1.1 An introduction to the cgs units

Transcription

1 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 this course. The cgs units mean cm, g, and second. Length: in cm = m Mass: in g = kg Time: in s Force: in dyn = N (hence computing the mass and acceleration in cgs units, the corresponding force will automatically be in dyn) Energy: in erg = J (hence computing mass and velocity in cgs units, the corresponding energy will automatically be in erg) Power: in erg s 1 Flux: in Pressure: in Charge: in the electrostatic unit of charge (esu), also known as the statcoulomb, so that Coulomb s law becomes F = q 1 q 2 /r 2 B-field: in gauss (G) = 10 4 T Field energy density: U = E 2 /8π or B 2 /8π 1

2 CHAPTER 1. INTRODUCTION What is a star? Stars are the building blocks of the visible Universe, so it is important to understand their structure and evolution. Why do they glow? How did they formed? What would they become after? Before we get into these, let us give a clear definition of stars. Roughly speaking, we define a star to be a gravitationally bounded object which has a nuclear energy source in the form of fusion in the core. Thus, Jupiter and Comet Halley are not stars. But this definition is not sufficient in this level of complexity because it also includes brown dwarf in the family of stars. Recall that brown dwarfs are objects that are too massive to be planets but too light to be stars. More precisely, in this context, a planet is a self-gravitating body without any nuclear fusion in the core although it may have other means of energy generation (such as radioactive decay). Thus Jupiter is a planet although Jupiter has an internal heating mechanism. (Do you know what internal heating mechanism it has?) A brown dwarf, in contrast, is a self-gravitating object that burns (or has burnt) D, Li, Be and B via nuclear fusion. Since the amount of D, Li, Be and B in such object is small, such nuclear burning cannot sustain over a long time. More importantly, the energy released in such a burning alone is not sufficient to stop the brown dwarf from further gravitational collapse. Thus, astronomers naturally want to exclude brown dwarfs from the rank of stars. (We shall study brown dwarf in detail as a special topic if we have time.) In addition, the above crude definition of star also excludes certain types of objects which burn nuclear fuel in a shell rather than at the center. Hence, we use the following more refined definition of star: A star is a gravitationally bounded object which has a sustainable nuclear fusion energy source in the interior. Using this refined definition, brown dwarf is not a star. Neutron star is not a star either. Red giant is a star for it has a hydrogen-burning shell although its core is not burning any nuclear fuel. An self-gravitating object whose sustainable nuclear fusion energy source in the interior no longer exists is called a dead star. Thus, neutron star is a dead star. So in the straightest sense, a neutron star is not a star even though it was a star. Note that the above definition says nothing about the shape of a star. In fact, stars may be spherical, ellipsoidal, egg-shaped and so on.

3 CHAPTER 1. INTRODUCTION Assumptions in our study of stellar structure and evolution 1. The star is isolated. Although most stars in the universe are not isolated, it makes sense to study the structure and evolution of an isolated star before going on to study the more complicated multiple stellar systems. 2. Right at the moment of formation of a star, its chemical composition is uniform. Why this is a good approximation? I shall further discuss the validity of this assumption when we study main sequence stars. 3. The star is irrotational and is spherically symmetric. Exercise: For a star with mass M, radius R, and angular velocity ω, calculate the ratio of rotational energy to gravitational energy. What is the ratio for the Sun? This ratio is very small for most stars. Thus, rotational distortion can be safely neglected. In a similar way, the ratio of magnetic energy to the gravitational potential energy of a star is B 2 R 4 /8πGM 2 < R 4 /M 2 where M and R are in units of solar mass and solar radius, respectively. Thus, the magnetic energy is very small compared with the gravitational potential energy. Hence, both rotation and magnetic field for almost all stars do not significantly break their spherical symmetry. Therefore, a spherically symmetric star serves as a very good approximation in the study of most isolated stars. 4. Mass loss or mass gain of a star is neglected. A star, like our Sun, may loss its mass through ejection of stellar wind. Some stars may gain mass by accretion. These processes are important in the study of certain stars such as white dwarfs, AGB stars and binary stars. Nonetheless, the mass change rate for most main sequence stars is small and can be neglected. 5. The star is in local thermodynamic equilibrium (LTE). In other words, every microscopically large but macroscopically small region is very close to and hence can be well approximated by a region in thermodynamic equilibrium. Thus, it makes sense to talk about thermodynamic variables such as temperature, density and pressure in these regions. This approximation greatly simplifies our analysis and serves as the starting point to the study of more refined calculations. How good these approximations

4 CHAPTER 1. INTRODUCTION 4 are in the study of main sequence stars? To answer this question, let me tell you that the photon mean free path near the core of our Sun is about a few cm, which is very short compared with the radius of the Sun. The mean free path for electron and nucleus is even shorter near the solar core. The typical equilibration time for photons, nuclei and electrons in the solar core is also very short compared with, say, the lifetime of the Sun. (I will justify these claims later on in this course.) Hence, LTE is a very good approximation for studying the overall structure of the Sun and most other stars. In contrast, LTE is no longer a very good approximation to study stellar atmosphere such as the chromosphere of our Sun. (Do you know why?) Fortunately, the atmosphere of many main sequence stars has a small mass compared with the star itself. Therefore, LTE still serves as a very good simplification in our first course of stellar structure. (Another exception is when we study neutrinos emitted by a star. Do you know why?) Consequence of our assumptions Since we assume that a star is spherically symmetric, we can forget about its angular velocity as well as the magnetic field strength. Thus, we may characterize the properties of a star if we know the following: density ρ, pressure P, temperature T and chemical composition of the star as functions of distance from the stellar core r and time since the birth of the star t. 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 g, which is about 1000 times more massive than Jupiter, or about times more massive than the Earth. Radius: Ranges from about 0.1 to 1000 solar radii for most living stars. The symbol for solar radius is R. Actually, 1R cm, which is about 100 times that of the Earth. Radiation: Ranges from about 0.01 to 10 6 solar luminosity for most living stars. The symbol for solar luminosity is L and 1 L erg s 1. (We shall talk more about luminosity and radiation later in this course.) Chemical composition: Most young stars are made up of primarily H and He plus a trace amount (which is about a few percent by mass)

5 CHAPTER 1. INTRODUCTION 5 of heavy elements. In astronomy, heavy elements (or some books call them by the misleading term heavy metals ) are defined as elements with atomic number greater than 2. Surface temperature: Ranges from about 4000 K to K for most main sequence stars. Magnetic field strength: Of the order of 1 G over most of the solar surface (although it can go up to about 3000 G in sunspots). But some living stars may have field strength as high as several thousand G. Rotation: Observation suggests that most stars rotate rather slowly. As for the Sun, the rotational period is about 1 month. 1.5 The Hertzsprung-Russell diagram As the name suggests, this diagram was first studied by E. Hertzsprung and H. N. Russell in the 1910 s. In modern language, Hertzsprung-Russell (H-R) diagram is a plot (usually a log-log or a semi-log one) of surface temperature vs. luminosity of stars. Conventionally, the luminosity is plotted in the usual direction while the surface temperature is plotted in the reverse direction. That is, luminosity increases as we move upward while temperature decreases as we move rightward in the H-R diagram. H-R diagrams for stars in star clusters give strong evidence for stellar evolution. Recall that a star cluster consists of a collection of gravitationally bounded stars within a space of radius of the order of 100 light years. Thus, we can safely assume that stars in a single star cluster are formed from roughly the same material at about the same time under roughly the same physical conditions. Thus, mass of a star is perhaps the most important parameter to determine the properties and evolution of that star within the same star cluster. It turns out that stars are not located randomly in the H-R diagram; they cluster around a few regions in the diagram instead. (See figure 1.1.) Besides, these regions differ from one star cluster to the next. It is reasonable to assume that the chemical composition for different stellar clusters just before their formation is more or less the same. Thus, the difference between the location of stars of different clusters on the H-R diagram should be mostly determined by their age. It is rather hard to determine the absolute age of a star cluster. (By absolute age, we mean how old the stars in that cluster.)

6 CHAPTER 1. INTRODUCTION Supergiants Luminosity (L ) White Dwarfs Main Sequence Gaints Temperature (K) Figure 1.1: The H-R diagram (schematic) Nevertheless, it is quite easy to determine the approximate age of star clusters since old star clusters tend to have condensed core. (Do you know why?) 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 strongly on the degree of core condensation and hence the age of the corresponding star cluster. (See figure 1.2.) Most stars in a young cluster locate in a band called the main sequence. Consequently, we call a star lying on the main sequence a main sequence star. For older clusters, a significant number of

7 CHAPTER 1. INTRODUCTION 7 stars occupies the red giant branch. This is a strong evidence that stars evolve. More importantly, there is a turnoff point on the main sequence above which stars have been evolved away from the main sequence. This suggests that different stars on the main sequence evolve at different rates. Specifically, a hot and bright main sequence star evolves faster than a cold and dim one. In fact, nowadays, astronomers turn things around and use the location of the turnoff point on the H-R diagram to determine the age of a star cluster more precisely. 1.6 Mass-luminosity relationship For those main sequence stars with known masses, A. Eddington found a relation between the mass and the luminosity in (Do you know how they find the mass of some main sequence stars?) The relationship goes like L M α. (See figure 1.3.) This is an approximate relationship. Actually, after Eddington, different people put slightly different values for α. The value of α is about 3 to 5 over the whole range of stellar masses M. Yet, a closer look reveals that the exponent α changes for different values of M. Specifically, α is about 2.5 to 3.0 for M < 0.5M or when M > 2.5M while α is about 3.5 to 4.5 for 0.5M < M < 2.5M. Any theory of stellar structure must be able to reproduce the mass-luminosity relationship Luminosity (L ) αm αm 3 α 4 M Mass (M ) Figure 1.3: The mass-luminosity relationship of main-sequence stars.

8 Chapter 2 Stellar Structure Equations To understand the stellar structure, we need to determine many physical parameters inside a star, including mass / density, temperature, pressure, and energy production rate, by solving the following equations invoking conservation laws. Remember that it is important to specify the boundary conditions otherwise a differential equation cannot be solved. 2.1 Mass conservation equation For a spherically symmetric star, let r be the distance from the star center and m = m(r) be the mass enclosed by the sphere of radius r centered at the center of the star. (Note: Some authors use the notation M(r) instead of m(r).) Clearly, m(0) = 0 and m(r) = M where M and R denote the mass and the radius of the star. More importantly, due to spherical symmetry, r 4πr 2 ρ(r) dr = m(r), (2.1) 0 where ρ(r) denotes the density of the star at radius r. (Obviously, ρ(r) > 0 for 0 r < R and ρ(r) = 0 for r > R.) Alternatively, Eq. (2.1) can be rewritten as the following differential form dm dr =. (2.2) To summarize, conservation of mass relates ρ and m. So, we have introduced one equation and yet one new unknown. So, we have to find out more equations relating ρ, P, T, m, etc, before we can solve the problem of stellar structure. 8

9 CHAPTER 2. STELLAR STRUCTURE EQUATIONS Momentum conservation and the equation of hydrostatic equilibrium The next equation we introduce comes from the conservation of momentum; or more precisely, in LTE the net force acting on any microscopically large but macroscopically small region is zero. Inside a star in LTE, the forces involve are gravity and pressure gradient due to gas or photon. Specifically, dp dr = Gm(r)ρ(r) r 2. (2.3) We denote the central pressure P (0) by P c. And clearly, P (R) can be approximated by 0. Using Eq. (2.2), Eq. (2.3) can be rewritten as dp dm =. (2.4) Exercise: Using the equation above, show that in any star. (Hint: r(m) < R.) P c > GM 2 8πR 4 (2.5) In particular, this inequality tells us that the central pressure of our Sun is at least dyn cm atm. 2.3 Virial theorem Multiplying Eq. (2.4) by the volume V (r) = 4πr 3 /3, we have P (R) P (0) V (r) dp = 1 3 M 0 Gm(r) r dm. (2.6)

10 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 10 Note that the R.H.S. above is Ω/3 where Ω is the gravitational potential energy of the entire star. Using integration by parts, the L.H.S. above equals V (R) 0 P dv = M 0 P/ρ dm. Hence, we arrive at M Ω = 3 0 P ρ dm. (2.7) This equation is one form of the virial theorem (sometimes known as the global form of the virial theorem). (An alternative form of the virial theorem, sometimes called the local form of the virial theorem, is obtained by integrating from r = 0 to r = R s (< R). The result is Ms P P s V s 0 ρ dm = Ω s 3, (2.8) where Ω s is the gravitational potential energy for the sphere centered at the star and with radius R s.) Exercise: It can be shown that the gravitational potential energy of a star Ω < GM 2. From the ideal gas law P = nkt = ρkt/µm 2R A, where µ is the mean molecular weight and m A is the unit atomic mass, and the virial theorem, show that the average temperature For the Sun, show that T 1 M M 0 T dm > GMµm A 6kR. (2.9) T > ( µ)k (2.10) Exercise: For non-relativistic classical ideal gas, P = nkt where n is the number density of the gas; and the average K.E. per molecule is 3kT/2. Show that 2U + Ω = 0 where U is the thermal energy of a star.

11 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 11 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 gravitational contraction, the star becomes hotter, more tightly bound and has to radiate some energy to space. It is kind of an analog to a system with negative heat capacity. In contrast, if the star is made up of extremely relativistic particles, K.E. per particle is 3kT, i.e., the pressure equals one-third the energy density. Hence, in this case, virial theorem implies that U + Ω = 0. In other words, a star making up of extremely relativistic particles can be in LTE only when its total energy is 0. So, such a star is not bounded. Another consequence of the virial theorem concerns the conversation law. Consider a star making up of non-relativistic classical ideal gas. Suppose further that the timescale concerned is small enough that the total energy of the star is roughly a constant. Since 2U + Ω = 0, the total energy of the star can be expressed as a function of U (or Ω) alone. Thus, the thermal energy and the gravitational potential energy of the star are also conserved. In turns out that this consequence of the virial theorem is useful to understand several properties of late stage stellar evolution. 2.4 Simple stellar models So far, we have 3 unknowns, namely, m(r), ρ(r) and P (r) but only two equations, namely, the mass conservation equation and equation of hydrostatic equilibrium. So, we need another equation to relate m, ρ and P. Astronomers have proposed a number of extremely simplified equations to fulfill this task. Since all these proposed equations are not derived from first principle, they are nothing more than crude approximations of the realistic situation. Yet, these approximations are sometimes quite useful to investigate the approximate structure of a star. The first such model is called constant density model, namely, we assume ρ is a constant inside the entire star. The second one is called the linear density model, namely, ρ(r) = ρ c (1 r/r) where ρ c is a constant. Exercise: Solve m(r) and P (r) in the above two models.

12 CHAPTER 2. STELLAR STRUCTURE EQUATIONS Polytrope model The last such model I am going to introduce, which is the most important model of this kind, is called the polytrope model. The is motivated by adiabatic expansion of ideal gas, P V γ = constant, (2.11) where γ is the adiabatic index (or the heat capacity ratio). We therefore assume P = Kρ γ, (2.12) for some constants K. In the classical case, it can be shown that γ is related to the degrees of freedom f by γ = 1 + 2/f, i.e. γ = 5/3 for monatomic gas and γ = 7/5 for diatomic gas. As we will show later in this chapter, Eq also applies to a photon gas, classical and relativistic degenerate gas with different values of γ. Exercise: Combining the mass conservation and hydrostatic equilibrium equations, show that ( ) 1 d r 2 dp = 4Gπρ. (2.13) r 2 dr ρ dr Exercise: Using the polytrope equation (Eq. 2.12) and write γ = 1 + 1/n, the above equation becomes (n + 1)K 4πGnr 2 [ d r 2 ρ dr ] (1 n)/n dρ dr = ρ. (2.14) The boundary conditions for this differential equations are ρ(0) = ρ c, ρ(r) = 0, and dρ(0)/dr = 0. (Do you know why?)

13 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 13 Exercise: Let ρ = ρ c θ n where ρ c is the central density of the star, show that ( 1 d ξ 2 dθ ) = θ n, (2.15) ξ 2 dξ dξ where r = [(n + 1)Kρ c (1 n)/n /4πG] 1/2 ξ. The corresponding boundary conditions are θ = 1 and dθ/dξ = 0 at ξ = 0. This equation is called Lane-Emden Equation and can be solved numerically. The significance of the polytrope model is that Lane-Emden equation is independent of the mass M, radius R and central density ρ c of a star. So, once you have numerically solved the Lane-Emden equation for a given value of n, the numerical solution can be used to deduce the solution of any star with the same polytropic index n (or γ). For the special cases of n = 0, n = 1 and n = 5, Lane-Emden equation can be solved exactly. Exercise: For n = 0, show that θ(ξ) = 1 ξ 2 /6 and ρ = ρ c whenever ξ 6. (2.16) The case of n = 1, Lane-Emden equation can be rewritten as d 2 (ξθ) = ξθ (2.17) dξ2 whose solution that matches the correct boundary conditions above is θ(ξ) = sin ξ/ξ. Hence, sin ξ ρ = ρ c. (2.18) ξ

14 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 14 Exercise: Show that Eq. (2.18) is a solution to Eq. (2.17). For the case of n = 5, Lane-Emden equation can be rewritten as d 2 z dt = z(1 z4 ) 2 4, (2.19) where ξ = e t and θ = z/(2ξ) 1/2. Multiplying Eq. (2.19) by dz/dt, we obtain 1 2 Boundary conditions demand that C = 0 and hence ( ) 2 dz = 1 dt 8 z z6 + C. (2.20) ( ) 1/2 dz dt = z 1 z4. (2.21) 2 3 After making the substitution z 4 /3 = sin 2 ζ and upon integration, we have e t = C tan(ζ/2) where C is a constant of integration. Applying the boundary conditions and after simplification, we obtain ( ) 5/2 ρ = ρ c 1 + ξ2. (2.22) 3 (Could you express ρ c as a function of M and R? Besides, could you express ρ and P as a function of r?) It is straightforward to check that the smallest positive root for the solution θ(ξ) of the Lane-Emden equation equals ξ = ξ 1 6 and ξ = ξ 1 π when n = 0 and 1 respectively. This value of ξ 1 corresponds to the boundary of a star. On the other hand, θ 0 for any real-valued ξ for n = 5. In other words, the solution of the Lane-Emden equation for n = 5 does not correspond to a physical star. Exerxise: From the definition of ξ, show that R = [ (n + 1)K 4πG ] 1/2 ρ (1 n)/2n c ξ 1, (2.23)

