Stellar Physics
1. Introduction
We need to figure out how stars work and how stars change. Stars do not stay static
throughout their lives, but vary their properties slowly over time. But how do stars vary
over time? To understand this we need to understand the structure of stars and how all of
the structures are defined. Sometimes we need to make some assumptions about how
things work, sometimes we are still rather uncertain about the physics that are operating
in stars, and sometimes we understand the physics, but we are unable to make all of the
required calculations.
First a very important assumptions to make life simple – we’re going to assume that stars
are spherically symmetric, or put basically, they are perfect spheres. This isn’t really
true, but it makes life simple to say this. If we operate under this assumption that stars
are the same at the same distance from the center, we need only use one parameter to
define our location in the star, usually distance from the center. If we wanted to have
non-spherical stars, then we’d need 2 or 3 dimensional parameters to follow which makes
the problem much more complex.
With one dimension to worry about we can also describe everything relative to that
distance. What are all of the values that we need to keep track of? The main physical
characteristics to keep track of include Mass (M), Temperature (T), density (), Pressure
(P), Flux –energy flow (F), energy production (q), opacity () and composition (X, Y, Z).
Fortunately there are all sorts of relationships between these parameters so that it is
possible to figure out how stars work (at least on a computer).
Even though it is often easiest to use radius as the way of marking your location inside of
a star, sometimes using radii is not good. Some of the parameters listed above can
change quickly over a short distance so marking the pressure or density or temperature
over distance doesn’t provide a fine enough resolution to take into account these rapid
changes. Sometimes mass (M) is the parameter used to measure the change in the other
characteristics, in which the radii, temperature or pressure would be defined at positions
in the star that are fractions of the entire mass. Generally speaking it doesn’t matter how
you do it, but in some instances one scale is better than the other. So let’s start going
over all of the fun relationships involved in stellar physics.
2. Conservation of Mass
The simplest relation is between mass, density and radius. You are usually familiar with
the idea that mass = density x volume. But you can’t do that here. Why? Mainly the
density of a star is not uniform as you go from the center to the surface. In fact densities
go from 20 times denser than iron to less than the density of the air in this room. You
therefore need to use a relationship that shows how these thing relate regardless of the
actual values of mass, density and volume.
Notes 2 - 1
The Conservation of mass relation is
dm
 4r 2 
dr
2-1
And yes, this formula is actually what you’d see in calculus, but don’t get freaked out.
The dm and dr are calculus ways for saying “small changes in mass, m” and “small
changes in radius, r”, so when you see a “d” in this part of the notes, it is probably a
calculus “d” indicating a difference or change, not a value for distance.
And of course M or m refers to mass here, not magnitudes. Confusing, but remember the
context of the variables so that will help you keep them straight.
Basically this equation shows how mass varies over distance. You have to remember that
the density () is not a constant but will change with radius (varies within the star).
Technically I should have some notation there that indicates that  varies with radius, but
that’s just extra text.
Of course if you do have a constant density then the formula could be solved simply by
using calculus and showing the variation of the mass over the entire radius of the star.
You would go from the center (r=0, m=0 ), to the surface (r=R, m=M). This gives
M
R
 dm  4r
0
M
2
 (dr)
0
4R3 
3
This is just the basic mass=volume x density relationship that you’d have for a constant
density. Since the density in a star in reality isn’t constant you have to gradually go from
the center to the surface and find how much the mass change2 (dm) as you go a little
distance in radius (dr).
You need to view the “d” notation as a “difference” or “change” in values. There are
