Astro 160: The Physics of Stars

Astro 160: The Physics of StarsServed by Roger GriffithNutritional Facts:Serving size: 1 Semester (16 weeks)Servings per container: many problems and solutionsProblem set 2Problem # 1In class, I derived the relationship between the luminosity and mass of stars under the assumption thatenergy is transported by radiative diffusion and that the opacity is due to Thomson scattering. We willcarry out many related estimates so it is important to become familiar with this process. Consider a starin hydrostatic equilibrium in which energy transport is by radiative diffusion. The star is composed ofionozed 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 theopacity is due to free-free absorption, fo which κ ≈ 10 23 ρT −7/2 cm 2 g −1 (ρ is in cgs).We know that the radiation flux is given byF rad ∼ caT 3κρ ∇Twhere we know that a is the radiation constant, c is the speed of light, T is the temperature, κ is theopacity, which in our case is given by free-free absorption, ρ is the mass density and ∇T is the temperaturegradient. We have the following relationshipsρ ∝ M R 3∇T ∝ dTdR ∝ T R − T cR R − R C∝ − T CR κ ∝ 1023 ρT −7/2given these relationships we can findF rad ∝ caT 15/2 R 5M 2we also know that the luminosity can be written aswhich gives usL = 4πr 2 F rad⇒ F rad = L4πr 2L ∝ caT 15/2 R 7M 2we can find the temperature by using the virial theorem which can be written asT ≈ GMm pµ3Rk1

where k is now the boltzman constant. Substituting this expression into the above equation yields( ) 15/2L ∝ caR −1/2 M 11/2 Gmpkthis 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 isthe mass-luminosity scaling (proportianality) relationship for stars?we now know that the luminosity scales asL ∝ M 11/2 R −1/2we can find the relationship between the mass M and the radius R of a star by using hydrostaticequilibrium.dPdr= − GMr 2 ρP c ∝ M R ρρT ∝ M R ρM ∝ Rsince T is constant, substituting this into the luminosity relationship yieldsL ∝ M 5(c). Give a quantitative argument as to whether free-free opacity dominates electron scattering opacityin 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, whichcan be written as with T constantκ ∝ ρ ρ ∝ M R 3 M ∝ Rthus we findκ ∝ 1 M 2this expression tells us that the lower the mass of the star the higher the opacity, thus in lower massstars the free-free opacity dominates.Problem # 2The central density and temperature of the sun are ρ c ≃ 150 g cm 3 and T c ≃ 1.5 × 10 7 K. For theconditions at the center of the sun, answer the following questions. Assume that the sun is composedsolely of ionized hydrogen.(a). What is the mean free path of an electron due to electron-electron Coulomb collisions? What isthe typical time between collisions?2

We know that the mean free path is given byl = 1n e σwe know that for a completely ionized hydrogen gas thatand the interaction cross section is given byn e ∼ n p ∼ ρ m pσ = πr 2where r is the Coulomb radius found comparing the thermal energy to the Coulomb energye 2r ∼ kTr ∼ e2kTusing these relationships we find the mean free path to beand the collision time is given byand the velocity can be found by usingσ ∼l ∼ m ( )p kT 2ρ c π e 2t col = lv e32 kT = 1 2 m ev 2 v =πe4(kT) 2√3kTm ethus the time is given ast e =√ ( )me m p kT 23kT ρ c π e 2(b). What is the mean free path of an proton due to proton-proton Coulomb collisions? What is thetypical time between collisions? Which occurs more rapidly, electron-electron or proton-proton Coulombcollisions?The mean free path of proton-proton collisions would be the same as for the electron-electron collosionbecause the gas is completely ionized. The mean free path is given byl ∼ m ( )p kT 2ρ c π e 2The collision time would be the same except now that the mass is the mass of the proton not theelectron. i.et p =√ ( )mp m p kT 23kT ρ c π e 23

