Department of Physics, HKU - Chapter 1 Introductionphys4650/notes.pdf · 2020-01-21 · CHAPTER 1. INTRODUCTION 6 40000 20000 10000 5000 2500 106 104 102-2 10-4 Temperature (K) Luminosity

Chapter 1

Introduction

1.1 An introduction to the cgs units

The cgs units, instead of the MKS units, are commonly used by astrophysicists.We will therefore follow the convention to adopt the former throughout thiscourse. The cgs units mean cm, g, and second.

Length: in cm = mMass: in g = kgTime: in sForce: in dyn = N (hence computing the mass and

acceleration in cgs units, the corresponding forcewill automatically be in dyn)

Energy: in erg = J (hence computing mass and ve-locity in cgs units, the corresponding energy willautomatically be in erg)

Power: in erg s−1

Flux: inPressure: inCharge: in the electrostatic unit of charge (esu), also known

as the statcoulomb, so that Coulomb’s law be-comes F = q1q2/r

2

B-field: in gauss (G) = 10−4 TField energy density: U = E2/8π or B2/8π

1

CHAPTER 1. INTRODUCTION 2

1.2 What is a star?

Stars are the building blocks of the visible Universe, so it is important to under-stand 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 cleardefinition of stars. Roughly speaking, we define a star to be a gravitationallybounded object which has a nuclear energy source in the form of fusion inthe core. Thus, Jupiter and Comet Halley are not stars. But this definition isnot sufficient in this level of complexity because it also includes brown dwarf inthe family of stars.

Recall that brown dwarfs are objects that are too massive to be planets but toolight to be stars. More precisely, in this context, a planet is a self-gravitatingbody without any nuclear fusion in the core although it may have other means ofenergy generation (such as radioactive decay). Thus Jupiter is a planet althoughJupiter has an internal heating mechanism. (Do you know what internal heatingmechanism it has?) A brown dwarf, in contrast, is a self-gravitating object thatburns (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 along time. More importantly, the energy released in such a burning alone is notsufficient 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 ofobjects which burn nuclear fuel in a shell rather than at the center.

Hence, we use the following more refined definition of star:

A star is a gravitationally bounded object which has a sustainablenuclear fusion energy source in the interior.

Using this refined definition, brown dwarf is not a star. Neutron star is not a stareither. Red giant is a star for it has a hydrogen-burning shell although its coreis not burning any nuclear fuel. An self-gravitating object whose sustainablenuclear 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 isnot 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.


1.3 Assumptions in our study of stellar struc-

ture and evolution

1. The star is isolated. Although most stars in the Universe are not iso-lated, it makes sense to study the structure and evolution of an isolatedstar before going on to study the more complicated multiple stellar sys-tems.

2. Right at the moment of formation of a star, its chemical com-position is uniform. Why this is a good approximation? I shall furtherdiscuss the validity of this assumption when we study main sequence stars.

3. The star is irrotational and is spherically symmetric.

Exercise: For a star with mass M , radius R, and angular velocity ω,calculate the ratio of rotational energy to gravitational energy. What isthe ratio for the Sun?

This ratio is very small for most stars. Thus, rotational distortion canbe safely neglected. In a similar way, the ratio of magnetic energy to thegravitational potential energy of a star is B2R4/8πGM2 <∼ 10−11R4/M2

where M and R are in units of solar mass and solar radius, respectively.Thus, the magnetic energy is very small compared with the gravitationalpotential energy. Hence, both rotation and magnetic field for almost allstars do not significantly break their spherical symmetry. Therefore, aspherically symmetric star serves as a very good approximation in thestudy 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 gainmass by accretion. These processes are important in the study of certainstars such as white dwarfs, AGB stars and binary stars. Nonetheless,the mass change rate for most main sequence stars is small and can beneglected.

5. The star is in local thermodynamic equilibrium (LTE). In otherwords, every microscopically large but macroscopically small region is veryclose to and hence can be well approximated by a region in thermodynamicequilibrium. Thus, it makes sense to talk about thermodynamic variablessuch as temperature, density and pressure in these regions. This approx-imation greatly simplifies our analysis and serves as the starting point tothe study of more refined calculations. How good these approximations


are in the study of main sequence stars? To answer this question, letme tell you that the photon mean free path near the core of our Sun isabout 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 solarcore. The typical equilibration time for photons, nuclei and electrons inthe solar core is also very short compared with, say, the lifetime of theSun. (I will justify these claims later on in this course.) Hence, LTE isa very good approximation for studying the overall structure of the Sunand most other stars. In contrast, LTE is no longer a very good approxi-mation to study stellar atmosphere such as the chromosphere of our Sun.(Do you know why?) Fortunately, the atmosphere of many main sequencestars has a small mass compared with the star itself. Therefore, LTE stillserves as a very good simplification in our first course of stellar structure.(Another exception is when we study neutrinos emitted by a star. Do youknow why?)

Consequence of our assumptions

Since we assume that a star is spherically symmetric, we can forget about itsangular velocity as well as the magnetic field strength. Thus, we may charac-terize 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 distancefrom the stellar core r and time since the birth of the star t.

1.4 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 isabout 1000 times more massive than Jupiter, or about 3×105 times moremassive 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 isabout 100 times that of the Earth.

• Radiation: Ranges from about 0.01 to 106 solar luminosity for most livingstars. The symbol for solar luminosity is L and 1L ≈ 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 primarilyH and He plus a trace amount (which is about a few percent by mass)


of heavy elements. In astronomy, heavy elements (or some books callthem by the misleading term “heavy metals”) are defined as elementswith atomic number greater than 2.

• Surface temperature: Ranges from about 4000 K to 30000 K for mostmain sequence stars.

• Magnetic field strength: Of the order of 1 G over most of the solarsurface (although it can go up to about 3000 G in sunspots). But someliving stars may have field strength as high as several thousand G.

• Rotation: Observation suggests that most stars rotate rather slowly. Asfor the Sun, the rotational period is about 1 month.

1.5 The Hertzsprung-Russell diagram

As the name suggests, this diagram was first studied by E. Hertzsprung andH. 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 temperaturevs. 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 decreasesas we move rightward in the H-R diagram.

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 starswithin a space of radius of the order of 100 light years. Thus, we can safelyassume that stars in a single star cluster are formed from roughly the samematerial at about the same time under roughly the same physical conditions.Thus, mass of a star is perhaps the most important parameter to determine theproperties and evolution of that star within the same star cluster.

It turns out that stars are not located randomly in the H-R diagram; theycluster 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 stellarclusters just before their formation is more or less the same. Thus, the differencebetween the location of stars of different clusters on the H-R diagram should bemostly determined by their age. It is rather hard to determine the absolute ageof a star cluster. (By absolute age, we mean how old the stars in that cluster.)


20000 250050001000040000

610

410

210

−2

−410

Temperature (K)

Lu

min

osi

ty (

L )

10

1

Giants

Main Sequence

Supergiants

White Dwarfs

Figure 1.1: The H-R diagram (schematic)

Nevertheless, it is quite easy to determine the approximate age of star clusterssince old star clusters tend to have condensed core. (Do you know why?)

Figure 1.2: A comparison of H-R diagram between young and old clusters

Astronomers find that the location of stars in the H-R diagram depends stronglyon the degree of core condensation and hence the age of the correspondingstar cluster. (See figure 1.2.) Most stars in a young cluster locate in a bandcalled the main sequence. Consequently, we call a star lying on the mainsequence 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 whichstars have been evolved away from the main sequence. This suggests thatdifferent stars on the main sequence evolve at different rates. Specifically, a hotand 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 turnoffpoint on the H-R diagram to determine the age of a star cluster more precisely.

1.6 Mass-luminosity relationship

For those main sequence stars with known masses, A. Eddington found a relationbetween the mass and the luminosity in 1924. (Do you know how they find themass of some main sequence stars?) The relationship goes like L ≈ Mα. (Seefigure 1.3.) This is an approximate relationship. Actually, after Eddington,different people put slightly different values for α. The value of α is about3 to 5 over the whole range of stellar masses M . Yet, a closer look revealsthat the exponent α changes for different values of M . Specifically, α is about2.5 to 3.0 for M <∼ 0.5M or when M >∼ 2.5M while α is about 3.5 to 4.5for 0.5M <∼ M <∼ 2.5M. Any theory of stellar structure must be able toreproduce the mass-luminosity relationship.

Lum

inosi

ty (

L )

Mass (M )

10

10

10

10

1

0.1 1 10

−2

6

4

2

M3

M4

M3

α

α

α

Figure 1.3: The mass-luminosity relationship of main-sequence stars.

Chapter 2

Stellar Structure Equations

To understand the stellar structure, we need to determine many physical param-eters inside a star, including mass / density, temperature, pressure, and energyproduction rate, by solving the following equations invoking conservation laws.Remember that it is important to specify the boundary conditions otherwise adifferential equation cannot be solved.

2.1 Mass conservation equation

For a spherically symmetric star, let r be the distance from the star centerand m = m(r) be the mass enclosed by the sphere of radius r centered at thecenter 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 theradius of the star. More importantly, due to spherical symmetry,∫ r

04πr2ρ(r) dr = m(r) , (2.1)

where ρ(r) denotes the density of the star at radius r. (Obviously, ρ(r) > 0 for0 ≤ 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 introducedone equation and yet one new unknown. So, we have to find out more equationsrelating ρ, P , T , m, etc, before we can solve the problem of stellar structure.

8

CHAPTER 2. STELLAR STRUCTURE EQUATIONS 9

2.2 Momentum conservation and the equation

of hydrostatic equilibrium

The next equation we introduce comes from the conservation of momentum; ormore precisely, in LTE the net force acting on any microscopically large butmacroscopically small region is zero. Inside a star in LTE, the forces involveare gravity and pressure gradient due to gas or photon. Specifically,

dP

dr= −Gm(r)ρ(r)

r2. (2.3)

We denote the central pressure P (0) by Pc. And clearly, P (R) can be approxi-mated 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 >GM2

8πR4(2.5)

in any star. (Hint: r(m) < R.)

In particular, this inequality tells us that the central pressure of our Sun is atleast 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)V (r) dP = −1

3

∫ M

0

Gm(r)

rdm . (2.6)


Note that the R.H.S. above is Ω/3 where Ω is the gravitational potential en-ergy of the entire star. Using integration by parts, the L.H.S. above equals−∫ V (R)

0 P dV = −∫M

0 P/ρ dm. Hence, we arrive at

Ω = −3∫ M

0

P

ρdm . (2.7)

This equation is one form of the virial theorem (sometimes known as theglobal form of the virial theorem).

(An alternative form of the virial theorem, sometimes called the local form ofthe virial theorem, is obtained by integrating from r = 0 to r = Rs(< R). Theresult is

PsVs −∫ Ms

0

P

ρdm =

Ωs

3, (2.8)

where Ωs is the gravitational potential energy for the sphere centered at thestar and with radius Rs.)

Exercise: It can be shown that the gravitational potential energy of a starΩ < −GM2

2R. From the ideal gas law P = nkT = ρkT/µmA, where µ is the mean

molecular weight and mA is the unit atomic mass, and the virial theorem, showthat the average temperature

T ≡ 1

M

∫ M

0T dm >

GMµmA

6kR. (2.9)

For the Sun, show thatT > (4× 106µ)K (2.10)

Exercise: For non-relativistic classical ideal gas, P = nkT where n is the numberdensity of the gas; and the average K.E. per molecule is 3kT/2. Show that2U + Ω = 0 where U is the thermal energy of a star.


Since the total energy of the star is U + Ω < 0, it is in a bound state.

One interesting consequence of the virial theorem is that upon gravitationalcontraction, the star becomes hotter, more tightly bound and has to radiatesome energy to space. It is kind of an analog to a system with negative heatcapacity.

In contrast, if the star is made up of extremely relativistic particles, K.E. perparticle is 3kT , i.e., the pressure equals one-third the energy density. Hence, inthis case, virial theorem implies that U + Ω = 0. In other words, a star makingup of extremely relativistic particles can be in LTE only when its total energyis 0. So, such a star is not bounded.

Another consequence of the virial theorem concerns the conversation law. Con-sider a star making up of non-relativistic classical ideal gas. Suppose furtherthat the timescale concerned is small enough that the total energy of the staris roughly a constant. Since 2U + Ω = 0, the total energy of the star can beexpressed as a function of U (or Ω) alone. Thus, the thermal energy and thegravitational potential energy of the star are also conserved. In turns out thatthis consequence of the virial theorem is useful to understand several propertiesof 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 equa-tions, namely, the mass conservation equation and equation of hydrostatic equi-librium. So, we need another equation to relate m, ρ and P . Astronomers haveproposed a number of extremely simplified equations to fulfill this task. Since allthese proposed equations are not derived from first principle, they are nothingmore than crude approximations of the realistic situation. Yet, these approxi-mations are sometimes quite useful to investigate the approximate structure ofa star.

The first such model is called constant density model, namely, we assume ρ isa constant inside the entire star. The second one is called the linear densitymodel, namely, ρ(r) = ρc(1− r/R) where ρc is a constant.

Exercise: Solve m(r) and P (r) in the above two models.


2.5 Polytrope model

The last such model I am going to introduce, which is the most importantmodel of this kind, is called the polytrope model. The is motivated by adiabaticexpansion of ideal gas,

PV γ = constant, (2.11)

where γ is the adiabatic index (or the heat capacity ratio). We thereforeassume

P = Kργ , (2.12)

for some constants K. In the classical case, it can be shown that γ is related tothe degrees of freedom f by γ = 1 + 2/f , i.e. γ = 5/3 for monatomic gas andγ = 7/5 for diatomic gas. As we will show later in this chapter, Eq. 2.12 alsoapplies to a photon gas, classical and relativistic degenerate gas with differentvalues of γ.

Exercise: Combining the mass conservation and hydrostatic equilibrium equa-tions, show that

1

r2

d

dr

(r2

ρ

dP

dr

)= −4Gπρ . (2.13)

Exercise: Using the polytrope equation (Eq. 2.12) and write γ = 1 + 1/n, theabove equation becomes

(n+ 1)K

4πGnr2

d

dr

[r2ρ(1−n)/ndρ

dr

]= −ρ . (2.14)

The boundary conditions for this differential equations are ρ(0) = ρc, ρ(R) = 0,and dρ(0)/dr = 0. (Do you know why?)


Exercise: Let ρ = ρcθn where ρc is the central density of the star, show that

1

ξ2

d

dξ

(ξ2 dθ

dξ

)= −θn , (2.15)

where r = [(n+ 1)Kρ(1−n)/nc /4πG]1/2ξ. The corresponding boundary conditions

are θ = 1 and dθ/dξ = 0 at ξ = 0. This equation is called Lane-EmdenEquation and can be solved numerically.

The significance of the polytrope model is that Lane-Emden equation is inde-pendent of the mass M , radius R and central density ρc of a star. So, once youhave numerically solved the Lane-Emden equation for a given value of n, thenumerical solution can be used to deduce the solution of any star with the samepolytropic index n (or γ).

For the special cases of n = 0, n = 1 and n = 5, Lane-Emden equation can besolved exactly.

Exercise: For n = 0, show that θ(ξ) = 1− ξ2/6 and

ρ = ρc whenever ξ 6=√

6 . (2.16)

The case of n = 1, Lane-Emden equation can be rewritten as

d2

dξ2(ξθ) = −ξθ (2.17)

whose solution that matches the correct boundary conditions above is θ(ξ) =sin ξ/ξ. Hence,

ρ = ρcsin ξ

ξ. (2.18)


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

d2z

dt2=z(1− z4)

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

8z2 − 1

24z6 + C . (2.20)

Boundary conditions demand that C = 0 and hence

dz

dt= −z

2

(1− z4

3

)1/2

. (2.21)

After making the substitution z4/3 = sin2 ζ and upon integration, we havee−t = C ′ tan(ζ/2) where C ′ is a constant of integration. Applying the boundaryconditions and after simplification, we obtain

ρ = ρc

(1 +

ξ2

3

)−5/2

. (2.22)

(Could you express ρc as a function of M and R? Besides, could you express ρand P as a function of r?)

It is straightforward to check that the smallest positive root for the solutionθ(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 astar. On the other hand, θ 6= 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 aphysical star.

Exerxise: From the definition of ξ, show that

R =

[(n+ 1)K

4πG

]1/2

ρ(1−n)/2nc ξ1 , (2.23)


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 massconservation and equation of hydrostatic equilibrium, we obtain

M = 4π

[(n+ 1)K

4πG

]3/2

ρ(3−n)/2nc

[−ξ2 dθ

dξ

]∣∣∣∣∣ξ=ξ1

, (2.24)

Pc =GM2

R4

4π(n+ 1)

(dθ

dξ

)2−1

∣∣∣∣∣∣∣ξ=ξ1

, (2.25)

ρ = ρc

[−3

ξ

dθ

dξ

]∣∣∣∣∣ξ=ξ1

, (2.26)

and

Ω = − 3

5− nGM2

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 forn = 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 ρ is 0 for case of n = 5?)

Interestingly, for n = 3, the mass of the star is independent of its centraldensity. This particular polytropic index is sometimes also called the Eddingtonstandard model. (We shall say more about the Eddington standard model laterin the next chapter.)

2.6 Equation of state

We have used the conservation of mass, energy and momentum to constructthree equations for stellar structure. Yet, we have P , T , m, ρ, L, ε, r as well asthe chemical composition as our variables. To further our analysis, we have tolook for equations that tell us specific properties of the material in a star. One


such equation is the equation of state (EOS), namely, an equation expressingthe pressure P as a function of T , ρ and chemical compositions.

