In this course, we won t talk much about earlier/later phases:

Transcription

1 PHYSICS OF THE SUN The Sun is a main-sequence star... still burning its initial supply of hydrogen. In this course, we won t talk much about earlier/later phases: Mostly, we ll concentrate on how the Sun currently maintains itself in a quasi-steady state, with Solar mass... M = kg Solar radius... R = m Solar luminosity... L = W Solar core temperature... T core = 15.7 million K Solar effective temperature... T eff = 5770 K Age of Sun & solar system... t 4.6 Gyr For context, main sequence stars have M between about 0.1 and 100 M. Brown dwarfs (not massive enough to burn hydrogen) have M between about 0.01 to 0.1 M. Less than that: planets! Also, two key derived quantities: {surface flux} = σt 4 eff = L 4πR 2, {surface gravity} = g = GM R

2 What we will do: In this part of the course, we ll try to answer two basic questions about the basic physics of a stellar interior: Where does the energy come from? Gravitational contraction Nuclear burning (fusion) Radioactive decay (fission)? Tidal heating (if it has a massive neighbor & eccentric orbit)? How is energy transported out through the interior? radiation (slow diffusion of photons) heat conduction (slow diffusion of particles) convection (unstable bubbling) After that, we ll talk about the solar atmosphere, which is the part we actually see (photons stop random walking; start freely escaping). Sources of Energy First, gravity. Sir Arthur Eddington, in early 1900s, postulated that stars generated their L by slowly contracting, and converting gravitational energy into heat & light. We now know that this isn t how the Sun generates energy, but this basic process should still be explored. Thus, let s look at the effects of gravity on the solar interior. Remember our MHD momentum equation, for B = 0, u t +u u = g 1 ρ P A static star is held together via hydrostatic equilibrium, in which all u terms = 0. We solved for ρ(r) for the simple case of both g and T being constant, but that s not true in general. 5.2

3 For spherical symmetry, a point a distance r from the origin only cares about the mass interior to r, which we ll call M r (r). (Effects of mass outside r cancel out.) We can treat the interior mass as if it s a point-mass at the origin. Thus, g = ê r GM r r 2 and P r = GM rρ r 2 We also write a conservation equation for M r by looking at thin, infinitesimal shells. It s kind of just bookkeeping: Mass contained: dm r Volume: 4πr 2 dr Alternate estimate of mass contained: ρ(4πr 2 dr) Thus, Enclosed mass: M r = M r r Mr 0 = 4πr 2 ρ dm r = r 0 dr 4πr 2 ρ(r ) This equation essentially takes the place of the mass conservation equation (again, in the limit of u = 0 and / t = 0). These two equations are basic building blocks for much of what we ll do regarding models of stellar interiors. 5.3

4 However... first, we d like to know something even more basic than how static objects are maintained. Let s posit some thought experiments.... If the pressure gradient support got turned off somehow, how fast would gravity alone be able to make changes to the system and reconfigure the mass? We can derive the stellar dynamical time t D in several ways. Here s a simple way to estimate the free-fall time: Imagine a test particle at distance R from a point-mass M at the center. Here, let s assume that a = GM /R 2 stays constant over the time we re considering. It s Physics d 2 r dt = dv r 2 dt Integrating again, r r 0 = at2 2 = a, so v = at (starting from rest at v 0 = 0, t 0 = 0) (r r 0 = R ), and thus t D 2R 3 GM We could do better by considering the actual collapse of a spherical distribution of gas... i.e., as it collapses (r ), the gravitational force increases. However, the end-result is only off by a bit... replace factor of 2 above by (π 2 /8) For the Sun, t D 30 minutes.... The above is nice to know, but in most situations we know that spherical, self-gravitating bodies are supported by gas pressure, too. The time it would take gravity to fight against the real reservoir of thermal energy is going to really be much longer than t D. 5.4

