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PHYS377: A six week marathon through the firmament. by Orsola De Marco [email protected] Office: E7A 316 Phone: 9850 4241. Week 3, May 10-12, 2009. Overview of the course. Where and what are the stars. How we perceive them, how we measure them. - PowerPoint PPT Presentation
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PHYS377:A six week marathon
through the firmament
by
Orsola De [email protected]
Office: E7A 316Phone: 9850 4241
Week 3, May 10-12, 2009
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.
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USun ≈−GM 2
λRwhere
λ ≈ 0.5
tKH =USun
LSun≈ 50Myr
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 m = 0.0458 x 10-27 gr
E = (m) 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 E = 4.1 x 10-12 J.• Total energy E: NH/4 E = 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.
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 probability of nuclearfusion.
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E ≈e2
4πε0
⎛
⎝ ⎜
⎞
⎠ ⎟
22mp
h2≈ 2keV ≈
3
2kTc
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.
Equilibrium abundances
• Start with the CNO cycle:12C +1H -> 13N + 106 years)13N -> 13C + e+ + (14 min)13C +1H -> 14N + (3x105 years)14N +1H ->15O + (3x 108 years)15O -> 15N + e+ + (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.