The polytrope model assumes adiabatic ideal gas. To model a star, astrophysi-cists may use some extremely accurate and hence complicated EOSs. I shall notdo so in this introductory course for the following reason. Although those EOSsmore accurately model the material behavior, their predictions on the interiorstructure of a star are in most cases not significantly different from the simpleEOS we will use in this course. The simplest EOS, which at the same timeturns out to be the one we will use in this course, is the ideal gas law that youhave learned in high school. That is, P = nkT where n is the number densityof 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 considerion, 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 astar. 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 aparticle (or system of particles), the probability of a particle is in state s isproportional to gse

−Es/kT where gs is the degeneracy of state s and Es is theenergy of state s relative to some fixed reference energy level (sometimes takento be the ground state). More importantly, the proportionality constant is thesame for all the states. Thus, the ratio of the probability that the particle is instate (s+ 1) to the probability that it is in state s is given by the Boltzmannformula

gs+1

gse−(Es+1−Es)/kT . (2.28)

Exercise: At what temperature a gas of neutral hydrogen will have equal numberof atoms in the ground and first excited states? (Hint: for hydrogen atoms, thedegeneracy is gn = 2n2, i.e., the ground state is n = 1 and the first excited state


is n = 2.)

This is even higher than the surface of the Sun. However, we know from ob-servations that some hydrogen atoms are ionized in the Sun. How can that bepossible? (The answer lies in the Saha Equations, which will be discussed laterin this chapter.)

The Statistical Mechanics and Thermodynamics course also tells us that thethermodynamic properties of a system with fixed number of particles N andfixed temperature T and fixed volume V is encoded in the so-called partitionfunction

Z ≡∑i

gi exp(− EikT

), (2.29)

where the sum is over all possible energy states of the system. Furthermore, theratio between the mean number of particles Ns in state s and the total numberof particles N ≡ ∑Nj in a sample is given by

Ns

N=

gse−Es/kT∑

j gje−Ej/kT

≡ gse−Es/kT

Z. (2.30)

For an ideal classical non-relativistic gas particle with ground state energy E0

and degeneracy g, its partition function is given by

Z(1, V, T ) =g

(2πh)3

∫ ∫e−(E0+p2/2m)/kT d3~x d3~p

=gV

(2πh)3e−E0/kT

∫ +∞

04πp2e−p

2/2mkT dp

=gV

λ3e−E0/kT (2.31)

where m is the mass of the particle and

λ ≡ h√

2π

mkT. (2.32)

Note that we have assumed that the momentum of the gas particle is isotrop-ically distributed in order to obtain the above expression for Z(1, V, T ). Thepartition function of a system of N indistinguishable classical non-relativisticideal gas particles each with ground state energy E0 and degeneracy g is thengiven by

Z(N, V, T ) =Z(1, V, T )N

N !. (2.33)


Recall again from the Statistical Mechanics and Thermodynamics course thatthe most convenient method to study the situation in which the particle numberin the system may change is to use the so-called grand partition function

Q(µ, V, T ) ≡+∞∑N=0

Z(N, V, T ) exp(µN

kT

), (2.34)

where µ is the chemical potential. In the thermodynamic limit, µ can be in-terpreted as the work needed to add one extra particle to the system at constantV and T . The grand partition function encodes the thermodynamical informa-tion of a system with fixed chemical potential, volume and temperature. Sinceex =

∑∞0 xn/n!, the grand partition function of an ideal classical gas in the

non-relativistic limit equals

Q(µ, V, T ) = exp[gV

λ3e(µ−E0)/kT

]. (2.35)

More generally, the grand partition function of a system of interacting particlesof various different species such that each species can be approximated by aclassical non-relativistic ideal gas is given by

Q(µs, V, T ) = exp

[∑s

gsV

λ3s

e(µs−E0,s)/kT

], (2.36)

where the s is the species label. Hence, the corresponding grand potentialequals

G(µs, V, T ) ≡ −kT lnQ = −kT∑s

gsV

λ3s

exp(µs − E0,s

kT

). (2.37)

Note that G is minimized in chemical equilibrium. Since G = E−TS−∑s µsNs,we have

Ns = − ∂G∂µs

∣∣∣∣∣T,V,µii 6=s

=gsV

λ3s

exp(µs − E0,s

kT

). (2.38)

By rearranging terms, the above equation becomes

µs = kT ln

(Nsλ

3s

gsV

)+ E0,s = kT ln

(nsλ

3s

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 sthionized state in its ground level. For example, the label 0, 0 of H refers to the


state of hydrogen atom with its electron in the lowest energy level. And I usethe label e to denote electron. Clearly, at non-zero temperature, reactions suchas

Xs+ X(s+1)+ + e− (2.40)

may occur, where X denotes the atomic species. Hence, at thermodynamicequilibrium,

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+1,0neλ3s+1,0λ

3egs,0

ns,0λ3s,0gs+1,0ge

= e−χs+1/kT . (2.42)

From Eq. (2.32), ge = 2 and the approximation that the masses of the sth andthe (s+ 1)th ionized atom are about the same, the above equation becomes

ns+1,0

ns,0=gs+1,0

gs,0nefs+1(T ) , (2.43)

where

fs+1(T ) =2(2πmekT )3/2

h3exp

(−χs+1

kT

), (2.44)

In general, not all the sth ionized atoms in a thermalized system are in theground state. From Eq. (2.30), Eq. (2.43) can be re-written as

ns+1

ns=Zs+1

Zsnefs+1(T ) , (2.45)

where

Zs =∑i

gs,i exp

[−(Es,i − Es,0)

kT

](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 equationis useful in astronomy. For instance, if we know the electron partial pressurePe, then the electron number density ne can be well approximated by Pe/kT inmany cases. We now apply Saha equation to estimate the degree of ionizationof our Sun. For simplicity, we consider only the form of Saha equation inEq. (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 analysisby 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 Sahaequation demands that

n1

n0

=x

1− x=f1(T )

ne≈ mHf1(T )

ρx(2.47)

and hence x ≈ 98.8%.

From the result above, almost all hydrogen in our Sun is ionized. By the sametoken, it is straight-forward to check that most atoms in a main sequence starare ionized. On the other hand, in the solar atmosphere where T ≈ 6000 K andne ≈ 1013 cm−3. Saha equation tells us that the degree of ionization of hydrogenis about 10−4. (What about other atoms, such as Na, in the atmosphere, andHe and other heavy elements in the core?

Note however that Saha equation has its limitations. It works for thermalizedregion 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 coreof 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 arethe three most important constituents of a star. The electron pressure is givenby

Pe = nekT (2.48)

where ne is the electron number density, and the cation pressure is given by

Pi = nikT (2.49)

where ni is the cation number density.

What are the values of ne and ni? To answer this question, we denote themass fraction of H, He and “heavy elements” (that is, those elements heavierthan H and He) in a star by X, Y and Z = 1 − X − Y respectively. (Moreprecisely, X and Y are the mass fraction of 1H and 4He respectively. Do youknow why we can forget about the contribution of 2H, 3H, 3He?) Then, numberdensity of hydrogen and helium nuclei equal Xρ/mH and Y ρ/4mH, respectively.Therefore,

ni =ρ

mH

(X +

Y

4+

Z

〈A〉

), (2.50)

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 averageweight of a particle in units of hydrogen mass, i.e.

µ ≡ 〈m〉mH

(2.51)

such that

µmH = 〈m〉 =total mass of gas

total number of particles. (2.52)

This gives

nµmH = n〈m〉 = ρ

n =ρ

µmH

(2.53)

A direct comparison with Eq. (2.50) gives

1

µi=

(X +

Y

4+

Z

〈A〉

). (2.54)

Next, we consider electrons. By virial theorem, we know that the thermal energyof a star is of the order of GM2/R. That is,

GM2/R ∼ kTM/mH , (2.55)


where T ≈ 106 K is the average temperature of a star. Hence, for most partof 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.Nonetheless, this assumption has little effect in determining the overall structureof 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 atypical main sequence star.)

Similar to Eq. (2.50) above, but a hydrogen atom gives one electron, an heliumatom gives two, and a heavy element atom with atomic number Z gives Zelectrons. Hence, the total number of electrons per unit volume is

ne =ρ

mH

(X + 2

Y

4+ Z

⟨ZA

⟩)=

ρ

mHµe. (2.56)

(Do not confuse the mass fraction of heavy elements Z with Z.)

For fully ionized gas, the total gas pressure law is contributed by ions andelectrons

Ptot = Pi + Pe = nikT + nekT (2.57)

=ρ

mH

(1

µi+

1

µe

)kT ≡ ρkT

µmH

. (2.58)

Here we define1

µ≡ 1

µI+

1

µe. (2.59)

From Eqs. (2.54) and (2.56), we have

1

µ= 2X +

3

4Y + Z

⟨Z + 1

A

⟩(2.60)

for fully ionized gas. As most nuclei contain about the same number of protonsand neutrons, the last term can be approximated by Z/2 and this can be furtherreduced to

1

µ≈ 2X +

3

4Y +

Z

2

= 2X +3

4Y +

1

2(1−X − Y ) . (2.61)

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 isµ for neutral gas (assume 〈A〉 = 15.5)? For fully ionized gas, show thatµ ≈ 0.6. Hence,

Ptot =ρkT

0.6mH

. (2.62)

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 radiationpressure. Recall from the course Statistical Mechanics and Thermodynamicsthat photons are bosons. Therefore, in thermal equilibrium, photons obeyBose-Einstein statistics. That is, the distribution of photons is isotropic andthe number density of photons with frequencies between ν and ν + ∆ν equals

n(ν) dν =8πν2

c3

dν

exp(hν/kT )− 1. (2.63)

To see why photon number density follows Eq. (2.63), one recall that the energyof having s photons each with momentum ~p equals spc. As photon is bosonic,s can take on any natural number. Moreover, the probability P (s) that thereare exactly s photons each with momentum ~p follows the constraint

P (s+ 1)

P (s)= e−pc/kT = e−hν/kT , (2.64)

where ν is the frequency of the photon. Consequently,

P (s) =e−shν/kT∑+∞

s′=0 e−s′hν/kT = e−shν/kT (1− e−hν/kT ) (2.65)


provided that ~p is an allowed momentum of a photon. Thus, the expectednumber of photons with an allowed momentum ~p 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 polar-izations.) As a result, the total expected number of photons equals

〈Nγ〉 =1

h3

∫ ∫〈Nγ(~x)〉 dV d3~p

=1

h3

∫dV

∫ 2

ehν/kT − 1d3~p

= V∫ 8πν2

c3(ehν/kT − 1)dν . (2.67)

Hence, Eq. (2.63) is valid.

Note that the first fraction in Eq. (2.63) is called phase space factor whilethe second fraction is the Bose-Einstein distribution factor. Moreover, thisdistribution n(ν) is sometimes called the blackbody spectrum.

For a photon of momentum ~p hitting a wall and then reflected back elastically,the magnitude of the change in momentum equals 2p cos θ = 2hν cos θ/c whereθ is the angle between ~p and the normal of the wall. The radiation pressure duesimply the momentum transfer to the wall per unit time per unit surface area.That is,

Prad =1

h3

∫ +∞

0

∫ π/2

0

∫ 2π

0c cos θ

2hν cos θ

c

2

ehν/kT − 1

(hν

c

)2

sin θ dφdθd

(hν

c

)

=∫ +∞

0

hν

3n(ν) dν

=4σ

3cT 4 , (2.68)

where σ = 2π5k4/15c2h3 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 givenby

urad =∫ +∞

0hνn(ν)dν =

4σ

cT 4 . (2.70)

Hence, urad = 3Prad.

Exercise:


1. For a photon gas with volume V and temperature T , what is the internalenergy U?

2. Given that the entropy S of the gas is related to U by dU = TdS and Sremains a constant for an adiabatic process, show that a photon gas hasan adiabatic index γ = 4/3.

2.11 Other forces

As a final note, we estimate the contribution of other forces to the EOS. Mostof the stellar material should be ionized, therefore, the dominant interactionbetween particles in stellar interior (besides gravity) is electrostatic in nature.The typical distance between particles in stellar interior d ≈ (AmH/ρ)1/3 whereA is the average atomic weight of particles, mH is the mass of hydrogen atom,and ρ is the average density of a star. Consequently, the typical Coulombenergy per particle is about Z2e2/d (recall that 4πε0 → 1 in cgs), where Z isthe average atomic number of particles. In contrast, virial theorem tells us thatthe gravitational potential energy and internal energy of a star are of the sameorder. Consequently, the ratio of Coulombic energy to internal energy of a staris about

Z2e2

G(AmH)4/3M2/3. (2.71)

Note that Z and A are of order of 1. Therefore, the above ratio is about 10−2,which is much less than 1. Thus, electrostatic potential energy contribute onlya small fraction of the internal energy of a star; and most of the internal en-ergy are in the form of heat. More specifically, the thermal K.E. of a typicalparticle inside a star is much greater than the electrostatic potential energy itexperiences.

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 AmH. 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 completelydifferent for dead stars such as white dwarfs and neutron stars. Because thedensity of dead stars is high, ∆x∆p for some particle species inside these starsare 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 theinterior of most living stars. (However, quantum mechanical effect is importantin some other “stars” such as brown dwarfs and supernovae. Radiation pressurealso plays an important role in the structure and stability of a super-massivestar. We shall come across them later on in this course.)

2.12 Degenerate gas equation of state

For completeness, we will now derive the EOS for degenerate gas, in which thequantum effect dominates. This is important in some cases, such as browndwarf, 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 extremedensities, 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 inan increase in the gas pressure. Fundamentally, it is the exchange interactionthat creates the “force”, although this is merely a quantum mechanical effectrather than a real force.

Consider a particle in a box (i.e. an infinite potential well in 3D) of lengthL, solution of the Schrodinger equation requires the particle wavelength tosatisfy

λx =2L

lx, λy =

2L

ly, and λz =

2L

lz, (2.72)

where lx, ly, and lz are positive integers. From the de Broglie equationp = h/λ, the K.E. is

E =p2

2m=

h2

8mL2(l2x + l2y + l2z) ≡

h2l2

8mL2, (2.73)

i.e. it is proportional to l2 ≡ l2x + l2y + l2z . At low temperature, we can assumethat all low-energy states are occupied up to the Fermi level lF . The numberof states within a sphere of radius lF is 4πl3F/3. But since lx ly, and lz takesonly positive numbers, we have to divide this by 8 and each state can have twoparticles when we take the spin into account. Therefore, the total number of


particles is

Ntot = 2(

1

8

)(4

3πl3F

)=πl3F3

. (2.74)

This gives lF = (3Ntot/π)1/3 and hence the Fermi energy of

EF =h2l2F

8mL2=

h2

8mL2

(3Ntot

π

)2/3

=h2

8m

(3n

π

)2/3

, (2.75)

where n = Ntot/L3 (not to be confused with the polytropic index) is the particle

number density. The total energy is

Etot =∫ lF

0E dN =

∫ lF

0

(h2l2

8mL2

)d(π

3l3)

=

=3

5NtotEF ∝ V −2/3 . (2.76)

Exercise: Show that the gas pressure is

P ≡ −∂Etot

∂V=

2

5nEF ∝ V −5/3 , (2.77)

and hence the non-relativistic degenerate gas EOS is a polytrope. Whatis γ and the polytropic index?

When the density is extremely large, e.g., inside a neutron star, EF becomescomparable to the rest mass of the Fermions, we need to consider relativisticeffects and modify Eq. 2.73 with E =

√m2c4 + p2c2 ≈ pc.

Exercise: What is E in terms of l? Hence, derive Etot and P . Show that therelativistic degenerate gas EOS is a polytrope and find the index.

Chapter 3

Energy Production and TransferProcesses

3.1 Energy conservation and the energy pro-

duction equation

We have the equations that govern P and ρ (i.e. m) as functions of radialdistance. So far, our simple stellar model is only a giant ball of gas that isbounded by self gravity and obeys the adiabatic gas law. Now, we want to gofurther by considering a more realistic equation of state, i.e., P (ρ) and differentchemical compositions. In this chapter, we will talk about energy generationinside 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 gasis zero. Let L(r) denotes the luminosity at the spherical shell with radius rcentered at the core of the star. (That is, L(r) is the energy per unit timemoves out of the spherical shell of radius r.) Energy conservation demands that

dL

dr= 4πr2ρ(r)ε(r) , (3.1)

where ε is the rate of energy production per unit mass of material at the radiusr 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 notationq 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

dm= ε(r) . (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 derivativeφ, τ ≡ φ/φ, such that it gives us an idea on the characteristic time for thatvariable to change significantly.

• The free-fall / dynamical timescale: For a star with mass M and ra-

dius R, the escape velocity is of order of√GM/R. The free-fall timescale

is, therefore, of the order of τff ≈√R3/GM ≈ 1/

√Gρ where ρ denotes the

mean density of the star. For our Sun, the free-fall timescale is of order ofan hour. In short, any force that is unbalanced inside a star occurs at thefree-fall timescale. Thus, if we are interested in the processes of a stablestar 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 astar to radiate its thermal energy away at constant luminosity. By virialtheorem, τKH ≈ GM2/RL. For our Sun, τKH is about 3 × 107 yr. Fora time much longer than the Kelvin-Helmholtz timescale, we may safelyassume that the star is in thermal equilibrium.

• The nuclear burning timescale: This is the timescale for burning allthe nuclear fuel of a star at constant luminosity and is therefore equalto τnucl ≈ Mc2∆/L, where ∆ ≈ 10−3 is the typical binding energy of anucleon 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 ofa star. (You may notice that the nuclear burning timescale overestimatesthe lifespan of a star. Do you know why?)

Since τff τKH τnucl, we know once again that our assumption of LTEis a good approximation to study the interior structure and evolution ofa star.


