1. The physics of stellar interiors
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
Thus far, we have assumed that stars are composed of ideal gases and hence the pressure is given by the ideal gas equation of state we have just derived.

2. The physics of stellar interiors
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,
Having derived the four differential equations of stellar structure, we got to find three associated relations for P, ,  to T, , and chemical compositions Xi, in order to study the evolution of the stellar structure. Therefore, we will have to look at the physics of stellar interiors, including nuclear physics, atomic physics and thermodynamics. This is a complex subject and hence only the basic physical processes which determine P, ,  will be described here.
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 
. An exact calculation of  is very complicated as it depends on the fractional ionisation of all the elements in all parts of a star.
Fortunately, we can simplify things enormously and only introduce a small error in the calculation of 
This is a valid assumption because hydrogen and helium are very much more abundant than all of the other elements and they are certainly fully ionised in stellar interiors.
the mass densities for H, He and the other heavier elements are : X  , Y , and Z  of heavier elements respectively.

4. In a in a fully ionised gas:
No of parti./ per p.or n. mH
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,
Since = nmH 
 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
One might have thought that electrostatic effects are responsible for the first departures from an ideal gas,
In fact, the first such departures are due to interactions between the free electrons in the highly ionised gas because electrons have to obey Pauli's exclusion principle.
It is not possible, however, to know both the position and momentum of an electron with complete accuracy
 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
The ions in such a gas have higher momenta than the electrons, they are less likely to violate Pauli's exclusion principle. The pressure due to the ions can then be treated as an ideal gas, but the pressure due to the degenerate electrons is much larger and hence the gas obeys a different equation of state, which we shall now derive.
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,
To obtain the equation of state, we need to have an expression of pressure.
A completely degenerate gas is one in which all of the momentum states up to some critical value p0 are filled, while the states with momentum greater than p0 are empty.
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
Suppose we have particles inside a box. Let consider the wall with its normal along the positive x-direction. The wall has a surface area of A. Now pressure is simply the force divided by are P=F/A
These particles will be bounced back with the same momentum but in the opposite direction, so the momentum change is 2p for each particle,
This is our final expression for the pressure, valid at all velocities and for all momentum distribution, as long as the particle distribution is isotropic.
Also the particles may have different momentum, so we need to integrate over the momentum distribution

12. which can be rearranged to give,
To evaluate this integral, we cannot simply use the the relation p=mvp because 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)
We will not evaluate this integral here , Instead we will consider two limiting cases:
Recalling that N = 8 p03V / 3 h3 and defining the electron density, ne = N / V

13. * 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 .
Using
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
This occurs when the velocity of an electron approaches that of light, in which case its momentum approaches infinity.
Our aim is to obtain an equation of state for a degenerate gas.
Hence in a degenerate gas, the pressure depends only on the density and chemical composition and is independent of temperature.

14. The average distance between electron is
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
We could find some the regions where the degenerated electrons can be treated as ideal gas.
The electron’s number density must satisfy:

15. e) What types of stars are the above equations applicable to?
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
We will see later in the course that there are stars in which ….
We will also see that many stars temporarily develop degenerate cores as they evolve off the main-sequence.
electrons
neutrons
degeneracy pressure
the force of gravity

16. 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.
Fig. 2. different absorption mechanisms
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
The first three are known as true absorption processes because they involve the disappearance of a photon, whereas the fourth process only alters the direction of a photon.
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.
Note that, in both free-free and bound-free absorption, low energy photons are more likely to be absorbed than high energy photons.
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. A photon is scattered by an electron or an atom.
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:
Our aim is to derive an expression for the opacity which can be used to solve the equations of stellar structure.
We have seen that, as stars…
A discussion of the detailed form for opacity is outside the scope of this course - it involves complicated numerical calculations in atomic physics.

