas law is introduced in PHYS1211:
PV = NkT,
where P is the gas pressure, V its volume, N the number of
particles in the volume, T is the temperature (on the absolute
scale), and k = 1.3807 × 10–16 ergs/K is the Boltzmann constant.
It is also expressed using the gas constant R as:
PV = nRT, where n is the number of moles of gas.
The relationship can be derived from first principles through the
kinetic theory of gases, which specifies that:
(i) gas consists of small particles (molecules, atoms) that are
smaller than the distances separating them,
(ii) particles are in constant motion and make perfectly elastic
collisions with container walls,
(iii) the motions of particles are random, i.e. ⅓ are moving in any
specific direction.
Consider a box of N particles of mass m, where ⅓N are moving in
the x-direction. The average
force on the right-hand wall
(shaded) is given by the rate
of change of momentum:
Favg = Δ(mv)/Δt
A particle takes time Δt to
complete a trip from one wall
of the box back to the same
wall, i.e. Δt = 2l/v, where l
is the dimension of the box.
Δ(mv) = mv(before) – mv(after) = mv – (–mv) = 2mv
The resulting pressure (force/unit area) on the shaded wall is
given by the total force exerted by all particles in the box on the
end wall, i.e.:
P  13 N
Favg
l2
2

mv
N
v
mv
 13 N 2  2 2mv  13 N 3
l t 3l
2l
l
But l 3 = V, the volume of the box, so:
2
m
v
P  13 N
or PV  13 Nm v2
V
Since Nm = ρV (density × volume):
PV  13 Vv 2 or P  13 v2
But the average kinetic energy of a particle ½mv2 = 3⁄2 kT, so:
PV  13 Nm v2  13 N 3kT  NkT
or P 
 kT
m
, since  
Nm
V
The average kinetic energy of a gas molecule at room
temperature is:
16
1
.
3807

10
erg / K
3
3
1
KE  2 kT  2
300
K

0
.
0388
eV

eV
25
12
1.6022 10 erg / eV
Pressure Equation of State
The ideal gas law is, once again:
PV = NkT, where
P = gas pressure
V = volume of the gas
N = number of particles in the gas
T = temperature (K)
k = 1.3807 × 10–16 ergs/K is the Boltzmann constant.
The number density for the gas can be written as n = N/V, so:
P = nkT,
and since the number density can also be written as n = ρ/μmH,
where ρ is the actual density (gm/cm3), mH is the mass of a
hydrogen atom, and μ is the mean molecular weight. So the gas
pressure can be written as:
 kT
Pg  n kT 
 mH
Mean Molecular Weight
As the name implies, the mean molecular weight is the average
mass of a gas particle, but in units of the mass of a hydrogen
atom, i.e.:
m

mH
where m is the average mass of a gas particle.
Consider possible examples:
Hydrogen: neutral,

m
m
 H 1
mH mH
ionized,
1
m
2 mH



mH
mH
molecular,
m
2mH


2
mH
mH
1
2
Helium:
neutral,
ionized,

m
4mH

4
mH
mH
4
m
m

 3 H 
mH
mH
4
3
Heavy element: neutral,
2 Aj mH
m


 2 Aj
mH
mH
ionized,
2 Aj mH
m


2
mH Aj  1mH
where Aj is the atomic number.
In stellar interiors the value of μ for a fully-ionized gas is desired.
How does one take into consideration the contributions from all
elements?
Let X = fractional abundance by mass of hydrogen,
Y = fractional abundance by mass of helium, and
Z = fractional abundance by mass of all heavy elements,
where X + Y + Z ≡ 1, by definition.
In stellar interiors the value of μ for a fully-ionized gas is desired.
How does one take into consideration the contributions from all
elements?
In a cubic centimeter of gas of density ρ there is Xρ of hydrogen,
Yρ of helium, and Zρ of heavy elements by mass. Each of the
elements contributes different numbers of electrons to the mix,
under the assumption of complete ionization: 1 for hydrogen, 2
for helium, and 3, 4, 5… for the heavy elements. The number of
particles per cubic centimeter is:
Xρ/mH × 2 = 2Xρ/mH for hydrogen
Yρ/4mH × 3 = 3Yρ/4mH for helium, and
Zρ/2AjmH × (Aj + 1) ≈ Zρ/2mH for the heavy elements.
The total number of particles per cubic centimeter is therefore
given by:

3
1
No. of particles 2 X  4 Y  2 Z 
mH


1
1
So:  



2 X  43 Y  12 Z 
n mH mH 2 X  3 Y  1 Z  
4
2
mH
1
i 
or:
for fully ionized gas,
3
1
2 X  4 Y  2 Z 
and: n 
X 
1
1
4
Y
1
Aj
Z

for neutral gas,
or, since X + Y + Z ≡ 1:
i 
4
6 X  Y  2
When Z is negligible:
i 
4
5 X  3
Example:
What is the mean molecular weight for gas in the Sun, where X =
0.75, Y = 0.23, and Z = 0.02?
Solution (instructor):
Consider the case for gas that is fully ionized, which gives:
1
i 
2 X  43 Y  12 Z 
1