Exercise: Estimate these timescales for the Sun. Based on the results, what isthe 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 absorbedand remitted in a random direction by the surrounding materials. In this way,energy carried by the photons is transferred to the surrounding material andheats it up. Therefore, the trajectories of photons inside a star can be regardedas 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 beingabsorbed or scattered. When a light ray travels in a medium, its intensity willbe reduced with distance x by I(x) = I0 exp(−κρx), where κ is the specificopacity or opacity coefficient and ρ is the density of the medium. Note thatκ in general depends on the frequency of light to be absorbed. In fact, the valueof κ used here is carefully averaged over frequency. absorbed.

Exercise: What should be the unit of κ? (Hint: what is the unit of κρ?)

In addition to energy transfer between photon and the medium, absorptionand/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 sphereper unit area per unit time) equals L(r)/4πr2. Therefore, the light energyabsorbed or scattered on this sphere per unit area per unit time per unit pathlength equals L(r)κρ/4πr2. Consequently, the momentum transferred to thesurrounding medium on this sphere per unit area per unit time per unit path


length is L(r)κρ/4πr2c. This number is equal to −1 times the radiation pressuregradient on that sphere (do you know why?) i.e.

dPrad

dr= −L(r)κ(r)ρ(r)

4πr2c.

Exercise: Recall that inside a star, LTE is a good approximation. Hence,radiation there follows a blackbody spectrum. Using P = 4σT 4/3c (Eq. 2.68),we have

dPrad

dr=

( )dT

dr.

Putting these together, we have

L(r) =

( )dT

dr. (3.3)

This equation is sometimes known as the radiative energy transport equa-tion. Clearly, T (0) = Tc (the core temperature) while T (R) = Ts ≈ 0 (thesurface temperature).

Note that the radiative transfer equation is valid if radiation is the principle heattransfer mechanism. If convection or conduction is important, this equation hasto be modified.

3.4 Physical processes leading to opacity and

Kramers opacity Approximation

The interactions between photons and matter (primarily with electrons andnuclei in most part of a star) can be classified below.

1. Electron scattering: Compton/Thomson scattering, that is, scatteringof photon by a free electron.

2. Free-free absorption: Absorption of a photon by a free electron whichmakes a transition to a higher energy level by briefly interacting with anucleus or ion.

3. Bound-free absorption: Also known as photon-ionization, namely, theionization of an electron from an atom or ion after absorbing a photon.


4. Bound-bound absorption: Exciting an electron from a bounded stateof a lower energy level to another bounded state of a higher energy levelby absorbing a photon.

Close to the stellar core, temperature is so high that atoms are almost com-pletely ionized. Thus, electron scattering and free-free absorption are the dom-inant process to create opacity in stellar core. In the outer stellar atmosphere,in contrast, the situation is much more complex. Bound-free and bound-boundabsorptions can be important. This is the reason why we see absorption lineson the surface of stars.

The specific opacity κ can be calculated by taking into account all the possibleinteractions between photon and the medium. This is a very complicated andinvolved task, requiring the work of a whole generation of atomic physicists.Fortunately, the resultant κ can usually be approximated by relatively simpleformula in the form κ = κ0ρ

aT b. (Once again, we shall only use the approximateform of κ in this course as the approximation is reasonably good for most partof the stellar interior.)

In particular, the electron scattering opacity, which is important in the hightemperature core of a star, is given by

κes ≈ 0.2(1 +X) cm2g−1 , (3.4)

while the free-free absorption, which may be important near the surface of astar, can be approximated within an accuracy of about 20% by the so-calledKramers opacity

κff ≈ 4× 1022(1 +X)(X + Y )ρT−7/2 cm2g−1 . (3.5)

Clearly, free-free absorption produces a higher opacity than electron scatter-ing at low temperature. In fact, the cross-over temperature between free-freeabsorption and electron scattering is about 106 K. Whereas when the tempera-ture is less than about 104 K, bound-free and bound-bound absorption becomemore important than Kramers opacity is no longer valid. In fact, at such a lowtemperature, κ starts to decrease as T decreases.

Interestingly, the presence of opacity places an upper limit on the luminosityof a star. Recall that P = Pgas + Prad. Clearly, dPgas/dr ≤ 0. So, from theequation of hydrostatic equilibrium and equation of radiative transfer, we have

dPrad

dr≥ dPrad

dr+

dPgas

dr= (3.6)


Therefore,

L ≤ ≡ LEdd . (3.7)

If this inequality is violated somewhere inside a star, then either radiative trans-port is no longer valid (such as in the core of certain stars) or hydrostatic equi-librium can no longer be maintained (such as near the surface of certain stars).In particular, near the surface of a star, m = M and the limiting luminositygiven by Eq. (3.7) becomes the so-called the Eddington luminosity

LEdd = 3.2× 104

(M

M

)(κ

κes

)−1

L . (3.8)

3.5 The Eddington standard model

Let us define a function η by L(r)/m(r) = L∗η/M where L is the luminosityacross a sphere of radius r centered at the star and L∗ is the total luminosityof a star.

Exercise: Using the stellar structure equation, show that dPrad/dr can be writ-ten as

dPrad

dP=

L∗κη

4πcGM. (3.9)

Assuming that the location where nuclear burning occurs is confined in a smallregion 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 isconstant over the entire star. Surely, this assumption is in no way close to thetruth, it leads to very interesting and realistic predictions about structure andevolution of stars.

Integrating Eq. (3.9) gives

Prad =κsL∗

4πcGMP ≡ P (1− β) . (3.10)


That is to say, the ratio of Prad to P is a constant throughout a star in the Ed-dington standard model. Historically, the ratio between gas and total pressureis defined as β, i.e. Pgas/P = β, such that Prad/P = 1− β. Thus, the radiativetransfer equation becomes

L∗ =4πcGM

κs(1− β) ≡ LEdd(1− β) . (3.11)

The ideal gas law demands that

T 4

1− β∝ Prad

1− β= P =

Pgas

β=ιρT

β, (3.12)

where ι = k/mHµ (see Eq. 2.58). Hence, T ∝ ρ1/3. Note that for gas, theequation of state may be written as P = Kρ4/3 for some constant K ∝ [(1 −β)ι4/β4]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

1− β = 0.003

(β

ι

)4 (M

M

)2

. (3.13)

This is known as the Eddington quartic equation. This equation has thefollowing predictions. First, for a given composition (fixed ι), β decreases as Mincreases. So, radiation pressure is more important in high mass stars. Second,combining with Eq. (3.11) gives L∗ ∝M3. So, the mass-luminosity relationis obtained. Actually, Eddington made this prediction before the confirmationby observations. Third, for a given M , nuclear burning implies that ι willdecrease gradually. Consequently, β decreases. Therefore, it is expected thatradiation pressure plays an increasingly important role as a star gets older.Combined with the Eddington limit, it strongly hints that some stars may ejectsome of its materials in its late stage of evolution.

Exercise: Combining Eqs. (3.11) and (3.13), show that L∗ ∝M3.


3.6 Nuclear energy generation

The energy generation mechanism of a star cannot be chemical in nature. Forchemical 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 about104 yr. In contrast, the energy generation by nuclear reaction is of the order of1 MeV per nucleus. This allows the Sun to shine at the present rate for about1010 yr. Thus, nuclear energy looks like a promising way to power the Sun andother stars. (Still recall that energy is released by fusion of light nuclei or fissionof heavy nuclei? Do you know that 56Fe is the most stable nucleus in the sensethat it has the highest binding energy per nucleon?)

But we have a problem. Virial theorem tells us that the thermal energy andgravitational potential energy of a star are of the same order. Thus, usingideal gas approximation, the average temperature of our Sun turns out to beabout 106 K. However, the typical K.E. of two 106 K hydrogen nuclei cannotovercome the Coulombic repulsion between them. Specifically, the Coulombenergy between two hydrogen nuclei placed 2 hydrogen nuclear radii apart isabout three orders of magnitude larger than the typical K.E. of the two hydrogennuclei. Thus, classically, there is no way for two hydrogen nuclei to fuse togethermaking a bigger nucleus and hence releasing the required nuclear energy to makethe star to shine. Fortunately, quantum mechanics saves us for hydrogen nucleusmay tunnel through the Coulombic energy barrier (see figure 3.1).

Consider a nucleus of charge Z1e moving towards another nucleus of chargeZ2e. Denote µ ≡ (m1 + m2)/m1m2 the reduced mass of the system (not to beconfused with the mean molecular weight). The Schrodinger equation describingthis system is

− h2

2µ∇2ψ +

Z1Z2e2

rψ = Eψ , (3.14)

for r > rN (the nuclear radius), where E is the K.E. of the first nucleus relativeto the second one when they are infinitely apart.

By separation of variables, (that is, by writing ψ = R(r)Y (θ, φ)), we concludethat the radial wavefunction of the above equation satisfies

− h2

2µ

d2χ

dr2+

Z1Z2e2

rχ(r) +

h2l(l + 1)

2µr2χ(r) = E χ(r) , (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,


Potential

E

r

r

1/r

Coulombic repulsion

rcrn

Wavefunction

Figure 3.1: An illustration of overcoming the Coulombic energy barrier throughtunneling.

Eq. (3.15) can be re-written as

d2χ

dr2=

2µ

h2

(Z1Z2e

2

r− E

)χ . (3.16)

Nonetheless, the above equation is too complicated to solve analytically. For-tunately, an approximate solution is available using W.K.B. approximation.Specifically, we assume that the radial wavefunction (in other words, the solu-tion of Eq. 3.16) to be in the form χ(r) = AeiS(r)/h for some constant A.

Exercise: Using Eq. (3.16), show that S(r) satisfies

ihS ′′ − S ′2 − h2k(r)2 = 0 , (3.17)

where

k(r)2 =2µ

h2

(Z1Z2e

2

r− E

). (3.18)

Note that in the quantum tunneling region, where Z1Z2e2/r > E, we can pick


k(r) ≥ 0.

Exercise: Expanding S(r) in powers of h, i.e. S =∑∞k=0 h

kSk(r) and keepingonly the zeroth order term (note that the term h2k2 is h independent as k isproportional to 1/h), show that

χ(r) = A exp[−∫ r

rck(r) dr

](3.19)

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 betweenthe two nuclei when the classical K.E. of the first nucleus is zero. That is,

Z1Z2e2

rc=

1

2µv2 . (3.20)

Besides, E = µv2/2 where v is the relative speed between the two nuclei.

Since rc rN and hence Z1Z2e2/r E, upon integration, we find that the

tunneling probability is about

|R(rN)|2 =

∣∣∣∣∣χ(rN)

rN

∣∣∣∣∣2

≈∣∣∣∣ ArN

∣∣∣∣2 exp

(−2πZ1Z2e

2

hv

). (3.21)


(Can you fill in the details of the calculation?) This result, which laid an im-portant foundation to the study of nuclear reactions in stars, was first obtainedby 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 nuclearforce to be effective is small but non-zero.

We expect that nuclear force is effective only within a de Broglie wavelengthλ = h/µv. Thus, we also expect the nuclear reaction cross section to be in theform

σ(v) ≈ πλ2 exp

(−2πZ1Z2e

2

hv

)≈ S(v)

v2exp

(−2πZ1Z2e

2

hv

), (3.22)

where S(v) is some expression that is weakly dependent on v.

Note further that the velocity distribution of nuclei follows the Maxwelliandistribution. Specifically, the probability of finding a particle with velocity ~v isproportional to (µ/2πkT )3/2 exp(−µv2/2kT ) where v = |~v|.

Consequently, the probability of nuclear reaction to occur between the twonuclei per unit time is proportional to

∫ +∞

0

v

rN

(µ

2πkT

)3/2

exp

(− µv

2

2kT

)S(v)

v2exp

(−2πZ1Z2e

2

hv

)4πv2 dv . (3.23)

For simplicity, we assume that S is a function that does not depend stronglyon v so that vS(v) can be assumed to be almost a constant throughout thespeed range considered. Further observe that the first term in the exponen-tial increases with v (and hence the initial K.E.) while the second term in theexponential decreases with v (and hence the initial K.E.). By Taylor series ex-pansion on the terms inside the two exponentials, we conclude that the resultantreaction probability has a peak around

vGamow =

(2πZ1Z2e

2kT

hµ

)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)


e−b/E

Probability

Energy

Maxwell−Boltzmann

TunnelingGamow Peak1/2

e−E/kT

Figure 3.2: An illustration of the dependency of reaction probability on energies.

Consequently, the reaction rate per unit volume is approximately proportionalto

n1n2vGamowS(vGamow)(

µ

2πkT

)3/2

exp

(−2πZ1Z2e

2

hvGamow

− µv2Gamow

2kT

)∆vwidth

∝ n1n2(kT )−2/3 exp

−3

2

(2πZ1Z2e

2

h

)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 chargeof the nucleus decreases. This is the reason why fusion of heavy nuclei requireshigh temperature.

However, there is an exception to this trend. Some reactions known as resonantreactions occur when the energy of the interacting particles correspond toenergy level of the intermediate unstable compound nuclei. (Still rememberwhat you have learned in high school chemistry?) Resonant reaction probability,therefore, is sharply peaked at the resonant energy. (That is, S(v) is sharplypeaked at certain values of v.) Such a probability may be several order ofmagnitudes 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 ≈ AρaT 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 reactionrate and hence a and b are very involved and complicated. The above discussionsonly serve as an overly simplified approximation.

3.7 Some specific nuclear reactions occurring

in stars

3.7.1 Proton-proton chain (p-p chain)

The simplest possible reaction is p + p −→ 2He. Unfortunately, 2He is a highlyunstable nucleus which quickly decomposes back into two protons. However,if a weak interaction that turns a proton into a neutron occurs during thecombination of two protons, then we will get some stable product. Namely,p + p −→ 2H +νe + e+. The reaction rate is very slow since weak interaction, asthe name suggests, is much weaker than strong and EM interactions. (Actually,experiments find that the relative interaction strength between strong, EM andweak 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 2Hwill undergo further nuclear reaction in the core of a main sequence star. Forexample:

p-p I chain:

p + p −→ 2H + νe + e+

2H + p −→ 3He + γ3He + 3He −→ 4He + 2p

Besides p-p I chain, a star may also react viap-p II chain:

3He + 4He −→ 7Be + γ7Be + e− −→ 7Li + νe

7Li + p −→ 2 4He


orp-p III chain:

7Be + p −→ 8B + γ8B −→ 8Be + e+ + νe

8Be −→ 2 4He

These three sets of reactions, whose foundation was first proposed by H. A.Bethe in 1939, operate simultaneously although their relative significance de-pend on the physical conditions at the stellar core.

But in any case, the net result is to convert 4 hydrogen nuclei into 1 heliumnucleus. The nuclear energy released per each conversion is experimentallymeasured to be about Q = 4M(1H) − M(4He) = 26.73 MeV. The energy isreleased in the form of mainly photons and neutrinos. Photons will later oninteract with the electrons in the star through electron opacity. Neutrinos, onthe other hand, are extremely weakly interacting particles. They leave the staressentially without interacting with the surrounding materials. Thus, neutrinosemitted from the star in this way are not thermalized. In other words, LTEdoes not apply to neutrinos in a star.

Finally, notice that in p-p I chain only the first reaction involves weak in-teraction, all other reactions involve strong and EM interactions. Hence, thebottle-neck for p-p chain is the first reaction. In fact, more accurate calculationstell us that the reaction rate for the first reaction is of the order of once every1010 years. Hence, it occurs in a massive scale in the core of a star because thereare numerous supply of protons in the stellar core. To summarize, the nuclearenergy generation rate per unit mass ε for p-p chain is determined by the rateof p + p −→ 2H + νe + e+; and it can be shown that the particular reaction rateis given by

εp−p = 2.4× 104ρX2e−3.38T−1/3

T−2/39 , (3.28)

where T9 is temperature in units of 109 K. This can be approximated by

εp−p ≈ AρX2T 4 (3.29)

for some constant A > 0, where X is the mass fraction of 1H.

3.7.2 Carbon-nitrogen-oxygen (CNO) cycle

H. A. Bethe and C. von Weizsacker independently pointed out in the late 1930’sthat 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:

12C + p −→ 13N + γ13N −→ 13C + e+ + νe

13C + p −→ 14N + γ14N + p −→ 15O + γ

15O −→ 15N + e+ + νe

15N + p −→ 12C + 4He

cycle 2:

14N + p −→ 15O + γ15O −→ 15N + e+ + νe

15N + p −→ 16O + γ16O + p −→ 17F + γ

17F −→ 17O + e+ + νe

17O + p −→ 14N + 4He

Eventually, H. A. Bethe received the Nobel Prize in Physics in 1967 “for hiscontributions to the theory of nuclear reactions, especially his discoveries con-cerning the energy production of stars.”

Although these reactions are very complicated, it is relatively easy to estimatethe overall rate of reaction for we need only to look closely at the slowest ratedeterminate reaction. Recall that β decay is temperature and density inde-pendent (over conditions inside a star). In contrast, proton capture reactionsare sensitively dependent on temperature. (Do you know why?) Therefore, atrelatively low temperature, it is the proton capture that sets the pace of thereaction. While at relatively high temperature, it is the β decay that determinesthe 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× 1025ρXZe−15.2T−1/3

T−2/39 , (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 thanthat 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 theonset of hydrogen burning:

2H + p −→ 3He + γ6Li + p −→ 3He + 4He7Li + p −→ 2 4He

9Be + p −→ 4He + 6Li10B + p −→ 4He + 7Be11B + p −→ 12C + γ11B + p −→ 3 4He.

(Note that 7Be formed in the above reaction decays via electron capture to7Li + νe with a half-life of about 53 d.)

These reactions are effective to destroy light elements such as 2H, Li, Be, Bwhenever the temperature is higher than about 106 K. This can be used to dis-tinguish young brown dwarfs from main-sequence stars, known as the Lithiumtest.