15 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 15 where ξ 1 is the smallest positive value of ξ having θ(ξ) = 0. By substituting θ and ξ in the Lane-Emden equation back our equation of mass conservation and equation of hydrostatic equilibrium, we obtain M = 4π [ ] 3/2 [ (n + 1)K ρ (3 n)/2n c ξ 2 dθ ], (2.24) 4πG dξ ξ=ξ1 and P c = GM 2 R 4 4π(n + 1) ( ) 1 2 dθ, (2.25) dξ ξ=ξ1 [ ρ = ρ c 3 ] dθ, (2.26) ξ dξ ξ=ξ1 Ω = 3 GM 2 5 n R. (2.27) Question: What physical quantities do θ and ξ correspond to? Note that R is independent of ρ c for n = 1, Ω > 0 for n > 5, Ω is undefined for n = 5. Therefore, the polytropic models for n = 1 and n 5 are not physical. (Can you verify that M is finite, P c is infinite and ρ is 0 for case of n = 5?) Interestingly, for n = 3, the mass of the star is independent of its central density. This particular polytropic index is sometimes also called the Eddington standard model. (We shall say more about the Eddington standard model later in the next chapter.) 2.6 Equation of state We have used the conservation of mass, energy and momentum to construct three equations for stellar structure. Yet, we have P, T, m, ρ, L, ϵ, r as well as the chemical composition as our variables. To further our analysis, we have to look for equations that tell us specific properties of the material in a star. One

16 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 16 such equation is the equation of state (EOS), namely, an equation expressing the pressure P as a function of T, ρ and chemical compositions. The polytrope model assumes adiabatic ideal gas. To model a star, astrophysicists may use some extremely accurate and hence complicated EOSs. I shall not do so in this introductory course for the following reason. Although those EOSs more accurately model the material behavior, their predictions on the interior structure of a star are in most cases not significantly different from the simple EOS we will use in this course. The simplest EOS, which at the same time turns out to be the one we will use in this course, is the ideal gas law that you have learned in high school. That is, P = nkt where n is the number density of gas particles. In the rest of this chapter, I will give a brief review of statistical mechanics, then derive Saha equation to show that the interior of a star is mostly ionized. This means that we will need to introduce mean molecular weight and consider ion, electron, and photon pressure in the EOS. 2.7 Brief review of statistical mechanics Ideal gas law is a good approximation to describe the properties of matter in a star. To see why, I first have to introduce the Saha equation discovered by M. Saha in Recall from the course Statistical Mechanics and Thermodynamics that for a particle (or system of particles), the probability of a particle is in state s is proportional to g s e Es/kT where g s is the degeneracy of state s and E s is the energy of state s relative to some fixed reference energy level (sometimes taken to be the ground state). More importantly, the proportionality constant is the same for all the states. Thus, the ratio of the probability that the particle is in state (s + 1) to the probability that it is in state s is given by the Boltzmann formula g s+1 e (E s+1 E s )/kt. (2.28) g s Exercise: At what temperature a gas of neutral hydrogen will have equal number of atoms in the ground and first excited states? (Hint: for hydrogen atoms, the degeneracy is g n = 2n 2, i.e., the ground state is n = 1 and the first excited state

17 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 17 is n = 2.) This is even higher than the surface of the Sun. However, we know from observations that some hydrogen atoms are ionized in the Sun. How can that be possible? (The answer lies in the Saha Equations, which will be discussed later in this chapter.) The Statistical Mechanics and Thermodynamics course also tells us that the thermodynamic properties of a system with fixed number of particles N and fixed temperature T and fixed volume V is encoded in the so-called partition function Z ( g i exp E ) i, (2.29) i kt where the sum is over all possible energy states of the system. Furthermore, the ratio between the mean number of particles N s in state s and the total number of particles N N j in a sample is given by N s N = g se Es/kT j g j e E j/kt g se Es/kT Z. (2.30) For an ideal classical non-relativistic gas particle with ground state energy E 0 and degeneracy g, its partition function is given by g Z(1, V, T ) = e (E 0+p 2 /2m)/kT d 3 x d 3 p (2π h) 3 gv + = (2π h) 3 e E 0/kT 4πp 2 e p2 /2mkT dp 0 = gv (2.31) λ 3 e E 0/kT where m is the mass of the particle and 2π λ h mkt. (2.32) Note that we have assumed that the momentum of the gas particle is isotropically distributed in order to obtain the above expression for Z(1, V, T ). The partition function of a system of N indistinguishable classical non-relativistic ideal gas particles each with ground state energy E 0 and degeneracy g is then given by Z(1, V, T )N Z(N, V, T ) =. (2.33) N!

18 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 18 Recall again from the Statistical Mechanics and Thermodynamics course that the most convenient method to study the situation in which the particle number in the system may change is to use the so-called grand partition function 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 interpreted as the work needed to add one extra particle to the system at constant V and T. The grand partition function encodes the thermodynamical information of a system with fixed chemical potential, volume and temperature. Since e x = 0 x n /n!, the grand partition function of an ideal classical gas in the non-relativistic limit equals [ ] gv Q(µ, V, T ) = exp λ 3 e(µ E 0)/kT. (2.35) More generally, the grand partition function of a system of interacting particles of various different species such that each species can be approximated by a classical non-relativistic ideal gas is given by Q({µ s }, V, T ) = exp [ s g s V λ 3 s e (µs E 0,s)/kT ], (2.36) where the s is the species label. Hence, the corresponding grand potential equals G({µ s }, V, T ) kt ln Q = kt ( ) g s V µs E 0,s exp. (2.37) s λ 3 s kt Note that G is minimized in chemical equilibrium. Since G = E T S s µ s N s, we have N s = G = g ( ) sv µs E 0,s exp. (2.38) µ s T,V,{µi } i s kt By rearranging terms, the above equation becomes ( Ns λ 3 ) ( s ns λ 3 ) s µ s = kt ln + E 0,s = kt ln + E 0,s, (2.39) g s V where n s is the number density of species s. λ 3 s g s 2.8 Saha equation Now consider a specific atomic species and use the label s, 0 to denote its sth ionized state in its ground level. For example, the label 0, 0 of H refers to the

19 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 19 state of hydrogen atom with its electron in the lowest energy level. And I use the label e to denote electron. Clearly, at non-zero temperature, reactions such as X s+ X (s+1)+ + e (2.40) may occur, where X denotes the atomic species. Hence, at thermodynamic equilibrium, µ s,0 = µ s+1,0 + µ e. (2.41) Exercise: From Eq. (2.39) and the fact that E 0,(s+1,0) + E 0,e E 0,(s,0) = χ s+1, the (s + 1)th ionization potential of the atomic species, show that n s+1,0 n e λ 3 s+1,0λ 3 eg s,0 n s,0 λ 3 s,0g s+1,0 g e = e χ s+1/kt. (2.42) From Eq. (2.32), g e = 2 and the approximation that the masses of the sth and the (s + 1)th ionized atom are about the same, the above equation becomes where n s+1,0 n s,0 f s+1 (T ) = 2(2πm ekt ) 3/2 h 3 = g s+1,0 g s,0 n e f s+1 (T ), (2.43) ( exp χ ) s+1, (2.44) kt In general, not all the sth ionized atoms in a thermalized system are in the ground state. From Eq. (2.30), Eq. (2.43) can be re-written as where Z s = i n s+1 n s = Z s+1 Z s n e f s+1 (T ), (2.45) g s,i exp [ ] (Es,i E s,0 ) is the partition function of those sth ionized atoms in various energy levels. kt (2.46) Eqs. (2.43) and (2.45) are known as the Saha equation. The Saha equation is useful in astronomy. For instance, if we know the electron partial pressure P e, then the electron number density n e can be well approximated by P e /kt in many cases. We now apply Saha equation to estimate the degree of ionization

20 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 20 of our Sun. Eq. (2.43). For simplicity, we consider only the form of Saha equation in 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 K. We simplify our analysis by assuming that the Sun is made up of entirely hydrogen. Thus, χ 1 = 13.6 ev. Let us define the degree of ionization by n 1 /(n 0 + n 1 ) x. Show that Saha equation demands that and hence x 98.8%. n 1 n 0 = x 1 x = f 1(T ) m Hf 1 (T ) n e ρx (2.47) From the result above, almost all hydrogen in our Sun is ionized. By the same token, it is straight-forward to check that most atoms in a main sequence star are ionized. On the other hand, in the solar atmosphere where T 6000 K and n e cm 3. Saha equation tells us that the degree of ionization of hydrogen is about (What about other atoms, such as Na, in the atmosphere, and He and other heavy elements in the core? Note however that Saha equation has its limitations. It works for thermalized region of a star. Hence, it may not be applicable to stellar atmosphere. Also, it assumes classical non-relativistic ideal gas, so it may not work near the core of some stars. 2.9 Gas law From now on, we model the EOS of a star by ideal gas law. More precisely, we assume that the gas pressure of a star is given by P gas = nkt. Since the temperature and luminosity of a star are high, we conclude that electrons, cations (in the form of partially or completely ionized atoms), and photons are the three most important constituents of a star. The electron pressure is given by P e = n e kt (2.48)

21 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 21 where n e is the electron number density, and the cation pressure is given by where n i is the cation number density. P i = n i kt (2.49) What are the values of n e and n i? To answer this question, we denote the mass fraction of H, He and heavy elements (that is, those elements heavier than H and He) in a star by X, Y and Z = 1 X Y respectively. (More precisely, X and Y are the mass fraction of 1 H and 4 He respectively. Do you know why we can forget about the contribution of 2 H, 3 H, 3 He?) Then, number density of hydrogen and helium nuclei equal Xρ/m H and Y ρ/4m H, respectively. Therefore, n i = ρ ( X + Y 4 + Z ), (2.50) A m H where A denotes the average mass number of heavy elements in a star. We introduce the mean molecular weight µ, which is defined as the average weight of a particle in units of hydrogen mass, i.e. µ m m H (2.51) such that This gives µm H = m = total mass of gas total number of particles. (2.52) A direct comparison with Eq. (2.50) gives nµm H = n m = ρ n = ρ µm H (2.53) ( 1 = X + Y µ i 4 + Z ) A. (2.54) Next, we consider electrons. By virial theorem, we know that the thermal energy of a star is of the order of GM 2 /R. That is, GM 2 /R k T M/m H, (2.55) where T 10 6 K is the average temperature of a star. Hence, for most part of a star, the temperature is so high that most atoms are completely ionized. (As we have already seen, this is not a valid assumption for stellar atmospheres.

22 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 22 Nonetheless, this assumption has little effect in determining the overall structure of a star. Similarly, this assumption is not valid for high atomic number atoms, but their numbers are small compared to that of hydrogen and helium for a typical main sequence star.) Similar to Eq. (2.50) above, but a hydrogen atom gives one electron, an helium atom gives two, and a heavy element atom with atomic number Z gives Z electrons. Hence, the total number of electrons per unit volume is n e = ρ m H ( X + 2 Y 4 + Z Z A (Do not confuse the mass fraction of heavy elements Z with Z.) ) = ρ m H µ e. (2.56) For fully ionized gas, the total gas pressure law is contributed by ions and electrons P tot = P i + P e = n i kt + n e kt (2.57) = ρ ( ) kt ρkt. (2.58) m H µ i µ e µm H Here we define 1 µ (2.59) µ I µ e From Eqs. (2.54) and (2.56), we have 1 µ = 2X + 3 Z Y + Z (2.60) A for fully ionized gas. As most nuclei contain about the same number of protons and neutrons, the last term can be approximated by Z/2 and this can be further reduced to 1 µ 2X Y + Z 2 = 2X Y + 1 (1 X Y ). (2.61) 2 Exercises: 1. What is µ for fully ionized hydrogen? 2. What about fully ionized pure helium gas? 3. For Solar abundances, X = 0.747, Y = 0.236, and Z = 0.017, what is µ for neutral gas (assume A = 15.5)? For fully ionized gas, show that µ 0.6. Hence, P tot = ρkt. (2.62) 0.6m H

23 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 23 Question: Why can the mean molecular weight be smaller than 1? 2.10 Photon pressure We have not finished the discussion on EOS yet since we leave out radiation pressure. Recall from the course Statistical Mechanics and Thermodynamics that photons are bosons. Therefore, in thermal equilibrium, photons obey Bose-Einstein statistics. That is, the distribution of photons is isotropic and the number density of photons with frequencies between ν and ν + ν equals n(ν) dν = 8πν2 c 3 dν exp(hν/kt ) 1. (2.63) To see why photon number density follows Eq. (2.63), one recall that the energy of having s photons each with momentum p equals spc. As photon is bosonic, s can take on any natural number. Moreover, the probability P (s) that there are exactly s photons each with momentum p follows the constraint P (s + 1) P (s) where ν is the frequency of the photon. Consequently, P (s) = = e pc/kt = e hν/kt, (2.64) 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 expected number of photons with an allowed momentum p is given by N γ ( p) = 2 + s=0 sp (s) = 2 e hν/kt 1. (2.66) (Note that the 2 above reflects the fact that a photon has two possible polarizations.) As a result, the total expected number of photons equals N γ = 1 N h 3 γ ( x) dv d 3 p = 1 2 dv h 3 e hν/kt 1 d3 p = V 8πν 2 dν. (2.67) c 3 (e hν/kt 1)

24 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 24 Hence, Eq. (2.63) is valid. Note that the first fraction in Eq. (2.63) is called phase space factor while the second fraction is the Bose-Einstein distribution factor. Moreover, this distribution n(ν) is sometimes called the blackbody spectrum. For a photon of momentum p hitting a wall and then reflected back elastically, the magnitude of the change in momentum equals 2p cos θ = 2hν cos θ/c where θ is the angle between p and the normal of the wall. The radiation pressure due simply the momentum transfer to the wall per unit time per unit surface area. That is, P rad = 1 h 3 + = hν 3 π/2 2π 0 0 n(ν) dν c cos θ 2hν cos θ c 2 e hν/kt 1 ( ) 2 hν sin θ dϕdθd c ( ) hν c = 4σ 3c T 4, (2.68) where σ = 2π 5 k 4 /15c 2 h 3 is the Stefan s constant. Thus, our ideal gas EOS for stellar matter is P = P i + P e + P rad. (2.69) Finally, note that the energy density of a photon gas at temperature T is given by Hence, u rad = 3P rad. Exercise: u rad = + 0 hνn(ν)dν = 4σ c T 4. (2.70) 1. For a photon gas with volume V and temperature T, what is the internal energy U? 2. Given that the entropy S of the gas is related to U by du = T ds and S remains a constant for an adiabatic process, show that a photon gas has an adiabatic index γ = 4/3.

25 CHAPTER 2. STELLAR STRUCTURE EQUATIONS Other forces As a final note, we estimate the contribution of other forces to the EOS. Most of the stellar material should be ionized, therefore, the dominant interaction between particles in stellar interior (besides gravity) is electrostatic in nature. The typical distance between particles in stellar interior d (Ām H / ρ) 1/3 where Ā is the average atomic weight of particles, m H is the mass of hydrogen atom, and ρ is the average density of a star. Consequently, the typical Coulomb energy per particle is about Z 2 e 2 /d (recall that 4πϵ 0 1 in cgs), where Z is the average atomic number of particles. In contrast, virial theorem tells us that the gravitational potential energy and internal energy of a star are of the same order. Consequently, the ratio of Coulombic energy to internal energy of a star is about Z 2 e 2. (2.71) G(Ām H ) 4/3 M 2/3 Note that Z and Ā are of order of 1. Therefore, the above ratio is about 10 2, which is much less than 1. Thus, electrostatic potential energy contribute only a small fraction of the internal energy of a star; and most of the internal energy are in the form of heat. More specifically, the thermal K.E. of a typical particle inside a star is much greater than the electrostatic potential energy it experiences. Well, you may also ask if quantum effects may alter the ideal gas EOS in a star. We may estimate the uncertainty in position x to be of order of d. Similarly, the uncertainty in momentum p is of the order of kt Ām H. Thus, for all living stars, x p is much larger than h. Hence, quantum effects play little role in affecting the ideal gas EOS of a star. (Note: the situation is completely different for dead stars such as white dwarfs and neutron stars. Because the density of dead stars is high, x p for some particle species inside these stars are of order of h and hence quantum effect drastically changes their EOS.) To summarize, ideal gas EOS is a good approximation in the study of the interior of most living stars. (However, quantum mechanical effect is important in some other stars such as brown dwarfs and supernovae. Radiation pressure also plays an important role in the structure and stability of a super-massive star. We shall come across them later on in this course.)

26 CHAPTER 2. STELLAR STRUCTURE EQUATIONS Degenerate gas equation of state For completeness, we will now derive the EOS for degenerate gas, in which the quantum effect dominates. This is important in some cases, such as brown dwarf, late stage of stellar evolution, white dwarfs, and neutron stars. Due to Pauli s exclusion principle, two indistinguishable Fermions (e.g. electrons, protons, neutrons, etc) cannot occupy the same state. At extreme densities, all the low-energy states are occupied. To further increase the density, it takes extra energy to force a Fermion to a higher energy level. This results in an increase in the gas pressure. Fundamentally, it is the exchange interaction that creates the force, although this is merely a quantum mechanical effect rather than a real force. Consider a particle in a box (i.e. an infinite potential well in 3D) of length L, solution of the Schrödinger equation requires the particle wavelength to satisfy λ x = 2L, λ y = 2L, and λ z = 2L, (2.72) l x l y l z where l x, l y, and l z are positive integers. From the de Broglie equation p = h/λ, the K.E. is E = p2 2m = h2 8mL 2 (l2 x + ly 2 + lz) 2 h2 l 2 8mL, (2.73) 2 i.e. it is proportional to l 2 lx 2 + ly 2 + lz. 2 At low temperature, we can assume that all low-energy states are occupied up to the Fermi level l F. The number of states within a sphere of radius l F is 4πlF 3 /3. But since l x l y, and l z takes only positive numbers, we have to divide this by 8 and each state can have two particles when we take the spin into account. Therefore, the total number of particles is ( ) ( ) 1 4 N tot = πl3 F = πl3 F 3. (2.74) This gives l F = (3N tot /π) 1/3 and hence the Fermi energy of E F = h2 l 2 F 8mL 2 = h2 8mL 2 ( 3Ntot π ) 2/3 = h2 8m ( 3n π ) 2/3, (2.75) where n = N tot /L 3 (not to be confused with the polytropic index) is the particle number density. The total energy is E tot = lf 0 E dn = lf 0 ( h 2 l 2 8mL 2 ) ( ) π d 3 l3

27 CHAPTER 2. STELLAR STRUCTURE EQUATIONS 27 = = 3 5 N tote F V 2/3. (2.76) Exercise: Show that the gas pressure is P E tot V = 2 5 ne F V 5/3, (2.77) and hence the non-relativistic degenerate gas EOS is a polytrope. What is γ and the polytropic index? When the density is extremely large, e.g., inside a neutron star, E F becomes comparable to the rest mass of the Fermions, we need to consider relativistic effects and modify Eq with E = m 2 c 4 + p 2 c 2 pc. Exercise: What is E in terms of l? Hence, derive E tot and P. Show that the relativistic degenerate gas EOS is a polytrope and find the index.

