q = F If we integrate this equation over all the mass in a star, we have q dm = F (M) F (0)

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1 Astronomy 112: The Physics of Stars Class 4 Notes: Energy an Chemical Balance in Stars In the last class we introuce the iea of hyrostatic balance in stars, an showe that we coul use this concept to erive crue limits on their internal properties even without constructing a etaile moel. In this class we will apply the same sort of analysis to the energy an chemical balance in stars in effect examining the principles governing the energy an chemical content of stars rather than their mechanical equilibrium. We will see that this leas to similar non-obvious conclusions. It also sets us up to begin making etaile moels in the next few weeks. I. The Energy Equation A. Static Stars We ene last class by writing own the first law of thermoynamics for a given shell within a star. This enables us to make some interesting statements about the total energy content of a star. First consier the simplest case of a star in equilibrium, so that each shell s volume an specific internal energy are constant in time. In this case the left-han sie of the first law of thermoynamics is exactly zero because nothing is changing with time, an we have q = F m. If we integrate this equation over all the mass in a star, we have q m = F m m q m = F (M) F () Consier the physical meaning of this equation. The left-han sie is the total rate of nuclear energy generation in the star, summing over all the star s mass. We call this quantity the nuclear luminosity L nuc a luminosity because it has units of energy per time; i.e., it is the total rate at which nuclear reactions in the star release energy. When we integrate the right-han sie, we win up with the ifference between the flux passing through the last mass shell, F (M), an the flux entering the star at m =, F (). The latter is obviously zero, unless there is a magical energy source at the center of the star. The former, F (M) is just the energy per unit time leaving the stellar surface. Thus, F (M) must be the star s total luminous output, which we see as light, an enote L.

2 Thus the equation we have erive simply states L nuc = L, i.e. for a state in equilibrium, the total energy leaving the stellar surface must be equal to the total rate at which nuclear reactions within the star release energy. This isn t exactly a shocking conclusion, but the machinery we use to erive it will be prove extremely useful when we consier stars that are not exactly in equilibrium. B. Time-Variable Stars The calculation we just performe can be generalize to the case of a star that is not exactly in equilibrium, so that the time erivatives are not zero. If we retain these terms an integrate over mass again, we have u M t m + P t ( ) 1 M m = q m F (M) + F () = L nuc L. ρ For the first term on the left-han sie, since m oes not epen on t, we can interchange the integral an the time erivative. Thus, we have u t m = u m = t t U, where U is the total internal energy of the star. For the secon term, it is convenient to rewrite the time erivative of 1/ρ in a slightly ifferent form: ( ) 1 = ( ) V = ( ) V, t ρ t m m t where we have use the fact that m oes not epen on t to interchange the orer of the erivatives. Here V is the volume of the material insie mass shell m. If this shell is at raius r, then V = (4/3)πr 3, an V t = 4πr2 r t. In other wors, the rate at which the volume occupie by a given mass of gas changes is equal to the surface area of its outer bounary (4πr 2 ) multiplie by the rate at which it expans or contracts (r/t). Putting this into the integral, we have P ( ) 1 M m = P ( 4πr 2 r ) m. t ρ m t

3 We can evaluate this integral by parts. Doing so gives P ( ) [ 1 m = 4πr 2 P r ] M 4πr 2 r P t ρ t t m m. For the first term, note that at m =, r/t =. This is because the innermost shell must stay at the center, unless a vacuum appears aroun the origin. Similarly, P = at m = M, because the pressure rops to zero at the ege of the star. (Strictly speaking it is not exactly zero, but it is negligibly small.) Thus, the term in square brackets must be zero. For the secon term, we can evaluate using the equation of motion that we erive in the last class. To remin you, we erive the equation If we re-arrange this we get r = Gm r 2 1 P ρ r. P r = GMρ ρ r. r 2 Converting to Lagrangian coorinates, we have P m = 1 P 4πr 2 ρ r = Gm 4πr r 4 4πr 2 If we substitute this into our integral, we have P ( ) 1 M Gm m = t ρ r ṙ m + 2 ṙ r m The meanings of the terms are a bit clearer if we rewrite them a bit. In particular, we can write Gm r ṙ = ( ) Gm 2 t r an so that we have P t ( ) 1 m = ρ t ṙ r = 1 2 t (ṙ2 ), Gm r m t ṙ 2 m = Ω + T. As earlier, we have use the fact that mass oes not epen on time to interchange the integral with the time erivative. The first term, which I have written Ω, is clearly just minus the time erivative of the gravitational potential energy; Gm/r is the gravitational potential experience by the shell at mass m, so the integral of Gm m/r is just the total

