PHYS 231 Lecture Notes Week 3

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1 PHYS 231 Lecture Notes Week 3 Reading from Maoz (2 nd edition): Chapter 2, Sec. 3.1, 3.2 A lot of the material presented in class this week is well covered in Maoz, and we simply reference the book, with additional comments and derivations as needed. References to slides on the web page are in the format slidesx.y/nn, where x is week, y is lecture, and nn is slide number. 3.1 Some Thermodynamics and Statistical Mechanics In order to discuss how stellar spectra depend on temperature, we need to say something about temperature determines the speeds, excitation levels, and ionization states of atoms in a star. There are many different physical processes going on, involving collisions among atoms and electrons, and interactions between matter and radiation. All of these processes transform energy from one form to another, and in principle trying to follow them all can get very complicated. Fortunately, there is a huge simplifying assumption we can make thermodynamic equilibrium. The idea is that, in a completely closed system, we expect all processes to come into equilibrium, in the sense that all physical properties of the material temperature, density, atomic excitation state, ionization fraction don t fluctuate in time. In other words, although atoms are banging into one another and bumping electrons up and down the atomic energy states, we can assume that for every process causing a transition from 1 to 2, there is another from 2 to 1, and they are in balance, meaning that the total numbers of atoms in state 1 or state 2 stay constant. The same applies to the speeds of atoms in the gas, and to interactions between atoms and radiation. There is a well-defined temperature T, and this in essence controls the equilibrium of all processes. In fact, stars are not closed systems energy is constantly leaking out, and their properties depend on location in the star but these deviations deviations from perfect thermodynamic equilibrium are generally small, and we can assume Local Thermodynamic Equilibrium in discussing the properties of the gas The Maxwell-Boltzmann Distribution With this assumption we can use the techniques statistical mechanics to determine the most probable distribution of physical properties. For example, in an ideal classical (non-relativistic) gas made up of particles (atoms) of mass m, the temperature T may be defined in terms of the mean kinetic energy of the gas particles: 1 2 mv2 = 3 2 kt. Statistical mechanics tells us how the speeds v are distributed. The probability of an atom having speed v in the range [v, v + δv) is p(v)δv, where p(v) = ( ) m 3/2 4πv 2 e mv2 /2kT. 2πkT The first factor simply ensures that p(v) dv = 1. Even though atoms are bouncing around and colliding with one another and an individual atom may change its speed billions of time per second, on average the distribution of speeds stays constant. The above expression is called the Maxwell- Boltzmann distribution. Note that since atoms are moving in all directions with many different speeds, the light they emit or absorb will be Doppler shifted slightly due to their motion relative to an observer. Thus, 1

2 for an example, a emission line that is very narrow in the rest frame of an atom will be broadened due to the thermal motion of the atoms (see slides3.1/3). Since the rms speed is v rms = v 2 1/2 = 3kT m, the line broadening will be on the order of λ λ v rms c 3kT mc 2. Thus measuring the widths of spectral lines gives us direct information on the temperature of the gas producing them The Boltzmann Formula Similar reasoning applies to the energy distribution of electrons in atoms. Imagine two electron states A and B, with E 2 > E 1 (slides3.1/4). As atoms collide and energy is exchanged, the electron will be bumped up and down between the states many times, but averaged over many atoms, statistical mechanics tells us that the ratio of the number of atoms in the upper state, n 2 to the number in the lower state, n 1 is given by the Boltzmann formula n 2 n 1 = g 2 g 1 e (E 2 E 1 )/kt. Here, g 2 and g 1 are the degeneracies of the two states, which simply means the number of different quantum states that have the same energy, or the number of different ways the electron can have that energy. The degeneracy of states in hydrogen can be directly calculated from quantum mechanics: g n = 2n 2, the n 2 coming from all allowed angular momentum states and the 2 from the two spin states of the electron. Note that, as T, the exponential term goes to zero and n 2 = in other words, all atoms are in the lower state. This immediately explains why low-temperature M stars don t have any hydrogen absorption lines. The visible (Balmer) hydrogen lines correspond to transitions starting at the first excited state (n = 2; see slides2.2/31). But for such a transition to occur, we must have some atoms in the n = 2 state. A transition between n = 1 and n = 2 has energy E 21 = E 2 E 1 = 1.2 ev. Thus, for this transition, n 2 n 1 = 4 e E 21/kT. For T = 5 K, this ratio is , so negligibly many hydrogen atoms are in the excited state. Increasing T to 1, K increases the ratio by a factor of more than 1 4, to still a small number, but large enough that enough hydrogen is in the excited state to create a strong line The Saha Equation As T in the Boltzmann formula, the exponential factor goes to 1, so n 2 n 1 = g 2 g 1, 2