we can now see the collision times for the electron-electron collision occurs more rapidly due to themass being so much smaller.t e ≪ t p(c). Which opacity is more important for photons, Thomson scattering or free-free absorption?We know thatandκ T = n eσ Tρ c= 2σ Tm p∼ 0.80κ F = 10 23 ρT −7/2 ∼ 1.15free-free absorption dominates the opacity for photons in this case? not sure why this is. We know thatThomson 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 anelectron (this should give you a feel for why photons are far more effective at moving energy around instars)? What is the typical time between photon absorptions/scattering?we know that the mean free path of a photon is given byl = 1 where σ T = 8π [e 2 ] 2n e σ T 3 4πε 0 m e c 2 = 6.65 × 10 −25 cm 2which yieldsl photon = m p∼ 8.3 × 10 −3 cm2ρσ Tl electron = m pρ c πThe typical time for a photon collision is given by( kTe 2 ) 2∼ 8.9 × 10 −7 cmt = l pc ∼ 2.8 × 10−13 s(e). For a photon undergoing a random walk because absorption/scattering, how long would it take tomove a distance R sun given the results in (d)? For comparison, it would take 2.3 seconds moving at thespeed of light to travel a distance R sun in the absence of scattering/absorption.We know that the diffusion time can be acquired witht di f f =thermal energyL∼ R2 nkTlc aT 4 ∼ R2 nklcaT 3 ∼we know that the average time for a photon to leave the star is given byt di f f ∼ R2 sunl ph c ∼ 104 yr4R2 2ρkm p lcaT 3

Problem # 3How old is the sun? In this problem we illustrate how the naturally occuring radioactive isotopes ofuranium, U 235 and U 238 can be used to determine the age of the rocks. Both isotopes decay via a sequenceof α-decays and β-decays to form stabel isotopes of lead: the decay chain of U 235 ends up with Pb 207 , andthe decay chain of U 238 ends up with Pb 206 . As a result, the number of uranium nuclei in a rock decaysexponetially with time in accord with:N 5 (t) = N 5 (0)e −λ 5tand N 8 (t) = N 8 (0)e −λ 8tTo avoid clutter, the last digit of the mass number of the isotope has been used as a subscript label. Thedecay constants λ 5 and λ 8 for the two isotopes corresponds to half-lives ofT 5 = ln2 = 0.7 × 10 9 yrs T 8 = ln2 = 4.5 × 10 9 yrsλ 5 λ 8The magnitudes of these half-lives are ideally suitable to the determination of the ages of the rockswhich are over a billion years old. Now consider a set of rock samples which were formed at the sametime, but with different chemical compositions. They differ in chemical composition because differentchemical elements are affected differently by the processes of rock formation. However rock formationprocesses do not favour one isotope over another. For example, on formation, the relative abundances ofU 235 and U 238 should be the same in every sample. But these abundances will change with time as thedeacy 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 206nuclei. Show that this ratio is the same for all rock samples which were formed at the same time,and that it is given byN 7 (t) − N 7 (0)N 6 (t) − N 6 (0) = N 5(t) e λ5t − 1N 8 (t) e λ8t − 1We know that the ratio of the two isotopes can be written asN 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 asN 5 (0) = N 5 (t)e λ 5tand N 8 (0) = N 8 (t)e λ 8tsubstituting this into our previous expression yieldswhich is what we were asked to show.N 7 (t) − N 7 (0)N 6 (t) − N 6 (0) = N 5(t) e λ5t − 1N 8 (t) e λ8t − 1• Consider a graph in which the measured abundances in the rock samples of Pb 207 and Pb 206 areplotted, N 7 (t) along the y-axis and N 6 (t) on the x-axis. Show that a straight line will be obtained ifall the samples were formed at the same time.5

