1. January 19, 2006 Lecture 1 - The Solar Interior oTopics to be covered: oSolar interior oCore oRadiative zone oConvection zone

2. January 19, 2006 Lecture 1 - The Solar Interior The Solar Interior - “The Standard Model” oCore oEnergy generated by nuclear fusion (the proton-proton chain). oRadiative Zone oEnergy transport by radiation. oConvective Zone oEnergy transport by convection.

3. January 19, 2006 Lecture 1 - The Solar Interior The Solar Interior oChristensen-Dalsgaard, J. et al., Science, 272, 1286 - 1292, (1996).

4. January 19, 2006 Lecture 1 - The Solar Interior The Solar Core oR: 0.0 - 0.25 R sun oT(r): 15 - 8 MK o  (r): 150 - 10 g cm -3 oTemperatures and densities sufficiently high to drive hydrogen burning (H->He). oUltimate source of energy in the Sun and Sun-like stars.

5. January 19, 2006 Lecture 1 - The Solar Interior The Solar Core oWhat is the temperature and pressure in the core? oAssume hydrostatic equilibrium: and mass conservation: oDivide to cancel  ’s => oTherefore, LHS => and RHS => P C = pressure at core P S = pressure at surface

6. January 19, 2006 Lecture 1 - The Solar Interior The Solar Core oAssuming P S << P C and setting r = R, oUsing the Ideal Gas Law k = Boltzmann’s const n = number density atoms/cm 3  = density = M/4  R 3 oThe core temperature is therefore o Which gives T c ~ 2.7 x 10 7 K (actual value is ~1.5 x 10 7 K).

7. January 19, 2006 Lecture 1 - The Solar Interior The Solar Core oCoulomb barrier between protons must be overcome for fusion to occur. oTo overcome Coulomb barrier, particles must have sufficient thermal kinetic energy to exceed Coulomb repulsion: oParticles have Maxwell-Boltzmann distribution: oThere is a high-energy tail, but not sufficient … need quantum mechanics.

8. January 19, 2006 Lecture 1 - The Solar Interior The Solar Core oFrom Heisenberg Uncertainty Principle a proton of a given (insufficient) energy may be located within nucleus of neighbouring proton. oCombined with high-energy M-B tail, we get the Gamow Peak. oSo protons in 3-10 keV energy range can overcome the Coulomb barrier (i.e., T>15MK). oFusion can therefore occur.

9. January 19, 2006 Lecture 1 - The Solar Interior Proton-proton cycle oThe p-p cycle occurs in three main steps. Step 1: 1 H + 1 H  2 H + e + + (Q = 1.44 MeV) oMight then expect a 2 H + 2 H reaction, but because of the large numbers of 1 H, the following is more probable: Step 2: 2 H + 1 H  3 He +  (Q = 5.49 MeV) o 3 He can then react with 1 H, but the resultant 4 Li is unstable (i.e. 3 He + 1 H  4 Li  3 He + 1 H). oThe final step is then: Step 3: 3 He + 3 He  4 He + 2 1 H +  (Q = 12.86 MeV) oThe net result is: 4 1 H  4 He + 2e + + 2 (Q = 26.7 MeV)

10. 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. oThe remaining ~1% is produced by the Carbon-Nitrogen-Oxygen (CNO) cycle. oCNO cycle is more important in more massive stars.

11. January 19, 2006 Lecture 1 - The Solar Interior Proton-proton vs. CNO

12. January 19, 2006 Lecture 1 - The Solar Interior The Radiative Zone oR: 0.25 - 0.8 R sun oT(r): 8 - 0.5 MK o  (r): 10 - 0.01 g cm -3 oHydrogen burning cuts off abruptly at r ~ 0.25 R sun. oInterior becomes optically thin or transparent as density decreases. oEnergy transported radiatively. oPhotons 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.

13. January 19, 2006 Lecture 1 - The Solar Interior The Radiative Zone oFor T = 15MK Wien’s displacement law implies max = 0.19 nm i.e., the center of the Sun is full of X-rays. oPhotons do 3D random walk out of Sun. oAssume photon moves l between interactions (mean free path) and takes a total number of steps N. oOn average it will have moved a distance oAs t difusion = N l / c and => t diffusion >10 4 yrs!

14. January 19, 2006 Lecture 1 - The Solar Interior Solar Interior oTotal radiative energy inside Sun is: J where a = 4  /c is the radiation constant. oCan thus estimate solar luminosity from, W oWhich gives, L ~ 3 x 10 26 W. oActual value is actually 4 x 10 26 W.

15. January 19, 2006 Lecture 1 - The Solar Interior The Convective Zone oR: 0.8 - 1 R sun oT(r): 0.5 MK - 6000 K. o  <0.01 g cm -3 oPhotons now absorbed as temperature is sufficiently low to allow atoms to form. Gas is optically thick or opaque. oContinuous absorption of photons by lower layers causes a temperature gradient to build up between the lower and upper layers. oPlasma become convectively unstable, and large convective motions become the dominant transport mechanism. T H > T C THTH TCTC r

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