3 meaning that, in the absence of other effects, all states would be populated according to their degeneracy. However, as the temperature increases, it becomes more and more likely that the atom will be ionized. An atom (or ion) becomes ionized then a bound electron gains enough energy to escape. This can occur radiatively (via the absorption of a photon) or collisionally (when two atoms run into one another). The reverse process is called recombination, where an electron and an ion encounter one another, lose some energy in the form of a photon, and become bound. This is a reversible process: X i X i+1 + e, where X i represents the i-th ionization state of element X. Spectroscopists often use the notation XI (roman numeral I to refer to the neutral atom, XII the singly ionized atom, etc., so some concrete examples might be OI OII + e NII NIII + e HI HII + e. In ionization equilibrium, the forward and backward rates are equal, and statistical mechanics tells us that the number densities of state i, state i + 1, and electrons must satisfy the Saha equation n i+1 n e n i = 2 Λ 3 g i+1 g i e χ i/kt. Here, g i is the degeneracy of ionization state i, χ is the energy needed to ionize state i to state i + 1, and ( ) h 2 1/2 Λ = 2πm e kt is a length in fact, the de Broglie wavelength of an electron. For typical stellar temperatures, Λ = m ( ) T 1/ K Let s focus on the ionization state of hydrogen. Let the total number density of hydrogen nuclei (atoms plus protons), hydrogen atoms, protons, and electrons be n, n H, n p, and n e, respectively, and define the ionization fraction as X = n p n. Clearly, n = n h + n p and, if all electrons were produced by ionization of hydrogen (not a bad assumption for most stars), n e = n p. In that case, g i = 4 (2 proton spin states times 2 electron spins), g i+1 = 2) (2 proton spin states), and the Saha equation becomes (with n H = (1 X)n) X 2 1 X = 1 nλ 3 e χ /kt, where χ = 13.6 ev. Thus, although we can t write down an explicit solution for X in terms of n and T, once n and T are specified, solving for X is just a matter of solving a quadratic equation X 2 + AX A =, where A is the right-hand side of the above equation. 3

