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Sources of energy and the origin of the chemical elements Origin of the elements Could the Sun be powered purely by gravitational contraction? Sphere of uniform density (underestimate?) has gravitational potential energy: 3 GM 2 Ug = − 5 R Change in radius liberates potential energy, but 50% of this (from Virial Theorem) is required to maintain pressure support (e.g. heat): Available energy = 0.5 U g ~ 10 41 J LSun ~ 4 ×10 26 W Timescale to maintain solar luminosity: t KH ~ 0.5 U g Lsun ~ 107 yr Note that every factor of 2 collapse would release as much energy again Gravity as source ASIDE: Note the curious behaviour of a self-gravitating body under its own pressure support: it has negative heat capacity: The Virial Condition for equilibrium state of self-gravitating system is: 2K + W = 0, i.e. E = -K = 0.5W K = kinetic energy = 3/2 NkT W = potential energy E = total energy 3 E = − NkT 2 dE 3 = − Nk dT 2 Self-gravitating system heats up when it loses energy. Same effect as an orbiting satellite speeding up under drag forces Virial theorem Could the Sun be powered purely by gravitational contraction? Number of particles M sun n~ ~ 1057 mp Chemical energies are typically eV (10-19 J) per particle, i.e. 1038 J available. Gives lifetime of order only 104 years! Nuclear reactions are typically MeV (10-13 J) per particle, i.e. 1044 J available. Gives lifetime of order 1010 years Chemistry as source Fusion energetics mp = 1.6726 × 10-27 kg mn = 1.6749 × 10-27 kg a.m.u.= 1.6605 × 10-27 kg (= 931 MeV) mHe = 4.0026 a.m.u. Making 4He out of 2n + 2p results in 0.7% of the mass (i.e. 6.5MeV) being lost, due to increased negative binding energy of the nucleus. Nuclear fusion reactions (a) provide the energy source in the Sun, and (b) produce the diversity of chemical elements out of initial H+He produced in the Big Bang. Fusion as source Nuclear energy levels • Pauli exclusion principle ensures np ~ nn • Proton levels are more widely spaced because of electrostatic repulsion • Gross shape of binding energy curve is a result of the finite range of the strong force (attraction saturates as n increase too far). Fusion as source Fusion energetics fusion releases energy Fusion as source Formation requires energy, fission releases energy Fusion reaction rates: Difficulty of fusion arises from coulomb repulsion between positively charged nuclei Potential energy of two charges separated by r Point of closest approach if initial velocity is v Ue = rclose q1q2 4πε 0 r 2q1q2 = 4πε 0 mv 2 Naively setting r ~ 10-15 m and mv2 = 3kT would require T ~ 1010K >> Tsun Coulomb repulsion • Quantum tunneling through potential barrier • Velocity distribution at a given T Velocity in center of mass frame of two particles of reduced m = m1m2/(m1+m2) & mv 2 # !! dv n(v) dv ∝ exp$$ − % 2kT " Tunneling probability through distance r given by de Broglie λ: & 4π 2 q1q2 # & 2π 2 rclose # !! !! ∝ exp$$ − P ∝ exp$$ − λ " % % 4πε 0 hv " Probability of reaction given by product of these two, plus usual number of encounters vσdt, where σ is cross section Tunneling and Maxwell distribution ( mv 2 πq1q2 % ## dv dt dN ∝ n1n2σ v exp && − − ' 2kT ε 0 hv $ ( ) ∝ exp − αE − βE −1 / 2 dEdt Highly peaked function with maximum dN/dE = 0 when the two terms are equal (plus a factor of 2 from differentiating) v max $ πq1q2 kT '1/ 3 =& ) ε hm % 0 ( Rate at this peak is given by 2/3 1/ 3 # 1/ 3 € & & # 3 π q q m T & # & # dN / dT ∝ exp$ − $$ 1 2 !! $ ! ! ∝ exp − $ 0 ! $ 2 % hε 0 " % kT " ! %T " % " '3$ T0 = % " &2# Tunneling and Maxwell distribution 3 2 ' πq1q2 $ ' m $ %% "" % " ~ 4 ×1010 K & hε 0 # & k # Very strong T dependence Steep dependence of fusion reaction rate coupled with negative heat capacity of gas provides natural thermostat This is why stars do not explode like Hbombs T4 at ~ 107K Later stages of fusion require higher temperatures because