28 Chapter 3 Energy Production and Transfer Processes 3.1 Energy conservation and the energy production equation We have the equations that govern P and ρ (i.e. m) as functions of radial distance. So far, our simple stellar model is only a giant ball of gas that is bounded by self gravity and obeys the adiabatic gas law. Now, we want to go further by considering a more realistic equation of state, i.e., P (ρ) and different chemical compositions. In this chapter, we will talk about energy generation inside a star. This will change the temperature, and hence P and ρ. With our LTE assumption, matter in the star is in thermal equilibrium. Thus, the gas inside a star will not expand or contract; the work done by the gas is zero. Let L(r) denotes the luminosity at the spherical shell with radius r centered at the core of the star. (That is, L(r) is the energy per unit time moves out of the spherical shell of radius r.) Energy conservation demands that dl dr = 4πr2 ρ(r)ϵ(r), (3.1) where ϵ is the rate of energy production per unit mass of material at the radius r from the stellar center. Note that ϵ is an implicit function of density ρ, temperature T and chemical composition. (Note: a few authors use the notation q instead of ϵ.) 28

29 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 derivative ϕ, τ ϕ/ ϕ, such that it gives us an idea on the characteristic time for that variable to change significantly. The free-fall / dynamical timescale: For a star with mass M and radius R, the escape velocity is of order of GM/R. The free-fall timescale is, therefore, of the order of τ ff R 3 /GM 1/ G ρ where ρ denotes the mean density of the star. For our Sun, the free-fall timescale is of order of an hour. In short, any force that is unbalanced inside a star occurs at the free-fall timescale. Thus, if we are interested in the processes of a stable star over a time span which is much longer than the free-fall timescale, then we are safe to assume that the star is in mechanical equilibrium. The Kelvin-Helmholtz timescale: This is the timescale τ KH for a star to radiate its thermal energy away at constant luminosity. By virial theorem, τ KH GM 2 /RL. For our Sun, τ KH is about yr. For a time much longer than the Kelvin-Helmholtz timescale, we may safely assume that the star is in thermal equilibrium. The nuclear burning timescale: This is the timescale for burning all the nuclear fuel of a star at constant luminosity and is therefore equal to τ nucl Mc 2 /L, where 10 3 is the typical binding energy of a nucleon divided by the rest energy of the nucleon. For our Sun, τ nucl yr. The nuclear burning timescale is a rough estimate of the lifespan of a star. (You may notice that the nuclear burning timescale overestimates the lifespan of a star. Do you know why?) Since τ ff τ KH τ nucl, we know once again that our assumption of LTE is a good approximation to study the interior structure and evolution of a star.

30 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 30 Exercise: Estimate these timescales for the Sun. Based on the results, what is the main energy source of the Sun? 3.3 Coupling between radiation and matter and the radiative transfer equation A photon in a star travels in straight line until it is scattered or absorbed and remitted in a random direction by the surrounding materials. In this way, energy carried by the photons is transferred to the surrounding material and heats it up. Therefore, the trajectories of photons inside a star can be regarded as random walks. Opacity refers to the degree of radiation scattering and absorption in a medium. It is related to the probability per unit path length that a photon is being absorbed or scattered. When a light ray travels in a medium, its intensity will be reduced with distance x by I(x) = I 0 exp( κρx), where κ is the specific opacity or opacity coefficient and ρ is the density of the medium. By definition, κρ is of unit cm 1 and hence κ is of unit cm 2 g 1. (To tell you the truth, κ in general depends on the frequency of light to be absorbed. In fact, the κ used here is carefully averaged over the frequency of light to be absorbed.) In addition to energy transfer between photon and the medium, absorption and/or scattering of radiation in the medium also lead to momentum transfer. Now consider the sphere of radius r from the center of a star. The luminosity, namely, light energy passes through this sphere per unit time, is L(r). Thus, the corresponding radiation flux (that is, light energy passes through this sphere per unit area per unit time) equals L(r)/4πr 2. Therefore, the light energy absorbed or scattered on this sphere per unit area per unit time per unit path length equals L(r)κρ/4πr 2. Consequently, the momentum transferred to the surrounding medium on this sphere per unit area per unit time per unit path length is L(r)κρ/4πr 2 c. This number is equal to 1 times the radiation pressure gradient on that sphere (do you know why?) i.e. dp rad dr = L(r)κ(r)ρ(r) 4πr 2 c.

31 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 31 Recall that inside a star, LTE is a good approximation. Hence, radiation there follows a blackbody spectrum. So, from Eq. (2.68), we have dp rad dr Putting these together, we have = 16σT 3 3c dt dr. L(r) = 64πσr2 T 3 3κρ dt dr. (3.3) This equation is sometimes known as the radiative transfer equation. Clearly, T (0) = T c (the core temperature) while T (R) = T s 0 (the surface temperature). Note that the radiative transfer equation is valid if radiation is the principle heat transfer mechanism. If convection or conduction is important, this equation has to be modified. 3.4 Physical processes leading to opacity and Kramers opacity Approximation The interactions between photons and matter (primarily with electrons and nuclei in most part of a star) can be classified below. 1. Electron scattering: Compton/Thomson scattering, that is, scattering of photon by a free electron. 2. Free-free absorption: Absorption of a photon by a free electron which makes a transition to a higher energy level by briefly interacting with a nucleus or ion. 3. Bound-free absorption: Also known as photon-ionization, namely, the ionization of an electron from an atom or ion after absorbing a photon. 4. Bound-bound absorption: Exciting an electron from a bounded state of a lower energy level to another bounded state of a higher energy level by absorbing a photon. Close to the stellar core, temperature is so high that atoms are almost completely ionized. Thus, electron scattering and free-free absorption are the dominant process to create opacity in stellar core. In the outer stellar atmosphere,

32 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 32 in contrast, the situation is much more complex. Bound-free and bound-bound absorptions can be important. This is the reason why we see absorption lines on the surface of stars. The specific opacity κ can be calculated by taking into account all the possible interactions between photon and the medium. This is a very complicated and involved task, requiring the work of a whole generation of atomic physicists. Fortunately, the resultant κ can usually be approximated by relatively simple formula in the form κ = κ 0 ρ a T b. (Once again, we shall only use the approximate form of κ in this course as the approximation is reasonably good for most part of the stellar interior.) In particular, the electron scattering opacity, which is important in the high temperature core of a star, is given by κ es 0.2(1 + X) cm 2 g 1, (3.4) while the free-free absorption, which may be important near the surface of a star, can be approximated within an accuracy of about 20% by the so-called Kramers opacity κ ff (1 + X)(X + Y )ρt 7/2 cm 2 g 1. (3.5) Clearly, free-free absorption produces a higher opacity than electron scattering at low temperature. In fact, the cross-over temperature between free-free absorption and electron scattering is about 10 6 K. Whereas when the temperature is less than about 10 4 K, bound-free and bound-bound absorption become more important than Kramers opacity is no longer valid. In fact, at such a low temperature, κ starts to decrease as T decreases. Interestingly, the presence of opacity places an upper limit on the luminosity of a star. Recall that P = P gas + P rad. Clearly, dp gas /dr 0. So, from the equation of hydrostatic equilibrium and equation of radiative transfer, we have Therefore, dp rad dr dp rad dr + dp gas dr = (3.6) L L Edd. (3.7) If this inequality is violated somewhere inside a star, then either radiative transport is no longer valid (such as in the core of certain stars) or hydrostatic equilibrium can no longer be maintained (such as near the surface of certain stars).

33 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 33 In particular, near the surface of a star, m = M and the limiting luminosity given by Eq. (3.7) becomes the so-called the Eddington luminosity ( ) ( ) M κ 1 L Edd = L. (3.8) M κ es 3.5 The Eddington standard model Let us define a function η by L(r)/m(r) = L η/m where L is the luminosity across a sphere of radius r centered at the star and L is the total luminosity of a star. Exercise: Using the stellar structure equation, show that dp rad /dr can be written as dp rad dp = L κη 4πcGM. (3.9) Assuming that the location where nuclear burning occurs is confined in a small region in the core, then over most part of a star, η decreases as r increases. On the other hand, κ increases with r (do you know why?). At this point, A. Eddington somehow made the bold assumption that the product κη κ s is constant over the entire star. Surely, this assumption is in no way close to the truth, it leads to very interesting and realistic predictions about structure and evolution of stars. Integrating Eq. (3.9) gives P rad = κ sl P P (1 β). (3.10) 4πcGM That is to say, the ratio of P rad to P is a constant throughout a star in the Eddington standard model. Historically, the ratio between gas and total pressure is defined as β, i.e. P gas /P = β, such that P rad /P = 1 β. Thus, the radiative transfer equation becomes L = 4πcGM κ s (1 β) L Edd (1 β). (3.11)

34 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 34 The ideal gas law demands that T 4 1 β P rad 1 β = P = P gas β = ιρt β, (3.12) where ι = k/m H µ (see Eq. 2.58). Hence, T ρ 1/3. Note that for gas, the equation of state may be written as P = Kρ 4/3 for some constant K [(1 β)ι 4 /β 4 ] 1/3. In other words, the equation of state is a polytrope with index 3. So, there is a unique relation between K and M. From Eq. (2.24) with n = 3, we have 1 β = ( ) 4 ( ) 2 β M. (3.13) ι M This is known as the Eddington quartic equation. This equation has the following predictions. First, for a given composition (fixed ι), β decreases as M increases. So, radiation pressure is more important in high mass stars. Second, combining with Eq. (3.11) gives L M 3. So, the mass-luminosity relation is obtained. Actually, Eddington made this prediction before the confirmation by observations. Third, for a given M, nuclear burning implies that ι will decrease gradually. Consequently, β decreases. Therefore, it is expected that radiation pressure plays an increasingly important role as a star gets older. Combined with the Eddington limit, it strongly hints that some stars may eject some of its materials in its late stage of evolution. Exercise: Combining Eqs. (3.11) and (3.13), show that L M Nuclear energy generation The energy generation mechanism of a star cannot be chemical in nature. For chemical reaction leads to an energy release per molecule of the order of 1 ev. Thus, chemical reaction can only support the present solar luminosity for about 10 4 yr. In contrast, the energy generation by nuclear reaction is of the order of 1 MeV per nucleus. This allows the Sun to shine at the present rate for about

35 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER yr. Thus, nuclear energy looks like a promising way to power the Sun and other stars. (Still recall that energy is released by fusion of light nuclei or fission of heavy nuclei? Do you know that 56 Fe is the most stable nucleus in the sense that it has the highest binding energy per nucleon?) But we have a problem. Virial theorem tells us that the thermal energy and gravitational potential energy of a star are of the same order. Thus, using ideal gas approximation, the average temperature of our Sun turns out to be about 10 6 K. However, the typical K.E. of two 10 6 K hydrogen nuclei cannot overcome the Coulombic repulsion between them. Specifically, the Coulomb energy between two hydrogen nuclei placed 2 hydrogen nuclear radii apart is about three orders of magnitude larger than the typical K.E. of the two hydrogen nuclei. Thus, classically, there is no way for two hydrogen nuclei to fuse together making a bigger nucleus and hence releasing the required nuclear energy to make the star to shine. Fortunately, quantum mechanics saves us for hydrogen nucleus may tunnel through the Coulombic energy barrier (see figure 3.1). Wavefunction r Potential Coulombic repulsion 1/r E r r n r c Figure 3.1: An illustration of overcoming the Coulombic energy barrier through tunneling. Consider a nucleus of charge Z 1 e moving towards another nucleus of charge Z 2 e. Denote µ (m 1 + m 2 )/m 1 m 2 the reduced mass of the system (not to be confused with the mean molecular weight). The Schrödinger equation describing

36 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 36 this system is h2 2µ 2 ψ + Z 1Z 2 e 2 ψ = Eψ, (3.14) r for r > r N (the nuclear radius), where E is the K.E. of the first nucleus relative to the second one when they are infinitely apart. By separation of variables, (that is, by writing ψ = R(r)Y (θ, ϕ)), we conclude that the radial wavefunction of the above equation satisfies h2 d 2 χ 2µ dr + Z 1Z 2 e 2 2 r χ(r) + h2 l(l + 1) 2µr 2 χ(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, Eq. (3.15) can be re-written as d 2 χ dr = 2µ ( Z1 Z 2 e 2 ) 2 h 2 E χ. (3.16) r Nonetheless, the above equation is too complicated to solve analytically. Fortunately, an approximate solution is available using W.K.B. approximation. Specifically, we assume that the radial wavefunction (in other words, the solution of Eq. 3.16) to be in the form χ(r) = Ae is(r)/ h for some constant A. Exercise: Using Eq. (3.16), show that S(r) satisfies where i hs S 2 h 2 k(r) 2 = 0, (3.17) k(r) 2 = 2µ ( Z1 Z 2 e 2 ) h 2 E r. (3.18) Note that in the quantum tunneling region, where Z 1 Z 2 e 2 /r > E, we can pick k(r) 0. Exercise: Expanding S(r) in powers of h, i.e. S = k=0 h k S k (r) and keeping only the zeroth order term (note that the term h 2 k 2 is h independent as k is

37 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 37 proportional to 1/ h), show that [ r ] χ(r) = A exp k(r) dr r c for some constant A whenever r is in the quantum tunneling region. (3.19) Note that the integral above is from r c to r N where r c is the distance between the two nuclei when the classical K.E. of the first nucleus is zero. That is, Z 1 Z 2 e 2 r c = 1 2 µv2. (3.20) Besides, E = µv 2 /2 where v is the relative speed between the two nuclei. Since r c r N and hence Z 1 Z 2 e 2 /r E, upon integration, we find that the tunneling probability is about R(r N ) 2 χ(r N ) 2 ( 2 A = r N exp 2πZ 1Z 2 e 2 ). (3.21) r N hv (Can you fill in the details of the calculation?) This result, which laid an important foundation to the study of nuclear reactions in stars, was first obtained by G. A. Gamow in the study of radioactivity in R. Atkinson and F. Houtermans quickly applied it to the case of energy generation of stars in In summary, the probability for two nuclei coming close enough for their nuclear force to be effective is small but non-zero. We expect that nuclear force is effective only within a de Broglie wavelength λ = h/µv. Thus, we also expect the nuclear reaction cross section to be in the form σ(v) πλ 2 exp ( 2πZ 1Z 2 e 2 hv ) S(v) v 2 exp ( 2πZ 1Z 2 e 2 hv ), (3.22)

38 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 38 where S(v) is some expression that is weakly dependent on v. Note further that the velocity distribution of nuclei follows the Maxwellian distribution. Specifically, the probability of finding a particle with velocity v is proportional to (µ/2πkt ) 3/2 exp( µv 2 /2kT ) where v = v. Consequently, the probability of nuclear reaction to occur between the two nuclei per unit time is proportional to + ( ) ) ( v µ 3/2 exp ( µv2 S(v) exp 2πZ 1Z 2 e 2 ) 4πv 2 dv. (3.23) 0 2πkT 2kT v 2 hv r N For simplicity, we assume that S is a function that does not depend strongly on v so that vs(v) can be assumed to be almost a constant throughout the speed range considered. Further observe that the first term in the exponential increases with v (and hence the initial K.E.) while the second term in the exponential decreases with v (and hence the initial K.E.). By Taylor series expansion on the terms inside the two exponentials, we conclude that the resultant reaction probability has a peak around v Gamow = ( 2πZ1 Z 2 e 2 ) 1/3 kt (3.24) known as the Gamow peak. In addition, the width of this peak is about (See figure 3.2) v width hµ ( ) 1/2 kt. (3.25) 3µ Consequently, the reaction rate per unit volume is approximately proportional to ( ) ( µ 3/2 2πZ1 Z 2 e 2 ) n 1 n 2 v Gamow S(v Gamow ) exp n 1 n 2 (kt ) 2/3 exp 3 2 2πkT ( 2πZ1 Z 2 e 2 h ) 2/3 ( µ kt hv Gamow ) 1/3 where n i is the number density of nuclear species i. µv2 Gamow 2kT v width, (3.26) Thus, the reaction rate per unit volume increases as T increases or as the charge of the nucleus decreases. This is the reason why fusion of heavy nuclei requires high temperature. However, there is an exception to this trend. Some reactions known as resonant reactions occur when the energy of the interacting particles correspond to

39 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 39 Probability Maxwell Boltzmann e E/kT Gamow Peak Tunneling e b/e 1/2 Energy Figure 3.2: An illustration of the dependency of reaction probability on energies. energy level of the intermediate unstable compound nuclei. (Still remember what you have learned in high school chemistry?) Resonant reaction probability, therefore, is sharply peaked at the resonant energy. (That is, S(v) is sharply peaked at certain values of v.) Such a probability may be several order of magnitudes higher than that of an ordinary non-resonant reaction. From Eq. (3.26), the nuclear energy generation rate ϵ depends on T 2/3 exp(t 1/3 ). To simplify the calculation, we can parameterize ϵ by ϵ Aρ a T b (3.27) for some constants A, a and b over a restricted range of temperature T. Finally, I must stress that the actual calculation of a specific nuclear reaction rate and hence a and b are very involved and complicated. The above discussions only serve as an overly simplified approximation.

