Chapter two The physics of stellar interiors

1. Chapter two The physics of stellar interiors 1. The pressures: --- equation of state of an ideal gas --- equation of state of a degenerate gas 2. What is Opacity and its approximate form? 3. Nuclear reactions in stars? -- binding energy -- Hydrogen and helium burning -- advanced burning

2. Chapter two The physics of stellar interiors 1.1 Equation of state of an ideal gas—The pressure Because temperature is so high in the interior of a star, matter can be regarded as an ideal gas, with every atom fully ionised. Pgas = n k T (1) --- m is the mean mass of the particles in the gas in terms of the mass of the hydrogen atom, mH --  is known as the mean molecular weight of the stellar material -- R = k / mH is defined as define the gas constant If radiation pressure can not be neglected, where a is the radiation density constant. How to calculate mean molecular weight?

3. Mean molecular weight calculation: Assuming that all of the material in a star is fully ionised -- Near the cool stellar surface, however, where even hydrogen and helium are not fully ionised, the assumption breaks down. X: fraction of material by mass in form of hydrogen Y : fraction of material by mass in form of helium Z: fraction of material by mass in form of heavier elements (metals ) X + Y + Z = 1. (5) If the mean mass density for the gas is  the mass densities for H, He and the other heavier elements are : X  , Y , and Z  of heavier elements respectively.

4. No of parti./ per p.or n. mH mH Since = nmH In a in a fully ionised gas: The total number of particles per cubic metre is then given by the sum of the above n = (2X / mH) + (3Y / 4mH) + (Z / 2mH). n = (/ 4mH) (8X + 3Y + 2Z). X + Y + Z = 1, and hence Z = 1 - X - Y,  n = (/ 4mH) (6X + Y + 2).  = 4 / (6X + Y + 2) (6) which is a good approximation to except in the cool outer regions of stars For solar composition, X=0.747, Y=0.236 and Z=0.017, resulting in  ~ 0.6 i.e. the mean mass of the particles in the Sun is a little over half the mass of a proton.

5. Same calculation on the mean molecular weights per ion: (1.24 for the Sun) Same calculation on the mean molecular weights per electron (1.144 for the Sun) Furthermore: you can show that

6. 1.2 Equation of state of a degenerate gas At sufficiently high densities, the gas particles in a star are packed so closely together that the interactions between them cannot be neglected. a) What kind of interaction is dominated for its departure from ideal gas? Electrostatic effects between electrons? Pauli's exclusive effects between electron? Or a) Pauli's exclusion principle No more than two electrons (of opposite spin) can occupy the same quantum state. The quantum state of an electron is given by the 6 values x, y, z, px, py, pz. There is an uncertainty x in any position coordinate and p in the corresponding momentum coordinate, such that  instead of thinking of a quantum state as a point in 6-D phase space, we can think of a quantum state as a volume h3 of phase space

7. b) What happens at the centre of a star as the e is increased The electrons are so crowded in position space (r ) that it is not possible to squeeze in an additional electron to this region of position space, unless its momentum is significantly different so that it occupies a different region of phase space (r, p) the additional e will therefore pose a higher P than it would have had at the same temperature in an ideal gas. Higher momentum mean higher kinetic energy  the electrons in such a gas will exert a greater pressure than predicted by the ideal gas equation of state.

8. The way in which the Maxwellian distribution of electron momenta is modified by the Pauli exclusion principle is shown in the figure Curve A: the Maxwellian distribution of electron momentum in an ideal gas At sufficiently high densities, the Maxwellian distribution begins to violate Pauli's exclusion principle ---horizontal line. These electrons must then occupy higher momentum states than predicted by a Maxwellian distribution.

9. In curve B, where dashed portions of the non-degenerate distribution above the value defined by the Pauli exclusion principle are transferred to higher momentasolid line portion As the density increases still further, more and more of the low momenta electrons are transferred to higher momentum states, as shown by curve C. A gas in which the Pauli exclusion principle is important is known as a degenerate gas.