3.7.4 Helium burning

At higher temperature, helium may fuse via

2 4He −→ 8Be . (3.32)

Unfortunately, 8Be 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 waslater on refined by F. Hoyle. They pointed out that if the temperature is higherthan about 108 K, the K.E. of helium nuclei will be very high. In this case, themean collision time between the unstable Be nucleus and a He nucleus is shorterthan the half-life of the unstable Be nucleus. Hence, it is possible to produce a12C by capturing a He nucleus by a 8Be nucleus. The net result is

3 4He −→ 12C (3.33)

known as the triple alpha process. The energy released per reaction is about7.28 MeV. Clearly, the net reaction rate is determined by the rate of reaction ofthe capture of He by Be. It turns out that the reaction rate per unit mass is

ε3α ≈ Aρ2Y 3T 40 , (3.34)


for some A > 0. Some of the carbon produced could further fuse into oxygenby

12C +4 He −→ 16O . (3.35)

3.7.5 Heavy elements burning and photodisintegration

At higher temperature, carbon and oxygen can overcome their Coulombic bar-riers and fuse together. The resultant nuclei are Mg, Na, Ne, S, P, Si, etc. Ateven higher temperature, Si etc. may also begin to fuse. Nonetheless, the situa-tion is not as simple as He, C or O burning. At temperature higher than about109 K, photodisintegration begins to take place. In photodisintegration, anucleus divides into two or more fragments after absorbing a high energy pho-ton. Thus, photodisintegration is, in general, an endothermic process similar tothe photo-ionization of an atom. Therefore, photodisintegration is effective onlyat sufficiently high temperature (so that there are enough high energy photonto “ionize” the nucleus). An example of photodisintegration process is shownbelow:

20Ne + γ −→ 16O + 4He . (3.36)

Due to the presence of heavier element fusion and photodisintegration, reactionsinvolving the burning of Si and other heavier nuclei occur in an almost dynamicequilibrium fashion. The net result is a gradual buildup of heavy elements suchas Ni, Fe, Co provided that the temperature is lower than about 2×109 K. If thetemperature is higher than about 3 × 109 K, photodisintegration becomes thedominant reaction and hence the star will cool itself down. (Do you know why?)In other words, photodisintegration sets an upper limit in the core temperatureof a star.

3.7.6 Neutron capture and β decay

Another type of reaction that are of interest is neutron capture. Since neutronis electrically neutral, it can come close to a nucleus without worrying aboutthe Coulombic barrier. Nevertheless, free neutron outside a nucleus is a rarespecies. A small amount of neutron is generated in a minor reaction involvingthe burning of carbon such as

13C + 4He −→ 16O + n . (3.37)

These neutrons may be captured by a heavy nucleus resulting in a heavier iso-tope. Upon a few neutron captures, the resultant nucleus may be too neutron-


rich so that it will β decay into a new nucleus with higher atomic number.Neutron capture and the subsequent β decay, therefore, are effective mecha-nism to produce a variety of heavy elements and isotopes. In particular, theyplay an important role in the synthesis of elements with atomic number higherthan about 60.

( n γ ),p

p

s,r

s,r

s,r

s,r

s,r

s,r

s,rs only

onlys

onlys

r

r

(Neutron−rich matter)

N

Z

ν−−

β−

β

r

r

s only

Figure 3.3: Schematic representation of the effects of s- and r-processes.

Astronomers generally divide neutron capture processes into two types. Inthe event that the neutron capture reaction rate is much slower than the β-decay rate, we call it the s-process. In an s-process, which is a more commonneutron capture process in astronomy, successive neutron capture leads to theproduction 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 thatthe neutron capture rate is much faster than the β-decay rate, known as ther-process. The r-process is effective in synthesizing a few highly neutron-richstable isotopes. The r-process is likely to be important in supernova explosionor in neutrino-heated atmosphere surrounding a new-born neutron star. (Seefigure 3.3.) Note, however, that both processes cannot synthesize the so-calledstable p-nuclei (p for proton-rich). Thus, we expect that p-nuclei are rare ina star. This prediction generally agrees with observations. (See figures 3.3and 3.4.) Astronomers believe that these proton-rich nuclei are formed by pro-ton capture at very high temperature or by the so-called (γ, n) reaction duringsupernova explosions.


4.8%

p process s only

N

112 114 115 116 117 118 119 120

121 123

122 124

0.38%

113 115

13s

110 111 112 113 114

Sb, 51

Sn, 50

In, 49

Cd, 48

63 64 65 66 67 68 70 71 72 73 7462

1.02%

12.4%

112d

12.8% 24.0%

0.69%

12.3% 28.8% 7.6%

116

7.6% 24.1% 8.5% 32.5% 6.1%

54h

69

95.8%

14.3% 27h

4.2%

Slow process path

only

Rapid process

43%57% 2.8d

Rapid Process

Z

Figure 3.4: Schematic representation of the effects of s- and r-processes.

3.7.7 Pair production

In the event that the temperature of certain region of a star exceeds≈ 2mec2/k ≈

1010 K, electron-positron pair production from thermal photon becomesimportant. (Actually, the effect cannot be neglected when the temperature isabout 109 K because a few highly energetic photons from the tail of the Planckdistribution can already undergo pair production.) Similar to photodisintegra-tion, pair production is endothermic and in effect placing a limit on the max-imum temperature of a star. It has been proposed that this mechanism couldgive rise to pair-instability supernova in very massive stars above 130M.

The work of stellar nucleosythnesis culminated in the publication of the articleby E. M. Burbidge, G. Burbidge, W. A. Fowler and F. Hoyle, Reviews of ModernPhysics 29, 547 (1957). This article introduced all the essential ideas involvingin the synthesis of elements from carbon to uranium in a star starting with thehydrogen and helium produced in the big bang. W. A. Fowler was awarded theNobel Prize in Physics in 1983 “for his theoretical and experimental studies ofthe nuclear reactions of importance in the formation of the chemical elements inthe universe.” (Actually, a similar idea was also pointed out independently inthe article A. G. W. Cameron, in Chalk River Report Number CRL-41, ChalkRiver 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 non-relativistic ideal gas. That is to say, the total energy of the star E obeys

E = −U =1

2Ω , (4.1)

where U is internal energy and Ω is potential energy. Suppose that the nu-clear energy generation rate of the star somehow increases a little bit. Then,E > 0 implying that U < 0. For ideal gas, that means T < 0. In other words,the average temperature of the star decreases. Consequently, the nuclear en-ergy generation rate will decrease. By the same argument, a slight decreasein nuclear energy generation rate will increase the internal energy of the starmaking the star a little bit hotter, therefore increasing the resultant nuclearenergy generation rate. This negative feedback makes the star thermally sta-ble against small perturbation in nuclear energy generation rate. We call thissecular 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 astar. A detail treatment of dynamical stability is very complicated. So, I onlyillustrate 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 pointr(m), the pressure is equal to the weight per unit area of layers between m andM . Equation of hydrostatic equilibrium (Eq. 2.4) implies that

P (m) =∫ M

m

Gm

4πr4dm . (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

4πr2ρdr = dm = 4πr′2ρ′ dr′

=

= . (4.4)

ρ′(m) =

≈ ρ(1 + 3∆) . (4.5)

Hence, the new hydrostatic pressure is given by

P ′(m) =∫ M

m

Gm

4πr4(1−∆)4dm

≈ P (1 + 4∆) . (4.6)

Assume further that the change in radius is adiabatic (i.e. PV γ =const.), then

P ′gas = P (1 + 3∆)γa ≈ P (1 + 3γa∆) (4.7)

where γa is the adiabatic exponent. Hence, in order to attain stable equilibrium,P ′gas > P ′ for all m if ∆ > 0 (in other words, a compression), such that the gascould “bounce back”. That is,

P (1 + 3γa∆) > P (1 + 4∆) .

This condition is equivalent to

γa >4

3. (4.8)

Note that the same condition is arrived for ∆ < 0 (in other words, an expansion).

We can show that the adiabatic exponent of a photon gas is 4/3. Hence, ifsome part of a star is dominated by radiation pressure, that part of the starwill become dynamically unstable against compression.


4.3 Convective instability

Recall in our earlier discussions that the radiative transfer equation holds onlywhen radiation is the principal energy transfer mechanism. Hence, the radiativetransfer 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).

P

P

1

2ρnew

ρ

ρ1

2

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, re-spectively. When displaced upward, its pressure and density will change. Wedenote its new pressure and density by P2 and ρnew, respectively. (See fig-ure 4.1.) Clearly, P2 equals to the pressure of the gas surrounding the newlocation of the lump. Denote also the density of gas surrounding the new loca-tion of the lump by ρ2. Obviously, the lump of gas is stable against convectionif ρnew > ρ2. Otherwise, the system will be convectively unstable. We mayassume that the change is adiabatic. (Why is this a good approximation?) Inthis case, P ∝ ργa and hence the convective stability condition becomes(

dP

dρ

)star

<

(dP

dρ

)adiabatic

. (4.9)

This is known as Schwarzschild’s criterion. This can be understood usingfigure 4.2: when pressure changes from P1 to P2, a stable configuration should


have ρ′2 < ρ∗ and hence a shallower slope than the adiabatic case. This givesEq. (4.9) above. Multiplying both sides by ρ/P , we have

γ ≡ ρ

P

(dP

dρ

)star

< γa . (4.10)

Figure 4.2: Schwarzschild’s criterion for convective instability.

Exercise: For an ideal gas with negligible radiation pressure, recall the ideal gaslaw

P =ρ

µmH

kT ∝ ρT , (4.11)

show that,dP

P=

dρ

ρ+

dT

T(4.12)

and henceP

T

(dT

dP

)star

<γa − 1

γa(4.13)


and ∣∣∣∣∣dTdr∣∣∣∣∣star

<

(γa − 1

γa

)T

P

∣∣∣∣∣dPdr∣∣∣∣∣star

. (4.14)

The above equation tells us that there is an upper limit for the temperaturegradient 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 thestar. Hence, the radiative transfer equation must be modified. Unfortunately,a comprehensive theory of convection is still lacking. This makes our analysisdifficult and complex.

However, if we assume that convection is extremely efficient in transportingenergy, then the temperature gradient dT/dr in the convective region of a starshould 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,

dT

dr= −γa − 1

γa

ρTGm

Pr2. (4.16)

Chapter 5

Evolution of Main SequenceStars

5.1 The complete set of stellar structure equa-

tions

After all our detail discussions, we finally have a complete set of stellar structureequations. They are the four first order coupled ordinary differential equations:

dr

dm=

1

4πr2ρ, (5.1)

dP

dm= − Gm

4πr4, (5.2)

dL

dm= ε , (5.3)

dT

dm=

− 3κL

256π2σr4T 3if convectively stable,

−γa − 1

γa

TGm

4πr4Pif convectively unstable,

(5.4)

κ = κ0ρaT b , (5.5)

ε = A(X, Y, Z)ρmT n , (5.6)

P = (ne + ni)kT +4σ

3cT 4 , (5.7)

where ni =ρ

mH

(X +

Y

4+

Z

〈A〉

),

52

CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS 53

ne =ρ

mH

(X +

Y

2+ Z

⟨ZA

⟩), and

X + Y + Z = 1 .

Exercise: Derive dT/dm from the radiative energy transport 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 sufficientlyshort time interval compared with the nuclear burning timescale, we know thatthe 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 whenthe time interval concerned is much shorter than the nuclear burning timescale.Therefore, we have a complete set of equations to determine the structure of astar.

These equations must satisfy the following boundary conditions:

r = 0 at m = 0 ,

P ≈ 0 at m = M ,

L = 0 at m = 0 , and

T ≈ 0 at m = M .

One may solve this set of equations to determine the stellar structure. Un-fortunately, the system of differential equations we are working with is toocomplicated to solve analytically. Indeed, explicitly (and numerically) solvingthis set of equations is the only way to accurately determine the structure of aspherically symmetric star.

A mathematical-oriented student may be doubtful about the existence anduniqueness of the solution to the stellar structure equations for any bound-ary conditions as well as any given total stellar mass and chemical compositionof the star. This is a very complicated mathematical problem. Nonetheless,theoretical astrophysicists have cooked up an extremely artificial situation thatthe solution of the stellar structure equations is not unique! Moreover, thereare situations that the solution may not exist at all. There is a good physicalargument for the non-existence of solution in some cases, too. Consider, forexample, the situation when the total stellar mass is, say, 1 g. The gravita-tional force will be so weak that any reasonable nuclear energy generation will


blow this object up. Thus, that object is likely to be supported by EM forcerather than gravity. Hence, we expect that no solution for the stellar structureequations 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 nu-merical 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 simpleinitial value problem for a system of first order coupled O.D.E.s. Numerical so-lution of the stellar structure equations, therefore, must be obtained with greatcare.

There are many proposed ways to numerically solve the stellar structure equa-tions. The first way is called the shooting method. Basically, the idea is thatwe start from the boundary at m = 0. At this boundary, we know for sure thatL = 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 thesolution until it hits the other boundary at m = M . We check if T and P atm = M are really equal to 0. If not, we carefully make another guess at them = 0 boundary for T and P until a solution is found. This method turns outto be not very accurate since T and P varies greatly between the interior andthe surface of a star leading to a great lost of numerical accuracy.

Another method is called the fitting method. Essentially, we guess the values ofP , T in the inner boundary and R, L in the outer boundary. Then we numer-ically integrate the equations and see if they matches continuously somewherein 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-Henyeymethod. This method is so involved that I will not discuss it in any detail inthis 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 boundaryconditions by a system of equations. And then, we carefully solve the systemof equations by Newton-Raphson iteration. If you want to know the detail, youmay 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


m(r)/M

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

r

X

L

ρ

T

Figure 5.1: A 1 M model during main-sequence hydrogen burning at time4.2699 × 109 yr, showing radius, density, temperature, total luminosity, andhydrogen abundance versus mass fraction.

Y

m/M0 1

L

X

Figure 5.2: Model of a 1M star just after it leaves the main sequence, at time10.31× 109 yr.


m(r)/M

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

L

X

P

T

rρ

Figure 5.3: Model of a 5M star just after it leaves the main sequence at time6.84461× 107 yr.

m(r)/M

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

r

L

L

ρ

P

T

X4

X4

Figure 5.4: Model of a 5M star during the giant star stage at time 7.04×107 yr.


a number of situations. (Note that the y-axis in all the figures below are inlinear 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 evolu-

tion

A zero age main sequence star (ZAMSS) is defined as a star which justbegins its hydrogen burning. In other words, ZAMSS is a new born star. Thelocation of ZAMSSs in the H-R diagram is known as the zero age main se-quence (ZAMS). Since a star is formed from the gradual collapse of a nebula,a spherically symmetric non-rotating ZAMSS is characterized by only two setsof 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 timefor the stellar matter to mix. Hence, X, Y , and Z should be uniform over theentire ZAMSS.

By numerically solving the stellar structure equations above, we know that thestructure of a ZAMSS can be summarized below: if the mass M is below about0.1M, thermal hydrogen fusion does not occur in the core and hence we haveeither a brown dwarf or a planet. We may talk about more about them in thespecial topics in this course. If the mass M is between about 0.1 to 0.3M, theentire star is convective and p-p chain is the major nuclear reaction occurringin the stellar core.

When M is between about 0.3M and 1.2M, p-p chain is the dominant re-action. The core is radiative and the surface layer is highly convective. WhenM >∼ 1.2M, CNO cycle is the dominant reaction and the core is convective.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 30M fortwo reasons. First, its core will be dominated by radiation pressure which leadsto instability. Second, the surface luminosity will be so high that mass ejectionis serious.

The above results from ZAMSS can be understood as follows. First, althoughthe energy production rate for CNO cycle scales like T 16 while that of the p-pchain scales like T 4, the corresponding coefficients (which we both call them Ain this note) for CNO cycle is much smaller than that of the p-p chain. (Doyou know why?) Hence, at temperature <∼ Tcrit, the p-p chain is the dominantreaction. This is the reason why CNO cycle is only important for ZAMSS withmass >∼ 1.2M.


Second, the sensitive temperature dependence of CNO cycle (namely, T 16) isthe origin of a convective core for high mass ZAMSS. For convective stability,Eq. (5.4) gives

d2T

dm2= − 3κ

256π2σr4T 3

[dL

dm+ L

d

dm

(1

r4T 3

)]

≈ − 3κ

256π2σr4T 3

dL

dm, when r ≈ 0, L ≈ 0

≈ −3κAρT n−3

256π2σr4. (5.8)

That is, for regions sufficiently close to the stellar core, |dT/dm| increases asn 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 thecore becomes convective.

Third, when M <∼ 1.2M, the temperature close to stellar surface is so lowthat free-free absorption (hence, Kramer’s opacity) is important. Such a largeopacity makes the stellar surface a good photon absorbing medium. Hence, theonly effective way to transfer these absorbed energy out to the stellar surfaceis convection. In contrast, when M >∼ 1.2M, the temperature close to stel-lar surface is high enough that electron scattering opacity becomes important.Hence, in this case, the stellar surface is radiative.

Finally, when 0.1M <∼ M <∼ 0.3M, temperature in most part of the staris so low that Kramer’s opacity is important. Hence, the whole star is fullyconvective.

