What makes a star shine

1
Energy Sources in Stars (10.3)

• What makes a star shine?

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Energy Sources in Stars (10.3)
What makes a star shine?

• Suns Energy Output
• Lsun 4 x 1026 W (J s-1)
• tsun 4.5 x 109 yr (1.4 x 1017 s) - oldest rocks
  - radioactive dating
• Lsun constant over geological timescale
  (fossil evidence)
• Therefore total energy output Etot Lsun x tsun
  6 x 1043 J
• How might the Sun have generated this energy?

3
Energy Sources in Stars

• Energy Source 1Gravitational contraction
• We believe Sun collapsed from a large gas cloud
  (R 8) to its present size

Potential energy released is ?Egrav
-(Enow - Einitial) P.E. is obtained from
integrating grav. potential over all points in
the sphere
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Energy Sources in Stars
Energy Source 1Gravitational contraction

• However, from Virial Theorem (Eth -1/2 Egrav),
  only 1/2 of the gravitational energy is available
  for energy release, the other half heats the
  star.
• So, ?Egrav -(Enow -Einitial)
•                   3/10(GMsun2)(1/Rsun - 1/?)
•                   1041 J
•                   1/600 Etot needed over Suns
  lifetime
• By this process, Sun could only radiate for
  t
  ?Egrav/Lsun 107 years
• t ?Egrav/Lsun is Kelvin-Helmholtz timescale
• Conclusion Gravitational Contraction not the
  main energy source in Sun

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Energy Sources in Stars
Energy Source 2Chemical Reactions

• e.g. How much energy would be released if Sun was
  completely ionized and then all gas recombined
  into neutral atoms? (binding energy of the H atom
  13.6 ev 1 ev 1.6 x 10-19 J ? 13.6 ev 2 x
  10-18 J)
• Assume Sun is 100 H (X 1)
• Total H atoms NH Msun/mH 1057
• ? Etot ionization 1057 x (2 x 10-18) J 2 x
  1039 J
• This is only 1/30,000 Etotal of Suns energy
  output
• Suns energy output burning 7000 kg coal each
  hour on every sq. metre Suns surface!
• Conclusion Chemical Reactions cannot be source
  Suns energy

6
Energy Sources in Stars
Energy Source 3Nuclear Energy

• Fission?
• Z 0.02 (2 Sun consists of heavy elements).
• Fissionable isotopes like 235U are rare
  (isotopes are nucleii with the same protons but
  different neutrons)
• Fusion?
• e.g. 4 protons get converted into 2p 2n i.e.
  4H ? 1 4He
• H is very abundant in the Sun (X 0.73)
• 4 H nuclei mass 4 x mp 6.693 x 10-27 kg
• 1 4He nucleus mass 6.645 x 10-27 kg (binding
  energy included)
• Difference is 0.048 x 10-27 kg 0.7 of original
  mass
• Where does this mass go?

7
Energy Sources in Stars
Energy Source 3Nuclear Energy The lost mass
is converted into energy according to Einsteins
equation for the rest energy of matter E mc2
(E energy, m mass, c speed light)

• Example
• Suppose that Sun was originally 100 H and only
  10 of that was available for fusion. Thus
• MH available 0.1 Msun
• ?MH destroyed through fusion 0.1 Msun x 0.007
• total nuclear energy available Etot (7 x 10-4
  Msun)c2 1.3 x 1044 J
• Suns total lifetime energy output 6 x 1043 J
  
• Etot 2 x the total energy Sun has emitted!!
• ? Sun could shine for at least another 5 x 109 yr
  at its current luminosity
• Hence tnuclear for Sun 1010 yrs

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Energy Sources in Stars
Energy Source 3Nuclear Energy

• Fission?
• Z 0.02 (2 Sun consists of heavy elements).
• Fissionable isotopes like 235U are rare
  (isotopes are nucleii with the same protons but
  different neutrons)
• Fusion?
• e.g. 4 protons get converted into 2p 2n i.e.
  4H ? 1 4He
• H is very abundant in the Sun (X 0.73)
• 4 H nuclei mass 4 x mp 6.693 x 10-27 kg
• 1 4He nucleus mass 6.645 x 10-27 kg (binding
  energy included)
• Difference is 0.048 x 10-27 kg 0.7 of original
  mass
• Where does this mass go?

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Can Fusion Occur in Suns Core?

• Classical Approach
• Ekin proton gt Ecoulomb
• 1/2 mpv2 gt e2/r (cgs form)
• LHS is thermal gas energy, thus, 3/2 kT gte2/r
• At Tcentral 1.6 x 107 K ? Ekin 3.3 x 10-16
  J
• For successful fusion, r
• 10-15 m (1 fm) ?

• Ecoulomb 2.3 x 10-13 J
• Ekin 10-3 Ecoulomb
• Classically, would need T1010 to overcome
  Coulomb barrier
• So, no Fusion??

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Can Fusion Occur in Suns Core?

• Can we be helped by considering the distribution
  of energies? - not all protons will have just Eth
  3/2 kT
• Proton velocities are distributed according to
  the Maxwellian equation P(v) 4?(m/2?kT)3/2 v2
  e-mv2/2kT
• At the high energy tail of the Maxwellian
  distribution, the relative number of protons,
  with E gt 103 x Eth is
• N(Ecoulomb 103 ltEthermalgt)/N(ltEthermalgt)
  e-?E/kt e-1000 10-430!!

Number protons in Sun Msun/MH 1057 so 1 in
10430 of these will have enough energy to
overcome Coulomb barrier. So again, no nuclear
reactions??
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Can Fusion Occur in Suns Core?

