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Chapter 1
Introduction
1.1
What is a star?
Stars are the building blocks of the visible Universe, so it is important to understand their structure and evolution. Why do they glow? How did they formed?
What would they become after? Before we get into these, let us give a clear
definition of stars. Roughly speaking, we define a star to be a gravitationally
bounded object which has a nuclear energy source in the form of fusion in the
core. Thus, Jupiter and Comet Halley are not stars. But this definition is not
sufficient in this level of complexity because it also includes brown dwarf in the
family of stars.
Recall that brown dwarfs are objects that are too massive to be planets but too
light to be stars. More precisely, in this context, a planet is a self-gravitating
body without any nuclear fusion in the core although it may have other means of
energy generation (such as radioactive decay). Thus Jupiter is a planet although
Jupiter has an internal heating mechanism. (Do you know what internal heating
mechanism it has?) A brown dwarf, in contrast, is a self-gravitating object that
burns (or has burnt) D, Li, Be and B via nuclear fusion. Since the amount of D,
Li, Be and B in such object is small, such nuclear burning cannot sustain over a
long time. More importantly, the energy released in such a burning alone is not
sufficient to stop the brown dwarf from further gravitational collapse. Thus,
astronomers naturally want to exclude brown dwarfs from the rank of stars.
(We shall study brown dwarf in detail as a special topic if we have time.)
In addition, the above crude definition of star also excludes certain types of
objects which burn nuclear fuel in a shell rather than at the center.
1
CHAPTER 1. INTRODUCTION
2
Hence, we use the following more refined definition of star:
A star is a gravitationally bounded object which has a sustainable
nuclear fusion energy source in the interior.
Using this refined definition, brown dwarf is not a star. Neutron star is not a star
either. Red giant is a star for it has a hydrogen-burning shell although its core
is not burning any nuclear fuel. An self-gravitating object whose sustainable
nuclear fusion energy source in the interior no longer exists is called a dead star.
Thus, neutron star is a dead star. So in the straightest sense, a neutron star is
not a star even though it was a star.
Note that the above definition says nothing about the shape of a star. In fact,
stars may be spherical, ellipsoidal, egg-shaped and so on.
1.2
An introduction to the cgs units
The cgs units mean cm, g, and second.
Length:
in cm =
m
Mass:
in g =
Time:
in s
Energy:
in erg =
J (hence computing mass and velocity in cgs units,
the corresponding energy will automatically be in erg)
Force:
in dyn =
N (hence computing the mass and acceleration in cgs
units, the corresponding force will automatically be in dyn)
kg
Pressure: ?
Power:
in erg s−1
Charge:
in the electrostatic unit of charge (esu), also known as the statcoulomb, so that Coulomb’s law becomes F = q1 q2 /r2 .
CHAPTER 1. INTRODUCTION
1.3
3
Some ideas of the physical system we are
talking about
• Mass: Ranges from about 0.1 to 30 solar masses for most living stars.
The symbol for solar mass is M⊙ . In fact, 1M⊙ ≈ 2 × 1033 g, which is
about 1000 times more massive than Jupiter, or about 3 × 105 times more
massive than the Earth.
• Radius: Ranges from about 0.1 to 1000 solar radii for most living stars.
The symbol for solar radius is R⊙ . Actually, 1R⊙ ≈ 7 × 1010 cm, which is
about 100 times that of the Earth.
• Radiation: Ranges from about 0.01 to 106 solar luminosity for most living
stars. The symbol for solar luminosity is L⊙ and 1 L⊙ ≈ 3.8 × 1033 erg s−1 .
(We shall talk more about luminosity and radiation later in this course.)
• Chemical composition: Most young stars are made up of primarily
H and He plus a trace amount (which is about a few percent by mass)
of heavy elements. In astronomy, heavy elements (or some books call
them by the misleading term “heavy metals”) are defined as elements
with atomic number greater than 2.
• Surface temperature: Ranges from about 4000 K to 30000 K for most
main sequence stars.
• Magnetic field strength: Of the order of 1 G over most of the solar
surface (although it can go up to about 3000 G in sunspots). But some
living stars may have field strength as high as several thousand G.
• Rotation: Observation suggests that most stars rotate rather slowly. As
for the Sun, the rotational period is about 1 month.
1.4
The Hertzsprung-Russell diagram
As the name suggests, this diagram was first studied by E. Hertzsprung and
H. N. Russell in the 1910’s. In modern language, Hertzsprung-Russell (H-R)
diagram is a plot (usually a log-log or a semi-log one) of surface temperature
vs. luminosity of stars. Conventionally, the luminosity is plotted in the “usual”
direction while the surface temperature is plotted in the “reverse” direction.
That is, luminosity increases as we move upward while temperature decreases
as we move rightward in the H-R diagram.
CHAPTER 1. INTRODUCTION
4
106
Supergiants
Luminosity (L )
104
102
Gaints
Ma
1
in
Se
q
ue
−2
nc
e
10
White Dwarfs
10−4
40000
20000
10000
5000
2500
Temperature (K)
Figure 1.1: The H-R diagram (schematic)
H-R diagrams for stars in star clusters give strong evidence for stellar evolution.
Recall that a star cluster consists of a collection of gravitationally bounded stars
within a space of radius of the order of 100 light years. Thus, we can safely
assume that stars in a single star cluster are formed from roughly the same
material at about the same time under roughly the same physical conditions.
Thus, mass of a star is perhaps the most important parameter to determine the
properties and evolution of that star within the same star cluster.
It turns out that stars are not located randomly in the H-R diagram; they
cluster around a few regions in the diagram instead. (See figure 1.1.) Besides,
these regions differ from one star cluster to the next.
It is reasonable to assume that the chemical composition for different stellar
clusters just before their formation is more or less the same. Thus, the difference
between the location of stars of different clusters on the H-R diagram should be
mostly determined by their age. It is rather hard to determine the absolute age
of a star cluster. (By absolute age, we mean how old the stars in that cluster.)
Nevertheless, it is quite easy to determine the approximate age of star clusters
since old star clusters tend to have condensed core. (Do you know why?)
Astronomers find that the location of stars in the H-R diagram depends strongly
on the degree of core condensation and hence the age of the corresponding star
cluster. (See figure 1.2.) Most stars in a young cluster locate in a band called
CHAPTER 1. INTRODUCTION
5
Figure 1.2: A comparison of H-R diagram between young cluster and old cluster
the main sequence. Consequently, we call a star lying on the main sequence a
main sequence star. For older clusters, a significant number of stars occupies
the red giant branch. This is a strong evidence that stars evolve. More importantly, there is a turnoff point on the main sequence above which stars have
been evolved away from the main sequence. This suggests that different stars
on the main sequence evolve at different rates. Specifically, a hot and bright
main sequence star evolves faster than a cold and dim one. In fact, nowadays,
astronomers turn things around and use the location of the turnoff point on the
H-R diagram to determine the age of a star cluster more precisely.
1.5
Mass-luminosity relationship
For those main sequence stars with known masses, A. Eddington found a relation
between the mass and the luminosity in 1924. (Do you know how they find the
mass of some main sequence stars?) The relationship goes like L ≈ M α . (See
figure 1.3.) This is an approximate relationship. Actually, after Eddington,
different people put slightly different values for α. The value of α is about
3 to 5 over the whole range of stellar masses M . Yet, a closer look reveals
that the exponent α changes for different values of M . Specifically, α is about
< 0.5M
>
2.5 to 3.0 for M ∼
⊙ or when M ∼ 2.5M⊙ while α is about 3.5 to 4.5
< M < 2.5M . Any theory of stellar structure must be able to
for 0.5M⊙ ∼
⊙
∼
reproduce the mass-luminosity relationship.
CHAPTER 1. INTRODUCTION
6
Luminosity (L )
106
104
α
2
10
1
α
α
M3
M4
M3
10−2
0.1
1
10
Mass (M )
Figure 1.3: The mass-luminosity relationship of main-sequence stars.
1.6
Assumptions in our study of stellar structure and evolution
1. The star is isolated. Although most stars in the universe are not isolated, it makes sense to study the structure and evolution of an isolated
star before going on to study the more complicated multiple stellar systems.
2. Right at the moment of formation of a star, its chemical composition is uniform. Why this is a good approximation? I shall further
discuss the validity of this assumption when we study main sequence stars.
3. The star is irrotational and is spherically symmetric. The rotational energy of a star is of order of M R2 ω 2 while the gravitational
potential energy of a star is about GM 2 /R. Hence, the ratio of rotational energy to gravitational energy is about R3 ω 2 /GM ≈ 2 × 10−5 R3 /M
where M and R are expressed in units of solar mass and solar radius,
respectively. This ratio is, therefore, very small for most stars. Thus,
rotational distortion can be safely neglected. In a similar way, the ratio of magnetic energy to the gravitational potential energy of a star is
< 10−11 R4 /M 2 where M and R are in units of solar mass
B 2 R4 /8πGM 2 ∼
and solar radius, respectively. Thus, the magnetic energy is very small
CHAPTER 1. INTRODUCTION
7
compared with the gravitational potential energy. Hence, both rotation
and magnetic field for almost all stars do not significantly break their
spherical symmetry. Therefore, a spherically symmetric star serves as a
very good approximation in the study of most isolated stars.
4. Mass loss or mass gain of a star is neglected. A star, like our Sun,
may loss its mass through ejection of stellar wind. Some stars may gain
mass by accretion. These processes are important in the study of certain
stars such as white dwarfs, AGB stars and binary stars. Nonetheless,
the mass change rate for most main sequence stars is small and can be
neglected.
5. The star is in local thermodynamic equilibrium (LTE). In other
words, every microscopically large but macroscopically small region is very
close to and hence can be well approximated by a region in thermodynamic
equilibrium. Thus, it makes sense to talk about thermodynamic variables
such as temperature, density and pressure in these regions. This approximation greatly simplifies our analysis and serves as the starting point to
the study of more refined calculations. How good these approximations
are in the study of main sequence stars? To answer this question, let
me tell you that the photon mean free path near the core of our Sun is
about a few cm, which is very short compared with the radius of the Sun.
The mean free path for electron and nucleus is even shorter near the solar
core. The typical equilibration time for photons, nuclei and electrons in
the solar core is also very short compared with, say, the lifetime of the
Sun. (I will justify these claims later on in this course.) Hence, LTE is
a very good approximation for studying the overall structure of the Sun
and most other stars. In contrast, LTE is no longer a very good approximation to study stellar atmosphere such as the chromosphere of our Sun.
(Do you know why?) Fortunately, the atmosphere of many main sequence
stars has a small mass compared with the star itself. Therefore, LTE still
serves as a very good simplification in our first course of stellar structure.
(Another exception is when we study neutrinos emitted by a star. Do you
know why?)
Consequence of our assumptions
Since we assume that a star is spherically symmetric, we can forget about its
angular velocity as well as the magnetic field strength. Thus, we may characterize the properties of a star if we know the following: density ρ, pressure P ,
temperature T and chemical composition of the star as functions of distance
from the stellar core r and time since the birth of the star t.
Chapter 2
Stellar Structure Equations
To understand the stellar structure, we need to determine many physical parameters inside a star, including mass / density, temperature, pressure, and energy
production rate, by solving the following equations invoking conservation laws.
Remember that it is important to specify the boundary conditions otherwise a
differential equation cannot be solved.
2.1
Mass conservation equation
For a spherically symmetric star, let r be the distance from the star center
and m = m(r) be the mass enclosed by the sphere of radius r centered at the
center of the star. (Note: Some authors use the notation M (r) instead of m(r).)
Clearly, m(0) = 0 and m(R) = M where M and R denote the mass and the
radius of the star. More importantly, due to spherical symmetry,
∫
r
4πr2 ρ(r) dr = m(r) ,
(2.1)
0
where ρ(r) denotes the density of the star at radius r. (Obviously, ρ(r) > 0 for
0 ≤ r < R and ρ(r) = 0 for r > R.)
Alternatively, Eq. (2.1) can be rewritten as the following differential form
dm
=
dr
.
(2.2)
To summarize, conservation of mass relates ρ and m. So, we have introduced
one equation and yet one new unknown. So, we have to find out more equations
relating ρ, P , T , m, etc, before we can solve the problem of stellar structure.
8
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
2.2
9
Momentum conservation and the equation
of hydrostatic equilibrium
The next equation we introduce comes from the conservation of momentum; or
more precisely, in LTE the net force acting on any microscopically large but
macroscopically small region is zero. Inside a star in LTE, the forces involve
are gravity and pressure gradient due to gas or photon. Specifically,
dP
Gm(r)ρ(r)
=−
.
dr
r2
(2.3)
We denote the central pressure P (0) by Pc . And clearly, P (R) can be approximated by 0.
Using Eq. (2.2), Eq. (2.3) can be rewritten as
dP
=
dm
.
(2.4)
Exercise: Using the equation above, show that
Pc >
GM 2
8πR4
(2.5)
in any star. (Hint: r(m) < R.)
In particular, this inequality tells us that the central pressure of our Sun is at
least 4.4 × 1014 dyn cm−2 ≈ 4 × 108 atm.
2.3
Virial theorem
Multiplying Eq. (2.4) by the volume V (r) = 4πr3 /3, we have
∫
P (R)
P (0)
1 ∫ M Gm(r)
V (r) dP = −
dm .
3 0
r
(2.6)
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
10
Note that the R.H.S. above is Ω/3 where Ω is the gravitational potential energy
of the entire ∫star. Using integration by parts, the L.H.S. above equals
∫ V (R)
− 0
P dV = − 0M P/ρ dm. Hence, we arrive at
Ω = −3
∫
M
0
P
dm .
ρ
(2.7)
This equation is one form of the virial theorem (sometimes known as the
global form of the virial theorem).
(An alternative form of the virial theorem, sometimes called the local form of
the virial theorem, is obtained by integrating from r = 0 to r = Rs (< R). The
result is
∫ Ms
P
Ωs
Ps Vs −
dm =
,
(2.8)
ρ
3
0
where Ωs is the gravitational potential energy for the sphere centered at the
star and with radius Rs .)
Exercise: It can be shown that the gravitational potential energy of a star
2
Ω < − GM
. From the ideal gas law P = nkT = ρkT /µmA , where µ is the mean
2R
molecular weight and mA is the unit atomic mass, and the virial theorem, show
that the average temperature
1 ∫M
GM µmA
T̄ ≡
.
T dm >
M 0
6kR
(2.9)
T̄⊙ = (4 × 106 µ)K
(2.10)
For the Sun, show that
Exercise: For non-relativistic classical ideal gas, P = nkT where n is the number
density of the gas; and the average K.E. per molecule is 3kT /2. Show that
2U + Ω = 0 where U is the thermal energy of a star.
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
11
Consequently, the total energy of the star is U + Ω < 0. In other words, the
star is in a bound state.
One interesting consequence of the virial theorem is that upon gravitational
contraction, the star becomes hotter, more tightly bound and has to radiate
some energy to space. It is kind of an analog to a system with negative heat
capacity.
In contrast, if the star is made up of extremely relativistic particles, K.E. per
particle is 3kT , i.e., the pressure equals one-third the energy density. Hence, in
this case, virial theorem implies that U + Ω = 0. In other words, a star making
up of extremely relativistic particles can be in LTE only when its total energy
is 0. So, such a star is not bounded.
Another consequence of the virial theorem concerns the conversation law. Consider a star making up of non-relativistic classical ideal gas. Suppose further
that the timescale concerned is small enough that the total energy of the star
is roughly a constant. Since 2U + Ω = 0, the total energy of the star can be
expressed as a function of U (or Ω) alone. Thus, the thermal energy and the
gravitational potential energy of the star are also conserved. In turns out that
this consequence of the virial theorem is useful to understand several properties
of late stage stellar evolution.
2.4
Simple stellar models
So far, we have 3 unknowns, namely, m(r), ρ(r) and P (r) but only two equations, namely, the mass conservation equation and equation of hydrostatic equilibrium. So, we need another equation to relate m, ρ and P . Astronomers have
proposed a number of extremely simplified equations to fulfill this task. Since all
these proposed equations are not derived from first principle, they are nothing
more than crude approximations of the realistic situation. Yet, these approximations are sometimes quite useful to investigate the approximate structure of
a star.
The first such model is called constant density model, namely, we assume ρ is
a constant inside the entire star. The second one is called the linear density
model, namely, ρ(r) = ρc (1 − r/R) where ρc is a constant.
Exercise: Solve m(r) and P (r) in the above two models.
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
2.5
12
Polytrope model
The last such model I am going to introduce, which is the most important
model of this kind, is called the polytrope model. The is motivated by adiabatic
expansion of ideal gas,
P V γ = constant,
(2.11)
where γ is the adiabatic index (or the heat capacity ratio). We therefore
assume
P = Kργ ,
(2.12)
for some constants K. In the classical case, it can be shown that γ is related to
the degrees of freedom f by γ = 1 + 2/f , i.e. γ = 5/3 for monatomic gas and
γ = 7/5 for diatomic gas. As we will show later in this chapter, Eq. 2.12 also
applies to a photon gas, classical and relativistic degenerate gas with different
values of γ.
Exercise: Combining the mass conservation and hydrostatic equilibrium equations, show that
(
)
1 d r2 dP
(2.13)
= −4Gπρ .
r2 dr ρ dr
Exercise: Write γ = 1 + 1/n, then the above equation can be written as
[
]
(n + 1)K d 2 (1−n)/n dρ
r ρ
= −ρ .
4πGnr2 dr
dr
(2.14)
The boundary conditions for this differential equations are ρ(R) = 0 and
dρ(0)/dr = 0. (Do you know why?)