5 To estimate it, we first need to derive a fundamental relationship between the two main energy components of a star: The Virial Theorem Again, there are several ways to go about this. We ll do it by building an analytic model of the solar interior that obeys the mass & momentum equations that we derived above. The big assumption in this model is that the gas maintains a constant density [ρ(r) = ρ 0 ] throughout the whole interior. This is a notoriously bad approximation, but amazingly the physics insight it gives us is huge... Start with mass conservation: dm r dr = 4πr 2 ρ 0 It s (relatively) easy to solve because it s separable: M 0 R dm r = 4πρ 0 dr r 2 ( ) R 3 M = 4πρ M 0 i.e., ρ 0 = 3 (4/3)πR 3 so if we know any two of {ρ 0,M,R } this gives the third. 0 = ρ If we stop the integration part-way, we get M r (r) = 4 3 πρ 0r 3 and this (black curve) often doesn t look very much like what one gets numerically for more centrally condensed stars (red curve). Putting aside that bit of incongruity, we can solve the hydrostatic equilibrium equation for P(r), dp dr = GM r ρ r 2 0 = 4πGρ2 0 r 3 (again, a separable diff. eqn.) 5.5

6 Keeping track of limits, P(r) dp = 4πGρ2 0 P c 3 r 0 dr r (though we don t know P c yet) P(r) P c = 2 3 πgρ2 0 r2 If we set P 0 at r = R, we can solve for the central pressure: P c = 2π 3 Gρ2 0R 2 = 3GM2 8πR 4 ( ) And also, P(r) = P c 1 r2 R 2 Note that these results are independent of the equation of state. It s true for an ideal gas... also true for weird crystalline matter in the cores of white dwarfs! However, we do know that the plasma in the solar interior obeys an ideal gas equation of state. So if we know P(r) and ρ 0, we can use the equation of state to solve for T(r), P(r) = ρ ) 0k B T(r), so... T(r) = T c (1 r2 P c = k BT c = GM µm H ρ 0 µm H 2R Note: for the Sun s ionized interior, µ 0.6, we get T c = 7 million K. A more realistic solar model would replace the 1/2 on the RHS by a factor closer to 1, to get the actual value of 15 million K. But we re close! Note that we ve solved for the core temperature of the Sun without knowing anything about nuclear fusion.... The above relationship looks very fundamental. Note that both sides have units of velocity squared: thermal speed escape speed in interior. R 2 5.6

7 Another way to express this is by looking at the total energy components of the star. If there s NO rotation, mass motions, magnetic fields, etc., Total energy: E tot = E K +E G (in actual Joules) E K is the thermal energy of an entire system, integrated over its volume. E G is the system s integrated gravitational potential energy. We d like to know how E K and E G relate to one another. For the thermal energy, E K = dv U = Also, dv P γ 1 for an ideal gas. V = 4 3 πr3, dv = 4πr 2 dr = d 3 r. For our constant-density star, made up of ideal gas, we can compute R E K = dv U = dr 4πr P(r) = = 3 GM 2 10 R 0 What about E G? For test particles of mass m in the field of a gravitating body of mass M, we know that E G = GM m r where r is the distance between the two bodies. So, for a star in its own potential well, it s analogous to write E G = dm r GM r r We can solve for this using our constant-density model, too: GM r E G = dm r = = 3 GM 2 = 2E K! r 5 R The general version of the virial theorem is:. and for an ideal gas with γ = 5/3, E G = 3(1 γ)e K E G = 2E K and E tot = ( E G E ) G 2 = E G 2 5.7

8 This actually holds not just for the constant-density model, but for more general stellar models (as long as it s made out of ideal gas). Note that since E G is negative, then the total energy E tot is negative, too. (The star is bound.)... Remember Sir Arthur Eddington and his theory about collapse generating energy. If R is decreasing, then E G decreases (i.e., becomes more negative) in time, and thus E tot keeps decreasing, too. Eddington assumed total energy is lost in the form of radiation; i.e., that L = de tot dt = 1 2 de G dt > 0. If all that s changing is R, then we can take the derivative, L = 1 ( ) ( ) d 3GM2 = 3GM2 d 1 = 3GM2 dr 2 dt 5R 10 dt R 10R 2 dt Note that L > 0 because both E G < 0 and dr /dt < 0. = E G dr 2R dt. How long could a star survive before collapsing to zero? i.e., we want to solve dr dt R t R t K H for the so-called Kelvin-Helmholtz timescale t K H. Using the above, we get t K H E G 2L This is also roughly the same as the time it would take for the star to lose all of its thermal energy by radiating it away at a given luminosity (t K H E K /L ). Thus, t K H is sometimes called the thermal diffusion timescale. 5.8