5.4 Homology relation

Let us consider our stellar structure equations supplemented by the simplifiedopacity, energy production equations as well as the equation of state. To illus-trate the idea, we only consider the case of (a) the opacity κ is temperatureand pressure independent; (b) the energy production equation is a power lawof 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 (mainsequence) star is radiative, the temperature in most part of the star is hotenough so that electron scattering is the dominant contribution to opacity, andthe mass of the star is less than about 10M so that contribution of radia-tion pressure is small. With these assumptions in mind, we write our stellar


structure equations below:

dP

dm= − Gm

4πr4, (5.9)

dr

dm=

1

4πr2ρ, (5.10)

dT

dm= − 3κL

256π2r2σr2T 3, (5.11)

dL

dm= AρT n , and (5.12)

P =ρkT

µmH

. (5.13)

Now, we define a dimensionless variable x by x = m/M . We denote

r = f1(x)R∗ , (5.14)

P = f2(x)P∗ , (5.15)

ρ = f3(x)ρ∗ , (5.16)

T = f4(x)T∗ , and (5.17)

L = f5(x)L∗ . (5.18)

Note that quantities on the right represent the maximum values inside a star.Exercise: Substitute Eqs. (5.14) and (5.15) into Eq. (5.9) to show that

P∗M

df2

dx= − GMx

4πf 41R

4∗. (5.19)

In a physical equation the dimensions on the two sides must match, and hencein Eq. (5.19), where x, f1, and f2 are dimensionless, P∗ must be proportional toGM2/R4

∗. Without loss of generality, we can take the proportionality constantto be unity. The equation above is then separated into

df2

dx= − x

4πf 41

and P∗ ∝GM2

R4∗

. (5.20)


Similarly, the above five equations can be decoupled into two sets. The firstset consists of four differential equations and one algebraic equation involvingf1(x) through f5(x) only:

df2

dx= − x

4πf 41

, (5.21)

df1

dx=

1

4πf 21 f3

, (5.22)

f2 = f3f4 , (5.23)

df4

dx= − 3f5

4f 34 (4πf 2

1 )2, and (5.24)

df5

dx= f3f

n4 . (5.25)

The second set of equations are algebraic equations relating ρ∗, T∗, R∗, P∗, L∗and M :

P∗ ∝ GM2/R4∗ , (5.26)

ρ∗ ∝ M/R3∗ , (5.27)

T∗ ∝ µP∗/ρ∗ , (5.28)

L∗ ∝ T 4∗R

4∗/M , and (5.29)

L∗ ∝ ρ∗Tn∗M . (5.30)

The importance of this re-expression of the set of stellar structure equations isthat the values of ρ∗, T∗, R∗, P∗ and L∗ can be expressed in terms of a singleparameter 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 valuesof stellar mass M . This similarity property is known as homology.

Exercise: Eliminating ρ∗, P∗, T∗, and R∗ from the above equations to show that

L∗ ∝M3 (5.31)

This mass luminosity relation agrees reasonably well with observations (seefigure 1.3.) A few more relations can be deduced as well. For example,

R∗ ∝ Mn−1n+3 , (5.32)


ρ∗ ∝ M2(3−n)n+3 , (5.33)

L1− 2(n−1)

3(n+3)∗ ∝ T 4

eff , and (5.34)

T∗ ∝ µM/R∗ , (5.35)

where L∗ = 4πR2∗σT

4eff 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 mainsequence stars, the radius increases with mass, mean density ρ∗ decreases withmass. For p-p chain burning stars, Eq. (5.34) gives

logL ≈ 5.6 log Teff + constant, (5.36)

and for CNO cycle burning stars,

logL ≈ 8.4 log Teff + constant. (5.37)

Again, these slopes are in rough agreement with the observation in the H-Rdiagram. Besides, the characteristic life time for a main sequence star can beestimated by τMS = M/L ∼ M−2. Thus, heavier main sequence star livesshorter 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 ∝M5+31/2n1+5/2n ≈M5.46 (5.38)

This results in a change of slope between high and low mass stars in the mass-luminosity relation (see Figure 1.3). Eq. (5.38) also gives

logL ≈ 4.12 log Teff + constant. (5.39)

5.5 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 ismuch shorter than the nuclear burning timescale, we know that the change inX, 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)

dt≈ −ε(r)

∆E, (5.40)


dY (r)

dt≈ ε(r)

∆E, and (5.41)

dZ(r)

dt= 0 (5.42)

to the evolution of a hydrogen burning star, where ∆E is the energy releasedper helium nucleus formed.

The effects of the change in mass fractions described by the above equationscan be summarized below: First, since hydrogen is burned only in the core,an inhomogeneous distribution of chemical elements in the star will eventuallyappear. Namely, the core becomes more and more hydrogen depleted while thesurface is still hydrogen rich. (More precisely, this statement is only true formain sequence stars with mass higher than about 0.2M in which the effectsof convection, diffusion and flow are not important in changing the distributionof particles inside them during its main sequence life.) Can you justify thisstatement? In addition, do you know why this statement is wrong for a 0.1Mmain sequence star?

Second, the star must evolve due to such a gradual chemical change. Sinceits core will be more and more hydrogen depleted and helium rich, hydrogenburning will eventually be unable to provide the gas and radiation pressuregradient, 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 sufficientsmall timescale, both the thermal and gravitational potential energies of a starare approximately conserved. So, as the core of the star contracts, we expectthe outer part of the star to expand. And since the temperature around the coreof the star increases due to core contraction, we also expect the temperature ofthe outer part of the star to decrease. So, the star is heading towards the redgiant phase.

Chapter 6

Evolution of Pre-Main SequenceStars

6.1 Collapse of interstellar cloud

The interstellar medium is not empty, but consists of gas, dust, clouds, andplasma. The densest part is interstellar clouds. If the mass is high enough,the interior can be shielded from UV radiation from stars, such that moleculescould exist. They are called molecular clouds and most of them contain H2 andCO. A star is formed from the collapse of molecular clouds. They can have highmasses of 103 to 107M and the typical temperature ranges from ∼ 1 K (mostlydark nebula) to ∼ 104 K (mostly bright emission nebula). Its density is about10−24–10−21 g cm−3. Clearly the EOS of a cloud can be well approximated bythe ideal gas law.

If the thermal pressure in a gas cloud cannot balance the gravity, part the cloudwill collapse under external perturbation (e.g., triggered by supernova shockwave). From the virial theorem, a cloud will collapse if −Ω > 2U . Assuming aspherical cloud of uniform temperature and density, the gravitational potentialenergy is

Ω = −3

5

GM2

R(6.1)

and the thermal energy is

U =3

2NkT =

3

2kT

M

µmH

. (6.2)

63

CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS 64

Hence, the condition to collapse is

3

5

GM2

R>

3MkT

µmH

. (6.3)

Using M = 4πρR3/3, we can rewrite the above equation in termal of ρ to obtain

M > ≡MJ , (6.4)

where MJ is called Jeans mass. It is the minimum mass of a cloud that cancollapse (i.e. upper limit for a stable cloud). Putting in the typical numbers,

MJ ≈ 105T 3/2M/√n . (6.5)

Thus, only giant gas cloud of mass of order of 105M or more will collapse whenperturbed.

Exercise: Similar to above, we can express the equation in terms of R and ρ,such that the Jeans length is given by

RJ =

(15kT

4πρµmHG

)1/2

. (6.6)

As the cloud collapse, its Jeans mass will change since T and n are changed. Ifthe 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 thefragmentation. This mechanism is believed to be the way for a large massivegas cloud to collapse into a number of solar mass stars, which forms a stellarcluster. As the cloud fragments get denser and hotter, they will eventuallybecome opaque and cooling becomes inefficient. The rise in temperature willincrease 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 gravitationalcollapse can be estimated by

Ω ∼ Ω

τff∼ GM2

R

√Gρ ∼

(3

4π

)1/2 G3/2M5/2

R5/2, (6.7)


where τff ≈√R3/GM ≈ 1/

√Gρ is the free-fall timescale (see Chapter 2.6).

This energy is carried away via blackbody radiation. For efficient cooling,

Ω < U = 4πR2σT 4 . (6.8)

This gives MJ,min ∼ 0.02MT1/4µ9/4.

We expect the collapsing process to take place within the free-fall timescale. Asρ is very low for a gas cloud, τff can be as long as a few million years. Theactual collapse is much longer, because rotation and magnetic field — whileneglected is this course — play an important role here. The conservation ofangular 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 lowopacity (i.e. mostly transparent to photons) since κ ∼ 0.01 g−1 cm2 and themean density ρ ∼ (3M/4πR

3) ∼ 2 × 10−16 g cm−3, giving an optical depthτ ≈ κρR 1. The gravitational potential energy is radiated away efficientlyand 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 acentral temperature of 2 × 104 K, hydrostatic equilibrium is achieved and aprotostar is formed. Numerical simulations suggest that the core has a massof 1.5×10−3M and a radius of 1.3R at this stage. A protostar is powered bygravitational energy and it is extremely luminous, can be ∼ 100 − 1000L fora 1M star, with surface temperature 0.6 times of the Sun, and a large radiusof 70R. However, protostars are embedded in clouds that make observationsdifficult. (How to observe a protostar?) Detailed structure of protostars fromnumerical simulations can be found in Lasson, R.B. (1969) MNRAS, 145, 271.

The protostar will continue to accrete, until the surrounding material is de-pleted. Once the infall and accretion cease, mass of the star is nearly fixedand the protostar is now known as a pre-main sequence (PMS) star. In thisstage, radiation can escape, hence, the PMS star cools down and contracts.The energy source is still gravitational contraction, but the evolution timescaleis governed by the Kelvin-Helmholtz timescale τKH , as compared to τff for pro-tostars. Note that τKH ∝ 1/R, meaning that it will increase during contraction,while τff ∝ 1/ρ decreases during contraction. (How to observe a PMS star?)


Exercise: Estimate the size and temperature of a protostar by assuming thatall gravitational energy released is used to dissociate molecular hydrogen andto 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.)

−Ω =3

5

GM2

R≈ M

mH

(X

2χH2 +XχH +

Y

4χHe

)(6.9)

and take Y ≈ 1−X, we obtain

R ≈ 50

1− 0.2X

M

MR . (6.10)

As hydrostatic equilibrium is established, the temperature can be estimatedfrom the virial theorem

3

2

kT

µmH

M = −1

2Ω , (6.11)

which gives T ≈ 6× 104 K.

6.3 Hayashi (convective) track

The core temperature of a PMS star is much higher than the surface tempera-ture due to the large opacity of hydrogen. This results in a large temperaturegradient, making the entire PMS star (up to the photosphere) fully convective.As the radius continues to shrink, the core heats up. However, the surfacetemperature remains stable, because the photosphere can adjust its thicknessto counteract the rise of the core temperature. As a result, the luminosity de-creases as R2, and the star moves down nearly vertically on the H-R diagramover 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 withindex n. Hence, from Eq. (2.24) the coefficient K is given by

Kn = Cn( ) . (6.13)

where the coefficient Cn depends on n only.


The value of R is determined by joining the fully convective interior to theradiative photosphere at the boundary r = R. Quasi-hydrostatic equilibriumcondition requires

dP

dr≈ −GMρ

R2. (6.14)

Hence,

PR ≈GM

R2

∫ +∞

Rρdr . (6.15)

In the photosphere, the optical depth of the photosphere is of order of 1,

κ∫ +∞

Rρdr ≈ 1 (6.16)

where κ is the mean opacity of the photosphere. Taking κ to be the opacity atR and approximate it as a power-law of ρ and the surface effective temperatureTeff , we have

κ0ρaRT

beff

∫ +∞

Rρdr ≈ 1 . (6.17)

Hence,

PR ≈GM

R2κ0

ρ−aR T−beff . (6.18)

Exercise: Using Eqs. (6.12), (6.13), (6.18), the ideal gas law, and the blackbodyluminosity, show that

logPR = logM − 2 logR− a log ρR − b log Teff + constant

n logPR = (n− 1) logM + (3− n) logR + (n+ 1) log ρR + constant

logPR = log ρR + log Teff + constant

logL = 2 logR + 4 log Teff + constant

We can writelogL ≈ A log Teff +B logM + constant . (6.19)


Figure 6.1: Evolution paths of pre-main sequence stars from numerical simu-lations. The nearly vertical lines are the Hayashi tracks, then followed by thenearly horizontal Henyey tracks.

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. Thisrepresents a very steep evolutionary path on the H-R diagram, i.e. a PMS starevolves by reducing its luminosity with more or less constant effective surfacetemperature (see figure 6.1).

The Hayashi track also sets the boundary for which a stable star can exist. Theregion 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 realPMS star does not evolve on the H-R diagram exactly on the Hayashi track asthe assumptions we used in the arguments above is overly simplistic. However,the Hayashi track is still reasonably close to reality. The relevant timescale atthis stage is the Kelvin-Helmholtz timescale which is very short. Contractionto main sequence stars takes up < 1% of the life time of a star, while mainsequence stage takes up 80%. This is part of the reasons why so few pre-mainsequence stars are detected. One example of PMS star is T Tauri stars, whichare less massive stars that evolve more slowly along the Hayashi track, and theyare highly variable due to mass ejection.

6.4 Henyey (radiative) track

Upon further contraction, the internal temperature rises to the point that opac-ity greatly decreases (Eq. 3.5: κff ∼ T−7/2). The core is in radiative equilib-rium. The convective zone recedes from the center, leaving only a convectiveenvelope. The core of the star contracts on the free-falling timescale, the grav-itational energy becomes more negative. By the virial theorem, the internalenergy, and hence the surface temperature, has to increase rapidly, while theluminosity increases gradually. Therefore, the PMS star will move away fromthe 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 veryearly. For example, the central temperature of stars above 10M can reach3 × 107 K in 1000 years, then CNO is ignited (but not in equilibrium) and aradiative core is formed due to the high temperature. Therefore, very massivestars almost left the Hayashi track and on the Henyey track instantaneously.On the other hand, stars < 0.5M remain in the convective phase throughouttheir lives and there is no Henyey track for them.

Most light elements are burned off in this stage (see Chapter 3.7.3), exceptdeuterium, which is ignited while the star is still on the Hayashi track), but theenergy production is negligible compared to the gravitational energy. Finally,the core temperature is high enough that stable hydrogen burning (p-p chain orCNO cycle) gradually take over the energy generation and gravitational collapseslows down. The star becomes a zero age main sequence (ZAMS) star. Ittakes solar-type stars 3 × 107 yr for the protostar and PMS phases, which isonly 1% of the entire life time. Massive stars above 10M can take less than105 yr and low-mass stars can spend more than 108 yr before hydrogen burningreaches 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 12C in thecore such that the core has to slightly contract to adjust the rate of hydrogenburning. If you are interested to learn more, can refer to a simulation paper byIcko Iben Jr. (1965) ApJ, 141, 993, or the book by the same author: StellarEvolution Physics, Volume 1 (2013) (Cambridge: CUP).

Chapter 7

Evolution of Post-MainSequence Stars

7.1 Red giants

As the core temperature keep on increasing due to contraction, the thin layer ofhydrogen-rich material located just on top of the helium-rich core will be heatedup. Eventually, this thin hydrogen shell will burn. This shell burning becomesthe 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 entirelyhelium. 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∫ Vcore

0P dV = PcoreVcore +

αGM2core

3Rcore

=4πPcoreR

3core

3+αGM2

core

3Rcore

, (7.1)

where the label core refers to the boundary of the core (i.e., r = Rcore), and αis a geometric factor depending on the mass distribution of the core.

For ideal isothermal classical non-relativistic gas with temperature Tc = Tcore,∫ Vcore

0P dV =

kTcµcoremH

∫ρ dV =

kTcMcore

µcoremH

, (7.2)

where µcore is the mean molecular weight of the material within a distance ofRcore from the stellar core. Hence,

Pcore(Rcore) =3kTcMcore

4πµcoremHR3core

− αGM2core

4πR4core

. (7.3)

71

CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 72

By considering dPcore/dRcore = 0, we know that for a fixed core mass Mcore,Pcore attains its maximum value of Pcore,max ∼ T 4

c /M2coreµ

4core. For a stable core,

this maximum pressure must be balanced by the pressure Penv exerted by theenvelope. By assuming that the core is very small, Penv can be estimated byEq. (2.5), namely, Penv ≥ GM2/8πR4. Hence, the condition for stability of anisothermal classical non-relativistic core is Pcore,max ≥ GM2/8πR4. Now, apply-ing one of the homology relations in Eq. (5.35), namely, Tc ∼ µenvmHGM/kR,the stability condition becomes

Mcore

M<∼ constant

(µenv

µcore

)2

. (7.4)

Actually, M. Schonberg and S. Chandrasekhar deduced this result with theconstant 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 non-degenerate and non-burning helium core is about 0.13M.

Now, let’s consider the case of a high mass star (say of mass >∼ 2M). Oneof the homology relations in Eq. (5.33) shows that the central density ρc of amain sequence star decreases as its mass increases. So, for high mass star, thecentral density is not very high so that the stellar core is still non-degenerateeven after it has exhausted its hydrogen fuel. The mass of this isothermalhelium core will gradually increase as hydrogen is burnt in the shell. As thecore mass exceeds the so-called Schonberg-Chandrasekhar limit, the corewill collapse. In time, the core will acquires the temperature gradient necessaryto balance gravity. But this temperature gradient causes loss of heat and hencethe core will further collapse leading to further temperature rise. This corecontraction process will take place in the Kelvin-Helmholtz timescale, which isabout GM2

core/RcoreLcore ≈ 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 bythe contracting core will be absorbed by the outer layer of the star. Thus, asthe 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 ared giant is formed.

Since the evolution from high mass main sequence to red giant is very rapidin stellar evolution timescale, very few stars are found in the H-R diagramimmediately outside the main sequence. This gap is called the Hertzsprunggap (see figure 7.1.)

There is an interesting consequence of the evolution towards red giants for high


B

C

A

Lum

inosi

ty

Temperature

Figure 7.1: A schematic illustration of post-main sequence evolution for a 5 Mstar on H-R diagram (the light solid curve). The heavy solid curve showsthe main sequence. Stars spend very short time when evolve from point Bto point C. Only a few of the stars are found in this region, known as theHertzsprung gap.

mass stars. It turns out that the convective zone of a high mass main se-quence star is rather extended. So, by the time when the hydrogen burningshell is formed and the outer layer of the high mass red giant becomes convec-tive, this convective region overlaps with the original main sequence convectivezone. Hence, some material originally located deep inside the stellar core willbe brought up to the stellar surface. This process is called the dredge-up (orthe first dredge-up since high mass star will undergo several dredge-ups in itslife and this is the first one). Do you know what kind of observational evidencesupport for this idea of dredge-up?