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
13
Exercise: Let ρ = ρc θn where ρc is the central density of the star, show that
(
1 d
2 dθ
ξ
ξ 2 dξ
dξ
)
= −θn ,
(2.15)
where r = [(n + 1)Kρc(1−n)/n /4πG]1/2 ξ. The corresponding boundary conditions
are θ = 1 and dθ/dξ = 0 at ξ = 0. This equation is called Lane-Emden
Equation and can be solved numerically.
The significance of the polytrope model is that Lane-Emden equation is independent of the mass M , radius R and central density ρc of a star. So, once you
have numerically solved the Lane-Emden equation for a given value of n, the
numerical solution can be used to deduce the solution of any star with the same
polytropic index n (or γ).
For the special cases of n = 0, n = 1 and n = 5, Lane-Emden equation can be
solved exactly.
Exercise: For n = 0, show that θ(ξ) = 1 − ξ 2 /6 and
√
ρ = ρc whenever ξ ̸= 6 .
(2.16)
The case of n = 1, Lane-Emden equation can be rewritten as
d2
(ξθ) = −ξθ
dξ 2
(2.17)
whose solution that matches the correct boundary conditions above is θ(ξ) =
sin ξ/ξ. Hence,
sin ξ
ρ = ρc
.
(2.18)
ξ
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
14
Exercise: Show that Eq. (2.18) is a solution to Eq. (2.17).
For the case of n = 5, Lane-Emden equation can be rewritten as
z(1 − z 4 )
d2 z
=
,
dt2
4
(2.19)
where ξ = e−t and θ = z/(2ξ)1/2 . Multiplying Eq. (2.19) by dz/dt, we obtain
1
2
(
dz
dt
)2
1
1
= z2 − z6 + C .
8
24
(2.20)
Boundary conditions demand that C = 0 and hence
(
dz
z4
z
1−
=−
dt
2
3
)1/2
.
(2.21)
After making the substitution z 4 /3 = sin2 ζ and upon integration, we have
e−t = C ′ tan(ζ/2) where C ′ is a constant of integration. Applying the boundary
conditions and after simplification, we obtain
(
ρ = ρc
ξ2
1+
3
)−5/2
.
(2.22)
(Could you express ρc as a function of M and R? Besides, could you express ρ
and P as a function of r?)
It is straightforward to check that the smallest positive
root for the solution
√
θ(ξ) of the Lane-Emden equation equals ξ = ξ1 ≡ 6 and ξ = ξ1 ≡ π when
n = 0 and 1 respectively. This value of ξ1 corresponds to the boundary of a
star. On the other hand, θ ̸= 0 for any real-valued ξ for n = 5. In other words,
the solution of the Lane-Emden equation for n = 5 does not correspond to a
physical star.
Exerxise: From the definition of ξ, show that
[
(n + 1)K
R=
4πG
]1/2
ξ1 ,
ρ(1−n)/2n
c
(2.23)
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
15
where ξ1 is the smallest positive value of ξ having θ(ξ) = 0.
By substituting θ and ξ in the Lane-Emden equation back our equation of mass
conservation and equation of hydrostatic equilibrium, we obtain
[
(n + 1)K
M = 4π
4πG
[
]3/2
ρ(3−n)/2n
c

(
GM 2 
dθ
Pc =
4π(n
+
1)
R4
dξ
[
ρ̄ = ρc
dθ
−ξ 2
dξ
)2 −1  ]
3 dθ −
ξ dξ ]
,
(2.24)
ξ=ξ1
,
(2.25)
ξ=ξ1
,
(2.26)
ξ=ξ1
and
Ω=−
3 GM 2
.
5−n R
(2.27)
Question: What physical quantities do θ and ξ correspond to?
Note that R is independent of ρc for n = 1, Ω > 0 for n > 5, Ω is undefined for
n = 5. Therefore, the polytropic models for n = 1 and n ≥ 5 are not physical.
(Can you verify that M is finite, Pc is infinite and ρ̄ is 0 for case of n = 5?)
Interestingly, for n = 3, the mass of the star is independent of its central
density. This particular polytropic index is sometimes also called the Eddington
standard model. (We shall say more about the Eddington standard model later
in the next chapter.)
2.6
Equation of state
We have used the conservation of mass, energy and momentum to construct
three equations for stellar structure. Yet, we have P , T , m, ρ, L, ϵ, r as well as
the chemical composition as our variables. To further our analysis, we have to
look for equations that tell us specific properties of the material in a star. One
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
16
such equation is the equation of state (EOS), namely, an equation expressing
the pressure P as a function of T , ρ and chemical compositions.
The polytrope model assumes adiabatic ideal gas. To model a star, astrophysicists may use some extremely accurate and hence complicated EOSs. I shall not
do so in this introductory course for the following reason. Although those EOSs
more accurately model the material behavior, their predictions on the interior
structure of a star are in most cases not significantly different from the simple
EOS we will use in this course. The simplest EOS, which at the same time
turns out to be the one we will use in this course, is the ideal gas law that you
have learned in high school. That is, P = nkT where n is the number density
of gas particles.
In the rest of this chapter, I will give a brief review of statistical mechanics,
then derive Saha equation to show that the interior of a star is mostly ionized.
This means that we will need to introduce mean molecular weight and consider
ion, electron, and photon pressure in the EOS.
2.7
Brief review of statistical mechanics
Ideal gas law is a good approximation to describe the properties of matter in a
star. To see why, I first have to introduce the Saha equation discovered by M.
Saha in 1920.
Recall from the course Statistical Mechanics and Thermodynamics that for a
particle (or system of particles), the probability of a particle is in state s is
proportional to gs e−Es /kT where gs is the degeneracy of state s and Es is the
energy of state s relative to some fixed reference energy level (sometimes taken
to be the ground state). More importantly, the proportionality constant is the
same for all the states. Thus, the ratio of the probability that the particle is in
state (s + 1) to the probability that it is in state s is given by the Boltzmann
formula
gs+1 −(Es+1 −Es )/kT
e
.
(2.28)
gs
Exercise: At what temperature a gas of neutral hydrogen will have equal number
of atoms in the ground and first excited states? (Hint: for hydrogen atoms, the
degeneracy is gn = 2n2 , i.e., the ground state is n = 1 and the first excited state
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
17
is n = 2.)
This is even higher than the surface of the Sun. However, we know from observations that some hydrogen atoms are ionized in the Sun. How can that be
possible? (The answer lies in the Saha Equations, which will be discussed later
in this chapter.)
The Statistical Mechanics and Thermodynamics course also tells us that the
thermodynamic properties of a system with fixed number of particles N and
fixed temperature T and fixed volume V is encoded in the so-called partition
function
(
)
∑
Ei
gi exp −
Z≡
(2.29)
,
kT
i
where the sum is over all possible energy states of the system. Furthermore, the
ratio between the mean number of particles Ns in state s and the total number
∑
of particles N ≡ Nj in a sample is given by
Ns
gs e−Es /kT
gs e−Es /kT
≡
=∑
.
−Ej /kT
N
Z
j gj e
(2.30)
For an ideal classical non-relativistic gas particle with ground state energy E0
and degeneracy g, its partition function is given by
g ∫ ∫ −(E0 +p2 /2m)/kT 3 3
Z(1, V, T ) =
e
d ⃗x d p⃗
(2πh̄)3
∫ +∞
gV
−E0 /kT
2 −p2 /2mkT
=
e
4πp
e
dp
(2πh̄)3
0
gV −E0 /kT
e
(2.31)
=
λ3
where m is the mass of the particle and
√
λ ≡ h̄
2π
.
mkT
(2.32)
Note that we have assumed that the momentum of the gas particle is isotropically distributed in order to obtain the above expression for Z(1, V, T ). The
partition function of a system of N indistinguishable classical non-relativistic
ideal gas particles each with ground state energy E0 and degeneracy g is then
given by
Z(1, V, T )N
Z(N, V, T ) =
.
(2.33)
N!
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
18
Recall again from the Statistical Mechanics and Thermodynamics course that
the most convenient method to study the situation in which the particle number
in the system may change is to use the so-called grand partition function
(
+∞
∑
µN
Q(µ, V, T ) ≡
Z(N, V, T ) exp
kT
N =0
)
,
(2.34)
where µ is the chemical potential. In the thermodynamic limit, µ can be interpreted as the work needed to add one extra particle to the system at constant
V and T . The grand partition function encodes the thermodynamical information of a system with fixed chemical potential, volume and temperature. Since
∑
n
ex = ∞
0 x /n!, the grand partition function of an ideal classical gas in the
non-relativistic limit equals
Q(µ, V, T ) = exp
[
gV (µ−E0 )/kT
e
λ3
]
.
(2.35)
More generally, the grand partition function of a system of interacting particles
of various different species such that each species can be approximated by a
classical non-relativistic ideal gas is given by
Q({µs }, V, T ) = exp
[
∑ gs V
λ3s
s
]
e(µs −E0,s )/kT
,
(2.36)
where the s is the species label. Hence, the corresponding grand potential
equals
G({µs }, V, T ) ≡ −kT ln Q = −kT
∑ gs V
s
λ3s
(
µs − E0,s
exp
kT
)
.
Note that G is minimized in chemical equilibrium. Since G = E −T S −
we have
)
(
∂G µs − E0,s
gs V
Ns = −
= 3 exp
.
∂µs T,V,{µi }i̸=s
λs
kT
(2.37)
∑
s
µs Ns ,
(2.38)
By rearranging terms, the above equation becomes
(
Ns λ3s
µs = kT ln
gs V
)
(
+ E0,s
ns λ3s
= kT ln
gs
)
+ E0,s ,
(2.39)
where ns is the number density of species s.
2.8
Saha equation
Now consider a specific atomic species and use the label s, 0 to denote its sth
ionized state in its ground level. For example, the label 0, 0 of H refers to the
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
19
state of hydrogen atom with its electron in the lowest energy level. And I use
the label e to denote electron. Clearly, at non-zero temperature, reactions such
as
Xs+ ⇀
(2.40)
↽ X(s+1)+ + e−
may occur, where X denotes the atomic species. Hence, at thermodynamic
equilibrium,
µs,0 = µs+1,0 + µe .
(2.41)
Exercise: From Eq. (2.39) and the fact that E0,(s+1,0) + E0,e − E0,(s,0) = χs+1 ,
the (s + 1)th ionization potential of the atomic species, show that
ns+1,0 ne λ3s+1,0 λ3e gs,0
= e−χs+1 /kT .
3
ns,0 λs,0 gs+1,0 ge
(2.42)
From Eq. (2.32), ge = 2 and the approximation that the masses of the sth and
the (s + 1)th ionized atom are about the same, the above equation becomes
ns+1,0
gs+1,0
=
fs+1 (T ) ,
ns,0
gs,0 ne
where
(2.43)
(
2(2πme kT )3/2
χs+1
fs+1 (T ) =
exp −
3
h
kT
)
,
(2.44)
In general, not all the sth ionized atoms in a thermalized system are in the
ground state. From Eq. (2.30), Eq. (2.43) can be re-written as
Zs+1
ns+1
=
fs+1 (T ) ,
ns
Zs ne
where
Zs =
∑
i
[
−(Es,i − Es,0 )
gs,i exp
kT
(2.45)
]
(2.46)
is the partition function of those sth ionized atoms in various energy levels.
Eqs. (2.43) and (2.45) are known as the Saha equation. The Saha equation
is useful in astronomy. For instance, if we know the electron partial pressure
Pe , then the electron number density ne can be well approximated by Pe /kT in
many cases. We now apply Saha equation to estimate the degree of ionization
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
20
of our Sun. For simplicity, we consider only the form of Saha equation in
Eq. (2.43).
Exercise: The mean density of our Sun is about 1.4 g cm−3 . By virial theorem,
the mean temperature of our Sun is about 6 × 106 K. We simplify our analysis
by assuming that the Sun is made up of entirely hydrogen. Thus, χ1 = 13.6 eV.
Let us define the degree of ionization by n1 /(n0 + n1 ) ≡ x. Show that Saha
equation demands that
n1
x
f1 (T )
mH f1 (T )
=
≈
=
(2.47)
n0
1−x
ne
ρx
and hence x ≈ 98.8%.
From the result above, almost all hydrogen in our Sun is ionized. By the same
token, it is straight-forward to check that most atoms in a main sequence star
are ionized. On the other hand, in the solar atmosphere where T ≈ 6000 K and
ne ≈ 1013 cm−3 . Saha equation tells us that the degree of ionization of hydrogen
is about 10−4 . (What about other atoms, such as Na, in the atmosphere, and
He and other heavy elements in the core?
Note however that Saha equation has its limitations. It works for thermalized
region of a star. Hence, it may not be applicable to stellar atmosphere. Also,
it assumes classical non-relativistic ideal gas, so it may not work near the core
of some stars.
2.9
Gas law
From now on, we model the EOS of a star by ideal gas law. More precisely,
we assume that the gas pressure of a star is given by Pgas = nkT . Since the
temperature and luminosity of a star are high, we conclude that electrons,
cations (in the form of partially or completely ionized atoms), and photons are
the three most important constituents of a star. The electron pressure is given
by
Pe = ne kT
(2.48)
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
21
where ne is the electron number density, and the cation pressure is given by
Pi = ni kT
(2.49)
where ni is the cation number density.
What are the values of ne and ni ? To answer this question, we denote the
mass fraction of H, He and “heavy elements” (that is, those elements heavier
than H and He) in a star by X, Y and Z = 1 − X − Y respectively. (More
precisely, X and Y are the mass fraction of 1 H and 4 He respectively. Do you
know why we can forget about the contribution of 2 H, 3 H, 3 He?) Then, number
density of hydrogen and helium nuclei equal Xρ/mH and Y ρ/4mH , respectively.
Therefore,
(
)
ρ
Y
Z
ni =
X+ +
,
(2.50)
mH
4
⟨A⟩
where ⟨A⟩ denotes the average mass number of heavy elements in a star.
We introduce the mean molecular weight µ, which is defined as the average
weight of a particle in units of hydrogen mass, i.e.
µ≡
such that
µmH = ⟨m⟩ =
⟨m⟩
mH
(2.51)
total mass of gas
.
total number of particles
(2.52)
This gives
nµmH = n⟨m⟩ = ρ
ρ
n =
µmH
(2.53)
A direct comparison with Eq. (2.50) gives
(
Y
Z
1
= X+ +
µi
4
⟨A⟩
)
.
(2.54)
Next, we consider electrons. By virial theorem, we know that the thermal energy
of a star is of the order of GM 2 /R. That is,
GM 2 /R ∼ k T̄ M/mH ,
(2.55)
where T̄ ≈ 106 K is the average temperature of a star. Hence, for most part
of a star, the temperature is so high that most atoms are completely ionized.
(As we have already seen, this is not a valid assumption for stellar atmospheres.
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
22
Nonetheless, this assumption has little effect in determining the overall structure
of a star. Similarly, this assumption is not valid for high atomic number atoms,
but their numbers are small compared to that of hydrogen and helium for a
typical main sequence star.)
Similar to Eq. (2.50) above, but a hydrogen atom gives one electron, an helium
atom gives two, and a heavy element atom with atomic number Z gives Z
electrons. Hence, the total number of electrons per unit volume is
(
⟨
ρ
Y
Z
ne =
X +2 +Z
mH
4
A
⟩)
=
ρ
.
mH µe
(2.56)
(Do not confused the mass fraction of heavy elements Z with Z.)
For fully ionized gas, the total gas pressure law is contributed by ions and
electrons
Ptot = Pi + Pe = ni kT + ne kT
(
)
ρ
1
1
ρkT
=
+
kT ≡
.
mH µi µe
µmH
(2.57)
(2.58)
Here we define
1
1
1
≡
+
.
µ
µI µe
From Eqs. (2.54) and (2.56), we have
⟨
1
Z +1
3
= 2X + Y + Z
µ
4
A
(2.59)
⟩
(2.60)
for fully ionized gas. As most nuclei contain about the same number of protons
and neutrons, the last term can be approximated by Z/2 and this can be further
reduced to
1
3
Z
≈ 2X + Y +
µ
4
2
3
1
= 2X + Y + (1 − X − Y ) .
(2.61)
4
2
Exercises:
1. What is µ for fully ionized hydrogen?
2. What about fully ionized pure helium gas?
3. For Solar abundances, X = 0.747, Y = 0.236, and Z = 0.017, what is µ
for neutral gas? For fully ionized gas, show that µ ≈ 0.6. Hence,
Ptot =
ρkT
.
0.61mH
(2.62)
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
23
Question: Why can the mean molecular weight be smaller than 1?
2.10
Photon pressure
We have not finished the discussion on EOS yet since we leave out radiation
pressure. Recall from the course Statistical Mechanics and Thermodynamics
that photons are bosons. Therefore, in thermal equilibrium, photons obey
Bose-Einstein statistics. That is, the distribution of photons is isotropic and
the number density of photons with frequencies between ν and ν + ∆ν equals
8πν 2
dν
n(ν) dν = 3
.
c exp(hν/kT ) − 1
(2.63)
To see why photon number density follows Eq. (2.63), one recall that the energy
of having s photons each with momentum p⃗ equals spc. As photon is bosonic,
s can take on any natural number. Moreover, the probability P (s) that there
are exactly s photons each with momentum p⃗ follows the constraint
P (s + 1)
= e−pc/kT = e−hν/kT ,
P (s)
(2.64)
where ν is the frequency of the photon. Consequently,
e−shν/kT
P (s) = ∑+∞ −s′ hν/kT = e−shν/kT (1 − e−hν/kT )
s′ =0 e
(2.65)
provided that p⃗ is an allowed momentum of a photon. Thus, the expected
number of photons with an allowed momentum p⃗ is given by
⟨Nγ (⃗p)⟩ = 2
+∞
∑
s=0
sP (s) =
2
ehν/kT
−1
.