of larger nuclear charges H to He He to C, O C to O, Ne, Na, Mg Ne to O, Mg O to Mg – S Si to around Fe 1×107 K 1×108 K 5×108 K 1×108 K 2×108 K 3×109 K Note also that flattening of binding energy curve implies that energy release per reaction goes down in later stages Question: Are the interiors of stars hot enough for fusion? (c.f. Tsurface ~ 6000 K for the Sun) Quite general arguments indicate yes: GM 2 U grav = − f R with f of order unity Hydrostatic support at radius r dP Gm =− 2r ρ dr r R R Gmr 3 dP 3 4 π r dr = − 4 π r dr ∫ ∫ 2 dr 0 0 r R −3 ∫ P(r) 4π r 2 dr = −U grav 0 P Thermal pressure V =− 1 U grav 3 V ρ P ~ kTint m Tint ~ GMm ~ 3×10 6 K 3kR for Sun Question: How does it get that hot? Tint ~ GMm ~ 3×10 6 K 3kR Overall energetics of gravitational collapse: 3M GM 2 kT ~ 2m R 2 GMm T~ 3 kR So there is enough energy in liberation of gravitational PE (but without a lot to spare). What actually happens? Initial gravitational P.E. goes into dissociating H2 molecules (4.5eV) and ionizing H atoms (13.6eV) and proto-star collapses in free-fall. When interior is ionized plasma (when R ~ 100 Rsun and T ~ 30,000 K), mean free path of photons becomes very short and interior heats up, establishing temperature gradient and hydrostatic support governed by above equation. for Sun Are stars inevitable? Tint ~ GMm ~ 3×10 6 K 3kR for Sun Proto-star collapses in quasi-hydrostatic equilibrium with Tinterior rising as R shrinks. Negative heat capacity causes star to increase in temperature as energy is lost through radiation to exterior. Virial condition implies ½ of Ugrav goes into internal heat energy, rest is radiated. When Tinterior reaches point where fusion can balance heat loss at surface, a static stable structure (= star) is formed, with constant R. This ignition point will inevitably be reached at some R unless another source of pressure (e.g. degenerate electron pressure) can support the object before T gets high enough for fusion. Tint ~ Minimum mass of stars de Broglie wavelength of thermal electron λ~ GMm ~ 3×10 6 K 3kR h 1/2 ( mekT ) Critical density above which QM effects dominate m ( m kT ) ρ~ 3 ~m e 3 λ h We had before, for general hydrostatic support….. kT ~ 3/2 GMm ~ GmM 2/3ρ1/3 3R so the temperature increases as ρ1/3. What is maximum temperature reached before degeneracy kicks in? ! G 2 m8/3me $ 4/3 kT ~ # &M 2 h " % For the Sun, temperature could rise to 107K. Fusion will not occur for M < 0.08 Msun ( = “brown dwarfs”) Conclusion: for all more massive objects, stars are “inevitable” for Sun Mass-luminosity relation Details not critical Tint ~ GMm ~ 3×10 6 K 3kR for Sun Transport of energy out of the (opaque) star by diffusion due to random walk of photons Energy density and SHC of radiation Heat transport due to diffusion of photons in T gradient set by m.f.p.. Net luminosity passing out of radius r (assuming no fusion at this radius) Approximate for order of magnitude To get luminosity of star Uγ = aT 4 C = 4aT 3 jx = −K r dT dx 4 4 1 K r = caT 3 l = caT 3 3 3 κρ 4 1 dT L(r) = 4π r 2 jr = 4π r 2 caT 3 3 κρ dr dT Tint ~ dr R L = ( 4π ) 2 4 caG 4 1 3 M 4 4 3 k κ Lifetimes of stars L = ( 4π ) 2 4 caG 4 1 3 M 4 4 3 k κ This is a strong function of the mass of a star. Actually, putting in dependence of κ(T), it is more like M3.5 Implication: Since the amount of fuel for fusion will be (more or less) proportional to mass, expect lifetime of stars to strongly depend on M-2.5. e.g. 30 Msun star has lifetime < 107 years, c.f. 1010 years for Sun. Is there a maximum mass for a star? Yes. If internal pressure dominated by radiation pressure rather than thermal pressure then it is likely to be unstable. Mmax ~ 60-100 Msun Self-gravity vs. thermodynamics • Ordinary stars (like our Sun) are self-gravitating and therefore have to be hot inside to sustain the thermal pressure that resists the inward pull of gravity • The interstellar medium, however, is