Transcript Slides from the second lecture

Lecture derived largely from Michael Balogh’s course notes
The Research School of Astronomy and Astrophysics
Mount Stromlo Observatory
The tight relationship between
Mass and luminosity his a key to
the underlying physics.
Stellar interiors
• It is nearly impossible to learn anything directly about the
interior of stars
– The optical depth increases very quickly with distance from the
surface, so photons of all wavelengths originate near the surface
– An important exception are neutrinos, which interact very weakly
with matter and so can escape from the Sun unimpeded
– Also, solar oscillations give us a way to probe the interior by
measuring the sound speed at various depths
• Therefore most of what we know is obtained by
constructing a theoretical model containing all known
physical laws, and consistent with observable surface
features.
– This generally requires detailed computer programs that only
became feasible to run in the mid-60s (though most of the
physics was already understood).
Equations of Stellar Structure
• Hydrostatic
equilibrium:
dP
GM r 

dr
r2
• Mass
conservation:
dM r
 4r 2 
dr
• Equation of
state:
kT
P
mH
• 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
• The Solar luminosity is 3.8x1026 J/s
– From age-dating moon rocks, we know the solar system has been
around for ~4.6 billion years. If the luminosity of the Sun has been
approximately constant, the total energy released would be 4.8x1043J.
• The luminosity of a star can be related to the total amount of energy
produced per unit mass, e:
dL  edm
• For a spherically symmetric star, we can calculate the luminosity of a
thin radial shell:
dLr
2
 4r e
dr
• Where Lr is the luminosity due to all energy generated within radius r.
– This is the fourth fundamental equation of stellar structure
– So we need to know something about the energy generation mechanism
Gravitational potential
• So: how much energy can we get out of gravity?
• Assume the Sun was originally much larger than it is today, and
contracted. This releases gravitational potential energy.
• The initial potential energy is ~0 (since R is large). The change in
energy is therefore:
E  E f  Ei 
GM 2

R
 3.6 10 41 J
• The Solar luminosity is 3.8x1026 J/s, so this energy would be radiated
in only 1015 s, or ~107 years. This is the Kelvin-Helmholtz timescale
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
The Equation
•Einstein showed that mass and
energy are equivalent, and related
2
by:
E  mc
• Thus atomic masses1u
can be expressed as
energies
 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
• There is also a binding energy associated with the
nucleons themselves. Making a larger nucleus out
of smaller ones is a process known as fusion.
• For example:
H  H  H  H  He  ....
• 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!
– The … is other stuff like neutrinos, gamma rays, positrons…
– ~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 1044 J
• Assuming the Sun’s luminosity has been constant at 3.8x1026 J/s,
it would take ~10 billion years to radiate all this energy.
– Nuclear energy can sustain the solar luminosity over its likely
lifetime
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
• each of the above reactions occurs at its own rate. The
first step is the slowest because it operates by 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
2
He  24He  47 Be  
7
4
Be  e  37 Li   e
7
3
Li 11H 2 24 He
• 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
7
1
8
beryllium-7 nucleus
Be

H

4
1
5B  
B 48Be  e   e
8
5
Be 2 24He
8
4
• This reaction only occurs 0.3% of the time in the Sun.
Nuclear energy
• Since the rate equations can
be very complex, it is common
to approximate it as a power
law over a particular
rix  r0 X i X x T
temperature range
• Here, Xi and Xx are the mass fractions of the incident and target
particles, respectively
• The parameter  can have a wide range of values depending on the
set of reactions
• If each reaction releases an energy L, the amount of energy released
per unit mass is just
L
e ix   rix  e 0 X i X x   T 

• The sum over all reactions gives the nuclear
reaction contribution to e in our fundamental
equation
dLr
 4r 2 e
dr
The PP chain
• The nuclear energy generation rate for the PP chain,
including all three branches:
2 / 3 33.80T61 / 3
6
e pp  2.38 10 5 X T
4
5 
2
e