(2.66)
(Note that the 2 above reflects the fact that a photon has two possible polarizations.) As a result, the total expected number of photons equals
1 ∫ ∫
⟨Nγ (⃗x)⟩ dV d3 p⃗
h3 ∫
∫
1
2
= 3 dV
d3 p⃗
hν/kT
h
e
−1
∫
8πν 2
= V
dν .
c3 (ehν/kT − 1)
⟨Nγ ⟩ =
(2.67)
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
24
Hence, Eq. (2.63) is valid.
Note that the first fraction in Eq. (2.63) is called phase space factor while
the second fraction is the Bose-Einstein distribution factor. Moreover, this
distribution n(ν) is sometimes called the blackbody spectrum.
For a photon of momentum p⃗ hitting a wall and then reflected back elastically,
the magnitude of the change in momentum equals 2p cos θ = 2hν cos θ/c where
θ is the angle between p⃗ and the normal of the wall. The radiation pressure due
simply the momentum transfer to the wall per unit time per unit surface area.
That is,
(
)2
(
)
2hν cos θ
hν
hν
1 ∫ +∞ ∫ π/2 ∫ 2π
2
Prad = 3
c cos θ
sin θ dϕdθd
h 0
c
ehν/kT − 1
c
c
0
0
∫ +∞
hν
=
n(ν) dν
3
0
4σ 4
=
T ,
(2.68)
3c
where σ = 2π 5 k 4 /15c2 h3 is the Stefan’s constant.
Thus, our ideal gas EOS for stellar matter is
P = Pi + Pe + Prad .
(2.69)
Finally, note that the energy density of a photon gas at temperature T is given
by
∫ +∞
4σ 4
urad =
hνn(ν)dν =
T .
(2.70)
c
0
Hence, urad = 3Prad .
Exercise:
1. For a photon gas with volume V and temperature T , what is the internal
energy U ?
2. Given that the entropy S of the gas is related to U by dU = T dS and S
remains a constant for an adiabatic process, show that a photon gas has
an adiabatic index γ = 4/3.
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
2.11
25
Other forces
As a final note, we estimate the contribution of other forces to the EOS. Most
of the stellar material should be ionized, therefore, the dominant interaction
between particles in stellar interior (besides gravity) is electrostatic in nature.
The typical distance between particles in stellar interior d ≈ (ĀmH /ρ̄)1/3 where
Ā is the average atomic weight of particles, mH is the mass of hydrogen atom,
and ρ̄ is the average density of a star. Consequently, the typical Coulomb
energy per particle is about Z̄2 e2 /d (recall that 4πϵ0 → 1 in cgs), where Z̄ is
the average atomic number of particles. In contrast, virial theorem tells us that
the gravitational potential energy and internal energy of a star are of the same
order. Consequently, the ratio of Coulombic energy to internal energy of a star
is about
Z̄2 e2
.
(2.71)
G(ĀmH )4/3 M 2/3
Note that Z̄ and Ā are of order of 1. Therefore, the above ratio is about 10−2 ,
which is much less than 1. Thus, electrostatic potential energy contribute only
a small fraction of the internal energy of a star; and most of the internal energy are in the form of heat. More specifically, the thermal K.E. of a typical
particle inside a star is much greater than the electrostatic potential energy it
experiences.
Well, you may also ask if quantum effects may alter the ideal gas EOS in a star.
We may estimate the uncertainty in position ∆x to be√of order of d. Similarly,
the uncertainty in momentum ∆p is of the order of kT ĀmH . Thus, for all
living stars, ∆x ∆p is much larger than h. Hence, quantum effects play little
role in affecting the ideal gas EOS of a star. (Note: the situation is completely
different for dead stars such as white dwarfs and neutron stars. Because the
density of dead stars is high, ∆x ∆p for some particle species inside these stars
are of order of h and hence quantum effect drastically changes their EOS.)
To summarize, ideal gas EOS is a good approximation in the study of the
interior of most living stars. (However, quantum mechanical effect is important
in some other “stars” such as brown dwarfs and supernovae. Radiation pressure
also plays an important role in the structure and stability of a super-massive
star. We shall come across them later on in this course.)
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
2.12
26
Degenerate gas equation of state
For completeness, we will now derive the EOS for degenerate gas, in which the
quantum effect dominates. This is important in some cases, such as brown
dwarf, late stage of stellar evolution, white dwarfs, and neutron stars.
Due to Pauli’s exclusion principle, two indistinguishable Fermions (e.g.
electrons, protons, neutrons, etc) cannot occupy the same state. At extreme
densities, all the low-energy states are occupied. To further increase the density,
it takes extra energy to force a Fermion to a higher energy level. This results in
an increase in the gas pressure. Fundamentally, it is the exchange interaction
that creates the “force”, although this is merely a quantum mechanical effect
rather than a real force.
Consider a particle in a box (i.e. an infinite potential well in 3D) of length
L, solution of the Schrödinger equation requires the particle wavelength to
satisfy
2L
2L
2L
λx =
,
λy =
, and λz =
,
(2.72)
lx
ly
lz
where lx , ly , and lz are positive integers. From the de Broglie equation
p = h/λ, the total K.E. is
p2
h2
h2 l 2
2
2
2
=
(l
+
l
+
l
)
≡
,
(2.73)
y
z
2m
8mL2 x
8mL2
i.e. it is proportional to l2 ≡ lx2 + ly2 + lz2 . At low temperature, we can assume
that all low-energy states are occupied up to the Fermi level lF . The number
of states within a sphere of radius lF is 4πlF3 /3. But since lx ly , and lz takes
only positive numbers, we have to divide this by 8 and each state can have two
particles when we take the spin into account. Therefore, the total number of
particles is
( )(
)
1
4 3
πl3
Ntot = 2
πlF = F .
(2.74)
8
3
3
This gives lF = (3Ntot /π)1/3 and hence the Fermi energy of
E=
(
)
(
)
h2 lF2
h2
3Ntot 2/3
h2 3n 2/3
EF =
=
,
=
8mL2
8mL2
π
8m π
where n = Ntot /L3 is the particle number density. The total energy is
∫
Etot =
∫
lF
E dN =
0
0
lF
(
h2 l 2
8mL2
)
(
π 3
d
l
3
(2.75)
)
=
=
3
Ntot EF ∝ V −2/3 .
5
(2.76)
CHAPTER 2. STELLAR STRUCTURE EQUATIONS
27
Exercise: Show that the gas pressure is
P ≡−
∂Etot
2
= nEF ∝ V −5/3 ,
∂V
5
(2.77)
and hence the non-relativistic degenerate gas EOS is a polytrope. What
is γ and the polytropic index n?
When the density is extremely large, e.g., inside a neutron star, EF becomes
comparable to the rest mass of the Fermions,
to consider relativistic
√ 2 4 we2 need
2
effects and modify Eq. 2.73 with E = m c + p c ≈ pc.
Exercise: What is E in terms of l? Hence, derive Etot and P . Show that the
relativistic degenerate gas EOS is a polytrope and find the index.
Chapter 3
Energy Production and Transfer
Processes
3.1
Energy conservation and the energy production equation
We have the equations that govern P and ρ (i.e. m) as functions of radial
distance. So far, our simple stellar model is only a giant ball of gas that is
bounded by self gravity and obeys the adiabatic gas law. Now, we want to go
further by considering a more realistic equation of state, i.e., P (ρ) and different
chemical compositions. In this chapter, we will talk about energy generation
inside a star. This will change the temperature, and hence P and ρ.
With our LTE assumption, matter in the star is in thermal equilibrium. Thus,
the gas inside a star will not expand or contract; the work done by the gas
is zero. Let L(r) denotes the luminosity at the spherical shell with radius r
centered at the core of the star. (That is, L(r) is the energy per unit time
moves out of the spherical shell of radius r.) Energy conservation demands that
dL
= 4πr2 ρ(r)ϵ(r) ,
dr
(3.1)
where ϵ is the rate of energy production per unit mass of material at the radius
r from the stellar center. Note that ϵ is an implicit function of density ρ,
temperature T and chemical composition. (Note: a few authors use the notation
q instead of ϵ.)
28
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
29
Using Eq. (2.2), the above energy production equation can be written as
dL
= ϵ(r) .
dm
(3.2)
Note that L(0) = 0 and L(R) = L where L is the luminosity of the star. Besides,
the region with ϵ(r) > 0 is the region where nuclear fusion takes place.
3.2
Some important timescales
A timescale is defined as the ratio between a variable ϕ and its time derivative
ϕ̇, τ ≡ ϕ/ϕ̇, such that it gives us an idea on the characteristic time for that
variable to change significantly.
• The free-fall / dynamical timescale:√For a star with mass M and radius R, the escape velocity is of order of GM/R. The free-fall timescale
√
√
is, therefore, of the order of τff ≈ R3 /GM ≈ 1/ Gρ̄ where ρ̄ denotes the
mean density of the star. For our Sun, the free-fall timescale is of order of
an hour. In short, any force that is unbalanced inside a star occurs at the
free-fall timescale. Thus, if we are interested in the processes of a stable
star over a time span which is much longer than the free-fall timescale,
then we are safe to assume that the star is in mechanical equilibrium.
• The Kelvin-Helmholtz timescale: This is the timescale τKH for a
star to radiate its thermal energy away at constant luminosity. By virial
theorem, τKH ≈ GM 2 /RL. For our Sun, τKH is about 3 × 107 yr. For
a time much longer than the Kelvin-Helmholtz timescale, we may safely
assume that the star is in thermal equilibrium.
• The nuclear burning timescale: This is the timescale for burning all
the nuclear fuel of a star at constant luminosity and is therefore equal
to τnucl ≈ M c2 ∆/L, where ∆ ≈ 10−3 is the typical binding energy of a
nucleon divided by the rest energy of the nucleon. For our Sun, τnucl ≈
1011 yr. The nuclear burning timescale is a rough estimate of the lifespan of
a star. (You may notice that the nuclear burning timescale overestimates
the lifespan of a star. Do you know why?)
Since τff ≪ τKH ≪ τnucl , we know once again that our assumption of LTE
is a good approximation to study the interior structure and evolution of
a star.
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
30
Exercise: Estimate these timescales for the Sun. Based on the results, what is
the main energy source of the Sun?
3.3
Coupling between radiation and matter and
the radiative transfer equation
A photon in a star travels in straight line until it is scattered or absorbed
and remitted in a random direction by the surrounding materials. In this way,
energy carried by the photons is transferred to the surrounding material and
heats it up. Therefore, the trajectories of photons inside a star can be regarded
as random walks.
Opacity refers to the degree of radiation scattering and absorption in a medium.
It is related to the probability per unit path length that a photon is being absorbed or scattered. When a light ray travels in a medium, its intensity will be
reduced with distance x by I(x) = I0 exp(−κρx), where κ is the specific opacity or opacity coefficient and ρ is the density of the medium. By definition,
κρ is of unit cm−1 and hence κ is of unit cm2 g−1 . (To tell you the truth, κ in
general depends on the frequency of light to be absorbed. In fact, the κ used
here is carefully averaged over the frequency of light to be absorbed.)
In addition to energy transfer between photon and the medium, absorption
and/or scattering of radiation in the medium also lead to momentum transfer.
Now consider the sphere of radius r from the center of a star. The luminosity,
namely, light energy passes through this sphere per unit time, is L(r). Thus,
the corresponding radiation flux (that is, light energy passes through this sphere
per unit area per unit time) equals L(r)/4πr2 . Therefore, the light energy
absorbed or scattered on this sphere per unit area per unit time per unit path
length equals L(r)κρ/4πr2 . Consequently, the momentum transferred to the
surrounding medium on this sphere per unit area per unit time per unit path
length is L(r)κρ/4πr2 c. This number is equal to −1 times the radiation pressure
gradient on that sphere. (Do you know why?) Recall that inside a star, LTE
is a good approximation. Hence, radiation there follows a blackbody spectrum.
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
31
So, from Eq. (2.68), we have
L(r)κ(r)ρ(r)
dPrad
16σT 3 dT
=
−
=
−
.
4πr2 c
dr
3c dr
In other words,
L(r) = −
64πσr2 T 3 dT
.
3κρ
dr
(3.3)
This equation is sometimes known as the radiative transfer equation. Clearly,
T (0) = Tc (the core temperature) while T (R) = Ts ≈ 0 (the surface temperature).
Note that the radiative transfer equation is valid if radiation is the principle heat
transfer mechanism. If convection or conduction is important, this equation has
to be modified.
3.4
Physical processes leading to opacity and
Kramers opacity Approximation
The interactions between photons and matter (primarily with electrons and
nuclei in most part of a star) can be classified below.
1. Electron scattering: Compton/Thomson scattering, that is, scattering
of photon by a free electron.
2. Free-free absorption: Absorption of a photon by a free electron which
makes a transition to a higher energy level by briefly interacting with a
nucleus or ion.
3. Bound-free absorption: Also known as photon-ionization, namely, the
ionization of an electron from an atom or ion after absorbing a photon.
4. Bound-bound absorption: Exciting an electron from a bounded state
of a lower energy level to another bounded state of a higher energy level
by absorbing a photon.
Close to the stellar core, temperature is so high that atoms are almost completely ionized. Thus, electron scattering and free-free absorption are the dominant process to create opacity in stellar core. In the outer stellar atmosphere,
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
32
in contrast, the situation is much more complex. Bound-free and bound-bound
absorptions can be important. This is the reason why we see absorption lines
on the surface of stars.
The specific opacity κ can be calculated by taking into account all the possible
interactions between photon and the medium. This is a very complicated and
involved task, requiring the work of a whole generation of atomic physicists.
Fortunately, the resultant κ can usually be approximated by relatively simple
formula in the form κ = κ0 ρa T b . (Once again, we shall only use the approximate
form of κ in this course as the approximation is reasonably good for most part
of the stellar interior.)
In particular, the electron scattering opacity, which is important in the high
temperature core of a star, is given by
κes ≈ 0.2(1 + X) cm2 g−1 ,
(3.4)
while the free-free absorption, which may be important near the surface of a
star, can be approximated within an accuracy of about 20% by the so-called
Kramers opacity
κff ≈ 4 × 1022 (1 + X)(X + Y )ρT −7/2 cm2 g−1 .
(3.5)
Clearly, free-free absorption produces a higher opacity than electron scattering at low temperature. In fact, the cross-over temperature between free-free
absorption and electron scattering is about 106 K. Whereas when the temperature is less than about 104 K, bound-free and bound-bound absorption become
more important than Kramers opacity is no longer valid. In fact, at such a low
temperature, κ starts to decrease as T decreases.
Interestingly, the presence of opacity places an upper limit on the luminosity
of a star. Recall that P = Pgas + Prad . Clearly, dPgas /dr ≤ 0. So, from the
equation of hydrostatic equilibrium and equation of radiative transfer, we have
dPrad
dPrad dPgas
≥
+
=
dr
dr
dr
(3.6)
Therefore,
L≤
≡ LEdd .
(3.7)
If this inequality is violated somewhere inside a star, then either radiative transport is no longer valid (such as in the core of certain stars) or hydrostatic equilibrium can no longer be maintained (such as near the surface of certain stars).
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
33
In particular, near the surface of a star, m = M and the limiting luminosity
given by Eq. (3.7) becomes the so-called the Eddington luminosity
(
LEdd = 3.2 × 10
3.5
4
M
M⊙
)(
κ
κes
)−1
L⊙ .
(3.8)
The Eddington standard model
Let us define a function η by L(r)/m(r) = L∗ η/M where L is the luminosity
across a sphere of radius r centered at the star and L∗ is the total luminosity
of a star.
Exercise: Using the stellar structure equation, show that dPrad /dr can be written as
dPrad
L∗ κη
=
.
(3.9)
dP
4πcGM
Assuming that the location where nuclear burning occurs is confined in a small
region in the core, then over most part of a star, η decreases as r increases.
On the other hand, κ increases with r (do you know why?). At this point, A.
Eddington somehow made the bold assumption that the product κη ≡ κs is
constant over the entire star. Surely, this assumption is in no way close to the
truth, it leads to very interesting and realistic predictions about structure and
evolution of stars.
Integrating Eq. (3.9) gives
κs L∗
P ≡ P (1 − β) .
(3.10)
4πcGM
That is to say, the ratio of Prad to P is a constant throughout a star in the Eddington standard model. Historically, the ratio between gas and total pressure
is defined as β, i.e. Pgas /P = β, such that Prad /P = 1 − β. Thus, the radiative
transfer equation becomes
Prad =
L∗ =
4πcGM
(1 − β) ≡ LEdd (1 − β) .
κs
(3.11)
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
34
The ideal gas law demands that
T4
Prad
Pgas
ιρT
∝
=P =
=
,
1−β
1−β
β
β
(3.12)
where ι = k/mH µ (see Eq. 2.58). Hence, T ∝ ρ1/3 and the equation of state
may be written as P = Kρ4/3 for some constant K ∝ [(1 − β)ι4 /β 4 ]1/3 . In other
words, the equation of state is a polytrope with index 3. So, there is a unique
relation between K and M .
From Eq. (2.24) with n = 3, we have
( )4 (
β
1 − β = 0.003
ι
M
M⊙
)2
.
(3.13)
This is known as the Eddington quartic equation. This equation has the
following predictions. First, for a given composition (fixed ι), β decreases as M
increases. So, radiation pressure is more important in high mass stars. Second,
combining with Eq. (3.11) gives L∗ ∝ M 3 . So, the mass-luminosity relation
is obtained. Actually, Eddington made this prediction before the confirmation
by observations. Third, for a given M , nuclear burning implies that ι will
decrease gradually. Consequently, β decreases. Therefore, it is expected that
radiation pressure plays an increasingly important role as a star gets older.
Combined with the Eddington limit, it strongly hints that some stars may eject
some of its materials in its late stage of evolution.
Exercise: Combining Eqs. (3.11) and (3.13), show that L∗ ∝ M 3 .