40 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER Some specific nuclear reactions occurring in stars Proton-proton chain (p-p chain) The simplest possible reaction is p + p 2 He. Unfortunately, 2 He is a highly unstable nucleus which quickly decomposes back into two protons. However, if a weak interaction that turns a proton into a neutron occurs during the combination of two protons, then we will get some stable product. Namely, p + p 2 H + ν e + e +. The reaction rate is very slow since weak interaction, as the name suggests, is much weaker than strong and EM interactions. (Actually, experiments find that the relative interaction strength between strong, EM and weak forces is about 1 : 10 2 : 10 5 for all stellar astrophysical processes.) This is not the end of the story for main sequence stars as the resultant 2 H will undergo further nuclear reaction in the core of a main sequence star. For example: p-p I chain: p + p 2 H + ν e + e + 2 H + p 3 He + γ 3 He + 3 He 4 He + 2p Besides p-p I chain, a star may also react via p-p II chain: 3 He + 4 He 7 Be + γ 7 Be + e 7 Li + ν e 7 Li + p 2 4 He or p-p III chain: 7 Be + p 8 B + γ 8 B 8 Be + e + + ν e 8 Be 2 4 He

41 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 41 These three sets of reactions, whose foundation was first proposed by H. A. Bethe in 1939, operate simultaneously although their relative significance depend on the physical conditions at the stellar core. But in any case, the net result is to convert 4 hydrogen nuclei into 1 helium nucleus. The nuclear energy released per each conversion is experimentally measured to be about Q = 4M( 1 H) M( 4 He) = MeV. The energy is released in the form of mainly photons and neutrinos. Photons will later on interact with the electrons in the star through electron opacity. Neutrinos, on the other hand, are extremely weakly interacting particles. They leave the star essentially without interacting with the surrounding materials. Thus, neutrinos emitted from the star in this way are not thermalized. In other words, LTE does not apply to neutrinos in a star. Finally, notice that in p-p I chain only the first reaction involves weak interaction, all other reactions involve strong and EM interactions. Hence, the bottle-neck for p-p chain is the first reaction. In fact, more accurate calculations tell us that the reaction rate for the first reaction is of the order of once every years. Hence, it occurs in a massive scale in the core of a star because there are numerous supply of protons in the stellar core. To summarize, the nuclear energy generation rate per unit mass ϵ for p-p chain is determined by the rate of p + p 2 H + ν e + e + ; and it can be shown that the particular reaction rate is given by ϵ p p = ρx 2 e 3.38T 1/3 T 2/3 9, (3.28) where T 9 is temperature in units of 10 9 K. This can be approximated by ϵ p p AρX 2 T 4 (3.29) for some constant A > 0, where X is the mass fraction of 1 H Carbon-nitrogen-oxygen (CNO) cycle H. A. Bethe and C. von Weizsäcker independently pointed out in the late 1930 s that besides the p-p chain, hydrogen may burn via the CNO cycle using carbon, nitrogen and oxygen as catalysis as shown in the two cycles below: cycle 1: 12 C + p 13 N + γ 13 N 13 C + e + + ν e 13 C + p 14 N + γ 14 N + p 15 O + γ

42 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER O 15 N + e + + ν e 15 N + p 12 C + 4 He cycle 2: 14 N + p 15 O + γ 15 O 15 N + e + + ν e 15 N + p 16 O + γ 16 O + p 17 F + γ 17 F 17 O + e + + ν e 17 O + p 14 N + 4 He Eventually, H. A. Bethe received the Nobel Prize in Physics in 1967 for his contributions to the theory of nuclear reactions, especially his discoveries concerning the energy production of stars. Although these reactions are very complicated, it is relatively easy to estimate the overall rate of reaction for we need only to look closely at the slowest rate determinate reaction. Recall that β decay is temperature and density independent (over conditions inside a star). In contrast, proton capture reactions are sensitively dependent on temperature. (Do you know why?) Therefore, at relatively low temperature, it is the proton capture that sets the pace of the reaction. While at relatively high temperature, it is the β decay that determines the reaction rate. In fact, under the conditions of the interior of most stars, proton capture is the bottle-neck. The rate of energy release per unit mass is and it can be approximated by ϵ CNO = ρxze 15.2T 1/3 T 2/3 9, (3.30) ϵ CNO AρXZT 16, (3.31) for some constant A > 0. The constant A in Eq. (3.31) is much smaller than that in Eq. (3.29). Do you know the reason why? Pre-main sequence light element burning The following proton capture reactions already function effectively before the onset of hydrogen burning: 2 H + p 3 He + γ

43 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 43 6 Li + p 3 He + 4 He 7 Li + p 2 4 He 9 Be + p 4 He + 6 Li 10 B + p 4 He + 7 Be 11 B + p 12 C + γ 11 B + p 3 4 He. (Note that 7 Be formed in the above reaction decays via electron capture to 7 Li + ν e with a half-life of about 53 d.) These reactions are effective to destroy light elements such as 2 H, Li, Be, B whenever the temperature is higher than about 10 6 K. This can be used to distinguish young brown dwarfs from main-sequence stars, known as the Lithium test Helium burning At higher temperature, helium may fuse via 2 4 He 8 Be. (3.32) Unfortunately, 8 Be is highly unstable which decays back to 2 He in about s. The solution to this problem was given by E. Salpeter in 1952 and was later on refined by F. Hoyle. They pointed out that if the temperature is higher than about 10 8 K, the K.E. of helium nuclei will be very high. In this case, the mean collision time between the unstable Be nucleus and a He nucleus is shorter than the half-life of the unstable Be nucleus. Hence, it is possible to produce a 12 C by capturing a He nucleus by a 8 Be nucleus. The net result is 3 4 He 12 C (3.33) known as the triple alpha process. The energy released per reaction is about 7.28 MeV. Clearly, the net reaction rate is determined by the rate of reaction of the capture of He by Be. It turns out that the reaction rate per unit mass is ϵ 3α Aρ 2 Y 3 T 40, (3.34) for some A > 0. Some of the carbon produced could further fuse into oxygen by 12 C + 4 He 16 O. (3.35)

44 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER Heavy elements burning and photodisintegration At higher temperature, carbon and oxygen can overcome their Coulombic barriers and fuse together. The resultant nuclei are Mg, Na, Ne, S, P, Si, etc. At even higher temperature, Si etc. may also begin to fuse. Nonetheless, the situation is not as simple as He, C or O burning. At temperature higher than about 10 9 K, photodisintegration begins to take place. In photodisintegration, a nucleus divides into two or more fragments after absorbing a high energy photon. Thus, photodisintegration is, in general, an endothermic process similar to the photo-ionization of an atom. Therefore, photodisintegration is effective only at sufficiently high temperature (so that there are enough high energy photon to ionize the nucleus). An example of photodisintegration process is shown below: 20 Ne + γ 16 O + 4 He. (3.36) Due to the presence of heavier element fusion and photodisintegration, reactions involving the burning of Si and other heavier nuclei occur in an almost dynamic equilibrium fashion. The net result is a gradual buildup of heavy elements such as Ni, Fe, Co provided that the temperature is lower than about K. If the temperature is higher than about K, photodisintegration becomes the dominant reaction and hence the star will cool itself down. (Do you know why?) In other words, photodisintegration sets an upper limit in the core temperature of a star Neutron capture and β decay Another type of reaction that are of interest is neutron capture. Since neutron is electrically neutral, it can come close to a nucleus without worrying about the Coulombic barrier. Nevertheless, free neutron outside a nucleus is a rare species. A small amount of neutron is generated in a minor reaction involving the burning of carbon such as 13 C + 4 He 16 O + n. (3.37) These neutrons may be captured by a heavy nucleus resulting in a heavier isotope. Upon a few neutron captures, the resultant nucleus may be too neutronrich so that it will β decay into a new nucleus with higher atomic number. Neutron capture and the subsequent β decay, therefore, are effective mechanism to produce a variety of heavy elements and isotopes. In particular, they play an important role in the synthesis of elements with atomic number higher than about 60.

45 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 45 β ν ( n, γ ) p s only s,r s,r p s only s,r s only r Z s,r s,r s only s,r r r s,r r β (Neutron rich matter) N Figure 3.3: Schematic representation of the effects of s- and r-processes. Astronomers generally divide neutron capture processes into two types. In the event that the neutron capture reaction rate is much slower than the β- decay rate, we call it the s-process. In an s-process, which is a more common neutron capture process in astronomy, successive neutron capture leads to the production of a chain of stable isotopes until it reaches a radioactive species, at which point β-decay will occur. (See figure 3.3.) Another possibility is that the neutron capture rate is much faster than the β-decay rate, known as the r-process. The r-process is effective in synthesizing a few highly neutron-rich stable isotopes. The r-process is likely to be important in supernova explosion or in neutrino-heated atmosphere surrounding a new-born neutron star. (See figure 3.3.) Note, however, that both processes cannot synthesize the so-called stable p-nuclei (p for proton-rich). Thus, we expect that p-nuclei are rare in a star. This prediction generally agrees with observations. (See figures 3.3 and 3.4.) Astronomers believe that these proton-rich nuclei are formed by proton capture at very high temperature or by the so-called (γ, n) reaction during supernova explosions Pair production In the event that the temperature of certain region of a star exceeds 2m e c 2 /k K, electron-positron pair production from thermal photon becomes

46 CHAPTER 3. ENERGY PRODUCTION AND TRANSFER 46 Z Sb, 51 Sn, 50 In, 49 Cd, 48 p process s only % Slow process path % d 0.69% 4.2% 12.8% 24.0% 0.38% 14.3% 7.6% 24.1% 8.5% 32.5% 27h 4.8% 6.1% 95.8% 13s 12.3% 28.8% 54h 7.6% N % 2.8d Rapid Process Rapid process only % Figure 3.4: Schematic representation of the effects of s- and r-processes. important. (Actually, the effect cannot be neglected when the temperature is about 10 9 K because a few highly energetic photons from the tail of the Planck distribution can already undergo pair production.) Similar to photodisintegration, pair production is endothermic and in effect placing a limit on the maximum temperature of a star. It has been proposed that this mechanism could give rise to pair-instability supernova in very massive stars above 130 M. The work of stellar nucleosythnesis culminated in the publication of the article by E. M. Burbidge, G. Burbidge, W. A. Fowler and F. Hoyle, Reviews of Modern Physics 29, 547 (1957). This article introduced all the essential ideas involving in the synthesis of elements from carbon to uranium in a star starting with the hydrogen and helium produced in the big bang. W. A. Fowler was awarded the Nobel Prize in Physics in 1983 for his theoretical and experimental studies of the nuclear reactions of importance in the formation of the chemical elements in the universe. (Actually, a similar idea was also pointed out independently in the article A. G. W. Cameron, in Chalk River Report Number CRL-41, Chalk River Labs., Ontario (1957).)

47 Chapter 4 Stability of Stars 4.1 Secular thermal stability Recall from the virial theorem that 2U + Ω = 0 for stars modeled by nonrelativistic ideal gas. That is to say, the total energy of the star E obeys E = U = 1 2 Ω, (4.1) where U is internal energy and Ω is potential energy. Suppose that the nuclear energy generation rate of the star somehow increases a little bit. Then, Ė > 0 implying that U < 0. For ideal gas, that means T < 0. In other words, the average temperature of the star decreases. Consequently, the nuclear energy generation rate will decrease. By the same argument, a slight decrease in nuclear energy generation rate will increase the internal energy of the star making the star a little bit hotter, therefore increasing the resultant nuclear energy generation rate. This negative feedback makes the star thermally stable against small perturbation in nuclear energy generation rate. We call this secular thermal stability of a star. Does this argument work for degenerate stars or stars with degenerate cores? 4.2 Dynamical stability Dynamical stability deals with the stability of motion of mass parcels in a star. A detail treatment of dynamical stability is very complicated. So, I only illustrate the idea by a highly simplified example. 47

48 CHAPTER 4. STABILITY OF STARS 48 Consider a gaseous sphere of mass M in hydrostatic equilibrium. At any point r(m), the pressure is equal to the weight per unit area of layers between m and M. Equation of hydrostatic equilibrium (Eq. 2.4) implies that P (m) = M m Gm dm. (4.2) 4πr4 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 4πr 2 ρdr = dm = 4πr 2 ρ dr = ρ (m) = Hence, the new hydrostatic pressure is given by =. (4.4) ρ(1 + 3 ). (4.5) M P Gm (m) = m 4πr 4 (1 ) dm 4 P (1 + 4 ). (4.6) Assume further that the change in radius is adiabatic (i.e. P V γ =const.), then P gas = P (1 + 3 ) γ a P (1 + 3γ a ) (4.7) where γ a 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 gas could bounce back. That is, P (1 + 3γ a ) > P (1 + 4 ). This condition is equivalent to γ a > 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, if some part of a star is dominated by radiation pressure, that part of the star will become dynamically unstable against compression.

49 CHAPTER 4. STABILITY OF STARS Convective instability Recall in our earlier discussions that the radiative transfer equation holds only when radiation is the principal energy transfer mechanism. Hence, the radiative transfer equation does not apply when convection is important. It is, therefore, natural to ask the condition for a star to be convectively stable (or unstable). P2 ρ new ρ 2 P 1 ρ 1 Figure 4.1: The condition of a small lump of has rising. Consider a small lump of gas whose pressure and density are P 1 and ρ 1, respectively. When displaced upward, its pressure and density will change. We denote its new pressure and density by P 2 and ρ new, respectively. (See figure 4.1.) Clearly, P 2 equals to the pressure of the gas surrounding the new location of the lump. Denote also the density of gas surrounding the new location of the lump by ρ 2. Obviously, the lump of gas is stable against convection if ρ new > ρ 2. Otherwise, the system will be convectively unstable. We may assume that the change is adiabatic. (Why is this a good approximation?) In this case, P ρ γ a and hence the convective stability condition becomes ( ) ( ) dp dp < dρ dρ star adiabatic. (4.9) This is known as Schwarzschild s criterion. This can be understood using figure 4.2: when pressure changes from P 1 to P 2, a stable configuration should

50 CHAPTER 4. STABILITY OF STARS 50 have ρ 2 < ρ and hence a shallower slope than the adiabatic case. This gives Eq. (4.9) above. Multiplying both sides by ρ/p, we have γ ρ ( ) dp < γ a. (4.10) P dρ star Figure 4.2: Schwarzschild s criterion for convective instability. Exercise: For an ideal gas with negligible radiation pressure, recall the ideal gas law P = ρ kt ρt, µm H (4.11) show that, dp P = dρ ρ + dt T (4.12) and hence P T ( ) dt < γ a 1 (4.13) dp γ star a

51 CHAPTER 4. STABILITY OF STARS 51 and dt dr ( γa 1 < star γ a ) T P dp dr. (4.14) star The above equation tells us that there is an upper limit for the temperature gradient for a convectively stable star. You may ask what happens if a portion of a star becomes convectively unstable. The most important consequence is that convection (rather than radiation) becomes the most important energy transport mechanism in that part of the star. Hence, the radiative transfer equation must be modified. Unfortunately, a comprehensive theory of convection is still lacking. This makes our analysis difficult and complex. However, if we assume that convection is extremely efficient in transporting energy, then the temperature gradient dt/dr in the convective region of a star should be very close to the critical value given by Eq. (4.14). In other words, P T ( ) dt = γ a 1. (4.15) dp γ star a Exercise: Using the hydrostatic equilibrium equation, dt dr = γ a 1 ρt Gm. (4.16) γ a P r 2

52 Chapter 5 Evolution of Main Sequence Stars 5.1 The complete set of stellar structure equations After all our detail discussions, we finally have a complete set of stellar structure equations. They are the four first order coupled ordinary differential equations: dr dm = 1 4πr 2 ρ, (5.1) dp dm = Gm 4πr, (5.2) 4 dl dm = ϵ, (5.3) 3κL if convectively stable, dt dm = 256π 2 σr 4 T 3 γ (5.4) a 1 T Gm if convectively unstable, γ a 4πr 4 P κ = κ 0 ρ a T b, (5.5) ϵ = A(X, Y, Z)ρ m T n, (5.6) P = (n e + n i )kt + 4σ 3c T 4, (5.7) where n i = ρ ( X + Y m H 4 + Z ), A 52

53 CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS 53 n e = ρ m H X + Y + Z = 1. ( X + Y ) Z 2 + Z A, and Exercise: Derive dt/dm from the radiative transfer equation (Eq. 3.3). Here, we have 12 unknowns functions of r, namely, m, ρ, P, L, ϵ, T, κ, n e, n i, X, Y and Z. Yet, we only have 10 equations. Fortunately, at a sufficiently short time interval compared with the nuclear burning timescale, we know that the chemical composition X, Y and hence also Z are almost constants. Thus, we know X and Y as well. Besides, they are roughly time independent when the time interval concerned is much shorter than the nuclear burning timescale. Therefore, we have a complete set of equations to determine the structure of a star. These equations must satisfy the following boundary conditions: r = 0 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. Unfortunately, the system of differential equations we are working with is too complicated to solve analytically. Indeed, explicitly (and numerically) solving this set of equations is the only way to accurately determine the structure of a spherically symmetric star. A mathematical-oriented student may be doubtful about the existence and uniqueness of the solution to the stellar structure equations for any boundary conditions as well as any given total stellar mass and chemical composition of the star. This is a very complicated mathematical problem. Nonetheless, theoretical astrophysicists have cooked up an extremely artificial situation that the solution of the stellar structure equations is not unique! Moreover, there are situations that the solution may not exist at all. There is a good physical argument for the non-existence of solution in some cases, too. Consider, for example, the situation when the total stellar mass is, say, 1 g. The gravitational force will be so weak that any reasonable nuclear energy generation will

54 CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS 54 blow this object up. Thus, that object is likely to be supported by EM force rather than gravity. Hence, we expect that no solution for the stellar structure equations can exist for a 1 g object. 5.2 Introduction to the numerical solution of stellar structure equation All D.E.s require boundary and/or initial conditions in order to produce a numerical solution. We have a complication for the stellar structure equations. Namely, the boundary conditions we are imposing are at two different places, namely, at m = 0 and m = M. In other words, we are not dealing with a simple initial value problem for a system of first order coupled O.D.E.s. Numerical solution of the stellar structure equations, therefore, must be obtained with great care. There are many proposed ways to numerically solve the stellar structure equations. The first way is called the shooting method. Basically, the idea is that we start from the boundary at m = 0. At this boundary, we know for sure that L = r = 0. Then, we guess the values of T = T c and P = P c at m = 0. Now, we have a well-defined initial value problem, so we numerically integrate the solution until it hits the other boundary at m = M. We check if T and P at m = M are really equal to 0. If not, we carefully make another guess at the m = 0 boundary for T and P until a solution is found. This method turns out to be not very accurate since T and P varies greatly between the interior and the surface of a star leading to a great lost of numerical accuracy. Another method is called the fitting method. Essentially, we guess the values of P, T in the inner boundary and R, L in the outer boundary. Then we numerically integrate the equations and see if they matches continuously somewhere in the middle of the star. This was a method of choice in the 1960 s and 1970 s. The most popular method to date is the so-called Newton-Raphson-Henyey method. This method is so involved that I will not discuss it in any detail in this course. I just want to give you an idea what this method is all about. Essentially, it approximates the coupled O.D.E. together with the boundary conditions by a system of equations. And then, we carefully solve the system of equations by Newton-Raphson iteration. If you want to know the detail, you may refer to Chap. 7 of C. J. Hansen and S. D. Kawaler, Stellar Interiors: Physical Principles, Structure, and Evolution. Figures below show the numerical solution of the stellar structure equations in

55 CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS 55 X L T ρ r m(r)/m Figure 5.1: A 1 M model during main-sequence hydrogen burning at time yr, showing radius, density, temperature, total luminosity, and hydrogen abundance versus mass fraction. Y L X 0 1 m/m Figure 5.2: Model of a 1 M star just after it leaves the main sequence, at time yr.

56 CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS 56 L ρ r X T P m(r)/m Figure 5.3: Model of a 5 M star just after it leaves the main sequence at time yr. X 4 L T r P ρ L X m(r)/m Figure 5.4: Model of a 5 M star during the giant star stage at time yr.