10. c) The number density of the degenerate electrons Consider a group of electrons occupying a volume V of position space which have momenta lying in the range p and p + p. The volume of the phase space occupied by these electrons : 4  p2 V p. The number of quantum states in this volume of phase space is (4 p2V / h3) p.  2 * (4 p2V / h3) p. If Npp is the number of electrons in V with momentum in the range p and p +p, Pauli's exclusion principle tells us that: A completely degenerate gas Strictly speaking, this occurs at absolute zero temperature. Hence, The total number of electrons N in volume V

11. d) What is the pressure of the degenerate electrons? The pressure P on the surface A We can consider the simplest case: The particles all move in the +x direction, and have momentum p, the number density is n(r,p). Over a period of dt, the particle within volume Av(p)dt will collide with the surface, the total number is The total momentum change will be For an isotropic particle distribution, there will be on average 1/6 of particle moving in the positive x direction, hence Also the particles may have different momentum, so we need to integrate over the momentum distribution

12. To evaluate this integral, we cannot simply use the the relation p=mvpbecause at high density it is possible that p0 >> mec. We must use the relation between p and vp given by the special theory of relativity. This is which can be rearranged to give, * For a non-relativistically degenerate gas (i.e. po << mec) Recalling that N = 8 p03V / 3 h3 and defining the electron density, ne = N / V

13. Using * A relativistically degenerate gas (i.e. p0 >> mec). In order to derive the equation of state for degenerated electrons, we must convert the electron density ne to mass density . where X is the mass fraction of H, we can finally write: the equation of state of an NR degenerate gas the equation of state of a UR degenerate gas Hence in a degenerate gas, the pressure depends only on the density and chemical composition and is independent of temperature.

14. There is no a sharp transition between relativistically degenerate and non-relativistically degenerate gases. Similarly, there is not a sharp transition between the ideal gas equation of state and the corresponding degenerate formulae Can we find the condition that the electron number density ne must satisfy for a degenerate electron gas to be considered perfect ? For a degenerate electron gas to be considered perfect, the Coulomb energy per particle, must be smaller than the maximum kinetic energy, i.e. where p0 is given by The average distance between electron is The electron’s number density must satisfy:

15. electrons neutrons There is a region of temperature and density in which some intermediate and much more complicated equation of state must be used, a situation known as partial degeneracy. e) What types of stars are the above equations applicable to? In the stars, in which no nuclear fusion is occurring -- it is the outward-acting force due to degeneracy pressure that balances the inward-acting gravitational force. White dwarfs, brown dwarfs and neutron stars are examples of such stars degeneracy pressure the force of gravity

16. Fig. 2. different absorption mechanisms 2. What is Opacity and its approximate form? Opacity-- which is the resistance of material to the flow of heat, which in most stellar interiors is determined by all the following processes which scatter and absorb photons. a) Bound-bound absorption an electron is moved from one orbit in an atom or ion into another orbit of higher energy due to the absorption of a photon. E2 - E1 = hbb Bound-bound processes are responsible for the spectral lines visible in stellar spectra, which are formed in the atmospheres of stars But B-B processes are not of great importance in stellar interiors; -- as most of the atoms are highly ionised and only a small fraction contain electrons in bound orbits. -- the photons in stellar interiors are so energetic that they are more likely to cause bound-free absorptions

17. b) bound-free absorption the ejection of an electron from a bound orbit around an atom or ion into a free hyperbolic orbit due to the absorption of a photon E3 - E1 = hbf. Bound-free processes hence lead to continuous absorption in stellar atmospheres. In stellar interiors, however, the importance of bound-free processes is reduced due to the rarity of bound electrons c) Free-free absorption when a free electron of energy E3 absorbs a photon of frequency ff and moves to a state with energy E4, where E4 - E3 = hff. There is no restriction on the energy of a photon which can induce a free-free transition and hence free-free absorption is a continuous absorption process which operates in both stellar atmospheres and stellar interiors.