other things that are small increments of change, like dm and dr (mass and radius), and
when those items change, other things change as well. For example since we are
considering spherical stars, the mass interval, dm, is actually a shell of thickness dr and a
volume of size dV = 4  r2 dr. So a change in mass and radius also impacts a change in
volume and other parameters. Of course if stars were not spherical, you’d have to use a
different relationship.
3. Thermal Equilibrium
So we have small mass intervals, in small shell shaped volumes, with a density that varies
with distance from the center of the star. What else do we have? The volume of
material also has an internal energy, a temperature (T) and pressure (P). But of course
these things can change due to various things Heat can be absorbed by the material
Heat can be emitted (lost) by the material
Work can be done on the material (you could compress it or expand it)
Notes 2 - 2
The measure of energy or heat flow through the material is the flux (F). For a star to be
stable (“happy”), each layer must not create or expel more energy than comes out or into
it. Basically it has to remain in equilibrium – it can’t have too much energy and it can’t
have too little. The best situation is for it to remain constant in energy flow over time.
This is where we introduce the concept of Thermal Equilibrium – the balance of energy
in each layer. If a star is creating energy, q, that energy has to go through the material but
it can’t just stay in any particular layer, but it has to flow out or away from that layer, so
there is a flux through each layer. How much flux? Depends upon the amount of energy
and the material it is going through.
Relation for Thermal Equilibrium is the following –
dF
 4r 2 q
dr
2-2
And remember, density and energy both will be functions of radius, and will vary as you
go from one part of the star to another. Basically all energy that is produced is also given
off. You should (to remain a happy, stable star) have the same rate of energy emission as
energy production.
The value of the energy creation isn’t one that you have probably encountered previously.
q is the rate of nuclear energy released per unit mass of material, and it will vary
depending upon how you are making energy and the conditions under which the energy is
produced. We’ll go over the details about this later. It is worth noting that the values for
the energy flux have conditions at the center and surface of the star. Since a radius=0
has no mass, there is no energy production there (F=0), while at the surface of the star,
the flux has to equal the luminosity of the star (F=L).
4. Hydrostatic Equilibrium
What are the other attributes of a small chunk of the star?
Let’s look at the pressure variation. First we’ll start off
with one small piece of the star’s interior with the
following characteristics (also shown in the diagram)
Mass = dm
Height = dr
Surface area = dS (it is a cylinder shape here, but that’s
not important)
Density=
Volume = height x area = dr dS
So mass= density x volume or dm =  dr dS
What are the various forces that are acting on this chunk? Since it is a very small chunk,
we’ll assume that all sideways forces are cancelled out. This is okay since things change
up and down here, not side to side – which we can use since we’re only using radius as an
Notes 2 - 3
important direction of how things change. Well we have the forces on the chunk from
the bottom and the top, as well as the mass of the chunk itself.
Pressure at the bottom of the chunk, P(r), is the pressure defined at that distance r from
the center of the star, while the pressure at the top, P(r+dr), is the pressure defined at that
location from the star’s center, r+dr. There is also the force of gravity due to the mass of
the chunk, and this is a downward force. By definition, F=ma or force=mass x
acceleration. And this should equal all of the forces up and down. This would give you
F = dm a = - gravity (down) +pressure (up) – pressure (down)
We’re using the basic assumption that down is negative and up is positive. We’ll also use
the relation that P(r+dr) = P(r) + dP , or the pressure at the top equals the pressure at the
bottom and a small change in pressure. We’ll substitute that in the relationship in the
steps below-
F  (dm)a  
Gm(dm)
 P(r )dS  P(r  dr)dS
r2
Gm(dm)
 P(r )dS  ( P(r )  dP)dS
r2
Gm(dm)