20.75  43 0.23 

1
2
0.02
1
1.5000 0.1725 0.0100
1

 0.5944
1.6825
i.e., slightly larger than ½, the value for a pure hydrogen gas.
Example:
Obtain an estimate for the temperature at the centre of the Sun.
Assume, for simplicity, μi = 0.5944, i.e. fully-ionized gas.
Solution (instructor):
We can use the perfect gas law if we know the density at the centre
of the Sun. Approximate the value using estimates for conditions
at r = ½R (as before), where we assume that ρ =  . Assume also
that P = ½Pc at that point, that k ≈ 4⁄3 × 10–16 erg/K, mH ≈ 3⁄2 × 10–24
gm, and μi = 0.5944. Then:
 kT
P
so
 mH
3
 mH P 0.5944mH 12 Pc 43  RSun
T 

k
k M Sun


3
2
6 10  7 10 
 10 2 10 
 24
10
4
3
10 3
15
16
33
 1.2 107 K
If the average temperature = ½Tc , then Tc = 2.4 × 107 K, fairly
close to the actual value of 1.6 × 107 K.
The central density is found from:
c 
 mH Pc
kTc
0.59441.67261024 6 1015 
3


1
.
8
gm/cm
1.36071016 2.4 107 
whereas the actual value is ~82 gm/cm3.
Since the mass of a hydrogen atom is 1.6726 × 10–24 gm, a density
of 1.8 gm/cm3 corresponds to a number density of n = ρ/μmH =
1.81 × 1024 /cm3. The volume occupied by a particle is 4⁄3πr3 = 1/n,
so the average particle radius is r = 5.1 × 10–9 cm = 0.51 Å,
compared with the Bohr radius of 0.523 Å.
With an actual density for the Sun of 82 gm/cm3, the average
particle radius is r = 1.4 × 10–9 cm = 0.14 Å, compared with the
Bohr radius of 0.523 Å.
Such highly compressed matter cannot exist as bound atoms, and
is referred to as pressure-ionized gas.
Example (from Mechanics):
Calculate the gravitational self-energy (energy of assembly piecewise from infinity) of a uniform sphere of mass M and radius R.
Solution:
Think of assembling the sphere a shell at a time (r = 0 to r = R).
For a shell of radius r the incremental potential energy is dV =
φdm, where dm is the mass of the shell and φ is the gravitational
potential for the mass already assembled, which is a sphere. Since
the mass being assembled forms a spherical shell, we have:
2


M
3
Mr
dr
2
2
dm   4 r dr   4
4r dr 
and
3
3
R
3R 
Gm
4 r 3
4 r 3  M 
r3
 
, where m 

M 3
4
3
r
3
3 3R 
R
So the potential self-energy of the mass is:
V   dV
where:
m dm
G 4 r 3   3Mr 2 dr 


dV   dm   G

3
r
r
3  R

GMr 4  4 M 
dr
  3  4
3 
R  3R 
3GM 2 r 4

dr
6
R
So the potential self-energy of the sphere is:
V 
3GM
R6
2 R
4
r
 dr  
0
2
5 R
3GM r
R6 5
3GM 2  R 5 
3 GM 2
   

6
R  5 
5 R
0
Stellar Energy Sources
As noted the gravitational potential energy required to contract a
star to its present size is given by:
3 GM 2
V 
5 R
But, of the potential energy lost by a star, according to the Virial
Theorem (ASTR 2100), one half is transformed into an increase
in the kinetic energy of the gas (heat) and the remainder is
radiated into space. The radiation lost by a star upon contraction
to the main sequence is therefore given by:
3 GM 2
E 
10 R
For the Sun, at its present mass (1.9891  1033 gm) and radius
(6.9598  1010 cm), the amount of energy radiated through
contraction is:
8
33 2
3 6.67210 1.989110
48
E 

1
.
138

10
ergs
10
10
6.959810





The present luminosity of the Sun is L = 3.851  1033 ergs/s, so if
it had been shining at the same luminosity for the entire duration
of its contraction (clearly erroneous) then the time scale for
contraction is given by the following:
Energy Lost  PresentLuminosity T imeScale for Contraction
or E  LSun  tKH
where tKH is the Kelvin-Helmholtz time scale (more properly the
Helmholtz-Kelvin time scale, although it originated with the
Scottish physicist John James Waterston, 1811-1883, whose
papers on the subject were rejected by the Royal Society of
London!). Here:
E
1.1381048 ergs
2.9551014 s
6
tKH Sun  



