PHYS377:A six week marathon

through the firmament

by

Orsola De Marcoorsola@science.mq.edu.au

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.

€

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.

The main sequence lifetime

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τ ≈10GyrM

M•

⎛

⎝ ⎜

⎞

⎠ ⎟

−3

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 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.

pp or CNO?

csep10.phys.utk.edu/astr162/lect/energy/cno-pp.html

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.

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?