The structure and evolution of
stars
1
Introduction
What are the main physical processes which determine the structure of stars ?
•
•
•
•
•
Stars are held together by gravitation – attraction exerted on each part
of the star by all other parts
Collapse is resisted by internal thermal pressure.
These two forces play the principal role in determining stellar structure
– they must be (at least almost) in balance
Thermal properties of stars – continually radiating into space. If
thermal properties are constant, continual energy source must exist
Theory must describe - origin of energy and transport to surface
We make two fundamental assumptions :
1) Neglect the rate of change of properties – assume constant with
time
2) All stars are spherical and symmetric about their centres
We will start with these assumptions and later reconsider their validity
2
For our stars – which are isolated, static, and spherically symmetric –
there are four basic equations to describe structure. All physical
quantities depend on the distance from the centre of the star alone
1) Equation of hydrostatic equilibrium: at each radius, forces due to
pressure differences balance gravity
2) Conservation of mass
3) Conservation of energy : at each radius, the change in the energy
flux = local rate of energy release
4) Equation of energy transport : relation between the energy flux and
the local gradient of temperature
These basic equations supplemented with
•
Equation of state (pressure of a gas as a function of its density
and temperature)
•
Opacity (how opaque the gas is to the radiation field)
•
Core nuclear energy generation rate
3

Equation of hydrostatic support
Balance between gravity and internal pressure is known as hydrostatic
equilibrium
Mass of element
m  (r)sr
where (r)=density at r
Consider forces acting in radial direction
1. Outward force: pressure exerted by stellar material
on the lower face:
P(r)s
2. Inward force: pressure exerted by stellar material
on the upper face, and gravitational attraction of all
stellar material lying within r

GM(r)
P(r  r)s 
m
2
r
GM(r)
 P(r  r)s 
(r)sr
2
r
4
In hydrostatic equilibrium:
GM(r)
P(r)s  P(r  r)s 
(r)sr
2
r
GM(r)
 P(r  r)  P(r)  
(r)r
2
r
If we consider an infinitesimal element, we write
P(r  r)  P(r) dP(r)

r
dr
for r0
Hence rearranging above we get
dP(r)
GM(r)(r)

dr
r2
The equation of hydrostatic support
5
Equation of mass conservation
Mass M(r) contained within a star of radius r is determined by the density of
the gas ( r).
Consider a thin shell inside the star with
radius r and outer radius r+r
V  4r 2r
 M  V(r)  4 r 2r(r)
dM(r)

 4 r 2 (r)
dr
In the limit where r  0
This the equation of mass conservation
6
Accuracy of hydrostatic assumption
We have assumed that the gravity and pressure forces are balanced how valid is that ?
Consider the case where the outward and inward forces are not equal,
there will be a resultant force acting on the element which will give rise to
an acceleration a
GM(r)
(r)sr  P(r)s  (r)sra
2
r
dP(r) GM(r)


(r)  (r)a
2
dr
r
P(r  r)s 
Now acceleration due to gravity is g=GM(r)/r2

dP(r)
 g(r)  (r)a
dr
Which is the generalised form of the equation of hydrostatic support
7
Accuracy of hydrostatic assumption
Now suppose there is a resultant force on the element (LHS 0).
Suppose their sum is small fraction of gravitational term ()
(r)g  (r)a
Hence there is an inward acceleration of
a  g
Assuming it begins at rest, the spatial displacement d after a time t is
1 2 1
d  at  gt 2
2
2
Class Tasks
1. Estimate the timescale for the Sun’s radius to change by an observable
amount (as a function of ). Assume  is small, is the timescale likely ?
(r=7x108 m ; g=2.5x102 ms-2)
2.
We know from geological and fossil records that it is unlikely to have
changed its flux output significantly over the last 109 . Hence find an upper
limit for . What does this imply about the assumption of hydrostatic
equilibrium ?
8
9
The dynamical timescale
If we allowed the star to collapse i.e. set d=r and substitute g=GM/r2
1
2
1 2r 3 
t