9.4

10
yr
33
7
LSun 3.85110 ergs/s 3.155710 s/yr
or ~107 years. The actual value should be smaller because the
Sun’s luminosity was greater during the contraction phase, but
the main point is that the estimate is much shorter than the
estimated age of the solar system of ~4.6  109 years.
Nuclear Energy Sources
Consider the rest masses of the fundamental nuclear particles:
Proton: 1.672623 × 10–24 gm
Neutron: 1.674929 × 10–24 gm
Electron: 9.109390 × 10–28 gm
Atomic mass unit, 1 u = 1.660540 × 10–24 gm = 931.49432 MeV,
for E = mc2.
The original nucleon symbolism was:
where
i.e.,
1
1
H
A
Z
X
A = mass number = number of nucleons
Z = number of protons (usually omitted)
X = chemical symbol of the element as specified by Z.
is redundant, since
1
H indicates the same thing.
Typical masses:
1H
= 1.007825 u = 938.78326 MeV
2H = 2.014102 u = 1876.12457 MeV
4He = 4.002603 u = 3728.40196 MeV
5Li = 5.0125 u = 4669.115279 MeV
8Be = 8.005305 u = 7456.89614 MeV
The major reaction in astronomy converts 4 hydrogen nuclei
(protons) into a helium nucleus (4He).
But 4 1H = 1.007825 u × 4
and 1 4He = 4.002603 u
Difference
= 4.031280 u
= 4.002603 u
= 0.028677 u = 0.0071 of 4 1H
The energy released = mc2 = 0.028677 u × 1.660540 × 10–24 gm c2
= 26.71 MeV.
The lifetime of a star via nuclear reactions depends upon how
much of its hydrogen content is converted to energy via nuclear
reactions.
For the Sun we can estimate:
Enuclear  0.10 0.0071 MSun c2  1.31051 ergs
At L = 3.851 × 1033 ergs/s,
Enuclear
1.31051 ergs
3.381017 s
10
tnuclear Sun  



10
yr
33
7
LSun
3.85110 ergs/s 3.155710 s/yr
Also, if tnuclear = Enuclear/L* ≈ XM*c2/M*4
then tnuclear = Xc2/M*3 ≈ 1010 yrs/(M*/M)3 .
1 M, tnuclear = 1010 years
2 M, tnuclear = 109 years (A-star)
4.6 M, tnuclear = 108 years
10 M, tnuclear = 107 years (B-star)
21.5 M, tnuclear = 106 years (O-star)
0.5 M, tnuclear = 1011 years > 1/H0 (estimated age of the universe)
The lifetime of the Sun and stars via nuclear reactions is
consistent with the nuclear ages of meteorites, as well as the oldest
rocks on the Earth and the Moon.
Properties of nuclear particles:
Particle
Baryon
Lepton
proton


neutron


electron


positron


neutrino


muon


Other baryons, the hyperons:
xi (Ξ)


sigma (Σ)


lamda (Λ)


Middle family, the mesons:
κ-mesons
π-mesons
photons


Spin
±½ћ
±½ћ
±½ћ
±½ћ
±½ћ
±½ћ
Charge
+1
0
–1
+1
0
±1
±½ћ
±½ћ
±½ћ
0,
0, ±1
0
0 or ±1ћ
0 or ±1ћ
0
0, ±1
0, ±1
0
Nuclear reactions conserve: (i) number of nucleons, (ii) number of
leptons, (iii) electronic/nuclear charge, (iv) particle spin.
Example:
What is the missing particle in the following reaction?
37Cl + ν ↔ 37Ar + ?
e
The collision of an electron neutrino with a chlorine-37 nucleus
produces an argon-37 nucleus plus a to-be-identified particle.
Solution:
37Cl
is element 17 with 17 protons, 20 neutrons. 37Ar is element 18
with 18 protons, 19 neutrons.
Nucleons in = 17 + 20 = 37. Nucleons out = 18 + 19 = 37. 
Leptons in = 1. Leptons out = 0, so missing particle is a lepton. 
Charge in = +17 (17 protons). Charge out = +18 (18 protons), so
missing particle has a charge of –1. 
Spin in = 19ћ (17 protons, 20 neutrons, 1 neutrino). Spin out =
18½ћ (18 protons, 19 neutrons), so missing particle has spin ½ћ. 
The only lepton with spin ½ћ and charge of –1 is the electron.
i.e.,
37Cl
+ νe ↔
37Ar
+ e–
The proton-proton reaction:
The first step in the proton-proton cycle is stymied by the fact
that the colliding particles are small and both positively charged.
Although the nuclear strong force takes over at small separations
of the particles, at larger distances they are blocked from
interacting by mutual Coulomb repulsion.
How is the Coulomb barrier overcome? Consider the potential
energy of a Coulomb barrier:
Z 1Z 2e2
PotentialEnergy 
r
From statistical mechanics we know that kinetic energy and
thermal energy are related through the reduced mass of a
particle:
2
1

v
 23 kTclassical
so:
2 m
2Z1Z 2e
2 11 4.80310 esu 
10
Tclassical 


1
.
114

10
K
16
13
3k r
3  1.38110 erg/K110 cm
2
10
2
which is much higher than the Sun’s central temperature.
But, by the Heisenberg uncertainty principle, the uncertainty in a
particle’s momentum, Δpx, is related to the uncertainty in its
position, Δx, via:
ΔpxΔx ≥ ½ћ = h/4π
Reaction Rates:
The rates of individual nuclear reactions depend upon a variety of
factors:
1. Atomic nuclei must have sufficient energy to penetrate the
Coulomb barrier of the target nucleus, only nuclei with specific
energies specified by the high velocity “tail” of the MaxwellBoltzmann distribution can react with the target nuclei.
2. The cross-section for the reaction, σ(E), must be substantial.
The restriction from (1) implies that energetic nuclei capable of
reacting with the target nuclei are described by the MB
3
distribution, i.e.,
2
E


