Transcript Document

Lecture 11
Energy transport
Review: Nuclear energy
• If each reaction releases an energy L, the amount of energy
released per unit mass is just
L
 ix   rix   0 X i X x   T 

• The sum over all reactions gives the nuclear reaction contribution
to  in our fifth fundamental equation:
dLr
 4r 2 
dr
Proton-proton chain (PPI)
1
1
H 11H 12H  e    e
2
1
H 11H  23He  
3
2
He  23He  24He  211H
The net reaction is:
411H 24He  2e   2e  2
But each of the above reactions occurs at its own rate. The first
step is the slowest because it requires a proton to change into a
neutron:


Energy  p  n  e  e
This occurs via the weak force. The rate of this reaction
determines the rate of Helium production
Proton-proton chain (PPII and PPIII)
Alternatively, helium-3 can react with
helium-4 directly:
3
4
7
He

He

2
2
4 Be  
7
4
Be  e  37 Li   e
7
3
Li 11H 2 24He
In the Sun, this reaction occurs 31% of the time; PPI
occurs 69% of the time.
Yet another route is via the collision
between a proton and the beryllium-7
nucleus 7 Be  1H  8B  
4
1
5
B 48Be  e    e
8
5
Be 2 24He
8
4
This reaction only occurs 0.3% of the
time in the Sun.
The PP chain
The nuclear energy generation rate for the PP chain, including all
three branches:
2 / 3 33.80T61 / 3
6
 pp  2.38 10 5 X T
4
5 

5
10 kg / m
3
2
e
W / kg
T6  T / 106 K
Near T~1.5x107 K (i.e. the central temperature of the Sun):
 pp  1.07 107 5 X 2T64 W/kg
Example
 pp  1.07 10 5 X T W/kg
7
2
4
6
If we imagine a core containing 10% of the Sun’s mass,
composed entirely of hydrogen (X=1), calculate the total
energy produced by the PP reaction.
The CNO cycle
There is a second, independent cycle in
which carbon, nitrogen and oxygen act
as catalysts. The main branch
(accounting for 99.6% of CNO
reactions) is:
C 11H 137N  
12
6
13
7
N 136C  e    e
C 11H 147N  
13
6
14
7
N 11H 158O  
O 157N  e    e
15
8
15
7
N 11H 126C  24He
 CNO  8.67 10 5 XX
25
at T~1.5x107 K
2 / 3 152.28T61 / 3
CNO 6
T
e
W / kg
 CNO  8.24 1027 5 XX CNOT619.9 W/kg
Helium collisions
• As hydrogen is converted into helium, the mean
P
molecular weight increases.
• To keep the star in approximate pressure equilibrium,
the density and temperature of the core must rise
kT

mH
Recall that the temperature at which quantum tunneling becomes
possible is:
T
1
40 2
4 Z12 Z 22 e 4
3 kh2
 
 1.9 10 
 mH
7
 2 2
Z1 Z 2 K

As H burning progresses, the temperature increases and eventually He
burning becomes possible
The triple-alpha process
The burning of helium occurs via the triple alpha
process:
4
2
He  24He  48Be
8
4
Be  24He 126C  
The intermediate product 8-beryllium is very unstable, and will decay if
not immediately struck by another Helium. Thus, this is almost a 3body interaction
11 2 3 3 44.027T81
3
5
8

 5.09 10  Y T e
 3  3.85 10  Y T
8
2
5
3
41.0
8
W / kg
W/kg
Note the very strong temperature dependence. A 10% increase in
T increases the energy generation by a factor 50.
Nucleosynthesis
At the temperatures conducive to helium burning, other reactions
can take place by the capturing of -particles (He atoms).
C  24He 168O  
12
6
20
O 24He 10
Ne  
16
8
Nucleosynthesis
The binding energy per nucleon describes the stability of a nucleus. It is easier to break
up a nucleus with a low binding energy.
Break
Summary
We have now established four important equations:
Hydrostatic
equilibrium:
Mass conservation:
dP
GM r 

2
dr
r
dM r
 4r 2 
dr
Equation of state:
kT
P
mH
Energy production
dLr
2
 4r 
dr
There are 5 variables (P,,Mr, T and Lr) and 4 equations. To solve
the stellar structure we will need to know something about the
energy transportation.
Energy transport
 Radiation: the photons carry the energy as
they move through the star, and are
absorbed at a rate that depends on the
opacity.
 Convection: buoyant, hot mass will rise
 Conduction: collisions between particles
transfer kinetic energy of particles. This
is usually not important because gas
densities are too low.
Radiation transport
When we considered the properties of radiation, we found an
equation relating the pressure gradient to the radiative flux:
dPrad


Frad
dr
c
From this we can derive an expression for the temperature
gradient, assuming a blackbody.
dT
3 Lr

dr
64 r 2T 3
In regions of high opacity, or high
radiative flux, the temperature
gradient must be steep to
transport the energy outward.