lecture 1 - the solar interior
DESCRIPTION
Lecture 1 - The Solar Interior. Topics to be covered: Solar interior Core Radiative zone Convection zone. The Solar Interior - “The Standard Model”. Core Energy generated by nuclear fusion (the proton-proton chain). Radiative Zone Energy transport by radiation. Convective Zone - PowerPoint PPT PresentationTRANSCRIPT
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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
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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.
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January 19, 2006 Lecture 1 - The Solar Interior
The Solar InteriorThe Solar Interior
o Christensen-Dalsgaard, J. et al., Science, 272, 1286 - 1292, (1996).
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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.
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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 =>
€
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
€
∴PC = PS +GM 2
8πr4
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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
€
PC ~GM 2
8πR4
€
TC ~GMmH
kR
€
PC = nkT =ρkT
mH
o Which gives Tc ~ 2.7 x 107 K (actual value is ~1.5 x 107 K).
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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
€
P(E)dE ∝ Ee−
E
kT dE
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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.
€
(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)
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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.
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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.
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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!
€
d = l N
€
R = l N => tdiffusion = R2 / lc
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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.
€
E = aT 4 4
3πR3 ⎛
⎝ ⎜
⎞
⎠ ⎟
€
L =E
tdiffusion
=16π
3σT 4Rl
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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
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January 19, 2006 Lecture 1 - The Solar Interior
The Convective ZoneThe Convective Zone
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