B GM 
Assuming 1
1
2
2r 3 
t d  

GM 
td is known as the dynamical time. What is it a measure of ?
r=7x108
m
M =1.99x1030 kg
10
Accuracy of spherical symmetry assumption
Stars are rotating gaseous bodies – to what
extent are they flattened at the poles ?
If so, departures from spherical symmetry
must be accounted for
Consider mass m near the surface of star of
mass M and radius r
Element will be acted on by additional
inwardly acting force to provide circular
motion.
Centripetal force is given by:
m r
2
Where  = angular velocity of star
There will be no departure from spherical symmetry provided that
GMm
m r
1 or
2
r
2
GM
  3
r
2
11
Accuracy of spherical symmetry assumption
Note the RHS of this equation is similar to td
1
2
2r 3 
t d  
 or
GM 
2
2
   2
td
GM 2
 2
3
r
td
And as =2/P ; where P=rotation period
If spherical symmetry is to hold then P >> td
For example td(sun)~2000s and P~1 month
For the majority of stars, departures from spherical symmetry can be
ignored.
Some stars do rotate rapidly and rotational effects must be included in the
structure equations - can change the output of models
12
Before deriving the relations for (3) and (4) we will consider several
applications of our current knowledge. You will derive mathematical formulae
for the following
1) Minimum value for central pressure of a star
2) The Virial theorem
3) Minimum mean temperature of a star
4) State of stellar material
In doing this you will learn important assumptions and approximations that
allow the values for minimum central pressure, mean temperature and the
physical state of stellar material to be derived
13
Minimum value for central pressure of star
We have only 2 of the 4 equations, and no knowledge yet of material
composition or physical state. But can deduce a minimum central pressure :
Why, in principle, do you think there needs to be a minimum value ? given what
we know, what is this likely to depend upon ?
dM(r)
 4r 2 (r)
dr
dP(r) dM(r) dP
GM


dr
dr
dM
4r 4
dP(r)
GM(r)(r)

dr
r2
Divide these two equations:

Can integrate this to give

Lower limit to RHS:


Ms
0
Pc  Ps 

GM
dM 
4
4r

Ms
0
Ms
0
GM
dM
4
4r
GM
GMS2
dM 
4
4rs
8rs4
14

Hence we have
GMs2
Pc  Ps 
8rs4
We can approximate the pressure at the surface of the star to be zero:
GMs2
Pc 
8rs4
For example for the Sun:
Pc=4.5  1013 Nm-2 = 4.5  108 atmospheres
This seems rather large for gaseous material – we shall see that this is not
an ordinary gas.
15
The Virial theorem
Again lets take the two equations of hydrostatic equilibrium and mass
conservation and divide them
dP(r) dM(r) dP
GM


dr
dr
dM
4r 4
Now multiply both sides by 4r2
GM
4r dP  
dM
r
And integrate over the whole star

3
3  V dP   
Ps
Ms
Pc
0
GM
dM
r
Where V = vol contained within radius r
Use integration by parts to integrate LHS
3PV c  3 
s
Vs
Vc
M s GM

PdV   
dM
0
r
At centre, Vc=0 and at surface Ps=0
16
Hence we have
3  PdV 
Vs
0

Ms
0
GM
dM  0
r
Now the right hand term = total gravitational potential energy of
the star or it is the energy released in forming the star from its
components dispersed to infinity.
Thus we can write the Virial Theorem :
3  PdV    0
Vs
0
Thisis of great importance in astrophysics and has many
applications. We shall see that it relates the gravitational energy
of a star to its thermal energy
17
Minimum mean temperature of a star
We have seen that pressure, P, is an important term in the equation of
hydrostatic equilibrium and the Virial theorem. We have derived a minimum value
for the central pressure (Pc>4.5  108 atmospheres)
What physical processes give rise to this pressure – which are the most
important ?
• Gas pressure Pg
• Radiation pressure Pr
• We shall show that Pr is negligible in stellar interiors and pressure is
dominated by Pg
To do this we first need to estimate the minimum mean temperature of a star
Consider the  term, which is the gravitational potential energy:
 

Ms
0
GM
dM
r
18
We can obtain a lower bound on the RHS by noting: at all points inside
the star r<rs and hence 1/r > 1/rs


Ms
0
GM
dM 
r

GM
GMs
dM 
rs
2rs
Ms
0
Now dM=dV and the Virial theorem can be written
  3  0 PdV  3  0
Vs
Ms
P

dM
Now pressure is sum of radiation pressure and gas pressure: P = Pg +Pr
Assume, for now, that stars are composed of ideal gas with negligible Pr
kT
m
where n  number of particles per m
P  nkT 
The eqn of state of ideal gas
3
m  average mass of particles
k  Boltzmann' s constant
19
Hence we have
  3  0
Ms
P

dM  3  0
Ms
kT
dM
m
And we may use the inequality derived above to write
  3  0
Ms