We know thatwhereN 7 (t) = N 5(t) e λ5t − 1N 8 (t) e λ8t − 1 · N 6(t)N 5 (t) e λ5t − 1N 8 (t) e λ8t − 1 = constant• Given that the current ratio of naturraly occurring U 235 to U 238 is 0.0071, evaluate the gradient ofthe 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 plugin numbers(a). t = 1 billion years.We know thatλ 5 ∼ 9.90 × 10 −10 yr −1 λ 8 ∼ 1.5 × 10 −10 yr −1given these and the fact that we know the ratio between U 235 and U 238 we can find the gradient, for 1bilion years we get0.0071 · eλ 5t − 1e λ 8t − 1 = 0.0715For 3 billion years we get0.0071 · eλ 5t − 1e λ 8t − 1 = .231and finally for 5 billion years we get0.0071 · eλ 5t − 1e λ 8t − 1 = .891Problem # 4 Radiative AtmospheresIn this problem we will solve for the structure of the outer part of a star assuming that energy istransported solely by radiative diffusion (which is not the case in the sun, but is the case in stars moremassive than the sun). The star has a mass M and a luminosity L. Assume that the luminosity and massare approximately constant at the large radii of interest, that gas pressure dominates, and that the opacity isdue to electron scattering. Do not assume that the atmosphere is thin (i.e even though M r ≈ constant = M,because rchanges, the gravitional acceleartion is not constant).Write down the equations for hydrostatic equlibrium and energy transport by radiative diffusion. Usethese to calculate dP rad /dP, the change in radiatio pressure with pressure in the atmosphere. What doesthis result imply for how the ratio of gas pressure to radiation pressure changes as a function of thedistance in the atmosphere? Show that your result for dP rad /dP implies that ρ ∝ T 3 and P ∝ ρ 4/3 forradiative atmospheres (in the language that we will use in the next week, this means that the radiative partof 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 rP rad = 1 3 aT 4dP raddr= 1 3 a d dr (T 4 ) = 4 3 T 4 ∇T6

and we know that the radiation flux is given byF = − 4 3caT 3κρ ∇Tthus we can writethe hydrostatic equilibrium equation isF = dP raddrcκρdP raddr= F κρ cdeviding these two expressions yielddPdr = −GM r 2 ρdP raddP= Fκr2cGMwe know that the Flux and luminosity are related bythus we findL = 4πr 2 FdP radF = L4πr 2dP = Lκ4πcGMthis result implies that the ratio of the gas pressure to radiation pressure is independent of the distancein the atmosphere. To show ρ ∝ T 3 we can just use scaling argumentsP radP g∝ L M ⇒ T 4ρT ∝ L Msince we assumed that L and M are constant than this givesT 3 ∝ ρTo show that P ∝ ρ 4/3 we can also use scaling argument, we also know that the radiation pressurescales as some constant times the gas pressureP radP g∝ L M P t = P g + P rad ⇒ P g = P t − P radthus we findbut we know thatthus we know thatP rP t − P r∝ 1 ⇒ P r ∝ λP gP g ∝ ρT T ∝ ρ 1/3 P g ∝ ρ 4/3λP g ∝ P t − λP gP t ∝ 2λP g ∝ ρ 4/37

Problem set 3Problem # 1(a). Show that heat transfer by radiative diffusion implies a non-zero gradient for the radiation pressurewhich is proportional to the radient heat flux. Bearing in mind that the magnitude of the force per unitvolume in a fluid due to the pressure is equal to the pressure gradient, find the radient heat flux densitywhich can, by itself, support the atmosphere of a star with surface gravity g. Hence show that a star ofmass M has a maximum luminosity given byL max = 4πcGMκwhere κ is the opacity near the surface. Obtain a numerical estimate for this luminosity by assumingthat the surface is hot enough for the opacity to be dominated by electron scattering. (This maximumluminosity is called the Eddington luminosity.To show that the heat transfer by radiative defusion implies a non-zero gradient we must begin withF r = − 4 aT 33 κρ ∇TF r ∝ P radknowing these relationships we can doP rad = 1 3 aT 4dP rdr = 4 3 dTaT3 drthus this implies that there is a non-zero gradient.⇒ dTdr = 3 1 dP r4 aT 3 drTo show thatL max = 4πcGMκwe must begin with the equation derived from problem 4 in the last problem set, i.ebut since we know thatdP rdP = Lκ4πcGMand we find thatP = P g + P r P g ≪ P r ⇒ P ≈ P rdP rdP r= 1L = 4πcGMκ≈ 3.3 × 10 4 L sun( MM sun)We cannot obtain a numerical estimate because we do not know tha mass. We could use M sun but thiswould not be correct.8

