Transcript Slide 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
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.
  nm  nmH
Pgas = n k T
(1)
(2)
Pgas
kT
kT
RT



m
mH 

(3)
--- 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
RT
If radiation pressure can not be neglected,
P

where a is the radiation density constant.
How to calculate mean molecular weight?
aT 4

3
(4)
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.
In a in a fully ionised gas:
No of parti./
per p.or n.
mH
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,
 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.
Same calculation on the mean molecular weights per ion:
ni 
X Y
Z


m H 4m H 12m H

i 
1
12

X  Y / 4  Z / 12 1  11X  2Y
(1.24 for the Sun)
Same calculation on the mean molecular weights per electron
ni 
X
Y
Z


m H 2m H 2m H
(1.144 for the Sun)

e 
1
2

X  Y / 2  Z / 2 1 X
Furthermore: you can show that
n  ni  ne
1


1
i

1
e
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
xp  h
(7)
 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.
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.
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
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.
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:
 p0 full
2
p
8p V

A completely degenerate gas
N pp 

p
 p0 empty
h3
Strictly speaking, this occurs at absolute
zero temperature. Hence,
The total number of electrons N in volume V
8V
N 3
h
p0
8p 2V
N ( p) 
, p  p0
h3
N ( p)  0,
p  p0
2
3
p
dp

8

Vp
/
3
h
(8)
0

3
0
d) What is the pressure of the degenerate electrons?
The pressure P on the surface A
P
F
A
F 
dp
dt
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
dN  nAv( p)dt
The total momentum change will be dp  (2 p)dN  (2 p)  nAv( p)dt
F
dp
 2 pnAv ( p )
dt
P
F
 2npv ( p )
A
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 P 
we need to integrate over the momentum distribution

P
1
npv ( p )
3
1
0 3 npv ( p)dp (9)
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
me v p
p 
(1 
which can be rearranged to give,
p
p 2  12
vp 
(1  2 2 )
me
me c
vp
2
c2
)
1
2

8
P
3me h3
1
P   npv ( p ) dp
3
0
N
8p 2
ndp  dp  3 dp
V
h
p0
p 4 dp
1
0
p2 2
(1  2 2 )
me c
* For a non-relativistically degenerate gas (i.e. po << mec)
1
p2
(1 
)2 1
2
2
me c
8
P
3h 3 me
p0

0
8p 5 0
p dp 
15me h 3
4
Recalling that N = 8 p03V / 3 h3 and defining the electron density, ne = N / V
2
1 3 3 h2 5
P  ( )( )
ne 3
20 
m
(10)
* A relativistically degenerate gas (i.e. p0 >> mec).
p 2 12
p
(1  2 2 ) 
me c
me c
8c
P 3
3h
1
3
4
2cp0
1 3 
0 p dp  3h3  8    hcne 3
p0
4
3
(11)
In order to derive the equation of state for degenerated electrons, we must
convert the electron density ne to mass density .
Using
ne 
X
mH

 (1  X )
2mH

 (1  X )
2mH
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
P  K1 
P  K2 
4
3
5
3
(12)
K1 
2
h
20me
2
3
 3   (1  X ) 
  
 ,

2
m
  
H 
1
3
(13)
5
3
4
 hc  3   (1  X )  3
K 2     

 8     2mH 
(14)
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.
p 02
2me
where p0 is given by
The average distance between electron is ne
1
e 2 ne 3
3ne 23
h2

(
)
4 0
2me 8
The electron’s number density must satisfy:
8 me e 2 3
ne 
(
)
2
9  0 h

6  1028 m 3
 3h 3 ne
p 0  
 8

1
3
1
3


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
electrons
degeneracy pressure
neutrons
the force of gravity
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.
Fig. 2. different absorption mechanisms
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
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 freefree 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:
   (  , T , chem icalcom position)
In restricted ranges of density and temperature, however, the results of
detailed calculations can be represented by simple power laws of the form
  0  T 
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.
Fig. 3. opacity over temperature
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
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,
Fig. 3. opacity over temperature
-- 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.
1
   0  2T 4
(16)
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.

0
T
3..5
(17 )
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:
Q(Z , N )  [Zmp  Nmn  m(Z , N )]c2
(18)
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
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.
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.
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
PPI : 1. p  p  d  e  e ;
2. d  p3He   ;
PPII : 3' 3He 4He7Be   ;
PPIII : 4".
Be  p8B   ;
7
3. 3He  3He4He  p  p
(19)
4'. 7Be  e 7Li  e ; 5'. 7Li  p4He4He
(20)
5". 8B8Be  e  e ; 6". 8Be  2 4He (21)
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:
p  n  e  e
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
PPI : 3He  3He4He
PPII : 3He 4He7Be  
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:
 PPchain   0 T 4
ii) The CNO cycle:
There are two different branches forming a bi-cycle, each with six reactions
1.
C  p 13N  
12
3.
13
2.
C  p 14 N  
3.
N 13C  e    e
2.
13
N  p 15 O  
4.
14
5.
15
6.
15
1. 14N  p 15N  
O N  e  e

15
N  p  C  He
12
4
O N  e  e
15
15

N  p 16 O  
* C, N act as catalysts in the
reactions
15
4. 16O  p 17 F  
5. 17F 17 O  e    e
6. 17O  p 14 O  4He
4 p  4He and two + decays + ves
The released energy: 25 MeV.~ 6×1018 J kg-1
* 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 rate of energy generation:
CNOcycle:    0 T 17
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:
4
He 4He8Be
8
Be 4He12 C  
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 :
   0  2T 40
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
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
n(12 C ) 
0.5m
12mH
n(16 O) 
0.5m
16mH
Hence, the total energy released per unit mass is :
n(12 C )Q(3 )  n(16 O)Q(4 ) Q(3 ) / 24  Q(4 ) / 32
Q

 7.3 1013 Jkg 1
m
mH
c) A comparison for energy release rates:
PPchain :    0 T 4
CNOcycle :    0 T 17
Triple  alpha
reaction :    0  2T 40
 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.
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
Ne1020 + He24  Mg24
Mg24 + He4  Si1428
4.7 MeV
9.3 MeV
10.0 MeV
(Ne also from C + C)
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.
At ~109 K
Summary:
1. The pressures:
kT kT RT


a) equation of state of an ideal gas Pgas 
m
mH 

P
(3)
 = 4 / (6X + Y + 2)
ni 
X
Y
Z


m H 4m H 12m H
ne 
X Y
Z


mH 2mH 2mH


i 
e 
RT

aT 4

3
(4)
(6)
1
12

X  Y / 4  Z / 12 1  11X  2Y
n  ni  ne
1
1
2

X  Y / 2  Z / 2 1 X


1
i

1
e
b) equation of state of a degenerate gas
P  K1 
P  K2 
4
3
5
3
2
3
5
3
h
 3   (1  X ) 
  
 ,
20me     2mH 
2
(12)
K1 
(13)
 hc  3  3  (1  X )  3
K2  
  

 8     2mH 
1
4
(14)
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
8 me e 2 3
ne 
(
)
2
9  0 h

6  1028 m 3
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?
PPchain :    0 T 4
CNOcycle :    0 T 17
Triple  alpha
reaction :    0  2T 40
Each step of further burning requires a further jump in central temperature and
thus progressively larger stellar masses