3.6
Nuclear energy generation
The energy generation mechanism of a star cannot be chemical in nature. For
chemical reaction leads to an energy release per molecule of the order of 1 eV.
Thus, chemical reaction can only support the present solar luminosity for about
104 yr. In contrast, the energy generation by nuclear reaction is of the order of
1 MeV per nucleus. This allows the Sun to shine at the present rate for about
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
35
1010 yr. Thus, nuclear energy looks like a promising way to power the Sun and
other stars. (Still recall that energy is released by fusion of light nuclei or fission
of heavy nuclei? Do you know that 56 Fe is the most stable nucleus in the sense
that it has the highest binding energy per nucleon?)
But we have a problem. Virial theorem tells us that the thermal energy and
gravitational potential energy of a star are of the same order. Thus, using
ideal gas approximation, the average temperature of our Sun turns out to be
about 106 K. However, the typical K.E. of two 106 K hydrogen nuclei cannot
overcome the Coulombic repulsion between them. Specifically, the Coulomb
energy between two hydrogen nuclei placed 2 hydrogen nuclear radii apart is
about three orders of magnitude larger than the typical K.E. of the two hydrogen
nuclei. Thus, classically, there is no way for two hydrogen nuclei to fuse together
making a bigger nucleus and hence releasing the required nuclear energy to make
the star to shine. Fortunately, quantum mechanics saves us for hydrogen nucleus
may tunnel through the Coulombic energy barrier (see figure 3.1).
Wavefunction
r
Potential
Coulombic repulsion
1/r
E
r
rn
rc
Figure 3.1: An illustration of overcoming the Coulombic energy barrier through
tunneling.
Consider a nucleus of charge Z1 e moving towards another nucleus of charge Z2 e.
Denote µ the reduced mass of the system. The Schrödinger equation describing
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
36
this system is
h̄2 2
Z1 Z2 e2
∇ ψ+
ψ = Eψ ,
(3.14)
2µ
r
for r > rN (the nuclear radius), where E is the K.E. of the first nucleus relative
to the second one when they are infinitely apart.
−
By separation of variables, (that is, by writing ψ = R(r)Y (θ, ϕ)), we conclude
that the radial wavefunction of the above equation satisfies
−
h̄2 d2 χ Z1 Z2 e2
h̄2 l(l + 1)
+
χ(r)
+
χ(r) = E χ(r) ,
2µ dr2
r
2µr2
(3.15)
where l is the angular momentum between the two nuclei and R(r) = χ(r)/r.
Now, I consider simplest case of a head-on collision so that l = 0. In this case,
Eq. (3.15) can be re-written as
d2 χ
2µ
= 2
2
dr
h̄
(
)
Z1 Z2 e2
−E χ .
r
(3.16)
Nonetheless, the above equation is too complicated to solve analytically. Fortunately, an approximate solution is available using W.K.B. approximation.
Specifically, we assume that the radial wavefunction (in other words, the solution of Eq. 3.16) to be in the form χ(r) = AeiS(r)/h̄ for some constant A.
Exercise: Using Eq. (3.16), show that S(r) satisfies
ih̄S ′′ − S ′2 − h̄2 k(r)2 = 0 ,
where
2µ
k(r) = 2
h̄
2
(
Z1 Z2 e2
−E
r
(3.17)
)
.
(3.18)
Note that in the quantum tunneling region, where Z1 Z2 e2 /r > E, we can pick
k(r) ≥ 0.
∑
k
Exercise: Expanding S(r) in powers of h̄, i.e. S = ∞
k=0 h̄ Sk (r) and keeping
only the zeroth order term (note that the term h̄2 k 2 is h̄ independent as k is
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
37
proportional to 1/h̄), show that
[
χ(r) = A exp −
∫
]
r
k(r) dr
(3.19)
rc
for some constant A whenever r is in the quantum tunneling region.
Note that the integral above is from rc to rN where rc is the distance between
the two nuclei when the classical K.E. of the first nucleus is zero. That is,
Z1 Z2 e2
1
= µv 2 .
rc
2
(3.20)
Besides, E = µv 2 /2 where v is the relative speed between the two nuclei.
Since rc ≫ rN and hence Z1 Z2 e2 /r ≫ E, upon integration, we find that the
tunneling probability is about
)
(
2
χ(r ) 2
A 2
2πZ
Z
e
N 1
2
exp −
|R(rN )| = ≈ .
rN rN h̄v
2
(3.21)
(Can you fill in the details of the calculation?) This result, which laid an important foundation to the study of nuclear reactions in stars, was first obtained
by G. A. Gamow in the study of radioactivity in 1928. R. Atkinson and F.
Houtermans quickly applied it to the case of energy generation of stars in 1929.
In summary, the probability for two nuclei coming close enough for their nuclear
force to be effective is small but non-zero.
We expect that nuclear force is effective only within a de Broglie wavelength
λ = h/µv. Thus, we also expect the nuclear reaction cross section to be in the
form
(
2πZ1 Z2 e2
σ(v) ≈ πλ exp −
h̄v
2
)
(
S(v)
2πZ1 Z2 e2
≈ 2 exp −
v
h̄v
)
where S(v) is some expression that is weakly dependent on v.
,
(3.22)
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
38
Note further that the velocity distribution of nuclei follows the Maxwellian
distribution. Specifically, the probability of finding a particle with velocity ⃗v is
proportional to (µ/2πkT )3/2 exp(−µv 2 /2kT ) where v = |⃗v |.
Consequently, the probability of nuclear reaction to occur between the two
nuclei per unit time is proportional to
∫
+∞
0
v
rN
(
µ
2πkT
)3/2
(
µv 2
exp −
2kT
)
(
)
S(v)
2πZ1 Z2 e2
exp
−
4πv 2 dv .
2
v
h̄v
(3.23)
For simplicity, we assume that S is a function that does not depend strongly
on v so that vS(v) can be assumed to be almost a constant throughout the
speed range considered. Further observe that the first term in the exponential increases with v (and hence the initial K.E.) while the second term in the
exponential decreases with v (and hence the initial K.E.). By Taylor series expansion on the terms inside the two exponentials, we conclude that the resultant
reaction probability has a peak around
(
vGamow =
2πZ1 Z2 e2 kT
h̄µ
)1/3
(3.24)
known as the Gamow peak. In addition, the width of this peak is about
(
∆vwidth ≈
kT
3µ
)1/2
.
(3.25)
(See figure 3.2)
Consequently, the reaction rate per unit volume is approximately proportional
to
(
µ
n1 n2 vGamow S(vGamow )
2πkT

3
∝ n1 n2 (kT )−2/3 exp −
2
(
)3/2
2πZ1 Z2 e2
h̄
(
)
2
−2πZ1 Z2 e2 µvGamow
exp
−
∆vwidth
h̄vGamow
2kT
)2/3 (
µ
kT
)1/3

 ,
(3.26)
where ni is the number density of nuclear species i.
Thus, the reaction rate per unit volume increases as T increases or as the charge
of the nucleus decreases. This is the reason why fusion of heavy nuclei requires
high temperature.
However, there is an exception to this trend. Some reactions known as resonant
reactions occur when the energy of the interacting particles correspond to
energy level of the intermediate unstable compound nuclei. (Still remember
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
39
Probability
−E/kT
Maxwell−Boltzmann e
1/2
Tunneling e−b/E
Gamow Peak
Energy
Figure 3.2: An illustration of the dependency of reaction probability on energies.
what you have learned in high school chemistry?) Resonant reaction probability,
therefore, is sharply peaked at the resonant energy. (That is, S(v) is sharply
peaked at certain values of v.) Such a probability may be several order of
magnitudes higher than that of an ordinary non-resonant reaction.
From Eq. (3.26), the nuclear energy generation rate ϵ depends on T −2/3 exp(T −1/3 ).
To simplify the calculation, we can parameterize ϵ by
ϵ ≈ Aρa T b
(3.27)
for some constants A, a and b over a restricted range of temperature T .
Finally, I must stress that the actual calculation of a specific nuclear reaction
rate and hence a and b are very involved and complicated. The above discussions
only serve as an overly simplified approximation.
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
3.7
3.7.1
40
Some specific nuclear reactions occurring
in stars
Proton-proton chain (p-p chain)
The simplest possible reaction is p + p −→ 2 He. Unfortunately, 2 He is a highly
unstable nucleus which quickly decomposes back into two protons. However,
if a weak interaction that turns a proton into a neutron occurs during the
combination of two protons, then we will get some stable product. Namely,
p + p −→ 2 H + νe + e+ . The reaction rate is very slow since weak interaction, as
the name suggests, is much weaker than strong and EM interactions. (Actually,
experiments find that the relative interaction strength between strong, EM and
weak forces is about 1 : 10−2 : 10−5 for all stellar astrophysical processes.)
This is not the end of the story for main sequence stars as the resultant 2 H
will undergo further nuclear reaction in the core of a main sequence star. For
example:
p-p I chain:
p + p −→
2
H + p −→
3
He + 3 He −→
2
H + νe + e+
3
He + γ
4
He + 2p
Besides p-p I chain, a star may also react via
p-p II chain:
3
He + 4 He −→ 7 Be + γ
7
Be + e− −→ 7 Li + νe
7
Li + p −→ 2 4 He
or
p-p III chain:
7
Be + p −→ 8 B + γ
8
B −→ 8 Be + e+ + νe
8
Be −→ 2 4 He
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
41
These three sets of reactions, whose foundation was first proposed by H. A.
Bethe in 1939, operate simultaneously although their relative significance depend on the physical conditions at the stellar core.
But in any case, the net result is to convert 4 hydrogen nuclei into 1 helium
nucleus. The nuclear energy released per each conversion is experimentally
measured to be about Q = 4M (1 H) − M (4 He) = 26.73 MeV. The energy is
released in the form of mainly photons and neutrinos. Photons will later on
interact with the electrons in the star through electron opacity. Neutrinos, on
the other hand, are extremely weakly interacting particles. They leave the star
essentially without interacting with the surrounding materials. Thus, neutrinos
emitted from the star in this way are not thermalized. In other words, LTE
does not apply to neutrinos in a star.
Finally, notice that in p-p I chain only the first reaction involves weak interaction, all other reactions involve strong and EM interactions. Hence, the
bottle-neck for p-p chain is the first reaction. In fact, more accurate calculations
tell us that the reaction rate for the first reaction is of the order of once every
1010 years. Hence, it occurs in a massive scale in the core of a star because there
are numerous supply of protons in the stellar core. To summarize, the nuclear
energy generation rate per unit mass ϵ for p-p chain is determined by the rate
of p + p −→ 2 H + νe + e+ ; and it can be shown that the particular reaction rate
is given by
−1/3
−2/3
ϵp−p = 2.4 × 104 ρX 2 e−3.38T
T9 ,
(3.28)
where T9 is temperature in units of 109 K. This can be approximated by
ϵp−p ≈ AρX 2 T 4
(3.29)
for some constant A > 0, where X is the mass function of 1 H.
3.7.2
Carbon-nitrogen-oxygen (CNO) cycle
H. A. Bethe and C. von Weizsäcker independently pointed out in the late 1930’s
that besides the p-p chain, hydrogen may burn via the CNO cycle using carbon,
nitrogen and oxygen as catalysis as shown in the two cycles below:
cycle 1:
12
C+p
13
N
13
C+p
14
N+p
−→
−→
−→
−→
13
N+γ
C + e+ + νe
14
N+γ
15
O+γ
13
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
O −→
15
N + p −→
15
42
15
N + e+ + νe
12
C + 4 He
cycle 2:
14
N+p
15
O
15
N+p
16
O+p
17
F
17
O+p
−→
−→
−→
−→
−→
−→
15
O+γ
N + e+ + νe
16
O+γ
17
F+γ
17
O + e+ + νe
14
N + 4 He
15
Eventually, H. A. Bethe received the Nobel Prize in Physics in 1967 “for his
contributions to the theory of nuclear reactions, especially his discoveries concerning the energy production of stars.”
Although these reactions are very complicated, it is relatively easy to estimate
the overall rate of reaction for we need only to look closely at the slowest rate
determinate reaction. Recall that β decay is temperature and density independent (over conditions inside a star). In contrast, proton capture reactions
are sensitively dependent on temperature. (Do you know why?) Therefore, at
relatively low temperature, it is the proton capture that sets the pace of the
reaction. While at relatively high temperature, it is the β decay that determines
the reaction rate. In fact, under the conditions of the interior of most stars,
proton capture is the bottle-neck. The rate of energy release per unit mass is
ϵCNO = 4.4 × 1025 ρXZe−15.2T
−1/3
−2/3
T9
,
(3.30)
and it can be approximated by
ϵCNO ≈ AρXZT 16 ,
(3.31)
for some constant A > 0. The constant A in Eq. (3.31) is much smaller than
that in Eq. (3.29). Do you know the reason why?
3.7.3
Pre-main sequence light element burning
The following proton capture reactions already function effectively before the
onset of hydrogen burning:
2
H + p −→
3
He + γ
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
6
Li + p
7
Li + p
9
Be + p
10
B+p
11
B+p
11
B+p
43
−→ 3 He + 4 He
−→ 2 4 He
−→ 4 He + 6 Li
−→ 4 He + 7 Be
−→ 12 C + γ
−→ 3 4 He.
(Note that 7 Be formed in the above reaction decays via electron capture to
7
Li + νe with a half-life of about 53 d.)
These reactions are effective to destroy light elements such as 2 H, Li, Be, B
whenever the temperature is higher than about 106 K. This can be used to distinguish young brown dwarfs from main-sequence stars, known as the Lithium
test.
3.7.4
Helium burning
At higher temperature, helium may fuse via
2 4 He −→ 8 Be .
(3.32)
Unfortunately, 8 Be is highly unstable which decays back to 2 He in about 2 ×
10−16 s. The solution to this problem was given by E. Salpeter in 1952 and was
later on refined by F. Hoyle. They pointed out that if the temperature is higher
than about 108 K, the K.E. of helium nuclei will be very high. In this case, the
mean collision time between the unstable Be nucleus and a He nucleus is shorter
than the half-life of the unstable Be nucleus. Hence, it is possible to produce a
12
C by capturing a He nucleus by a 8 Be nucleus. The net result is
3 4 He −→ 12 C
(3.33)
known as the triple alpha process. The energy released per reaction is about
7.28 MeV. Clearly, the net reaction rate is determined by the rate of reaction of
the capture of He by Be. It turns out that the reaction rate per unit mass is
ϵ3α ≈ Aρ2 Y 3 T 40 ,
(3.34)
for some A > 0. Some of the carbon produced could further fuse into oxygen
by
12
C +4 He −→ 16 O .
(3.35)
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
3.7.5
44
Heavy elements burning and photodisintegration
At higher temperature, carbon and oxygen can overcome their Coulombic barriers and fuse together. The resultant nuclei are Mg, Na, Ne, S, P, Si, etc. At
even higher temperature, Si etc. may also begin to fuse. Nonetheless, the situation is not as simple as He, C or O burning. At temperature higher than about
109 K, photodisintegration begins to take place. In photodisintegration, a
nucleus divides into two or more fragments after absorbing a high energy photon. Thus, photodisintegration is, in general, an endothermic process similar to
the photo-ionization of an atom. Therefore, photodisintegration is effective only
at sufficiently high temperature (so that there are enough high energy photon
to “ionize” the nucleus). An example of photodisintegration process is shown
below:
20
Ne + γ −→ 16 O + 4 He .
(3.36)
Due to the presence of heavier element fusion and photodisintegration, reactions
involving the burning of Si and other heavier nuclei occur in an almost dynamic
equilibrium fashion. The net result is a gradual buildup of heavy elements such
as Ni, Fe, Co provided that the temperature is lower than about 2×109 K. If the
temperature is higher than about 3 × 109 K, photodisintegration becomes the
dominant reaction and hence the star will cool itself down. (Do you know why?)
In other words, photodisintegration sets an upper limit in the core temperature
of a star.
3.7.6
Neutron capture and β decay
Another type of reaction that are of interest is neutron capture. Since neutron
is electrically neutral, it can come close to a nucleus without worrying about
the Coulombic barrier. Nevertheless, free neutron outside a nucleus is a rare
species. A small amount of neutron is generated in a minor reaction involving
the burning of carbon such as
13
C + 4 He −→ 16 O + n .
(3.37)
These neutrons may be captured by a heavy nucleus resulting in a heavier isotope. Upon a few neutron captures, the resultant nucleus may be too neutronrich so that it will β decay into a new nucleus with higher atomic number.
Neutron capture and the subsequent β decay, therefore, are effective mechanism to produce a variety of heavy elements and isotopes. In particular, they
play an important role in the synthesis of elements with atomic number higher
than about 60.
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
45
−
β−
ν
( n,γ )
s only
p
s,r
s,r
s only
p
s,r
s only
r
s,r
Z
s,r
s,r
s only
s,r
r
r
r
β
−
(Neutron−rich matter)
N
Figure 3.3: Schematic representation of the effects of s- and r-processes.
Astronomers generally divide neutron capture processes into two types. In
the event that the neutron capture reaction rate is much slower than the βdecay rate, we call it the s-process. In an s-process, which is a more common
neutron capture process in astronomy, successive neutron capture leads to the
production of a chain of stable isotopes until it reaches a radioactive species,
at which point β-decay will occur. (See figure 3.3.) Another possibility is that
the neutron capture rate is much faster than the β-decay rate, known as the
r-process. The r-process is effective in synthesizing a few highly neutron-rich
stable isotopes. The r-process is likely to be important in supernova explosion
or in neutrino-heated atmosphere surrounding a new-born neutron star. (See
figure 3.3.) Note, however, that both processes cannot synthesize the so-called
stable p-nuclei (p for proton-rich). Thus, we expect that p-nuclei are rare in
a star. This prediction generally agrees with observations. (See figures 3.3
and 3.4.) Astronomers believe that these proton-rich nuclei are formed by proton capture at very high temperature or by the so-called (γ, n) reaction during
supernova explosions.