18. d) Scattering A photon is scattered by an electron or an atom. One can think of scattering as a collision between two particles which bounce off one another. If the energy of the photon satisfies: h << mc2 The particle is scarcely moved by the collision. In this case the photon can be imagined to bounce off a stationary particle. Although this process does not lead to the true absorption of radiation, it does slow the rate at which energy escapes from a star because it continually changes the direction of the photons. e) Approximate form for Opacity As stars are nearly in thermodynamic equilibrium, with only a slow outward flow of energy, the opacity should have the form:

19. Fig. 3. opacity over temperature In restricted ranges of density and temperature, however, the results of detailed calculations can be represented by simple power laws of the form where  and  are slowly varying functions of density and temperature 0 is a constant for stars of a given chemical composition. ________:  as a function of T for a star of given density (10-1 kg m-3) and chemical composition. -----------: the approximate power-law forms for the opacity described below. i) is low at high T and remains constant with increasing temperature. -- This is because at high T most of the atoms are fully ionised and the photons have high energy and are free-free absorbed less easily than lower energy photons. In this regime the dominant opacity mechanism is electron scattering, which is independent of T, resulting in an approximate analytical form for the opacity given by = = 0, i.e curve c:

20. Fig. 3. opacity over temperature ii) is also low at low T and decreases with decreasing temperature. -- In this regime, most atoms are not ionised and there are few electrons available to scatter radiation or to take part in free-free absorption processes, -- while most photons have insufficient energy to ionise atoms via free-bound absorption. An approximate analytical form for the opacity at low temperatures is given by = ½ and  = 4, i.e. iii) has a maximum at intermediate T where b-f and f-f absorption are very important and thereafter decreases with increasing T. A reasonable analytical approximation to the opacity in this regime is given by = 1 and  = - 3.5, i.e.

21. 3. Nuclear reactions in stars 3.1 Binding Energy The total mass of a nucleus is less than the mass of its constituent nucleons. The binding energy, Q(Z,N), of a nucleus composed of Z protons and N neutrons is: The more stable the nucleus, the greater the energy that is released when it is formed. A more useful measure of stability is the binding energy per nucleon, Q(Z,N)/(Z+N). This is the energy needed to remove an average nucleon from the nucleus and is proportional to the fractional loss of mass when the compound nucleus is formed.

22. Fig.4. Binding energy per nucleon over atomic number 3.2 Nuclear fusion and fission reactions a) Fusion reaction If two nuclei lying to the left of the maximum in the figure fuse to form a compound which also lies to the left of the maximum, energy will be released. b) fission reaction If a heavy nucleus lying to the right of the maximum in the above splits into two or more fragments which also lie to the right of the maximum, energy will be released as well.

23. The maximum possible energy released per kg from fission reactions is much less than that from fusion reactions. Also very heavy nuclei do not appear to be very abundant in nature, we may conclude that nuclear fusion reactions are by far the most important source of energy generation in stars.  3.3 Hydrogen and helium burning We turn to look at the most important nuclear reactions which occur in stars a) Hydrogen burning reactions The reaction must proceed through a series of steps : There are many possibilities here, but we will be looking at the main two hydrogen-burning reaction chains The proton-proton (PP) chain and the carbon-nitrogen (CNO) cycle.

24. Fig7. The proton-proton reaction chain i) The PP chain It divides into three main branches, which are called the PPI, PPII and PPIII chains.

25. Fig7. The proton-proton reaction chain The reaction rate of the PP chain is set by the rate of the slowest step, which is the fusion of two protons to produce deuterium. This is because it is necessary for one of the protons to undergo an inverse  decay: It occurs via the weak nuclear force and the average proton in the Sun will undergo such a reaction approximately once in the lifetime of the Sun. The subsequent reactions occur much more quickly, with the second step of the PP chain taking approximately 6 seconds and the third step approximately 106 years in the Sun

26. The relative importance of the PPI and PPII chains depend on the relative importance of If T < 1.4x107K If T > 1.4x107K If T > 3x107K PPIII is dominant but it is never important for energy generation, since at this temperature, some other H burning process may favourably compete with the p-p chains. but it does generate abundant high energy neutrinos. The energy released by 4p 4He: Q0 = 26. 73 MeV But ( Pn +e+ + ve ) Release neutrinos, 0.73MeV Q(4p4He) ~ 26 MeV The rate of energy generation:

27. ii) The CNO cycle: There are two different branches forming a bi-cycle, each with six reactions * C, N act as catalysts in the reactions * The slowest reaction in is the capture of a proton by 14N in the left cycle, and capture of a proton by 16O in the right cycle. 4 p  4He and two + decays + ves * The rate of energy generation: The released energy: 25 MeV.~ 6×1018 J kg-1

28. b) He burning reactions When there is no longer any H left to burn in the central regions of a star, gravity compresses the core until the temperature reaches the point when helium burning reactions become possible. So two 4He nuclei fuse to form a 8Be nucleus, but this is very unstable to fission and rapidly decays to two 4He nuclei again Very rarely , however a third 4He can be added to 8Be before it decays, forming 12C by the so-called triple-alpha reaction: Total effect: 3 4He  12C Energy released: Q=7.275MeV ~ 5.8×1013 Jkg-1 The rate of the reaction chain is decided by the second step. The rate of energy generation is :

29. When a sufficient number of C nuclei have accumulated by 3 reactions,  Capture by these C nuclei is possible: i.e. 12C + 4He  16 O The energy released by this reaction is Q = 7.162 MeV ~ 4.3 ×1013 J kg-1 Once He is used up in the central regions of a star, further contraction and heating additional nuclear reactions such as the burning of C and heavier elements. In summery: 3 4He  12C Energy released: Q=7.275MeV ~ 5.8×1013 Jkg-1 4 4He  16O Energy released: Q = 7.275 + 7.162 MeV

30. Example: If helium burning produces equal amounts (mass fractions) of carbon and oxygen, what is energy generated per unit mass? Consider a mass element m of helium, half of which turns into carbon half into oxygen, by nuclear processes that can be expressed as 312C and 4 16O. The energy released in the first process is Q(3) = 7.275 MeV While energy released in the second is given by adding to it the energy released by  capture on a 12C nucleus, Q(4) = 7.275 +7.162 = 14.437 MeV The number of 12C nuclei produced is given by The number of 16O nuclei produced is given by Hence, the total energy released per unit mass is :

31. c) A comparison for energy release rates:  is the energy release per unit mass per unit time, The energy released by the PP chain and CNO cycle are smooth functions of temperature * the rate of fusion is a very sensitive function of temperature * fusion reactions involving successively heavier elements (in ascending order: the PP chain, the CNO cycle and the triple-alpha reaction) The fusion reaction become even more temperature dependent (and require higher temperatures to operate) in order to overcome the larger Coulomb barrier due to the heavier (and hence more positively charged) nuclei. *  also depends on the density of the stellar material. For two-particle reactions such as PP chain and CNO cycle reactions, the dependence on density is linear, whereas for three-particle reactions such as the triple-alpha process, the dependence is quadratic.

32. 3.3 Advanced burning Each step of further burning requires a further jump in central temperature and thus progressively larger stellar masses. E.g. Carbon burning requires > 4MS The following is by no means a complete list of all the possible reactions and the nuclear physics gets too complex to consider here. At 6 x 108 K: O816 + He24 Ne1020 4.7 MeV (Ne also from C + C) Ne1020 + He24 Mg24 9.3 MeV Mg24 + He4 Si1428 10.0 MeV At ~109 K C12 + C12 Mg24 14 MeV O16 + O16 S32 16 MeV Mg24 + S32 Fe56 END OF FUSION Old massive stars accumulate a series of shells as illustrated below just prior to the supernova stage.

33. Summary: 1. The pressures: a) equation of state of an ideal gas  = 4 / (6X + Y + 2) (6) b) equation of state of a degenerate gas Hence in a degenerate gas, the pressure depends only on the density and chemical composition and is independent of temperature. The condition that the electron number density ne must satisfy for a degenerate electron gas to be considered perfect is

34. 2. An approximate form for Opacity , , and o are dependent on temperature, density and chemical components of the stellar materials. 3. Nuclear reactions in stars? Each step of further burning requires a further jump in central temperature and thus progressively larger stellar masses