To summarize, within about 106 yr, a red giant star is formed. A thin layer ofspherical 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 willcontribute 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 thecentral density of this type of main sequence stars is rather high. In fact, it isso high that the electron there is close to degenerate. Hence, the Schonberg-


Chandrasekhar limit no longer applies. As the core contracts after it has usedup its hydrogen fuel, it will quickly go degenerate.

Provided that the temperature is much lower than the Fermi energy, the equa-tion of state of the helium core can be well-approximated by an electron Fermigas at almost zero temperature. Recall from Eq.( 2.77), the EOS is a polytropeof index n = 1.5.

Exercise: From Eqs. (2.23) and (2.24) concerning the mass, radius and centraldensity of a polytropic star, show that the mass and radius of the degeneratehelium core obey

Mcore ∼ R−3core . (7.5)

Most importantly, it turns out that the electron degenerate pressure of thehelium core is sufficient to support the stellar envelope. As a result, for lowmass stars, the core contraction is slow and hence its road to red giants isgradual. Thus, there is no Hertzsprung gap in the lower bottom part of the H-R diagram. Note also that there is no dredge-up at this phase of the evolutionof 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 arelocated at the so-called red giant branch in the H-R diagram characterizedby the high luminosity and low surface temperature. The structure of a typicalred giant star is shown in figure 7.2. The exception is when the mass of the staris less than about 0.5M. In this case, the expansion of the radius of the staris 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 isno longer valid. In fact, the most direct way to answer the question of what thestructure is and what evolution of a red giant star take places is by numericalsolution. Such analysis has been carried out since the 1960’s by various groupsof astrophysicists, and perhaps the most famous one was headed by I. Iben, Jr.


CoreHelium

Hydrogen−riched zone

Hydrogen burning shell

Figure 7.2: The structure of a typical red giant star.

As the star expands, the surface gravity is low and hence material ejection inthe form of stellar wind can be very significant. Hence, from this time on, massconservation of the whole star is no longer a good approximation. This furthermakes the modeling of stellar evolution for red giant and beyond difficult. Ourdescription of subsequent evolution of a star is, therefore, forced to be ratherdescriptive at the level of this course. To add further complications, it turnsout that the precise subsequent evolution of a star depends on the mass of thestar. (Not just the initial mass of the star, but how much mass is lost by stellarwind.)

7.2 Helium core burning and helium flash

As the core collapses further, the central temperature and pressure will be highenough to burn helium via the triple-alpha process. For stars with mass greaterthan about 2M so that the stellar cores are non-degenerate, the ignition oftriple-alpha process is relatively smooth. That is, when the central temperaturereaches 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 heliumburning will increase steadily with time while that due to hydrogen burningwill decrease steadily (although it will not stop completely). Due to the hightemperature sensitivity of helium burning (ε ∼ T 40), the core of such star isconvective.


The situation is slightly more complicated if the main sequence star is less thanabout 2M (but higher than about 0.5M). Recall that the core of this type ofstars is degenerate. Unlike ideal gas law, pressure of degenerate Fermi matter(in this case electrons) at low temperature is almost independent of temperatureT . Hence, for low mass star, electrons in the helium-rich core becomes stronglydegenerate before helium burning sets in.

We expect that the degenerate helium core of a low mass star is much moremassive than its envelope. So, homology relation (Eq. 5.35) holds approximatelyfor the stellar core. In particular, we have Tcore ∼Mcore/Rcore. Combining withEq. (7.5), we conclude that low mass stars ignite core helium burning whenevertheir core masses reach a threshold irrespective of their envelope masses. Detailcalculation shows that this threshold value is about 0.5M.

Since the degenerate matter is highly conductive, the degenerate helium coreis highly isothermal. So, the entire core burns simultaneously once the heliumis ignited. Furthermore, the weak dependence of pressure of degenerate matteron temperature implies that the rise in core temperature will not affect thecore pressure too much. Thus, the helium burning rate of the core will increasequickly leading to a thermonuclear runaway. So, helium ignites in an explosion,known as helium flash, in low mass stars. Actually, the peak luminosity (whichlasts for a few seconds at most) is about 1011 L which equals the luminosityof the entire galaxy! Nevertheless, we cannot see helium flash directly fromthe EM radiation emitted from the star as most of the energy released in thissudden explosion is absorbed by the stellar envelope. (Nevertheless, we caninfer 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 canbe well approximated by an ideal gas once more. Thus, the core expands andhelium burning becomes stable.

Finally, for extremely low mass stars with mass <∼ 0.5M, their degeneratehelium cores are never hot enough to ignite helium burning. The degeneratecores 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, thecentral helium burning leads to an increase in the nuclear energy generationrate. Consequently, the core expands and hence the envelope contracts. Andthe star becomes smaller, less luminous and its surface becomes hotter. (Doyou know why?) So, the star descends from the red giant branch by moving tothe lower or lower-left in the H-R diagram.

The locus of high-mass core helium burning star forms the so-called the helium


main sequence — a somewhat misleading term as this type of stars havehydrogen burning shells which still significantly contribute to nuclear energygeneration. In theory, the helium main sequence is a band similar to the highmass part of the (hydrogen) main sequence except that it is shifted to the rightin the H-R diagram. In practice, the helium main sequence is mixed up withthe thick red giant branch. Computer simulations show that for high massstars, the evolutionary track from red giant to helium main sequence is roughlyhorizontally to the left in the H-R diagram. (Interestingly, when the helium inthe core is used up at a later time, the star will again move roughly horizontallyto the right in the H-R diagram. That is to say, the combined evolution makethe star to form a loop in the H-R diagram.) Some stars on the helium mainsequence are Cepheid pulsating variables whose period is of order of days tomonths.

The locus of low-mass core helium burning star forms the so-called horizontalbranch, a roughly horizontal strip stretching between the main sequence andthe red giant branch. (See figure 7.3 below.) Clearly, these stars have aboutthe same core masses just before core helium ignition. So, it is not surprisingthat the location of a star on the horizontal branch is due to the envelopemass. Some stars on the horizontal branch are pulsating variables, known asRR Lyrae variables. Periods of RR Lyrae variables, which are about a fewhours, are much shorter than those of Cepheid variables. If time permits, I willtalk about pulsating variables in the special topics.

The lifetime of a core helium burning star, irrespective of its mass, is muchshorter than the corresponding core hydrogen burning main sequence star. It isbecause 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 sequencestar of the same mass.

7.3 AGB star

Evolution of a core helium burning star is in some sense quite similar to that ofa core hydrogen burning main sequence star. More precisely, a carbon-oxygen-rich core is gradually formed. Then the star further expands. Most of theenergy production now occurs in the helium burning shell and the hydrogenburning shell on top of the core. Once more, material from the core is dredgedup and mixed into the envelope. If the carbon oxygen core is degenerate, thestar is called an asymptotic giant branch (AGB) star (see figure 7.4.) A starmay leave the AGB phase by either complete removal of hydrogen envelope bystellar wind; or ignition of carbon in the core.


Lu

min

osi

ty L

/L

10000 5000 2500

10

102

1

eff

AB

Main Sequence

Red Giant

Horizontal Branch

Effective Temperature T (K)

Figure 7.3: A schematic H-R diagram showing the evolution of low-mass starsfrom red giant branch to horizontal branch. Point A corresponds to large lossof 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 non-burning layer. The outer hydrogen burning shell burns hydrogen thereby in-creasing the mass of the helium non-burning layer. Similarly, the helium burningshell consumes the helium and hence eating into the helium non-burning layer.Due to the great difference between rates of hydrogen and helium burning pro-cesses, 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-burninglayer changes periodically. During most time in the cycle, hydrogen is burnt inthe external shell while the helium burning shell is extinct. Thus, the heliumlayer separating the two shells increases in mass. The gradual contraction ofthis 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 thehelium shell burns, the outer layer of hydrogen burning shell expands and coolsdown. This leads to a great suppression of the hydrogen burning rate. As thehelium burning front advances, the temperature of the hydrogen burning shellincreases while that of the helium burning shell decreases. Thus, a new cycleof hydrogen and helium shell burning begins. To summarize, the AGB star is


core

CO riched

H riched zoneH burning shell

He riched zone

He burning shell

Figure 7.4: The typical structure of an AGB star.

burning hydrogen most of the time while only it is burning helium occasion-ally. In fact, astronomers sometimes called the helium-shell-burning phase of anAGB star the thermal pulse. This cycle of hydrogen and helium ignition withthermal pulse lead to sudden increase in radiation pressure in the interior of anAGB 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 10−5M yr−1.) A planetary nebula is formedeventually. Detail study of planetary nebulae will be studied as a special topicsif time permits.

An AGB star will undergo up to a few thermal pulses in its life. Each cycle ofhydrogen- and helium-shell burning also leads to dredge-up of materials. (Thereis a confusion of naming here as astronomers studying AGB stars also calledthe first time when materials in the core of an AGB star is transported to thestellar 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 pulsatingvariables, such as Mira variables, are AGB stars.

While most of the outer layer of an AGB star will eventually be ejected forminga planetary nebula, the core of an AGB star will collapse after using up all itshydrogen and helium fuel. The mass of an AGB star is not high enough so thateven when the electrons in the core become degenerate, the core temperature isnot 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 discussedin other astrophysics courses. Note that only a very small portion of starsare massive enough to have a non-degenerate C-O core while do not undergosupernova explosion.

CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 80L

um

inosi

ty L

/L

Effective Temperature T (K)eff

250040000 20000 10000 5000

10

10

10

10

10

10

10

10

1

4

5

3

2

−2

−3

−1

white dwarf

Main Sequence Star

Red Giant

Planetary Nebula

red supergiant

AGB Star

Horizontal Branch

Figure 7.5: An illustration of evolutionary path of a low mass star from mainsequence star to white dwarf in H-R diagram. Note that the path from AGBstar via planetary nebula to white dwarf is not fixed.

7.4 Evolution of high mass star after the for-

mation of carbon-oxygen-rich core

The evolution of a high mass star up to the formation of a carbon-oxygen-richcore is similar to that of a 3M star. But there are small differences. Due to itshigh mass, energy generation rate in the core is so high that radiation pressureis important even in the main sequence phase. This leads to mass loss by stellarwind when the initial mass of the star is greater than about 30M. In fact, thestellar wind is so strong in some of these stars that the entire hydrogen envelopeis blown away leaving the helium-rich core. Such stars — luminous, hydrogendepleted, 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


the main sequence phase is already close to the Eddington limit. Hence, theluminosity remain more or less constant in spite of internal changes. Thus, itsevolutionary track in the H-R diagram is more or less horizontal.

After the formation and collapse of a carbon-oxygen-rich core, the central tem-perature is hot enough to burn carbon and later on oxygen. At this stage, thecore temperature is so high that energy lost due to neutrino emission becomessignificant. In fact, all major core nuclear burning processes take place in rapidsuccession until the inner core is 56Fe rich. Surrounding the core are shells thatburn different types of nuclei.

Finally, the star is so massive that its core remains non-degenerate until thevery last stage of its evolution. (Do you know why?)

7.5 Supernova

The theory of supernova explosion is rather involved. Here, I shall only outlinethe basic idea behind.

As 56Fe is the most energetically favorable nucleus, no major exothermic nuclearreaction can take place in the 56Fe-rich core. When this core contracts, electronin the core eventually becomes degenerate. Unfortunately, since the star ismassive, 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. Thatis, the core can be approximated by a polytropic EOS with index n = 3. Butfor a polytropic stellar core, R3−n

core ∼ M1−ncore . Hence, as n tends to 3, there is

only one possible solution for Mcore. Thus, the core is unstable upon additionof mass. (One way to remember this fact is that dP/dρ is not big enough tocounterbalance gravity as ρ increases.)

Since the pressure of the relativistic electron degenerate core is not strongenough to oppose gravity, the core further collapses. Hence, the core is furtherheated up. The EOS of the degenerate core matter is insensitive to tempera-ture making the heat up more unrestrained. Now, the temperature is so hotthat photodisintegration of iron back to 4He and then further back to protonsand neutrons becomes important. This endothermic reaction quickly absorbsthe heat from the core and hence reduces the central pressure. With a suddendecrease in central pressure, the core contracts with an even higher rate. (Infact, at almost the free-fall rate.) Even worse, the core becomes so dense thatprotons can capture electrons and convert into neutrons plus neutrinos. Thisreaction is also endothermic. Besides, it reduces the number of particles in the


core. The combined effect is to reduce the core pressure further. Finally, theneutron gas in the core becomes degenerate and the neutron degenerate pres-sure is sufficient, at least temporarily, to stop the collapse. Such neutron coreis about 10 km in radius.

The gravitational energy released by the contracting core is of the order ofGM2

core/Rcore ≈ 1053 to 1054 erg. The energy absorbed during the disintegrationof nuclei is only about 1052 erg which is about one tenth to one hundredth that ofthe total gravitational energy released. Thus, most of the gravitational energyreleased is available to eject the material outside the core and for producingthe huge luminosity and neutrino flux we see during a type II supernovaexplosion (also called core-collapse supernova).

Recall that in the final phase of the collapse, a large number of neutrinos isproduced in the core within a short period of time. These neutrinos are re-sponsible for transferring the released gravitational potential energy out fromthe core. Although neutrino interacts very weakly with matter under most cir-cumstances, a non-negligible neutrino opacity builds up in the core because ofthe high density and high neutrino flux. Hence, some of the neutrino energy isabsorbed by the envelope. Consequently, the envelope of the star quickly ex-pands. This explosion is known as type II supernova (or simply supernova whenno confusion is possible). (In contrast, type I supernova is the explosion thatdestroy a white dwarf as it accrues too much material from its surrounding.Observationally, the spectrum of a type II supernova contains hydrogen lineswhile that of type I supernova does not. Do you know why?)

When observed far away from this dying star, the luminosity of the star suddenlygoes up. At first, most radiation is in the UV range as the effective surfacetemperature is high. But the surface temperature quickly drops and the objectradiates mainly in the optical range. The later time optical emission is thenpowered by radioactive decay of newly formed elements. Typical light curve ofa type II supernova is shown in figure 7.6 below.

There is an important prediction in this model, namely, that we should see theneutrino 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 becamevisible for SN1987A.

Heavy elements in the envelope are ejected into the interstellar space in su-pernova explosion. These elements are produced in two ways. They may beproduced in the core and shell burning prior to supernova. Alternatively, theymay be formed during the supernova. Since the peak temperature in a super-


Blu

e M

agn

itud

e B

elow

Max

imu

m L

ight

0 50 100 150 200 250 300 350 400

8

7

6

5

4

3

2

1

0

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 equilibriumis achieved on a timescale of seconds. In this way, different kinds of nuclei areproduced. In particular, the 56Ni formed will later on decay to 56Co with a half-life of 6.1 d. Then 56Co will decay to the stable 56Fe with a half-life of 77.1 d.The energy released in these two decays powers the supernova light curve afterthe initial decline from maximum. Signatures of the decay of these unstablenuclei are observed in supernova light curves.

Chapter 8

Special Topic: Introduction toStellar Pulsation

Throughout this course, we focus on the static structure of a star. Moreprecisely, we assume that the star is in mechanical equilibrium all the time.Nonetheless, under suitable conditions, the equilibrium configuration of certainstars is unstable against perturbation. Such perturbation may lead to stellarpulsation. Study of stellar pulsation is important for astronomers for the fol-lowing reasons: 1. Distance measurement. Certain stellar pulsation leads topulsation variable. The most famous one is the Cepheid variable, which iswell-known for its period-luminosity relationship that has been used asstandard candle to set the cosmological distance scale. 2. Stellar structure.Our Sun and other main-sequence stars also show pulsations, albeit with muchsmaller amplitudes. By observing different oscillation modes of a star, one caninfer its interior structure. This is known as asteroseismology (or helioseis-mology for the specific case of the Sun).

8.1 Some basic terminology

The instability strip is a narrow region in the Hertzsprung-Russell diagramwhere most of the pulsating variables lie (except the Mira variables), e.g., itintersects with the horizontal branch at RR Lyrae, crosses the supergiants atCepheids, and crosses the white dwarf region at ZZ Ceti.

The pulsations in Cepheids, Miras, and RR Lyrae are radial. They show peri-odic contraction and expansion with periods from 1 to 100 days. On the other

84

CHAPTER 8. STELLAR PULSATION 85

Figure 8.1: Location of pulsating variables on the H-R diagram.

hand, the Sun, δ Scuti, and white dwarfs are non-radial pulsators. Differ-ent regions on the stellar surface contract and expand differently. Non-radialoscillations can be described by the spherical harmonics

Y ml (θ, φ) = (−1)m

√√√√2l + 1

4π

(l −m)!

(l +m)!Pml (cos θ)eimφ. (8.1)

Stellar pulsation is similar to sound waves in musical instruments. They arestanding waves with the center of the star as a node and the stellar surface asan antinode. In the fundamental mode (l = 0), there is no other nodes and

Type Period ModeMira 100–1000 d radialCepheids 1–50 d radialW Virginis 10–20 d radialRR Lyrae 1 hr–1 d radialδ Scuti 1–3 hr rad. & non-rad.β Cephei 3–7 hr rad. & non-rad.ZZ Ceti 100–1000 s non-rad.

Table 8.1: Some common pulsating variables.


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 secondovertone (l = 2) has two, and so on. A star can exhibit several modes at thesame time. Cepheids are believed to oscillate in the fundamental mode, whileboth the fundamental mode and the first overtone can exist in RR Lyrae andMiras.