57 CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS 57 a number of situations. (Note that the y-axis in all the figures below are in linear scale and are in units of r/r, ρ/ρ c, T/T c, L/L, and X, respectively.) 5.3 Zero age main sequence star and its evolution A zero age main sequence star (ZAMSS) is defined as a star which just begins its hydrogen burning. In other words, ZAMSS is a new born star. The location of ZAMSSs in the H-R diagram is known as the zero age main sequence (ZAMS). Since a star is formed from the gradual collapse of a nebula, a spherically symmetric non-rotating ZAMSS is characterized by only two sets of parameters, namely, its total mass (M) and its chemical composition (X, Y, and Z). Note that the gradual collapse of the nebula should give enough time for the stellar matter to mix. Hence, X, Y, and Z should be uniform over the entire ZAMSS. By numerically solving the stellar structure equations above, we know that the structure of a ZAMSS can be summarized below: if the mass M is below about 0.1 M, thermal hydrogen fusion does not occur in the core and hence we have either a brown dwarf or a planet. We may talk about more about them in the special topics in this course. If the mass M is between about 0.1 to 0.3 M, the entire star is convective and p-p chain is the major nuclear reaction occurring in the stellar core. When M is between about 0.3 M and 1.2 M, p-p chain is the dominant reaction. The core is radiative and the surface layer is highly convective. When M > 1.2 M, 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 30 M for two reasons. First, its core will be dominated by radiation pressure which leads to instability. Second, the surface luminosity will be so high that mass ejection is serious. The above results from ZAMSS can be understood as follows. First, although the energy production rate for CNO cycle scales like T 16 while that of the p-p chain scales like T 4, the corresponding coefficients (which we both call them A in this note) for CNO cycle is much smaller than that of the p-p chain. (Do you know why?) Hence, at temperature < T crit, the p-p chain is the dominant reaction. This is the reason why CNO cycle is only important for ZAMSS with mass > 1.2M.

58 CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS 58 Second, the sensitive temperature dependence of CNO cycle (namely, T 16 ) is the origin of a convective core for high mass ZAMSS. For convective stability, Eq. (5.4) gives d 2 T 3κ = dm 2 256π 2 σr 4 T 3 [ dl dm + L d ( ) ] 1 dm r 4 T 3 3κ dl, when r 0, L 0 256π 2 σr 4 T 3 dm n 3 3κAρT 256π 2 σr. (5.8) 4 That is, for regions sufficiently close to the stellar core, dt/dm increases as n increases. It turns out that for p-p chain, n = 4 and the core is radiative. While for CNO cycle, n = 16 4, the temperature gradient is so high that the core becomes convective. Third, when M < 1.2 M, the temperature close to stellar surface is so low that free-free absorption (hence, Kramer s opacity) is important. Such a large opacity makes the stellar surface a good photon absorbing medium. Hence, the only effective way to transfer these absorbed energy out to the stellar surface is convection. In contrast, when M > 1.2 M, the temperature close to stellar surface is high enough that electron scattering opacity becomes important. Hence, in this case, the stellar surface is radiative. Finally, when 0.1 M < M < 0.3 M, temperature in most part of the star is so low that Kramer s opacity is important. Hence, the whole star is fully convective. 5.4 Homology relation Let us consider our stellar structure equations supplemented by the simplified opacity, energy production equations as well as the equation of state. To illustrate the idea, we only consider the case of (a) the opacity κ is temperature and pressure independent; (b) the energy production equation is a power law of temperature and density; and (c) the EOS is just the simple ideal gas law. These assumptions hold to give good approximation if most part of the (main sequence) star is radiative, the temperature in most part of the star is hot enough so that electron scattering is the dominant contribution to opacity, and the mass of the star is less than about 10M so that contribution of radiation pressure is small. With these assumptions in mind, we write our stellar

59 CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS 59 structure equations below: dp dm = Gm 4πr, 4 (5.9) dr dm = 1 4πr 2 ρ, (5.10) dt dm = 3κL 256π 2 r 2 σr 2 T, 3 (5.11) dl dm = AρT n, and (5.12) P = ρkt. µm H (5.13) Now, we define a dimensionless variable x by x = m/m. We denote r = f 1 (x)r, (5.14) P = f 2 (x)p, (5.15) ρ = f 3 (x)ρ, (5.16) T = f 4 (x)t, and (5.17) L = f 5 (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 df 2 M dx = GMx 4πf1 4 R 4. (5.19) In a physical equation the dimensions on the two sides must match, and hence in Eq. (5.19), where x, f 1, and f 2 are dimensionless, P must be proportional to GM 2 /R 4. Without loss of generality, we can take the proportionality constant to be unity. The equation above is then separated into df 2 dx = x 4πf1 4 and P GM 2. (5.20) R 4

60 CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS 60 Similarly, the above five equations can be decoupled into two sets. The first set consists of four differential equations and one algebraic equation involving f 1 (x) through f 5 (x) only: df 2 dx = x, (5.21) 4πf1 4 df 1 dx = 1, (5.22) 4πf1 2 f 3 f 2 = f 3 f 4, (5.23) df 4 dx = 3f 5, and (5.24) 4f4 3 (4πf1 2 ) 2 df 5 dx = f 3f n 4. (5.25) The second set of equations are algebraic equations relating ρ, T, R, P, L and M: P GM 2 /R 4, (5.26) ρ M/R 3, (5.27) T µp /ρ, (5.28) L T 4 R 4 /M, and (5.29) L ρ T n M. (5.30) The importance of this re-expression of the set of stellar structure equations is that the values of ρ, T, R, P and L can be expressed in terms of a single parameter M. Moreover, once we solve the set of equations that relates the f 1 through f 5, we solve a whole family of stellar structure equations for all values of stellar mass M. This similarity property is known as homology. Exercise: Eliminating ρ, P, T, and R from the above equations to show that L M 3 (5.31) This mass luminosity relation agrees reasonably well with observations (see figure 1.3.) A few more relations can be deduced as well. For example, R M n 1 n+3, (5.32)

61 CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS (n 1) ρ M 2(3 n) n+3, (5.33) 3(n+3) L Teff 4, and (5.34) T µm/r, (5.35) where L = 4πR 2 σt 4 eff and µ is the mean molecular weight. Since n is about 4 for p-p chain and 16 for CNO cycle, we know that for main sequence stars, the radius increases with mass, mean density ρ decreases with mass. For p-p chain burning stars, Eq. (5.34) gives and for CNO cycle burning stars, log L 5.6 log T eff + constant, (5.36) log L 8.4 log T eff + constant. (5.37) Again, these slopes are in rough agreement with the observation in the H-R diagram. Besides, the characteristic life time for a main sequence star can be estimated by τ MS = M/L M 2. Thus, heavier main sequence star lives shorter time on the main sequence. For low mass stars, constant opacity is no longer a good approximation, instead, the Kramers opacity law holds. In this case, it can be shown that L M 5+31/2n 1+5/2n M 5.46 (5.38) This results in a change of slope between high and low mass stars in the massluminosity relation (see Figure 1.3). Eq. (5.38) also gives log L 4.12 log T eff + 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 is much shorter than the nuclear burning timescale, we know that the change in X, Y, and Z and hence also the change in stellar structure can be neglected. This explains why main sequence star is so stable throughout most of its life. As time goes, we have to supplement the stellar structure equations by dx(r) dt ϵ(r) E, (5.40)

62 CHAPTER 5. EVOLUTION OF MAIN SEQUENCE STARS 62 dy (r) dt dz(r) dt ϵ(r) E, and (5.41) = 0 (5.42) to the evolution of a hydrogen burning star, where E is the energy released per helium nucleus formed. The effects of the change in mass fractions described by the above equations can be summarized below: First, since hydrogen is burned only in the core, an inhomogeneous distribution of chemical elements in the star will eventually appear. Namely, the core becomes more and more hydrogen depleted while the surface is still hydrogen rich. (More precisely, this statement is only true for main sequence stars with mass higher than about 0.2 M in which the effects of convection, diffusion and flow are not important in changing the distribution of particles inside them during its main sequence life.) Can you justify this statement? In addition, do you know why this statement is wrong for a 0.1 M main sequence star? Second, the star must evolve due to such a gradual chemical change. Since its core will be more and more hydrogen depleted and helium rich, hydrogen burning will eventually be unable to provide the gas and radiation pressure gradient, dp/dr, to counter balance gravity, namely Gmρ/r 2 near the core. At that time, the core will contract. Third, recall from the consequences of the virial theorem that for a sufficient small timescale, both the thermal and gravitational potential energies of a star are approximately conserved. So, as the core of the star contracts, we expect the outer part of the star to expand. And since the temperature around the core of the star increases due to core contraction, we also expect the temperature of the outer part of the star to decrease. So, the star is heading towards the red giant phase.

63 Chapter 6 Evolution of Pre-Main Sequence Stars 6.1 Collapse of interstellar cloud The interstellar medium is not empty, but consists of gas, dust, clouds, and plasma. The densest part is interstellar clouds. If the mass is high enough, the interior can be shielded from UV radiation from stars, such that molecules could exist. They are called molecular clouds and most of them contain H 2 and CO. A star is formed from the collapse of molecular clouds. They can have high masses of 10 3 to 10 7 M and the typical temperature ranges from 1 K (mostly dark nebula) to 10 4 K (mostly bright emission nebula). Its density is about g cm 3. Clearly the EOS of a cloud can be well approximated by the ideal gas law. If the thermal pressure in a gas cloud cannot balance the gravity, part the cloud will collapse under external perturbation (e.g., triggered by supernova shock wave). From the virial theorem, a cloud will collapse if Ω > 2U. Assuming a spherical cloud of uniform temperature and density, the gravitational potential energy is Ω = 3 GM 2 (6.1) 5 R and the thermal energy is U = 3 2 NkT = 3 2 kt M µm H. (6.2) 63

64 CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS 64 Hence, the condition to collapse is 3 GM 2 5 R > 3MkT µm H. (6.3) Using M = 4πρR 3 /3, we can rewrite the above equation in termal of ρ to obtain M > M J, (6.4) where M J is called Jeans mass. It is the minimum mass of a cloud that can collapse (i.e. upper limit for a stable cloud). Putting in the typical numbers, M J 10 5 T 3/2 M / n. (6.5) Thus, only giant gas cloud of mass of order of 10 5 M or more will collapse when perturbed. Exercise: Similar to above, we can express the equation in terms of R and ρ, such that the Jeans length is given by R J = ( ) 1/2 15kT. (6.6) 4πρµm H G As the cloud collapse, its Jeans mass will change since T and n are changed. If the cloud cools efficiently, the increase in n will lower the Jeans mass. Hence, the stability condition in part of the cloud will be violated, resulting in the fragmentation. This mechanism is believed to be the way for a large massive gas cloud to collapse into a number of solar mass stars, which forms a stellar cluster. As the cloud fragments get denser and hotter, they will eventually become opaque and cooling becomes inefficient. The rise in temperature will increase the Jeans mass. Once M J exceeds the physical mass of the cloud, fragmentation will stop. We can estimate when does this happen. The power released from gravitational collapse can be estimated by Ω Ω GM 2 ( ) 3 1/2 G 3/2 M 5/2 Gρ, (6.7) τ ff R 4π R 5/2

65 CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS 65 where τ ff R 3 /GM 1/ Gρ is the free-fall timescale (see Chapter 2.6). This energy is carried away via blackbody radiation. For efficient cooling, This gives M J,min 0.02M T 1/4 µ 9/4. Ω < U = 4πR 2 σt 4. (6.8) We expect the collapsing process to take place within the free-fall timescale. As ρ is very low for a gas cloud, τ ff can be as long as a few million years. The actual collapse is much longer, because rotation and magnetic field while neglected is this course play an important role here. The conservation of angular momentum and magnetic pressure slow down the collapse. 6.2 Protostars and pre-main sequence stars The initial temperature of a cloud is only a few 10 to 100 K. It has a low opacity (i.e. mostly transparent to photons) since κ 0.01 g 1 cm 2 and the mean density ρ (3M /4πR 3 ) g cm 3, giving an optical depth τ κρr 1. The gravitational potential energy is radiated away efficiently and the temperature of the cloud stays as a constant. The infalling matter gradually increases the opacity and hence the temperature. When the core temperature reaches 2000 K, molecular hydrogen disintegrates, then followed by hydrogen ionization and later on helium ionization. At a central temperature of K, hydrostatic equilibrium is achieved and a protostar is formed. Numerical simulations suggest that the core has a mass of M and a radius of 1.3R at this stage. A protostar is powered by gravitational energy and it is extremely luminous, can be L for a 1M star, with surface temperature 0.6 times of the Sun, and a large radius of 70R. However, protostars are embedded in clouds that make observations difficult. (How to observe a protostar?) Detailed structure of protostars from numerical simulations can be found in Lasson, R.B. (1969) MNRAS, 145, 271. The protostar will continue to accrete, until the surrounding material is depleted. Once the infall and accretion cease, mass of the star is nearly fixed and the protostar is now known as a pre-main sequence (PMS) star. In this stage, radiation can escape, hence, the PMS star cools down and contract. The energy source is still gravitational contraction, but the evolution timescale is governed by the Kelvin-Helmholtz timescale τ KH, as compared to τ ff for protostars. Note that τ KH 1/R, meaning that it will increase during contraction, while τ ff 1/ρ decreases during contraction. (How to observe a PMS star?)

66 CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS 66 Exercise: Estimate the size and temperature of a protostar by assuming that all gravitational energy released is used to dissociate molecular hydrogen and to ionize hydrogen and helium. (Energy to dissociate H 2 is χ H2 = 4.5 ev, ionization energy of hydrogen and helium are χ H = 13.6 ev and χ He =24.6 ev ev=79 ev, respectively.) Ω = 3 GM 2 5 R and take Y 1 X, we obtain M m H ( X 2 χ H2 + Xχ H + Y 4 χ He) (6.9) R 50 M R. (6.10) 1 0.2X M As hydrostatic equilibrium is established, the temperature can be estimated from the virial theorem 3 kt M = 1 2 µm H 2 Ω, (6.11) which gives T K. 6.3 Hayashi (convective) track The core temperature of a PMS star is much higher than the surface temperature due to the large opacity of hydrogen. This results in a large temperature gradient, making the entire PMS star (up to the photosphere) fully convective. As the radius continues to shrink, the core heats up. However, the surface temperature remains stable, because the photosphere can adjust its thickness to counteract the rise of the core temperature. As a result, the luminosity decreases as R 2, and the star moves down nearly vertically on the H-R diagram over the first million years, this is known as the Hayashi track. The pressure is related to the density by P = Kρ 1+1/n, (6.12) where n = (γ a 1) 1. In other words, we are solving a polytropic EOS with index n. Hence, from Eq. (2.24) the coefficient K is given by where the coefficient C n depends on n only. K n = C n ( ). (6.13)

67 CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS 67 The value of R is determined by joining the fully convective interior to the radiative photosphere at the boundary r = R. Quasi-hydrostatic equilibrium condition requires dp dr GMρ. (6.14) R 2 Hence, P R GM R 2 + R ρdr. (6.15) In the photosphere, the optical depth of the photosphere is of order of 1, + κ ρdr 1 (6.16) R where κ is the mean opacity of the photosphere. Taking κ to be the opacity at R and approximate it as a power-law of ρ and the surface effective temperature T eff, we have Hence, κ 0 ρ a RT b eff + R ρdr 1. (6.17) P R GM R 2 κ 0 ρ a R T b eff. (6.18) Exercise: Using Eqs. (6.12), (6.13), (6.18), the ideal gas law, and the blackbody luminosity, show that log P R = log M 2 log R a log ρ R b log T eff + constant n log P R = (n 1) log M + (3 n) log R + (n + 1) log ρ R + constant log P R = log ρ R + log T eff + constant log L = 2 log R + 4 log T eff + constant We can write log L A log T eff + B log M + constant. (6.19)

68 CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS 68 Figure 6.1: Evolution paths of pre-main sequence stars from numerical simulations. The nearly vertical lines are the Hayashi tracks, then followed by the nearly horizontal Henyey tracks. Using the results above, we found A = (7 n)(a + 1) 4 a + b 0.5(3 n)(a + 1) 1 and (n 1)(a + 1) + 1 B = 0.5(3 n)(a + 1) 1. For realistic values of n = 1.5, a 1 and b = 4, the slope A = 20. This represents a very steep evolutionary path on the H-R diagram, i.e. a PMS star evolves by reducing its luminosity with more or less constant effective surface temperature (see figure 6.1). The Hayashi track also sets the boundary for which a stable star can exist. The region on the H-R diagram to the right of the line is called the forbidden zone

69 CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS 69 such that no stars can be in hydrostatic equilibrium. Finally, note that a real PMS star does not evolve on the H-R diagram exactly on the Hayashi track as the assumptions we used in the arguments above is overly simplistic. However, the Hayashi track is still reasonably close to reality. The relevant timescale at this stage is the Kelvin-Helmholtz timescale which is very short. Contraction to main sequence stars takes up < 1% of the life time of a star, while main sequence stage takes up 80%. This is part of the reasons why so few pre-main sequence stars are detected. One example of PMS star is T Tauri stars, which are less massive stars that evolve more slowly along the Hayashi track, and they are highly variable due to mass ejection. 6.4 Henyey (radiative) track Upon further contraction, the internal temperature rises to the point that opacity greatly decreases (Eq. 3.5: κ ff T 7/2 ). The core is in radiative equilibrium. The convective zone recedes from the center, leaving only a convective envelope. The core of the star contracts on the free-falling timescale, the gravitational energy becomes more negative. By the virial theorem, the internal energy, and hence the surface temperature, has to increase rapidly, while the luminosity increases gradually. Therefore, the PMS star will move away from the Hayashi track on the H-R diagram, making a sharp turn to the upper left. This is known as the Henyey track. For high mass stars, this can happen very early. For example, the central temperature of stars above 10M can reach K in 1000 years, then CNO is ignited (but not in equilibrium) and a radiative core is formed due to the high temperature. Therefore, very massive stars almost left the Hayashi track and on the Henyey track instantaneously. On the other hand, stars < 0.5M remain in the convective phase throughout their lives and there is no Heyney track for them. Most light elements are burned off in this stage (see Chapter 3.7.3), except deuterium, which is ignited while the star is still on the Hayashi track), but the energy production is negligible compared to the gravitational energy. Finally, the core temperature is high enough that stable hydrogen burning (p-p chain or CNO cycle) gradually take over the energy generation and gravitational collapse slows down. The star becomes a zero age main sequence (ZAMS) star. It takes solar-type stars yr for the protostar and PMS phases, which is only 1% of the entire life time. Massive stars above 10M can take less than 10 5 yr and low-mass stars can spend more than 10 8 yr before hydrogen burning reaches equilibrium. You may notice that in Figure 6.1 there are some kinks near the end of the

70 CHAPTER 6. EVOLUTION OF PRE-MAIN SEQUENCE STARS 70 Heyney track before the ZAMS. This is due to the depletion of 12 C in the core such that the core has to slightly contract to adjust the rate of hydrogen burning. If you are interested to learn more, can refer to a simulation paper by Icko Iben Jr. (1965) ApJ, 141, 993, or the book by the same author: Stellar Evolution Physics, Volume 1 (2013) (Cambridge: CUP).