19. where  and  are slowly varying functions of density and 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.
Fig. 3. opacity over temperature
________:  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.
Fig.3 shows how the power law approximations fit the curve derived from detailed opacity calculations.
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. ii)  is also low at low T and decreases with decreasing temperature.
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.
In the next part of the course we will see how it is possible to use these simple approximate expressions for the opacity in conjunction with the equations of stellar structure obtain some important insights to the structure of stars.
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).
As mentioned when we looked at energy generations, it is now known that most of the energy radiated by stars must be released by nuclear reactions. In this section we will consider why energy can be released by nuclear reactions.
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. 3.2 Nuclear fusion and fission reactions
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. Also very heavy nuclei do not appear to be very abundant in nature,
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 most important series of fusion reactions are those converting hydrogen to helium in a process known as hydrogen burning
The chances of four protons fusing together to form helium in one go are completely negligible.
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. i) The PP chain
It divides into three main branches, which are called the PPI, PPII and PPIII chains.
Fig7. The proton-proton reaction chain
The first reaction is the interaction of two protons (p or 1H) to form a nucleus of heavy hydrogen (deuterium, d, or 2H), consisting of one proton and one neutron, with the emission of a positron (e+) and a neutrino (e). The deuteron then captures another proton and forms the light isotope of Helium with the emission of a  -ray. The 3He nucleus can then either interact with another 3He nucleus or with a nucleus of 4He (an  particle), which has either already been formed or has been present since the birth of the star. The former case is the last reaction of the PPI chain, whereas the latter reaction leads into either the PPII or the PPIII chain, as shown below:
It can be seen that there is another choice in the chain when 7Be(beryllium) either captures an electron to form 7Li(lithium) in the PPII chain or captures another proton to form 8B(boron) in the PPIII chain. At the end of the PPIII chain, the unstable nucleus of 8Be breaks up to form two 4He nuclei.

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. but it does generate abundant high energy neutrinos.
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 = MeV
The relative importance of the PPI and PPII chains depend on the relative importance of the reactions of 3He with 3He in PPI as compared to the reactions of 3He with 4He in PPII.
The other hydrogen burning reaction of importance is the CNO cycle:
But ( Pn +e+ + ve )
Release neutrinos, 0.73MeV
Q(4p4He) ~ 26 MeV
The rate of energy generation:

27. * C, N act as catalysts in the reactions
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.
There is a small percentage of the initial composition of any star consists of carbon, nitrogen, and oxygen (CNO) nuclei. These nucleimay induce a chain of reactions that transform H into He, in which they themselves act similarly to catalysts in chemical reactions: they are destroyed and reformed in a cyclic.
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
In such reactions, two 4He nuclei fuse to form a 8Be nucleus, but this is very unstable to fission and rapidly decays to two 4He nuclei again
It can be seen that the reaction leaps from helium to carbon in one go, by-passing lithium, beryllium and boron. It is no coincidence that these three elements are over 105 times less abundant by number than carbon.
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 = 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
We will not discuss these reactions here as the majority of the possible energy release by nuclear fusion reactions has occurred by the time that hydrogen and helium have been burnt.
Energy released: Q=7.275MeV ~ 5.8×1013 Jkg-1
4 4He  16O
Energy released: Q = 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) = MeV
While energy released in the second is given by adding to it the energy released by  capture on a 12C nucleus,
Q(4) = = 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. E.g. Carbon burning requires > 4MS
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  Ne MeV (Ne also from C + C)
Ne He24  Mg MeV
Mg24 + He4  Si MeV
At ~109 K C12 + C12  Mg MeV
O16 + O16  S MeV
Mg24 + S32  Fe END OF FUSION
Old massive stars accumulate a series of shells as illustrated below just prior to the supernova stage.
The physics of such “multiphase stars” becomes extremely complex.
We shall see in the next part of the course how the use of the above approximate expressions for the laws of energy production enables us to obtain useful qualitative information about the structures of stars.

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.
Thus far, we have assumed that stars are composed of ideal gases and hence the pressure is given by the ideal gas equation of state we have just derived.
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
Thus far, we have assumed that stars are composed of ideal gases and hence the pressure is given by the ideal gas equation of state we have just derived.