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STARS Level 1 2006-07 Sections C and D M. R. W. Masheder (Section C) The Internal Structure of Stars Stars are large spheres of hot gas in equilibrium between internal pressure and selfgravity. They are then governed by the law of hydrostatic equilibrium (see Astrophysical Concepts) Hydrostatic Equilibrium [Figs. 38-40: Hydrostatic Equilibrium] Assume that the star is spherically symmetric so that pressure P, density and temperature T are all functions of radius r only. If we consider a fluid element of cross-sectional area A and depth r at radial position r, then the outward force due to the pressure (gradient) will be just Fp = PA - (P + P)A = - PA = - r(dP/dr)A. (Note that the force is outward only if dP/dr is negative). The inward gravitational force on the element (mass m), due to the mass inside GM ( r )m radius r will be F g r2 But given the respective volumes and densities, the masses must be m = (r)rA and M ( r ) r ( s )4s 2 ds 0 For equilibrium Fp = Fg so dP GM ( r ) ( r ) 2 dr r To solve this explicitly, we need to know (or assume) (r) or have another relationship between P and . However, we can obtain some simple approximations. Purely on dimensional grounds it is clear that if we just approximate dP/dr by the total change in pressure from the centre of the star (P = Pc) to the surface (P = 0) divided by 2 the radius of the star (R), of total mass M, we must have P GM 3GM c R 4R 4 More accurately, assume a constant density 3M 4R 3 . This implies M(r)=(r/R)3M, so putting these into the hydrostatic equilibrium equation we have dP GMr 3 3M 1 dr R 3 4R 3 r 2 Integrating from the centre to the surface we then get R Pc 0 rdr 3GM 2 4R 6 3GM 2 8R 4 For the Sun, M = 2 x 1030kg, R=7 x 108m (so = 1400 kg m−3), giving Pc ≈1.4 x 1014 Nm−2. For a realistic (r) decreasing outwards, Pc is even higher. Such pressures must therefore exist at the centres of stars to support them against their own self gravity. We should next consider how these pressures are produced. Page 1 in tranche 3 of 8 pages. There were 18 previous pages STARS Level 1 2006-07 Sections C and D M. R. W. Masheder Thermal Gas Pressure In a gas at temperature T the particles (mass m) have mean kinetic energy (1/2)m <v2>= (3/2)kT. Momentum transfer through collisions then produces a pressure given, for an ideal gas, by the perfect gas law P=nkT=(k/m)T where n and are the particle number density and the mass density, respectively. The relationship between P and ρ (and in this case T) is called the equation of state. Provided we have an ionized gas (plasma), the perfect gas law still holds at the very high densities calculated above. Assuming a star to be made entirely of ionized hydrogen the mean particle mass m = (me + mp)/2 ≈ mp/2. (If we include helium this increases to about 0.62 mp). Using the central pressure calculated for a homogeneous sphere and the perfect gas law we get a central temperature Tc m Pc m GM 6 10 6 K k k 2R Note that at these temperatures (3/2)kT ~1 keV >> 13.6 eV, so the hydrogen is indeed ionized. They are also high enough for thermonuclear fusion reactions to occur. Radiation Pressure A photon with energy hν carries momentum hν/c and therefore exerts pressure through the transfer of this momentum when photons are absorbed or scattered. For a black body we know that the flux (of energy of photons) passing through unit area of a surface is σT4. Thus we would expect a pressure (flux of momentum) ~ σT4/c. Specifically, for an isotropic radiation field this becomes Prad 4 T 4 1 4 aT 3 c 3 At T~107 K. Prad ~3×1012 Nm-2 ~ Pc/50 However, as T increases, Prad may dominate. Page 2 in tranche 3 of 8 pages. There were 18 previous pages STARS Level 1 2006-07 Sections C and D M. R. W. Masheder Degeneracy Pressure Degeneracy pressure is a quantum mechanical effect, important at very high densities. It occurs because the Pauli exclusion principle allows only one electron per quantum state (or 2 per energy state, as the electron has a choice of two spin states). The number of available states per unit volume is limited for a confined set of particles. The exclusion principle thus prevents the particles being too close together and therefore acts as a pressure. The Heisenberg uncertainty principle relates position (x) and momentum (p) uncertainties (in 1-D) through x p> h. If a particle is constrained in a very small volume then its momentum must be large. In 3-D this means that we can only have 2 electrons in a 6-dimensional phase-space element defined by 4r2dr × 4p2dp = h3. Thus the (physical) space density of particles with momentum p to p + dp is limited to dn = 2/4r2dr = 8p2dp/h3. The total density of particles with momentum less than p is then just the integral of this, i.e. n=(8p3)/(3h3) for completely degenerate gas. As before, pressure is the same as rate of momentum transfer, pnv p2n for nonrelativistic particles or pnc pn for relativistic particles. But from above, n p3 so we have a degeneracy pressure Pd p5 in the nonrelativistic case or p4 for the relativistic one. Finally, since ρ n p3, this implies that non- relativistic degenerate