8.2 A simple model

In this introduction, we will focus on the theory of the simplest possible modelof stellar pulsation, namely, adiabatic radial pulsation in the small amplitudelimit. 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, AnnualReview 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 becomesmore opaque to radiation. Radiative diffusion slows down, heat, and hencepressure, build up. This eventually pushes the layer outward. The expansioncools down the material, and it becomes more transparent to radiation, suchthat 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 ofindividual particles in a star. (An alternative approach is that of Euler, whichfocuses on the evolution of thermodynamic variables such as density and pres-sure. Both approaches give the same physics but the later one is sometimesmore 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

∂r= 4πr2ρ(r, t) . (8.2)

In contrast, the equation of hydrostatic equilibrium (Eq. 2.3 in Chapter 2)becomes more complicated in the time dependent case. We assume that theoscillation is radial so that all physical parameters are functions of r andt only. In this case, the mass of a small spherical shell from radius r tor + ∆r equals 4πr2ρ(r, t)∆r. The gravitational force acting on this shell is−Gm(r, t)4πr2ρ(r, t)∆r/r2. The (gas plus radiation) pressure gradient acting


on this shell is ∂P/∂r. Hence, by Newton’s second law,

4πr2ρr = −G4πr2ρm

r2− 4πr2∂P

∂r. (8.3)

In other words,

r = −Gmr2− 4πr2 ∂P

∂m. (8.4)

Assuming that the oscillation is adiabatic and the star is made up of an idealgas, 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 ofoscillation 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 expansionabout the equilibrium configuration of the star. Besides, the oscillation of thestar can be well approximated by a S.H.M. In other words, using m and t arethe 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)

andρ(m, t) = ρ0(m)[1 + d(m)eiωt] , (8.8)

where P0(m), r0(m) and ρ0(m) are the equilibrium pressure, radius and densityon the sphere whose mass contained equals m.

Substituting the above equations into the three coupled partial differential equa-tions and keeping up to the first order terms, we obtain

P0

ρ0

∂p

∂r0

= ω2r0x+Gm(p+ 4x)

r20

, (8.9)

r0∂x

∂r0

= −3x− d , (8.10)

and

p = −3γax− γar0∂x

∂r0

. (8.11)

Eliminating p and d, we have

∂

∂r0

(γa∂x

∂r0

)+

4

r0

∂

∂r0

(γax)− Gmρ0γaP0r2

0

∂x

∂r0

+ρ0

P0

[Gm

r30

(4− 3γa) + ω2

]x = 0 .

(8.12)


So the problem of stellar oscillation becomes the problem of finding eigenvaluesω2 for above linear partial differential equation. If the eigenvalue ω2 > 0, thesolution is oscillatory which corresponds to an radial oscillation of the star. Ifω2 < 0, the solution contains an exponentially decaying or growing term whichindicates instability.

Eq. (8.12) is still too complicated to be solved analytically in general. Never-theless, we have one specific meaningful case in which this eigenvalue problemhas been investigated in detail. This is the case of (1) the adiabatic expansioncoefficient γa is a constant throughout the star; and (2) x is m independent. Inthis case, Eq. (8.12) becomes

ω2 = (3γa − 4)Gm

r30

= (3γa − 4)4πGρ

3, (8.13)

where ρ is the mean density of the star. In other words, the oscillation of thestar is stable if γa > 4/3; and the period of oscillation Π in this case is given by

Π = 2π

[3

4π(3γa − 4)Gρ

]1/2

. (8.14)

For non-relativistic ideal gas, γa = 5/3. Put this into the equation above, wehave:

Type PeriodRR Lyrae ≈ 0.5 dClassical Cepheids ≈ 7 dW Virinis ≈ 15 d

This agrees reasonably well with observations.

We have not finished our discussions yet. In reality, our linear analysis in theadiabatic regime is an overly simplified approximation. Actually, if we includehigher order effects, radial oscillation of a star will be damped out quickly unlessthere is an effective way to pump energy into the oscillator. Indeed, pulsatingvariables, such as Cepheids, occupy only a restricted strip on the H-R diagram,strongly suggesting that an effective excitation mechanism exists only underrather restrictive conditions.


8.3 κ-mechanism and the instability strip

Detail analysis of the excitation mechanism is rather involved. The most famousone is the so-called κ-mechanism. This mechanism assumes that the opacityincreases upon compression in some region of the star so that the radiativeluminosity is blocked in the compression phase of oscillation. Therefore, theregion gains thermal energy in compression phase and loses thermal energy inexpansion phase.

Recall that for most part of an “ordinary” star, κ is well approximated byκ0ρT

−7/2 (Kramer’s law). Hence, upon adiabatic compression, the opacity ac-tually decreases. Thus, the condition for κ-mechanism is not satisfied in moststars. Nevertheless, such a condition holds in regions where there is partialionization of H and Hei and/or Heii. In fact, κ-mechanism due to such par-tial ionization zone is responsible for the excitation of pulsation in a numberof pulsating variables including RR Lyrae, classical Cepheids and W Virginis.Because the condition of onset of the κ-mechanism due to partial ionization isvery stringent, pulsating variables are rare and can only be found in restrictedregions on the H-R diagram.

We will not go into the detail mathematics here, but only give a qualitativedescription of the κ-mechanism and its connection to the instability strip. Inthe partial ionization regions, part of the energy input from compression canbe turned into more ionization instead of heating. Therefore, it is possible toincrease in density more than in temperature. Hence, κ can be increased dueto compression. Later in the cycle, electron-ion recombination during decom-pression can release energy to lower the opacity.

Where is the κ-mechanism expected to occur? Location of the partial ionizationzone in a star depends strongly on the effective temperature. If the temperatureis too high, the ionization will be close to the surface, resulting in insufficientmass to drive sustainable oscillations. On the other hand, if the temperature istoo 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 beforethe κ-mechanism to explain stellar pulsations. It suggests that compression atthe center of a star can lead to higher density and temperature, thus, increases


the thermonuclear energy production rate. However, detailed calculations showthat the oscillation amplitude at the central regions of most stars are small, suchthat this effect is insignificant. However, this is still an important mechanismin massive stars.

8.5 Cepheid variables

Classical Cepheids (or Cepheids for short) are luminous supergiants. It is namedafter the first known example, δ-Cephei, which is still an important calibratorfor the period-luminosity relation (see below). Their immediate precursors aremassive young O or B type stars. Their periods are about 2 to 60 d, meanluminosities about 300 to 40000L, mean surface temperature about 4000 to8000 K, and peak-to-peak magnitude variation of about 0.4 to 1.4 mag. Itsstellar atmosphere is cold enough to have a layer of partially ionized Hei andHeii. Hence, κ-mechanism leads to stellar pulsation. (See Figure 8.2 belowfor a typical light curve, color change and radial velocity of a classical Cepheidvariable.)

0.0 0.5 1.0

0.0 0.5 1.0

V (kms )r−1 0.0 0.5 1.0

4.5

4.0

3.5

mv

Phase

6500

6000

5500

Teff

20

10

0

−10

−20

Figure 8.2: Location of δ Cephei in the sky, and the variation of visual magni-tude, effective temperature and radial velocity.

In 1912, Henrietta Leavitt accidentally found that the period and luminosityof classical Cepheid variables are related. In a more modern language, such a


period-luminosity relation is found to be

log

(L

L

)= 2.43 + 1.18 log Π . (8.15)

That is, the longer the period, the brighter the star is. This famous relationis also called the Leavitt Law after her. Note that the above relation is validplus or minus a few percent error.

We may understand the period-luminosity relation as follows. First, the peak-to-peak magnitude difference is relatively small, hence we may assume thatthe effective surface temperature of a classical Cepheid to be a constant. Theluminosity of the star L ≈ 4πσR2T 4 is therefore a function of stellar radius Ronly. Although classical Cepheids are supergaints, the homology relation formain sequence star still holds approximately. Hence, we have an approximatemass-luminosity relation L ∼ Mα for classical Cepheids. Finally, we have theperiod-mean density relation Π ∼ ρ−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 theempirical formula above, but we cannot derive the exact value since the meandensity approximation is oversimplified.

Indeed, the above argument is rather general and can be applied to quite anumber of pulsation variables including W Virginis and Mira type variables.A notable exception is the RR Lyrae variables. Observations suggest that theabsolute 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 areof 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 toother observables such as spectrum or position. The period-luminosity relationimmediately turned Cepheids into standard candles, offering a very powerfultool for distance measurements. Edwin Hubble observed Cephieds in othergalaxies to discover Hubble’s law. Even today, Cepheids play an importantrole in the construction and calibration of the cosmic distance ladder, up to theorder of 10 Mpc (see the Hubble Space Telescope Key Project).

Chapter 9

A Schematic Picture of StellarEvolution

In the previous chapters we have seen that the timescale of stellar evolution is setby the (slow) rate of consumption of nuclear fuel. The rate of nuclear burningincreases with density and rises sharply with temperature, and the structureequations of a star show that both the temperature and the density decreasefrom the center outwards. We may therefore conclude that the evolution of astar is led by the central region or core, with the outer regions lagging behind.Changes in composition first occur in the core, and as the core is graduallydepleted of each nuclear fuel, the evolution of the star progresses.

We can therefore learn a lot about the evolution of stars by studying the changesthat occur in its center. To do this, we consider the diagram of Tc and ρc. Fromthese quantities, together with the composition, we can calculate the wholeevolution of the star. In the diagram of Tc and ρc the time evolution of a starwill be a track. Since the only property that distinguishes the evolutionarytrack of a star from that of any other star of the same composition is its mass,we will get different lines in this plane for different masses.

All the processes that occur in a star have characteristic temperature and den-sity ranges, so the different combinations of temperature and density will deter-mine the state of the stellar material, and the dominant physical processes thatare expected to occur. In this way we can divide the (T, ρ) diagram into zonesrepresenting different physical states or processes. By looking at the positionof stars in this diagram as a function of time and mass, we should be able totrace the processes that make up the evolution of a star.

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CHAPTER 9. A SCHEMATIC PICTURE OF STELLAR EVOLUTION 93

9.1 Zones of the equation of state

Figure 9.1: Mapping the temperature-density diagram according to the equationof state.

The most common state of the gas that we find in stars is the ideal gas statefor both components: ions and electrons. We can then write the equation ofstate as:

P =R

µρT = K0ρT , (9.1)

where K0 is a constant. At high density and relatively low temperature theelectrons become degenerate, and, since their contribution to the pressure isdominant, we can then write

P = K1ρ5/3 . (9.2)

The transition from one state to another is gradual with the change in densityand temperature, but we can draw in 9.1 an approximate boundary in the (T, ρ)plane. This line has to be (from the equations above):

log ρ = 1.5 log T + constant, (9.3)

which is a straight line with slope 1.5 in fig. 9.1. Above it lies the electrondegeneracy zone, and below it the ideal gas zone. For still higher densities,relativistic effects play a role, and the equation of state becomes

P = K2ρ4/3 . (9.4)


The boundary between the ideal gas zone and the relativistic degeneracy zoneis then given by

log ρ = 3 log T + constant. (9.5)

This means that the boundary between the ideal gas zone and the electrondegeneracy zone becomes steeper as the density increases. Within this zone, thetransition from relativistic to non-relativistic occurs when the rise in pressurewith increasing density becomes constrained by the limiting velocity c. We arein this case when

K1ρ5/3 K2ρ

4/3 , or ρ(

K2

K1

)3

. (9.6)

This is a horizontal line in fig. 9.1. In the ideal gas zone radiation pressurehas been neglected. Its contribution to the total pressure becomes important,however, at high temperatures and low densities and SHOULD be added to thegas pressure. Eventually, radiation pressure will become dominant, with theequation of state changing to

P =1

3aT 4 . (9.7)

The boundary between this zone and the ideal gas zone, again, is of the form

log ρ = 3 log T + constant. (9.8)

9.2 Zones of nuclear burning

A nuclear burning process becomes important in a star whenever the rate ofenergy release by this process constitutes a significant fraction of the rate atwhich energy is radiated away (which is the stellar luminosity). Since nuclearreaction rates are very sensitive to the temperature, one can define thresholdlines in the (T, ρ) diagram: on one side of the threshold the rate of burningcan be considered negligible, while on the other side it is considerable. Thesethreshold lines are shown in fig. 9.2.

We consider six different stages of nuclear burning: the p-p cycle, the CNOcycle, helium burning into carbon through the 3α reaction, carbon burning,oxygen burning, and silicon burning. Consider

ε = ε0ρmT n , (9.9)

in most cases m = 1, and n 1, hence, the thresholds are almost verticallines (do you know why?). In reality, n depends on temperature, therefore thelines are curved. Nucleosynthesis ends with iron. Iron nuclei heated to very


Figure 9.2: Mapping the temperature-density diagram according to nuclearprocesses.

high temperatures are disintegrated by energetic photons into helium nuclei (α-particles). This energy absorbing process reaches equilibrium with the relativeabundance of iron to helium nuclei determined by the values of temperatureand density. A final threshold may be defined by requiring that the number ofhelium and iron nuclei are approximately equal (see fig. 9.2).

9.3 Unstable zones

We showed in chapter 4 that the condition for dynamical stability is γa > 4/3.So in regions with γa < 4/3 the stellar models might be expected to becomeunstable. These are the far extremes of the relativistic degeneracy zone III andthe radiation pressure dominated zone IV (and also the iron photodisintegra-tion zone). Another zone is unstable due to pair production (very high T andlow density, see fig. 9.3). All this means that there is a relatively small zoneremaining for possible evolutionary tracks of stars. To finish, nuclear burningis thermally unstable in degenerate gases, relativistic or non-relativistic. So forthat reason also the nuclear burning thresholds in fig. 9.2 have been discontinuedafter crossing the boundary into the degeneracy zone II.


Figure 9.3: Outline of the stable and unstable zones in the temperature-densitydiagram.

9.4 Evolutionary tracks in the (T, ρ) plane

The question is now, where a star of mass M can be found in the (T, ρ) plane.To answer this, we use a polytropic model. For a polytrope, we have seen thatwe have the following relation between central density and pressure

Pc = (4π)13BnGM

23ρ

43c . (9.10)

This relation depends only slightly on n, especially for stable polytropes, forwhich n lies between 1.5 and 3. Although a star in hydrostatic equilibriumis not a polytrope, this equation does provide a good approximation. Apartfrom this relation, the central pressure is related to the central density andtemperature by the equation of state. Within the different zones of the (Tc, ρc)plane we have different equations of state. Combining them with Eq. (9.10) wecan eliminate Pc to obtain a relation between ρc and Tc. For example, for anideal gas, we have

ρc =K3

0

4πB3nG

3

T 3c

M2∝ T 3

c . (9.11)

This means that we can put a number of parallel lines with slope 3 in ourdiagram for various values of M (see fig. 9.4).

In zone II (non-relativistic degenerate electrons) we have the following equation


Figure 9.4: Relation of central density to central temperature for stars of dif-ferent masses within the stable ideal gas and degenerate gas zones.

of state, P = K1ρ53 , leading to

ρc = 4π(B1.5G

K1

)3

M2 . (9.12)

Hence, ρc is independent of Tc , and we have for each mass a correspondinghorizontal line in the (Tc, ρc) plane. These lines can be joined, for low masses,to lines with slope 3 in the ideal gas part of the diagram. We have seen thatthe maximum mass a degenerate star can have is Mch = 1.44M, the Chan-drasekhar limit. One can see that the paths for masses smaller than this willeventually bend into the degenerate region, while paths for higher mass starswill remain straight. To conclude, a star of fixed mass M has its own distincttrack in the (Tc, ρc) plane. There are two distinct shapes: straight for M > Mch

and knee-shaped for M < Mch. We can understand the relationship betweentracks of different masses as follows: when mass is increased, gravity becomesmore important, so a higher pressure is required to counterbalance gravity. Inan ideal gas this can be done with a higher density or a higher temperature.A higher density, however, implies smaller distances between particles, causinglarger gravitational forces. Since the hydrostatic pressure is proportional to ahigher power of the density than the gas pressure (4/3 as opposed to 1) a higherdensity would only worsen the imbalance. So for a higher mass a lower densityor a higher temperature is required. In the case of a degenerate electron gas,the temperature is less important. Now the hydrostatic pressure is proportional


to a lower power of the density than the gas pressure (4/3 vs. 5/3) so that ahigher density is required for equilibrium in a more massive star.

9.5 The journey of a star

Figure 9.5: Schematic illustration of the evolution of stars according to theircentral temperature-density tracks.

In fig. 9.5 we have combined all the previous figures. We will now choose amass, identify its path and follow the journey of a star along it. Stars formin gaseous clouds, with much lower densities and temperatures than we areused to here. At the beginning, a star radiates energy without a central energysource, which means that it contracts and heats up (we will see this in the nextchapter). In the (Tc, ρc) plane this means that the stars ascends along its tracktowards higher temperatures and densities. Eventually, it will cross the firstnuclear burning threshold. At this point hydrogen is ignited in the centre, andthe star adjusts into thermal equilibrium. The star then stops for a long time,until hydrogen in the centre is exhausted. Note that for low masses the trackcrosses the p-p track, while for high masses the CNO-track is crossed. So starsare expected to burn hydrogen differently according to their masses.

For more massive stars radiation pressure becomes more and more important,eventually dominating gas pressure. Because of this, stars cannot be more


massive than about 100M (an example of such a massive star is η Car). The(Tc, ρc) diagram can also be used to infer a lower limit to the stellar mass range.Since the hydrogen burning threshold does not extend to temperatures below afew million K, the highest value of M for which M still touches this thresholdcan be regarded as the lower stellar mass limit. Objects of lower masses willnever ignite nuclear fuel. These objects (of mass < 0.1M) are called browndwarfs.