2n 1
1
 E 2 e k T dE
nE dE  1 
 2  kT 
where KE = ½μmv2 is the kinetic energy of the particle. The
restriction from (2) produces:
number of reactions/nucleus/time
 E  
number of incident particles/area/time
Denote dNE as the number of particles of velocity vE = (2E/μm)½
that can strike a target nucleus in time dt, i.e.:
dnE   E vE niE dE dt,
where niE is the number of incident particlesof energy E.
But
ni
niE dE  nE dE, i.e., some fractionof the totalflux
n


0
0
of incident particles, n   nE dE, and ni   niE dE.
So:
ni
# reactions/nucleus dNE  E vE niE dE dt


  E vE nE dE
timeinterval
dt
dt
n
For nx targets/unit volume, the total reaction rate per unit volume
per unit time is:

nE
ri , x   nx ni E vE
dE
n
0
But we still need to evaluate σ(E).
The radius of an atomic nucleus can be estimated as r ~ λ (= h/p),
the de Broglie wavelength.
2
h2 1
p
2
 E  ~    2 ~ , since KE  12 mv 
p
E
2 m
The size of the Coulomb barrier VC is also important, so:
2
 E  ~ e
But
2 2VC E
VC Z1Z 2 e 2 r 2 Z1Z 2 e 2 2 Z1Z 2 e 2



2
E
 m v 2 h p  pv
hv
and KE  12  m v 2 , so 
VC 2 Z1Z 2 e


E
h
i.e.,  E  ~
e
m
1
2
 bE
2KE
1
2 Z1Z 2 e  m2

h
1
2
1
2
2
2
1
2
E
Presumably the function describes the main type of variability,
although an additional term that is a slowly varying function of E
cannot be excluded. The reaction rate therefore becomes:
 2 
ri , x   
 kT 
3
2

nx ni
  m 
1
2
 S E  e
b E
1
2
e  E kT dE
0
Where the MB relation has been used as a substitute for nE. The
resulting functional dependence of the reaction rate resembles a
Gaussian, and is referred to as the Gamov peak.
eb
E
1
2
In actual cases there may be resonant peaks superposed because
of resonances with excited energy levels in the nucleus (analogous
to excited energy levels for atoms).
Examples of potential resonant cross-sections in the reaction rate
functional dependence.
Electrons can partially shield the positive charges of nuclei,
resulting in lower effective Coulomb barriers to reactions, i.e.:
Veff
Z1Z 2 e 2

 Ves r 
r
where Ves(r) < 0, is the contribution from electron screening.
When electron screening is ignored, the integration results in a
function that can be approximated as:
ri , x  r0 X i X x   'T 
where r0 is a constant,
Xi is the mass fraction of impacting particles for the reaction,
Xx is the mass fraction of target particles for the reaction, and
α' and β are exponents established by the integration using a
power-law expansion for the reaction rate equations, which have
messy integrals.
The energy released by nuclear reactions per gram of stellar
material is given by:
 i,x
 i,x
 0 
   rix or

 

 0 Xi X x  T
where α = α' – 1. The units are ergs/gm/s.
Energy generated through nuclear reactions is responsible for a
star’s luminosity, through the equation of continuity:
dL
2
 4 r  
dr
where ε = εnuclear + εgravity, where the latter term is not always
negligible.
Nuclear Reaction Chains:
All must conserve momentum, energy, spin, charge, etc.:
The Proton-Proton Reaction.
PPI: (69%)
H  H  H  e  e
2
H  1H  3He  
3
3
4
1
He  He  He  2 H
1
1

2
t½ = 7.9 × 109 yr
t½ = 4.4 × 10–8 yr
t½ = 2.4 × 105 yr
PPII: (30.85%)
He  He  Be  
7
Be  e  7 Li   e
3
7
4
7
Li  1H  4 He  4 He
t½ = 9.7 × 105 yr
t½ = 3.9 × 10–1 yr
t½ = 1.8 × 10–5 yr
PPIII: (0.15%)
He  He  Be  
7
1
8
Be  H  B *  
8
B *  8 Be  e   e
8
4
4
Be  He  He
3
4
7
t½ = 9.7 × 105 yr
t½ = 6.6 × 101 yr
instantaneous
t½ = 3.0 × 10–8 yr
Net Reaction:
4 H  He  2e  2 e  2
1
4

