pdfcoffee.com-astro160-problems-solutions-philipsquestion24pg40

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 know that N 7 ( t ) = N 5 ( t ) N 8 ( t ) e λ 5 t − 1 e λ 8 t − 1 · N 6 ( t ) where N 5 ( t ) N 8 ( t ) e λ 5 t − 1 e λ 8 t − 1 = constant • Given that the current ratio of naturraly occurring U 235 to U 238 is 0.0071, evaluate the gradient of the straight line for rock samples of age (a) 1 billion years, (b) 3 billion years and (c) 5 billion years. We know that the gradient of the straight line is just the constant in front of N 6 ( t ) so we just have to plug in numbers (a). t = 1 billion years. We know that λ 5 ∼ 9 . 90 × 10 − 10 yr − 1 λ 8 ∼ 1 . 5 × 10 − 10 yr − 1 given these and the fact that we know the ratio between U 235 and U 238 we can find the gradient, for 1 bilion years we get 0 . 0071 · e λ 5 t − 1 e λ 8 t − 1 = 0 . 0715 For 3 billion years we get 0 . 0071 · e λ 5 t − 1 e λ 8 t − 1 = . 231 and finally for 5 billion years we get 0 . 0071 · e λ 5 t − 1 e λ 8 t − 1 = . 891 Problem # 4 Radiative Atmospheres In this problem we will solve for the structure of the outer part of a star assuming that energy is transported solely by radiative diffusion (which is not the case in the sun, but is the case in stars more massive than the sun). The star has a mass M and a luminosity L . Assume that the luminosity and mass are approximately constant at the large radii of interest, that gas pressure dominates, and that the opacity is due to electron scattering. Do not assume that the atmosphere is thin (i.e even though M r ≈ constant = M , because r changes, the gravitional acceleartion is not constant). Write down the equations for hydrostatic equlibrium and energy transport by radiative diffusion. Use these to calculate dP rad / dP , the change in radiatio pressure with pressure in the atmosphere. What does this result imply for how the ratio of gas pressure to radiation pressure changes as a function of the distance in the atmosphere? Show that your result for dP rad / dP implies that ρ ∝ T 3 and P ∝ ρ 4 / 3 for radiative atmospheres (in the language that we will use in the next week, this means that the radiative part of the star is an n=3 polytrope). since we know what the radiation pressure is we can find what the change is with respect to r P rad = 1 3 aT 4 dP rad dr = 1 3 a d dr ( T 4 ) = 4 3 T 4 ∇ T 6

(b) . Assume that radiative diffusion dominates energy transport in stars and that the opacity is due to Thomson scattering. Use a scaling argument to estimate the mass M (in M sun ) at which the luminosity of a star is ≈ L edd . We can do an order of magnitude estimate with respect to the sun by L ∝ M 3 L L sun = parenleftbigg M M sun parenrightbigg 3 and substituting the Eddington luminosity for L we find that M is given by M = parenleftbigg 4 π cGM L sun κ T parenrightbigg 1 / 2 M 3 / 2 sun ≈ 180 M sun Problem # 2 The physical quantities near the center of a star are given in the following table. Neglecting radiation pressure and assuming the average gas particle mass ¯ m is 0.7 amu, determine whether energy transport is convective or radiative. r m ( r ) L r T r ρ ( r ) κ 0.1 R sun 0.028 M sun 24.2 L sun 2.2 × 10 7 K 3 . 1 × 10 4 kg m − 3 0.040 m 2 kg − 1 Using equation ?? from Phillips bracketleftbigg L ( r ) m ( r ) bracketrightbigg crit = γ − 1 γ 16 π Gc κ P r P and the following relationships P r = 1 3 aT 3 P ∼ P g = ρ ( r ) ¯ m k b T γ = 5 3 κ T = 0 . 04 m 2 / kg we find bracketleftbigg L ( r ) m ( r ) bracketrightbigg crit = 2 5 16 π Gc κ T aT 3 ¯ m 3 ρ ( r ) k b 0 . 175 W kg > . 07 W kg which implies that the energy transport of this star is primarily due to convection. Problem # 3 The surface of a star (the “photosphere”) is the place where the mean free path of the photons ℓ is comparable to the scale-height h of the atmosphere . At smaller radii (deeper in the star), the density is higher and ℓ ≪ h , which implies that the photons bounce around many times; at larger radii ρ is smaller, ℓ ≫ h , and the photons are rarely absorbed and so travel on straight lines to us. Thus ℓ ≈ h is a good approximation to the place in the atmosphere of a star where most of the light we see originates. a) The temperature at the photosphere of the sun is 5800 K. Estimate the mass density ρ in the photo- sphere. Assume that Thompson scattering dominates the opacity. 9

