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Astro 160: The Physics of Stars Served by Roger Griffith Nutritional Facts: Serving size: 1 Semester (16 weeks) Servings per container: many problems and solutions Problem set 2 Problem # 1 In class, I derived the relationship between the luminosity and mass of stars under the assumption that energy is transported by radiative diffusion and that the opacity is due to Thomson scattering. We will carry out many related estimates so it is important to become familiar with this process. Consider a star in hydrostatic equilibrium in which energy transport is by radiative diffusion. The star is composed of ionozed hydrogen and is supported primarily by gas pressure. (a) . Derive an order of magnitude estimate of the luminosity L of a star of mass M and radius R if the opacity is due to free-free absorption, fo which κ ≈ 10 23 ρ T − 7 / 2 cm 2 g − 1 ( ρ is in cgs). We know that the radiation flux is given by F rad ∼ caT 3 κρ ∇ T where we know that a is the radiation constant, c is the speed of light, T is the temperature, κ is the opacity, which in our case is given by free-free absorption, ρ is the mass density and ∇ T is the temperature gradient. We have the following relationships ρ ∝ M R 3 ∇ T ∝ dT dR ∝ T R − T c R R − R C ∝ − T C R κ ∝ 10 23 ρ T − 7 / 2 given these relationships we can find F rad ∝ caT 15 / 2 R 5 M 2 we also know that the luminosity can be written as L = 4 π r 2 F rad ⇒ F rad = L 4 π r 2 which gives us L ∝ caT 15 / 2 R 7 M 2 we can find the temperature by using the virial theorem which can be written as T ≈ GMm p μ 3 Rk 1

where k is now the boltzman constant. Substituting this expression into the above equation yields L ∝ caR − 1 / 2 M 11 / 2 parenleftbigg Gm p k parenrightbigg 15 / 2 this gives us an order of magnitude estimate of the luminosity of a star with mass M and radius R . (b) . If all stars have roughly the same central temperature, and are supported by gas pressure, what is the mass-luminosity scaling (proportianality) relationship for stars? we now know that the luminosity scales as L ∝ M 11 / 2 R − 1 / 2 we can find the relationship between the mass M and the radius R of a star by using hydrostatic equilibrium. dP dr = − GM r 2 ρ P c ∝ M R ρ ρ T ∝ M R ρ M ∝ R since T is constant, substituting this into the luminosity relationship yields L ∝ M 5 (c) . Give a quantitative argument as to whether free-free opacity dominates electron scattering opacity in stars more massive that the sun or in stars less massive that the sun. We can solve this problem by looking at the defenition for the opacity in free-free absorption, which can be written as with T constant κ ∝ ρ ρ ∝ M R 3 M ∝ R thus we find κ ∝ 1 M 2 this expression tells us that the lower the mass of the star the higher the opacity, thus in lower mass stars the free-free opacity dominates. Problem # 2 The central density and temperature of the sun are ρ c ≃ 150 g cm 3 and T c ≃ 1 . 5 × 10 7 K. For the conditions at the center of the sun, answer the following questions. Assume that the sun is composed solely of ionized hydrogen. (a) . What is the mean free path of an electron due to electron-electron Coulomb collisions? What is the typical time between collisions? 2

We know that the mean free path is given by l = 1 n e σ we know that for a completely ionized hydrogen gas that n e ∼ n p ∼ ρ m p and the interaction cross section is given by σ = π r 2 where r is the Coulomb radius found comparing the thermal energy to the Coulomb energy e 2 r ∼ kT r ∼ e 2 kT σ ∼ π e 4 ( kT ) 2 using these relationships we find the mean free path to be l ∼ m p ρ c π parenleftbigg kT e 2 parenrightbigg 2 and the collision time is given by t col = l v e and the velocity can be found by using 3 2 kT = 1 2 m e v 2 v = radicalBigg 3 kT m e thus the time is given as t e = radicalbigg m e 3 kT m p ρ c π parenleftbigg kT e 2 parenrightbigg 2 (b) . What is the mean free path of an proton due to proton-proton Coulomb collisions? What is the typical time between collisions? Which occurs more rapidly, electron-electron or proton-proton Coulomb collisions? The mean free path of proton-proton collisions would be the same as for the electron-electron collosion because the gas is completely ionized. The mean free path is given by l ∼ m p ρ c π parenleftbigg kT e 2 parenrightbigg 2 The collision time would be the same except now that the mass is the mass of the proton not the electron. i.e t p = radicalbigg m p 3 kT m p ρ c π parenleftbigg kT e 2 parenrightbigg 2 3