 P(r )dS  P(r )dS  dPdS
r2

Also remember that we have that dS=dm/(dr)
When you substitute that in you get the following, which can be further simplified by
dividing out the dm term….
Gm(dm) dm dP

 dr
r2
Gm 1 dP
a 2 
 dr
r
(dm)a  
This relation shows how the chunk of stuff would accelerate due to gravity and external
pressures. But we don’t want layers to be moving around – they should be sitting still!
We want stability. In that case a=0. If you put that in the formula and move some things
around you get -
dP
GM
  2
dr
r
2-3
This is the formula of Hydrostatic Equilibrium – very important for the stability of stars
as well as anything that needs gas or air pressure to stay intact.
This formula can also be written in terms of mass,
dP
Gm

dm
4r 4
This formula can be integrated from the surface to the center of the star to help us
approximate the pressure at the center. We assume the surface pressure =0 (not really
technically true but pretty close) which will give you a lower limit for the center pressure
Notes 2 - 4
for the star, not an actual value. We’ll also simplify things by setting r=R rather than
integrating the radius value.
P(surface)  P(M )  0
P(center)  P(0)  Pc
M
P(M )  P(0)  P(0)   
0
Pc 
Gm
Gm
dm   
dm
4
4r
4R4
2
GM
8R4
2-4
This is really just a lower limit for the central pressure, it can be much greater.
5. Energy Balance
Now let’s look at the overall energy inventory of the star and see how these things
interact with one another.
First of all there is the gravitational potential energy – basically the gravity that is used to
bring the whole star together. That’s pretty straightforward and makes use of the laws of
gravity to define it.
There is also the energy caused by the pressure exerted by the star. There are various
ways to define this energy using the mass, density and pressure values, mainly by using
various equations of state (I’ll get to those later). Typically for simplicity, the ideal gas
laws are used since they are simple and well defined. The energy that is included here is
mainly through the various internal gas pressures due to the motion of the particles.
There is also kinetic energy (energy of motion) due to the various parts of the star moving
around. This isn’t always noticeable, and may not be obvious since parts can be moving
up/down or in/out in a rather non-uniform manner.
And there is the energy from energy production, and energy lost (the luminosity of the
star).
So add up all of these energies and keep track of whether they add something to the
energy inventory or take something away and see how they influence the star over time.
For the star to be balanced thermally (so energy doesn’t stay bottled up in the star and
overheat things), the total energy of the star shouldn’t change over time.
In reality the various energy values will not remain the same value during the existence
of the star. As the star changes internally, the structure changes, and this will impact all
of the various forms of energy. However, they still need to balance out in spite of these
changes. Fortunately there are some simplifications that we can make use of. If a star is
Notes 2 - 5
in hydrostatic equilibrium, the kinetic energy should equal 0. And we also want thermal
equilibrium, so the amount of energy produced equals the amount that is given off.
With thermal and hydrostatic equilibrium, we only need to worry about contributions to
the energy balance of a star due to the gravitational energy, and the internal energy.
These are related (in complicated, hard to explain ways – in what is known as the Virial
Theorem), but here are the basic rules –
If the energy balance in the star is negative (more energy is given off than is produced)
what would you expect to happen? Think of it like a balloon – with less inside of it, it
will collapse in on itself since gravity is always pulling inwards. For the case of a star
this scenario indicates that the star’s internal energy isn’t high enough to counteract
gravity, since it has lost too much energy. Since there is less gas pressure exerted by the
inside of the star, the star’s self gravity allows the star to contract. But what happens
when you contract something? It heats up the interior and the end result is more energy
production inside takes place, which helps to increase the internal gas pressure, which
counteracts the inward pull of gravity. So the balance gets restored.
And if the opposite is true, that the energy balance is positive (more energy is produced
than is given off), what do you think will happen? Since the star’s internal energy is
excessive, the internal gas pressure will increase, and the star will expand – the internal
energy is too much for the gravity. And what happens when objects expand? They cool
down. This will decrease the internal temperature and reduce the energy production and
again restore the balance.
So there is a balancing act in control that helps to maintain the energy flow as well as the
internal temperature and pressure balance of the star.
The next step is to get some more formula to help us solve how these stars behave. The
relations given above aren’t enough.
6. Equation of State
The interrelationship between pressure, density and temperature (and also chemical
composition, but that doesn’t change much) is known as the Equation of State (EOS).
Often these are given as various ways to calculate pressure, and there are quite a few
different ways to do that. Let’s take a peek at these different relationships.
First you have to take into account that different atoms influence materials differently,
and now we’ll look at how that factors into the various EOS relations. You can have
material that is ionized, or unionized, so you need to account for the effect of atoms as a
whole, atomic nuclei and free electrons. Generally speaking stars are pretty hot, so there
is a lot of ionization going on. We’ll look at the ions first.
Notes 2 - 6
6.1 Ideal Gas Law
The composition of the gas influences the density and for the most part the density is
influenced by the mean atomic mass, . This term makes it easier to calculate density
since we can then just use a standardized unit for mass and not have to worry about all of
the composition details. Those all go into . But since material can be in various states
(ionized or unionized), that has to be considered.
The mean atomic mass of ions is given by I, which is given in terms of composition (X,
Y, Z) as
1
I
X
Y
Z

4  A
This is to say that hydrogen has a mass of 1 (X/1), helium has a mass of 4 (Y/4) and the
metals have an average mass of <A>. If we assume the typical composition of stars,
which have most of their metal mass comprised of C, N, O, Ne, and Fe, this gives an
average value of <A> ~20. So this relation can be approximated as
1
I
X
Y Z