and finally the temperature can be found by using P = nk b T = 2 ρ m p k b T ⇒ T ( r ) = m p P 2 ρ k b and substituting the P and ρ from the previous expressions we find T ( r ) = m p K 3 / 5 2 k b bracketleftbigg 2 5 K − 3 / 5 GM bracketleftbigg 1 R − 1 R c bracketrightbigg + P 2 / 5 c bracketrightbigg b) In detailed solar models, the pressure at the base of the convection zone is ≈ 5 . 2 × 10 13 dyne/cm 2 and the density is ρ ≈ 0 . 175 g cm − 3 . Using your solution from a), estimate the radius of the base of the convection zone R c . Compare this to the correct answer of R c ≈ 0 . 71 R sun If we solve the density equation for R c we find 1 R c = 1 R − bracketleftBig ( ρ K 3 / 2 ) 2 / 3 − P 2 / 5 c bracketrightBig · 5 K 3 / 5 2 GM and plugging in values we find that 1 R c = 1 . 998 × 10 − 9 m − 1 ⇒ R c ≈ 5 . 11 × 10 8 m = 0 . 72 R sun c) In your model, what is the temperature of the sun at 0.99 R sun , 0.9 R sun , and at the base of the solar convection zone. This gives you a good sense of how quickly the temperature rises from its surface value of ≈ 5800 K as one enters the interior of the sun. To find the temperature as a function of radius we would use the temperature equation derived from part (a). i.e T ( r = 0 . 99 R sun ) = m p K 3 / 5 2 k b bracketleftbigg − 2 3 K − 3 / 5 GM r 25 66 · R sun + P 2 / 5 c bracketrightbigg ≈ 4 . 1 × 10 4 K T ( r = 0 . 90 R sun ) = m p K 3 / 5 2 k b bracketleftbigg − 2 3 K − 3 / 5 GM r 15 18 · R sun + P 2 / 5 c bracketrightbigg ≈ 5 . 1 × 10 5 K T ( r = 0 . 72 R sun ) = m p K 3 / 5 2 k b P 2 / 5 c ≈ 1 . 8 × 10 6 K Problem set 4 Problem # 1 I mentioned in class that there are two ways to estimate the energy carried by convection. The first is that the energy flux is F c ≈ 1 / 2 ρ v 3 c ≡ F c , 1 where v c is the characteristic velocity of the convective motions. 12

Problem # 3 Polytropes (a). The mass M of a star is given by M = Z R 0 4 π r 2 ρ ( r ) dr Use the Lane-Emden equation for polytropes, and the dimensionless density and radius defined in lecture, to rewrite this in terms of the central density of the star as ρ c = ¯ ρ a n = parenleftbigg 3 M 4 π R 3 parenrightbigg a n where a n is a dimensionless number, the ratio of the central density to the mean density of the star. a n is a function that you should determine that depends only on the solution to the Lane-Emden equation (you cannot actual evaluate a n in general without numerically solving for θ [ ζ ] , so your answer will just be in terms of the solution to the Lane-Emden equation). Since we know that Θ = parenleftbigg ρ ( r ) ρ parenrightbigg 1 / n ξ = r a given these two relations we can find ρ ( r ) = Θ n ρ c r 2 = a 2 ξ 2 a = R dr = Rd ξ and from the Lane-Amden equation we know d d ξ parenleftbigg ξ 2 d Θ d ξ parenrightbigg = − ξ 2 Θ n given these following relationships we find that M = − Z 1 0 4 π R 3 ξ 2 Θ n ρ c d ξ = − 4 π R 3 ρ c Z 1 0 d d ξ parenleftbigg ξ 2 d Θ d ξ parenrightbigg d ξ thus we find that a n is given by a n = − 1 3 Z 1 0 d d ξ parenleftbigg ξ 2 d Θ d ξ parenrightbigg d ξ and we can finally show that ρ c = 3 M 4 π R 3 a n (b). Show that the central pressure of a polytrope can be written as P c = 4 π G ρ 2 c a 2 n + 1 where a negationslash = a n is the constant (with units of length) defined in lecture (note that the polytropic relation P = K ρ γ can be used to write K = P c ρ − γ c . Use this result and (a) to derive an expression for the central pressure of a polytropic model of the form P c = parenleftbigg GM 2 R 4 parenrightbigg c n 15

