Transcript PowerPoint Presentation - Research in observational

PHYS377:
A six week marathon
through the firmament
Week 3, May 10-12, 2009
by
Orsola De Marco
[email protected]
Office: E7A 316
Phone: 9850 4241
Overview of the course
1.
Where and what are the stars. How we perceive them, how we
measure them.
2.
(Almost) 8 things about stars: stellar structure equations.
3.
The stellar furnace and stellar change.
4.
Stars that lose themselves and stars that lose it:
stellar mass loss and explosions.
5.
Stellar death: stellar remnants.
6.
When it takes two to tango: binaries and binary interactions.
What powers a star?
• In the 1850s Helmholz thought
it might be conversion of
potential energy (i.e. the energy
of gravity) into photons.
GM 2
U Sun 
R
where
  0.5
t KH
U Sun

 50Myr
LSun
This is too short a
time! Why?
Eddington’s interpretation
• Sun powered by gravity?
Would last too short a time.
• 1896-1903: Discovery of
radioactivity: could the sun
be powered by radioactivity?
No: stars do not have many
radioactive elements, and
their energy is dependent on
temperature.
• 1905: Special Relativity
E=mc2.
• 1920: Aston measures mass
deficit. Eddington realizes!
Arthur Eddington 1882-1944 UK
Aston’s mass deficit
• Mass of 4 H nucleii = 6.6905 x 10-27 gr
• Mass of 1 He nucleus = 6.6447 x 10-27 gr
• Mass difference Dm = 0.0458 x 10-27 gr
DE = (Dm) c2 = 4.1 x 10-12 J
Is this a lot of energy?
NOTE: 4.1 x 10-12 J = 4.1 x 10-12 J / 1.60 x 10-19 J/eV = 26 MeV
There are a lot of H atoms
in the Sun….
•
•
•
•
•
•
NH = MSun/mp = 1.2 x 1057
Number of reactions are NH/4
Each yields DE = 4.1 x 10-12 J.
Total energy E: NH/4 DE = 1.2 x 1045 J
Lifetime of the sun = E/L = 100 Gyr
Why is this age about 10 times the total
lifetime of the Sun?
• We can now rewrite an equation for the
main sequence lifetimes of stars.
The main sequence lifetime
 M 3
  10Gyr 
M 

Also known as the nuclear timescale
Stellar birth: the Virial Theorem
• The Virial Theorem: the complex n-body
problem has a surprisingly simply
property: for a bound system, the total,
time averaged, kinetic energy (K) is
related to the total potential energy (U):
2K + U = 0
Stellar birth: consequences of
the Virial theorem
• For collapsing stars, equilibrium is
reached when ½ of the gravitational
energy is stored as thermal energy and
the other ½ is radiated away. The
collapse timescale of a star is therefore
the time it takes to radiate the energy
away, which is easily computed by U/L
(L is the luminosity of a star). This is the
Kelvin-Helmoltz timescale!
Stellar birth
• A molecular cloud is unstable against
collapse if the time sound takes to reach
the centre from the surface is longer than
the free-fall time. This condition defines
the Jeans length, the maximum radius
that a cloud can have before it collapses
to form a star.
Tunnelling!
• In order for protons to come close enough
to react, their kinetic energies would have
too be 1000 times higher than they are.
• Fortunately, there is a finite probability that
a proton can gain that energy temporarily
by quantum uncertainty.
This means that there
is a low but finite
2
2 

2m p
probability of nuclear
e
3
E  

2keV

kTc
 2
2
fusion.
4 0  h
Three fusion chains
for H and He “burning”
• H to He: The pp chain
• H to He at higher T: The CNO cycle
• He to C and O: The triple alpha reaction.
• Reaction rate is a function of probability of
a given reaction. Probability is a function of
temperature.
• Equilibrium abundances depend on the
reaction speed of the various reactions.
pp or CNO?
csep10.phys.utk.edu/astr162/lect/energy/cno-pp.html
Equilibrium abundances
• Start with the CNO cycle:
12C +1H -> 13N + g(106 years)
13N -> 13C + e+ + m(14 min)
13C +1H -> 14N + g (3x105 years)
14N +1H ->15O + g (3x 108 years)
15O -> 15N + e+ + m(82 sec)
15N + 1H -> 12C + 4He (104 years)
Equilibrium abundances
• Although the CNO cycle does not create new
C,N and O, their relative abundances are altered
by the process.
• But what are the equilibrium abundances? For,
e.g., 14N, one has to wait 3x108 years for its
equilibrium value to be established:
d14N/dt = 0 = 13C x reac. rate (13C + H) – 14N x reac. rate (14N + H)
14N
/ 13C = 3 x 108 / 3 x 105 = 1000; 12C/13C = 3.3; 14N/12C = 300
• Equilibrium abundances are expected after the first
dredge up.
On to He fusion
• Bottleneck: there are no stable elements of
mass number 5 or 8, so He + p, or He + He,
likely to happen in an H and He-rich
environment, end up in products that vanish
immediately.
• As the Tc rises 8Be, although unstable, can
be formed at a rate high enough to result in
a non zero abundance of this element which
can the react with another 4He.
The end of the line: Fe
• Massive stars keep fusing core elements till
the core is made of iron, which does not
liberate energy by fusion because it has the
lowest
binding
energy per
nucleon.
What
happens
then?