71 Chapter 7 Evolution of Post-Main Sequence Stars 7.1 Red giants As the core temperature keep on increasing due to contraction, the thin layer of hydrogen-rich material located just on top of the helium-rich core will be heated up. Eventually, this thin hydrogen shell will burn. This shell burning becomes the main source of energy for the star and the star becomes a red giant. Let s consider an idealized situation in which the core is made up of entirely helium. This core is probably be very close to isothermal. (Do you know why?) Nevertheless, an arbitrarily massive isothermal core is not stable. To see why, we use the local form of the virial theorem. Recall from Eq. (2.8) that Vcore 0 P dv = P core V core + αgm 2 core 3R core = 4πP corer 3 core 3 + αgm 2 core 3R core, (7.1) where the label core refers to the boundary of the core (i.e., r = R core ), and α is a geometric factor depending on the mass distribution of the core. For ideal isothermal classical non-relativistic gas with temperature T c = T core, Vcore P dv = kt c ρ dv = kt cm core, (7.2) 0 µ core m H µ core m H where µ core is the mean molecular weight of the material within a distance of R core from the stellar core. Hence, P core (R core ) = 3kT cm core 4πµ core m H R 3 core αgm 2 core 4πR 4 core. (7.3) 71

72 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 72 By considering dp core /dr core = 0, we know that for a fixed core mass M core, P core attains its maximum value of P core,max Tc 4 /Mcoreµ 2 4 core. For a stable core, this maximum pressure must be balanced by the pressure P env exerted by the envelope. By assuming that the core is very small, P env can be estimated by Eq. (2.5), namely, P env GM 2 /8πR 4. Hence, the condition for stability of an isothermal classical non-relativistic core is P core,max GM 2 /8πR 4. Now, applying one of the homology relations in Eq. (5.33), namely, T c µ env m H GM/kR, the stability condition becomes M core M < constant ( µenv µ core ) 2. (7.4) Actually, M. Schönberg and S. Chandrasekhar deduced this result with the constant equals 0.37 in For a pure helium core, µ core = 4/3. And for a solar composition envelope, µ env 0.6. This implies the maximum mass of a stable isothermal nondegenerate and non-burning helium core is about 0.13 M. Now, let s consider the case of a high mass star (say of mass > 2M ). One of the homology relations in Eq. (5.33) shows that the central density ρ c of a main sequence star decreases as its mass increases. So, for high mass star, the central density is not very high so that the stellar core is still non-degenerate even after it has exhausted its hydrogen fuel. The mass of this isothermal helium core will gradually increase as hydrogen is burnt in the shell. As the core mass exceeds the so-called Schönberg-Chandrasekhar limit, the core will collapse. In time, the core will acquires the temperature gradient necessary to balance gravity. But this temperature gradient causes loss of heat and hence the core will further collapse leading to further temperature rise. This core contraction process will take place in the Kelvin-Helmholtz timescale, which is about GMcore/R 2 core L core 10 6 yr. Such a contraction can be approximated by a quasi-static equilibrium process. Besides, we expect that most of the gravitational potential energy released by the contracting core will be absorbed by the outer layer of the star. Thus, as the core contracts and heats up, the outer layer will expand and cool down. Apart from the virial theorem, this is an alternative way to understand how a red giant is formed. Since the evolution from high mass main sequence to red giant is very rapid in stellar evolution timescale, very few stars are found in the H-R diagram immediately outside the main sequence. This gap is called the Hertzsprung gap (see figure 7.1.) There is an interesting consequence of the evolution towards red giants for high

73 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 73 Luminosity A B C Temperature Figure 7.1: A schematic illustration of post-main sequence evolution for a 5 M star on H-R diagram (the light solid curve). The heavy solid curve shows the main sequence. Stars spend very short time when evolve from point B to point C. Only a few of the stars are found in this region, known as the Hertzsprung gap. mass stars. It turns out that the convective zone of a high mass main sequence star is rather extended. So, by the time when the hydrogen burning shell is formed and the outer layer of the high mass red giant becomes convective, this convective region overlaps with the original main sequence convective zone. Hence, some material originally located deep inside the stellar core will be brought up to the stellar surface. This process is called the dredge-up (or the first dredge-up since high mass star will undergo several dredge-ups in its life and this is the first one). Do you know what kind of observational evidence support for this idea of dredge-up? To summarize, within about 10 6 yr, a red giant star is formed. A thin layer of spherical shell immediately outside the core now is hot enough to burn hydrogen. We call it the hydrogen burning shell. The core during this about 10 6 yr will contribute very little to the nuclear energy generation. The situation is slightly different in the case of low mass stars (with mass < 2M ). Again one of the homology relations in Eq. (5.33) tells us that the central density of this type of main sequence stars is rather high. In fact, it is so high that the electron there is close to degenerate. Hence, the Schönberg-

74 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 74 Chandrasekhar limit no longer applies. As the core contracts after it has used up its hydrogen fuel, it will quickly go degenerate. Provided that the temperature is much lower than the Fermi energy, the equation of state of the helium core can be well-approximated by an electron Fermi gas at almost zero temperature. Recall from Eq. 2.77, the EOS is a polytrope of index n = 1.5. Exercise: From Eqs. (2.23) and (2.24) concerning the mass, radius and central density of a polytropic star, show that the mass and radius of the degenerate helium core obey M core Rcore 3. (7.5) Most importantly, it turns out that the electron degenerate pressure of the helium core is sufficient to support the stellar envelope. As a result, for low mass stars, the core contraction is slow and hence its road to red giants is gradual. Thus, there is no Hertzsprung gap in the lower bottom part of the H- R diagram. Note also that there is no dredge-up at this phase of the evolution of low mass stars except for the fully convective extremely low mass stars. In both the low and high mass star cases, the onset of hydrogen shell burning (and before the helium shell burning that we are going to discuss), the stars are located at the so-called red giant branch in the H-R diagram characterized by the high luminosity and low surface temperature. The structure of a typical red giant star is shown in figure 7.2. The exception is when the mass of the star is less than about 0.5M. In this case, the expansion of the radius of the star is so small that it may not be able to reach the red giant branch at all. From now on, our highly simplified analysis of the stellar structure equations is no longer valid. In fact, the most direct way to answer the question of what the structure is and what evolution of a red giant star take places is by numerical solution. Such analysis has been carried out since the 1960 s by various groups of astrophysicists, and perhaps the most famous one was headed by I. Iben, Jr.

75 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 75 Hydrogen burning shell Helium Core Hydrogen riched zone Figure 7.2: The structure of a typical red giant star. As the star expands, the surface gravity is low and hence material ejection in the form of stellar wind can be very significant. Hence, from this time on, mass conservation of the whole star is no longer a good approximation. This further makes the modeling of stellar evolution for red giant and beyond difficult. Our description of subsequent evolution of a star is, therefore, forced to be rather descriptive at the level of this course. To add further complications, it turns out that the precise subsequent evolution of a star depends on the mass of the star. (Not just the initial mass of the star, but how much mass is lost by stellar wind.) 7.2 Helium core burning and helium flash As the core collapses further, the central temperature and pressure will be high enough to burn helium via the triple-alpha process. For stars with mass greater than about 2 M so that the stellar cores are non-degenerate, the ignition of triple-alpha process is relatively smooth. That is, when the central temperature reaches the helium ignition temperature, which is about 10 8 K, helium will burn. After that, the contribution of energy production of the entire star due to helium burning will increase steadily with time while that due to hydrogen burning will decrease steadily (although it will not stop completely). Due to the high temperature sensitivity of helium burning (ϵ T 40 ), the core of such star is convective.

76 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 76 The situation is slightly more complicated if the main sequence star is less than about 2 M (but higher than about 0.5 M ). Recall that the core of this type of stars is degenerate. Unlike ideal gas law, pressure of degenerate Fermi matter (in this case electrons) at low temperature is almost independent of temperature T. Hence, for low mass star, electrons in the helium-rich core becomes strongly degenerate before helium burning sets in. We expect that the degenerate helium core of a low mass star is much more massive than its envelope. So, homology relation (Eq. 5.33) holds approximately for the stellar core. In particular, we have T core M core /R core. Combining with Eq. (7.5), we conclude that low mass stars ignite core helium burning whenever their core masses reach a threshold irrespective of their envelope masses. Detail calculation shows that this threshold value is about 0.5 M. Since the degenerate matter is highly conductive, the degenerate helium core is highly isothermal. So, the entire core burns simultaneously once the helium is ignited. Furthermore, the weak dependence of pressure of degenerate matter on temperature implies that the rise in core temperature will not affect the core pressure too much. Thus, the helium burning rate of the core will increase quickly leading to a thermonuclear runaway. So, helium ignites in an explosion, known as helium flash, in low mass stars. Actually, the peak luminosity (which lasts for a few seconds at most) is about L which equals the luminosity of the entire galaxy! Nevertheless, we cannot see helium flash directly from the EM radiation emitted from the star as most of the energy released in this sudden explosion is absorbed by the stellar envelope. (Nevertheless, we can infer its onset in principle from the evolutionary track. Do you know how?) A few seconds after the flash, the core temperature is hot enough that it can be well approximated by an ideal gas once more. Thus, the core expands and helium burning becomes stable. Finally, for extremely low mass stars with mass < 0.5 M, their degenerate helium cores are never hot enough to ignite helium burning. The degenerate cores of these stars will keep on contracting, evolving directly from red giants (if you can still can them giants) to helium white dwarfs. No matter the star starts to burn core helium gradually or in a flash, the central helium burning leads to an increase in the nuclear energy generation rate. Consequently, the core expands and hence the envelope contracts. And the star becomes smaller, less luminous and its surface becomes hotter. (Do you know why?) So, the star descends from the red giant branch by moving to the lower or lower-left in the H-R diagram. The locus of high-mass core helium burning star forms the so-called the helium

77 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 77 main sequence a somewhat misleading term as this type of stars have hydrogen burning shells which still significantly contribute to nuclear energy generation. In theory, the helium main sequence is a band similar to the high mass part of the (hydrogen) main sequence except that it is shifted to the right in the H-R diagram. In practice, the helium main sequence is mixed up with the thick red giant branch. Computer simulations show that for high mass stars, the evolutionary track from red giant to helium main sequence is roughly horizontally to the left in the H-R diagram. (Interestingly, when the helium in the core is used up at a later time, the star will again move roughly horizontally to the right in the H-R diagram. That is to say, the combined evolution make the star to form a loop in the H-R diagram.) Some stars on the helium main sequence are Cepheid pulsating variables whose period is of order of days to months. The locus of low-mass core helium burning star forms the so-called horizontal branch, a roughly horizontal strip stretching between the main sequence and the red giant branch. (See figure 7.3 below.) Clearly, these stars have about the same core masses just before core helium ignition. So, it is not surprising that the location of a star on the horizontal branch is due to the envelope mass. Some stars on the horizontal branch are pulsating variables, known as RR Lyrae variables. Periods of RR Lyrae variables, which are about a few hours, are much shorter than those of Cepheid variables. If time permits, I will talk about pulsating variables in the special topics. The lifetime of a core helium burning star, irrespective of its mass, is much shorter than the corresponding core hydrogen burning main sequence star. It is because the energy released per nucleon burnt is lower for helium. In addition, the luminosity of a core helium burning star is much higher than a main sequence star of the same mass. 7.3 AGB star Evolution of a core helium burning star is in some sense quite similar to that of a core hydrogen burning main sequence star. More precisely, a carbon-oxygenrich core is gradually formed. Then the star further expands. Most of the energy production now occurs in the helium burning shell and the hydrogen burning shell on top of the core. Once more, material from the core is dredged up and mixed into the envelope. If the carbon oxygen core is degenerate, the star is called an asymptotic giant branch (AGB) star (see figure 7.4.) A star may leave the AGB phase by either complete removal of hydrogen envelope by stellar wind; or ignition of carbon in the core.

78 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS A B Red Giant Luminosity L /L 10 Horizontal Branch Main Sequence Effective Temperature T eff (K) Figure 7.3: A schematic H-R diagram showing the evolution of low-mass stars from red giant branch to horizontal branch. Point A corresponds to large loss of mass in helium flash while point B corresponds to small loss of mass. The two burning shells of an AGB star are separated by a helium-rich nonburning layer. The outer hydrogen burning shell burns hydrogen thereby increasing the mass of the helium non-burning layer. Similarly, the helium burning shell consumes the helium and hence eating into the helium non-burning layer. Due to the great difference between rates of hydrogen and helium burning processes, the two burning fronts do not advance at the same rate all the time. The two shells supply energy in turn and the mass of the helium non-burning layer changes periodically. During most time in the cycle, hydrogen is burnt in the external shell while the helium burning shell is extinct. Thus, the helium layer separating the two shells increases in mass. The gradual contraction of this helium layer increases the temperature of the helium shell until it ignites. Such ignition may be in the form of a shell flash for low mass star. As the helium shell burns, the outer layer of hydrogen burning shell expands and cools down. This leads to a great suppression of the hydrogen burning rate. As the helium burning front advances, the temperature of the hydrogen burning shell increases while that of the helium burning shell decreases. Thus, a new cycle of hydrogen and helium shell burning begins. To summarize, the AGB star is

79 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 79 H riched zone H burning shell CO riched core He burning shell He riched zone Figure 7.4: The typical structure of an AGB star. burning hydrogen most of the time while only it is burning helium occasionally. In fact, astronomers sometimes called the helium-shell-burning phase of an AGB star the thermal pulse. This cycle of hydrogen and helium ignition with thermal pulse lead to sudden increase in radiation pressure in the interior of an AGB star. It leads to a rapid mass loss in the star in the form of stellar wind. (Its mass loss rate is of order of 10 5 M yr 1.) A planetary nebula is formed eventually. Detail study of planetary nebulae will be studied as a special topics if time permits. An AGB star will undergo up to a few thermal pulses in its life. Each cycle of hydrogen- and helium-shell burning also leads to dredge-up of materials. (There is a confusion of naming here as astronomers studying AGB stars also called the first time when materials in the core of an AGB star is transported to the stellar surface the first dredge-up. So, be careful when you read the literature.) As expected, AGB stars are variables. In fact, a lot of long period pulsating variables, such as Mira variables, are AGB stars. While most of the outer layer of an AGB star will eventually be ejected forming a planetary nebula, the core of an AGB star will collapse after using up all its hydrogen and helium fuel. The mass of an AGB star is not high enough so that even when the electrons in the core become degenerate, the core temperature is not high enough to ignite carbon burning. The core now becomes a dead star, whose gravitational force is counter-balanced by electron degenerate pressure. This star is called a white dwarf. Properties of white dwarf will be discussed in other astrophysics courses. Note that only a very small portion of stars are massive enough to have a non-degenerate C-O core while do not undergo supernova explosion.

80 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS Planetary Nebula AGB Star red supergiant Red Giant Luminosity L/L Horizontal Branch Main Sequence Star white dwarf Effective Temperature T eff (K) 2500 Figure 7.5: An illustration of evolutionary path of a low mass star from main sequence star to white dwarf in H-R diagram. Note that the path from AGB star via planetary nebula to white dwarf is not fixed. 7.4 Evolution of high mass star after the formation of carbon-oxygen-rich core The evolution of a high mass star up to the formation of a carbon-oxygen-rich core is similar to that of a 3 M star. But there are small differences. Due to its high mass, energy generation rate in the core is so high that radiation pressure is important even in the main sequence phase. This leads to mass loss by stellar wind when the initial mass of the star is greater than about 30 M. In fact, the stellar wind is so strong in some of these stars that the entire hydrogen envelope is blown away leaving the helium-rich core. Such stars luminous, hydrogen depleted, and with high mass loss rate are known as Wolf-Rayet stars. Besides, the radiation pressure in the core is so strong that the luminosity in

81 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 81 the main sequence phase is already close to the Eddington limit. Hence, the luminosity remain more or less constant in spite of internal changes. Thus, its evolutionary track in the H-R diagram is more or less horizontal. After the formation and collapse of a carbon-oxygen-rich core, the central temperature is hot enough to burn carbon and later on oxygen. At this stage, the core temperature is so high that energy lost due to neutrino emission becomes significant. In fact, all major core nuclear burning processes take place in rapid succession until the inner core is 56 Fe rich. Surrounding the core are shells that burn different types of nuclei. Finally, the star is so massive that its core remains non-degenerate until the very last stage of its evolution. (Do you know why?) 7.5 Supernova The theory of supernova explosion is rather involved. Here, I shall only outline the basic idea behind. As 56 Fe is the most energetically favorable nucleus, no major exothermic nuclear reaction can take place in the 56 Fe-rich core. When this core contracts, electron in the core eventually becomes degenerate. Unfortunately, since the star is massive, the core mass is so high that degenerate electrons become relativistic. As discussed in Section 2.12, the pressure scales like ρ 4/3 rather than ρ 5/3. That is, the core can be approximated by a polytropic EOS with index n = 3. But for a polytropic stellar core, Rcore 3 n Mcore 1 n. Hence, as n tends to 3, there is only one possible solution for M core. Thus, the core is unstable upon addition of mass. (One way to remember this fact is that dp/dρ is not big enough to counterbalance gravity as ρ increases.) Since the pressure of the relativistic electron degenerate core is not strong enough to oppose gravity, the core further collapses. Hence, the core is further heated up. The EOS of the degenerate core matter is insensitive to temperature making the heat up more unrestrained. Now, the temperature is so hot that photodisintegration of iron back to 4 He and then further back to protons and neutrons becomes important. This endothermic reaction quickly absorbs the heat from the core and hence reduces the central pressure. With a sudden decrease in central pressure, the core contracts with an even higher rate. (In fact, at almost the free-fall rate.) Even worse, the core becomes so dense that protons can capture electrons and convert into neutrons plus neutrinos. This reaction is also endothermic. Besides, it reduces the number of particles in the

82 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 82 core. The combined effect is to reduce the core pressure further. Finally, the neutron gas in the core becomes degenerate and the neutron degenerate pressure is sufficient, at least temporarily, to stop the collapse. Such neutron core is about 10 km in radius. The gravitational energy released by the contracting core is of the order of GM 2 core/r core to erg. The energy absorbed during the disintegration of nuclei is only about erg which is about one tenth to one hundredth that of the total gravitational energy released. Thus, most of the gravitational energy released is available to eject the material outside the core and for producing the huge luminosity and neutrino flux we see during a type II supernova explosion (also called core-collapse supernova). Recall that in the final phase of the collapse, a large number of neutrinos is produced in the core within a short period of time. These neutrinos are responsible for transferring the released gravitational potential energy out from the core. Although neutrino interacts very weakly with matter under most circumstances, a non-negligible neutrino opacity builds up in the core because of the high density and high neutrino flux. Hence, some of the neutrino energy is absorbed by the envelope. Consequently, the envelope of the star quickly expands. This explosion is known as type II supernova (or simply supernova when no confusion is possible). (In contrast, type I supernova is the explosion that destroy a white dwarf as it accrues too much material from its surrounding. Observationally, the spectrum of a type II supernova contains hydrogen lines while that of type I supernova does not. Do you know why?) When observed far away from this dying star, the luminosity of the star suddenly goes up. At first, most radiation is in the UV range as the effective surface temperature is high. But the surface temperature quickly drops and the object radiates mainly in the optical range. The later time optical emission is then powered by radioactive decay of newly formed elements. Typical light curve of a type II supernova is shown in figure 7.6 below. There is an important prediction in this model, namely, that we should see the neutrino arriving the Earth faster than the brightening up of the supernova. This prediction was confirmed in the famous supernova explosion SN1987A. Actually, neutrinos were observed a few hours before the supernova became visible for SN1987A. Heavy elements in the envelope are ejected into the interstellar space in supernova explosion. These elements are produced in two ways. They may be produced in the core and shell burning prior to supernova. Alternatively, they may be formed during the supernova. Since the peak temperature in a super-

83 CHAPTER 7. EVOLUTION OF POST-MAIN SEQUENCE STARS 83 0 Blue Magnitude Below Maximum Light Days after Maximum Light Figure 7.6: A typical light curve of a type II supernova. nova explosion can be as high as K locally, nuclear statistical equilibrium is achieved on a timescale of seconds. In this way, different kinds of nuclei are produced. In particular, the 56 Ni formed will later on decay to 56 Co with a halflife of 6.1 d. Then 56 Co will decay to the stable 56 Fe with a half-life of 77.1 d. The energy released in these two decays powers the supernova light curve after the initial decline from maximum. Signatures of the decay of these unstable nuclei are observed in supernova light curves.