9 For the Sun s present-day values of M, R, L, we get t K H 10 7 years. ( only 10 million years) Of course, carbon dating, moon rocks, and lots of other evidence point to the age of the solar system as years, so there must be some other major energy source besides gravity! We know this other source is nuclear burning. However, the Sun s core temperature is NOT set by nuclear burning! We saw earlier how it s set by the balance between gravity & gas pressure. Fusion allows this high T c to be maintained for much longer than would otherwise be possible! t D 30 minutes t K H 10 7 years t MS life years Stellar interiors are thus one big balance... If the momentum equation is unbalanced (i.e., gravity only), the star collapses almost immediately. If the momentum equation is balanced ( P = ρg), but the energy equation is unbalanced (i.e., cooling via L without anything else on the RHS), the star takes millions of years to collapse. However, if we balance the energy equation with heating from nuclear burning, we get something much more stable! 5.9

10 Nuclear energy generation Let s delve into the different types of nuclear reactions that can occur when nuclei are bombarding one another. Basic ingredients of nuclear reactions: Baryons/Nucleons: protons (p + ), neutrons (n 0 ) Leptons: electrons (e ), electron neutrinos (ν e ) Anti-particles: Photons: p, n 0, e +, ν e γ ( raw energy, makes up for E = mc 2 imbalances) Nuclear reactions are similar to chemical reactions: ingredients come together and get transformed into products. In all these reactions, the conserved quantities (i.e., equal on left and right sides) are: total mass-energy (E = mc 2 ) total charge baryon number (i.e., # of nucleons) lepton number (leptons = 1, anti-leptons = 1) (In strong interactions, the #s of protons & neutrons tend to be conserved separately, too which helps automatically conserve charge.) There are 2 main types of reactions: Strong interactions: e.g., N 1 +N 2 N 3 +γ These involve nucleons grouping up or splitting apart. Usually, N 1 is an arbitrary nucleus, and N 2 is often just a p +, n 0, or α particle. For two positively charged nuclei, they overcome Coulomb repulsion only if they get very close to one another. The above is an exergonic (energy releasing) fusion reaction. If the γ is on the left side, it s endergonic (energy consuming). If N 3 N 1 +N 2, it s fission. 5.10

11 Weak interactions: e.g., radioactive beta decay... n 0 p + +e +ν e These usually occur within an existing nucleus. There s usually no γ produced. We care about them because they help get a nucleus into a state in which strong reactions can then happen. All sorts of other combinations (of what goes on the left/right sides) are possible, as long as the conserved quantities are conserved... Thus, some other possible weak interactions are: p + n 0 +e + +ν e p + +e n 0 +ν e (positron emission in proton-rich nuclei) (inverse beta decay; electron capture in NS) n 0 +ν e p + +e (used in chlorine neutrino detectors; 37 Cl+ν e 37 Ar+e 37 Ar has short half-life; detect its decay!) A good way to think about weak reactions is to realize that a nucleus with Z protons has an ideal number of neutrons needed to make it a stable isotope. Too many neutrons, and it will undergo n p beta decay. Too few neutrons, and it will undergo p n inverse beta decay. Mini-chemistry refresher: (most stable isotopes of example elements) Element Z (# protons) N (# neutrons) Helium 2 2 Carbon 6 6 Oxygen 8 8 Sulfur Chlorine Iron Silver Gold Uranium Each proton wants to repel away from its neighbor protons. The strong force keeps nuclei together, but nuclei with more protons need even more neutrons to help prevent the Coulomb repulsion due to all those positive charges