3.7.7
Pair production
In the event that the temperature of certain region of a star exceeds ≈ 2me c2 /k ≈
1010 K, electron-positron pair production from thermal photon becomes
CHAPTER 3. ENERGY PRODUCTION AND TRANSFER
121
Sb, 51
57%
114
123
s only
p process
112
46
115
116
117
118
119
2.8d
120
43%
122
124
27h 4.8%
6.1%
Sn, 50
1.02% 112d
Z
0.69%
0.38% 14.3% 7.6%
113
Slow process path
In, 49
4.2%
110
111
112
32.5%
Rapid Process
95.8%
113
24.1% 8.5%
115
Rapid process
only
13s
116
114
Cd, 48
12.4% 12.8% 24.0% 12.3% 28.8%
62
63
64
65
66
54h
7.6%
67
68
69
70
71
72
73
74
N
Figure 3.4: Schematic representation of the effects of s- and r-processes.
important. (Actually, the effect cannot be neglected when the temperature is
about 109 K because a few highly energetic photons from the tail of the Planck
distribution can already undergo pair production.) Similar to photodisintegration, pair production is endothermic and in effect placing a limit on the maximum temperature of a star. It has been proposed that this mechanism could
give rise to pair-instability supernova in very massive stars above 130 M⊙ .
The work of stellar nucleosythnesis culminated in the publication of the article
by E. M. Burbidge, G. Burbidge, W. A. Fowler and F. Hoyle, Reviews of Modern
Physics 29, 547 (1957). This article introduced all the essential ideas involving
in the synthesis of elements from carbon to uranium in a star starting with the
hydrogen and helium produced in the big bang. W. A. Fowler was awarded the
Nobel Prize in Physics in 1983 “for his theoretical and experimental studies of
the nuclear reactions of importance in the formation of the chemical elements in
the universe.” (Actually, a similar idea was also pointed out independently in
the article A. G. W. Cameron, in Chalk River Report Number CRL-41, Chalk
River Labs., Ontario (1957).)
Chapter 4
Stability of Stars
4.1
Secular thermal stability
Recall from the virial theorem that 2U + Ω = 0 for stars modeled by nonrelativistic ideal gas. That is to say, the total energy of the star E obeys
1
E = −U = Ω ,
(4.1)
2
where U is internal energy and Ω is potential energy. Suppose that the nuclear energy generation rate of the star somehow increases a little bit. Then,
Ė > 0 implying that U̇ < 0. For ideal gas, that means Ṫ < 0. In other words,
the average temperature of the star decreases. Consequently, the nuclear energy generation rate will decrease. By the same argument, a slight decrease
in nuclear energy generation rate will increase the internal energy of the star
making the star a little bit hotter, therefore increasing the resultant nuclear
energy generation rate. This negative feedback makes the star thermally stable against small perturbation in nuclear energy generation rate. We call this
secular thermal stability of a star.
Does this argument work for degenerate stars or stars with degenerate cores?
4.2
Dynamical stability
Dynamical stability deals with the stability of motion of mass parcels in a
star. A detail treatment of dynamical stability is very complicated. So, I only
illustrate the idea by a highly simplified example.
47
CHAPTER 4. STABILITY OF STARS
48
Consider a gaseous sphere of mass M in hydrostatic equilibrium. At any point
r(m), the pressure is equal to the weight per unit area of layers between m and
M . Equation of hydrostatic equilibrium (Eq. 2.4) implies that
∫
M
P (m) =
m
Gm
dm .
4πr4
(4.2)
Now, we consider a small uniform radial change in the radius
r −→ r′ = r(1 − ∆) .
(4.3)
The mass dm is conserved. If ∆ is sufficiently small, then the new density ρ′ (m)
will be given by
4πr2 ρdr = dm = 4πr′2 ρ′ dr′
= 4πr2 (1 − ∆)2 ρ′ d [(1 − ∆)r]
= 4πr2 (1 − ∆)3 ρ′ dr .
ρ
ρ′ (m) =
(1 − ∆)3
≈ ρ(1 + 3∆) .
(4.4)
(4.5)
Hence, the new hydrostatic pressure is given by
∫
M
Gm
dm
− ∆)4
m
≈ P (1 + 4∆) .
′
P (m) =
4πr4 (1
(4.6)
Assume further that the change in radius is adiabatic (i.e. P V γ =const.), then
′
Pgas
= P (1 + 3∆)γa ≈ P (1 + 3γa ∆)
(4.7)
where γa is the adiabatic exponent. Hence, in order to attain stable equilibrium,
′
Pgas
> P ′ for all m if ∆ > 0 (in other words, a compression), such that the gas
could “bounce back”. That is,
P (1 + 3γa ∆) > P (1 + 4∆) .
This condition is equivalent to
4
.
(4.8)
3
Note that the same condition is arrived for ∆ < 0 (in other words, an expansion).
γa >
We can show that the adiabatic exponent of a photon gas is 4/3. Hence, if
some part of a star is dominated by radiation pressure, that part of the star
will become dynamically unstable against compression.
CHAPTER 4. STABILITY OF STARS
4.3
49
Convective instability
Recall in our earlier discussions that the radiative transfer equation holds only
when radiation is the principal energy transfer mechanism. Hence, the radiative
transfer equation does not apply when convection is important. It is, therefore,
natural to ask the condition for a star to be convectively stable (or unstable).
P2 ρ
new
ρ2
P1 ρ1
Figure 4.1: The condition of a small lump of has rising.
Consider a small lump of gas whose pressure and density are P1 and ρ1 , respectively. When displaced upward, its pressure and density will change. We
denote its new pressure and density by P2 and ρnew , respectively. (See figure 4.1.) Clearly, P2 equals to the pressure of the gas surrounding the new
location of the lump. Denote also the density of gas surrounding the new location of the lump by ρ2 . Obviously, the lump of gas is stable against convection
if ρnew > ρ2 . Otherwise, the system will be convectively unstable. We may
assume that the change is adiabatic. (Why is this a good approximation?) In
this case, P ∝ ργa and hence the convective stability condition becomes
(
dP
dρ
)
(
<
star
dP
dρ
)
.
(4.9)
adiabatic
This is known as Schwarzschild’s criterion. This can be understood using
figure 4.2: when pressure changes from P1 to P2 , a stable configuration should
CHAPTER 4. STABILITY OF STARS
50
have ρ′2 < ρ∗ and hence a shallower slope than the adiabatic case. This gives
Eq. (4.9) above. Multiplying both sides by ρ/P , we have
ρ
γ≡
P
(
dP
dρ
)
< γa .
(4.10)
star
Figure 4.2: Schwarzschild’s criterion for convective instability.
Exercise: For an ideal gas with negligible radiation pressure, recall the ideal gas
law
ρ
P =
kT ∝ ρT ,
(4.11)
µmH
show that,
dP
dρ dT
=
+
(4.12)
P
ρ
T
and hence
P
T
(
dT
dP
)
<
star
γa − 1
γa
(4.13)
CHAPTER 4. STABILITY OF STARS
and
dT dr (
<
star
γa − 1
γa
51
)
T
P
dP dr .
(4.14)
star
The above equation tells us that there is an upper limit for the temperature
gradient for a convectively stable star.
You may ask what happens if a portion of a star becomes convectively unstable.
The most important consequence is that convection (rather than radiation)
becomes the most important energy transport mechanism in that part of the
star. Hence, the radiative transfer equation must be modified. Unfortunately,
a comprehensive theory of convection is still lacking. This makes our analysis
difficult and complex.
However, if we assume that convection is extremely efficient in transporting
energy, then the temperature gradient dT /dr in the convective region of a star
should be very close to the critical value given by Eq. (4.14). In other words,
P
T
(
dT
dP
)
=
star
γa − 1
.
γa
(4.15)
Exercise: Using the hydrostatic equilibrium equation,
γa − 1 ρT Gm
dT
=−
.
dr
γa
P r2
(4.16)
Chapter 5
Evolution of Main Sequence
Stars
5.1
The complete set of stellar structure equations
After all our detail discussions, we finally have a complete set of stellar structure
equations. They are the four first order coupled ordinary differential equations:
dr
1
=
,
dm
4πr2 ρ
dP
Gm
= −
,
dm
4πr4
dL
= ϵ,
dm

(5.1)
(5.2)
(5.3)
3κL



−


2

256π r2 σr2 T 3
dT
=

dm


γa − 1 T Gm


 −
4
κ =
ϵ =
P =
where ni =
if convectively stable,
if convectively unstable,
γa 4πr P
κ0 ρ a T b ,
A(X, Y, Z)ρm T n ,
4σ 4
(ne + ni )kT +
T ,
3c )
(
ρ
Y
Z
X+ +
,
mH
4
⟨A⟩
52
(5.4)
(5.5)
(5.6)
(5.7)
CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS
(
⟨
ρ
Y
Z
ne =
X + +Z
mH
2
A
X +Y +Z = 1 .
53
⟩)
, and
Exercise: Derive dT /dm from the radiative transfer equation (Eq. 3.3).
Here, we have 12 unknowns functions of r, namely, m, ρ, P , L, ϵ, T , κ, ne ,
ni , X, Y and Z. Yet, we only have 10 equations. Fortunately, at a sufficiently
short time interval compared with the nuclear burning timescale, we know that
the chemical composition X, Y and hence also Z are almost constants. Thus,
we know X and Y as well. Besides, they are roughly time independent when
the time interval concerned is much shorter than the nuclear burning timescale.
Therefore, we have a complete set of equations to determine the structure of a
star.
These equations must satisfy the following boundary conditions:
r=0
P ≈0
L=0
T ≈0
at
at
at
at
m=0,
m=M ,
m = 0 , and
m=M .
One may solve this set of equations to determine the stellar structure. Unfortunately, the system of differential equations we are working with is too
complicated to solve analytically. Indeed, explicitly (and numerically) solving
this set of equations is the only way to accurately determine the structure of a
spherically symmetric star.
A mathematical-oriented student may be doubtful about the existence and
uniqueness of the solution to the stellar structure equations for any boundary conditions as well as any given total stellar mass and chemical composition
of the star. This is a very complicated mathematical problem. Nonetheless,
theoretical astrophysicists have cooked up an extremely artificial situation that
the solution of the stellar structure equations is not unique! Moreover, there
are situations that the solution may not exist at all. There is a good physical
argument for the non-existence of solution in some cases, too. Consider, for
example, the situation when the total stellar mass is, say, 1 g. The gravitational force will be so weak that any reasonable nuclear energy generation will
CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS
54
blow this object up. Thus, that object is likely to be supported by EM force
rather than gravity. Hence, we expect that no solution for the stellar structure
equations can exist for a 1 g object.
5.2
Introduction to the numerical solution of
stellar structure equation
All D.E.s require boundary and/or initial conditions in order to produce a numerical solution. We have a complication for the stellar structure equations.
Namely, the boundary conditions we are imposing are at two different places,
namely, at m = 0 and m = M . In other words, we are not dealing with a simple
initial value problem for a system of first order coupled O.D.E.s. Numerical solution of the stellar structure equations, therefore, must be obtained with great
care.
There are many proposed ways to numerically solve the stellar structure equations. The first way is called the shooting method. Basically, the idea is that
we start from the boundary at m = 0. At this boundary, we know for sure that
L = r = 0. Then, we guess the values of T = Tc and P = Pc at m = 0. Now,
we have a well-defined initial value problem, so we numerically integrate the
solution until it hits the other boundary at m = M . We check if T and P at
m = M are really equal to 0. If not, we carefully make another guess at the
m = 0 boundary for T and P until a solution is found. This method turns out
to be not very accurate since T and P varies greatly between the interior and
the surface of a star leading to a great lost of numerical accuracy.
Another method is called the fitting method. Essentially, we guess the values of
P , T in the inner boundary and R, L in the outer boundary. Then we numerically integrate the equations and see if they matches continuously somewhere
in the middle of the star. This was a method of choice in the 1960’s and 1970’s.
The most popular method to date is the so-called Newton-Raphson-Henyey
method. This method is so involved that I will not discuss it in any detail in
this course. I just want to give you an idea what this method is all about.
Essentially, it approximates the coupled O.D.E. together with the boundary
conditions by a system of equations. And then, we carefully solve the system
of equations by Newton-Raphson iteration. If you want to know the detail, you
may refer to Chap. 7 of C. J. Hansen and S. D. Kawaler, “Stellar Interiors:
Physical Principles, Structure, and Evolution”.
Figures below show the numerical solution of the stellar structure equations in
CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS
55
a number of situations. (Note that the y-axis in all the figures below are in
linear scale and are in units of r/R, ρ/ρc , T /Tc , L/L∗ , and X, respectively.)
5.3
Zero age main sequence star and its evolution
A zero age main sequence star (ZAMSS) is defined as a star which just
begins its hydrogen burning. In other words, ZAMSS is a new born star. The
location of ZAMSSs in the H-R diagram is known as the zero age main sequence (ZAMS). Since a star is formed from the gradual collapse of a nebula,
a spherically symmetric non-rotating ZAMSS is characterized by only two sets
of parameters, namely, its total mass (M ) and its chemical composition (X, Y ,
and Z). Note that the gradual collapse of the nebula should give enough time
for the stellar matter to mix. Hence, X, Y , and Z should be uniform over the
entire ZAMSS.
By numerically solving the stellar structure equations above, we know that the
structure of a ZAMSS can be summarized below: if the mass M is below about
0.1 M⊙ , thermal hydrogen fusion does not occur in the core and hence we have
either a brown dwarf or a planet. We may talk about more about them in the
special topics in this course. If the mass M is between about 0.1 to 0.3 M⊙ , the
entire star is convective and p-p chain is the major nuclear reaction occurring
in the stellar core.
When M is between about 0.3 M⊙ and 1.2 M⊙ , p-p chain is the dominant reaction. The core is radiative and the surface layer is highly convective. When
> 1.2 M , CNO cycle is the dominant reaction and the core is convective.
M ∼
⊙
In addition, the remaining part of the star (except for a thin outermost layer)
is radiative. A stable ZAMSS cannot have mass greater than about 30 M⊙ for
two reasons. First, its core will be dominated by radiation pressure which leads
to instability. Second, the surface luminosity will be so high that mass ejection
is serious.
The above results from ZAMSS can be understood as follows. First, although
the energy production rate for CNO cycle scales like T 16 while that of the p-p
chain scales like T 4 , the corresponding coefficients (which we both call them A
in this note) for CNO cycle is much smaller than that of the p-p chain. (Do
< T
you know why?) Hence, at temperature ∼
crit , the p-p chain is the dominant
reaction. This is the reason why CNO cycle is only important for ZAMSS with
> 1.2M .
mass ∼
⊙
CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS
56
X
L
T
ρ
0.0
0.1
0.2
r
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
m(r)/M
Figure 5.1: A 1 M⊙ model during main-sequence hydrogen burning at time
4.2699 × 109 yr, showing radius, density, temperature, total luminosity, and
hydrogen abundance versus mass fraction.
L
Y
0
X
1
m/M
Figure 5.2: Model of a 1 M⊙ star just after it leaves the main sequence, at time
10.31 × 109 yr.
CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS
57
L
ρ
r
T
X
P
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
m(r)/M
Figure 5.3: Model of a 5 M⊙ star just after it leaves the main sequence at time
6.84461 × 107 yr.
L
X4
T
r
ρ
L
X4
P
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
m(r)/M
Figure 5.4: Model of a 5 M⊙ star during the giant star stage at time 7.04×107 yr.
CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS
58
Second, the sensitive temperature dependence of CNO cycle (namely, T 16 ) is
the origin of a convective core for high mass ZAMSS. For convective stability,
Eq. (5.4) gives
[
(
)]
3κ
dL
d
1
d2 T
= −
+L
2
2
4
3
4
dm
256π σr T dm
dm r T 3
3κ
dL
≈ −
, when r ≈ 0, L ≈ 0
2
4
3
256π σr T dm
3κAρT n−3
≈ −
.
256π 2 σr4
(5.8)
That is, for regions sufficiently close to the stellar core, |dT /dm| increases as
n increases. It turns out that for p-p chain, n = 4 and the core is radiative.
While for CNO cycle, n = 16 ≫ 4, the temperature gradient is so high that the
core becomes convective.
< 1.2 M , the temperature close to stellar surface is so low
Third, when M ∼
⊙
that free-free absorption (hence, Kramer’s opacity) is important. Such a large
opacity makes the stellar surface a good photon absorbing medium. Hence, the
only effective way to transfer these absorbed energy out to the stellar surface
> 1.2 M , the temperature close to stelis convection. In contrast, when M ∼
⊙
lar surface is high enough that electron scattering opacity becomes important.
Hence, in this case, the stellar surface is radiative.
< M < 0.3 M , temperature in most part of the star
Finally, when 0.1 M⊙ ∼
⊙
∼
is so low that Kramer’s opacity is important. Hence, the whole star is fully
convective.
5.4
Homology relation
Let us consider our stellar structure equations supplemented by the simplified
opacity, energy production equations as well as the equation of state. To illustrate the idea, we only consider the case of (a) the opacity κ is temperature
and pressure independent; (b) the energy production equation is a power law
of temperature and density; and (c) the EOS is just the simple ideal gas law.
These assumptions hold to give good approximation if most part of the (main
sequence) star is radiative, the temperature in most part of the star is hot
enough so that electron scattering is the dominant contribution to opacity, and
the mass of the star is less than about 10M⊙ so that contribution of radiation pressure is small. With these assumptions in mind, we write our stellar
CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS
59
structure equations below:
dP
dm
dr
dm
dT
dm
dL
dm
Gm
,
4πr4
1
=
,
4πr2 ρ
3κL
= −
,
2
256π r2 σr2 T 3
(5.10)
= AρT n ,
(5.12)
= −
P =
and
ρkT
.