26.73 MeV of energy
Energies of Neutrino Products:
From 1H:
Maximum Energy = 0.42 MeV
From 7Be:
Maximum Energy = 0.86 MeV
From 8B:
Maximum Energy = 14.0 MeV
The nuclear energy generation rate for the process can be written
as:
 2 3 33.80T6
6
 PP  2.3810  X f PP PP cPP T e
6
2
1
3
where T6 = units of temperature in 106 K,
fPP = fPP(X, Y, ρ, T) ≈ 1 is the electron screening factor,
φPP = φPP(X, Y, T) ≈ 1 is a correction factor to account for the
various branches of the PP chain, and
cPP ≈ 1 is a correction factor for higher order terms.
In simple form the relationship is written as:
 PP
2
4

  0,PP  X f PP PP cPP T6
for temperatures near 1.5  107 K. In other words, the energy
generation rate for the proton-proton chain varies as the local
density and the temperature to the fourth power, ε ~ ρT4.
The CNO Bi-Cycle.
C  1H  13N  
13
13

N  C  e  e
13
C  1H  14 N  
14
1
14
N  H O
15
O  15N  e   e
12
15
N  1H  12 C  4 He
t½ = 1.3 × 107 yr
t½ = 2.8 × 10–5 yr
t½ = 2.7 × 106 yr
t½ = 3.2 × 108 yr
t½ = 5.6 × 10–6 yr
t½ = 1.1 × 105 yr
99.96% of the time, or 0.04% of the time
N  1H  16 O  
16
1
17
O H F
17
F  17 O  e   e
17
1
14
4
O  H  N  He
15
t½ = 3.0 × 10–6 yr
In the CNO cycle, discovered by Hans Bethe in 1938, isotopes of
carbon (C), nitrogen (N), and oxygen (O) act as catalysts for the
reaction. The element fluorine (F) is also involved. Although the
elements are not destroyed in the reactions, they proceed at such
different rates that the isotopes of nitrogen (N) increase in
abundance while those of carbon (C) and oxygen (O) decrease.
The proton-proton chain dominates for cool stars like the Sun, the
CNO bi-cycle for stars hotter than the Sun, which have higher
core temperatures.
The Triple-Alpha Process.
He  4 He  8Be *  
4
8
12
He  Be *  C  
4
t½ = 1.3 × 107 yr
t½ < 8.2 × 10–24 yr
Summary:
Reaction
ρ-dependence X-dependence T-dependence
PP Chain
ρ1
X2
T64
CNO Bi-Cycle
ρ1
XXCNO
T619.9
Triple-Alpha
ρ2
Y3
T841.0
Note the higher temperature dependence of the CNO cycle. It is
dependent upon the CNO abundances, but dominates over the PP
chain for stars somewhat more massive than the Sun where the
central temperatures are higher.
The triple-alpha reaction involves an unstable isotope in the
production of 8Be. The reaction proceeds because, under the high
density conditions at the centers of evolved stars, a third alpha
particle (4He nucleus) can collide with 8Be before it has time to
decay. The resulting production of energy has an extremely strong
temperature dependence.
The Helium Flash.
When stellar core material is electron degenerate, the local
pressure does not depend upon T. Thus, when He-burning is
initiated, the high T-dependence of the 3α process means the
energy is generated, raises T locally, thereby increasing the
reaction rate, but does not produce a pressure or density decrease
to moderate the reaction. The result is known as a helium flash. It
only occurs in red giants for stars roughly as massive as the Sun
or less, and in more advanced stages of other stars, typically in
the white dwarf stage. It is best pictured using the pressure
equation:
 kT
Pg  n kT 
 mH
When H-burning or He-burning occurs, the result is a gradual
increase in the mean molecular weight μ. If T and ρ remain
unchanged, P decreases and unbalances the previously-existing
hydrostatic equilibrium. The core of the star collapses so that
both P and ρ increase. That enhances the energy generation rate,
making the star more luminous. If Teff does not change, the star
becomes larger, since L = 4πR2σTeff4.
Other Reactions.
More advanced reactions involve fusion of 12C to 16O, 23Mg
(endothermic) or 20Ne, 23Na, 24Mg (exothermic), as well as fusion
of 16O to 24Mg (endothermic) or 28Si, 31P, 31S, 32S, (exothermic).
Various reactions are possible. Consider the binding energy per
nucleon:
Eb m c2 Zmp  ( A  Z )mn  mnucleus 