12 We care about the energy released in all of the above types of reactions, both by photons (γ) which are rapidly thermalized and by neutrinos. Why is energy released? The output nuclei are in a lower mass-energy state than the input nuclei. The relative energies are measured in terms of B, the nuclear binding energy. It s energy that needs to be added to break a nucleus apart into its component protons & neutrons: m nucleus + B c 2 = m p,m n If the reaction is going, that B energy is released! Expressing masses in units of MeV/c 2, we can compare ( mp = MeV/c 2,m n = MeV/c 2,m e = MeV/c 2) to some example binding energies: For 1 H, B = 0 (though the bound electron contrib s 13.6 ev) For 2 D, B 2.2 MeV ( 1.1 MeV/nucleon) For 4 He, B 27 MeV ( 7 MeV/nucleon) For 56 Fe, B some big # ( 8.8 MeV/nucleon) Handout: Binding energies per nucleon (B/A) are shown as a function of atomic mass A. Unstable isotopes have lower binding energies than the stable ones, so they re more apt to break apart. Being higher up on this diagram means it s more stable. Thus, more input energy would be needed to break it apart. Favorable reactions go from lower to higher binding energy. ր is fusion... տ is fission. Note: One gets the most bang for the buck out of converting H to He. Less from further stages of fusion? This is why a star s main sequence (H-burning) lifetime is so long, compared to successive stages of nuclear burning. 5.12

13 Top: Binding energies of atomic nuclei (B/A) in MeV, shown as a function of mass number A. Data for all known isotopes (black points) come from Audi & Wapstra (1993, Nucl. Phys., A565, 1). The red curve connects the stable isotopes. Bottom: Solar elemental abundances, from Asplund et al. (2009, Ann. Rev. Astron. Astrophys., 47, 481). 5.13

14 Anyway, if we manage to fuse 4 1 H 4 He, there s 27 MeV left over (i.e., released as γ). Is it plausible to power a star in this way? In other words, how much 1 H would need to fuse into 4 He in order to support the Sun over it s lifetime so far? L = E t, solve for E = L t For L W, and t years ( seconds/year) s Thus, E J MeV. If it s 30 MeV per reaction, then reactions are needed. 4 (m H kg) kg go into each reaction Thus, we need kg of hydrogen to power the Sun over its lifetime so far. Since M kg, this is 1/20 of a solar mass. This is a plausible order of magnitude, since we know it s just the core that undergoes nuclear fusion. Problem: How can these reactions ever really occur? Strong interactions involve two positively charged nuclei... But here s where the strong nuclear force comes into play. It turns attractive for tiny distances: 5.14

15 V 12 (r) is the potential energy of nucleon 1 as seen by nucleon 2. (V 12 = V 21 ) Positive potential: repulsive force. Negative potential: attractive force (like a gravity well ). In the outer part, for two positive charges, V 12 (r) k eq 1 q 2 r > 0. But, if the two + charges manage to get close enough together, the potential goes negative and they stick! (This inner part doesn t care about charge: pp, pn, nn are similar.) To penetrate the barrier, a particle s kinetic energy, E = mv 2 /2, must exceed max(v 12 ). (I used nonrelativistic E here because even though the electrons may be relativistic in stellar cores, the ions & nuclei rarely go so fast.) Let s compare E and V 12 with one another... Because N 1 and N 2 can have unequal masses (m 1,m 2 ) and charges (q 1 = Z 1 e,q 2 = Z 2 e), it makes the most sense to use a center-of-mass coordinate system. This is handy; you ve probably seen from classical mechanics how this makes the calculations easier. Total energy: E = 1 2 m 1 v m 2 v 2 2 = 1 2 m 12 v

