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Chapter 10

Chapter 10. Stellar Interiors. Revised 10/8/2012. The laws of stellar structure that lead to L=M 3. Hydrostatic Equilibrium Ideal gas law Energy transport Mass conservation Energy conservation. Hydrostatic Equilibrium. Mathematically…….

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Chapter 10

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  1. Chapter 10 Stellar Interiors Revised 10/8/2012

  2. The laws of stellar structure that lead to L=M3 • Hydrostatic Equilibrium • Ideal gas law • Energy transport • Mass conservation • Energy conservation

  3. Hydrostatic Equilibrium

  4. Mathematically…….. The pressure gradient across a slab of area A supports the weight, mg, of the layer so that mg = (Pl – Pu ) A For a spherical shell, A = 4pr2, and the mass of a shell dm = 4pr2rdr Thus, dP/dr = -gr Where g is the “local value” of g inside the star g = GM(r)/r2 Where M(r) is the mass enclosed by the radius r

  5. Continuing …. So, dP/dr = - GM(r) r(r) /r2 “equation of Hydrostatic equilibrium” which may be integrated to determine Pc, exactly, with some simplifying assumptions; P(0) = Pc, and P(R) = 0 and constant density r(r) ~ r ~ Mo/ (4pR3o /3) Then, Pc ~ G Mo r/2 Ro

  6. Which upon substituting standard values for the constants yields, Pc ~ 1.4 x 1014 N/m2 Which is the wrong answer! More detailed calculations, that allow for the fact that the density increases towards the center of the Sun, yield Pc ~ 2.7 x 1016 N/m2 The problem here is that the Sun’s density is not constant!

  7. Ideal Gas Law What provides the Pressure? - the thermal energy of the hot gas, which we assume is a “perfect gas” governed by the ideal gas law; P = n(r) k T(r) Where n(r) is the number density (atoms/m3), k is Boltzmann’s constant and T(r) is the temperature gradient. We can express n(r) in terms of the density r(r) since n(r) = r(r) /mH where mH is the mass of a hydrogen atom

  8. P(r) = r(r) k T(r) /mH which may be differentiated to get the temperature gradient, or, we can use the same trick as before to get an estimate for the central temperature since re-arranging, Tc = Pc mH / r k Which upon substituting the values for the constants, (including the wrong Pressure from the previous example!), yields Tc ~ 1.1 x 107 K (which is close to the correct answer !) At these temperatures hydrogen is dissociated (ionized) into protons and electrons, an electrically neutral mixture called a plasma.

  9. This version of the ideal gas law, P(r) = r(r) k T(r) /mH, only works for a star composed entirely of Hydrogen, which, of course, they are not. So, to accommodate different chemical compositions, astronomers introduce the concept of mean molecular weight, denoted by the symbol . Then the gas law becomes, P(r) = r(r) k T(r) /  mH, With the understanding that  is the average mass of an atom, in units of the hydrogen mass, so that the average mass of a stellar atom is m, where m =  mH Mean Molecular Weight

  10. Energy Generation & Transport The temperature is only high enough for fusion reactions to occur in the very center of the Sun, a region enclosing ~ 10% of the Sun’s total mass called the “core”. The primary mode of energy transport is radiation.

  11. Radiative Transfer Equation The Sun’s central temperature is much higher than the surface temperature. Thus, heat flows from hot to cold, down a temperature gradient. The fact that the surface temperature of the Sun is much lower than The core temperature means that the photon’s are loosing a lot of energy on their journey out of the star. Thus, there must be some source of opacity.

  12. The primary source of opacity is ionization. Protons and electrons will recombine into H atoms unless they are continually ionized by photons. A secondary source of opacity is electron scattering, where photons are scattered by electrons. (see Chapter 9) dPrad = F/c (Note there are now 2 pressures; gas and dradiation. Prad is the latter). where  =   dr, is the radial optical depth and F = L / 4 r2 is the outward flux, expressed in terms of the luminosity, L. Also, from the second moment of the R.T.E., we have Prad = 4 T4/ 3c which can be differentiated w.r.t.r. Following some algebra one finds, L = 16  T3 4 r2 dT 3  dr

  13. Important note In this and similar equations, it is important to note that T, r, r are “interior” values, themselves a function of r, so that T = T(r), r = r(r). L=L(r), etc.