4 Slides3.1/7 shows the solution to this equation as a function of T for a range of choices of hydrogen density (left to right): n = 1 6, n = 1 9, n = 1 2, n = 1 24, n = 1 26, and n = 1 28 m 3, corresponding to densities found in interstellar space, near the solar surface, and in the solar interior. Note that (unsurprisingly) X increases as T increases, and (less obviously) decreases as n increases, because recombination is faster at higher densities. For densities typical of the solar interior close to the surface, X >.5 for T > 17 K. The outer layers of hot O and B stars are almost completely ionized, which explains why no hydrogen absorption lines are seen there is no atomic hydrogen to produce them. If we simply estimate the temperature T at which the typical atomic energy 3 2kT equals the ionization energy of hydrogen, 13.6 ev, we find T = K, a temperature consistent with the above discussion of the Boltzmann formula (in that the exponential term is close to 1), but much hotter than any known star. Why then, is the hydrogen near the solar surface ionized? The answer is that the ionization fraction is determined by a balance between ionization and recombination. As the density decreases, recombination becomes less effective, and the equilibrium moves toward full ionization. In the outer layers of the Sun, ionization occurs due to relatively rare interactions that lie far out on the high-energy exponential tail of the Maxwell-Boltzmann (or Planck) distribution, with E kt, but if recombination is ineffective in opposing this process, they can nevertheless lead to a large ionization fraction. 3.2 Atoms and Radiation Before we leave the subject, a few more remarks on the interactions between matter and radiation, and the resulting emission and absorption spectra we see. This section goes considerably beyond the syllabus of the course, but it is intended to give some flavor of how emission and absorption processes at the atomic level relate to the large-scale radiation field in a star. We Consider a system in thermal equilibrium at temperature T. That means that all aspects of all atoms and ions speeds, internal excitation state, and ionization state are all determined by T. Now consider, as above, an idealized two-level atom, with levels 1 and 2, with E 2 > E 1, in that environment Emission Let s first consider the atom s emission. An atom in the excited 2 state, left to its own devices, will eventually spontaneously emit a photon and drop to the lower 1 state, where E 2 E 1 = E 21 = hν, the frequency associated with the transition. The probability per unit time of such an occurrence is the Einstein coefficient A 21 for the two levels. it is a quantum-mechanical property of the atom, and can in principle be calculated for any given system. The emission is isotropic, and we can easily write down an expression for the resulting emissivity (unit W m 3 ster 1 ): j ν = hν A 21 n 2 φ(ν)/4π. Here, hν is the line energy (J), A 21 is the transition probability per unit time (s 1 ), n 2 is the density of atoms in the upper state (m 3 ), and the final 4π means that j ν is emission per steradian. The dimensionless factor φ(ν) is the line profile a delta-function spike for an idealized line, but in reality a sharply peaked, but broadened, function centered on ν, with φ(ν)dν = 1 (see slides3.1/3). This process gives rise to the emission lines we see. Notice that observations of emission lines from different transitions in the same gas (for example, the hydrogen lines Hα and Hβ) can provide accurate information on the gas temperature. Let s compare emission lines from the 2 1 and the i k transitions in the same atom. Temporarily 4

5 ignoring the line profiles, the ratio of the line strengths is j 21 = E 21 A 21 n 2 j ik E ik A ik n i ( ) E21 A 21 = e (E 2 E i )/kt, E ik A ik where we have assumed thermodynamic equilibrium in the second equation. The factors in parenthesis are atomic, or quantum mechanical, and hence in principle known. Thus, relative emission line strengths provide another accurate means of determining the temperature of stellar gas. Similar considerations apply to absorption lines Absorption Now let s turn to absorption lines. Our two-level atom can interact with an incoming photon in two ways: (1) it may absorb the photon and move from the lower 1 state to the upper 2 state, or (2) an atom in the upper 2 state may drop to the lower 1 state and emit another photon identical to the first. The first transition is what we usually think of as absorption; the second is called stimulated emission. Both are occurring in stars. In the presence of a radiation field of energy density (per frequency interval) u ν, the probability per unit time of an atom in the lower state absorbing a photon and jumping to the upper state is B 12 u ν. The probability per unit time of the reverse transition is B 21 u ν. The B ij are more Einstein coefficients, again characteristic of the atom, not the radiation field. Thus, if n 1 and n 2 are the number densities of atoms in the lower and upper states, respectively, and taking all radiative processes into account, we can write down a simple differential equation for the rate of change of n 2 : dn 2 = A 21 n 2 + B 12 u ν n 1 B 21 u ν n 2. dt Once again, thermodynamic equilibrium simplifies things a lot, and provides us with an important connection between the Einstein coefficients. The above equation is true in all cases, and in particular in the case of thermodynamic equilibrium, where the left-hand side is zero, u ν is given by the Planck expression u ν = 4π c B ν = 8πhν3 ( ) 1 ( 1 c 3 e hν/kt 1 F ν e hν/kt 1), and n 1 and n 2 are related by the Boltzmann formula n 2 = g 2 e hν/kt. n 1 g 1 Thus g 2 A 21 n 1 e hν/kt g 2 = B 12 u ν n 1 B 21 u ν n 1 e hν/kt g 1 g ) 1 A 21 g 2 (e hν/kt 1 = B 12 F ν g 1 e hν/kt B 21 F ν g 2. This latter relation must be true for any temperature T, so looking separately at the temperaturedependent and temperature-independent terms, we must have A 21 = F ν B 21 B 21 = g 1. B 12 g 2 5