16 where m 12 = m 1m 2 m 1 +m 2 (reduced mass), v 12 = v 1 v 2 (relative velocity). The Coulomb potential (for r > r n ) is given by V 12 (r) = k e Z 1 Z 2 e 2 /r. In order for random thermal motions to overcome the barrier, we want 1 2 m 12v12 2 = 3 2 k BT = k ez 1 Z 2 e 2 which means T must exceed 2k ez 1 Z 2 e 2. r n 3k B r n Big question: what s the value of r n? A typical nucleus has a size of 1 fm m. (Five orders of magnitude smaller than an atom; 1 Å m.) For protons (Z 1 = Z 2 = 1) and the assumption of r n 1 fm, we get T K. This is kind of high for stellar interiors. It s telling us that we re missing something... Remember, however, that weird quantum stuff (wave-particle duality) happens when we get down to the size of an object s de Broglie wavelength λ db. It doesn t make much sense to talk about smaller sizes. Recall that, in quantum mechanics, all particles exhibit wave-like characteristics. de Broglie showed that a kind of uncertainty principle is obeyed by particles, x p > h If we know a particle s momentum p, it will exhibit matter waves with wavelength λ db = h p and particles will produce diffraction patterns when shot at targets, flow through narrow slits, etc. 5.16

17 One minor complication: the most probable momentum p depends on T, so λ db depends on T also. In an ideal gas, 3 2 k BT = p2 2m 12, so λ db = h 3m12 k B T If λ db is bigger than the actual size of the nucleus, then the precise position of the nucleus is fuzzy. Thus, let s try assuming r n = λ db. We get T = 2k ez 1 Z 2 e 2 3m12 k B T 3k B h which requires us to solve for T 1/2 and square it... T = 4 3 k 2 e Z2 1 Z2 2 e4 m 12 k B h K which is much more realistic. Here, λ db r n m, between 1 fm & 1 Å. Thus, the interactions between nuclei are inherently quantum mechanical (i.e., probabilistic and fuzzy ). 5.17

18 Okay, we ve shown that nuclear fusion can supply enough energy, and that barrier penetration is plausible. Computing the probability of barrier penetration is a nasty quantum problem... George Gamow worked it out in the 1940s. We need to know how often the reactions actually occur in a plasma of given ρ, T, P,... i.e., we want the reaction rate between input species 1 & 2, Total # of fusion reactions that r 12 = occur per unit volume, per unit (in units of #/m3 /s) time The reaction rate tells us how number densities change vs. time. r 12 ends up being a function of ρ, T, and composition (the mixture of elements in the star). For example, if the system contained nothing but the (N 1 +N 2 N 3 ) fusion reaction, and we write the rate as r 12,3, then n 1 t = r 12,3, n 2 t = r 12,3, n 3 t = +r 12,3 In general, n i t = j r ij,k + m,n r mn,i and for a hypothetical single reaction in which 4 protons converge simultaneously to turn into a helium nucleus, we have n p t = 4r, n 4He t = +r 5.18

19 And, conservation of total baryon number is approximately the same as total mass conservation, so one can assume ρ/ t = 0 and use the mass fractions X = m pn p ρ, Y = 4m pn 4He ρ i.e., as X goes down, Y goes up. to show that X t = Y t All that quantum mechanics (combined with assumption of a M B distribution of nuclei in the hot plasma) gives r 12 ρ 2 {complicated function of T } and the ρ 2 is really n 1 n 2... i.e., it depends on how many particles of both input species are present. The T dependence is really a v dependence (or momentum p dependence), and all the de Broglie, barrier penetration stuff is in there. For stellar interiors, we will care about how much power is released, per kg of material in the core, per second: ǫ = L r r 12B net M r ρ (make sure units work out) where B net is the net binding energy released by one full reaction. Astronomers like to fit power-laws to things, and it s often seen that ǫ ǫ 0 ρt ν works pretty well. For the present-day Sun, ν