Ms
0
kT
GM s2
dM 
m
2rs
GM s2 m
TdM 
6krs
We can think of the LHS as the sum of the temperatures of all the mass
elements dM which make up the star
–
The mean temperature of the star T is then just the integral divided
by the total mass of the star Ms
 MsT 
T
GM sm
6krs

Ms
0
TdM
20
Minimum mean temperature of the sun
As an example for the sun we have
m
T  4 10
K
mH
6
where m H 1.67 1027 kg
Now we know that H is the most abundant element in stars and for a
fully ionised hydrogen star m/mH=1/2 (as there are two particles, p + e–,
for each H atom). And for any other element m/mH is greater
–
T

> 2  106 K
21
Physical state of stellar material
We can also estimate the mean density of the Sun using:
av 
3M
3
-3
1.4
10
kgm
4r3
Mean density of the sun is only a little higher than water and other ordainary
–
liquids. We know such liquids become gaseous at T much lower thanT
–
Also the average K.E. of particles at T is much higher than the ionisation
potential of H. Thus the gas must be highly ionised, i.e. is a plasma.
It can thus withstand greater compression without deviating from an ideal gas.
Note that an ideal gas demands that the distances between the particles are
much greater than their sizes, and nuclear dimension is 10-15 m compared to
atomic dimension of 10-10 m
Lets revisit the issue of radiation vs gas pressure. We assumed that the
radiation pressure was negligible. The pressure exerted by photons on the
particles in a gas is:
Prad
aT 4

3
(See Prialnik Section 3.4)
Where a = radiation density constant
22
Now compare gas and radiation pressure at a typical point in the Sun
Pr aT 4

Pg
3
kT maT 3

m
3k
27
1.67
10
Taking T ~ Tav  2 10 6 K,  ~  av  1.4 10 3 kgm3 and m 
kg
2
Pr
Gives
~ 104
Pg
Hence radiation pressure appears to be negligible at a typical (average) point
in the Sun. In summary, with no knowledge of how energy is generated in
stars we have been able to derive a value for the Sun’s internal temperature
and deduce that it is composed of a near ideal gas plasma with negligible
radiation pressure
23
Mass dependency of radiation to gas pressure
However we shall later see that Pr does become significant in higher mass
stars. To give a basic idea of this dependency: replace  in the ratio equation
above:
maT 3
4 ma rs3T 3

3M s 
9k M s
3k 3 
4 rs 
Ms
And from the Virial theorem : T ~
rs
Pr

 M s2
Pg
Pr

Pg
i.e. Pr becomes more significant in higher mass stars.
24
Summary and next lecture
With only two of the four equations of stellar structure, we have
derived important relations for Pc and mean T
We have derived and used the Virial theorem – this is an important
formula and concept in this course, and astrophysics in general.
You should be comfortable with the derivation and application of
this theorem.
In the next lecture we will explore the energy generation and
energy transport in stars to provide the four equations that can be
simultaneously solved to provide structural models of stars.
25
Energy generation in stars
So far we have only considered the dynamical properties of the star, and the
state of the stellar material. We need to consider the source of the stellar
energy.
Let’s consider the origin of the energy i.e. the conversion of energy from some
form in which it is not immediately available into some form that it can radiate.
How much energy does the sun need to generate in order to shine with it’s
measured flux ?
L0  4 10 26 W  4 10 26 Js -1
Sun has not changed flux in 10
9
 Sun has radiated 1.2 x10
J
43
yr (3x10 6 s)
E  mc 2
 mlost  10 26 kg  104 M 0
26
Source of energy generation:
What is the source of this energy ? Four possibilities :
• Cooling or contraction
• Chemical Reactions
• Nuclear Reactions
Cooling and contraction
These are closely related, so we consider them together. Cooling is simplest
idea of all. Suppose the radiative energy of Sun is due to the Sun being much
hotter when it was formed, and has since been cooling down .We can test how
plausible this is.
Or is sun slowly contracting with consequent release of gravitational potential
energy, which is converted to radiation ?
27
Source of energy generation:
In an ideal gas, the thermal energy of a particle (where nf=number of degrees
of freedom = 3)
kT

nf
2
3kT

2
3knT
Total thermal energy per
 unit volume
2 volume
N = number of particles per unit
Now, Virial theorem :
3  0 PdV    0
Vs