4 20
2-5
remember, typically X=0.7, Y=.28, Z=.02, so I=1.3
To determine the pressure in a gas caused by ions, you would then use the relation
Ion Pressure PI 
T
2-6
I
Where  is the gas constant.
Now you also have to look at the contribution of free electrons to pressure and density.
We define another mean mass for the electrons, e = mean number of free electrons per
nucleon. In this case hydrogen has 1 free electron, helium has 2, and the metals average
out to 2 free electrons in many cases. The formula is then
1
e
X
Y Z
  X  (Y  Z ) / 2  X  (1  X ) / 2  (1  X ) / 2
2 2
2-7
And the corresponding electron pressure relation is
Pe 
T
2-8
e
The overall contribution of electrons and ions is then
 1
1
T
Pgas  PI  Pe    T 

  I e 
2-9
where we define
 1
1
3Y Z
     2 X  
   I e 
4 2
1
2-10
Notes 2 - 7
This is the law for an “ideal” gas – one that behaves nicely. This is pretty reasonable for
most stars and most of a star’s interior.
6.2 Degenerate Material
But there are situations where this law doesn’t work. These would commonly be when
the pressures are incredibly high. Under these conditions the rules of quantum physics
take effect. In these situations atoms are limited on how close they can be to one another
by the Pauli Exclusion principle. This says that electrons cannot be in the same quantum
state. This only happens at very high densities. When a material gets to this extreme
condition, it is considered to be electron degenerate. There are two types of EOS for
electron degenerate material – first the non-relativistic case.
Pe,deg
h2

20me
3
 
 
2/3
  


 mH e 
5/3

 K1 '  
 e 
5/3
2-11
With K1’ = 107 (mks units).
At higher pressures, the electrons become relativistic, so another formula is used for that
(still electron degenerate, but relativistic electron degenerate)
1/ 3
Pe,rel deg
hc  3 
  