A
A
A
At low atomic weights the most stable nuclei are 1H, 2H, 3He, 6Li,
4He, 12C, 16O, 24Mg, 40Ca, and 56Fe. At high atomic weights the
most stable nuclei are 86Kr, 107Ag, 127I, 174Yb, 208Pb, and 238U. Such
unusually stable nuclei are called magic nuclei. The maximum
binding energy per nucleus occurs at the iron peak, and all other
fusion reactions producing heavier nuclei are endothermic.
The problem of explaining
the existence of heavy
elements in the universe
can be restricted to
explaining the existence of
nuclear reactions that are
endothermic in stars. The
solution is advanced stages
of evolution in massive
stars (proton and alpha
capture) and supernova
explosions
(neutron
capture).
Energy Transport and Thermodynamics:
So far we have developed the following equations of stellar
structure:
Equation of Continuity:
dM ( r )
2
 4 r 
dr
Hydrostatic Equilibrium:
dP  G M ( r ) 

dr
r2
Energy Generation:
dL
 4 r 2  
dr
But what about the temperature gradient, dT/dr?
Radiative Transport:
Energy can be transported through a star by radiation,
convection, or conduction. Conduction is unimportant in most
gaseous stellar interiors, but the other two processes are
important. In stellar atmospheres radiative energy transport is
described by:
dPrad   

Frad
dr
c
But Prad = ⅓aT4, so:
Thus:
dPrad 4aT 3 dT   


Frad
dr
3 dr
c
dT
3  

Frad
3
dr 4ac T
But Frad = Lr/4πr2, giving, for radiative energy transport:
dT  3   L r

3
2
dr 4ac T 4 r
Convective Transport:
Convection is a three-dimensional process that must be
approximated by one-dimensional equations for simple stellar
interior models. It is a process that is difficult to model correctly,
but begins with certain assumptions.
1
 1 dP
The pressure scale height HP is defined as:

HP
P dr
r H P
or, if HP = constant,
P  P0 e
HP is the distance over which the gas pressure P decreases by a
factor of 1/e.
Example:
Estimate a typical value for the pressure scale height in the Sun.
Solution:
At the midpoint of the Sun we estimated ρ = 4M/3πR3.
But dP/dr = –GMρ/r2 ≈ –GM2/2R5 = –Pc/R.
So the pressure scale height can be estimated as:
HP = –P/dP/dr = –½Pc/(–Pc/R) = ½R
More typical values in the Sun are of order HP ≈ 0.1R.
Thermodynamics:
According to the first law of thermodynamics, the change in
internal energy of a mass element, per unit mass, is the difference
between the amount of heat added and the work done by the
element on its surroundings:
dU = dQ – dW.
The internal energy of the mass element, U, is:
internalenergy of the gas
U
unit mass
average energy # particles


particle
mass
1
 23 kT 
 23 nRT
 mH
where nR = k/μmH is the universal gas constant.
The change in heat of a gaseous mass element is usually expressed
in terms of the specific heat C of the gas, where C is the amount of
heat required to raise the temperature of a unit mass by 1K, i.e.:
dQ
CP 
at constantpressure
dT P
dQ
CV 
at constantvolume
dT V
The amount of work done per unit mass by a gas on its
surroundings is dW. The usual expression for work done is:
dW = PdV,
so
dU = dQ – PdV.
When the volume does not change, i.e., dV = 0, we have:
dQ
dU  dQ 
dT  CV dT
dT V
But:
dU  d  23 nRT  23 nR dT, so CV  23 nR
If the pressure is held constant instead so that the volume changes
as heat is added, then:

dQ
dV
dV
dU 
dT  P
dT   CP  P
dT P
dT P
dT

But
so:
and
Define:

 dT  CV dT
P
PV = nRT,
dV
P
 nR and
dT
CP  nR  CV
CP  CV  nR  23 nR  nR  25 nR
CP

CV
as the ratio of specific heats.
CP 25 nR 5

 3

CV 2 nR 3
For a monatomic gas:
When a gas undergoes ionization, some of the heat dU that would
normally increase the average kinetic energy of the gas is used
instead for ionization. The temperature of the gas therefore
increases less rapidly so dT is lower than otherwise. In such
instances:
dQ
CP 
is larger than for an ideal gas
dT P
dQ
CV 
dT V
so:
is also larger than for an ideal gas
CP

1
CV
in such instances.
The effect is particularly pronounced in stellar ionization zones.
For adiabatic processes there is no net heat flow into or out of a
mass element, so dQ = 0 and dU = –dW = –PdV. From PV = nRT
we have:
PdV + VdP = nRdT and
dU = CVdT, so:
dT 
dU  P dV

CV
CV
 nR 
and PdV  VdP  nRdT     PdV
 CV 
From the definition of specific heats we have:
nR CP  CV CP


1 1
CV
CV
CV
so:
PdV + VdP = –(γ–1)PdV
(1+γ–1)PdV = –VdP
That gives:
γPdV = –VdP
dV
dP


V
P
or:
or:
The result is the adiabatic gas law: PVγ = K, where K is a constant.
But the perfect gas law also gives: PV = nRT, so:
P V  nR T 



P  V  nR T 



PV
K

 P  K T
   1




nR
or P   1  
 K
 nR
, where K   
 K

 
T


1   1




The sound speed in a gas is related to the incompressibility and
inertia of the gas, specifically:
 dP 
 where B   V  dV ad
For adiabatic sound waves:
vs  B
is the bulk modulus.
 dP 
  P 
V 