matter will have Pd = A1ρ5/3 while relativistic degenerate matter has Pd = A2ρ4/3 for some constants A1 and A2. Notice particularly that there is no dependence on temperature. Thus degenerate stars will not change their size as they cool. Electron degeneracy pressure is important in the very dense white dwarf stars. Since collisions tend to equalise the energy (mv2) of particles in a gas, the much more massive nuclei will have higher momentum (mv) by a factor (mn/me). They will therefore occupy a (mn/me)3/2 times larger volume in momentum space than the electrons, so do not contribute much to the pressure. At very high densities, proton and electrons combine to form neutrons (inverse beta-decay). Neutron degeneracy can set in when the number density is about 105 times higher than for electron degeneracy. This occurs in neutron stars. [Fig. 41 : Condensed Objects] Page 3 in tranche 3 of 8 pages. There were 18 previous pages STARS Level 1 2006-07 Sections C and D M. R. W. Masheder (Section D) Energy Sources in Stars As a star shines it is losing energy at a rate equal to its luminosity L. If it has a lifetime τ it must have available an energy source E=τL. From estimates of the age of the Earth, we know that the Sun must be at least 4.6 Gyr old (1.5 × 1017s). This requires E > 6 × 1043J. Consider first gravitational energy. As the separation of two masses m1 and m2 is reduced from infinity to r, gravitational potential energy is released :- EG GM 1M 2 r In the same way, if a star of mass M is formed by the collapse of a very large cloud down to radius R the energy released is E GM G r constant which depends on the density distribution of the star. 2 where is a If the density is uniform, = 3/5. Thus for the Sun we have an approximate energy supply :- 3 (6.67 10 11 )( 2 10 30 ) 2 EG 2 10 41 J 8 5 7 10 This is about 300 times too small. Or, put another way, the lifetime due only to gravitational contraction (the KelvinHelmholtz timescale) is 300 times too short. Nuclear Energy A general atomic nucleus consists of Z protons and N neutrons. Z is the atomic number and the number of nucleons, A = Z + N, is the mass number. For light nuclei, N ≈ Z. In the nuclear shell model, the protons and neutrons fill up parallel sets of energy levels, in a similar way to the electrons in shells outside the nucleus. The binding energy of a nucleus is A B ( Zm Nm M A )c 2 where M Z is the total mass of the nucleus. B is usually p n Z given in MeV and is the amount of energy needed to pull the nucleus apart. For comparing the stability of nuclei, the binding energy per nucleon B/A is more useful. [Fig. 42: Binding Energy] Two forces are important in nuclei, the strong nuclear force and the electrostatic force. The strong force is attractive and applies between all nucleons. As its name implies it is stronger than other forces on nuclear scales, but is very short range,~10−15m (smaller than one large nucleus). The strong force also saturates, in the sense that a nucleon can only bind to a limited number of others around it. In dense nuclear matter all bonds will be saturated so we would expect B/A to tend to a constant value, as seen for heavy nuclei. On the other hand, for light nuclei, there are Page 4 in tranche 3 of 8 pages. There were 18 previous pages STARS Level 1 2006-07 Sections C and D M. R. W. Masheder unused bonds so additional nucleons can be pulled in closer. B/A should therefore increase with A. The electrostatic (Coulomb) force is repulsive (all positive charges) but affects only the protons. It is weak but extends to all scales. Its contribution to the binding energy works like gravitational potential, 3 Z 2 e 2 for a nucleus of size R. B 5 4 0 R The density of nucleons is roughly constant so R A1/3. Also, until we reach very heavy nuclides Z ≈A/2. Therefore B Z2 4 / 3 A2 / 3 A A In heavy nuclei Z does not increase as fast as A, so B/A is reduced. The opposing effects of the two forces produce a peak in B/A for moderately heavy nuclei, specifically at A = 56, i.e. Iron. Fe56 is the most stable nucleus. This implies that energy will be released if nuclei up to Fe are synthesised from lighter ones (nuclear fusion) and will also be released if nuclei heavier than Fe are split into two lighter ones (nuclear fission). [Fig. 43: Binding Energy] Stars are mostly hydrogen and helium, so fusion is potentially important for supplying their energy output but fission is not. For fusion to occur, the electrostatic repulsion between the two light nuclei must be overcome. This is possible in stars because of the high thermal energies of the nuclei. This leads to thermonuclear fusion. (Actually, quantum tunnelling through the coulomb barrier is crucial to this process which would otherwise require temperatures some 1000 times higher.) Bigger nuclei have more coulomb repulsion between them, so higher temperatures are required for fusion. This is important for understanding stellar evolution. The most important fusion reactions in stars are as follows. Hydrogen Burning The net effect of hydrogen burning is to turn four protons into a helium nucleus 4H1 1→ He42+ 2e+ + 2νe where e+ represents