4 gravitational potential energy over the entire star. Similarly, the term I have labelle T is just the time erivative of the total kinetic energy of the clou; ṙ 2 /2 is the kinetic energy per unit mass of a shell, so the integral of (1/2)ṙ 2 m is just the total kinetic energy of the star. Putting it all together, we arrive at the total energy equation for the star: U + Ω + T = L nuc L. This just represents the total energy equation for the star, an it is fairly intuitively obvious. It just says that the time rate of change of internal energy plus gravitational potential energy plus kinetic energy is equal to the rate at which nuclear reactions a energy minus the rate at which energy is raiate away from the stellar surface. We can get something slightly more interesting if we consier a star that is expaning or contracting extremely slowly, so that its very close to hyrostatic balance. In this case we can make two simplifications: (1) we can rop the term T, because the star s kinetic energy is tiny compare to its internal energy or gravitational potential energy; (2) we can use the virial theorem for hyrostatic objects, which says that U = Ω/2. In this case the energy equation reas 1 2 Ω = L nuc L. This equation is eciely non-obvious. It tells us how quickly the gravitational potential energy of the star changes in response to energy generation by nuclear reactions an energy loss by raiation. C. The Kelvin-Helmholtz Timescale It is useful to consier an orer-of-magnitue version of the energy equation, because it tells us something important about the nature of stars. Consier how long will it take the gravitational potential energy (an thus the stellar raius) to change by a significant amount in a star without any nuclear energy generation (i.e. where L nuc = ). The gravitational potential energy is Ω = α GM 2 R, where α is a number of orer unity that epens on the ensity istribution within the star, an R is the star s raius. If we efine t as the time require to alter the gravitational potential energy significantly, then at the orer-of-magnitue level the equation we have written reas 1 ( ) GM 2 L t R t GM 2 RL.

5 We efine the quantity t KH = GM 2 /(RL) as the Kelvin-Helmholtz timescale, name after the 19th century physicists Kelvin (of the Kelvin temperature scale) an Helmholtz, who first pointe out its importance. The meaning of t KH is that it is the time for which a star coul be powere by gravity alone without its raius changing very much. Similarly, if we have a star that is not in energy balance for some reason, so that L nuc L, then t KH tells us about the time that will be require for the star to reach energy equilibrium. If we plug in numerical values for the Sun, we fin t KH = GM /(R 2 L ) = 3 Myr. This number is the answer to the Star Trek problem of what woul happen if you somehow shut off all nuclear reactions insie a star. The answer is: absolutely nothing for about 3 Myr. You coul turn off all the nuclear reactions in the Sun, an unless you were observing neutrinos (which get out immeiately), you wouln t notice anything change for tens of millions of years. In fact, before the iscovery of nuclear energy, it was believe that gravity was the main process causing the Sun to shine. This playe an important role in the history of science, because, if gravity really i power the Sun, it woul imply that the Sun s properties coul only have been constant for 3 Myr. This woul be a natural limit for the age of the Earth, or at least for the time for which life as we know it coul have existe on Earth. This was an important argument against the Darwinian theory of evolution, which require billions of years to explain the evelopment of life. Of course, as we ll iscuss in a few minutes, in this case the biologists were right an the astronomers were wrong. In reality L nuc is very close to L, so the real timescale for the Sun s evolution is vastly longer than 3 Myr. II. Nuclear Energy an the Nuclear Timescale There is one more important timescale for star s, which comes from consiering how long L nuc L can be maintaine. This epens a bit on nuclear chemistry, which will be our topic in a few weeks. For now, we ll simply take on faith that the main nuclear reaction that occurs in the Sun is burning hyrogen into helium. You can figure out how much energy this yiels by comparing the masses of hyrogen an helium. The starting point of the reaction is 4 hyrogen nuclei an the final point is 1 helium nucleus. The mass of 1 hyrogen nucleus is g, while the mass of a helium nucleus is The ifference in mass is m = 4m H m He = g. If we phrase this as the change in mass per hyrogen atom we starte with, this is ɛ = m 4m H =.66. In other wors, this reaction converts.66% of the mass of each proton into energy. Einstein s relativity then tell us that the excess energy release per hyrogen atom by this reaction is E = ɛm H c 2 = erg = 6.2 MeV.