typically much colder and energy flows continuously from the star to the exterior. • No thermodynamic equilibrium is possible under this circumstances, losing energy the star will be forced to contract and to get hotter and hotter • Thermonuclear production of energy keeps the star structure stable for a limited amount of time. When it runs out of fuel a major structural reassessment has to happen. What supports a star against self-gravity? X mass fraction of H Y mass fraction of He Z mass fraction of metals Available sources of pressure • Non-degenerate plasma: P = nkT = (ρkT/mH)(2 X+0.75 Y+0.5 Z) • Radiation pressure: P=aT4/3 • Non-relativistic degenerate plasma: P∝ [0.5 (1+X) ρ]5/3 • Ultra-relativistic degenerate plasma: P∝ [0.5 (1+X) ρ]4/3 • Atomic and molecular interactions • Quark and hadronic interactions • Rotation and magnetic fields The Hertzsprung-Russell diagram Brown Dwarfs A brown dwarf is a self-gravitating gaseous object composed mainly of hydrogen and helium, whose mass is too small to ignite stable thermonuclear hydrogen fusion in its interior (but they do produce some helium isotopes by fusing protons with deuterium, lithium and berillium). Theoretical expectations: M < 0.08 M¤. Incapable of generating substantial amounts of nuclear energy, the gravitational contraction of a brown dwarf takes place until the pressure of the degenerated electrons in its interior stops it. Young BDs can release substantial amounts of gravitational energy, while older BDs radiate remnant internal heat only. Gliese 229a: cool, main sequence red dwarf (0.46 M¤, 0.53 R¤, 0.016 L¤), approximately 3 Gyr old. Gliese 229b: brown dwarf (25-65 MJ, 0.9-1.1 RJ, T~1100 K) orbiting at about 39 AU from Gliese 229a and showing abundant methane in its atmosphere. Main-sequence stars Objects with M > 0.08 M¤ ignite stable thermonuclear reactions in their centres and become stars. In the initial phase of their lives they are characterized by chemical homogeneity and core-hydrogen burning. Stars with M < 2 M¤ fuse hydrogen into helium by the p-p chain, while more massive stars do it by the CNO cycle. In the HR diagram they populate an approximately one-dimensional locus along which positions are mainly determined by the stellar mass. Vogt-Russell Theorem: The mass and chemical composition of a star uniquely determine its radius, luminosity, and internal structure, as well as its subsequent evolution. Gives approximate scaling relations: L ∝ M3.5 E∝M t ∝ M-2.5 R∝M Te ∝ M0.5 The Hertzsprung-Russell diagram What happens when Hydrogen fuel is exhausted in the core? Inert He core surrounded by shell still burning H to He. Changes structure of star: x1000+ increase in L, plus 100x increase in size (with decrease in temperature) à Red Giant Star (N.B. Sun will consume Earth) Temperature rise in core may ignite He to C fusion à Horizontal Branch Star Followed by He shell burning and so on à Asymptotic Giant Branch Star If ever the core gets dense enough to be supported by degenerate electron pressure, it will cease contracting and raising its temperature. Therefore no new fusion reactions will occur. e.g. the Sun will never fuse C because it will stabilize its core below the ignition temperature of C. White Dwarfs When the triple-α process in a red-giant is complete, the star core starts contracting. For a main sequence mass of M < 4 M¤ there is not enough energy to ignite the carbon fusion process, thus the collapse keeps going until it is halted by the pressure arising from electron degeneracy. From the stage of PN the star loses its extended envelope. The resulting hot and small (relative to the Sun) star is called a White Dwarf. Stars with M< 0.4 M¤ will not become red giants because they do not become hot enough for He burning, but their lifetimes will be longer than the present age of the Universe. Typical C,O white dwarf specs: • mass of about 1 M¤ (on average) • size