Download - Lecture 1 - The Solar Interior
January 19, 2006 Lecture 1 - The Solar Interior
Lecture 1 - The Solar InteriorLecture 1 - The Solar Interior
o Topics to be covered:
o Solar interioro Core
o Radiative zone
o Convection zone
January 19, 2006 Lecture 1 - The Solar Interior
The Solar Interior - “The Standard Model”The Solar Interior - “The Standard Model”
o Coreo Energy generated by
nuclear fusion (the proton-proton chain).
o Radiative Zoneo Energy transport by
radiation.
o Convective Zoneo Energy transport by
convection.
January 19, 2006 Lecture 1 - The Solar Interior
The Solar InteriorThe Solar Interior
o Christensen-Dalsgaard, J. et al., Science, 272, 1286 - 1292, (1996).
January 19, 2006 Lecture 1 - The Solar Interior
The Solar CoreThe Solar Core
o R: 0.0 - 0.25 Rsun
o T(r): 15 - 8 MK
(r): 150 - 10 g cm-3
o Temperatures and densities sufficiently high to drive hydrogen burning (H->He).
o Ultimate source of energy in the Sun and Sun-like stars.
January 19, 2006 Lecture 1 - The Solar Interior
The Solar CoreThe Solar Core
o What is the temperature and pressure in the core?
o Assume hydrostatic equilibrium:
and mass conservation:
o Divide to cancel ’s =>
o Therefore, LHS =>
and RHS =>
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dP
dr= −
GMρ
r2
€
dM
dr= −4πr2ρ
€
dP
dr/dM
dr=
dP
dM= −
GM
4πr4
€
−dP
dMdM = PC − PS0
M
∫
€
GM
4πr4dM
0
M
∫ =GM 2
8πr4
PC = pressure at core PS = pressure at surface
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∴PC = PS +GM 2
8πr4
January 19, 2006 Lecture 1 - The Solar Interior
The Solar CoreThe Solar Core
o Assuming PS << PC and setting r = R,
o Using the Ideal Gas Law
k = Boltzmann’s const
n = number density atoms/cm3
= density = M/4R3
o The core temperature is therefore
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PC ~GM 2
8πR4
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TC ~GMmH
kR
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PC = nkT =ρkT
mH
o Which gives Tc ~ 2.7 x 107 K (actual value is ~1.5 x 107 K).
January 19, 2006 Lecture 1 - The Solar Interior
The Solar CoreThe Solar Core
o Coulomb barrier between protons must be overcome for fusion to occur.
o To overcome Coulomb barrier, particles must have sufficient thermal kinetic energy to exceed Coulomb repulsion:
o Particles have Maxwell-Boltzmann distribution:
o There is a high-energy tail, but not sufficient … need quantum mechanics.
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=1010K!€
3
2kT >
e2
rnuc
€
=>T >2e2
3krnuc
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P(E)dE ∝ Ee−
E
kT dE
January 19, 2006 Lecture 1 - The Solar Interior
The Solar CoreThe Solar Core
o From Heisenberg Uncertainty Principle a proton of a given (insufficient) energy may be located within nucleus of neighbouring proton.
o Combined with high-energy M-B tail, we get the Gamow Peak.
o So protons in 3-10 keV energy
range can overcome the Coulomb
barrier (i.e., T>15MK).
o Fusion can therefore occur.
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(ΔxΔp ≥ h /2)
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January 19, 2006 Lecture 1 - The Solar Interior
Proton-proton cycleProton-proton cycle
o The p-p cycle occurs in three main steps.
Step 1: 1H + 1H 2H + e+ + (Q = 1.44 MeV)
o Might then expect a 2H + 2H reaction, but because of the large numbers of 1H, the following is more probable:
Step 2: 2H + 1H 3He + (Q = 5.49 MeV)
o 3He can then react with 1H, but the resultant 4Li is unstable (i.e. 3He + 1H 4Li 3He + 1H).
o The final step is then:
Step 3: 3He + 3He 4He + 21H + (Q = 12.86 MeV)
o The net result is: 4 1H 4He + 2e+ + 2 (Q = 26.7 MeV)
January 19, 2006 Lecture 1 - The Solar Interior
Proton-proton cycle (cont.)Proton-proton cycle (cont.)
o ~99% of the Sun’s energy is produced via the p-p cycle. o The remaining ~1% is produced by the Carbon-Nitrogen-Oxygen (CNO) cycle.o CNO cycle is more important in more massive stars.
January 19, 2006 Lecture 1 - The Solar Interior
Proton-proton vs. CNOProton-proton vs. CNO
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January 19, 2006 Lecture 1 - The Solar Interior
The Radiative ZoneThe Radiative Zone
o R: 0.25 - 0.8 Rsun
o T(r): 8 - 0.5 MK
(r): 10 - 0.01 g cm-3
o Hydrogen burning cuts off abruptly at r ~ 0.25 Rsun
.
o Interior becomes optically thin or transparent as density decreases.
o Energy transported radiatively.
o Photons cannot be absorbed in the radiative zone as the temperature are too high to allow atoms to form. Therefore no mechanism for the absorption of photons.
January 19, 2006 Lecture 1 - The Solar Interior
The Radiative ZoneThe Radiative Zone
o For T = 15MK Wien’s displacement law implies max = 0.19 nm i.e., the center of the Sun is full of X-rays.
o Photons do 3D random walk out of Sun.
o Assume photon moves l between interactions (mean free path) and takes a total number of steps N.
o On average it will have moved a distance
o As tdifusion = N l / c and
=> tdiffusion >104 yrs!
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d = l N
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R = l N => tdiffusion = R2 / lc
January 19, 2006 Lecture 1 - The Solar Interior
Solar InteriorSolar Interior
o Total radiative energy inside Sun is: J
where a = 4/c is the radiation constant.
o Can thus estimate solar luminosity from, W
o Which gives, L ~ 3 x 1026 W.
o Actual value is actually 4 x 1026 W.
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E = aT 4 4
3πR3 ⎛
⎝ ⎜
⎞
⎠ ⎟
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L =E
tdiffusion
=16π
3σT 4Rl
January 19, 2006 Lecture 1 - The Solar Interior
The Convective ZoneThe Convective Zone
o R: 0.8 - 1 Rsun
o T(r): 0.5 MK - 6000 K.
<0.01 g cm-3
o Photons now absorbed as temperature is sufficiently low to allow atoms to form. Gas is optically thick or opaque.
o Continuous absorption of photons by lower layers causes a temperature gradient to build up between the lower and upper layers.
o Plasma become convectively unstable, and large convective motions become the dominant transport mechanism.
TH > TC
TH
TC
r
January 19, 2006 Lecture 1 - The Solar Interior
The Convective ZoneThe Convective Zone
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