6 Thus, as a ray of light of intensity I ν passes through distance δx of the medium, the combination of the two absorption ( B ) terms changes the intensity by an amount δi ν = hνφ(ν) c (n 1 B 12 n 2 B 21 ) I ν δx, where φ(ν) is the line profile centered on hν, as before. Using the above equations to write everything in terms of A 21, and after a little algebra, we find δi ν = α ν I ν δx, where is the absorption coefficient. α ν = c2 g 2 8πν 2 n 1 A 21 φ(ν) (1 ) e hν/kt g Radiation Transfer in Stars Setting aside the nitty gritty details of the previous section, we can say that, as a beam of light of intensity I ν passes through medium, we can say that the intensity changes due to emission and absorption according to the transfer equation di ν dx = α νi ν + j ν. The quantities j ν and α ν contain all necessary information about emission and absorption processes in the medium. Both are functions of density, temperature, composition, and frequency, and both, in general, are very complex. Note in particular from the preceding discussion that both are proportional to the local density of the medium. It is common to write j ν = ε ν ρ α ν = κ ν ρ, where ε ν and κ ν are, respectively, the emissivity and the opacity of the gas Mean Free Path We won t solve the transfer equation in detail, but we can use it to draw some important general conclusions about radiation in stars. Let s focus first on radiation moving through a purely absorbing medium (j ν = ), so the transfer equation simplifies to We can easily solve this by writing so di ν dx = α νi ν. di ν I ν = α ν dx I ν = I ν e α νdx = I ν e τν. 6

7 The incoming beam is simply attenuated as it moves through the medium. The quantity τ ν = α ν dx which determines the amount of attenuation is called the optical depth. As a rule of thumb, we can think of a photon as moving through the medium until τ 1, at which point it is absorbed and subsequently re-emitted. The mean free path l ν is the distance the photon travels before this occurs. In a uniform medium, α ν is constant with respect to x, so τ ν = α ν l ν and l ν = 1 α ν. Now consider an optically thin medium, in which absorption may be neglected. In that case, the transfer equation is even simpler: di ν dx = j ν, and the solution is I ν = j ν dx. In general, optically thin media are expected to produce emission lines (see the discussion of Kirchhoff s laws in week 2) Limb Darkening and Stellar Absorption Lines In the deep interior of the Sun, the opacity is so high that photons are effectively reabsorbed very close to where they were emitted, and light is effectively trapped. Put another way, the optical depth from any given location to the surface is very large, τ ν 1. We will discuss in more detail in a moment how solar energy works its way outward to the surface. Eventually, however, near the surface, both the opacity and the distance a photon must travel to escape decrease, and eventually we reach a point where the optical depth to the surface equals 1 a photon has a good chance of escaping without further interaction with solar matter. The light we see therefore comes from the part of the Sun lying above the depth R p, where R R p α ν (r) dr 1. This region is called the photosphere, and given conditions in the outer layers of the Sun, it is about 5 km deep, much less than the radius of the Sun, which is why the Sun (in visible light, at least) appears to have a very sharp edge. How deep we see into the Sun depends on the angle between our line of sight and the radial direction at the solar surface. Imagine for simplicity that α ν is constant, α. Then if we look straight down, we see down to a depth l = 1/α (see slides3.2/3). However, if we look at an angle θ to the radial direction, the light we see started closer to the surface, at depth l cos θ. Since the temperature decreases with radius, we effectively see a cooler part of the Sun when we observe at an oblique angle, and the Sun looks darker. This phenomenon, called limb darkening is evident in slides3.2/2. In addition, because α ν depends on frequency, R p does too, so the depth we can see also depends on the wavelength of the radiation. Near the center of an absorption line, where α ν is greatest, l ν is least; conversely, near the edge, α ν is smaller and l ν is larger (see slides3.2/4)). Thus the light we see in different parts of the line actually comes from different parts of the star. In particular, at line center we are seeing higher, cooler parts of the star, which is the real reason why the center is darker than the edges. 7