6 We can use this to estimate how long the nuclear reaction that burns hyrogen into helium can keep a star going. The star raiates energy at a rate L, an thus the rate at which hyrogen atoms must be burne is L/ E = L/(ɛm H c 2 ). To estimate how long this can keep going, we simply ivie the total number of hyrogen atoms in the star, roughly M/m H, but the rate at which they are burne. This efines the nuclear timescale t nuc = M/m H L/(ɛm H c 2 ) = ɛmc2 L (In making this estimate we have implicitly neglecte the fact that not all of the Sun s mass is hyrogen, but that s a fairly small correction.) Evaluating this for the Sun gives 1 Gyr, a staggeringly long time almost a factor of 1 larger than the age of the universe. In fact, we ll see later in the course that the true time for which nuclear reactions can ol up the Sun is about a factor of 1 smaller, because the Sun can t actually use all of its hyrogen as fuel. Nonetheless, this result emonstrates that nuclear energy is able to hol up the Sun for much, much longer than the Kelvin-Helmholtz timescale. III. The Hierarchy of Timescales an Evolutionary Moels A. General Iea The three timescales we ve compute this week tell us a great eal about the ingreients we nee to make a moel of stars. Putting them in orer, we have t nuc t KH t yn, an this is true not just for the Sun, but for all stars. This has the important implication that, on timescales comparable to t nuc, we can assume that stars are in nearly perfect mechanical an thermal equilibrium. This enables a great simplification in making moels of stars. Our approach to making stellar moels for the rest of the course will therefore procee through a series of steps: 1. Assume that the star is in perfect mechanical equilibrium (since t yn t KH t nuc ), an compute the resulting luminosity. 2. Assume that the star is in perfect energy equilibrium (since t KH t nuc ), an compute the reaction rate require to provie this luminosity. 3. Evolve the chemical makeup of the star using the erive reaction rates. For a star that is powere by hyrogen burning into helium, steps 2 an 3 are extraorinarily simple. If we want to keep track of a star s evolution, we just nee to keep track of the mass of hyrogen an the mass of helium within it. Let M be the total stellar mass, an let M H an M He be the mass of hyrogen an helium. If we neglect the mass of other elements (a reasonable first approximation) an any change in mass ue to raiation (which is tiny) an ue to stellar wins (which is small), then we can write M = M H + M He = constant. Steps 2 an 3 therefore