(b). Assume that radiative diffusion dominates energy transport in stars and that the opacity is due toThomson scattering. Use a scaling argument to estimate the mass M (in M sun ) at which the luminosity ofa star is ≈ L edd .We can do an order of magnitude estimate with respect to the sun byL ∝ M 3( )L M 3=L sun M sunand substituting the Eddington luminosity for L we find that M is given byM =( 4πcGML sun κ T) 1/2M 3/2sun ≈ 180 M sunProblem # 2The physical quantities near the center of a star are given in the following table. Neglecting radiationpressure and assuming the average gas particle mass ¯m is 0.7 amu, determine whether energy transport isconvective or radiative.r m(r) L r T r ρ(r) κ0.1R sun 0.028M sun 24.2L sun 2.2×10 7 K 3.1 × 10 4 kg m −3 0.040 m 2 kg −1Using equation ?? from Phillipsand the following relationships[ ] L(r)m(r)crit= γ − 1γ16πGc P rκ Pwe findP r = 1 3 aT 3P ∼ P g = ρ(r)¯m k bT γ = 5 3 κ T = 0.04 m 2 /kg[ ] L(r)m(r)crit= 2 516πGcκ T0.175 W kg > .07 W kgaT 3 ¯m3ρ(r)k bwhich implies that the energy transport of this star is primarily due to convection.Problem # 3The surface of a star (the “photosphere”) is the place where the mean free path of the photons l iscomparable to the scale-height h of the atmosphere . At smaller radii (deeper in the star), the density ishigher 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 goodapproximation 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 photosphere.Assume that Thompson scattering dominates the opacity.9

Knowing that l∼ h we can derive the following relationshipnσ = κρ ρ = nσ Tl = 1κ T nσ = h = k bT¯mgthus we find that the density is given by and assuming ¯m ≈ m pρ =¯mgk b κ T T ≈ 8.26 × 10−4 g/cm 3b) In reality, the surface of the sun is so low that hydrogen is primarily neutral. There are thus not thatmany free electrons to Thompson scatter off of. The opacity at the surface of the sun is instead due to theH − ion and is given by κ ≈ 2.5 × 10 −31 ρ 1/2 T 9 cm 2 g −1 . Using this (correct) opacity, repeat the estimatefrom a) of the density at the photosphere of the sun.Substituting the opacity given into the above expression yieldsρ 3/2 =¯mg2.5 × 10 −13 kT 10b()¯mg 2/3⇒ ρ =2.5 × 10 −31 k b T 10 ≈ 9.8 × 10 −8 g/cm 3c) Just beneath the photosphere, energy is transported by convection, not radiation, for the reasonsdiscussed in class (in fact, the photosphere is the place where photons travel so freely out of the star thatenergy transport by radiation finally dominates over convection). Estimate the convective velocity nearthe photosphere given your density from b).The convective heat flux is given byF c = 1 ( 2Fc2 ρv3 c v c =ρand knowing thatF c = L4πr 2we find that the convective velocity is given byv c =) 1/3( ) 2L 1/34πr 2 ≈ 1.09 × 10 6 cm/sρd) What is the characteristic timescale for convective ”blobs” to move around near the pho-tosphere?How does this compare to the observed timescale for granulation on the surface of the sun, which was afew min in the movie we watched in class?since we know that the characteristic time scale is given byt blob = l v c≈ h v cwe know that h which is the scale height of the sun is given byh = kTg ¯m = 1.7 × 107 cm10