J / kg / s
T6  T / 106 K
105 kg / m3
• Near T~1.5x107 K (i.e. the central temperature of the
Sun):
e pp  1.07 10 5 X T J/kg/s
7
2
4
6
Example
e pp  1.07 10 5 X T J/kg/s
7
2
4
6
• At the centre of the Sun, 5~1.5
and T6~15 so
e pp  8.110 X J/kg/s
3
2
• If we imagine a core containing 10% of the
Sun’s mass, composed entirely of 75%
hydrogen (X=0.75), the total energy
produced by the PP reaction is:
L pp  8.1103  0.1 0.752  MSun J/s
 9 1026 J / s
• Recall the solar luminosity is 3.8x1026 J/s
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:
12
1
13
C

H

6
1
7N 
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
e CNO  8.24 1027 5 XX CNOT619.9 J/kg/s
• at T~1.5x107 K
The CNO cycle
• Compared with the PP cycle, the CNO cycle is weaker
at ~106 K, but much more strongly temperature
dependent.
e pp  1.07 107 5 X 2T64 J/kg/s
e CNO  8.24 1027 5 XX CNOT619.9 J/kg/s
• Assuming XCNO~0.03X, the two cycles are approximately equally
energetic when
• which is just a little hotter than the central solar temperature.
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 3-body interaction
44.027T81
3
8
3
2 3 41.0
5
8
e 3  5.09 10  Y T f e
J / kg / s
e 3  3.85 10  Y T J/kg/s
11
2
5
8
3
• Note the very strong temperature dependence. A 10%
increase in T increases the energy generation by a
Nucleosynthesis
• At the temperatures conducive to helium burning, other
reactions can take place by the capturing of -particles
12
4
16
(He atoms).
C

