Chapter twoThe 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?

-- Hydrogen and helium burning

-- binding energy

-- advanced burning

Chapter twoThe 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) )3( Pgas

RT

m

kT

m

kT

H

)2(Hmnnm

--- 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, )4(3

4aTTRP

where a is the radiation density constant.

How to calculate mean molecular weight?

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.

The total number of particles per cubic metre is then given by the sum of the above

n = (2X / mH) + (3Y / 4mH) + (Z / 2mH).

In a in a fully ionised gas:

n = (/ 4mH) (8X + 3Y + 2Z).

X + Y + Z = 1, and hence Z = 1 - X - Y,

n = (/ 4mH) (6X + Y + 2). Since = nmH

= 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.

No of parti./ per p.or n.

mH mH

Same calculation on the mean molecular weights per ion:

YXZYXm

Z

m

Y

m

Xn i

HHHi 2111

12

12/4/

1

124

(1.24 for the Sun)

Same calculation on the mean molecular weights per electron

XZYXm

Z

m

Y

m

Xn e

HHHi

1

2

2/2/

1

22

(1.144 for the Sun)

Furthermore: you can show that

ei nnn ei

111

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?

Or Pauli's exclusive effects between electron?

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

)7(hpx

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

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.

the electrons in such a gas will exert a greater pressure than predicted by the ideal gas equation of state.

Higher momentum mean higher kinetic energy

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.

In curve B, where dashed portions of the non-degenerate distribution above the value defined by the Pauli exclusion principle are transferred to higher momentasolid 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.

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.

If Npp is the number of electrons in V with momentum in the range p and p +p, Pauli's exclusion principle tells us that:

ph

VppN p

3

28

2 * (4 p2V / h3) p.

Strictly speaking, this occurs at absolute zero temperature. Hence,

A completely degenerate gas

empty

full

0

0

p

pp

0

03

2

,0)(

,8

)(

pppN

pph

VppN

The total number of electrons N in volume V

)8(3/88 33

0

0

23

0

hVpdpph

VN

p

dt

dpF

A

FP

dtpnAvdN )(

dtpnAvpdNpdp )()2()2(

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

)(2 ppnAvdt

dpF )(2 pnpv

A

FP

)(3

1pnpvP

)9()(3

1

0

dppnpvP

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

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

2

1

2

2

)1(c

v

vmp

p

pe

which can be rearranged to give,

2

1

22

2

)1(

cm

p

m

pv

eep

dph

pdp

V

Nndp

3

28

0

)(3

1dppnpvP

0

0 2

1

22

2

4

3

)1(3

8p

e

e

cm

p

dpp

hmP

* For a non-relativistically degenerate gas (i.e. po << mec)

1)1( 2

1

22

2

cm

p

e3

05

0

43 15

8

3

8 0

hm

pdpp

mhP

e

p

e

Recalling that N = 8 p03V / 3 h3 and defining the electron density, ne = N / V

)10()3

)(20

1( 3

523

2

enm

hP

* A relativistically degenerate gas (i.e. p0 >> mec).

cm

p

cm

p

ee

2

1

22

2

)1( )11(3

8

1

3

2

3

83

43

1

3

40

0

33

0

e

p

hcnh

cpdpp

h

cP

In order to derive the equation of state for degenerated electrons, we must convert the electron density ne to mass density .

HHHe m

X

m

X

m

Xn

2

)1(

2

)1(

Using

where X is the mass fraction of H, we can finally write:

)12(3

5

1KP the equation of state of an NR degenerate gas

)13(3

4

2KP the equation of state of a UR degenerate gas

)14(2

)1(3

8

,2

)1(3

20

3

4

3

1

2

3

5

3

22

1

H

He

m

XhcK

m

X

m

hK

Hence in a degenerate gas, the pressure depends only on the density and chemical composition and is independent of temperature.

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.

em

p

2

20

where p0 is given by 3

13

0 8

3

enhp

The average distance between electron is 3

1en

3

22

0

312

)8

3(

24 e

e

e n

m

hne

The electron’s number density must satisfy:

3283

20

2

106)(9

8 mh

emn ee

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

electrons neutrons

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

Fig. 2. different absorption mechanisms

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 = hbb

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;

-- the photons in stellar interiors are so energetic that they are more likely to cause bound-free absorptions

-- as most of the atoms are highly ionised and only a small fraction contain electrons in bound orbits.