7 are entirely emboie by the equation Ṁ He = ṀH = L ɛc 2. This is a bit of an oversimplification, since in reality we nee to o this on a shellby-shell basis, an to worry about chemical mixing between the shells. Nonetheless, it conveys the basic iea. Of course the har part of this is in step 1: compute the luminosity of a star that is in mechanical equilibrium. This luminosity, it turns out, will epen on nothing but the star s mass which is what observations have alreay tol us, if we recall the mass luminosity relation. Deriving that mass-luminosity relation from first principles is going to consume the next four weeks of the course. Once we ve one that, however, what this timescale analysis tells us is that we will essentially have a full theory for stellar evolution. B. The Chemical Evolution Equations To make this more precise, we nee to write own the equations governing change in the composition of the star. To o this, we nee to introuce some notation. A star is mae of many ifferent elements the vast majority are hyrogen an helium, but there are others, an it turns out that they can play important roles, particularly in evolve stars. If we examine some volume of gas in a star, we can iscuss its ensity ρ, but we coul also count only the hyrogen atoms, only the helium atoms, etc., an compute the ensity for that species only. For convenience we number these species, so that we might write the ensity of hyrogen as ρ 1, the ensity of helium as ρ 2., etc. Depening on the level of sophistication of our moel, we might also istinguish ifferent isotopes of the same atom, so that we woul count orinary hyrogen an euterium (which has an extra neutron) separately. It is often convenient to work with quantities other than mass ensities, so we efine some alternatives. The mass fraction of species i is written X i = ρ i ρ. We often also want to count the number of atoms, instea of measuring their mass. To o this, we have to take into account the ifferences in their atomic weights. For example, helium atoms have four times the mass of hyrogen atoms, so if there are four hyrogen atoms for every helium atom, then they both have the same mass fraction. We write the atomic mass number for species i as an integer A i, which means that each atom of that species has an atomic mass of approximately A i m p, where m p = is the proton mass. Hyrogen has A = 1, so m H = m p, an we frequently write masses in terms of m H rather than m p. Given this efinition, it is clear that the number ensity of atoms is relate to their mass ensity by n i = ρ i, A i m H

8 an substituting this into the mass fraction X i immeiately gives X i = n i A i ρ m H. We can escribe any species in terms of its atomic mass A i an also its atomic number, Z i. The atomic number gives the number of protons, an thus the charge of the nucleus. For example for the stable isotopes of the most common elements we have Element Name Z A 1 H Hyrogen H Deuterium He Helium He Helium C Carbon C Carbon Some nuclear reactions also involve electrons an positrons. This have Z = ±1 an A = (to goo approximation we can generally neglect the mass of electrons compare to protons an neutrons). The mass fractions X i can be altere by nuclear reactions, which leave the total mass ensity fixe (to very goo approximation), but convert atoms of one species into atoms of another. We can write one species as I(Z i, A i ), another as J(Z j, A j ), an so forth. In this notation, any chemical reaction is I(Z i, A i ) + J(Z j, A j ) K(Z k, A k ) + L(Z l, A l ), where this can obviously be extene to more elements as neee if the reaction involves more species. Chemical reactions always have to conserve mass an charge, so we have two conservation laws: Z i + Z j = Z k + Z l A i + A j = A k + A l. If we want to know how the star s composition changes in time, we nee to know the rate at which reactions between two species occur. Computing these reactions rates is a problem we ll efer for now, but we can begin to think about them by noting that the reaction rate will always be proportional to the rate at which atoms of the two reactant species run into one another. To see what this implies, consier a given volume of space within which a chemical reaction is taking place, for example 2 H + 1 H 3 He, one of the steps in the energy-generating reaction chain in the Sun. What woul happen to the rate at which this reaction occurre in that volume if I were to