we find ρ c ≈ 77 . 26 g/cm 3 P c ≈ 1 . 24 × 10 15 T c ≈ 1 . 02 × 10 7 K thus we can see that the γ = 4 / 3 polytrope best represents the values observed in the sun, this is mainly due to the fact that the center of the sun is radiative and not convective. Since we now know that P ∝ ρ 4 / 3 radiative P ∝ ρ 5 / 3 convective Problem # 4 Consider a pre-main sequence “star” (gas cloud) of mass M undergoing Kelvin-Helmholz contraction. In class, we showed that fully convective stars move down the Hayashi line with T ef f ≈ constant. But stars with M > 0 . 3 M sun do not end up fully convective on the main sequence and so must go through a phase of KH contraction in which energy transport is dominated by photons. Assume throughout this problem that gas pressure dominates and that free-free ab- sorption dominates the opacity (because the temperature is lower during KH contraction than on the main sequence, free-free absorption tends to be even more important). Motivated by HW #2 Problem 1, assume that the luminosity of a star in which photons carry the energy out and the opacity is dominated by free-free absorption is given by L ≈ L sun ( M / M sun ) 11 / 2 ( R / R sun ) − 1 / 2 . (a). Determine how the radius, luminosity, and effective temperature vary as a function of time and mass M for a radiative star undergoing KH contraction. Don’t worry about the constants in these relations; all you need to calculate are proportionalities (i.e., how do the various quantities depend on time and mass M ). Do the luminosity and effective temperature increase or decrease as the star contracts? since we know that P g > P r κ = κ f f thus we know that L f f ≈ L sun parenleftbigg M M sun parenrightbigg 11 / 2 parenleftbigg R R sun parenrightbigg − 1 / 2 ≈ L rad ≈− 1 2 GM 2 R 2 dR dt since we are doing proportionalities we find M 2 R 2 t R T ∝ M 11 / 2 R − 1 / 2 so we find that the radius scales as R ∝ 1 M 7 t 2 plugging this into L ∝ M 11 / 2 R − 1 / 2 ∝ M 9 t thus L ∝ M 9 t and to find the temperature L ∝ R 2 T 4 ef f T ef f ∝ parenleftbigg L R 2 parenrightbigg 1 / 4 ∝ M 23 / 4 t 5 / 4 18

froma a purely physical argument we know that things cannot fall faster than gravity can pull it. Thus t KH ≫ t f f and saying t f f ≫ t KH would be an unphysical statement due to the knowledge we have about gravity. We also know that when things collapse that the radius gets smaller, hence collapse and from the relation- ships of time in both free-fall and Kelvin-Helmholtz contraction we can see that as R ↓ that t f f ↓ thus the only possible solution is that t KH ≫ t f f . ( b). Estimate the critical radius R c ( inR sun ) at which t KH ≈ t f f for a given cloud of mass M (in M sun ). Assume, as we did in class, that the cloud is fully convective at early times. Show that for R < R c , the cloud undergoes KH contraction according to your criterion from a). Recall that the luminosity of a fully convective star is L ≈ 0 . 2 L sun ( M / M sun ) 4 / 7 ( R / R sun ) 2 . We can find the critical radius by setting the above expression equal to each other, i.e 1 radicalbig G ( ρ ) = bracketleftBigg parenleftbigg M M sun parenrightbigg 1 / 2 R sun R c bracketrightBigg 3 ( 2 × 10 7 yrs ) re-arranging this expression for R c we find R c = parenleftbigg M M sun parenrightbigg 1 / 2 R sun ( G ( ρ ) ) 1 / 6 ( 2 × 10 7 yrs ) 1 / 3 and letting ( ρ ) = 3 M 4 π R 3 c ≈ M R 3 c we find R c ≈ M 4 / 9 R 2 / 3 sun ( 2 × 10 7 yrs ) 1 / 3 G 1 / 9 M 1 / 3 sun ≈ M 4 / 9 7 . 6 × 10 − 3 if we put the mass off the sun we get R c ≈ 4 . 79 × 10 10 cm Since we know that t f f ∝ R 3 / 2 t KH ∝ 1 R 3 we can see that R ↑ t KH ↓ t f f ↑ R ↓ t KH ↑ t f f ↓ from these two relationships we can see that for KH contraction to occur that t KH ≫ t f f and also that R < R c . (c). What is the central temperature of the (fully convective) cloud (in K) as a function of its mass M (in M sun ) when R = R c ? We know that the temperature of a fully convective object is given by T c = d n GM Rk b parenleftbigg 3 a n 4 π parenrightbigg 1 / 3 m p 2 21