He

6
2
8O  
20
O 24He 10
Ne  
16
8
• Further atom building in solar-type stars
is prohibited by the increasingly strong
Coulomb barrier (Not enough T and
density to get an alpha into a nucleus)
– However, in more massive stars, the
central temperature may be
significantly higher, allowing further
reactions to occur.
– Iron is the most stable nucleus. Thus,
iron will be the ultimate product of
nucleosynthesis, if temperatures are
high enough to overcome the Coulomb
barrier and fuse together.
Energy transport
• To solve for T, P, M and L as a function of radius r, we
need one more equation.
• We need a differential equation relating temperature to
radius, which is related to the transport of energy from
the core to the surface.
– Radiation: the photons carry the
energy as they move through the star,
and are absorbed at a rate that
depends on the opacity. Slow
– Convection: buoyant, hot mass will
rise. Very efficient where it works.
– Conduction: collisions between
particles transfer kinetic energy of
particles. This is usually not
important because gas densities are
too low. Slow. Slow.
•
The
Solar
model
In this way we can
build up a model of
the interior structure of
the Sun
•
•
In general the
differential equations
are solved numerically
Boundary conditions:
–
–
in the simplest case,
, P and T =0 at r=R
M,L=0 at r=0
Convection in the Sun
• For the solar model we can plot dlnP/dlnT as a function of
radius. When this is <2.5, convection will occur. This is the
region where a boyant blob will continue to remain boyant
as it rises inside the star.
The Solar interior
•
The interior can be
divided into three regions:
1.
Core: In the very centre, the
temperature is high enough
to sustain nuclear reactions,
via the PP chain
The radiative zone: energy is
transported via radiation
The convective zone: the
temperature here is lower,
and the opacity is higher due
to recombination. Thus
convection takes over as the
main mechanism for energy
transport.
2.
3.
The solar model: evolution
• In the core, the mass fraction of H
has decreased from X=0.75 to
X=0.34 after 4.6Gyr
– Accordingly, the mass fraction of
He has increased from 0.24 to 0.64
• At the surface, the fraction of H has
also increased by ~0.03, due to the
diffusive settling of more massive
elements.
• The change in core H fraction has a
direct effect on the luminosity and
temperature, by changing the
nuclear reaction rates and the
amount of pressure support
Energy production
• Although nuclear reaction rates are higher where the
temperature is higher, most of the energy is not produced
at the centre of the Sun, because:
– The amount of mass in a shell at radius r is
dM  4r 2 dr
• i.e. there is more mass per unit volume at large radius (assuming
constant density)
– The mass fraction of hydrogen (X) at the centre has been
depleted due to fusion, and the rate equations depend on X2.
The Evolutionary Cycle of a Star
like the Sun
•
•
•
•
•
•
The Main Sequence
The Sub-Giant Branch
The Red Giant Branch
The Horizontal Branch
The Asymptotic Giant Branch.
Planetary Nebulae and White Dwarf stars
8.1
Evolutionary
Tracks on the
H-R Diagram
Plot how the
observations of a
star change
during its
lifetime.
8.2 Main Sequence Evolution
Stars start their lives on the Main Sequence
with approximately 75% H and 25% He.
H-He envelope
Core burning
H to He
In the center: as H burns to He:
• the fraction of He (and mean molecular
weight ) goes up
• to maintain the pressure support the
core contracts
• the temperature & Pressure increase
• the rate of energy generation increases
due to the sensitivity of energy
generation to Temperature.
At the surface:
• the increased energy
generation means that
the luminosity goes up
• the envelope is heated
and expands so the
radius of the star
increases
Duration: about 10
billion years for a 1
solar mass star
8.3 Sub-Giant Branch
When the core runs out of hydrogen it
loses its source of pressure support.
In the center:
• the core contracts and heats up
H-He
envelope • a new shell-source of burning
H-burning
hydrogen is ignited just outside the
shell
core
He core, gradually
contracting and
growing from
“ash” from the
shell
• He “ash” from the shell source
increases the mass of the contracting
core
• the energy generation rate slowly
increases, heating the envelope
At the surface:
• the luminosity
increases slightly
• radius of the star
continues to increase
• the effective
temperature decreases.
Duration: about 1
billion years for a 1
solar mass star
Rapidly
expanding
H-He
envelope
8.4 Red Giant Branch:
In the center:
• the core becomes dense enough to
be supported by electron
degeneracy pressure
Increasingly
• it continues to grow and contract
rapidly-burning as it gains He-ash from the shell
H- shell
source
• the rate of He-burning increases
as the shell is heated by sinking
into the center
He core supported by
electron degeneracy
pressure, still growing
• Not all the energy from the shell
from He ash, contracting source can be carried by radiation and heating up
the envelope becomes convective.
“Forbidden region”
At the surface:
• The luminosity increases
dramatically
• the effective temperature
decreases slightly
• the stellar radius increases
dramatically.
• The star loses some mass
via stellar winds
Duration: about 1 billion
years for a 1 solar mass star
8.5 Horizontal Branch
Once the temperature and density of the core are high
enough Helium-burning via the triple-alpha process is
ignited. This ignition process is known as the helium
flash.
In the center:
• the core first expands and
cools
• then the star burns helium
steadily in the core
• the energy generation rate
is lower than on the RGB
Steady envelope
H-burning shell
He-burning core
Helium flash
At the surface:
• the luminosity is
steady and much
less than on the
RGB
• the radius is
smaller than the
RGB
Duration: about 0.1
billion years.
(the “Heliumburning Main
Sequence”)
Rapidly
expanding
envelope
H-burning
shell
He-burning shell
Carbon core
8.6 Asymptotic Giant
Branch
When all the Helium in the
core is burnt:
In the center:
• the core contracts until
supported by electron
degeneracy pressure
• a He-burning shell is
ignited
(Like an RGB star but with
He-burning shell).
At the surface:
• The luminosity increases
• the effective temperature
stays (approximately) the
same
• the stellar radius increases
dramatically.
• The star loses mass via
pulsations
8.7 Planetary Nebulae
In low-mass stars the core never gets hot enough to ignite
Carbon-burning. However, the rate of He-burning in the
shell source is unstable, so the star pulsates and the
envelope of the star is ejected in a planetary nebula.
The core of the star remains behind as a white dwarf star
that gradually cools.
The gas in planetary nebulae is heated by UV-photons
from hot, young white dwarf stars. They are typically
expanding at speeds of 10-30km/s. They will glow, giving
off optical photons for 10,000-50,000 years until they are
too diffuse and too far from the White Dwarf star to be
heated.
Direct observations of the core:
neutrinos
• Because of their low cross section, neutrinos can escape from the
Sun’s core to the Earth without further interaction
• One type of neutrino detector on Earth uses an isotope of chlorine,
which will (rarely) interact with a neutrino to produce a radioactive
37
isotope of argon.
Cl    37Ar  e 
17
e
18
• This reaction requires the neutrino to have
an energy of 0.814 MeV or more, and can
only detect neutrinos from the “sidereactions” in the PP chain:
• PPI
3
4
7
2 He  2 HeI 4 Be  
7
4
Be  e  37 Li   e
7
3
Li  H 2 He
1
1
4
2
• PPI
Be 11H 58B  
II8
8

