Transcript Document

Lecture 10
Energy production
Summary
We have now established three important equations:
Hydrostatic
equilibrium:
Mass conservation:
Equation of state:
dP
GM r 

dr
r2
dM r
 4r 2 
dr
kT
P
mH
•
•
Can be modified to include
radiation pressure
Breaks down at high
temperatures and at high
densities.
There are 4 variables (P,,Mr and T) and 3 equations. To solve the
stellar structure we will need to know something about the
energy production and transportation
Stellar luminosity
Where does the energy come from?
Possibilities:
• Gravitational potential energy (energy is
released as star contracts)
• Chemical energy (energy released when
atoms combine)
• Nuclear energy (energy released when
atoms form)
Gravitational potential
The gravitational potential energy for a
system of two particles:
Mm
U  G
r
As M and m are brought closer together, the potential
energy becomes more negative.
Assume the Sun was originally much
larger than it is today, and contracted.
How much gravitational potential
energy is released?.
Chemical energy
Chemical reactions are based on the interactions
of orbital electrons in atoms. Typical energy
differences between atomic orbitals are ~10
eV.
e.g. assume the Sun is pure hydrogen. The total number of atoms is
therefore
M Sun 1.99 1030
57
n


1
.
19

10
mH
1.67 1027
If each atom releases 10 eV of energy due to chemical reactions,
this means the total amount of chemical energy available is
 1058 eV  1039 J
This is ~100 times less than the gravitational potential energy
available, and would be radiated in only 100,000 years at the
present solar luminosity
Atomic nuclei
Usually we express the mass of nuclei in atomic mass
units u, defined to be 1/12 the mass of 12C (the 12 is
the mass number A. 12C has 6 protons, 6 neutrons
and 6 electrons)
1u=1.66054x10-27kg
m p  1.00728u
mn  1.00866u
me  0.0005486u
• The mass of 1 proton + 1 electron is 1.0078285u
• Note 6 p + 6 n + 6 e- = 12.099. Does this make
sense?
The Equation
Einstein showed that mass and energy are
equivalent, and related by:
E  mc
Thus atomic masses can be
expressed as energies
2
1u  931.49432 MeV/c
2
When equating rest masses to energies, it is customary to omit the factor c2 and
leave it implicitly assumed.
Binding energy
The mass of an atom (protons+neutrons+electrons) is not equal to
the mass of the individual particles.
e.g. the hydrogen atom is less massive than the sum of the proton
and electron mass by 1.46x10-8 u
This mass difference corresponds to an energy:
E  13.6eV
This energy is the binding energy. It is the energy released when
an electron and proton are brought together.
Binding energy
There is also a binding energy associated with the
nucleons themselves.
• Recall Carbon-12: 6 p + 6 n + 6 e- = 12.099.
• The mass difference is 0.099u, equivalent to 92.22 MeV!
• This is the binding energy of the C-12 atom
Fusion
Making a larger nucleus out of smaller
ones is a process known as fusion.
For example:
H  H  H  H  He  low mass remnants
The mass of 4 H atoms is 4.031280u. The mass of He is 4.002603u.
• The mass difference is 0.028677u, equivalent to 26.71 MeV.
 As we will see, the energy of the typical low-mass remnants amounts
to only ~1 MeV.
 ~0.7% of the H mass is converted into energy
Nuclear energy: fusion
In contrast with chemical reactions, nuclear
reactions (which change one type of nucleus
into another) typically release energies in the
MeV range, 1 million times larger.
E.g. Assume the Sun was originally 100% hydrogen, and
converted the central 10% of H into helium.
This would release an energy:
E  0.1 0.007  M Sunc 2
 1.3 10 44 J
Assuming the Sun’s luminosity has been constant at 3.8x1026 W,
it would take ~10 billion years to radiate all this energy.
 Nuclear energy can sustain the solar luminosity over its likely
lifetime
Nuclear energy: fusion
Nuclear energy is sufficient to sustain the Sun’s
luminosity. But fusion is difficult to achieve! Can it
actually occur naturally in the Sun?
Break
Coulomb repulsion
The repulsive force between like-charged
particles results in a potential barrier
that gets stronger as the particles get
closer:
•
•
The strong nuclear force
becomes dominant on very small
scales, 10-15 m.
What temperature is required to
overcome the Coulomb barrier?
Z1Z 2 e 2
UC 
40
r
1
Statistical mechanics
• If the gas is in thermal equilibrium with
temperature T, the atoms have a range of
velocities described by the Maxwell-
Boltzmann distribution function.
• The number density of gas particles with
speed between v and v+dv is:
 m 
nv dv  n

 2kT 
3/ 2
e
The most probable velocity:
The average kinetic energy:
mv 2

2 kT
4v 2 dv
v2  2
kT
m
1
3
m v 2  kT
2
2
Quantum tunneling
The answer lies in quantum physics. The uncertainty principle
states that momentum and position are not precisely defined:

xp x 
2
• The uncertainty in the position
means that if two protons can
get close enough to each other,
there is some probability that
they will be found within the
Coulomb barrier.
 This is known as tunneling.
 The effectiveness of this
process depends on the
momentum of the particle
Quantum tunneling
Approximately: tunneling is possible if the
protons come within 1 deBroglie wavelength
of each other:
For two protons,
at T~107 K
h h
h
 

p v
3kT
1.13 103

 3.57 107 m
T
• So the protons don’t need to get anywhere near 10-15m before they can
begin to tunnel past the barrier
•
Without this quantum effect, fusion would not be possible in the Sun and
such high luminosities could never be achieved.
Quantum tunneling
What temperature is required for two protons to come
within one de Broglie wavelength of each other?
deBroglie 
h
3kT
Review: Nuclear fusion
• Fusion is possible if the particle energy (3/2 kT) is
equal to or greater than the Coulomb potential energy:
Z1Z 2 e 2
UC 
40
r
1
where the distance r can be taken to be the de Broglie
wavelength:
h

3kT
Fusion is therefore possible at temperatures
 
T  1.9  10 
 mH
7
 2 2
 Z1 Z 2 K

Nuclear energy
• The energy production rate must be related to the rate of
collisions.
• Since the full rate equation can be very complex, it is common to
approximate it as a power-law over a particular temperature
range
rix  r0 X i X x  1 T 
•
•
•
Here, Xi and Xx are the mass fractions of the incident and target
particles, respectively
For two-body interactions (i.e. p+p collisions), ~1
The parameter  can have a wide range of values
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
Nuclear reactions
So – what are the specific reactions we’re talking about??
 The probability that four H atoms will collide at once to form a single He
atom is exceedingly small. Even this simple fusion reaction must occur via a
number of steps.