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 = hbf.

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 = hff.

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.

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:

),,( ncompositiochemicalT

In restricted ranges of density and temperature, however, the results of detailed calculations can be represented by simple power laws of the form

T0where 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.

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: )15(0

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.

)16(42

1

0 T

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.

)17(5..3

0

T

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:

)18()],([),( 2cNZmNmZmNZQ np

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.

Fig.4. Binding energy per nucleon over atomic number

a) Fusion reaction

3.2 Nuclear fusion and fission reactions

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.

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 :

The proton-proton (PP) chain and the carbon-nitrogen (CNO) cycle.

There are many possibilities here, but we will be looking at the main two hydrogen-burning reaction chains

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.

)19(.3;.2;.1: 4333 ppHeHeHeHepdedppPPI e

)20(.5;.4;3: 447'77'743' HeHepLiLieBeBeHeHePPII e

)21(2".6;".5;".4: 488887 HeBeeBeBBpBePPIII e

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:

Fig7. The proton-proton reaction chain

eenp

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

The relative importance of the PPI and PPII chains depend on the relative importance of

HeHeHePPI 433: BeHeHePPII 743:

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 ( Pn +e+ + ve ) Release neutrinos, 0.73MeV

40 TPPchain The rate of energy generation:

Q(4p4He) ~ 26 MeV

ii) The CNO cycle:

HeCpN

eNO

OpN

NpC

eCN

NpC

e

e

41215

1515

1514

1413

1313

1312

.6

.5

.4

.3

.2

.1

HeOpO

eOF

FpO

OpN

eNO

NpN

e

e

41417

1717

1716

1615

1515

1514

.6

.5

.4

.3

.2

.1

There are two different branches forming a bi-cycle, each with six reactions

4 p 4He and two + decays + ves

* 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.

The released energy: 25 MeV.~ 6×1018 J kg-1 170: TCNOcycle

* The rate of energy generation:

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:

CHeBe

BeHeHe1248

844

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 : 4020 T

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

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 16O nuclei produced is given by

Hm

mCn

12

5.0)(12

The number of 12C nuclei produced is given by

Hm

mOn

16

5.0)(16

Hence, the total energy released per unit mass is :

1131612

103.732/)4(24/)3()4()()3()(

Jkgm

QQ

m

QOnQCnQ

H

Example:

If helium burning produces equal amounts (mass fractions) of carbon and oxygen, what is energy generated per unit mass?

c) A comparison for energy release rates:

4020

170

40

:

:

:

TreactionalphaTriple

TCNOcycle

TPPchain

The energy released by the PP chain and CNO cycle are smooth functions of temperature

is the energy release per unit mass per unit time,

* 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.

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 + He2

4 Ne1020 4.7 MeV (Ne also from C + C)

Ne1020 + He2

4 Mg24 9.3 MeV Mg24 + He4 Si14

28 10.0 MeV

At ~109 K C12 + C12 Mg24 14 MeVO16 + O16 S32 16 MeVMg24 + S32 Fe56 END OF FUSION

Old massive stars accumulate a series of shells as illustrated below just prior to the supernova stage.

1. The pressures:

Summary:

a) equation of state of an ideal gas

b) equation of state of a degenerate gas

)3( Pgas

RT

m

kT

m

kT

H

)4(3

4aTTRP

YXZYXm

Z

m

Y

m

Xn i

HHHi 2111

12

12/4/

1

124

= 4 / (6X + Y + 2) (6)

XZYXm

Z

m

Y

m

Xn e

HHH

e

1

2

2/2/

1

22

ei nnn

ei 111

)12(3

5

1KP

)13(3

4

2KP )14(2

)1(3

8

,2

)1(3

20

3

4

3

1

2

3

5

3

22

1

H

He

m

XhcK

m

X

m

hK

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

3283

20

2

106)(9

8 mh

emn ee

2. An approximate form for Opacity

3. Nuclear reactions in stars?

, , and o are dependent on temperature, density and chemical components of the stellar materials.

4020

170

40

:

:

:

TreactionalphaTriple

TCNOcycle

TPPchain

Each step of further burning requires a further jump in central temperature and thus progressively larger stellar masses