µmH
(5.9)
(5.11)
(5.13)
Now, we define a dimensionless variable x by x = m/M . We denote
r
P
ρ
T
L
=
=
=
=
=
f1 (x)R∗ ,
f2 (x)P∗ ,
f3 (x)ρ∗ ,
f4 (x)T∗ , and
f5 (x)L∗ .
(5.14)
(5.15)
(5.16)
(5.17)
(5.18)
Exercise: Substitute Eqs. (5.14) and (5.15) into Eq. (5.9), we obtain
P∗ df2
GM x
=−
.
M dx
4πf14 R∗4
(5.19)
In a physical equation the dimensions on the two sides must match, and hence
in Eq. (5.19), where x, f1 , and f2 are dimensionless, P∗ must be proportional to
GM 2 /R∗4 . Without loss of generality, we can take the proportionality constant
to be unity. The equation above is then separated into
x
df2
=−
dx
4πf14
and P∗ =
GM 2
.
R∗4
(5.20)
Similarly, the above five equations can be decoupled into two sets. The first
set consists of four differential equations and one algebraic equation involving
CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS
60
f1 (x) through f5 (x) only:
df2
dx
df1
dx
f2
df4
dx
df5
dx
x
,
4πf14
1
=
,
4πf12 f3
= f3 f4 ,
3f5
,
= − 3
4f4 (4πf12 )2
= −
(5.21)
(5.22)
(5.23)
and
= f3 f4n .
(5.24)
(5.25)
The second set of equations are algebraic equations relating ρ∗ , T∗ , R∗ , P∗ , L∗
and M :
P∗
ρ∗
T∗
L∗
L∗
=
=
=
=
=
GM 2 /R∗4 ,
M/R∗3 ,
µP∗ /ρ∗ ,
T∗4 R∗4 /M ,
ρ∗ T∗n M .
and
(5.26)
(5.27)
(5.28)
(5.29)
(5.30)
The importance of this re-expression of the set of stellar structure equations is
that the values of ρ∗ , T∗ , R∗ , P∗ and L∗ can be expressed in terms of a single
parameter M . Moreover, once we solve the set of equations that relates the f1
through f5 , we solve a whole family of stellar structure equations for all values
of stellar mass M . This similarity property is known as homology.
Exercise: Eliminating ρ∗ , P∗ , T∗ , and R∗ from the above equations, we get
L∗ ∝ M 3
(5.31)
This mass luminosity relation agrees reasonably well with observations (see
figure 1.3.) A few more relations can be deduced as well. For example,
n−1
R∗ ∝ M n+3 ,
(5.32)
CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS
ρ∗ ∝ M
2(n−1)
1− 3(n+3)
L∗
T∗
2(3−n)
n+3
,
4
∝ Teff
, and
∝ µM/R∗ ,
61
(5.33)
(5.34)
(5.35)
4
where L∗ = 4πR∗2 σTeff
and µ is the mean molecular weight.
Since n is about 4 for p-p chain and 16 for CNO cycle, we know that for main
sequence stars, the radius increases with mass, mean density ρ∗ decreases with
mass. For p-p chain burning stars, Eq. (5.34) gives
log L ≈ 5.6 log Teff + constant,
(5.36)
and for CNO cycle burning stars,
log L ≈ 8.4 log Teff + constant.
(5.37)
Again, these slopes are in rough agreement with the observation in the H-R
diagram. Besides, the characteristic life time for a main sequence star can be
estimated by τMS = M/L ∼ M −2 . Thus, heavier main sequence star lives
shorter time on the main sequence.
For low mass stars, constant opacity is no longer a good approximation, instead,
the Kramers opacity law holds. In this case, it can be shown that
L∝M
5+31/2n
1+5/2n
≈ M 5.46
(5.38)
This results in a change of slope between high and low mass stars in the massluminosity relation (see Figure 1.3). Eq. (5.38) also gives
log L ≈ 4.12 log Teff + constant.
5.5
(5.39)
Evolution away from the main sequence
So far, we have omitted one fact in the evolution of a star. As a star burns H,
its mass fraction X changes. Provided that the time we are talking about is
much shorter than the nuclear burning timescale, we know that the change in
X, Y , and Z and hence also the change in stellar structure can be neglected.
This explains why main sequence star is so stable throughout most of its life.
As time goes, we have to supplement the stellar structure equations by
dX(r)
ϵ(r)
≈ −
,
dt
∆E
(5.40)
CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS
dY (r)
ϵ(r)
≈
,
dt
∆E
dZ(r)
= 0
dt
and
62
(5.41)
(5.42)
to the evolution of a hydrogen burning star, where ∆E is the energy released
per helium nucleus formed.
The effects of the change in mass fractions described by the above equations
can be summarized below: First, since hydrogen is burned only in the core,
an inhomogeneous distribution of chemical elements in the star will eventually
appear. Namely, the core becomes more and more hydrogen depleted while the
surface is still hydrogen rich. (More precisely, this statement is only true for
main sequence stars with mass higher than about 0.2 M⊙ in which the effects
of convection, diffusion and flow are not important in changing the distribution
of particles inside them during its main sequence life.) Can you justify this
statement? In addition, do you know why this statement is wrong for a 0.1 M⊙
main sequence star?
Second, the star must evolve due to such a gradual chemical change. Since
its core will be more and more hydrogen depleted and helium rich, hydrogen
burning will eventually be unable to provide the gas and radiation pressure
gradient, dP/dr, to counter balance gravity, namely −Gmρ/r2 near the core.
At that time, the core will contract.
Third, recall from the consequences of the virial theorem that for a sufficient
small timescale, both the thermal and gravitational potential energies of a star
are approximately conserved. So, as the core of the star contracts, we expect
the outer part of the star to expand. And since the temperature around the core
of the star increases due to core contraction, we also expect the temperature of
the outer part of the star to decrease. So, the star is heading towards the red
giant phase.
Chapter 6
Evolution of Pre-Main Sequence
Stars
6.1
Collapse of interstellar cloud
The interstellar medium is not empty, but consists of gas, dust, clouds, and
plasma. The densest part is interstellar clouds. If the mass is high enough,
the interior can be shielded from UV radiation from stars, such that molecules
could exist. They are called molecular clouds and most of them contain H2 and
CO. A star is formed from the collapse of molecular clouds. They can have high
masses of 103 to 107 M⊙ and the typical temperature ranges from ∼ 1 K (mostly
dark nebula) to ∼ 104 K (mostly bright emission nebula). Its density is about
10−24 –10−21 g cm−3 . Clearly the EOS of a cloud can be well approximated by
the ideal gas law.
If the thermal pressure in a gas cloud cannot balance the gravity, part the cloud
will collapse under external perturbation (e.g., triggered by supernova shock
wave). From the virial theorem, a cloud will collapse if −Ω > 2U . Assuming a
spherical cloud of uniform temperature and density, the gravitational potential
energy is
3 GM 2
Ω=−
(6.1)
5 R
and the thermal energy is
3
M
3
.
U = N kT = kT
2
2 µmH
63
(6.2)
CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS
64
Hence, the condition to collapse is
3M kT
3 GM 2
>
.
5 R
µmH
(6.3)
Put in M = 4πρR3 /3, we obtain
≡ MJ ,
M>
(6.4)
where MJ is called Jeans mass. It is the minimum mass of a cloud that can
collapse (i.e. upper limit for a stable cloud). Putting in the typical numbers,
√
MJ ≈ 105 T 3/2 M⊙ / n .
(6.5)
Thus, only giant gas cloud of mass of order of 105 M⊙ or more will collapse when
perturbed.
Exercise: Show that the Jeans length is given by
(
RJ =
15kT
4πρµmH G
)1/2
.
(6.6)
As the cloud collapse, its Jeans mass will change since T and n are changed. If
the cloud cools efficiently, the increase in n will lower the Jeans mass. Hence,
the stability condition in part of the cloud will be violated, resulting in the
fragmentation. This mechanism is believed to be the way for a large massive
gas cloud to collapse into a number of solar mass stars, which forms a stellar
cluster. As the cloud fragments get denser and hotter, they will eventually
become opaque and cooling becomes inefficient. The rise in temperature will
increase the Jeans mass. Once MJ exceeds the physical mass of the cloud,
fragmentation will stop.
We can estimate when does this happen. The power released from gravitational
collapse can be estimated by
(
Ω
GM 2 √
3
Ω̇ ∼
∼
Gρ ∼
τf f
R
4π
)1/2
G3/2 M 5/2
,
R5/2
(6.7)
CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS
65
√
√
where τf f ≈ R3 /GM ≈ 1/ Gρ is the free-fall timescale (see Chapter 2.6).
This energy is carried away via blackbody radiation. For efficient cooling,
Ω̇ < U̇ = 4πR2 σT 4 .
(6.8)
This gives MJ,min ∼ 0.02M⊙ T 1/4 µ9/4 .
We expect the collapsing process to take place within the free-fall timescale. As
ρ is very low for a gas cloud, τf f can be as long as a few million years. The
actual collapse is much longer, because rotation and magnetic field — while
neglected is this course — play an important role here. The conservation of
angular momentum and magnetic pressure slow down the collapse.
6.2
Protostars and pre-main sequence stars
The initial temperature of a cloud is only a few 10 to 100 K. It has a low
opacity (i.e. mostly transparent to photons) since κ ∼ 0.01 g−1 cm2 and the
mean density ρ ∼ (3M⊙ /4πR3 ) ∼ 2 × 10−16 g cm−3 , giving an optical depth
τ ≈ κρR ≪ 1. The gravitational potential energy is radiated away efficiently
and the temperature of the cloud stays as a constant.
The infalling matter gradually increases the opacity and hence the temperature.
When the core temperature reaches ∼ 2000 K, molecular hydrogen disintegrates,
then followed by hydrogen ionization and later on helium ionization. At a
central temperature of 2 × 104 K, hydrostatic equilibrium is achieved and a
protostar is formed. Numerical simulations suggest that the core has a mass
of 1.5 × 10−3 M⊙ and a radius of 1.3R⊙ at this stage. A protostar is powered by
gravitational energy and it is extremely luminous, can be ∼ 100 − 1000L⊙ for
a 1M⊙ star, with surface temperature 0.6 times of the Sun, and a large radius
of 70R⊙ . However, protostars are embedded in clouds that make observations
difficult. (How to observe a protostar?) Detailed structure of protostars from
numerical simulations can be found in Lasson, R.B. (1969) MNRAS, 145, 271.
The protostar will continue to accrete, until the surrounding material is depleted. Once the infall and accretion cease, mass of the star is nearly fixed
and the protostar is now known as a pre-main sequence (PMS) star. In this
stage, radiation can escape, hence, the PMS star cools down and contract. The
energy source is still gravitational contraction, but the evolution timescale is
governed by the Kelvin-Helmholtz timescale τKH , as compared to τf f for protostars. Note that τKH ∝ 1/R, meaning that it will increase during contraction,
while τf f ∝ 1/ρ decreases during contraction. (How to observe a PMS star?)
CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS
66
Exercise: Estimate the size and temperature of a protostar by assuming that
all gravitational energy released is used to dissociate molecular hydrogen and
to ionize hydrogen and helium. (Energy to dissociate H2 is χH2 = 4.5 eV,
ionization energy of hydrogen and helium are χH = 13.6 eV and χHe =24.6 eV+
54.4 eV=79 eV, respectively.)
3 GM 2
M
−Ω =
≈
5 R
mH
and take Y ≈ 1 − X, we obtain
R≈
(
X
Y
χH2 + XχH + χHe
2
4
50
M
R⊙ .
1 − 0.2X M⊙
)
(6.9)
(6.10)
As hydrostatic equilibrium is established, the temperature can be estimated
from the virial theorem
3 kT
1
M =− Ω,
(6.11)
2 µmH
2
which gives T ≈ 6 × 104 K.
6.3
Hayashi (convective) track
The core temperature of a PMS star is much higher than the surface temperature due to the large opacity of hydrogen. This results in a large temperature
gradient, making the entire PMS star (up to the photosphere) fully convective.
As the radius continues to shrink, the core heats up. However, the surface
temperature remains stable, because the photosphere can adjust its thickness
to counteract the rise of the core temperature. As a result, the luminosity decreases as R2 , and the star moves down nearly vertically on the H-R diagram
over the first million years, this is known as the Hayashi track.
The pressure is related to the density by
P = Kρ1+1/n ,
(6.12)
where n = (γa − 1)−1 . In other words, we are solving a polytropic EOS with
index n. Hence, from Eq. (2.24) the coefficient K is given by
K n = Cn Gn M n−1 R3−n .
Note that the coefficient Cn depends on n only.
(6.13)
CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS
67
The value of R is determined by joining the fully convective interior to the
radiative photosphere at the boundary r = R. Quasi-hydrostatic equilibrium
condition requires
dP
GM ρ
≈− 2 .
(6.14)
dr
R
Hence,
∫
GM +∞
PR ≈ 2
ρdr .
(6.15)
R R
In the photosphere, the optical depth of the photosphere is of order of 1,
∫
+∞
κ̄
R
ρdr ≈ 1
(6.16)
where κ̄ is the mean opacity of the photosphere. Taking κ̄ to be the opacity at
R and approximate it as a power-law of ρ and the surface effective temperature
Teff , we have
∫
b
κ0 ρaR Teff
+∞
R
ρdr ≈ 1 .
(6.17)
Hence,
PR ≈
GM −a −b
ρ T .
R2 κ0 R eff
(6.18)
Exercise: Using Eqs. (6.12), (6.13), (6.18), the ideal gas law, and the blackbody
luminosity, show that
log PR
n log PR
log PR
log L
=
=
=
=
log M − 2 log R − a log ρR − b log Teff + constant
(n − 1) log M + (3 − n) log R + (n + 1) log ρR + constant
log ρR + log Teff + constant
2 log R + 4 log Teff + constant
We can write
log L ≈ A log Teff + B log M + constant .
(6.19)
Using the results above, we found
A=
(7 − n)(a + 1) − 4 − a + b
0.5(3 − n)(a + 1) − 1
and
B=−
(n − 1)(a + 1) + 1
.
0.5(3 − n)(a + 1) − 1
For realistic values of n = 1.5, a ≈ 1 and b = 4, the slope A = 20. This
represents a very steep evolutionary path on the H-R diagram, i.e. a PMS star
CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS
68
Figure 6.1: Evolution paths of pre-main sequence stars from numerical simulations. The nearly vertical lines are the Hayashi tracks, then followed by the
nearly horizontal Henyey tracks.
evolves by reducing its luminosity with more or less constant effective surface
temperature (see figure 6.1).
The Hayashi track also sets the boundary for which a stable star can exist. The
region on the H-R diagram to the right of the line is called the forbidden zone
such that no stars can be in hydrostatic equilibrium. Finally, note that a real
PMS star does not evolve on the H-R diagram exactly on the Hayashi track as
the assumptions we used in the arguments above is overly simplistic. However,
the Hayashi track is still reasonably close to reality. The relevant timescale at
this stage is the Kelvin-Helmholtz timescale which is very short. Contraction
to main sequence stars takes up < 1% of the life time of a star, while main
sequence stage takes up 80%. This is part of the reasons why so few pre-main
sequence stars are detected. One example of PMS star is T Tauri stars, which
are less massive stars that evolve more slowly along the Hayashi track, and they
CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS
69
are highly variable due to mass ejection.
6.4
Henyey (radiative) track
Upon further contraction, the internal temperature rises to the point that opacity greatly decreases (Eq. 3.5: κf f ∼ T −7/2 ). The core is in radiative equilibrium. The convective zone recedes from the center, leaving only a convective
envelope. The core of the star contracts on the free-falling timescale, the gravitational energy becomes more negative. By the virial theorem, the internal
energy, and hence the surface temperature, has to increase rapidly, while the
luminosity increases gradually. Therefore, the PMS star will move away from
the Hayashi track on the H-R diagram, making a sharp turn to the upper left.
This is known as the Henyey track. For high mass stars, this can happen very
early. For example, the central temperature of stars above 10M⊙ can reach
3 × 107 K in 1000 years, then CNO is ignited (but not in equilibrium) and a
radiative core is formed due to the high temperature. Therefore, very massive
stars almost left the Hayashi track and on the Henyey track instantaneously.
On the other hand, stars < 0.5M⊙ remain in the convective phase throughout
their lives and there is no Heyney track for them.
Most light elements are burned off in this stage (see Chapter 3.7.3), except
deuterium, which is ignited while the star is still on the Hayashi track), but the
energy production is negligible compared to the gravitational energy. Finally,
the core temperature is high enough that stable hydrogen burning (p-p chain or
CNO cycle) gradually take over the energy generation and gravitational collapse
slows down. The star becomes a zero age main sequence (ZAMS) star. It
takes solar-type stars 3 × 107 yr for the protostar and PMS phases, which is
only 1% of the entire life time. Massive stars above 10M⊙ can take less than
105 yr and low-mass stars can spend more than 108 yr before hydrogen burning
reaches equilibrium.
You may notice that in Figure 6.1 there are some kinks near the end of the
Heyney track before the ZAMS. This is due to the depletion of 12 C in the
core such that the core has to slightly contract to adjust the rate of hydrogen
burning. If you are interested to learn more, can refer to a simulation paper by
Icko Iben Jr. (1965) ApJ, 141, 993, or the book by the same author: Stellar
Evolution Physics, Volume 1 (2013) (Cambridge: CUP).