8  
  


 mH e 
4/3

 K 2 '  
 e 
4/3
2-12
With K2’ = 1.24 x 1010 (mks units)
NOTE: Electron degenerate pressure relations depends only on DENSITY, not
temperature! This is a very important aspect that we’ll get to later.
6.3 Radiation Pressure
While matter can exert pressure, it is also possible to have radiation exerting pressure.
This is especially important for very low density areas, and plays a role in very high
luminosity stars. Photons will exert a pressure on material (they have energy so they can
do that). The amount of pressure depends only on temperature (one of those black body
rules)
a
Prad  T 4
3
2-13
To be accurate, if you want to determine the total pressure of a star you should calculate
the influence of both gas pressure (due to ions, electrons) and radiation pressure P = Pgas + Prad
And it might be the case that Pgas can be ideal, or electron degenerate.
I’ll make a slight detour here to note a special case of gas pressure, that of an adiabatic
system – one that has no exchange of heat with the environment. While this may not
seem realistic, it does exist under certain circumstances, and the relationships are often
Notes 2 - 8
used when special situations arise. This is something that was seen with the degenerate
gases.
2-14
P  Ka  
Where the value of  = 4/3 is used for a relativistic degenerate gas and  = 5/3 works for
an ideal gas, or a non-relativistic degenerate gas. The value for the constant will depend
upon the situation that you are in as well. In general  will vary greatly depending upon
the environment, the amount of ionization, the effects of ions, electrons, atoms, etc.
How ionized a gas is depends on the type of gas, the temperature and the density.
Different gases have different potentials for ionization and some ionize more easily than
others or it is more difficult.
7. Radiative Transfer
7.1 Opacity
How does energy travel through a star? Easiest way to deal with it is through the
radiative transfer laws. These depend upon how much energy goes into a slab of
material, which is the energy flux rate (energy/area/time) H, as well as the density of the
slab, its thickness and the material’s ability to absorb or scatter/deflect energy which is
what we define as the opacity of the star (). So what does the opacity of a material
depends upon? This is based upon how radiation and matter interact.
There are several things that contribute to the opacity, and some of these can cause
energy to be absorbed by matter, and some cause energy to be emitted by matter. There
are often reverse processes at work here where one absorbs the energy and the other gives
it off. Depending upon the conditions (temperature and density), the amount of energy
absorption or scattering and the amount of energy emission may or may not occur.
Electron scattering – basically how an electron can alter the direction of a photon, but it
doesn’t reduce the energy. Not really an absorption process, but does alter the flow of
energy. For non-relativistic electrons, the process is known as Thomson scattering,
which is what we usually run into in the interiors of stars.
Free-Free Absorption and Bremsstrahlung Radiation – the first is where an electron will
absorb a photon which gives the electron more energy. This happens when the electron is
free (not part of an atom). Eventually the electron will give off some energy, in which
case it gives off the energy as Bremsstrahlung Radiation.
Bound-Free Absorption and Radiative Recombination – this is the one that you are most
likely to hear about in introductory astronomy courses. A photon with a specific energy
is absorbed by an electron, and the electron escapes from the atom, leaving the atom
ionized. For the electron to go back to the atom, it must give off energy via radiative
recombination.
Bound-Bound Absorption (Excitation) and De-Excitation – the electron in an atom again
absorbs the photon, but in this case it doesn’t leave the atom. Instead it just goes to a
Notes 2 - 9
higher energy level. It will eventually go back to a lower energy level, once it emits the
energy it needs to make that transition.
For regions with high levels of ionization (often very high T) you commonly see electron
scattering and free-free absorption at work. For these situations, the calculations to
determine the value for the opacity are pretty straightforward. Bound-Free and BoundBound (and their counterparts) tend to be seen in lower temperature regions, such as the
atmospheres of stars. Accurately calculating the opacity in these situations are very,
very difficult since the excitation or ionization of atoms can happen in so many ways and
result in so many different configurations of the atoms. Since elements such as iron,
carbon, silicon, oxygen, and others have multiple ways in which they interact with
matter, accounting for all of these different atom configurations requires supercomputers.
There is quite a bit of variation of the value of opacity with temperature and density for
the lower temperature areas due to all of the various excitation/de-excitation scenarios
that would exist with all of these elements.
We’ll look at the simple situation where the temperatures are very high, and there is full
ionization. When electron scattering and free-free absorption are dominant, the opacity is
nearly constant.
2-15a
  1
Usually this opacity is based upon electron scattering and so it mainly depends upon the
electron mean mass, e, in which case it would be
  1 ≈ (1/2) 0.4 (1+X)
At more moderate temperatures, and lower densities, the Bound-Free effects start to also
come into play. In these situations, the opacity can be approximated by
   2 T 3.5
2-15b
In this case the constant depends upon the chemical composition of the material, so
1
2
  (7.5 1022 )(1  X )
Z2
T 3.5
A
Where A is the average atomic mass of the metals, and this would have to be a sum of all
metals (Z) which makes this a very rough approximation..
At the lowest temperature seen usually in the outer layers of a star, bound-bound
transitions are dominant and the relation that approximates them is
   3  0.5T 4
2-15c
Calculating opacities is very difficult to do. The graph of opacities shown below is from
calculations done by different theoretical physics groups with different results. While
these differences do not seem significant, they do have an influence on our calculations
of the characteristics of stars, and these errors can add up. In fact it wasn’t until the early
1990’s that revised opacity calculations were able to explain a long standing
observational problem with how we calculated the masses of certain stars. Once the new
opacity values were calculated, the problem no longer existed! Typically values for
Notes 2 - 10
opacity are calculated for different compositions (X, Y, Z) for various values of
temperature and density. Pretty much all of this data can be obtained freely on-line for
theoreticians to use in computer models.
You should note that in the
graph shown here the
opacity values are
logarithmic, so a small
variation on this graph
indicates a relatively large
variation in the values. For
example, note the range of
values found at Log T=6.
Typically, the variation of opacity in a star can be approximated by a function of the form
2-16
   i  aT b
Where i, a, b are constants for the particular situation or location in the star.
7.2 Radiative Transfer Relation
Let’s assume that you have a way of getting values for opacity at various parts of the star.
Now how much energy does a slab of material actually absorb? This is the same as the
amount of how much the energy flow (flux) changes, which is basically how much
momentum the slab absorbs due to the photon’s energy transferring into the material as
the photons are absorbed. You have to remember that when material absorbs light it is
absorbing energy – that’s the concept behind radiation pressure.
This can be given by the following – the flux rate (H) combined with the opacity (rate of
absorption), the density, and the speed, influences the change in the radiation pressure
over time.
1