V


  P
 dV ad
 V 
so : vs 
P

For convective energy transport one assumes that the gas bubbles
are adiabatic. Also the specific volume V = 1/ρ refers to the
volume per unit mass.
So if P = KV–γ , we must also have P = Kργ .
The formula for the pressure gradient in stellar interiors
therefore becomes:
dP
K  d
P d
  1 d
  K


dr
dr
 dr
 dr
With the perfect gas law:
so:
 kT
Pg  n kT 
 mH
dP  P d P d P dT



dr
 dr  dr T dr
If there is no gradient in mean molecular weight in a star (not
necessarily a valid assumption!), then:
dP P d P dT  P d



dr  dr T dr
 dr
from the adiabatic expression for P and ρ. The resulting equation
gives an expression for the adiabatic temperature gradient.

dT
T
P
d

T
K

 dP


Namely: 
   1
    1
P  dr
P   P dr
 dr  ad


or:
But:
 1 T K   dP 
1  T dP

1





P K  dr 
  P dr

  1 T dP   1  mH T
 dT 


 
 P dr

 kT
 dr ad
   1  mH GM ( r )


k
r2
  GM ( r )  


2
r


k
CP
 nR,  
, CP  CV  nR
 mH
CV
 CP


1

 1 C  C
P
V
so:    1   mH   CV



   k
 C
C
nR
P


P



C
V


 1  nR  1
1

  

 nR  CP  nR CP
The resulting temperature gradient is:
 1 GM ( r )  g
GM ( r )
 dT 

, where g 

 
2
CP
r2
 dr ad CP r
If:
dT
dT

dr actual
dr
temperature gradient is superadiabatic.
ad
The temperature gradient is a negative quantity since the
temperature decreases with increasing radius inside a star.
dT
dT
If:
heat will be transported by convection.

dr actual dr ad
Otherwise the heat is transported outwards by radiation. The
deep interiors of stars are relatively simple to understand. They
have either convective or radiative cores depending upon the
temperature gradient. They may also be semiconvective in the
region immediately outside a convective core. Stellar atmospheres
are more complicated since heat may be transported by both
methods.
The test for whether or not convective or radiative transport
describes the temperature gradient inside a star is therefore to
test whether or not a displaced bubble of gas rises or falls. It will
rise, i.e. convection applies if ρi(bubble) < ρi(surroundings).
The condition is:
dT
dT

dr actual
dr
ad
or:
T dP   

 
P dT    1 
or:
d ln P   

 
d ln T    1 
See textbook, also for mixing length model.
Stellar Models:
The complete set of differential equations describing the interiors
of stars is therefore:
Equation of Continuity:
dM ( r )
 4 r 2 
dr
Hydrostatic Equilibrium:
dP  G M ( r ) 

