Energy generation in stars
So far, we have only considered the dynamical properties of the star,
and the state of the stellar material. But what is 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?
26 26 -1
9 6
43
2
26 4
( ) 4 10 W 4 10 Js
Sun has not changed flux in 10 yr (3x10 s)
Sun has radiated 1.2 x10 J
10 kg 10 ( )lost
L Sun
E mc
m M Sun−
=  = 

=
 = =
Source of energy generation
What is the source of this energy? We can think of four possibilities:
• Cooling
• 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 the Sun is due
to it being much hotter when it was formed, and has since been
cooling down.
Or is the Sun slowly contracting with consequent release of
gravitational potential energy, which is converted to radiation?
Source of energy generation
In an ideal gas, the thermal energy of a particle; where 𝑛 𝑓 … number
of degrees of freedom (= 3)
Total thermal energy per unit volume
𝑛 … number of particles per unit volume
Assume that stellar material is ideal gas (negligible 𝑃𝑟)
3
2 2
f
kT kT
n= =
3
2
knT
=
0
3 0
sV
PdV +  =
0
3 0
sV
P nkT
nkTdV
 =
+  =
Now Virial Theorem:
Source of energy generation
Now lets define 𝑈 … integral over volume of the thermal energy per
unit volume
Thermal energy per unit volume
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.
Negative gravitational potential energy of a star is related by the
inequality
3
2 0
2
knT
U=  +  =
2 2
as an approximation assume ~
2 2
s s
s s
GM GM
r r
−  − 
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 𝐿 𝑠, for a time
2
~ s
th
s s
GM
t
L r
For the values of the Sun: 𝑡𝑡ℎ 𝑆𝑢𝑛 = 3 × 107 years.
𝑡𝑡ℎ… thermal timescale (or Kelvin-Helmholtz timescale)
Hence if the Sun is 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.
Chemical Reactions
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.
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).
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 + 𝛿𝑟) … rate of energy flow
across sphere of radius 𝑟 + 𝛿𝑟
Because the shell is thin: 2
2
( ) 4
and ( ) 4 ( )
V r r r
m r r r r
  
   
=
=
We define 𝜀 ... energy release per unit mass per unit volume (Wkg-1)
Hence energy release in a shell is written as:
Conservation of energy leads us to
2
2
2
( ) ( ) 4 ( )
( )
( ) 4 ( )
and for 0
( )
4 ( )
L r r L r r r r
L r r
L r r r r
r
r
dL r
r r
dr
    

   


  
+ = +

+
− =
→
=
2
4 ( )r r r   
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 , 𝜌 𝑟 , 𝜀(𝑟). The next step is
to investigate the energy transport inside stars.
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 ~3 Τ𝑘𝑇 2 is comparable to energy carried by
typical photon ~ Τℎ𝑐 𝜆
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 ~ 10-10m
Photons can move across temperature gradients more easily, hence larger
transport of energy. Conduction is negligible, radiation transport in dominant
Convection
Convection is the mass motion of gas
elements which 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 + 𝛿𝑟. It
expands, 𝑃(𝑟) and 𝜌 𝑟 are reduced to
𝑃 − 𝛿𝑃 and 𝜌 − 𝛿𝜌
Convective element of stellar material
But these may not be the same as the new surrounding gas conditions. Define
those as 𝑃 − 𝛿𝑃 and 𝜌 − 𝛿𝜌
If gas element is denser than surroundings at 𝑟 + 𝛿𝑟 then will sink (i.e. stable)
If it is less dense then it will keep on rising – convectively unstable
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
Let’s make two assumptions
1. The element rises adiabatically
2. 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 𝛿𝑃 = ∆𝑃 at all times
The first assumption means that the element must obey the adiabatic relation
between pressure and volume
Where 𝛾 = Τ𝑐 𝑝 𝑐 𝑣 is the specific heat (i.e. the energy in J to raise temperature of
1kg of material by 1K) at constant pressure 𝑐 𝑝, divided by specific heat at constant
volume 𝑐 𝑣
   −  − 
constantPV
=
Given that 𝑉 is inversely proportional to 𝜌, we can write
Hence equating the term at 𝑟 and 𝑟 + 𝛿𝑟:
If 𝛿𝜌 is small we can expand (𝜌 − 𝛿𝜌) using the binomial theorem
as follows
Now we need to evaluate the change in density of the surroundings,
𝛿𝜌
Lets consider an infinitesimal rise of 𝛿𝑟
constant
P


=
( )
P P P
 

  
−
=
−
-1
( ) ~ and combining last two expressions
P
P
  
    

 

− −
=
r
d
dr

  =
And substituting these expressions for 𝛿𝜌 and ∆𝜌 into the condition
for convective instability derived above:
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
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 Τ𝑑𝑃 𝑑𝑟. Note
that Τ𝑑𝑃 𝑑𝑟 is negative, hence the inequality sign must change.
r
d
P
P dr
 
 


0,
P dP
r
r dr



→ =
dP d
P dr dr
 


And for an ideal gas in which radiation pressure is negligible (where
m is the mean mass of particles in the stellar material)
And can differentiate to give
1
d dP
dr drP
d
P dP
P d
dP
 

 


 

 
 
 
 
ln ln ln constant
kT
P
m
P T


=
= + +
dP d dT
P T


= +
And combining this with the equation above gives ….
Condition for occurrence of convection
1P dT
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.
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 𝛾~1.
In such a case, a star can have a cool outer convective layer.
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.