Chapter 7
Evolution of Post-Main
Sequence Stars
7.1
Red giants
As the core temperature keep on increasing due to contraction, the thin layer of
hydrogen-rich material located just on top of the helium-rich core will be heated
up. Eventually, this thin hydrogen shell will burn. This shell burning becomes
the main source of energy for the star and the star becomes a red giant.
Let’s consider an idealized situation in which the core is made up of entirely
helium. This core is probably be very close to isothermal. (Do you know why?)
Nevertheless, an arbitrarily massive isothermal core is not stable. To see why,
we use the local form of the virial theorem. Recall from Eq. (2.8) that
∫
0
Vcore
P dV = Pcore Vcore +
2
3
2
αGMcore
4πPcore Rcore
αGMcore
=
+
,
3Rcore
3
3Rcore
(7.1)
where the label core refers to the boundary of the core (i.e., r = Rcore ), and α
is a geometric factor depending on the mass distribution of the core.
For ideal isothermal classical non-relativistic gas with temperature Tc = Tcore ,
∫
Vcore
P dV =
0
kTc ∫
kTc Mcore
ρ dV =
,
µcore mH
µcore mH
(7.2)
where µcore is the mean molecular weight of the material within a distance of
Rcore from the stellar core. Hence,
Pcore (Rcore ) =
2
3kTc Mcore
αGMcore
−
.
3
4
4πµcore mH Rcore
4πRcore
70
(7.3)
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
71
By considering dPcore /dRcore = 0, we know that for a fixed core mass Mcore ,
2
Pcore attains its maximum value of Pcore,max ∼ Tc4 /Mcore
µ4core . For a stable core,
this maximum pressure must be balanced by the pressure Penv exerted by the
envelope. By assuming that the core is very small, Penv can be estimated by
Eq. (2.5), namely, Penv ≥ GM 2 /8πR4 . Hence, the condition for stability of an
isothermal classical non-relativistic core is Pcore,max ≥ GM 2 /8πR4 . Now, applying one of the homology relations in Eq. (5.33), namely, Tc ∼ µenv mH GM/kR,
the stability condition becomes
Mcore <
constant
M ∼
(
µenv
µcore
)2
.
(7.4)
Actually, M. Schönberg and S. Chandrasekhar deduced this result with the
constant equals 0.37 in 1942.
For a pure helium core, µcore = 4/3. And for a solar composition envelope,
µenv ≈ 0.6. This implies the maximum mass of a stable isothermal nondegenerate and non-burning helium core is about 0.13 M⊙ .
> 2M ). One
Now, let’s consider the case of a high mass star (say of mass ∼
⊙
of the homology relations in Eq. (5.33) shows that the central density ρc of a
main sequence star decreases as its mass increases. So, for high mass star, the
central density is not very high so that the stellar core is still non-degenerate
even after it has exhausted its hydrogen fuel. The mass of this isothermal
helium core will gradually increase as hydrogen is burnt in the shell. As the
core mass exceeds the so-called Schönberg-Chandrasekhar limit, the core
will collapse. In time, the core will acquires the temperature gradient necessary
to balance gravity. But this temperature gradient causes loss of heat and hence
the core will further collapse leading to further temperature rise. This core
contraction process will take place in the Kelvin-Helmholtz timescale, which is
2
about GMcore
/Rcore Lcore ≈ 106 yr.
Such a contraction can be approximated by a quasi-static equilibrium process.
Besides, we expect that most of the gravitational potential energy released by
the contracting core will be absorbed by the outer layer of the star. Thus, as
the core contracts and heats up, the outer layer will expand and cool down.
Apart from the virial theorem, this is an alternative way to understand how a
red giant is formed.
Since the evolution from high mass main sequence to red giant is very rapid
in stellar evolution timescale, very few stars are found in the H-R diagram
immediately outside the main sequence. This gap is called the Hertzsprung
gap (see figure 7.1.)
There is an interesting consequence of the evolution towards red giants for high
Luminosity
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
72
B
C
A
Temperature
Figure 7.1: A schematic illustration of post-main sequence evolution for a 5 M⊙
star on H-R diagram (the light solid curve). The heavy solid curve shows
the main sequence. Stars spend very short time when evolve from point B
to point C. Only a few of the stars are found in this region, known as the
Hertzsprung gap.
mass stars. It turns out that the convective zone of a high mass main sequence star is rather extended. So, by the time when the hydrogen burning
shell is formed and the outer layer of the high mass red giant becomes convective, this convective region overlaps with the original main sequence convective
zone. Hence, some material originally located deep inside the stellar core will
be brought up to the stellar surface. This process is called the dredge-up (or
the first dredge-up since high mass star will undergo several dredge-ups in its
life and this is the first one). Do you know what kind of observational evidence
support for this idea of dredge-up?
To summarize, within about 106 yr, a red giant star is formed. A thin layer of
spherical shell immediately outside the core now is hot enough to burn hydrogen.
We call it the hydrogen burning shell. The core during this about 106 yr will
contribute very little to the nuclear energy generation.
The situation is slightly different in the case of low mass stars (with mass
<
∼ 2M⊙ ). Again one of the homology relations in Eq. (5.33) tells us that the
central density of this type of main sequence stars is rather high. In fact, it is
so high that the electron there is close to degenerate. Hence, the Schönberg-
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
73
Chandrasekhar limit no longer applies. As the core contracts after it has used
up its hydrogen fuel, it will quickly go degenerate.
Provided that the temperature is much lower than the Fermi energy, the equation of state of the helium core can be well-approximated by an electron Fermi
gas at almost zero temperature. Recall from Eq. 2.77, the EOS is a polytrope
of index n = 1.5.
Exercise: From Eqs. (2.23) and (2.24) concerning the mass, radius and central
density of a polytropic star, show that the mass and radius of the degenerate
helium core obey
−3
.
Mcore ∼ Rcore
(7.5)
Most importantly, it turns out that the electron degenerate pressure of the
helium core is sufficient to support the stellar envelope. As a result, for low
mass stars, the core contraction is slow and hence its road to red giants is
gradual. Thus, there is no Hertzsprung gap in the lower bottom part of the HR diagram. Note also that there is no dredge-up at this phase of the evolution
of low mass stars except for the fully convective extremely low mass stars.
In both the low and high mass star cases, the onset of hydrogen shell burning
(and before the helium shell burning that we are going to discuss), the stars are
located at the so-called red giant branch in the H-R diagram characterized
by the high luminosity and low surface temperature. The structure of a typical
red giant star is shown in the figure 7.2. The exception is when the mass of the
star is less than about 0.5M⊙ . In this case, the expansion of the radius of the
star is so small that it may not be able to reach the red giant branch at all.
From now on, our highly simplified analysis of the stellar structure equations is
no longer valid. In fact, the most direct way to answer the question of what the
structure is and what evolution of a red giant star take places is by numerical
solution. Such analysis has been carried out since the 1960’s by various groups
of astrophysicists, and perhaps the most famous one was headed by I. Iben, Jr.
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
74
Hydrogen burning shell
HeliumCore
Hydrogen−riched zone
Figure 7.2: The structure of a typical red giant star.
As the star expands, the surface gravity is low and hence material ejection in
the form of stellar wind can be very significant. Hence, from this time on, mass
conservation of the whole star is no longer a good approximation. This further
makes the modeling of stellar evolution for red giant and beyond difficult. Our
description of subsequent evolution of a star is, therefore, forced to be rather
descriptive at the level of this course. To add further complications, it turns
out that the precise subsequent evolution of a star depends on the mass of the
star. (Not just the initial mass of the star, but how much mass is lost by stellar
wind.)
7.2
Helium core burning and helium flash
As the core collapses further, the central temperature and pressure will be high
enough to burn helium via the triple-alpha process. For stars with mass greater
than about 2 M⊙ so that the stellar cores are non-degenerate, the ignition of
triple-alpha process is relatively smooth. That is, when the central temperature
reaches the helium ignition temperature, which is about 108 K, helium will burn.
After that, the contribution of energy production of the entire star due to helium
burning will increase steadily with time while that due to hydrogen burning
will decrease steadily (although it will not stop completely). Due to the high
temperature sensitivity of helium burning (ϵ ∼ T 40 ), the core of such star is
convective.
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
75
The situation is slightly more complicated if the main sequence star is less than
about 2 M⊙ (but higher than about 0.5 M⊙ ). Recall that the core of this type of
stars is degenerate. Unlike ideal gas law, pressure of degenerate Fermi matter
(in this case electrons) at low temperature is almost independent of temperature
T . Hence, for low mass star, electrons in the helium-rich core becomes strongly
degenerate before helium burning sets in.
We expect that the degenerate helium core of a low mass star is much more
massive than its envelope. So, homology relation (Eq. 5.33) holds approximately
for the stellar core. In particular, we have Tcore ∼ Mcore /Rcore . Combining with
Eq. (7.5), we conclude that low mass stars ignite core helium burning whenever
their core masses reach a threshold irrespective of their envelope masses. Detail
calculation shows that this threshold value is about 0.5 M⊙ .
Since the degenerate matter is highly conductive, the degenerate helium core
is highly isothermal. So, the entire core burns simultaneously once the helium
is ignited. Furthermore, the weak dependence of pressure of degenerate matter
on temperature implies that the rise in core temperature will not affect the
core pressure too much. Thus, the helium burning rate of the core will increase
quickly leading to a thermonuclear runaway. So, helium ignites in an explosion,
known as helium flash, in low mass stars. Actually, the peak luminosity (which
lasts for a few seconds at most) is about 1011 L⊙ which equals the luminosity
of the entire galaxy! Nevertheless, we cannot see helium flash directly from
the EM radiation emitted from the star as most of the energy released in this
sudden explosion is absorbed by the stellar envelope. (Nevertheless, we can
infer its onset in principle from the evolutionary track. Do you know how?)
A few seconds after the flash, the core temperature is hot enough that it can
be well approximated by an ideal gas once more. Thus, the core expands and
helium burning becomes stable.
< 0.5 M , their degenerate
Finally, for extremely low mass stars with mass ∼
⊙
helium cores are never hot enough to ignite helium burning. The degenerate
cores of these stars will keep on contracting, evolving directly from red giants
(if you can still can them giants) to helium white dwarfs.
No matter the star starts to burn core helium gradually or in a flash, the
central helium burning leads to an increase in the nuclear energy generation
rate. Consequently, the core expands and hence the envelope contracts. And
the star becomes smaller, less luminous and its surface becomes hotter. (Do
you know why?) So, the star descends from the red giant branch by moving to
the lower or lower-left in the H-R diagram.
The locus of high-mass core helium burning star forms the so-called the helium
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
76
main sequence — a somewhat misleading term as this type of stars have
hydrogen burning shells which still significantly contribute to nuclear energy
generation. In theory, the helium main sequence is a band similar to the high
mass part of the (hydrogen) main sequence except that it is shifted to the right
in the H-R diagram. In practice, the helium main sequence is mixed up with
the thick red giant branch. Computer simulations show that for high mass
stars, the evolutionary track from red giant to helium main sequence is roughly
horizontally to the left in the H-R diagram. (Interestingly, when the helium in
the core is used up at a later time, the star will again move roughly horizontally
to the right in the H-R diagram. That is to say, the combined evolution make
the star to form a loop in the H-R diagram.) Some stars on the helium main
sequence are Cepheid pulsating variables whose period is of order of days to
months.
The locus of low-mass core helium burning star forms the so-called horizontal
branch, a roughly horizontal strip stretching between the main sequence and
the red giant branch. (See figure 7.3 below.) Clearly, these stars have about
the same core masses just before core helium ignition. So, it is not surprising
that the location of a star on the horizontal branch is due to the envelope
mass. Some stars on the horizontal branch are pulsating variables, known as
RR Lyrae variables. Periods of RR Lyrae variables, which are about a few
hours, are much shorter than those of Cepheid variables. If time permits, I will
talk about pulsating variables in the special topics.
The lifetime of a core helium burning star, irrespective of its mass, is much
shorter than the corresponding core hydrogen burning main sequence star. It is
because the energy released per nucleon burnt is lower for helium. In addition,
the luminosity of a core helium burning star is much higher than a main sequence
star of the same mass.
7.3
AGB star
Evolution of a core helium burning star is in some sense quite similar to that of
a core hydrogen burning main sequence star. More precisely, a carbon-oxygenrich core is gradually formed. Then the star further expands. Most of the
energy production now occurs in the helium burning shell and the hydrogen
burning shell on top of the core. Once more, material from the core is dredged
up and mixed into the envelope. If the carbon oxygen core is degenerate, the
star is called an asymptotic giant branch (AGB) star (see figure 7.4.) A star
may leave the AGB phase by either complete removal of hydrogen envelope by
stellar wind; or ignition of carbon in the core.
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
102
B
Luminosity L /L
A
77
Red Giant
Horizontal Branch
10
Main Sequence
1
10000
5000
2500
Effective Temperature T eff (K)
Figure 7.3: A schematic H-R diagram showing the evolution of low-mass stars
from red giant branch to horizontal branch. Point A corresponds to large loss
of mass in helium flash while point B corresponds to small loss of mass.
The two burning shells of an AGB star are separated by a helium-rich nonburning layer. The outer hydrogen burning shell burns hydrogen thereby increasing the mass of the helium non-burning layer. Similarly, the helium burning
shell consumes the helium and hence eating into the helium non-burning layer.
Due to the great difference between rates of hydrogen and helium burning processes, the two burning fronts do not advance at the same rate all the time.
The two shells supply energy in turn and the mass of the helium non-burning
layer changes periodically. During most time in the cycle, hydrogen is burnt in
the external shell while the helium burning shell is extinct. Thus, the helium
layer separating the two shells increases in mass. The gradual contraction of
this helium layer increases the temperature of the helium shell until it ignites.
Such ignition may be in the form of a shell flash for low mass star. As the
helium shell burns, the outer layer of hydrogen burning shell expands and cools
down. This leads to a great suppression of the hydrogen burning rate. As the
helium burning front advances, the temperature of the hydrogen burning shell
increases while that of the helium burning shell decreases. Thus, a new cycle
of hydrogen and helium shell burning begins. To summarize, the AGB star is
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
78
H burning shell
H riched zone
CO riched
core
He burning shell
He riched zone
Figure 7.4: The typical structure of an AGB star.
burning hydrogen most of the time while only it is burning helium occasionally. In fact, astronomers sometimes called the helium-shell-burning phase of an
AGB star the thermal pulse. This cycle of hydrogen and helium ignition with
thermal pulse lead to sudden increase in radiation pressure in the interior of an
AGB star. It leads to a rapid mass loss in the star in the form of stellar wind.
(Its mass loss rate is of order of 10−5 M⊙ yr−1 .) A planetary nebula is formed
eventually. Detail study of planetary nebulae will be studied as a special topics
if time permits.
An AGB star will undergo up to a few thermal pulses in its life. Each cycle of
hydrogen- and helium-shell burning also leads to dredge-up of materials. (There
is a confusion of naming here as astronomers studying AGB stars also called
the first time when materials in the core of an AGB star is transported to the
stellar surface the first dredge-up. So, be careful when you read the literature.)
As expected, AGB stars are variables. In fact, a lot of long period pulsating
variables, such as Mira variables, are AGB stars.
While most of the outer layer of an AGB star will eventually be ejected forming
a planetary nebula, the core of an AGB star will collapse after using up all its
hydrogen and helium fuel. The mass of an AGB star is not high enough so that
even when the electrons in the core become degenerate, the core temperature is
not high enough to ignite carbon burning. The core now becomes a dead star,
whose gravitational force is counter-balanced by electron degenerate pressure.
This star is called a white dwarf. Properties of white dwarf will be discussed
in other astrophysics courses. Note that only a very small portion of stars
are massive enough to have a non-degenerate C-O core while do not undergo
supernova explosion.
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
105
Planetary Nebula
red supergiant
104
103
Luminosity L/L
79
AGB Star
102
Red Giant
Horizontal Branch
10
1
Main Sequence Star
−1
10
10−2
white dwarf
−3
10
40000
20000
10000
5000
2500
Effective Temperature T eff (K)
Figure 7.5: An illustration of evolutionary path of a low mass star from main
sequence star to white dwarf in H-R diagram. Note that the path from AGB
star via planetary nebula to white dwarf is not fixed.
7.4
Evolution of high mass star after the formation of carbon-oxygen-rich core
The evolution of a high mass star up to the formation of a carbon-oxygen-rich
core is similar to that of a 3 M⊙ star. But there are small differences. Due to its
high mass, energy generation rate in the core is so high that radiation pressure
is important even in the main sequence phase. This leads to mass loss by stellar
wind when the initial mass of the star is greater than about 30 M⊙ . In fact, the
stellar wind is so strong in some of these stars that the entire hydrogen envelope
is blown away leaving the helium-rich core. Such stars — luminous, hydrogen
depleted, and with high mass loss rate — are known as Wolf-Rayet stars.
Besides, the radiation pressure in the core is so strong that the luminosity in
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
80
the main sequence phase is already close to the Eddington limit. Hence, the
luminosity remain more or less constant in spite of internal changes. Thus, its
evolutionary track in the H-R diagram is more or less horizontal.
After the formation and collapse of a carbon-oxygen-rich core, the central temperature is hot enough to burn carbon and later on oxygen. At this stage, the
core temperature is so high that energy lost due to neutrino emission becomes
significant. In fact, all major core nuclear burning processes take place in rapid
succession until the inner core is 56 Fe rich. Surrounding the core are shells that
burn different types of nuclei.
Finally, the star is so massive that its core remains non-degenerate until the
very last stage of its evolution. (Do you know why?)
7.5
Supernova
The theory of supernova explosion is rather involved. Here, I shall only outline
the basic idea behind.
As 56 Fe is the most energetically favorable nucleus, no major exothermic nuclear
reaction can take place in the 56 Fe-rich core. When this core contracts, electron
in the core eventually becomes degenerate. Unfortunately, since the star is
massive, the core mass is so high that degenerate electrons become relativistic.