thus we find that the blob timescale ist blob ≈ 15.6 swhich is a lot shorter than the timescale given by the movie which was approximately 2 minutes.e) Is the assumption ds/dr ≈ 0 valid near the surface of the sun? Why or why not?Since we know that the temperature gradient near the surface of the sun is very high and energy ismostly transported by photons impies that we cannot make the assumption ds/dr ≈ 0 .Problem # 4 Convective atmospheresIn HW 2, you calculated the structure of a stellar atmosphere in which energy is transported by radiativediffusion; you showed that such an atmosphere satisfies P ∝ ρ 4/3 . Here we will consider the problem ofa convective atmosphere, which is much more relevant to sun-like stars. For simplicity, assume that theatmosphere is composed of fully ionized hydrogen. The solar convection zone contains very little mass(only ≈ 2of the mass of the sun). Thus, let’s consider a model in which we neglect the mass of theconvection zone in comparison to the rest of the sun. For the reasons discussed in class, we can modelthe convection zone as having P = Kρ γ with γ = 5/3 and K a constant. R c is the radius of the base of theconvection zone.a) Solve for the density, temperature, and pressure as a function of radius in the convection zone. Donot assume that the convection zone is thin (i.e., even though M r = constant = M , because r changessignificantly in the convection zone, do not assume that the gravitational acceleration is constant).To solve for the Pressure we can begin withthus we findand integrating over the following limits we findthis integral yields( )dPP 1/γdr = −ρGM r 2 P = Kρ γ ⇒ ρ =KZ P( ) P −3/5dP = − GMKr 2 drP rc( PKthus we find that the pressure is given by5[ ]2K 3/5 P 2/5 − Pc2/5) −3/5 Z R GMdP = −R c r 2 dr[ 1= GMR − 1 ]R c( [ 2 1P =5 K−3/5 GMR − 1 ] ) 5/2+ Pc2/5R cTo solve for the density we can just plug this solution into( ) P 3/5ρ = = 1 ( [ 2 1K K 3/5 5 K−3/5 GMR − 1 ] ) 3/2+ Pc2/5R c11

and finally the temperature can be found by usingP = nk b T = 2ρm pk b T ⇒ T(r) = m pP2ρk band substituting the P and ρ from the previous expressions we findT(r) = m pK 3/52k b[ [ 2 15 K−3/5 GMR − 1 ] ]+ Pc2/5R cb) In detailed solar models, the pressure at the base of the convection zone is ≈ 5.2 × 10 13 dyne/cm 2and the density is ρ ≈ 0.175 g cm −3 . Using your solution from a), estimate the radius of the base of theconvection zone R c . Compare this to the correct answer of R c ≈ 0.71R sunIf we solve the density equation for R c we findand plugging in values we find that1= 1 [R c R −1R c= 1.998 × 10 −9 m −1(ρK 3/2 ) 2/3 − P 2/5c]· 5K3/52GM⇒ R c ≈ 5.11 × 10 8 m = 0.72R sunc) In your model, what is the temperature of the sun at 0.99R sun , 0.9R sun , and at the base of the solarconvection zone. This gives you a good sense of how quickly the temperature rises from its surface valueof ≈ 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 frompart (a). i.eT(r = 0.99R sun ) = m pK 3/5 [− 2 ]252k b 3 K−3/5 GM r + Pc2/5 ≈ 4.1 × 10 4 K66 · R sunT(r = 0.90R sun ) = m pK 3/5 [− 2 ]152k b 3 K−3/5 GM r + Pc2/5 ≈ 5.1 × 10 5 K18 · R sunT(r = 0.72R sun ) = m pK 3/5Pc2/5 ≈ 1.8 × 10 6 K2k bProblem set 4Problem # 1I mentioned in class that there are two ways to estimate the energy carried by convection. The first isthat the energy flux is F c ≈ 1/2ρv 3 c ≡ F c,1 where v c is the characteristic velocity of the convective motions.12