Assume that stellar material is ideal gas (negligible Pr)
 P  nkT
3  0 nkTdV    0
Vs
28
Source of energy generation:
Now lets define U= integral over volume of the thermal energy per unit volume
thermal energy per unit volume

3knT
2
 2U    0
The negative gravitational energy of a star is equal to twice its thermal energy.
This means that the time for which the present thermal energy of the Sun can
supply its radiation and the time for which the past release of gravitational
potential energy could have supplied its present rate of radiation differ by only
a factor two. We can estimate the later:
Negative gravitational potential energy of a star is related by the inequality
GMs2
 
2rs
as an approximation assume
GM s2
- ~
2rs
29
Source of energy generation:
Total release of gravitational potential energy would have been sufficient to
provide radiant energy at a rate given by the luminosity of the star Ls , for a time
GMs2
t th ~
Lsrs
Putting in values for the Sun: tth=3107 years.
Hence if sun where powered by either contraction or cooling, it would have
changed substantially in the last 10 million years. A factor of ~100 too short to
account for the constraints on age of the Sun imposed by fossil and geological
records.
Definition: tth is defined as the thermal timescale (or Kelvin-Helmholtz timescale)
Chemical Reactions
Can quickly rule these out as possible energy sources for the Sun. We calculated
above that we need to find a process that can produce at least 10-4 of the rest
mass energy of the Sun. Chemical reactions such as the combustion of fossil
fuels release ~ 510-10 of the rest mass energy of the fuel.
30
Source of energy generation:
Nuclear Reactions
Hence the only known way of producing sufficiently large amounts of energy is
through nuclear reactions. There are two types of nuclear reactions, fission and
fusion. Fission reactions, such as those that occur in nuclear reactors, or atomic
weapons can release ~ 510-4 of rest mass energy through fission of heavy nuclei
(uranium or plutonium).
Class task
Can you show that the fusion reactions can release enough energy to feasibly
power a star ?
Assume atomic weight of H=1.008172 and He4=4.003875
Hence we can see that both fusion and fission could in principle power the Sun.
Which is the more likely ?
As light elements are much more abundant in the solar system that heavy ones,
we would expect nuclear fusion to be the dominant source.
Given the limits on P(r) and T(r) that we have just obtained - are the central
31
conditions suitable for fusion ? We will return to this later.
Equation of energy production
The third equation of stellar structure:
relation between energy release and the
rate of energy transport
Consider a spherically symmetric star in
which energy transport is radial and in
which time variations are unimportant.
L(r)=rate of energy flow across sphere of
radius r
L(r+r)=rate of energy flow across
sphere of radius r +r
Because shell is thin:
V (r)  4 r 2r
and m(r)  4r 2 (r)r
32
We define  = energy release per unit mass per unit volume (Wkg-1)
Hence energy release in shell is written:
4 r 2 (r)r
Conservation of energy leads us to
L(r  r)  L(r)  4r 2 (r)r

L(r  r)
 L(r)  4r 2 (r)r
r
and for r  0
dL(r)
 4 r 2  (r)
dr
This is the equation of energy production.
We now have three of the equations of stellar structure. However we
have five unknowns P(r), M(r), L(r), (r) ,(r) . In order to make further
progress we need to consider energy transport in stars.
33
Method of energy transport
There are three ways energy can be transported in stars
• Convection – energy transport by mass motions of the gas
• Conduction – by exchange of energy during collisions of gas particles (usually e-)
• Radiation – energy transport by the emission and absorption of photons
Conduction and radiation are similar processes – they both involve transfer of
energy by direct interaction, either between particles or between photons and
particles.
Which is the more dominant in stars ?
Energy carried by a typical particle ~ 3kT/2 is comparable to energy carried by
typical photon ~ hc/
But number density of particles is much greater than that of photons. This would
imply conduction is more important than radiation.
Mean free path of photon ~ 10-2m
Mean free path of particle ~ 10-10 m
Photons can move across temperature gradients more easily, hence
larger transport of energy. Conduction is negligible, radiation transport
in dominant
34
Solar surface from Swedish Solar Telescope
Resolution ~100km
Granule size ~1000km
35
Convection
Convective element of stellar material
Convection is the mass motion of
gas elements – only occurs when
temperature gradient exceeds some
critical value. We can derive an
expression for this.
Consider a convective element at
distance r from centre of star.
Element is in equilibrium with
surroundings
Now let’s suppose it rises to r+r. It
expands, P(r) and (r) are reduced
to P- P and  - 
But these may not be the same as the same as the new surrounding gas
conditions. Define those as P- P and  - 
If gas element is denser than surroundings at r+r then will sink (i.e. stable)
If it is less dense then it will keep on rising – convectively unstable 36
The condition for instability is therefore
      