and since we know that fully convective stars have a polytropic index of γ = 5 / 3. Knowing this we find n = 3 a n = 5 . 99 d n = 0 . 477 thus T c ≈ 2 . 16 × 10 − 15 M R c ≈ 2 . 79 × 10 − 15 M 5 / 9 and if we want the temperature of a collapsing gas cloud with respect to M sun we get T c ≈ 1 . 9 × 10 4 K Problem # 2 The globular cluster M13 in Hercules contains about 0.5 million stars with an average mass of about half the solar mass. Use Jeans criteria to check whether this cluster could have formed in the early universe just after the time when the universe was cool enough for the electrons and nuclie to form neutral atoms; at this time the density of the universe was ρ ≈ 10 − 27 kg m − 3 and the temperature was T ≈ 10 4 K . Using the Jeans mass equation M j = parenleftbigg k b Gm p parenrightbigg 3 / 2 T 3 / 2 √ ρ using the values given we find M j ≈ 1 . 37 × 10 42 g and the mass of M13 is M 13 ≈ 0 . 5 × 10 6 · M sun ≈ 9 . 95 × 10 38 g and we can see that M j ≫ M 13 which means that this cluster could not have formed in the early universe. Things only collapse if the mass is greater than the Jeans mass Problem # 3 The binding energy per nucleon for 56 Fe is 8.8 MeV per nucleon. Estimate the energy released per kilogram of matter by the sequnce of reactions which fuse hydrogen to iron. We know that the enery released will be given as E tot = E b nucleon × N nucleon and the number of nucleons are given by N nucleon = M m p ≈ 1 kg m p ≈ 5 . 98 × 10 26 nucleon thus E tot ≈ 5 . 27 × 10 27 MeV (b). Consider two hypothetical stars of the same mass M and the same luminosity L (that is constant in time). The stars are initially pure hydrogen. In star A, fusion proceeds until the entire star is converted into 22

He. In star B, fusion proceeds until the entire star is converted into Fe. Which star has a longer lifetime, and by how much? We know that L He = E He t He L Fe = E Fe t Fe and since we know that these two luminosities are theoretically equal E He t Fe = E Fe t He which gives t He = E He E Fe t Fe = 6 . 4 8 . 5 t Fe ≈ 0 . 72 t Fe thus we see that t He < t Fe we can see that the time for all of the hydrogen to fuse into helium is less then the time for all of the hydrogen to fuse into iron so the star that is converted to iron has a longer lifetime. Problem # 4 (a) What is the classical distance of closest approach for two protons with an energy of 2 keV (the mean thermal energy at the center of the sun)? Estimate the probability that the protons tunnel through the Coulomb barrier trying to keep them apart. Answer the same two questions for two 4 He nuclei and for a proton and a 4 He nucleus with the same energy of 2 keV. The classical distance of closest approach is given by r c = e 2 Z 1 Z 2 E 0 ≈ 7 . 2 × 10 − 11 cm We know that the propbability for a particle-particle interaction is given by P = e − parenleftBig E G E parenrightBig 1 / 2 where E G = 2 π 2 α 2 Z 2 1 Z 2 2 ( m r c 2 ) is the Gamow energy. For a proton-proton interaction we find E G ≈ . 493 MeV thus the probability is given as P ≈ 1 . 51 × 10 − 7 for a He-He interaction we find r c = e 2 Z 1 Z 2 E 0 ≈ 2 . 8 × 10 − 10 cm E G = 32 π 2 α 2 2 m p c 2 ≈ 31 . 6 MeV and the probabilty is given by P ≈ 2 . 50 × 10 − 55 23