a positron and νe is an electron neutrino. This occurs at a temperature TH ~ 107K and is the energy source for main sequence stars; the net result in each case is the release of 26.7 MeV per He4 nucleus produced. [Fig. 44: Fusion: 4p->He] However, it does not proceed by the direct fusion of four protons (as it is unlikely that four will all arrive at the same point at the same time), but by successive addition. Several chains of reactions are important. The first are the PP (proton-proton) chains. In this case He4 is built up via deuterium and He3 (PP I), or via the production of Be7 and then either Li7 (PP II) or B8 (PP III). Page 5 in tranche 3 of 8 pages. There were 18 previous pages STARS Level 1 2006-07 Sections C and D M. R. W. Masheder [Fig. 45: Fusion details] [Fig. 46: Fusion reactions] Fig. 47: More PP reactions – see below] + 1 H + 1H 2H + e + e 2 H + 1H 3He + 3 He + 3He 4He +1H + 1H (PP I) OR 3 He + 4He 7Be + (PPII and PPIII) PP Chain 3 He + 4He 7Be + 7 Be + e 7Li + 7 Li + 1H 4He +4He - PPII reactions 7 Be + 1H 8B + 8 B 8Be + e + 8 Be 4He + 4He + PPIII reactions Page 6 in tranche 3 of 8 pages. There were 18 previous pages STARS Level 1 2006-07 Sections C and D M. R. W. Masheder PP I dominates at T < 1.4 × 107K and only 2% of the available energy is lost to the neutrinos (average energy 0.26 MeV). PPII dominates at 1.4 to 2.7 × 107K and has losses of 4% to the neutrinos (the average energy of the neutrino arising from the production of Li7 is 0.8 MeV). Finally, the PP III chain dominates above 2.3 × 107K, but with 28% of the energy going into neutrinos (7.2 MeV for the neutrino emitted by the Be8 decay). The CNO cycle (sometimes called the CNO bicycle) is rather different in that it uses existing carbon nuclei in the build up of the helium, though they are not ‘used up’, being returned at the end after processing through oxygen and nitrogen isotopes. Neutrinos produced at various stages all have energies ~1 MeV so the net energy loss is 6.4%. [Fig. 48: CNO cycle] [Fig. 49: CNO cycle reactions] Since 26.7 MeV is produced per 4 protons, this amounts to 0.7% of the original rest mass 4mpc2. The energy available from 1 M is therefore Enuc = M/4mp ×26.7 × 1.6 × 10-13 ≈ 1.3 × 1045J. Even if only 10% of the mass of a star like the Sun is used up in hydrogen burning on the main sequence this is still enough to fuel it for ≈1010 years, easily covering the known age of the Solar System. Helium Burning The next basic process is the conversion of three helium nuclei to one carbon. The first two He42 give a Be84 and although most of these decay back (half-life ≈ 2.6×10-16 sec), a few live long enough to fuse with another He42 to give C126 . This is the ‘Triple-alpha’ reaction. This operates at temperatures T ~ 2 × 108K and is a main energy source in red giants, producing 7.65 MeV per C nucleus formed. [Fig. 50: Helium and Carbon burning] Carbon Burning The simplest reaction is just the fusion of two C12 6 to give Mg24 12. It requires much higher temperatures, T~2 × 109K in order to instigate. It therefore only occurs in massive stars, late in their lives. There are also carbon burning processes at slightly lower temperatures (> 8 × 108K) which produce oxygen, neon and sodium. Note that these differ from the CNO cycle in that the carbon is used up as fuel, not merely used as a catalyst. Even heavier elements (up to iron) can be produced in nuclear fusion at yet higher temperatures in higher mass stars, viz. Carbon burning (C producing O, Ne) at 5 × 108K (M> 8M) Neon burning (Ne producing O, Mg) at 1.5 × 109K (M > 5M) Oxygen burning (producing elements up to S) at 2 × 109K (M> 10M) Silicon burning (producing elements up to Fe) at 3 × 109K (M > 11M) Elements heavier than iron are produced by successive neutron capture in red giants and supernova explosions. Page 7 in tranche 3 of 8 pages. There were 18 previous pages STARS Level 1 2006-07 Sections C and D M. R. W. Masheder Stability of Energy Generation In equilibrium, we require that the power generated in the core is equal to that radiated from the surface. But the energy generation rate is very temperature dependent, so the temperature in the core must self-regulate. In normal stars (supported by thermal pressure), if the energy generation rate rises above the equilibrium value, the core temperature rises, so the pressure rises. The core therefore expands and cools, lowering the energy generation rate again. Hence the equilibrium is restored (stable). The thermostat is very sensitive as dE/dt depends strongly on temperature, e.g. p-p chain 2T4 and the CNO cycle 2T16 In degenerate stars, if the energy generation rises, the core temperature again rises, but this has no effect on the degeneracy pressure. Thus there is no expansion or cooling. However, the increased temperature will lead to more energy generation, so there is a runaway thermonuclear explosion (e.g. the ‘helium flash’ at the onset of helium burning in the degenerate cores of low mass post-main sequence stars). Stable thermonuclear energy generation is therefore not possible in degenerate stars. Page 8 in tranche 3 of 8 pages. There were 18 previous pages