8 3.4 Free-Fall Time See Maoz Hydrostatic Equilibrium See Maoz Virial Theorem See Maoz 3.1. Here s an alternative derivation of the virial theorem. Starting from the equation of hydrostatic equilibrium dp dr = GMρ r 2. Multiplying by 4πr 3 and integrating, we have R 4πr 3 dp dr dr = R GM(r) 4πr 3 dr r 2 GM(r) = dm r = E gr, where the right side of the equation is the total gravitational potential energy of the star. The left side, as discussed in the text and in class, is R 3 4πr 2 P (r)dr. Now let s write down some thermodynamic expressions for the pressure P and the internal energy u of the (assumed ideal and nonrelativistic) gas so P = 2 3 u and P = nkt u = 3 2 nkt, R R 3 4πr 2 P (r)dr = 2 u(r) dv = 2T th, where E th is the total thermal energy. Thus, we recover the virial theorem E th = 1 2 E gr. This equation is important because it relates the total thermal energy of a star to the total gravitational energy two quantities determined by very different physical processes. The total energy of the star is E tot = E th + E gr. 8

9 Using the above relations, the virial theorem implies that E tot = 1 2 E gr. One obvious takeaway from this is that the total energy is negative. As a result, since the star is radiating energy into space, the total energy is decreasing, so the gravitational energy E gr becomes more negative and the thermal energy E th increases. Thus, as the star loses energy, it gets hotter. In a sense, it has negative heat capacity, and this spells long-term disaster as the loss of energy drives long-term evolution. 3.7 Mass Continuity Equation See Maoz Cross Sections We saw above that the absorption coefficient α ν depends on the local density, and may be written α ν = κ ν ρ, where κ ν contains a collection of quantum-mechanical and statistical factors that describe the detailed physics of the interaction. Alternatively, we can write α ν in terms of the local number density n. As we have just seen, α ν has a dimension of inverse length (1/l ν ). Factoring out the number density, the dimension of the other factor is length squared an area. Thus we can also write α ν = nσ ν, where σ ν is called the cross-section for the physical process of interest. In effect, it defines the area an absorber presents to the incoming radiation field for the given process to occur, and encapsulates the essential physics into a single number. Cross sections give us a convenient geometric means of quantifying interaction rates. Imagine that atoms actually have some small area that they present to the radiation. A photon striking that area interacts with the atom; otherwise, no interaction occurs. We can easily calculate the probability of an interaction, as follows. Imagine again a beam of light of intensity I ν moving through a medium of number density n and cross-section σ. Photons pass through a volume of area A perpendicular to the beam and length δx along the beam (see Maoz Fig. 3.2 and surrounding discussion). The total number of atoms in this volume Aδx is naδx, and the total area they present to the beam is A int = nσaδx. Thus, as seen by photons moving through the volume, the fraction of the area filled by interacting atoms, and hence the probability of an interaction, is δp = A int A = nσ νaδx A = nσ ν δx, and the above relation follows. The bottom line is that the absorption coefficient α can be rewritten in terms of the cross sections for various contributing physical processes: α ν = i n i σ i,ν = κ ν ρ Opacity and cross-sections are just different ways of expressing the same thing. 9