for the proton-He interaction we find r c = e 2 Z 1 Z 2 E 0 ≈ 1 . 44 × 10 − 10 cm E G ≈ 3 . 15 MeV thus the probability is given by P ≈ 5 . 8 × 10 − 18 (b) What energy E would be required for i) the two 4 He nuclei, ii) the proton and the 4 He nucleus, and iii) two 12 C nuclei to have the same probability of penetrating the Coulomb barrier as the two protons? For particles with energies equal to the mean thermal energy of the plasma, what temperatures do these correspond to? Since we know that E = E G ( ln P ) 2 for the He-He interaction we find E He − He ≈ 0 . 125 MeV for the proton-He interaction we find E p − He ≈ . 013 MeV for the carbon-carbon interaction the Gamow energy is given by E G ≈ 2592 π 2 α 2 6 m p c 2 ≈ 7 . 7 GeV and thus E c − c ≈ 31 MeV we know that T ≈ E k b so T He − He ≈ 1 . 45 × 10 9 K T p − He ≈ 1 . 5 × 10 8 K T c − c ≈ 3 . 6 × 10 11 K Problem # 5 Calculations of nuclear reaction rates are done in the center of mass (COM) frame, so it is useful to remember a few results about the COM. Consider two particles of mass m 1 and m 2 with positions x 1 and x 2 and velocities v 1 and v 2 . (a) .What is the velocity of the COM? We kbnow that the center of mass is given by com = m 1 r 1 + m 2 r 2 m 1 + m 2 24

so the velocity would be v com = m 1 v 1 + m 2 v 2 m 1 + m 2 (b). What are the velocities of each of the two particles in the COM reference frame (i.e., in the frame for which the COM is at the origin)? We know that the relative velocities are given by v rel − 1 = v 1 − v com v rel − 2 = v 2 − v com a bit of algebra yields v rel − 1 = m 2 m 1 + m 2 ( v 1 − v 2 ) v rel − 2 = m 1 m 1 + m 2 ( v 2 − v 1 ) (c). What is the total KE of the two particles in the COM frame? Show that this is equal to the KE of the reduced mass moving at the relative velocity, as claimed in class. We know that the total kinetic energy is given by K tot = 1 2 m 1 v 2 rel − 1 + 1 2 m 2 v 2 rel − 2 = m 1 m 2 2 ( m 1 + m 2 ) 2 ( m 2 ( v 1 − v 2 ) 2 + m 1 ( v 2 − v 1 ) 2 ) = m 1 m 2 2 ( m 1 + m 2 ) ( v 1 − v 2 ) 2 K tot = 1 2 m r ( v 1 − v 2 ) 2 where m r is the reduced mass. Problem set 7 Problem # 1 The Main Sequence for Fully Convective Stars In this problem we will determine the main sequence for fully convective low mass stars. We showed in lecture that fully convective stars have T ef f ≈ 4000 ( L / L sun ) 1 / 102 ( M / M sun ) 7 / 51 K (I actually derived a coefficient of 2600 K in lecture but commented that more detailed calculations get something similar but with the coefficient closer to the value of 4000 K used here). We can also write this result as 25

L ≈ 0 . 2 ( M / M sun ) 4 / 7 ( R / R ) 2 sun L sun ≡ Lconv I called this luminosity L conv since it is derived from the properties of energy transport alone (convec- tive interior + radiative atmosphere with H − opacity). The luminosity of a star is also given by L f usion = 4 π r 2 ρε ( T , ρ ) dr where ε is due to the proton-proton chain for low mass stars (this was given in lecture). As discussed in class, the main sequence is determined by the requirement that the energy escaping the star (in this case by convection) is equal to the energy generated in the star (in this case by pp fusion), i.e., that L conv = L f usion . a) Use scaling arguments to derive the power-law relations R ( M ) , L ( M ) , T c ( M ) , and L ( T ef f ) (the HR diagram) for fully convective stars, like we did for other examples in lecture. Approximate ε ∝ ρ T β with an appropriate choice of β (recall that low mass stars will have somewhat lower central temperatures than the sun, closer to ≃× 10 6 K, as you will see in part b). We know that L conv ∝ M 4 / 7 R 2 L f us ∝ R 3 ε ( ρ , T ) ∝ R 3 ρ 2 T β we can find what β is by β = − 2 3 + parenleftbigg E G 4 kT parenrightbigg 1 / 3 given that we know what the temperature is and also what E G for p-p reaction E G ≈ 500 keV T ≈ 5 × 10 6 K we find that β = 5 . 92 ≈ 6 . 0 we also know ρ ∝ M R 3 we know that in steady state L f usion = L conv thus we can find M 4 / 7 R 2 ∝ R 3 parenleftbigg M 2 R 6 parenrightbigg T 6 c ⇒ T 6 ∝ M − 5 / 21 R 5 / 6 we know from the Virial temperature, assuming gas pressure dominates T ∝ M R thus we find T c ∝ M 25 / 77 knowing this we can now find R ∝ M 52 / 77 with this and the relationship for the convective luminosity we find L ∝ M 4 / 7 R 2 ∝ M 148 / 77 26