d  aT 4 
dP
H
3
   4 aT 3 dT
  rad   
c
dr
dr
3
dr
Let’s assume the radiation is black body, in which case the value for the radiation
pressure is pretty simple. Multiply both sides by 4  r2 and you get the total flux in all
directions since F=4 r2 H.
Re-arrange the constants and such and you’ll get the Radiative Transfer relation -
Notes 2 - 11
F
4
dT
  4r 2 aT 3
c
3
dr
dT
3F

dr
4acT 3 4r 2
2-17
This relation gives us a way to measure the flow of energy through a substance. Even
thought this may have originally been an illustration of how energy flows through a layer
of material, but it does illustrate how the temperature varies with radius through the star –
and how it depends upon the opacity, density and distance from the center.
8. Fusion
8.1 Coulomb Barrier
Where does the energy come from in a star? Via the process of nuclear fusion. This is a
very high yield process of energy production, where one element is fused into another
and the end result is that energy is given off, along with some other items like neutrinos,
positrons, and other particles. Usually in a star there are multiple fusion reactions
occurring at a given time, not just one single process. Each of these reactions have
different energy outputs, and different probabilities of occurring. The likelihood of a
reaction occurring depends upon the binding energy of the atom (basically the energy that
is needed to hold it together).
The graph shows the
binding energy for
all elements from
hydrogen to
uranium. The
binding energy in
atoms increases as
you go to larger
atoms until you get
to iron, and after
iron the binding
energy decreases.
The effect of this is
that as you fuse
elements together to
create more massive
elements, you gain energy, but if you try to fuse elements heavier than iron, you lose
energy. That’s why fusion in stars is only productive to stars up to the point of iron
fusion. That’s also why you can only gain energy from very massive elements by fission,
breaking them apart. Also notice the differences between some of the energies – the
bigger the difference, the more energy is given off in the reaction. The difference
between hydrogen (H1) and helium (He4) is the largest, and this is the largest energy
Notes 2 - 12
fusion process. And as you get closer to iron, the difference in levels is less, so there is
less energy given off per reaction – or put another way, you get less bang-for-the-buck.
The only problem with nuclear fusion is that atoms don’t want to fuse – they must
overcome their own repulsive forces since the cores of atoms are made up of positively
charged particles. This is similar to trying to bring two magnetic poles together that are
the same polarity. However in the case of atoms, this is a very large repulsion, so large
that it would appear impossible for atoms to overcome this barrier which is known as the
Coulomb barrier.
One way for protons to overcome this barrier is to have very high velocities, but even the
highest velocities are not enough to get past the barrier. This is illustrated in the diagram
below where the proton can only get so far “up the hill” (which is a representation of the
Coulomb barrier). So how is it possible for there to even be fusion if there is no way to
get the particle past the barrier?
This is another fun part of quantum
mechanics – the effect known as
“tunneling”. Under normal
circumstances a particle can never
have enough energy to overcome the
Coulomb barrier, sort of like having a
ball roll back down the hill since it
isn’t going fast enough to get over the
hill (it doesn’t have enough energy).
But in the case of quantum tunneling,
the rules are different. Since particles
are more wave-like at the quantum
level, there is a finite probability that
the object can be on the other side of
the barrier. This apparent disregard
for normal barriers seems to make no
sense, but that’s part of quantum
mechanics – what may be “normally”
impossible, can be possible.
Tunneling occurs under very specific conditions, typically when the particle has a certain
velocity and that depends upon the temperature and density of the material (remember
hotter stuff moves faster).
8.2 Nuclear Reactions – Main Sequence
The nuclear reaction rates depend upon the temperature and density of the material, and
the type of material. This is because more massive particles will have a greater electrical
charge with more protons, and like-charged particles don’t like to get together. In those
cases you would need much higher temperatures to overcome their larger barriers.
Notes 2 - 13
The energy (q) that is given off by a fusion reaction is pretty much a function of
temperature and density, and can be approximated by
q  q0 T n
2-18
Where q0, and n depend upon the particular reaction.
The most common fusion reactions are those that convert hydrogen into helium. There
are actually 3 different versions of this reaction.
The first is the PP I (proton-proton I), which occurs usually when temperatures are below
13 million K. Each step is outlined below along with the time scale for each reaction to
occur. Since there are so many atoms in the core of a star, these reactions happen at a
regular pace, not as infrequently as you would think based upon the time scales.
1.
1 billion years
p  p  D2  e  
2.
3.
D 2  p3He  
3
He 3He4He  p  p
6 seconds
1 million years
The PP II is the next reaction. For temperatures between 13 and 23 million K, both PP I
and PP II will occur, while for temperatures between 23 and 30 million, PP II will
dominate. The first two steps are the same as steps 1 and 2 in the PP I reaction and then
you have
3
He 4He7 Be  
3’.
7
Be  e  7 Li  
4’.
7
5.’
Li  p4He 4He
For temperatures above 30 million K, PP III is the dominant reaction. It uses steps
1, 2 and 3’, and then
7
Be  p8B  
4’’
8
B8Be  e   
5’’
8
Be4 He 4He
6’’
Basically there is energy produced during the transition from H to He (or 4 protons are
converted into one helium nuclei). But there is also energy in the neutrinos, of which