we can now see the collision times for the electron-electron collision occurs more rapidly due to the mass being so much smaller. t e ≪ t p (c) . Which opacity is more important for photons, Thomson scattering or free-free absorption? We know that κ T = n e σ T ρ c = 2 σ T m p ∼ 0 . 80 and κ F = 10 23 ρ T − 7 / 2 ∼ 1 . 15 free-free absorption dominates the opacity for photons in this case? not sure why this is. We know that Thomson scattering is the primary way that photons move the energy out. (d) . What is the mean free path of a photon? How does this compare to the mean free path of an electron (this should give you a feel for why photons are far more effective at moving energy around in stars)? What is the typical time between photon absorptions/scattering? we know that the mean free path of a photon is given by l = 1 n e σ T where σ T = 8 π 3 bracketleftbigg e 2 4 πε 0 m e c 2 bracketrightbigg 2 = 6 . 65 × 10 − 25 cm 2 which yields l photon = m p 2 ρσ T ∼ 8 . 3 × 10 − 3 cm l electron = m p ρ c π parenleftbigg kT e 2 parenrightbigg 2 ∼ 8 . 9 × 10 − 7 cm The typical time for a photon collision is given by t = l p c ∼ 2 . 8 × 10 − 13 s (e) . For a photon undergoing a random walk because absorption/scattering, how long would it take to move a distance R sun given the results in (d)? For comparison, it would take 2.3 seconds moving at the speed of light to travel a distance R sun in the absence of scattering/absorption. We know that the diffusion time can be acquired with t di f f = thermal energy L ∼ R 2 lc nkT aT 4 ∼ R 2 nk lcaT 3 ∼ R 2 2 ρ k m p lcaT 3 we know that the average time for a photon to leave the star is given by t di f f ∼ R 2 sun l ph c ∼ 10 4 yr 4

Problem # 3 How old is the sun? In this problem we illustrate how the naturally occuring radioactive isotopes of uranium, U 235 and U 238 can be used to determine the age of the rocks. Both isotopes decay via a sequence of α -decays and β -decays to form stabel isotopes of lead: the decay chain of U 235 ends up with Pb 207 , and the decay chain of U 238 ends up with Pb 206 . As a result, the number of uranium nuclei in a rock decays exponetially with time in accord with: N 5 ( t ) = N 5 ( 0 ) e − λ 5 t and N 8 ( t ) = N 8 ( 0 ) e − λ 8 t To avoid clutter, the last digit of the mass number of the isotope has been used as a subscript label. The decay constants λ 5 and λ 8 for the two isotopes corresponds to half-lives of T 5 = ln2 λ 5 = 0 . 7 × 10 9 yrs T 8 = ln2 λ 8 = 4 . 5 × 10 9 yrs The magnitudes of these half-lives are ideally suitable to the determination of the ages of the rocks which are over a billion years old. Now consider a set of rock samples which were formed at the same time, but with different chemical compositions. They differ in chemical composition because different chemical elements are affected differently by the processes of rock formation. However rock formation processes do not favour one isotope over another. For example, on formation, the relative abundances of U 235 and U 238 should be the same in every sample. But these abundances will change with time as the deacy of U 235 and U 238 produce nuclei of Pb 207 and Pb 206 . • Consider the ratio of the increase in the number of Pb 207 nuclei relative to the increase of Pb 206 nuclei. Show that this ratio is the same for all rock samples which were formed at the same time, and that it is given by N 7 ( t ) − N 7 ( 0 ) N 6 ( t ) − N 6 ( 0 ) = N 5 ( t ) N 8 ( t ) e λ 5 t − 1 e λ 8 t − 1 We know that the ratio of the two isotopes can be written as N 7 ( t ) − N 7 ( 0 ) N 6 ( t ) − N 6 ( 0 ) = N 5 ( t ) − N 5 ( 0 ) N 8 ( t ) − N 8 ( 0 ) and given the first expression given in this problem, which can also be written as N 5 ( 0 ) = N 5 ( t ) e λ 5 t and N 8 ( 0 ) = N 8 ( t ) e λ 8 t substituting this into our previous expression yields N 7 ( t ) − N 7 ( 0 ) N 6 ( t ) − N 6 ( 0 ) = N 5 ( t ) N 8 ( t ) e λ 5 t − 1 e λ 8 t − 1 which is what we were asked to show. • Consider a graph in which the measured abundances in the rock samples of Pb 207 and Pb 206 are plotted, N 7 ( t ) along the y -axis and N 6 ( t ) on the x -axis. Show that a straight line will be obtained if all the samples were formed at the same time. 5