Let us continue with stars that do burn nuclear fuel. When the hydrogen inthe center is finally exhausted the star will lose energy again, and will have tocontract its core to compensate for it. This means that it will heat up. Forlow-mass stars, the tracks will cross into the degeneracy zone and bend to theleft. The pressure exerted by the degenerate electrons is enough to counteractgravity. Contraction slows down, and the star cools while radiating the accu-mulated thermal energy, tending to a constant density and radius, determinedby M. The higher the mass, the higher the final density and the lower the finalradius. For higher mass stars, the tracks will cross the next nuclear burningthreshold. Helium will now ignite the core, and we have another phase of ther-mal equilibrium. After exhausting helium, we have more contraction, and somestars develop degenerate cores, moving towards the left. This means that thereare two kinds of degenerate stars: some with helium cores, and others withcores of carbon and oxygen.

A star of mass Mch can in principle continue burning fuel without becomingdegenerate. However, since its track is very close to the degenerate zone insta-bilities might easily occur. For higher masses, more nuclear burning thresholdswill be crossed. These massive stars undergo contraction and heating cyclesalternating with thermal equilibrium burning of heavier and heavier nuclear fu-els, until their cores consist of iron. Further heating of iron inevitably leads toits photodisintegration, a highly unstable process. These stars will end as su-pernovae. Stars of very large masses have tracks that enter the instability zonebefore crossing the burning thresholds of heavy elements. They are thereforeexpected to be extremely short-lived, developing pair-production instabilitiesthat will result in a catastrophic event, like a Supernova explosion.

We can summarise by stating that stars can exist with masses between about 0.1and 100 solar masses. All start with hydrogen burning in their centres. When,however, hydrogen is exhausted in the centre, evolution will proceed differentlyfor different masses. Those under M = Mch contract and cool off after com-pletion of hydrogen or helium burning. Stars above this critical mass undergoall the nuclear burning processes, ending with iron synthesis. Subsequent heat-ing of the core develops into a highly unstable state, ending in a catastrophiccollapse or explosion. Stars of very high mass may reach dynamical instability


sooner, due to pair production.

9.6 The late evolutionary stages in the (Tc, ρc)

diagram

Figure 9.6: Schematic illustration of the stellar configuration in different evolu-tionary phases for a 10 M star (A,B,C,D,E) and a white dwarf.

Instead of using the (Tc, ρc) diagram just for the centre, one can also use it todescribe the structure of a star at a given evolutionary stage. A star can berepresented as a line connecting the center with the photosphere. The exactshape of these lines can be complicated — only polytropes are straight lines inthe diagram. In fig. 9.6 a number of lines are plotted with the start of all lines onthe track from previous section. These lines may be considered as the evolvingstructure of the star. Line A can be considered as the main sequence. The nextline, B, describes the star at a later stage when the hydrogen has been depletedin the core. On track B hydrogen is burning at the point where it crossesthe hydrogen burning threshold. This indicates that hydrogen burns in a shelloutside the helium core. The region interior to it, consisting of homogeneoushelium, is contracting and heating up.

Assuming that the contraction of the core occurs quasi-statically, on a timescalemuch larger than the dynamical timescale, we can assume that the virial the-


orem holds. If the amount of energy gained during this phase is negligiblecompared to the total energy, we may assume that energy is constant. In sucha case we may assume that both gravitational potential energy and thermalenergy are conserved. For this reason contraction of the core must be accom-panied by expansion of the envelope. Heating of the core will result in coolingof the envelope. On track B the surface temperature drops, so that the starbecomes redder. It becomes a red giant. One can show that a moderate amountof core contraction has to be compensated by a large amount of expansion ofthe envelope. If the total energy does not remain constant but increases, thenthe core will heat up even more, so there will be more expansion. If the to-tal energy of the star decreases the envelope might either contract, expand orremain unchanged.

When the He ignition temperature is finally reached, core contraction stops (lineC). Now there is helium burning in the core and hydrogen burning in a shell.This is a stable phase, and is called the Horizontal Branch. When helium in thecore itself is exhausted, another phase of contraction and envelope expansionsets in. Since the core is now more condensed, envelope expansion is even morepronounced. Now a star will evolve into a Supergiant (this is the AGB phase,the Asymptotic Giant Branch, line D). Now there are 2 shells burning. The starlooks like an onion with a central region of C, N, and O, a shell of helium, andthe hydrogen-rich envelope. The hydrogen burning shell feeds fresh fuel to thehelium burning one, so both move outwards. This process is quite complicatedand often unstable, leading to variability. Some of the most massive stars willeven reach phase E, burning C, N, and O.

9.7 Problems with this simple picture

In this chapter, we have given a schematic picture of the evolution of stars. Oneshould not forget that this is only schematic — real models will have to rely oncomplicated numerical calculations. There are however, as expected, a numberof problems with this picture. For example, stars as massive as 7–9M endup as white dwarf, i.e., degenerate stars. We mentioned before that this wouldonly happen for stars with mass M < Mch = 1.44M . How can these starslose almost all their mass? Another indication comes the fact that in the solarneighbourhood white dwarfs are found with masses of M = 0.4M. Stars withsuch low masses evolve very slowly, and would never have been able to reachthis stage in a Hubble time, i.e., in the time since the Big Bang.

As it turns out, stars lose a significant fraction of mass by stellar winds, in whichgas and dust is blown away from the star. Mass loss for massive stars can be


as high as 10−4M yr−1 for very massive stars. This means that stars do notstay on the same track, but move to different tracks, where they will behave asstars of mass M ′. Since no simple model of mass loss is available, this cannotbe very easily implemented in a schematic picture.

Another process that was neglected in the previous picture was neutrino emis-sion in dense cores, which has the effect of cooling the stars. As the rise intemperature between late burning stages is impeded by neutrino cooling, theslope of the tracks should become somewhat steeper than 3. Nonetheless, thisdoes not change anything qualitative of the picture.

What we have not been able to predict are: 1. time scales 2. detailed predictionof the outer appearance of stars. For that reason, we cannot compare thesemodels with observations; to do so we need to go to more detailed models.

Chapter 10

Special Topic: SubstellarObjects

In Chapter 5, we derived the minimum mass of a main sequence star to be∼ 0.1M. In this chapter, we briefly describe some topics related to stars nearand below the main-sequence edge, namely red dwarfs and brown dwarfs, allthe way down to planets.

10.1 Brown Dwarfs – Failed Stars

Figure 10.1: Size and surface temperature comparison between stars, browndwawfs, and planets.

Recall that a star is formed from the collapse of a nebula. The type of object

103

CHAPTER 10. SPECIAL TOPIC: SUBSTELLAR OBJECTS 104

eventually formed depends on the final mass. If it is greater than about 0.1M,sustainable hydrogen burning will occur in the core (due to sufficiently high coretemperature) before the core becomes degenerate. This is a star. More precisecalculations show that the hydrogen-burning main-sequence edge is 0.075M ≈75MJ (Burrows & Liebert 1993, Rev. Mod. Phys. 65, 2). This also slightlydepends on the metallicity: it can reach 90MJ for zero metallicity.

If the mass of an object is less than about 0.01M (≈ 10MJ , i.e. 10 Jupitermass), the temperature of the core will not be hot enough to burn any nuclearfuel. The result will be a “planet”. A planet is supported by mainly EM forcesbetween atoms or molecules against gravity.

Figure 10.2: Luminosity, core temperature, and radius evolution of low-massstars (blue), brown dwarfs (green), and planets (red) (Burrows et al. 2001).

An object with mass between 0.01M and 0.1M is called a brown dwarf.The IAU currently adopts a cutoff of 13MJ between brown dwarfs and planets,since the core temperature of an object above 13MJ is sufficiently hot to fuseD. A brown dwarf above 65MJ can even burn Li. Similar to a very low massmain sequence star, a brown dwarf is fully convective. However, D and Li arerare and hence this burning takes only a relatively short time (at most 108 yr).In addition, it turns out that the gas pressure produced by D and Li burningis not strong enough to halt the collapse. Eventually, almost all D and Li inthe star is used up, but the core temperature is still not high enough to burn


hydrogen. The brown dwarf is now supported by electron degenerate pressurein the core. Unlike a white dwarf, the temperature of a brown dwarf is stillrather high compared with the Fermi temperature of the electrons in the core(why?). Hence, electrons in the core of a brown dwarf is not fully degenerate,resulting in a rather complex EOS. By definition, a brown dwarf is not a star,due to the lack of sustainable hydrogen burning.

Although the idea of brown dwarf has been suggested since 1963, the first firmdetection (Gliese 229B) was only made in 1995 by Kulkarni et al. Observation-ally, it is very difficult to identify brown dwarfs because of their low surfacetemperature and hence low luminosity. The main search methods are directobservations and indirect mass estimate from binary systems.

Figure 10.3: Brown dwarfs are faint.

Figure 10.4: Images of GL 229B.

• Old visible brown dwarfs: direct search for cool stars whose temperatureand luminosity are below the minimum possible values for main-sequencestars, either by looking at stellar companions or in the field. The first con-firmed brown dwarf, Gliese 229B, was detected this way. It is a companionto a nearby (5.8 pc) red dwarf (i.e. M star) Gl 229A.

• Dynamical brown dwarfs: from the orbital information of binary systems,one can deduce the companion mass even it is invisible. However, this ismodulated by the inclination angle, therefore, is only a lower limit, unlessthe orbit is spatially resolved.

• Young brown dwarfs in clusters: brown dwarfs are at their brightest inyoung age, hence, easiest to see. Nevertheless, they have very similartemperature and luminosity as very low-mass stars. This can be solvedwith additional information such as age and chemical abundances. Theformer could be estimated from the colour-magnitude diagram for stars


in a cluster. Among the first brown dwarfs observed, PPL 15 and Teide1 in the Pleiades, are young objects.

10.2 Distinguishing Brown Dwarfs from Stars

To confirm brown dwarf candidates, we need some good ways to distinguishbrown dwarfs from stars. The most famous one is the Lithium test.

Figure 10.5: The lithium test.

• Lithium test: If lithium lines are observed in the spectrum, then the objectis very likely to be substellar, because low-mass stars are hot enough toburn Li and they are fully convective, so all Li will be depleted in a shorttime scale within 108 yr. Problems: 1. Heavier stars like the sun canretain Li in their outer layers. (This is generally not a problem since wecan rule out heavy stars from the luminosity or temperature). 2. Youngstars do not have enough time to deplete all the Li. 3. Brown dwarfsheavier than 65MJ can also burn Li.

• Methane test: stars are too hot to have methane in the atmosphere, butbrown dwarfs and planets are cool enough (< 1400 K) that methane gath-ered from the interstellar medium over time can stay in the atmosphere.Therefore, if the spectrum shows methane bands at 2 microns, then theobject cannot be a star.


10.3 Distinguishing Brown Dwarfs from Plan-

ets

It is generally difficult to distinguish between a brown dwarf and a giant planetand the 13MJ cutoff by IAU should be considered as a rule of thumb ratherthan a physical limit.

• Size (?): Massive brown dwarfs are supported by degenerated pressure,therefore, the variation of radius with mass is small and many of themhave similar sizes as Jupiter.

• Mass: If an object is much more massive than Jupiter, e.g., 20MJ , wecan rule out a planet.

• Spectrum: X-ray and radio emission have been detected from browndwarfs (e.g., LP 944−20). X-ray flares found in brown dwarfs are believedto be very similar to those in very low-mass stars: both due to changingmagnetic fields as a result of rapid rotation and convective interior.

10.4 Planets: Energy Sources

Moving down the mass scale, planets, by definition, have no nuclear reactionin the interior. However, it is found that Jupiter and Saturn radiate about 2–3times the energy as they absorb from the sun. Where does the energy comefrom?

• Residual heat: One idea is that the energy source could be residual heatleft over from the primordial nebula that collapsed into the planet. But asthe planets are formed for a long time, this is likely relatively unimportant.

• Kelvin-Helmholtz mechanism: The contraction of a star/planet can gen-erate heat. This was first proposed to explain the energy source of thesun. Recall that from the virial theorem, the total energy Utot, which isthe sum of the potential energy and the thermal energy, and

Utot = Ω + UT = −UT < 0 . (10.1)

So if a star loses energy (e.g., due to radiation), the thermal energy, andhence the temperature, actually increase! For this reason a star (or aplanet) appears to have “negative heat capacity”. (Does it violate any


laws in thermodynamics?) For the simplest case of a constant densityplanet, it can be shown that

Ω = −∫ R

0

Gm

rdm = −3M2G

5R. (10.2)

Hence, the temperature increases when a planet shrinks. (How aboutbrown dwarfs?)

• Helium rain: The Kelvin-Helmholtz mechanism alone is not enough toexplain the emission of Saturn. The atmosphere of Saturn is cool enoughfor helium to condense. The helium droplets then “rain out” and sinkto the core through the liquid hydrogen surrounding. The friction inthis process releases heat in the planet interior. Observation found a lowhelium abundance in the atmosphere of Saturn, providing support to thistheory. For Jupiter, this is less clear. Due to high temperature, it isgenerally believed that helium is still mix with hydrogen in the interior.However, recent studies suggest that Jupiter may also have Helium raintoo.

Figure 10.6: Helium rain in Saturn.Figure 10.7: Earth orbit as seenfrom the nearest star.

10.5 Extrasolar Planets: Detection

Let us move to extrasolar planets. The problem: the Earth is 1 AU from thesun, even viewed from the nearest star at ' 1 pc, the angular separation is 1′′

(by the definition of parsec). How small is this? Just like viewing the diameterof a coin over the Victoria Harbour. Also a star is much brighter than a planet.All these post a great challenge to planet detection.


• Pulsar timing: pulsars are extremely precise clocks due to their stablerotation. Any small change in their position or motion due to the presenceof a planet will result in detectable timing anomaly. This method is by farthe most sensitive down to 0.1 Earth mass (0.1M⊕). However, the harshpulsar environment is not habitable because of strong wind of particleswinds and EM radiation streaming out from the neutron star. Also, theplanet is formed from the fall-back materials in a supernova remnant,which could have very different properties than those collapsed from gasclouds. The first exoplanet was detected this way, around PSR B1257+12.

• Radial velocity: The motion of a planet would causes the parent star towobble. Unless the system is viewed edge-on, this will result in periodicchange of the radial velocity of the star. Due to the Doppler effect, thespectral lines will be redshifted and blueshifted periodically. This is themost efficient detection method since the chance of having a face-on sys-tem is rare. However, only the minimum mass of a planet can be deduced(why). The first planet found around a main-sequence star, 51 Pegasi,was found by this method. The current technology can measure velocityshift to 0.3 m/s level (is it enough to detect the Earth?).

Figure 10.8: Planet detection by radio velocity. The right panel shows thewobbling of the sun due to Jupiter, compared to the solar radius.

Figure 10.9: Planet detection by transit.

• Transit: To determine the EOS (or composition) of a planet. It is es-sential to know both the mass and radius. Transit is the only methodthat can provide reliable radius measurements. However, this requires a


nearly perfect alignment between the parent and the planet as viewed fromEarth, for which the probability is very small. NASA’s Kepler mission isdedicated to this kind of measurement with very precise photometry.

• Direct imaging: While the small (arcsecond scale) angular resolution men-tioned above can be achieved using the Hubble Space Telescope or ground-based optical telescopes with adaptive optics, it is still non-trivial and thecontrast between the star and the planet is often higher than 1 : 10−10.

Figure 10.10: Direct imaging of planets.

• Gravitational microlensing: When a star passes in front of a backgroundobject, the light from the latter is bended due to the gravitational poten-tial. The presence of a planet in the foreground lensing star will result ina slightly different gravitational potential, giving detectable effect. How-ever, the chance alignment of a star with a background source is veryrare, and since all stars have proper motions, the alignment only last forhours to days and almost never happen again. This makes confirmationor follow-up studies of the planets impossible.


Figure 10.11: Planet detection by microlensing.

10.6 Habitability of Exoplanets and Red Dwarfs

To look for lives in exoplanets, there are some important characteristics for theparent stars:

• Long-lived: at least a few billion years for life to evolve. Therefore, O, B,and A stars are not good candidates.

• Low variability: to provide a stable energy source

• UV radiation: this can trigger chemical reactions in the atmosphere andevolution of life. But too much is harmful.

And for the planets:

• Rocky planets: it is generally believed that life could not exist in gasgiants that has no solid surface.

• Massive: a too low mass planet cannot retain an atmosphere, which makeslife impossible.

• Rotation: if the rotation is too slow or tidally locked to the orbit, one sidewould be much hotter than the other, life could be difficult to form.

• Distance from star: this is perhaps the most critical factor for liquid waterto exist on surface. The habitable zone depends on the star’s luminosity.For our solar system, a conservative estimate suggests about 0.95–1.4 AU,and more optimistic estimate gives 0.84–1.7 AU. Lower mass (K or M)stars obviously have habitable zones closer than that of the sun. Anotherconsideration is tidal locking, which depends on the planet-star distance


and mass of the star. The tidal force constantly squeeze and stretch theplanet, converting its rotational energy into heat, until it is tidally lockedinto synchronous rotation in a time scale of

(a

AU

)6(M

M

)−2

× 1012 yr. (10.3)

Figure 10.12: Illustration of tidal locking. Left: Synchronous rotation. Right:Non-synchronous rotation. The planet is continuously deformed such that therotational energy is dissipated via friction.

Red dwarfs are stars at the low-end of the main sequence. They are very lowmass K or M-type stars ranging from 0.075M (= mass limit of a brown dwarf)to 0.6M and have surface temperature of 2000–4000 K. They are the mostcommon stars in the Milky Way and can live for very long time (T ∝M−2.5 ≈6 × 1011 yr). These are advantages to harbour life. However, their habitablezones are much smaller than that of the sun. This also makes it easier for aplanet to be tidally locked (although some recent studies suggest that the darkside may not be totally frozen). Like brown dwarfs, some red dwarfs show flaresin X-rays and UV during the early stage (first 1.2 × 109 yr) of their lives. Forlife to develop, some protection from radiation is needed or the planet has tomigrate into the habitable zone later to avoid radiation and tidal locking.


Figure 10.13: Habitable zone of red dwarfs compared to that of other stars.

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