with this we can now find what the effective temperature as a function of mass is, i.e L ∝ R 2 T 4 ef f thus T 4 ef f ∝ L R 2 ∝ M 4 / 7 which yields T ef f ∝ M 1 / 7 to find what the luminosity as a function of the effective temperature is (HR diagram) M ∝ T 7 ef f which yields L ∝ T 148 / 11 ef f In a) you just determined a scaling relation between stars of different mass, but not the absolute values of L , T ef f , etc. In class, we did the latter by scaling to the sun. Note, however, that it is not reasonable to estimate the properties of low mass stars by scaling from the properties of the sun, since the sun is not a fully convective star! Instead we need to actually determine the structure of some fully convective star. This is what we will do in the rest of the problem. We can significantly improve on the above scaling arguments by using the fact that fully convective stars are n = 3 / 2 polytropes. It turns out that for a polytrope, in equation (1) can be Taylor expanded near the center to yield L f usion ≃ 2 . 4 ε c M ( 3 + β ) 3 / 2 where I have again approximated ε ∝ ρ T β and where ε c is evaluated at the center of the star. I am not asking you to prove equation (2). You will have to trust me. Note that for a typical value of β for the pp chain, equation (2) says that L f usion ≃ 0 . 1 ε c M . This makes sense because fusion only takes place at the center of the star (not all of the mass participates). b) Use the results for n = 3 / 2 polytropes from HW 4, Problem # 3, to write the central temperature of the star T c , central density ρ c , and pp energy generation at the center of the star ε c in terms of the mass M and radius R . Assume X = 0 . 7 and μ = 0.6 (typical for stars just reaching the main sequence). Note that you should give expressions for T c , ρ c , and ε c here, with constants and real units, not just scaling relationships. So that the constants in front of your expressions are reasonable, please normalize M to M sun and R to R sun . The general expressions given by HW 4 problem #3 are T c = d n GM Rk b parenleftbigg 3 a n 4 π parenrightbigg 1 / 3 μm p P c = GM 2 R 4 c n ρ c = 3 M 4 π R 3 a n we found that for a n = 3 / 2 polytrope a n = 5 . 99 c n = 0 . 77 d n = 0 . 477 27

thus we find that T c = 0 . 322 m p GM sun k b R sun parenleftbigg M / M sun R / R sun parenrightbigg ρ c = 1 . 43 M sun R 3 sun parenleftbigg M / M sun ( R / R sun ) 3 parenrightbigg plugging in all the constants yields T c ≈ 7 . 43 × 10 6 K parenleftbigg M / M sun R / R sun parenrightbigg ρ c ≈ 8 . 41 parenleftbigg M / M sun ( R / R sun ) 3 parenrightbigg we also know ε c = 5 × 10 5 ρ c X 2 T − 2 / 3 7 e − 15 . 7 T − 1 / 3 7 we also know that we can approximate this as ε c = A ρ T β 7 X 2 = A ρ T 6 7 X 2 setting this two expressions equal to each other we can find what A is, i.e letting T ≈ 10 7 K we find A = 5 × 10 5 e − 15 . 7 ≈ 0 . 076 thus we find ε c ≈ 0 . 076 ρ c T 6 7 X 2 ≈ 0 . 037 ρ c T 6 7 substituting T c and ρ c gives ε c ≈ 0 . 053 parenleftbigg M M sun parenrightbigg 7 parenleftbigg R sun R parenrightbigg 9 c) Use equation (2), the results of b), and L conv = L f usion on the main sequence to determine the R ( M ) , L ( M ) , T c ( M ) , and L ( T ef f ) relations for fully convective stars. If you use the same β , your expressions here should be the same as in a) except that you should now be able to determine the absolute normalization for R ( M ) , L ( M ) , etc., i.e., you have determined the true luminosity and radius of a ful ly convective star from first principles. In doing this problem, remember that β is temperature dependent so make sure you check that your value of β is reasonable given the resulting central temperature that you calculate. Using the results from b and also L f usion ≃ 2 . 4 ε c M ( 3 + β ) 3 / 2 ≈ . 09 ε c M ( β ≈ 6 ) thus we know that L con = L f usion 0 . 2 parenleftbigg M M sun parenrightbigg 4 / 7 parenleftbigg R R sun parenrightbigg 2 L sun = 0 . 0047 parenleftbigg M M sun parenrightbigg 7 parenleftbigg R sun R parenrightbigg 9 parenleftbigg M M sun parenrightbigg M sun rearranging this we find parenleftbigg R R sun parenrightbigg 11 = 0 . 023 M sun L sun parenleftbigg M M sun parenrightbigg 52 / 7 which yield R R sun ≈ 0 . 67 parenleftbigg M M sun parenrightbigg 52 / 77 28

