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Home , Convection zone, Solar core

Lecture 1 - The Solar Interior o Topics to be covered:

o Solar interior o Core o Radiative zone o zone

January 19, 2006 Lecture 1 - The Solar Interior

The Solar Interior - “The Standard Model” o Core o Energy generated by (the -proton chain). o Radiative Zone o Energy transport by . o Convective Zone o Energy transport by convection.

January 19, 2006 Lecture 1 - The Solar Interior The Solar Interior

o Christensen-Dalsgaard, J. et al., Science, 272, 1286 - 1292, (1996).

January 19, 2006 Lecture 1 - The Solar Interior

The

o R: 0.0 - 0.25 Rsun o T(r): 15 - 8 MK o !(r): 150 - 10 g cm-3 o and densities sufficiently high to drive burning (H->He). o Ultimate source of energy in the and Sun-like .

January 19, 2006 Lecture 1 - The Solar Interior The Solar Core

o What is the and pressure in the core?

dP GM# o Assume hydrostatic equilibrium: = " dr r2

dM and mass conservation: = "4#r2$ dr ! dP dM dP GM o Divide to cancel !’s => / = = " dr dr dM 4#r4 ! M dP o Therefore, LHS => " dM = P " P PC = pressure at core #0 dM C S ! PS = pressure at surface 2 M GM GM and RHS => dM = #0 4"r4 8"r4 ! 2 GM " P = P + C S 8#r4 ! January 19, 2006 Lecture 1 - The Solar Interior

!

The Solar Core

GM 2 o Assuming P << P and setting r = R, P ~ S C C 8"R4 "kT o Using the Ideal Gas Law PC = nkT = mH ! k = Boltzmann’s const n = number density atoms/cm3 ! = density = M/4"!R 3

GMmH o The core temperature is therefore TC ~ kR

7 7 o Which gives Tc ~ 2.7 x 10 K (actual value is ~1.5 x 10 K). !

January 19, 2006 Lecture 1 - The Solar Interior The Solar Core o Coulomb barrier between must be overcome for fusion to occur. o To overcome Coulomb barrier, particles must have sufficient thermal kinetic energy to exceed Coulomb repulsion: 3 e2 kT > 2 rnuc 2e2 => T > 3krnuc ! =1010K! o Particles have Maxwell-Boltzmann distribution:

E ! # kT !P (E)dE " Ee dE o There is a high-energy tail, but not sufficient … need quantum mechanics. !

January 19, 2006 Lecture 1 - The Solar Interior

The Solar Core

o From Heisenberg Uncertainty Principle ( " x " p # / 2 ) a proton of a given h (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.

January 19, 2006 Lecture 1 - The Solar Interior Proton-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.)

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. CNO

January 19, 2006 Lecture 1 - The Solar Interior

The Radiative Zone o R: 0.25 - 0.8 Rsun o T(r): 8 - 0.5 MK o !(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 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 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 moves l between interactions (mean free path) and takes a total number of steps N. o On average it will have moved a distance d = l N o As t = N l / c and 2 difusion R = l N => tdiffusion = R /lc ! 4 => tdiffusion >10 yrs!

January 19, 2006 ! Lecture 1 - The Solar Interior

Solar Interior

# 4 & o Total radiative energy inside Sun is: E = aT 4% "R3 ( J $ 3 '

where a = 4$/c is the radiation constant.

! E 16" o Can thus estimate solar from, L = = #T 4 Rl W tdiffusion 3 o Which gives, L ~ 3 x 1026 W.

! o Actual value is actually 4 x 1026 W.

January 19, 2006 Lecture 1 - The Solar Interior The Convective Zone

o R: 0.8 - 1 Rsun o T(r): 0.5 MK - 6000 K. o ! <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.

TC o Continuous absorption of photons by lower layers causes a temperature gradient to build up between the lower and upper layers. r o become convectively unstable, and large convective motions become the dominant transport mechanism. TH

TH > TC

January 19, 2006 Lecture 1 - The Solar Interior

The Convective Zone

January 19, 2006 Lecture 1 - The Solar Interior Advanced Stellar Physics o Email: [email protected] / [email protected] o Office: 3.17A. o 80%: Final exam. o 20%: 2000 word essay and 10-min presentation. o Deadline and presentation: Last lecture (March 9) o Claire: “The Coronal Heating Debate” o Brian: “The Solar Activity Cycle”

January 19, 2006 Lecture 1 - The Solar Interior