there are more produced in the PP III cycle, less in the PP II cycle, and the least produced
in the PP I cycle. Also the positrons will produce energy as well since they will give off
energy when they encounter an electron and annihilate.
In the sun, most of the energy is generated by PPI (86%), while the remainder is
produced by PP II (14%) and even less by PP III (0.11%).
Total energy given off by the PP chain is 26.73 MeV, which includes the energy from the
destruction of the positrons, the emission of the neutrinos and the creation of photons.
Notes 2 - 14
For higher mass Main Sequence stars the core temperature is greater and the hydrogen
fusion reaction process is the CNO cycle. This is dominant for temperature greater than
20 million K.
C  p13N  
12
N 13C  e  
13
C  p14N  
13
14
N  p15O  
O15N  e  
15
N  p12C 4He
15
There is another reaction that can operate along with this one and might be called the
FNO cycle (but isn’t)
14
N  p15O  
O15N  e  
15
N  p16O  
15
O  p17F  
16
17
F 17O  e  
O  p14N  4He
17
Generally speaking the amount of energy produced by the PP or CNO cycles varies with
temperature and conditions. The general relations for the amount of energy generation
for the PP cycle and the CNO cycle respectively are
q pp  T 4
qcno  T
2-19a
2-19b
16
8.3 Nuclear Reactions – Post Main Sequence
At later stages of stellar evolution you have other fusion reactions in operation..
The easiest and most common form of fusion after the Main Sequence phase is helium
fusion. Helium fusion is also called the triple alpha reaction. The first step is
4
He 4He8Be
But 8Be is very unstable and it decays quickly (in about 2.6 x 10-16 of a second) – too
quick for it to react with another helium nuclei for the next step! Fortunately the high
density of the core gives a high likelihood of a collision occurring so we can get to the
next step in the process before the 8Be decays.
8
Be 4He12C
And there you have it! The energy given off here is only 7.28 MeV, so it isn’t as
powerful as the PP reaction.
Notes 2 - 15
The corresponding relation for the triple-alpha is
qTriple   2T 40
2-19c
To have helium fusion going you need a temperature of 100 million K and densities of
around 100,000 gm/cc (however, this may not be correct as you’ll see).
After helium fusion, it is pretty much a game of temperatures and densities as to which
fusion process occurs. Generally speaking fusion processes can be linked to stellar
masses since the gravitation forces exerted on the core drive fusion, and more mass
allows for higher temperatures and densities, and therefore more fusion processes.
For T above 500 million K, and densities above 200,000 gm/cc, carbon fusion will occur.
This is usually seen in stars of 4 times the mass of the sun or greater.
For temperatures above 1.2 billion K, and densities above 4 million gm/cc, neon fusion
can occur. The minimum mass star that this happens in is around 8 solar masses or
greater.
And at temperatures above 1.5 billion K, and densities above 10 million gm/cc, oxygen
fusion occurs. This is also seen in masses of about 8 solar masses or more.
At temperatures around 2.7-3.5 billion K, silicon fusion takes place. This happens in
stars above 8 solar masses, and the exact temperature at which it occurs will depend upon
the mass of the star.
At high temperatures you can also have elements that photodisintegrate. Basically they
form into other elements and then break down nearly simultaneously. So oxygen will
form into neon, which will form into oxygen – back and forth and back and forth…..
Carbon can fuse into various things like 23Mg, 24Mg, 23Na, 20Ne, 16O
12
C + 12C → 24Mg + γ
→ 23Mg + n
→ 23Na + 1H
→ 20Ne + 4He
→ 16O + 24He
Neon can produce (through photodisintegration) O, and 24Mg.
20
Ne + γ → 16O + 4He
20
Ne + 4He → 24Mg + γ
Oxygen fuses into 32S, 31S, 31P, 28Si, 24Mg
16
O + 16O → 32S + γ
→ 31S + n
→ 31P + 1H
→ 28Si + 4He
Notes 2 - 16
→ 24Mg + 24He
Silicon fusion goes by adding Helium at each step (alpha process).
Silicon -> sulfur -> argon -> calcium -> titanium-> chromium->iron->nickel
These reactions are rather poorly understood, and often physicists will revise the rates at
which these reactions occur and how much energy is produced. In 2009 there was a
revision of the triple-alpha nuclear reaction process, which indicated that it is possible for
it to occur at lower temperatures – perhaps as low as 30 million K. If this is correct, then
all calculations involving stars, particularly during post-Main Sequence evolution, will
need to be revised. At the moment these results are being studied and it will take some
time to determine if we need to re-write the textbooks (but I’ll wait a while before I
change my notes, so let’s just stick with the temperature of 100 million K for the triple
alpha cycle).
The basic effect of all of this fusion, apart from energy generation, is the change in the
composition of the core of the star. And once you change X, Y and Z, you change ,
which changes pressure, density,….well, since everything is all connected, everything
changes!
9. Time Scales
Looking at the previous section, it is obvious that stars can change their compositions due
to fusion, but how quickly does that happen? What other possible changes can a star
have during its life? We’ll look at how quickly (or not so quickly) a star can change.
Some of this involves gross generalizations, but they do provide some information about
stars that is useful.
First there is the Dynamical Time Scale, involves motions within a star. We’ll start by
taking a look at the surface of the star, R. If you start from this location and freefall to the
center how fast would you be going? This is also the same as the velocity needed to
escape from the star when you are at the surface. And the escape velocity is a pretty
simple relationship
vesc 
2GM
R
How long would it take for the star to fall in? That’s the dynamical time scale.
R
R3
 dyn 