to find for the temperature we can use T c ≈ 7 . 43 × 10 6 K parenleftbigg M / M sun R / R sun parenrightbigg using our previous resulst gives T c ≈ 1 . 1 × 10 7 K parenleftbigg M M sun parenrightbigg 25 / 77 to find for the luminosity we use L conv = 0 . 2 parenleftbigg M M sun parenrightbigg 4 / 7 parenleftbigg R R sun parenrightbigg 2 L sun plugging in for the radius we find L conv = 0 . 09 parenleftbigg M M sun parenrightbigg 148 / 77 L sun for the effective temperature we find L = 4 π R 2 σ T 4 ef f ⇒ 0 . 09 parenleftbigg M M sun parenrightbigg 148 / 77 L sun = 4 π R 2 σ T 4 ef f which can be simplified to 0 . 09 parenleftbigg M M sun parenrightbigg 148 / 77 L sun = 1 . 56 × 10 18 parenleftbigg M M sun parenrightbigg 104 / 77 T 4 ef f thus T ef f = 3868 K parenleftbigg M M sun parenrightbigg 1 / 7 which can also be written as M M sun = parenleftbigg T ef f 3868 K parenrightbigg 7 L ( T ef f ) = 4 . 85 × 10 − 50 L sun T 148 / 11 ef f d) What are your predicted luminosities, radii, and effective temperatures for main sequence stars with M = 0 . 1 and 0 . 3 M sun ? Compare your values to the values of L = 0 . 01 L sun , R = 0 . 3 R sun , and T ef f = 3450 K for M = 0 . 3 M sun and L = 10 − 3 L sun , R = 0 . 11 R sun , and T ef f = 3000 K for M = 0 . 1 M sun that I found in a graduate textbook (based on detailed models). Given our relationships we find for M = 0 . 1 M sun L = 1 . 1 × 10 − 3 L sun R = 0 . 14 R sun T ef f = 2783 K and for M = 0 . 3 M sun L = . 009 L sun R = 0 . 298 R sun T ef f = 3256 K 29

Problem # 2 Very Massive Stars Consider very massive stars with M ∼ 50 − 100 M sun . Recall that I showed in lecture and you showed on HW 3, Problem # 1, that in such stars, radiation pressure due to photons (a relativistic particle) is more important than gas pressure. Fusion is by the CNO cycle. Assume for now that energy is transported primarily by photons and that the opacity is due to Thomson scattering (reasonable for hot massive stars). a) Use scaling arguments to derive the power-law relations R ( M ) , L ( M ) , T c ( M ) , and L ( T ef f ) (the HR diagram) for very massive stars, like we did for other examples in lecture. Using radiative diffusion along with P rad = 1 3 aT 4 dP r dT = 4 3 aT 3 we also know dP dR = dP dT dT dR = 4 3 aT 3 dT dR = − ρ GM R 2 thus dT dR ∝ ρ M T 3 R 2 and radiative diffusion says L ρ R 2 T 3 ∝ dT dR ∝ ρ M T 3 R 2 which gives us L ∝ M using the Virial theorem, where P rad dominates rather than P gas we find T 4 ρ ∝ M R ⇒ T c ∝ M 1 / 2 R where the left hand term is from the radiation pressure, but since we know that is an energy density we must devide by the density to find what the energy is per particle. Now using the steady state for luminosity we find L ∝ M ρ T 18 where we chose β = 18 as a more appropriate value rather than the value given for the sun β = 20, this is motivated by the fact that more massive stars have somewhat higher temperatures, thus reducing β . We find T 18 ∝ 1 ρ ∝ R 3 M and using the result from Virial temperature we find M 9 R 18 ∝ R 3 M thus we find R ∝ M 10 / 21 and we also find for the central temperature T c ∝ M 1 / 42 30