• Quantum Approach
• Heisenberg Uncertainty Principle - ?p?x gt h/2?
• If ?p small, then ?x may be large enough that
  protons have non-negligible probability of being
  located within 1 fm of another proton inside
  Coulomb barrier

• Called Quantum Tunnelling
• Particles have wavelength, (? h/p ), associated
  with them (like photons) called de Broglie
  wavelength
• Proved in many experiments - for example
  diffraction electrons (wave phenomenon)
• Eg. ? free electron (3 x 106 m/s) 0.242 nm
  (size atom)
• ? person (70 k gm) jogging at 3 m/s 3 x 10-16
  m (negligible - person wont diffract!)
• Increased wavelength reflects loss momentum.

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Can Fusion Occur in Suns Core?

• Even with tunnelling, are stars hot enough for
  nuclear reactions to proceed?
• ? h/p is the wavelength associated with a
  massive particle (p momentum)
• In terms momentum, the kinetic energy of a proton
  is
• 1/2 mpv2 p2/2mp
• Assuming proton must be within 1 de Broglie ? of
  its target to tunnel
• set distance, r, of closest approach to ?,
  (where barrier height original K.E.
• incoming particle) gives
• e2/r e2/? p2/2mp (h/?)2/2mp
• Solving for ? (h2/2mpe2) and substituting r ?
  into 3/2 kT e2/r, we get the QM estimate of the
  temperature required for a nuclear reaction to
  occur
• TQM 4/3(e4mp/kh2) - putting in numbers
• TQM 107 K which is comparable to Tcore.
  Therefore, fusion is feasible at centre of Sun

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Proton-Proton Chain
Basic particles involved in nuclear reactions

• Conservation Laws in nuclear reactions
• (1) mass-energy
• (2) charge
• (3) difference between number particles and
  anti-particles conserved - ie particle cannot be
  created from anti-particle or vice-versa but a
  pair can be formed or destroyed without violating
  this rule

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Proton-Proton Chain

• Types of nuclear reactions
• Beta Decay n ? p e- ?- proceeds
  spontaneously - also for neutron inside nucleus
• eg Z-1, A ? Z, A e- ?- (Z p, N
  n, A ZN)
• Inverse Beta Decay p e- ? n ?
• eg 13N ? 13C e ?
• (p, ?) process AZ p ? A1(Z1) ?
• eg 12C p ? 13N ?
• (?, ?) process ? particle (4He) added to nucleus
  to make heavier particle AZ 4He ? A4(Z2)
  ?
• eg. 8Be 4He ? 12C ?

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Proton-Proton Chain

• Since the Suns mass consists mostly of H and He,
  we anticipate nuclear reactions involving these
  two elements
• eg p p ? 2He
• p 4He ? 5Li
• 4He 4He ? 8Be
• But there is a problem here - what is it?
• All these reactions produce unstable particles
• eg  p p ? 2He ? p p
• p 4He ? 5Li ? p 4He
• 4He 4He ? 8Be ? 4He 4He
• So among light elements there are no two-particle
  exothermic reactions producing stable particles.
  We have to look to more peculiar reactions or
  those involving rarer particles

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Proton-Proton Chain

• Using considerations of
• minimizing Coulomb barriers,
• crossections,
• and making sure conservation laws are obeyed
• the nuclear reaction chain at right produces the
  energy observed in the Sun.

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Proton-Proton Chain

• Summary of proton-proton chain
• 6p ? 4He 2p 2e 2? 2? or 4p ? 4He
  2e 2? 2?
• Mass difference between 4p and 1 4He 26.7 Mev
  x 6.424 x 1018 ev/J       4.2 x 10-12 J
• 3 of this energy (0.8 Mev) is carried off by
  neutrinos and does not      contribute to the
  Suns luminosity
• 2 e immediately annihilate with 2 e- and add 2 x
  0.511 Mev ( 1.02 Mev)
• So, the total energy available for the Suns
  luminosity per 4He formed      is (26.7 - 0.8
  1.02) Mev 26.9 Mev 4.2 x 10-12 J
• 4He formed/sec Lsun/4.2 x 10-12J 3.9 x1026
  J/s / 4.2x10-12 J       9.3 x 1037
  4He/s
• Increase of 4He mass/time dmHe/dt 9.3 x 1037
  4He/s x 6.68 x 10-27      kg/4He 6.2 x 1011 kg
  4He /s
• After 1010 years (3 x 1017 sec), M(4He) 1.9 x
  1029 kg 10 mass Sun

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Proton-Proton Chain

• The previous nuclear reaction chain (PPI) is not
  the only way to convert H into He.
• eg last step 3He 3He ?
  4He 2p can proceed differently if there is
  appreciable 4He present

PPI PPII  PPIII operate simultaneously
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Proton-Proton Chain Summary
69                               31
(PPI)
99.7                               0.3
(PPII)
(PPIII)
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Alternate Fusion Reactions H ? 4He

• First step in PP chain has very low reaction
  rate (weak interaction).
• 12C can act as a catalyst in fusion reaction.
• Net result
  12C 4p ? 12C 4He 2e 2n 3g
• Note 12C neither created nor destroyed. Also
  isotopes of N O are temporarily produced.
• CNO cycle dominates over PP chain if Tcore gt 1.8
  x 107 (slightly hotter than Sun)

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Other Fusion Reactions
We believe Universe began with only H, He, (Li,
Be) All other elements created in core of stars
(stellar nucleosynthesis) We are all made of
STARSTUFF