  14. The Standard Solar Model Numerically integrating the 5 equations of stellar structure reveal that the pressure, temperature, and density all increase rapidly towards the center of the Sun;

  15. The Physical Basis for L a M3 Can be deduced from the first 3 equations governing stellar structure; Pc ~ G Mo r/ Ro Pc ~ r k Tc /mH Lc ~ 16 sT3 4  r2 dT 3 kr dr From which we can derive the following proportionalities, Pc a Mo r/ Ro Pcar Tc Lca T4 r/r

  16. The Physical Basis for L a M3 (continued) Eliminating Pc , and T (since neither appear in the mass –luminosity relationship) and writing ra M/r3, one can show that L a M3 as observed ! Question: Using the proportionalities given on the previous page, show that L a M3.

  17. Energy Sources What source of energy makes the Sun shine? This was a mystery until quite recently when in 1938 Hans Bethe recognized that the temperature at the core of the Sun was high enough to support thermonuclear fusion reactions. Prior to 1938 there were all sorts of ideas floating around including The meteoric theory, whereby astronomers believed that the Sun was powered by in-falling comets.

  18. Comets do fall into the Sun as these pictures show!

  19. Plus, comets can yield a lot of energy…. The energy released by in-falling comets is that due to the gain in kinetic energy, E, of the comets as they fall into the gravitational potential well of the sun, where E = G M m /R and m is the mass of the comet, E r

  20. In-falling comets yield energy (continued) If all that energy were converted into heat and light then the rate of energy conversion, L, is related to the mass in-fall rate, dm/dt, so that L = G Mdm R dt Which, assuming a 100% conversion of mass to energy, the observed luminosity of the Sun, would require a mass in-fall rate of 6.3 x 1022 kg/yr. This may seem like a lot, but when expressed in terms of the mass of the moon, it corresponds to a rate of only 1 moon per year!!

  21. The Contraction Theory The meteoric theory was never disproved, but rather supplanted by the Contraction Theory. In this theory, the Sun has to contract in order to loose heat. The problem with this theory is that the Sun is not contracting, but back in the 19th Century, no one could tell. The amount of energy released by a contracting self-gravitating collection of particles may be calculated using the virial theorem, which states that one half of the gravitational potential energy, U, is radiated away, and the other half goes into heating up the star. So, now, U/2 = 3 G M2 /10 R, is released, as the Sun shrinks from infinity down to it’s current size.

  22. But, the problem with the Contraction theory that if the Sun had been radiating at it’s present rate, L, all the time, then the lifetime of the Sun, t, would be t ~ E/L = 3G M2 /10 L  R  which, upon substituting the appropriate constants, yields a lifetime of about 10 million years. But we know from carbon dating of terrestrial and lunar rocks that the earth is at least 4.5 billion years old ! and one would expect the Sun to be at least as old as the earth-moon system, if not older. So there’s a major time discrepancy.

  23. Nuclear Fusion We now know that the real energy source for the Sun’s luminosity is thermonuclear fusion reactions occurring in the Sun’s core. Although the temperatures in the Sun’s core are high, they are not high enough to overcome the coulomb repulsion force resulting from two positively charged nuclei colliding under the laws of classical physics. The solution to this problem is quantum mechanical tunneling. The essence of this energy generating process is that 4 1H are converted into one 4He, with the difference in mass-energy being released mostly as heat according to Einstein’s famous equation E = mc2. Here are the details;

  24. The total possible number of reactions, n = M  /4mH The energy released in each reaction, e, is e = (4 mH – mHe) c2 The total energy released, E, is E = n e E = M(4 mH – mHe) c2/4mH The quantity (4 mH – mHe) /4mH = 0.0071, thus E = 0.0071 M c2 But, only the core, 10% of the total mass, is involved in fusion, so actually E = 0.00071 M c2

  25. Solar Lifetime Substituting the appropriate constants yields E = 1.28 x 1044 J Which, at the rate of the Sun’s present luminosity, L = 4 x 1026 J/s, will last for a time, t t = E/L = 3.2 x 1017 s or 10 billion years !

  26. There are several reactions, in order of decreasing probability; The PPI chain 4 11H  42He + 2e+ + 2e + 2 The PPII chain The PPIII chain The CNO cycle

  27. PPI Chain

  28. Stellar Main Sequence Lifetimes We can use the mass luminosity relationship to predict the main sequence lifetimes for all other stars since t a E/L a M/L and L a M3 then, t a 1/M2 , ie. the most massive stars have the shortest lifetimes

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