Whether or not this condition is satisfied depends on two things:
•
The rate at which the element expands due to decreasing pressure
•
The rate at which the density of the surroundings decreases with
height
2.
Let’s make two assumptions
1. The element rises adiabatically
The element rises at a speed much less than the sound speed. During
motion, sound waves have time to smooth out the pressure differences
between the element and the surroundings. Hence P =P at all times
The first assumption means that the element must obey the adiabatic relation
between pressure and volume
PV   constant
Where =cp/cv is the specific heat (i.e. the energy in J to raise temperature
of 1kg of material by 1K) at constant pressure, divided by specific heat at
constant volume
37
Given that V is inversely proportional to  , we can write
P


 constant
Hence equating the term at r and r+r:
P  P
P
 

(  )

If  is small we can expand ( - ) using the binomial theorem as follows
( - ) ~  -  -1 
 
and combining last two expressions

P
P
Now we need to evaluate the change in density of the surroundings, 
Lets consider an infinitesimal rise of r
 
d
r
dr
38
And substituting these expressions for  and  into the condition for
convective instability derived above:

d
P  r
P
dr
And this can be rewritten by recalling our 2nd assumption that element will
remain at the same pressure as it surroundings, so that in the limit
P dP
r  0,

r dr
 dP d

P dr dr
The LHS is the density gradient that would exist in the surroundings if they
had an adiabatic relation between density and pressure. RHS is the actual
density in the surroundings. We can convert this to a more useful expression,
by first dividing both sides by dP/dr. Note that dP/dr is negative, hence the
inequality sign must change.
39
 d dP

P dr
dr

d


P
dP
P  d 1

 
  dP 
And for an ideal gas in which radiation pressure is negligible (where m is
the mean mass of particles in the stellar material)
P
kT
m
ln P  ln   ln T  constant
And can differentiate to give
dP d dT


P

T
And combining this with the equation above gives ….
40
Condition for occurrence of convection
P dT  1

T dP

Which is the condition for the occurrence of convection (in terms of the
temperature gradient). A gas is convectively unstable if the actual temperature
gradient is steeper than the adiabatic gradient.
If the condition is satisfied, then large scale rising and falling motions transport
energy upwards.
The criterion can be satisfied in two ways. The ratio of specific heats  is close
to unity or the temperature gradient is very steep.
For example if a large amount of energy is released at the centre of a star, it
may require a large temperature gradient to carry the energy away. Hence
where nuclear energy is being released, convection may occur.
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Condition for occurrence of convection
Alternatively in the cool outer layers of a star, gas may only be partially ionised,
hence much of the heat used to raise the temperature of the gas goes into
ionisation and hence the specific heat of the gas at constant V is nearly the same
as the specific heat at constant P , and ~1.
In such a case, a star can have a cool outer convective layer. We will come back
to the issues of convective cores and convective outer envelopes later.
Convection is an extremely complicated subject and it is true to say that the lack
of a good theory of convection is one of the worst defects in our present studies
of stellar structure and evolution. We know the conditions under which convection
is likely to occur but don’t know how much energy is carried by convection.
Fortunately we will see that we can often find occasions where we can manage
without this knowledge.
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Summary
Hence we have shown that the source of energy in the sun must be nuclear.
Presumably you all knew that anyway !
We have derived the third formula of the equations of stellar structure (the
equation of energy production). Next lecture we will derive the 4th equation – the
equation of radiative transport, and discuss how to solve this set of equations.
Before doing that we considered the mode of energy transport in stellar interiors,
and derived the condition for convection. We saw that convection may be
important in hot stellar cores and cool outer envelopes, but that a good
quantitative theory is lacking.
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