1. Structure of Stars Today’s Lecture: Structure of Stars Homework 4: Due Tuesday, March 4 Help Session: Wed., March 12, 5:00-6:30, LGRT 1033 Midterm Exam, in class Thursday, March 13 Reading for Today’s Lecture: Chapter 9.4-9.6 Reading for Next Lecture: Chapter 10.1-10.2 Gas Pressure/Hydrostatic Equilibrium Energy Transport Structure of the Sun and Stars Main Sequence Lifetimes

2. The Sun The Sun, like all stars, is a gaseous sphere composed of primarily hydrogen and helium. The Sun is about 100 times larger than Earth and about 300,000 times more massive. Average density about 1/3 that of the Earth. The visible surface is called the photosphere. The surface temperature of the Sun is about 6000 degrees K. The Sun is almost a perfect blackbody radiator.

3. What Supports a Star Against its own self Gravity ? Gas pressure – produced by the thermal motions of the gas atoms. Pressure is the force exerted by the gas per unit area and depends on the gas density (number of collisions) and its temperature (speed of gas atoms).

4. Gas Pressure: We start by computing the momentum transferred to area A in time t, which is just the total number of particles impacting A, multiplied by the average momentum transferred per particle. Consider the component of the velocity perpendicular to A (v x ), therefore in time t, all of the particles within a distance of v x t will impact A. The volume swept out is A v x t, therefore the total number of impacts in time t is: = number density · volume swept = n · A v x t, where n is the number of gas particles per unit volume. vxvx

5. The total momentum transferred is: = number density · volume swept · momentum/particle = n · A v x t · p x The gas particles have a Maxwell-Boltzmann velocity distribution, thus need to compute the average of v x p x or v x p x 〉. Note that v 2 = v x 2 + v y 2 + v z 2, however we also know that: v x 2 〉 = v y 2 〉 = v z 2 〉, so v x 2 〉 = 1/3 v 2 〉, or v x p x 〉 = v x 2 〉 m = 1/3 v 2 〉 m. The momentum transferred to A in time t is: n A t 1/3 v 2 〉 m. Force is momentum per unit time, so pressure is momentum transferred per unit time and per unit area, therefore: P = 1/3 n v 2 〉 m

6. For a thermal gas, the average energy per particle is: E 〉 = 3/2 kT = ½ m v 2 〉, therefore: v 2 〉 = 3 k T/m, Thus, the pressure of an ideal gas is given by: P = n k T, k is the Boltzmann constant, and T the gas temperature. Pressure has units of dynes cm -2, or ergs cm -3 or gm cm -1 s -2. If the particles have average mass m, this can be rewritten in terms of the mass density, ρ (units gm cm -3 ) = n m, as: P = ρ k T/m This is called the equation of state, and it relates pressure, density and temperature for an ideal gas.

7. Gas Pressure Force: How do we get an outward gas pressure force ?? A net force is produced only if there is a pressure difference. Imagine an area, A, with pressure P 1 on one side and P 2 on the other side. The net pressure force is then: F P = (P 1 – P 2 ) A, and has units of force which are dynes or gm cm s -2. For stars, the density and temperature (and therefore gas pressure) at their centers is much larger than at their surfaces. This decrease in gas pressure from center to surface, produces an net outward gas pressure force.

8. Stellar structure: Both ρ and T vary with stellar radius. We can first consider density and its relation to mass. Imagine a star is made up of numerous concentric, incrementally thin shells. Each shell has a mass density given by ρ(r). Then the incremental mass in a shell is: dM = ρ(r) ∗ volume The volume of a shell at radius r, with incremental radius dr is: volume = 4 π r 2 dr, therefore the incremental mass: dM = 4 π r 2 ρ(r) dr If we sum up the shells we get the mass out to radius r: M(r) = ∫ 0 r 4 π r 2 ρ(r) dr

9. Hydrostatic Equilibrium The stability of stars is due to the balance between the inward force of gravity and the outward force of gas pressure at every radius. We can compute at radius r, the gas pressure force and the force of gravity: Gravity Gas Pressure Consider a thin shell at radius r with mass dM, the force of gravity on the shell depends on the mass of the star interior to radius r, M(r): F g = -G dM M(r)/r 2 or F g = - 4π G ρ(r) M(r) dr

10. If the pressure difference from one side of the thin shell to the other side is given by dP, then the pressure force is just the surface area of the shell times the pressure difference, giving: F P = 4 π r 2 dP If the force of gravity equals the pressure force then: F P = 4 π r 2 dP = F g = - 4π G ρ(r) M(r) dr Rewriting this: dP/dr = -G ρ(r) M(r)/r 2 This is called the equation of hydrostatic equilibrium. If this equation is satisfied, then a star is stable and will not contract due to gravity, nor expand due to gas pressure.

11. Central Pressure and Temperature A crude estimate of a star's central pressure can be made by just considering the pressure difference needed to support the weight of the star. Start with: dP/dr = -G ρ(r) M(r)/r 2 Consider: dr = R (the radius of the star) and dP = [P(surface) – P(center)]. However the pressure at the surface is negligible compared to the pressure in the center, thus the hydrostatic equilibrium equation gives: dP/dr ~ - P center /R ~ - G 〈 ρ 〉 M/R 2 Therefore: P center ~ G 〈 ρ 〉 M/ R, average density of the star: 〈 ρ 〉 ~ M/R 3

12. Therefore: P center ~ G 〈 ρ 〉 M/ R ~ G M 2 /R 4 For the Sun: P center ~ 1x10 16 dynes cm -2 ( ~1x10 10 atm.) The equation of state allows us to relate pressure and temperature (remember that: P = ρ k T/m), therefore: T ~ m P/(ρ k) Substituting for P center we find: T center ~ m G M/(R k) For stars (dominated by fully ionized H & He): m = 0.6 m p For the Sun, find: T center ~ 14x10 6 K (the good agreement is somewhat fortuitous) Very crude estimates, but demonstrates need for the center of stars to have large temperatures and pressures.