dr
r2
Energy Generation:
dL
 4 r 2  
dr
Temperature Gradient:
 3   Lr
 dT 

 
3
2
dr
4
ac
T
4

r

rad
 1 GM ( r )
 dT 

 
2
 dr ad CP r
Stellar Models:
The results for a series of stellar models at the start of hydrogen
burning, i.e. with their initial composition unchanged, is depicted
in the following figures. In each case solid circles denote models
with a solar metallicity (X = 0.73, Y = 0.25, Z = 0.02), whereas
open circles denote models of extremely low metallicity (X =
0.749, Y = 0.25, Z = 0.001). The actual metallicity of the Sun is
presently under debate as a result of newer models describing
turbulence in the solar atmosphere.
Models are constructed numerically
using the differential equations as
difference equations. They are also
rearranged so that the dependent
variable is mass rather than radius.
Models are then denoted by the
number of mass cells they contain.
Spherical symmetry is also assumed,
although that assumption is relaxed
in more recent models.
20
log Pc
19
18
17
16
M/Msun
15
0.0
5.0
10.0
15.0
20.0
3
log rhoc
2
1
M/Msun
0
0.0
5.0
10.0
15.0
20.0
9
8
log Tc
7
6
M/Msun
5
0.0
5.0
10.0
15.0
20.0
14
log R
13
12
11
10
M/Msun
9
0.0
5.0
10.0
15.0
20.0
7
6
log L/Lsun
5
4
3
2
1
0
log Teff
-1
-2
4.5
4.0
3.5
One of the basic tenets of stellar evolutionary models is the VogtRussell Theorem, which states that the mass and chemical
composition of a star, and in particular how the chemical
composition varies within the star, uniquely determine its radius,
luminosity, and internal structure, as well as its subsequent
evolution. A consequence of the theorem is that it is possible to
uniquely describe all of the parameters for a star simply from its
location in the Hertzsprung-Russell Diagram. There is no proof for
the theorem, and in fact, it does fail in some special instances.
A prime example of where ambiguities arise occurs when one
compares models for two stars, one of which is spherically
symmetric and the other of which is flattened as a result of rapid
rotation. Both stars can occupy the same location in the
Hertzsprung-Russell diagram, at different evolutionary ages and
even for different masses.
Insights into Stellar Evolution:
Consider the sequence of events that happens as stars evolve:
During hydrogen burning, the primary 41H → 4He reaction
converts two protons and two electrons into two neutrons. The
reaction chain may suggest that two positrons are produced, but
they quickly self-annihilate through collisions with electrons, so
the net result is as stated.
The production of α-particles from protons and electrons has two
effects: (i) the mean molecular weight μ of the gas increases
slightly, and (ii) the gas opacity, which is dominated by electron
scattering in stellar cores, decreases because of the depleted
abundance of electrons.
Consequence (i) implies that the gas
pressure must decrease, since the mass
density of the gas is unaffected.
 kT
Pg 
 mH
Consequence (ii) implies that radiation escapes the core of the
star more easily.
Since the gas pressure is reduced, the core of the star responds to
the pressure imbalance by contracting. But according to the
Virial Theorem, a contracting sphere of gas converts half of the
decreased potential energy of the system into kinetic energy, i.e.
heat, while the remainder escapes as radiation.
The consequence of an increased temperature for the gas at the
stellar core is an increase in the nuclear reaction rate, so, along
with the radiation increase induced by contraction of the core, the
luminosity of the stellar core increases sharply, raising the photon
flux from the interior.
The decreased electron scattering opacity of the
gas near the stellar core means that radiation
escapes more easily into the stellar envelope,
where the main sources of opacity (from atoms
and ions) exist. The increased photon flux on
the gas transfers more outwards-directed
momentum to the gas particles, resulting in
an outwards expansion of the envelope gases.
So, as the core of the star contracts, the envelope expands!
Stellar evolution thus results in an increased luminosity as the
star evolves. The resulting change in the effective temperature of
the star is more complicated, since increased luminosity can be
accommodated by an increase in the stellar radius or an increase
in the effective temperature. Since the radius of the star must
increase because of envelope expansion, it is not clear what will
happen to the star’s effective temperature.
Stellar evolutionary models indicate that, for massive stars, the
effective temperature decreases as the star evolves. For low mass
stars like the Sun, however, the effective temperature actually
increases during the initial stages of hydrogen burning.
The difference presumably originates from the differences in how
energy is transmitted outwards in the two types of stars. Massive
stars have convective cores and radiative envelopes, whereas lowmass stars like the Sun have radiative cores and convective
envelopes. Convection mixes gas so that any changes in chemical
composition are transmitted throughout the convective region,
which is clearly more important for high mass stars: M > 1¼ M.
Question. Air is mostly (80%) composed of nitrogen molecules,
each of which consists of 28 nucleons (protons and neutrons),
where the mass of one nucleon is about one atomic mass unit
(1.6605402 × 10–24 g). The radius of a typical nitrogen molecule is
about 1 Å, and the density of air at sea level is roughly 1.2 × 10–3 g
cm–3. Calculate the mean free path for collisions between air
molecules under such conditions. The air is at room temperature,
i.e. about 300K, which allows you to estimate the root-meansquared speed, vrms, for air molecules. From that information,
calculate the average time between collisions of atoms.
Solution:
For nitrogen, σ = π(2r)2 = π(2 × 10–8)2 cm2 ≈ π × 4 × 10–16 cm2
The number density is n = ρ/m(N2) = 1.2 × 10–3 g cm–3/(28 ×
1.6605402 × 10–24 g) = 2.581 × 1019 cm–3
So the mean free path is l = 1/nσλ = 1/(2.581 × 1019 × π × 4 × 10–16 )
≈ 3 × 10–5 cm
vRMS = (3kT/m)½ = (3 × 1.38 × 10–16 × 300/4.65 × 10–23)½
≈ 5.17 × 104 cm/s
The average time between collisions is t = l/v = (1.25 × 10–4
cm)/(5.17 × 104 cm/s) ≈ 2 × 10–9 s
Question. According to a “standard model” for the Sun, the
central density is 162 g cm–3 and the Rosseland mean opacity κ is
1.16 cm2 g–1.
a. Calculate the mean free path for a photon at the centre of the
Sun.
b. If the mean free path remains constant for the photon’s
journey to the surface of the Sun, how long, on average, would it
take for photons to escape from the Sun?
Solution:
The mean free path for a photon is l = 1/κλρ = 1/(1.16 × 162) ≈ 5 ×
10–3 cm
The distance traveled by a photon between collisions is d2 = Nl2
In order to escape from the Sun a photon has to travel a distance
of 6.9598 × 1010 cm
Number of collisions is N = d2/l2 = (6.9598 × 1010)2/(5 × 10–3)2 =
1.9376 × 1026, each taking t = l/c = 5 × 10–3 cm/3 × 1010 cm/s = 1.7
× 10–13 s
Time to escape the Sun is Nt = 1.9376 × 1026 × 1.7 × 10–13 s) = 3.23
× 1013 s ≈ 106 years