9 remove half the euterium ( 2 H)? The remaining euterium atoms woul still encounter hyrogen atoms just as often, since their number woul be unchange, so the reaction rate woul just be reuce by a factor of 2. Similarly, the same argument shows that if I were, for example, ouble the number of hyrogen atoms, the reaction rate woul increase by a factor of 2. Clearly the reaction rate must be proportional to the number of members of each reactant species in the volume. Expressing this mathematically, the reaction rate per unit volume must be proportional to n i n j, where n i an n j are the number ensity of the species i an j involve in the reaction. If there are more species involve, then we just multiply by n k, n l, etc. We call the constant of proportionality the reaction rate, an write it R ijk, meaning the rate at which reactions between particles of species i an j occur, leaing to species k. If the reaction involves two members of the same species, the same argument applies, except that we have to be careful not to ouble-count. Thus rather than having the reaction rate be proportional to n i n j, it is proportional to n i (n i 1)/2 n 2 i /2, where the factor of 1/2 is to hanle the ouble-counting problem. You can convince yourself that this is right. Suppose there are 4 people in a room: Alice, Bob, Cathy, an Davi (A, B, C, an D). How many istinct couples can we make? Counting them is pretty easy, since we just pick one person, then pick another ifferent person. If we pick Alice, there are 3 possible partners: Bob, Cathy, an Davi. Similarly, if we pick Bob, there are three possible partners: Alice, Cathy, an Davi. We can write this out as AB AC AD BA BC BD CA CB CD DA DB DC The table has 3 4 = 12 entries. It s pretty clear, however, that half of the entries in this table are uplicates: we have both BA an AB. Thus to count the number of istinct couples, we have n i = 4 an n i (n i 1)/2 = 6. The argument is the same for chemical reactions: if we were to make a list of possible collisions, there woul be n i (n i 1)/2, which for large values of n i is approximately n 2 i /2. Thus the reaction rate is n 2 i R iik /2. With this notation out of the way, we are now prepare to write own the equations of chemical evolution. Suppose that we have a number ensity n i of species i, an that these atoms are estroye by a reaction with species j, which leas to species k. Clearly the rate of change in the number ensity of species i is just given by minus the rate at which reactions occur: t n i = n i n j R ijk. In this case we on t ivie by 2 when the reaction is species i with itself because each reaction estroys two atoms of species i, an the factors of 2 cancel. In

10 general there are multiple possible reactions with many possible partners, an we have to sum over all the reactions that estroy members of species i. Thus t n i = j,k n i n j R ijk. We must also take into account that reactions can create members of species i. Suppose we have a reaction between species l an species k that creates members of species i. In this case the rate at which members of species i are create is t n i = n l n k R lki if l an k are istinct, or t n i = 1 2 n ln k R lki if l an k are the same. We can unify the notation for these two cases by writing t n i = n ln k 1 + δ lk R lki, where δ lk is simply efine to be 1 if l an k are the same, an otherwise. This is nothing more than a notational convenience. Again, we nee to sum over all possible reactions that can create species i: t n i = l,k n l n k 1 + δ lk R lki, Finally, combining the rates of creation an estruction, we have t n i = l,k n l n k 1 + δ lk R lki j,k n i n j R ijk. If we prefer, we can also write this in terms of the composition fraction by substituting: ( ) t X i = ρ A i X l X k R lki X i X j R ijk m H A l A k 1 + δ lk A i A j l,k j,k C. The Evolution Equations We are now in a position to write own the basic equations governing a star s evolution. We envision a spherical star of fixe mass, whose chemical composition at some initial time is known. To figure out how its structure changes in time, our first step is to assume mechanical equilibrium, i.e. hyrostatic balance. We re alreay written own the equation for this: P m = Gm 4πr 4.

11 This equation gives us a relationship between the position of each shell of mass, r, an the graient in the pressure of the gas P. The solution tells us how the shells of gas must arrange themselves to maintain mechanical equilibrium. The secon step is to require thermal equilibrium, which means that we balance the energy being generate in the star against the energy it is raiating. We ve alreay written own the integrate version of this as L = L nuc, but if we re going to moel the shells iniviually, we want to use the non-integrate version we wrote own earlier: F m = q, where F is the flux passing through a mass shell, an q is the heat generate within it by nuclear burning. The final step is to figure out how it is changing chemically, which is escribe by the equations we have just written out: ( ) t X i = ρ A i X l X k R lki X i X j R ijk m H A l A k 1 + δ lk A i A j for each species i. l,k Of course these equations are all couple. The rates of nuclear reactions in the final step etermine the rate of heat generation q in the secon one, or vice versa. Similarly, the ensities, which come from r(m), also affect the chemical evolution rates. Thus we have a set of couple non-linear ifferential equations to solve. Moreover, our set of equations is not yet complete. Most obviously, we haven t yet written own the reaction rates R ijk or the way that nuclear energy generation rate q epens on them. Also, we have not specifie yet how the pressure epens on ensity, temperature, or anything else. To solve for the evolution of a star, we will nee to fill in these gaps. j,k