84 Chapter 8 Special Topic: Introduction to Stellar Pulsation Throughout this course, we focus on the static structure of a star. More precisely, we assume that the star is in mechanical equilibrium all the time. Nonetheless, under suitable conditions, the equilibrium configuration of certain stars is unstable against perturbation. Such perturbation may lead to stellar pulsation. Study of stellar pulsation is important for astronomers for the following reasons: 1. Distance measurement. Certain stellar pulsation leads to pulsation variable. The most famous one is the Cepheid variable, which is well-known for its period-luminosity relationship that has been used as standard candle to set the cosmological distance scale. 2. Stellar structure. Our Sun and other main-sequence stars also show pulsations, albeit with much smaller amplitudes. By observing different oscillation modes of a star, one can infer its interior structure. This is known as asteroseismology (or helioseismology for the specific case of the Sun). 8.1 Some basic terminology The instability strip is a narrow region in the Hertzsprung-Russell diagram where most of the pulsating variables lie (except the Mira variables), e.g., it intersects with the horizontal branch at RR Lyrae, crosses the supergiants at Cepheids, and crosses the white dwarf region at ZZ Ceti. The pulsations in Cepheids, Miras, and RR Lyrae are radial. They show periodic contraction and expansion with periods from 1 to 100 days. On the other 84

85 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. Different regions on the stellar surface contract and expand differently. Non-radial oscillations can be described by the spherical harmonics Yl m (θ, ϕ) = ( 1) m 2l + 1 (l m)! 4π (l + m)! P l m (cos θ)e imϕ. (8.1) Stellar pulsation is similar to sound waves in musical instruments. They are standing waves with the center of the star as a node and the stellar surface as an antinode. In the fundamental mode (l = 0), there is no other nodes and Type Period Mode Mira d radial Cepheids 1 50 d radial W Virginis d radial RR Lyrae 1 hr 1 d radial δ Scuti 1 3 hr rad. & non-rad. β Cephei 3 7 hr rad. & non-rad. ZZ Ceti s non-rad. Table 8.1: Some common pulsating variables.

86 CHAPTER 8. STELLAR PULSATION 86 the entire star pulse in a single direction at a given time. The first overtone (l = 1) correspond to one node between the center and the surface, the second overtone (l = 2) has two, and so on. A star can exhibit several modes at the same time. Cepheids are believed to oscillate in the fundamental mode, while both the fundamental mode and the first overtone can exist in RR Lyrae and Miras. 8.2 A simple model In this introduction, we will focus on the theory of the simplest possible model of stellar pulsation, namely, adiabatic radial pulsation in the small amplitude limit. Further analysis can be found in the following review articles: J. P. Cox, Reports on Progress in Physics 37, 563 (1974); A. Gautschy and H. Saio, Annual Review of Astronomy and Astrophysics 33, 75 (1995); A. Gautschy and H. Saio, Annual Review of Astronomy and Astrophysics 34, 551 (1996). The basic idea is simple, when a layer is compressed, it heats up and becomes more opaque to radiation. Radiative diffusion slows down, heat, and hence pressure, build up. This eventually pushes the layer outward. The expansion cools down the material, and it becomes more transparent to radiation, such that heat can escape more easily and the pressure subsequently drops. Finally, the layer falls back and the cycle starts over. Here we use the so-called Lagrange approach which focuses on the motion of individual particles in a star. (An alternative approach is that of Euler, which focuses on the evolution of thermodynamic variables such as density and pressure. Both approaches give the same physics but the later one is sometimes more effective in solving macroscopic problems involving fluid.) The conservation of mass (Eq. 2.2 in Chapter 2) in the time independent case) should be rewritten as m r = 4πr2 ρ(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 the oscillation is radial so that all physical parameters are functions of r and t only. In this case, the mass of a small spherical shell from radius r to r + r equals 4πr 2 ρ(r, t) r. The gravitational force acting on this shell is Gm(r, t)4πr 2 ρ(r, t) r/r 2. The (gas plus radiation) pressure gradient acting

87 CHAPTER 8. STELLAR PULSATION 87 on this shell is P/ r. Hence, by Newton s second law, In other words, 4πr 2 ρ r = G4πr2 ρm r 2 r = Gm r 2 4πr 2 P r. (8.3) 4πr 2 P m. (8.4) Assuming that the oscillation is adiabatic and the star is made up of an ideal gas, pressure is given by P = Kρ γ a (8.5) for some constant K. Analytical solution to these three partial differential equations, namely, Eqs. (8.2), (8.4) and (8.5) are difficult to find. Nonetheless, in most cases, the amplitude of oscillation of a star about its equilibrium position is small enough. In this case, we need only to keep up to the first order term in the Taylor series expansion about the equilibrium configuration of the star. Besides, the oscillation of the star can be well approximated by a S.H.M. In other words, using m and t are the two independent variables, we may write and P (m, t) = P 0 (m)[1 + p(m)e iωt ], (8.6) r(m, t) = r 0 (m)[1 + x(m)e iωt ], (8.7) ρ(m, t) = ρ 0 (m)[1 + d(m)e iωt ], (8.8) where P 0 (m), r 0 (m) and ρ 0 (m) are the equilibrium pressure, radius and density on the sphere whose mass contained equals m. Substituting the above equations into the three coupled partial differential equations and keeping up to the first order terms, we obtain and Eliminating p and d, we have ( ) x γ a r 0 r 0 P 0 p = ω 2 Gm(p + 4x) r 0 x + ρ 0 r 0 r r 0 r 0 (γ a x) Gmρ 0γ a P 0 r 2 0, (8.9) r 0 x r 0 = 3x d, (8.10) p = 3γ a x γ a r 0 x r 0. (8.11) x r 0 + ρ 0 P 0 [ Gm r 3 0 (4 3γ a ) + ω 2 ] x = 0. (8.12)

88 CHAPTER 8. STELLAR PULSATION 88 So the problem of stellar oscillation becomes the problem of finding eigenvalues ω 2 for above linear partial differential equation. If the eigenvalue ω 2 > 0, the solution is oscillatory which corresponds to an radial oscillation of the star. If ω 2 < 0, the solution contains an exponentially decaying or growing term which indicates instability. Eq. (8.12) is still too complicated to be solved analytically in general. Nevertheless, we have one specific meaningful case in which this eigenvalue problem has been investigated in detail. This is the case of (1) the adiabatic expansion coefficient γ a is a constant throughout the star; and (2) x is m independent. In this case, Eq. (8.12) becomes ω 2 = (3γ a 4) Gm r 3 0 = (3γ a 4) 4πG ρ 3, (8.13) where ρ is the mean density of the star. In other words, the oscillation of the star is stable if γ a > 4/3; and the period of oscillation Π in this case is given by [ ] 1/2 3 Π = 2π. (8.14) 4π(3γ a 4)G ρ For non-relativistic ideal gas, γ a = 5/3. Put this into the equation above, we have: Type RR Lyrae Classical Cepheids W Virinis Period 0.5 d 7 d 15 d This agrees reasonably well with observations. We have not finished our discussions yet. In reality, our linear analysis in the adiabatic regime is an overly simplified approximation. Actually, if we include higher order effects, radial oscillation of a star will be damped out quickly unless there is an effective way to pump energy into the oscillator. Indeed, pulsating variables, such as Cepheids, occupy only a restricted strip on the H-R diagram, strongly suggesting that an effective excitation mechanism exists only under rather restrictive conditions.

89 CHAPTER 8. STELLAR PULSATION κ-mechanism and the instability strip Detail analysis of the excitation mechanism is rather involved. The most famous one is the so-called κ-mechanism. This mechanism assumes that the opacity increases upon compression in some region of the star so that the radiative luminosity is blocked in the compression phase of oscillation. Therefore, the region gains thermal energy in compression phase and loses thermal energy in expansion phase. Recall that for most part of an ordinary star, κ is well approximated by κ 0 ρt 7/2 (Kramer s law). Hence, upon adiabatic compression, the opacity actually decreases. Thus, the condition for κ-mechanism is not satisfied in most stars. Nevertheless, such a condition holds in regions where there is partial ionization of H and Hei and/or Heii. In fact, κ-mechanism due to such partial ionization zone is responsible for the excitation of pulsation in a number of pulsating variables including RR Lyrae, classical Cepheids and W Virginis. Because the condition of onset of the κ-mechanism due to partial ionization is very stringent, pulsating variables are rare and can only be found in restricted regions on the H-R diagram. We will not go into the detail mathematics here, but only give a qualitative description of the κ-mechanism and its connection to the instability strip. In the partial ionization regions, part of the energy input from compression can be turned into more ionization instead of heating. Therefore, it is possible to increase in density more than in temperature. Hence, κ can be increased due to compression. Later in the cycle, electron-ion recombination during decompression can release energy to lower the opacity. Where is the κ-mechanism expected to occur? Location of the partial ionization zone in a star depends strongly on the effective temperature. If the temperature is too high, the ionization will be close to the surface, resulting in insufficient mass to drive sustainable oscillations. On the other hand, if the temperature is too low convection in the outer layer will strongly suppress the κ-mechanism. Therefore, we expect a narrow, near vertical strip in the H-R diagram. 8.4 ε-mechanism For completeness, we also mention the ε-mechanism. It was proposed before the κ-mechanism to explain stellar pulsations. It suggests that compression at the center of a star can lead to higher density and temperature, thus, increases

90 CHAPTER 8. STELLAR PULSATION 90 the thermonuclear energy production rate. However, detailed calculations show that the oscillation amplitude at the central regions of most stars are small, such that this effect is insignificant. However, this is still an important mechanism in massive stars. 8.5 Cepheid variables Classical Cepheids (or Cepheids for short) are luminous supergiants. It is named after the first known example, δ-cephei, which is still an important calibrator for the period-luminosity relation (see below). Their immediate precursors are massive young O or B type stars. Their periods are about 2 to 60 d, mean luminosities about 300 to L, mean surface temperature about 4000 to 8000 K, and peak-to-peak magnitude variation of about 0.4 to 1.4 mag. Its stellar atmosphere is cold enough to have a layer of partially ionized Hei and Heii. Hence, κ-mechanism leads to stellar pulsation. (See Figure 8.2 below for a typical light curve, color change and radial velocity of a classical Cepheid variable.) m v T eff V r (kms ) Phase Figure 8.2: Location of δ Cephei in the sky, and the variation of visual magnitude, effective temperature and radial velocity. In 1912, Henrietta Leavitt accidentally found that the period and luminosity of classical Cepheid variables are related. In a more modern language, such a

91 CHAPTER 8. STELLAR PULSATION 91 period-luminosity relation is found to be ( ) L log = log Π. (8.15) L That is, the longer the period, the brighter the star is. This famous relation is also called the Leavitt Law after her. Note that the above relation is valid plus or minus a few percent error. We may understand the period-luminosity relation as follows. First, the peakto-peak magnitude difference is relatively small, hence we may assume that the effective surface temperature of a classical Cepheid to be a constant. The luminosity of the star L 4πσR 2 T 4 is therefore a function of stellar radius R only. Although classical Cepheids are supergaints, the homology relation for main sequence star still holds approximately. Hence, we have an approximate mass-luminosity relation L M α for classical Cepheids. Finally, we have the period-mean density relation Π ρ 1/2. Combining these three constraints, we arrive at a period-luminosity relation for classical Cepheid variables L Π 4α/(3α 2). For α 3, we obtain L Π 1.7. This is about the same order as the empirical formula above, but we cannot derive the exact value since the mean density approximation is oversimplified. Indeed, the above argument is rather general and can be applied to quite a number of pulsation variables including W Virginis and Mira type variables. A notable exception is the RR Lyrae variables. Observations suggest that the absolute magnitudes of RR Lyrae variables fall into a narrow range of 0.6 ± 0.2. Hence, within about 10% accuracy, we may assume that all RR Lyrae stars are of the same luminosities and hence can be regarded as standard candles. 8.6 Cepheid variables in the cosmic distance ladder Since the period is much easier to measure to high precision, as compared to other observables such as spectrum or position. The period-luminosity relation immediately turned Cepheids into standard candles, offering a very powerful tool for distance measurements. Edwin Hubble observed Cephieds in other galaxies to discover Hubble s law. Even today, Cepheids play an important role in the construction and calibration of the cosmic distance ladder, up to the order of 10 Mpc (see the Hubble Space Telescope Key Project).

92 Chapter 9 A Schematic Picture of Stellar Evolution In the previous chapters we have seen that the timescale of stellar evolution is set by the (slow) rate of consumption of nuclear fuel. The rate of nuclear burning increases with density and rises sharply with temperature, and the structure equations of a star show that both the temperature and the density decrease from the center outwards. We may therefore conclude that the evolution of a star 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 gradually depleted of each nuclear fuel, the evolution of the star progresses. We can therefore learn a lot about the evolution of stars by studying the changes that occur in its center. To do this, we consider the diagram of T c and ρ c. From these quantities, together with the composition, we can calculate the whole evolution of the star. In the diagram of T c and ρ c the time evolution of a star will be a track. Since the only property that distinguishes the evolutionary track 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 density ranges, so the different combinations of temperature and density will determine the state of the stellar material, and the dominant physical processes that are expected to occur. In this way we can divide the (T, ρ) diagram into zones representing different physical states or processes. By looking at the position of stars in this diagram as a function of time and mass, we should be able to trace the processes that make up the evolution of a star. 92

93 CHAPTER 9. A SCHEMATIC PICTURE OF STELLAR EVOLUTION Zones of the equation of state Figure 9.1: Mapping the temperature-density diagram according to the equation of state. The most common state of the gas that we find in stars is the ideal gas state for both components: ions and electrons. We can then write the equation of state as: P = R µ ρt = K 0ρT, (9.1) where K 0 is a constant. At high density and relatively low temperature the electrons become degenerate, and, since their contribution to the pressure is dominant, we can then write P = K 1 ρ 5/3. (9.2) The transition from one state to another is gradual with the change in density and 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 Above it lies the electron degeneracy zone, and below it the ideal gas zone. For still higher densities, relativistic effects play a role, and the equation of state becomes P = K 2 ρ 4/3. (9.4)

94 CHAPTER 9. A SCHEMATIC PICTURE OF STELLAR EVOLUTION 94 The boundary between the ideal gas zone and the relativistic degeneracy zone is then given by log ρ = 3 log T + constant. (9.5) This means that the boundary between the ideal gas zone and the electron degeneracy zone becomes steeper as the density increases. Within this zone, the transition from relativistic to non-relativistic occurs when the rise in pressure with increasing density becomes constrained by the limiting velocity c. We are in this case when ( ) 3 K 1 ρ 5/3 K 2 ρ 4/3 K2, or ρ. (9.6) This is a horizontal line in fig In the ideal gas zone radiation pressure has been neglected. Its contribution to the total pressure becomes important, however, at high temperatures and low densities and SHOULD be added to the gas pressure. Eventually, radiation pressure will become dominant, with the equation of state changing to K 1 P = 1 3 at 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 of energy release by this process constitutes a significant fraction of the rate at which energy is radiated away (which is the stellar luminosity). Since nuclear reaction rates are very sensitive to the temperature, one can define threshold lines in the (T, ρ) diagram: on one side of the threshold the rate of burning can be considered negligible, while on the other side it is considerable. These threshold lines are shown in fig We consider six different stages of nuclear burning: the p-p cycle, the CNO cycle, helium burning into carbon through the 3α reaction, carbon burning, oxygen burning, and silicon burning. Consider ϵ = ϵ 0 ρ m T n, (9.9) in most cases m = 1, and n 1, hence, the thresholds are almost vertical lines (do you know why?). In reality, n depends on temperature, therefore the lines are curved. Nucleosynthesis ends with iron. Iron nuclei heated to very

95 CHAPTER 9. A SCHEMATIC PICTURE OF STELLAR EVOLUTION 95 Figure 9.2: Mapping the temperature-density diagram according to nuclear processes. high temperatures are disintegrated by energetic photons into helium nuclei (αparticles). This energy absorbing process reaches equilibrium with the relative abundance of iron to helium nuclei determined by the values of temperature and density. A final threshold may be defined by requiring that the number of helium 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 γ a > 4/3. So in regions with γ a < 4/3 the stellar models might be expected to become unstable. These are the far extremes of the relativistic degeneracy zone III and the radiation pressure dominated zone IV (and also the iron photodisintegration zone). Another zone is unstable due to pair production (very high T and low density, see fig. 9.3). All this means that there is a relatively small zone remaining for possible evolutionary tracks of stars. To finish, nuclear burning is thermally unstable in degenerate gases, relativistic or non-relativistic. So for that reason also the nuclear burning thresholds in fig. 9.2 have been discontinued after crossing the boundary into the degeneracy zone II.