20 Nuclear Burning Cycles Note that ǫ depends very strongly on T. Also, since V max Z 1 Z 2 m 12, we know that heavier nuclei have higher Coulomb barriers, so their reactions become enabled at higher temperatures. Thus, a star with a given core T will usually be undergoing only a small fraction of all possible reactions. Let s look at the most relevant sets of reactions for the Sun.... Hydrogen Burning What we want is: 4p + 2p + +2n 0 4 He + energy Having all 4 protons come together at once is hugely unlikely. Assembling heavier nuclei happens more efficiently in stages. For the present-day Sun, it happens in 3 steps. Problem: One might think the most basic building block would be the strong interaction, p + + p + 2 He but it is highly unstable. The di-proton 2 He breaks up into 2 protons pretty much instantaneously. All nuclei heavier than hydrogen need neutrons to be stable. However, there aren t any free neutrons flying around... In 1939, Hans Bethe realized that the only way to get the process started was with a weak interaction, (1) p + + p + D + + e + + ν e (i.e., one proton decays into a neutron, when near another proton, and then they can fuse). Note shorthand: p + is 1 H +, and D + is 2 H + (pn). This weak interaction is very slow & infrequent: its rate is only about times that of typical strong reactions. It is the bottleneck that determines how fast the whole process goes! 5.20

21 Once we have some D +, and the core T is at least 1 MK, there s deuterium burning: (2) p + + D + 3 He ++ + γ {strong reaction, so it s fast} Note how p + pn gives ppn. Some brown dwarfs (M = M ) can be kept warm for a few million years from this, but eventually they run out of their initial supply of D. In the Sun, the third step happens once there s a chance for there to be two 3 He nuclei in close proximity, (3) 3 He + 3 He 4 He + p + + p + + γ {also fast} Verify nucleon conservation for strong interactions: each 3 He is ppn, and the 4 He nucleus is ppnn. Note also that it would be unlikely for two 3 He nuclei to combine together to a single nucleus with atomic mass 6. There are no stable nuclei with A = 6. The 3 main boxed reactions together form the PP-I chain. Note there are two other dominant ways of doing this, but they both require an existing 4 He to come in as a catalyst

22 Involvement of heavier species in the PP-II and PP-III chains means that they start to dominate at higher temperatures. Using equations from Clayton s Chapter but note that above 20 MK, other (non-pp) reactions start to occur. For stars just a bit more massive than the Sun, these other reactions start to dominate over PP. When T core starts to exceed 20 MK, and there s some C, N, O present, they can act as catalysts for other avenues of hydrogen burning: the CNO cycle (or bi-cycle ). If we start with a 12 C nucleus (6p+6n)... The 12 C gradually captures 4 protons. 2 of them convert into neutrons (via weak interactions). Just before it settles into being a 16 O (8p+8n), an α particle (2p+2n) breaks off, and we re left with another 12 C again. Net result: 4 protons convert into one 4 He! 5.22

23 Alternatively, One 14 N (7p+7n) gradually captures 4 protons. 2 of them convert into neutrons (via weak interactions). Just before it settles into being a 18 F (9p+9n, which is unstable; the stable version has 10n), an α particle breaks off, and we re left with another 14 N again. For even hotter cores (T > 100 MK, i.e., not the present-day Sun!), there s helium burning: three 4 He s fusing into one 12 C... the triple alpha process. Even hotter (T > 500 MK), fusion proceeds by alpha capture reactions C 16 O 20 Ne 24 Mg 28 Si 32 S 36 Ar 40 Ca 44 Ti 48 Cr 52 Fe 56 Ni These sequences are the main reason for the so-called Oddo-Harkins rule: elements with even Z, which also are often those with A divisible by 4, are the most abundant. Once you go even, you never go back. 56 Ni undergoes 2 rapid weak (p n) beta decays: 56 Ni 56 Co 56 Fe. and we re at the stable peak of the binding energy curve. Fusing the elements beyond iron only happens in the most extreme environments (e.g., supernova explosions). That s it for energy generation by nuclear fusion in stars. At the beginning of the lecture on energy generation, I mentioned two other sources (besides gravity & fusion): Radioactive decay (fission) Tidal heating These are more important in planetary interiors than in stars; we won t discuss them further. 5.23