of about 1 Earth diameter • ~ 4% of solar luminosity • surface temperature of ~ 25000 K What about Life? Good news: Because they are so small (and have such a low L) they will shine for a very long time, even though they are not producing more energy from fusion Bad news: All the products of fusion are locked up in the interiors Mass Loss and “Planetary” Nebulae Shells of gas thrown off by Sun-like stars nearing the ends of their lives. The action of radiation pressure on small dust grains formed in the cool stellar atmospheres probably causes the mass loss. The UV radiation emitted by the remaining hot (but quickly inert) stellar core (white dwarf) causes the ejected gas to fluoresce as the planetary nebula. Stars with M < 8 M¤ loose most of their mass during this phase. But this is largely unprocessed material with a composition similar to that of the original gas cloud out of which the star formed. But more massive stars will go all the way to an Fe core. abundance Final structure of a 25 Msun star enclosed mass Onion-like structure with successive shells burning different fusion reactions But time is running out: Energy per reaction is dropping, required luminosity is increasing. For a 25 Msun star: – – – – – Hydrogen burns in the core for 7 million years Helium burns for 500,000 years Carbon burns for 600 years Oxygen burns for 6 months Silicon burns for 1 day What happens next? But first, back to that Gamow Peak……… Key point: Gamow peak means that most individual reactions have the same energy… i.e. the Δmc2 of the reaction plus the K.E. associated with the Gamow Peak. One consequence: The existence of well-placed energy levels in product nucleus leads to resonances, i.e. greatly increased σ at particular energies, leading to wide variation of reaction rates for different reactions. Reaction pathways for H → 4He fusion Extremely slow due to weak interaction τ ~ 1010 yr in Sun Slow, but less so than above due to spontaneous decay of unstable nuclei 13N (τ1/2 ~ 600s) and 15O (τ 1/2 ~ 120s) Also, requires higher T because of higher +ve charge on nuclei p-p chain 1H + 1H → 2H + e- + ν 2Η + 1Η → 3He + γ 3He + 3He → 4He + 1H + 1H CNO cycle 12C + 1H → 13N + γ 13N → 13C + e- + ν 13C + 1H → 14N + γ 14N + 1H → 15O + γ 15O → 15N + e- + ν 15N + 1H → 12C + 4He CNO cycle Helium burning and the interesting case of Carbon: Absence of more tightly bound nuclei between 4He and 12C Requires effectively a three particle reaction 4He + 4He + 4He → 12C + 7.7 MeV Which will be extremely rare statistically Carbon • Production of 8Be from two 4He requires 0.1 MeV of energy. • 8Be is unstable with half life of 10-17 s While it survives, 8Be can join with 4He to make an excited state of 12C. 8Be +4He has energy 7.33 MeV above ground state of 12C. Hoyle (1952) correctly predicted that 12C must have a resonant state just above 7.33 MeV for this reaction to occur often enough to produce significant amounts of 12C. Subsequently found at 7.65 MeV – a triumph of theoretical astrophysics. Most 12C* falls apart back to 8Be + 4He, but some de-excites to ground-state through emission of γ-ray photons. Carbon production Gamow peak C Resonance level Three states of 4He, 8Be and 12C* are so similar in energy that these three species will be in equilibrium, described by Saha equation. Production of 12C is effectively “leakage” out of equilibrium from 12C* excited state Equilibrium There is a potentially easy route to burn 12C: 12C + 4He → 16O + 7.16 MeV Absence of resonance (“near-miss”) for 16O in Gamow band hinders this reaction, allowing 12C to survive in reasonable abundance (~ 1/3 O) Destruction of 16O also hindered Energy levels Consequence: O and C are the most abundant elements after H and He – good for organics, water etc . The relatively high abundances of these two arise from coincidences in the nuclear energy levels of these and other nuclear species Abundances