96 CHAPTER 9. A SCHEMATIC PICTURE OF STELLAR EVOLUTION 96 Figure 9.3: Outline of the stable and unstable zones in the temperature-density diagram. 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 that we have the following relation between central density and pressure P c = (4π) 1 3 Bn GM 2 3 ρ 4 3 c. (9.10) This relation depends only slightly on n, especially for stable polytropes, for which n lies between 1.5 and 3. Although a star in hydrostatic equilibrium is not a polytrope, this equation does provide a good approximation. Apart from this relation, the central pressure is related to the central density and temperature by the equation of state. Within the different zones of the (T c, ρ c ) plane we have different equations of state. Combining them with Eq. (9.10) we can eliminate P c to obtain a relation between ρ c and T c. For example, for an ideal gas, we have ρ c = K3 0 4πB 3 ng 3 T 3 c M 2 T 3 c. (9.11) This means that we can put a number of parallel lines with slope 3 in our diagram for various values of M (see fig. 9.4). In zone II (non-relativistic degenerate electrons) we have the following equation

97 CHAPTER 9. A SCHEMATIC PICTURE OF STELLAR EVOLUTION 97 Figure 9.4: Relation of central density to central temperature for stars of different masses within the stable ideal gas and degenerate gas zones. of state, P = K 1 ρ 5 3, leading to ( ) B1.5 G 3 ρ c = 4π M 2. (9.12) K 1 Hence, ρ c is independent of T c, and we have for each mass a corresponding horizontal line in the (T c, ρ c ) 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 that the maximum mass a degenerate star can have is M ch = 1.44 M, the Chandrasekhar limit. One can see that the paths for masses smaller than this will eventually bend into the degenerate region, while paths for higher mass stars will remain straight. To conclude, a star of fixed mass M has its own distinct track in the (T c, ρ c ) plane. There are two distinct shapes: straight for M > M ch and knee-shaped for M < M ch. We can understand the relationship between tracks of different masses as follows: when mass is increased, gravity becomes more important, so a higher pressure is required to counterbalance gravity. In an ideal gas this can be done with a higher density or a higher temperature. A higher density, however, implies smaller distances between particles, causing larger gravitational forces. Since the hydrostatic pressure is proportional to a higher power of the density than the gas pressure (4/3 as opposed to 1) a higher density would only worsen the imbalance. So for a higher mass a lower density or a higher temperature is required. In the case of a degenerate electron gas, the temperature is less important. Now the hydrostatic pressure is proportional

98 CHAPTER 9. A SCHEMATIC PICTURE OF STELLAR EVOLUTION 98 to a lower power of the density than the gas pressure (4/3 vs. 5/3) so that a higher 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 their central temperature-density tracks. In fig. 9.5 we have combined all the previous figures. We will now choose a mass, identify its path and follow the journey of a star along it. Stars form in gaseous clouds, with much lower densities and temperatures than we are used to here. At the beginning, a star radiates energy without a central energy source, which means that it contracts and heats up (we will see this in the next chapter). In the (T c, ρ c ) plane this means that the stars ascends along its track towards higher temperatures and densities. Eventually, it will cross the first nuclear burning threshold. At this point hydrogen is ignited in the centre, and the 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 track crosses the p-p track, while for high masses the CNO-track is crossed. So stars are 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

99 CHAPTER 9. A SCHEMATIC PICTURE OF STELLAR EVOLUTION 99 massive than about 100 M (an example of such a massive star is η Car). The (T c, ρ c ) 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 a few million K, the highest value of M for which M still touches this threshold can be regarded as the lower stellar mass limit. Objects of lower masses will never ignite nuclear fuel. These objects (of mass < 0.1 M ) are called brown dwarfs. Let us continue with stars that do burn nuclear fuel. When the hydrogen in the center is finally exhausted the star will lose energy again, and will have to contract its core to compensate for it. This means that it will heat up. For low-mass stars, the tracks will cross into the degeneracy zone and bend to the left. The pressure exerted by the degenerate electrons is enough to counteract gravity. Contraction slows down, and the star cools while radiating the accumulated thermal energy, tending to a constant density and radius, determined by M. The higher the mass, the higher the final density and the lower the final radius. For higher mass stars, the tracks will cross the next nuclear burning threshold. Helium will now ignite the core, and we have another phase of thermal equilibrium. After exhausting helium, we have more contraction, and some stars develop degenerate cores, moving towards the left. This means that there are two kinds of degenerate stars: some with helium cores, and others with cores of carbon and oxygen. A star of mass M ch can in principle continue burning fuel without becoming degenerate. However, since its track is very close to the degenerate zone instabilities might easily occur. For higher masses, more nuclear burning thresholds will be crossed. These massive stars undergo contraction and heating cycles alternating with thermal equilibrium burning of heavier and heavier nuclear fuels, until their cores consist of iron. Further heating of iron inevitably leads to its photodisintegration, a highly unstable process. These stars will end as supernovae. Stars of very large masses have tracks that enter the instability zone before crossing the burning thresholds of heavy elements. They are therefore expected to be extremely short-lived, developing pair-production instabilities that will result in a catastrophic event, like a Supernova explosion. We can summarise by stating that stars can exist with masses between about 0.1 and 100 solar masses. All start with hydrogen burning in their centres. When, however, hydrogen is exhausted in the centre, evolution will proceed differently for different masses. Those under M = M ch contract and cool off after completion of hydrogen or helium burning. Stars above this critical mass undergo all the nuclear burning processes, ending with iron synthesis. Subsequent heating of the core develops into a highly unstable state, ending in a catastrophic collapse or explosion. Stars of very high mass may reach dynamical instability

100 CHAPTER 9. A SCHEMATIC PICTURE OF STELLAR EVOLUTION 100 sooner, due to pair production. 9.6 The late evolutionary stages in the (T c, ρ c ) diagram Figure 9.6: Schematic illustration of the stellar configuration in different evolutionary phases for a 10 M star (A,B,C,D,E) and a white dwarf. Instead of using the (T c, ρ c ) diagram just for the centre, one can also use it to describe the structure of a star at a given evolutionary stage. A star can be represented as a line connecting the center with the photosphere. The exact shape of these lines can be complicated only polytropes are straight lines in the diagram. In fig. 9.6 a number of lines are plotted with the start of all lines on the track from previous section. These lines may be considered as the evolving structure of the star. Line A can be considered as the main sequence. The next line, B, describes the star at a later stage when the hydrogen has been depleted in the core. On track B hydrogen is burning at the point where it crosses the hydrogen burning threshold. This indicates that hydrogen burns in a shell outside the helium core. The region interior to it, consisting of homogeneous helium, is contracting and heating up. Assuming that the contraction of the core occurs quasi-statically, on a timescale much larger than the dynamical timescale, we can assume that the virial the-

101 CHAPTER 9. A SCHEMATIC PICTURE OF STELLAR EVOLUTION 101 orem holds. If the amount of energy gained during this phase is negligible compared to the total energy, we may assume that energy is constant. In such a case we may assume that both gravitational potential energy and thermal energy are conserved. For this reason contraction of the core must be accompanied by expansion of the envelope. Heating of the core will result in cooling of the envelope. On track B the surface temperature drops, so that the star becomes redder. It becomes a red giant. One can show that a moderate amount of core contraction has to be compensated by a large amount of expansion of the envelope. If the total energy does not remain constant but increases, then the core will heat up even more, so there will be more expansion. If the total energy of the star decreases the envelope might either contract, expand or remain unchanged. When the He ignition temperature is finally reached, core contraction stops (line C). 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 the core itself is exhausted, another phase of contraction and envelope expansion sets in. Since the core is now more condensed, envelope expansion is even more pronounced. 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 star looks like an onion with a central region of C, N, and O, a shell of helium, and the hydrogen-rich envelope. The hydrogen burning shell feeds fresh fuel to the helium burning one, so both move outwards. This process is quite complicated and often unstable, leading to variability. Some of the most massive stars will even 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. One should not forget that this is only schematic real models will have to rely on complicated numerical calculations. There are however, as expected, a number of problems with this picture. For example, stars as massive as 7 9 M end up as white dwarf, i.e., degenerate stars. We mentioned before that this would only happen for stars with mass M < M ch = 1.44 M. How can these stars lose almost all their mass? Another indication comes the fact that in the solar neighbourhood white dwarfs are found with masses of M = 0.4 M. Stars with such low masses evolve very slowly, and would never have been able to reach this 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 which gas and dust is blown away from the star. Mass loss for massive stars can be

102 CHAPTER 9. A SCHEMATIC PICTURE OF STELLAR EVOLUTION 102 as high as 10 4 M yr 1 for very massive stars. This means that stars do not stay on the same track, but move to different tracks, where they will behave as stars of mass M. Since no simple model of mass loss is available, this cannot be very easily implemented in a schematic picture. Another process that was neglected in the previous picture was neutrino emission in dense cores, which has the effect of cooling the stars. As the rise in temperature between late burning stages is impeded by neutrino cooling, the slope of the tracks should become somewhat steeper than 3. Nonetheless, this does not change anything qualitative of the picture. What we have not been able to predict are: 1. time scales 2. detailed prediction of the outer appearance of stars. For that reason, we cannot compare these models with observations; to do so we need to go to more detailed models.

103 Chapter 10 Special Topic: Substellar Objects In Chapter 5, we derived the minimum mass of a main sequence star to be 0.1 M. In this chapter, we briefly describe some topics related to stars near and below the main-sequence edge, namely red dwarfs and brown dwarfs, all the way down to planets Brown Dwarfs Failed Stars Figure 10.1: Size and surface temperature comparison between stars, brown dwawfs, and planets. Recall that a star is formed from the collapse of a nebula. The type of object 103

104 CHAPTER 10. SPECIAL TOPIC: SUBSTELLAR OBJECTS 104 eventually formed depends on the final mass. If it is greater than about 0.1 M, sustainable hydrogen burning will occur in the core (due to sufficiently high core temperature) before the core becomes degenerate. This is a star. More precise calculations show that the hydrogen-burning main-sequence edge is M 75 M J (Burrows & Liebert 1993, Rev. Mod. Phys. 65, 2). This also slightly depends on the metallicity: it can reach 90 M J for zero metallicity. If the mass of an object is less than about 0.01 M ( 10 M J, i.e. 10 Jupiter mass), the temperature of the core will not be hot enough to burn any nuclear fuel. The result will be a planet. A planet is supported by mainly EM forces between atoms or molecules against gravity. Figure 10.2: Luminosity, core temperature, and radius evolution of low-mass stars (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 13 M J between brown dwarfs and planets, since the core temperature of an object above 13 M J is sufficiently hot to fuse D. A brown dwarf above 65 M J can even burn Li. Similar to a very low mass main sequence star, a brown dwarf is fully convective. However, D and Li are rare and hence this burning takes only a relatively short time (at most 10 8 yr). In addition, it turns out that the gas pressure produced by D and Li burning is not strong enough to halt the collapse. Eventually, almost all D and Li in the star is used up, but the core temperature is still not high enough to burn

105 CHAPTER 10. SPECIAL TOPIC: SUBSTELLAR OBJECTS 105 hydrogen. The brown dwarf is now supported by electron degenerate pressure in the core. Unlike a white dwarf, the temperature of a brown dwarf is still rather 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 firm detection (Gliese 229B) was only made in 1995 by Kulkarni et al. Observationally, it is very difficult to identify brown dwarfs because of their low surface temperature and hence low luminosity. The main search methods are direct observations 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 temperature and luminosity are below the minimum possible values for main-sequence stars, either by looking at stellar companions or in the field. The first confirmed brown dwarf, Gliese 229B, was detected this way. It is a companion to 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 is modulated by the inclination angle, therefore, is only a lower limit, unless the orbit is spatially resolved. Young brown dwarfs in clusters: brown dwarfs are at their brightest in young age, hence, easiest to see. Nevertheless, they have very similar temperature and luminosity as very low-mass stars. This can be solved with additional information such as age and chemical abundances. The former could be estimated from the colour-magnitude diagram for stars

106 CHAPTER 10. SPECIAL TOPIC: SUBSTELLAR OBJECTS 106 in a cluster. Among the first brown dwarfs observed, PPL 15 and Teide 1 in the Pleiades, are young objects Distinguishing Brown Dwarfs from Stars To confirm brown dwarf candidates, we need some good ways to distinguish brown 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 object is very likely to be substellar, because low-mass stars are hot enough to burn Li and they are fully convective, so all Li will be depleted in a short time scale within 10 8 yr. Problems: 1. Heavier stars like the sun can retain Li in their outer layers. (This is generally not a problem since we can rule out heavy stars from the luminosity or temperature). 2. Young stars do not have enough time to deplete all the Li. 3. Brown dwarfs heavier than 65 M J can also burn Li. Methane test: stars are too hot to have methane in the atmosphere, but brown dwarfs and planets are cool enough (< 1400 K) that methane gathered from the interstellar medium over time can stay in the atmosphere. Therefore, if the spectrum shows methane bands at 2 microns, then the object cannot be a star.

107 CHAPTER 10. SPECIAL TOPIC: SUBSTELLAR OBJECTS Distinguishing Brown Dwarfs from Planets It is generally difficult to distinguish between a brown dwarf and a giant planet and the 13 M J cutoff by IAU should be considered as a rule of thumb rather than a physical limit. Size (?): Massive brown dwarfs are supported by degenerated pressure, therefore, the variation of radius with mass is small and many of them have similar sizes as Jupiter. Mass: If an object is much more massive than Jupiter, e.g., 20 M J, we can rule out a planet. Spectrum: X-ray and radio emission have been detected from brown dwarfs (e.g., LP ). X-ray flares found in brown dwarfs are believed to be very similar to those in very low-mass stars: both due to changing magnetic fields as a result of rapid rotation and convective interior Planets: Energy Sources Moving down the mass scale, planets, by definition, have no nuclear reaction in the interior. However, it is found that Jupiter and Saturn radiate about 2 3 times the energy as they absorb from the sun. Where does the energy come from? Residual heat: One idea is that the energy source could be residual heat left over from the primordial nebula that collapsed into the planet. But as the planets are formed for a long time, this is likely relatively unimportant. Kelvin-Helmholtz mechanism: The contraction of a star/planet can generate heat. This was first proposed to explain the energy source of the sun. Recall that from the virial theorem, the total energy U tot, which is the sum of the potential energy and the thermal energy, and U tot = Ω + U T = U T < 0. (10.1) So if a star loses energy (e.g., due to radiation), the thermal energy, and hence the temperature, actually increase! For this reason a star (or a planet) appears to have negative heat capacity. (Does it violate any

108 CHAPTER 10. SPECIAL TOPIC: SUBSTELLAR OBJECTS 108 laws in thermodynamics?) For the simplest case of a constant density planet, it can be shown that Ω= 0 R Gm 3M 2 G dm =. r 5R (10.2) Hence, the temperature increases when a planet shrinks. (How about brown dwarfs?) Helium rain: The Kelvin-Helmholtz mechanism alone is not enough to explain the emission of Saturn. The atmosphere of Saturn is cool enough for helium to condense. The helium droplets then rain out and sink to the core through the liquid hydrogen surrounding. The friction in this process releases heat in the planet interior. Observation found a low helium abundance in the atmosphere of Saturn, providing support to this theory. For Jupiter, this is less clear. Due to high temperature, it is generally believed that helium is still mix with hydrogen in the interior. However, recent studies suggest that Jupiter may also have Helium rain too. Figure 10.6: Helium rain in Saturn Figure 10.7: Earth orbit as seen from the nearest star. Extrasolar Planets: Detection Let us move to extrasolar planets. The problem: the Earth is 1 AU from the sun, 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 diameter of a coin over the Victoria Harbour. Also a star is much brighter than a planet. All these post a great challenge to planet detection.

109 CHAPTER 10. SPECIAL TOPIC: SUBSTELLAR OBJECTS 109 Pulsar timing: pulsars are extremely precise clocks due to their stable rotation. Any small change in their position or motion due to the presence of a planet will result in detectable timing anomaly. This method is by far the most sensitive down to 0.1 Earth mass (0.1M ). However, the harsh pulsar environment is not habitable because of strong wind of particles winds and EM radiation streaming out from the neutron star. Also, the planet is formed from the fall-back materials in a supernova remnant, which could have very different properties than those collapsed from gas clouds. The first exoplanet was detected this way, around PSR B Radial velocity: The motion of a planet would causes the parent star to wobble. Unless the system is viewed edge-on, this will result in periodic change of the radial velocity of the star. Due to the Doppler effect, the spectral lines will be redshifted and blueshifted periodically. This is the most efficient detection method since the chance of having a face-on system 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 velocity shift 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 the wobbling 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 essential to know both the mass and radius. Transit is the only method that can provide reliable radius measurements. However, this requires a

110 CHAPTER 10. SPECIAL TOPIC: SUBSTELLAR OBJECTS 110 nearly perfect alignment between the parent and the planet as viewed from Earth, for which the probability is very small. NASA s Kepler mission is dedicated to this kind of measurement with very precise photometry. Direct imaging: While the small (arcsecond scale) angular resolution mentioned above can be achieved using the Hubble Space Telescope or groundbased optical telescopes with adaptive optics, it is still non-trivial and the contrast between the star and the planet is often higher than 1 : Figure 10.10: Direct imaging of planets. Gravitational microlensing: When a star passes in front of a background object, the light from the latter is bended due to the gravitational potential. The presence of a planet in the foreground lensing star will result in a slightly different gravitational potential, giving detectable effect. However, the chance alignment of a star with a background source is very rare, and since all stars have proper motions, the alignment only last for hours to days and almost never happen again. This makes confirmation or follow-up studies of the planets impossible.

111 CHAPTER 10. SPECIAL TOPIC: SUBSTELLAR OBJECTS 111 Figure 10.11: Planet detection by microlensing Habitability of Exoplanets and Red Dwarfs To look for lives in exoplanets, there are some important characteristics for the parent 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 and evolution of life. But too much is harmful. And for the planets: Rocky planets: it is generally believed that life could not exist in gas giants that has no solid surface. Massive: a too low mass planet cannot retain an atmosphere, which makes life impossible. Rotation: if the rotation is too slow or tidally locked to the orbit, one side would be much hotter than the other, life could be difficult to form. Distance from star: this is perhaps the most critical factor for liquid water to exist on surface. The habitable zone depends on the star s luminosity. For our solar system, a conservative estimate suggests about AU, and more optimistic estimate gives AU. Lower mass (K or M) stars obviously have habitable zones closer than that of the sun. Another consideration is tidal locking, which depends on the planet-star distance

112 CHAPTER 10. SPECIAL TOPIC: SUBSTELLAR OBJECTS 112 and mass of the star. The tidal force constantly squeeze and stretch the planet, converting its rotational energy into heat, until it is tidally locked into synchronous rotation in a time scale of ( a ) yr. (10.3) AU)6 ( M M Figure 10.12: Illustration of tidal locking. Left: Synchronous rotation. Right: Non-synchronous rotation. The planet is continuously deformed such that the rotational energy is dissipated via friction. Red dwarfs are stars at the low-end of the main sequence. They are very low mass K or M-type stars ranging from M (= mass limit of a brown dwarf) to 0.6 M and have surface temperature of K. They are the most common stars in the Milky Way and can live for very long time (T M yr). These are advantages to harbour life. However, their habitable zones are much smaller than that of the sun. This also makes it easier for a planet to be tidally locked (although some recent studies suggest that the dark side may not be totally frozen). Like brown dwarfs, some red dwarfs show flares in X-rays and UV during the early stage (first yr) of their lives. For life to develop, some protection from radiation is needed or the planet has to migrate into the habitable zone later to avoid radiation and tidal locking.

113 CHAPTER 10. SPECIAL TOPIC: SUBSTELLAR OBJECTS 113 Figure 10.13: Habitable zone of red dwarfs compared to that of other stars.