13. Note that in general, the central temperature for stars is crudely given as: T center = 1.4x10 7 (M/M ⊙ )/(R/R ⊙ ) Main Sequence Stars (data from Appendix E): Spectral Type M/M ⊙ R/R ⊙ T center M5 0.2 0.32 9x10 6 K M0 0.5 0.63 12x10 6 K G2 1.0 1.0 14x10 6 K A5 2.2 1.8 18x10 6 K B5 7.1 4.0 26x10 6 K B0 17.0 7.6 34x10 6 K O5 39.8 17.8 34x10 6 K More massive stars have hotter cores. O, B & A stars fuse H via the CNO cycle while G, K & M stars its via the p-p chain.

14. The CNO cycle and pp chain: Another reaction chain is called the CNO cycle, where carbon acts to catalyze the nuclear reactions. Removes the slow beta-decay, but increases the Coulomb barrier. The net result is the same as the pp chain, that 4 H nuclei are fused into 1 He nuclei.

15. The CNO cycle and pp chain: More massive stars have higher central temperatures when on the main sequence and CNO cycle is faster than the pp chain and is responsible for most of the energy production:

16. Energy Transport Heat energy flows from high temperature regions to low temperature regions. Thus, heat flows from center of star to the surface. Mechanisms of heat flow are conduction, convection and radiation, as illustrated below:

17. Energy Loss In the inner regions of stars like our Sun, heat is transported by radiation. Photons are produced by the hot gas. These photons are repeatedly scattered or absorbed by interactions with matter. This is a random walk process that requires of order 100,000 years for the photon energy to be transferred outward.

18. Energy Loss In the outer regions of stars like our Sun, heat is transported by convection. See the convective cells near the surface. At the surface the energy is radiated away.

19. Photosphere Visible surface of Sun where continuous spectrum originates. Solar granulation is result of convection of energy to surface – rising and falling columns of gas. The image to the left shows the solar granulation. The brighter areas are the rising columns of hot gas, and the darker regions are the falling cooler gas.

20. Stellar Models There a number of basic equations that need to be solved for a star's structure: Mass continuity equation [dM/dr = 4 π r 2 ρ(r)] Hydrostatic Equilibrium eq. [dP/dr = -G ρ(r) M(r)/r 2 ] Equation of State [P = ρ k T/m] Energy generation equation Energy transport equation These can be solved numerically to get a complete description of temperature and density as a function of radius. Once we pick a mass, then all of the properties of a star on the main sequence are determined (only very slight dependence on chemical composition).

21. Model of the Sun Model of the Sun

22. Photosphere The gas immediately above the photosphere is less dense and cooler (as low as 4200 K for the Sun) and is the origin of the Sun’s absorption line spectrum.

23. Corona Upper Atmosphere of the Sun 5800°K10,000°K 1,000,000 °K Outer atmosphere of the Sun heated by the transfer of kinetic energy in the convective motion of gas in photosphere upward by sound/magnetic waves. Kinetic energy converted to thermal energy in upper atmosphere.

24. Main Sequence Stars on the main sequence are fusing hydrogen into helium. These stars are in hydrostatic equilibrium, so the inward force of gravity is balanced by the outward force of gas pressure. The properties of main sequence stars are determined by mass – this is a mass sequence The mass of stars varies from 0.08 M ⊙ to larger than 100 M ⊙. Below 0.08 M ⊙, the central temperature is too low for fusion of hydrogen to occur.

25. Main Sequence Lifetimes No matter if a star is fusing hydrogen by the pp chain or CNO cycle, there is only a limited core hydrogen supply. The lifetime of a star on the main sequence must depend on the hydrogen supply and rate at which the hydrogen is being consumed: Lifetime = fuel supply/consumption of fuel Thus, the main sequence lifetime must scale approximately by: Mass divided by Luminosity

26. Main Sequence Lifetimes Assume the main sequence lifetime scales as M/L, we can use Sun as a calibration to derive: lifetime = 10 10 yrs (M/M ⊙ )/(L/L ⊙ ) Remember the mass-luminosity relation for stars on the main sequence (see figure). Note that the luminosity increases rapidly with increasing mass. Since the luminosity increases faster than mass, more massive stars have shorter main sequence lifetimes.

27. Main Sequence Lifetimes Using Stellar Properties from Appendix E and the following relationship: lifetime = 10 10 yrs (M/M ⊙ )/(L/L ⊙ ): Mass (M ⊙ ) Luminosity (L ⊙ ) M-S Lifetime 0.2 7.9x10 -3 3x10 12 yrs 0.5 0.063 8x10 10 yrs 1.0 1.0 1x10 10 yrs 2.2 20 1x10 9 yrs 3.6 80 5x10 8 yrs 7.1 630 1x10 8 yrs 17.0 1.3x10 4 1x10 7 yrs 39.8 3.2x10 5 1x10 6 yrs