vesc
2GM
Now let’s use the fact that average density = M/Volume, or

M
M 4
or 3  
4 3
R 3
R
3
Now put these densities into the dynamical time scale formula to get
Notes 2 - 17
 dyn 
3
R3

8G
2GM
2-20
Either way, this is a very short time span.
So what does this actually tell us? This is the time associated with stars not in hydrostatic
equilibrium, particularly pulsating stars, and indicates the time for their variation.
Therefore dyn gives a rough approximation to a star’s pulsation period.
Another time scale is the Thermal Time Scale – how long does it take a star to give off all
of its internal energy by contracting (gravitational energy). Since gravitation energy is
defined as GM2/R and energy emission is basically luminosity (L), you’d get
 th 
GM 2
RL
2-21
If you use typical values for this, you’d see that it isn’t a very long time at all. This was
one of the things that led people to find other causes of energy generation in stars, since
gravity would not produce enough energy in a star for very long.
There is also the Nuclear Time Scale – time for all energy generated by nuclear fusion to
be given off (emitted). And of course that’s where the fun formula E=mc2 comes into
play. In reality, the entire mass of a star isn’t converted to energy, but only a small
fraction, about 0.001 (=) of the mass becomes energy. The total energy produced by
fusion = Mc2. Combine this with the rate at which energy is given off by a star = L
 nuc 
Mc2
2-22
L
This is a very long time.
The order of these time scales is  dyn  th  nuc , so that stars’ evolution is much
longer than any other changes that occur in them. That’s why we usually ignore the
effects of pulsation or variability on a star since those are pretty short compared to the
changes that happen as stars evolve. Of course if you want to study the effects due to
pulsation on a star, you don’t have to worry about evolutionary changes since those are so
slow compared to the star’s pulsation rate.
Notes 2 - 18