As discussed in Section 2.12, the pressure scales like ρ4/3 rather than ρ5/3 . That
is, the core can be approximated by a polytropic EOS with index n = 3. But
3−n
1−n
for a polytropic stellar core, Rcore
∼ Mcore
. Hence, as n tends to 3, there is
only one possible solution for Mcore . Thus, the core is unstable upon addition
of mass. (One way to remember this fact is that dP/dρ is not big enough to
counterbalance gravity as ρ increases.)
Since the pressure of the relativistic electron degenerate core is not strong
enough to oppose gravity, the core further collapses. Hence, the core is further
heated up. The EOS of the degenerate core matter is insensitive to temperature making the heat up more unrestrained. Now, the temperature is so hot
that photodisintegration of iron back to 4 He and then further back to protons
and neutrons becomes important. This endothermic reaction quickly absorbs
the heat from the core and hence reduces the central pressure. With a sudden
decrease in central pressure, the core contracts with an even higher rate. (In
fact, at almost the free-fall rate.) Even worse, the core becomes so dense that
protons can capture electrons and convert into neutrons plus neutrinos. This
reaction is also endothermic. Besides, it reduces the number of particles in the
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
81
core. The combined effect is to reduce the core pressure further. Finally, the
neutron gas in the core becomes degenerate and the neutron degenerate pressure is sufficient, at least temporarily, to stop the collapse. Such neutron core
is about 10 km in radius.
The gravitational energy released by the contracting core is of the order of
2
GMcore
/Rcore ≈ 1053 to 1054 erg. The energy absorbed during the disintegration
of nuclei is only about 1052 erg which is about one tenth to one hundredth that of
the total gravitational energy released. Thus, most of the gravitational energy
released is available to eject the material outside the core and for producing
the huge luminosity and neutrino flux we see during a type II supernova
explosion (also called core-collapse supernova).
Recall that in the final phase of the collapse, a large number of neutrinos is
produced in the core within a short period of time. These neutrinos are responsible for transferring the released gravitational potential energy out from
the core. Although neutrino interacts very weakly with matter under most circumstances, a non-negligible neutrino opacity builds up in the core because of
the high density and high neutrino flux. Hence, some of the neutrino energy is
absorbed by the envelope. Consequently, the envelope of the star quickly expands. This explosion is known as type II supernova (or simply supernova when
no confusion is possible). (In contrast, type I supernova is the explosion that
destroy a white dwarf as it accrues too much material from its surrounding.
Observationally, the spectrum of a type II supernova contains hydrogen lines
while that of type I supernova does not. Do you know why?)
When observed far away from this dying star, the luminosity of the star suddenly
goes up. At first, most radiation is in the UV range as the effective surface
temperature is high. But the surface temperature quickly drops and the object
radiates mainly in the optical range. The later time optical emission is then
powered by radioactive decay of newly formed elements. Typical light curve of
a type II supernova is shown in figure 7.6 below.
There is an important prediction in this model, namely, that we should see the
neutrino arriving the Earth faster than the brightening up of the supernova.
This prediction was confirmed in the famous supernova explosion SN1987A.
Actually, neutrinos were observed a few hours before the supernova became
visible for SN1987A.
Heavy elements in the envelope are ejected into the interstellar space in supernova explosion. These elements are produced in two ways. They may be
produced in the core and shell burning prior to supernova. Alternatively, they
may be formed during the supernova. Since the peak temperature in a super-
CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS
82
Blue Magnitude Below Maximum Light
0
1
2
3
4
5
6
7
8
0
50
100
150
200
250
300
350
400
Days after Maximum Light
Figure 7.6: A typical light curve of a type II supernova.
nova explosion can be as high as 5×109 K locally, nuclear statistical equilibrium
is achieved on a timescale of seconds. In this way, different kinds of nuclei are
produced. In particular, the 56 Ni formed will later on decay to 56 Co with a halflife of 6.1 d. Then 56 Co will decay to the stable 56 Fe with a half-life of 77.1 d.
The energy released in these two decays powers the supernova light curve after
the initial decline from maximum. Signatures of the decay of these unstable
nuclei are observed in supernova light curves.
Chapter 8
Special Topic: Introduction to
Stellar Pulsation
Throughout this course, we focus on the static structure of a star. More
precisely, we assume that the star is in mechanical equilibrium all the time.
Nonetheless, under suitable conditions, the equilibrium configuration of certain
stars is unstable against perturbation. Such perturbation may lead to stellar
pulsation. Study of stellar pulsation is important for astronomers for the following reasons: 1. Distance measurement. Certain stellar pulsation leads to
pulsation variable. The most famous one is the Cepheid variable, which is
well-known for its period-luminosity relationship that has been used as
standard candle to set the cosmological distance scale. 2. Stellar structure.
Our Sun and other main-sequence stars also show pulsations, albeit with much
smaller amplitudes. By observing different oscillation modes of a star, one can
infer its interior structure. This is known as asteroseismology (or helioseismology for the specific case of the Sun).
8.1
Some basic terminology
The instability strip is a narrow region in the Hertzsprung-Russell diagram
where most of the pulsating variables lie (except the Mira variables), e.g., it
intersects with the horizontal branch at RR Lyrae, crosses the supergiants at
Cepheids, and crosses the white dwarf region at ZZ Ceti.
The pulsations in Cepheids, Miras, and RR Lyrae are radial. They show periodic contraction and expansion with periods from 1 to 100 days. On the other
83
CHAPTER 8. STELLAR PULSATION
84
Figure 8.1: Location of pulsating variables on the H-R diagram.
hand, the Sun, δ Scuti, and white dwarfs are non-radial pulsators. Different regions on the stellar surface contract and expand differently. Non-radial
oscillations can be described by the spherical harmonics
v
u
u
m
m t 2l + 1 (l − m)! m
Yl (θ, ϕ) = (−1)
Pl (cos θ)eimϕ .
4π (l + m)!
(8.1)
Stellar pulsation is similar to sound waves in musical instruments. They are
standing waves with the center of the star as a node and the stellar surface as
an antinode. In the fundamental mode (l = 0), there is no other nodes and
Type
Mira
Cepheids
W Virginis
RR Lyrae
δ Scuti
β Cephei
ZZ Ceti
Period
Mode
100–1000 d
radial
1–50 d
radial
10–20 d
radial
1 hr–1 d
radial
1–3 hr
rad. & non-rad.
3–7 hr
rad. & non-rad.
100–1000 s
non-rad.
Table 8.1: Some common pulsating variables.
CHAPTER 8. STELLAR PULSATION
85
the entire star pulse in a single direction at a given time. The first overtone
(l = 1) correspond to one node between the center and the surface, the second
overtone (l = 2) has two, and so on. A star can exhibit several modes at the
same time. Cepheids are believed to oscillate in the fundamental mode, while
both the fundamental mode and the first overtone can exist in RR Lyrae and
Miras.
8.2
A simple model
In this introduction, we will focus on the theory of the simplest possible model
of stellar pulsation, namely, adiabatic radial pulsation in the small amplitude
limit. Further analysis can be found in the following review articles: J. P. Cox,
Reports on Progress in Physics 37, 563 (1974); A. Gautschy and H. Saio, Annual
Review of Astronomy and Astrophysics 33, 75 (1995); A. Gautschy and H. Saio,
Annual Review of Astronomy and Astrophysics 34, 551 (1996).
The basic idea is simple, when a layer is compressed, it heats up and becomes
more opaque to radiation. Radiative diffusion slows down, heat, and hence
pressure, build up. This eventually pushes the layer outward. The expansion
cools down the material, and it becomes more transparent to radiation, such
that heat can escape more easily and the pressure subsequently drops. Finally,
the layer falls back and the cycle starts over.
Here we use the so-called Lagrange approach which focuses on the motion of
individual particles in a star. (An alternative approach is that of Euler, which
focuses on the evolution of thermodynamic variables such as density and pressure. Both approaches give the same physics but the later one is sometimes
more effective in solving macroscopic problems involving fluid.)
The conservation of mass (Eq. 2.2 in Chapter 2) in the time independent case)
should be rewritten as
∂m
= 4πr2 ρ(r, t) .
(8.2)
∂r
In contrast, the equation of hydrostatic equilibrium (Eq. 2.3 in Chapter 2)
becomes more complicated in the time dependent case. We assume that the
oscillation is radial so that all physical parameters are functions of r and
t only. In this case, the mass of a small spherical shell from radius r to
r + ∆r equals 4πr2 ρ(r, t)∆r. The gravitational force acting on this shell is
−Gm(r, t)4πr2 ρ(r, t)∆r/r2 . The (gas plus radiation) pressure gradient acting
CHAPTER 8. STELLAR PULSATION
86
on this shell is ∂P/∂r. Hence, by Newton’s second law,
G4πr2 ρm
∂P
4πr ρr̈ = −
.
− 4πr2
2
r
∂r
2
(8.3)
In other words,
r̈ = −
Gm
∂P
− 4πr2
.
2
r
∂m
(8.4)
Assuming that the oscillation is adiabatic and the star is made up of an ideal
gas, pressure is given by
P = Kργa
(8.5)
for some constant K.
Analytical solution to these three partial differential equations, namely, Eqs. (8.2),
(8.4) and (8.5) are difficult to find. Nonetheless, in most cases, the amplitude of
oscillation of a star about its equilibrium position is small enough. In this case,
we need only to keep up to the first order term in the Taylor series expansion
about the equilibrium configuration of the star. Besides, the oscillation of the
star can be well approximated by a S.H.M. In other words, using m and t are
the two independent variables, we may write
P (m, t) = P0 (m)[1 + p(m)eiωt ] ,
(8.6)
r(m, t) = r0 (m)[1 + x(m)eiωt ] ,
(8.7)
ρ(m, t) = ρ0 (m)[1 + d(m)eiωt ] ,
(8.8)
and
where P0 (m), r0 (m) and ρ0 (m) are the equilibrium pressure, radius and density
on the sphere whose mass contained equals m.
Substituting the above equations into the three coupled partial differential equations and keeping up to the first order terms, we obtain
Gm(p + 4x)
P0 ∂p
= ω 2 r0 x +
,
ρ0 ∂r0
r02
r0
(8.9)
∂x
= −3x − d ,
∂r0
and
p = −3γa x − γa r0
(8.10)
∂x
.
∂r0
(8.11)
Eliminating p and d, we have
(
∂x
∂
γa
∂r0
∂r0
)
[
]
4 ∂
Gmρ0 γa ∂x
ρ0 Gm
+
(γa x) −
+
(4 − 3γa ) + ω 2 x = 0 .
2
3
r0 ∂r0
P0 r0 ∂r0 P0 r0
(8.12)
CHAPTER 8. STELLAR PULSATION
87
So the problem of stellar oscillation becomes the problem of finding eigenvalues
ω 2 for above linear partial differential equation. If the eigenvalue ω 2 > 0, the
solution is oscillatory which corresponds to an radial oscillation of the star. If
ω 2 < 0, the solution contains an exponentially decaying or growing term which
indicates instability.
Eq. (8.12) is still too complicated to be solved analytically in general. Nevertheless, we have one specific meaningful case in which this eigenvalue problem
has been investigated in detail. This is the case of (1) the adiabatic expansion
coefficient γa is a constant throughout the star; and (2) x is m independent. In
this case, Eq. (8.12) becomes
Gm
r03
4πGρ̄
= (3γa − 4)
,
3
ω 2 = (3γa − 4)
(8.13)
where ρ̄ is the mean density of the star. In other words, the oscillation of the
star is stable if γa > 4/3; and the period of oscillation Π in this case is given by
[
3
Π = 2π
4π(3γa − 4)Gρ̄
]1/2
.
(8.14)
For non-relativistic ideal gas, γa = 5/3. Put this into the equation above, we
have:
Type
RR Lyrae
Classical Cepheids
W Virinis
Period
≈ 0.5 d
≈ 7d
≈ 15 d
This agrees reasonably well with observations.
We have not finished our discussions yet. In reality, our linear analysis in the
adiabatic regime is an overly simplified approximation. Actually, if we include
higher order effects, radial oscillation of a star will be damped out quickly unless
there is an effective way to pump energy into the oscillator. Indeed, pulsating
variables, such as Cepheids, occupy only a restricted strip on the H-R diagram,
strongly suggesting that an effective excitation mechanism exists only under
rather restrictive conditions.
CHAPTER 8. STELLAR PULSATION
8.3
88
κ-mechanism and the instability strip
Detail analysis of the excitation mechanism is rather involved. The most famous
one is the so-called κ-mechanism. This mechanism assumes that the opacity
increases upon compression in some region of the star so that the radiative
luminosity is blocked in the compression phase of oscillation. Therefore, the
region gains thermal energy in compression phase and loses thermal energy in
expansion phase.
Recall that for most part of an “ordinary” star, κ is well approximated by
κ0 ρT −7/2 (Kramer’s law). Hence, upon adiabatic compression, the opacity actually decreases. Thus, the condition for κ-mechanism is not satisfied in most
stars. Nevertheless, such a condition holds in regions where there is partial
ionization of H and Hei and/or Heii. In fact, κ-mechanism due to such partial ionization zone is responsible for the excitation of pulsation in a number
of pulsating variables including RR Lyrae, classical Cepheids and W Virginis.
Because the condition of onset of the κ-mechanism due to partial ionization is
very stringent, pulsating variables are rare and can only be found in restricted
regions on the H-R diagram.
We will not go into the detail mathematics here, but only give a qualitative
description of the κ-mechanism and its connection to the instability strip. In
the partial ionization regions, part of the energy input from compression can
be turned into more ionization instead of heating. Therefore, it is possible to
increase in density more than in temperature. Hence, κ can be increased due
to compression. Later in the cycle, electron-ion recombination during decompression can release energy to lower the opacity.
Where is the κ-mechanism expected to occur? Location of the partial ionization
zone in a star depends strongly on the effective temperature. If the temperature
is too high, the ionization will be close to the surface, resulting in insufficient
mass to drive sustainable oscillations. On the other hand, if the temperature is
too low convection in the outer layer will strongly suppress the κ-mechanism.
Therefore, we expect a narrow, near vertical strip in the H-R diagram.
8.4
ε-mechanism
For completeness, we also mention the ε-mechanism. It was proposed before
the κ-mechanism to explain stellar pulsations. It suggests that compression at
the center of a star can lead to higher density and temperature, thus, increases
CHAPTER 8. STELLAR PULSATION
89
the thermonuclear energy production rate. However, detailed calculations show
that the oscillation amplitude at the central regions of most stars are small, such
that this effect is insignificant. However, this is still an important mechanism
in massive stars.
8.5
Cepheid variables
Classical Cepheids (or Cepheids for short) are luminous supergiants. It is named
after the first known example, δ-Cephei, which is still an important calibrator
for the period-luminosity relation (see below). Their immediate precursors are
massive young O or B type stars. Their periods are about 2 to 60 d, mean
luminosities about 300 to 40000 L⊙ , mean surface temperature about 4000 to
8000 K, and peak-to-peak magnitude variation of about 0.4 to 1.4 mag. Its
stellar atmosphere is cold enough to have a layer of partially ionized Hei and
Heii. Hence, κ-mechanism leads to stellar pulsation. (See Figure 8.2 below
for a typical light curve, color change and radial velocity of a classical Cepheid
variable.)
mv
3.5
4.0
4.5
Teff
0.0
0.5
1.0
0.0
0.5
1.0
0.5
1.0
6500
6000
5500
Vr (kms −1)
20
10
0
−10
−20
0.0
Phase
Figure 8.2: Location of δ Cephei in the sky, and the variation of visual magnitude, effective temperature and radial velocity.
In 1912, Henrietta Leavitt accidentally found that the period and luminosity
of classical Cepheid variables are related. In a more modern language, such a
CHAPTER 8. STELLAR PULSATION
90
period-luminosity relation is found to be
(
L
log
L⊙
)
= 2.43 + 1.18 log Π .
(8.15)
That is, the longer the period, the brighter the star is. This famous relation
is also called the Leavitt Law after her. Note that the above relation is valid
plus or minus a few percent error.
We may understand the period-luminosity relation as follows. First, the peakto-peak magnitude difference is relatively small, hence we may assume that
the effective surface temperature of a classical Cepheid to be a constant. The
luminosity of the star L ≈ 4πσR2 T 4 is therefore a function of stellar radius R
only. Although classical Cepheids are supergaints, the homology relation for
main sequence star still holds approximately. Hence, we have an approximate
mass-luminosity relation L ∼ M α for classical Cepheids. Finally, we have the
period-mean density relation Π ∼ ρ̄−1/2 . Combining these three constraints,
we arrive at a period-luminosity relation for classical Cepheid variables L ∼
Π4α/(3α−2) . For α ≈ 3, we obtain L ∼ Π1.7 . This is about the same order as the
empirical formula above, but we cannot derive the exact value since the mean
density approximation is oversimplified.
Indeed, the above argument is rather general and can be applied to quite a
number of pulsation variables including W Virginis and Mira type variables.
A notable exception is the RR Lyrae variables. Observations suggest that the
absolute magnitudes of RR Lyrae variables fall into a narrow range of 0.6 ± 0.2.
Hence, within about 10% accuracy, we may assume that all RR Lyrae stars are
of the same luminosities and hence can be regarded as standard candles.
8.6
Cepheid variables in the cosmic distance
ladder
Since the period is much easier to measure to high precision, as compared to
other observables such as spectrum or position. The period-luminosity relation
immediately turned Cepheids into standard candles, offering a very powerful
tool for distance measurements. Edwin Hubble observed Cephieds in other
galaxies to discover Hubble’s law. Even today, Cepheids play an important
role in the construction and calibration of the cosmic distance ladder, up to the
order of 10 Mpc (see the Hubble Space Telescope Key Project).