7
4
5
B 4 Be  e  e
8
4
Be 2 24He
• The Homestake detector contains ~400,000
L of cleaning fluid
• 2x1030 atoms of Cl isotope
• Detect one Argon atom every 2-3 days.
Direct observations of the core:
neutrinos
• More recently, the GALLEX (also SAGE) experiments
uses 30 tons of natural gallium in a 100 ton aqueous
gallium chloride solution to detect neutrinos via:
Ga  e  3271Ge  e 
71
31
• This is sensitive to lower neutrino energies (0.233 MeV)
and can detect neutrinos from the main branch of the PP
chain
1
1
H 11H 12H  e    e
2
1
H 11H  23He  
3
2
He  23He  24He  211H
The Solar neutrino problem
• Both the Homestake and GALLEX experiments
detected fewer neutrinos (by a factor 2-3) than
were expected from the PP-chain reactions. This
problem existed for about 30 years.
• The solution to the
problem was suggested
by results from the SuperKamiokande detector in
Japan
– Uses 40,000 tons of
normal water. Neutrinos
scatter electrons in the
water to velocities
exceeding the speed of
light in water, producing a
flash of light.
– Results showed that
electron neutrinos
produced in the upper
atmosphere can change
into tau- or muonneutrinos
– This means neutrinos
must have some mass and
can therefore oscillate
The Solar neutrino problem solved
• The Sudbury Neutrino
Observatory uses heavy
water, and was able to
directly detect the flux of all
types of neutrinos from the
Sun.
• The flux of all neutrino
types is in agreement with
the predicted electron
neutrino production from
the PP reactions
• The results are completely
consistent with the
standard solar model.
Solar Oscillations also probe Core
• Stars ring like bells
with different modes,
and these are related
to the structure of the
star, and its
composition.
• Successful
comparison between
predicted mode
frequencies gives
sound speed, and
density as and
measured normal
• Until recently, the
model of the sun
seemed to reproduce
the solar oscillations.
– Can ‘see’ bottom of
Convective Zone.
– Accuracy has been
sufficient to correct
errors in opacity
codes.
• Recent Revision of
solar abundance now
causes some
problems…
Useful Stars- Cepheids
•
•
•
Massive Stars (3-20 Msun) (Young
Stars)
Pulsate from 3-100 days (Pulsation
rate depends on Mass)
Located in Instability Strip
LMC Cepheids
RR-Lyrae
Pulsate 4-14 hours – in an area
Known as the instability strip.
Stars which have low metal content
– old stars
Basically all the same brightness
Use as distance measurers
These Stars
Are Dead
Subgiants
Stars Larger than the Sun