and to find the effective temperature we know T 4 ef f ∝ L R 2 ∝ M R 2 ∝ M 1 / 21 simplifying gives T ef f ∝ M 1 / 84 and finally the luminosity as a function of T ef f is given by L ( T ef f ) ∝ T 84 ef f b) Estimate the fraction of the mass in the star that is undergoing convection (recall that fusion by the CNO cycle is very concentrated at small radii because of the strong temperature dependence). For comparison, detailed calculations show that the fraction of the mass that undergoes “core” convection increases from 10 % at 2 M sun to 75% at 60 M sun . The condition for convection is given by d ln T d ln P ≈ 1 4 P tot P rad L L Edd L r / L M r / M > γ − 1 γ since we know that γ = 4 3 P tot ≈ P rad gives us 1 4 L r L Edd M M r > 1 4 which simplyfies to L r L edd > M r M we know that in the limit that M → 150 M sun L r → L Edd , M r M < 1 which means that the fraction of the mass of the star that is undergoing convection approaches 1, which is 100% of the mass is undergoing convection. Its a little strange that stars that are much less massive than the sun and the stars that are much more massive than the sun are both almost fully convective. n=3/2 polytrope The previous case yielded a result for a n=3 polytrope, we find that for n = 3 / 2 polytrope γ = 5 3 M r M < 5 8 L r L Edd and in the limit where M → 150 M sun L r → L edd M r M < 5 8 31

This seems rather strange in the sense that stars that are approximately 60 M sun have a convective core that encompasses 75% of the mass, which means that the convective core decreases after M > 60 M sun ? c) Calculate the main sequence lifetime of a very massive star as a function of its mass M . Be sure to take into account the results of b). We know that the main sequence lifetime of a star is given by t MS ≈ E tot L Edd where E tot = NQ Q ≈ 7 MeV where that is the total energy per reaction, we also know N = M r m p = M m p ( n = 3 polytrope ) N = M r m p = 5 8 M m p ( n = 3 / 2 polytrope ) we also know that L Edd = 4 π cGM κ T so we find the main-sequence lifetime to be given as t MS ≈ κ T Q m p 4 π cG ( n = 3 polytrope ) t MS ≈ 5 κ T Q m p 32 π cG ( n = 3 / 2 polytrope ) we know that κ T ≈ 0 . 4 cm 2 / g Q ≈ 7 MeV ≈ 1 . 12 × 10 − 5 ergs thus we find that the main-sequence lifetime for both types of polytropes are given by 2 . 21 × 10 6 yr < t MS < 3 . 39 × 10 6 yr seems reasonable. Problem set 8 Problem # 1 Fermi gas 32

a) Above what density is a gas of room temperature fermions degenerate? Below what temperature would gas with the density of air be degenerate? We know that if the density of the gas is n g ≥ n Q where n Q is the quantum concentration, n Q is defined as n Q ≡ parenleftbigg 2 π ¯ mkT h 2 parenrightbigg 3 / 2 (1) and for a gas at room temperature to be degenerate n g ≥ n Q = parenleftbigg 2 π ¯ mkT h 2 parenrightbigg 3 / 2 ≈ 1 . 46 × 10 26 cm − 3 where we used T = 300K ¯ m = 28 m p due to the fact that air is mostly composed of N 2 . If we assume that the questio is only speaking about free electrons we get n g ≥ n Q = parenleftbigg 2 π m e kT h 2 parenrightbigg 3 / 2 ≈ 1 . 25 × 10 19 cm − 3 using the same temperature as before. To find the temperature at which gas with a density of air would be degenerate can by using the above expression, except now we must find what the density of air is at STP and use this, i.e n air = P kT = 2 . 52 × 10 19 cm − 3 = n Q and now using Equation 1 we find T = n 2 / 3 Q h 2 2 π ¯ mkT ≈ 9 . 2 × 10 − 3 K using ¯ m = 28 m p b) Compare the relative importance of the thermal energy, the electrostatic (Coulomb) energy between electrons and ions, and electron degeneracy (electron Fermi energy) in room temperature silver ( Z = 47; ρ ≃ 10g cm 3 ). Which dominates? We can write the thermal energy as E th ≈ 3 2 kT ≈ . 039 eV We can write the Coulomb energy as E coul ≈ Z 2 e 2 1 r we know r ∼ n − 1 / 3 ∼ parenleftBig ρ ¯ m parenrightBig − 1 / 3 33