This is the KE flux carried by moving blobs. The other estimate is that F c ≈ ρ∆Ev c where ∆E is thedifference in the thermal energy of a rising hot blob (or sinking cool blob) relative to the background star(where E is per unit mass). I claimed in lecture that these two expressions are equivalent, to order ofmagnitude (which is the accuracy of mixing length theory). In this problem, you will prove my claim.(a). Calculate the acceleration a due to buoyancy of a rising hot blob (or sinking cool blob) in terms ofthe fractional density difference ∆ρ/ρ relative to the background star. Don’t worry about the sign of theacceleration or ∆ρ?, just their magnitudes.We know that the accelaration of the blob due to bouyancy is given by( )ρba b = g − 1 = g ∆ρρ ∗ ρsince ρ b ≈ ρ ∗ .(b). Use (a) to calculate the convective velocity v c in terms of ∆ρ/ρ. Recall that in lecture we estimatedv c using the work done by the buoyancy force.We know that the work done by the bouyancy force can be found byW = 1 2 mv2 c =Z lthus we find that the convective velocity is given aswhich can also be expressed asv c =we can also write this as given that l ∼ H, thus√0F · dl = aml2g ∆ρρ lv 2 c2gl = ∆ρρv 2 c = 2g ∆ρ k b Tρmg = 2∆ρ ρk b Tm(c). Use (b) to calculate ∆E, the difference in the thermal energy (per unit mass) of a rising hot blob(or sinking cool blob) relative to the background star, in terms of v c .We can write the last expression asand from the equation of state, which is given asT = v 2 m ρc2k b ∆ρ∆E = 1 k b ∆T(∆ρ ·V)φ ¯mwhere (∆ρ ·V) is the mass. Using these two expression and what we found from part (b) we can seethat∆EρV = 12φ v2 c = ∆Ē ∆Em m ∝ v2 c213

d) Combine your previous results to show that F c,1 ≈ F c,2 .From (a) we know thatusing this and our solution we findF c,1 ≈ 1 2 ρv3 cF c,2 ≈ ρ ∆E m v c12 ρv3 c ≈ ρ ∆E m v c ⇒ 1 2 v2 c ≈ ∆E mProblem # 2Estimate the convective velocity v c and the dimensionless entropy gradient (ds/dr)(H/c p ) in the convectionzones of 0.1 and 10 M sun stars. Assume that the material undergoing convection is at about themean density of the star and that gas pressure dominates. You can either use a scaling argument to estimatethe density, temperature, luminosity, etc. of such stars or look up in a book (e.g., Carrol & Ostlie) anyproperties of 0.1 and 10M? stars that you need to make your estimate (e.g., radius and luminosity). Butyou can’t just look up v c and (ds/dr)(H/c p ).From class we know thatbut we know thatthuswhich reduces toF c = ρα 3 Cs3 H ds∣C p dr∣3/2F c ≈whereCs 2 = kTm pand for the convective velocity we findL4πR 2∣ ∣∣∣ H dsv c = C s C p dr∣∣ ( ) ∣∣∣ H dsL 2/3C p dr∣ = 14πR 2 ρC 2 s( ) H dsLR 2/3∣C p dr∣ = m p3M ∗ k b T∣ ∣∣∣ H dsv c = C s C p dr∣1/2ρ = 3M4πR 3=( ) LR 1/33Mfrom Carrol and Ostley we find that for 10M sun and .1M sun we find that the radius, and luminosity areapproximatelyM ≈ 10M sun R ≈ 6R sun L ≈ 5700L sun1/2M ≈ 0.1M sun R ≈ 0.2R sun L ≈ .0034L sungiven these values we find∣ ∣∣∣ H dsM = 10M sun C p dr∣ ≈ 3.61 × 10−6 v∣ ∣∣∣ H dsM = 0.1M sun C p dr∣ ≈ 5.8 × 10−10 v14c ≈ 5.3 × 10 4 cm/sc ≈ 698 cm/s

Problem # 3 Polytropes(a). The mass M of a star is given byM =Z R04πr 2 ρ(r)drUse 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( ) 3Mρ c = ¯ρa n =4πR 3 a nwhere a n is a dimensionless number, the ratio of the central density to the mean density of the star. a n isa function that you should determine that depends only on the solution to the Lane-Emden equation (youcannot actual evaluate a n in general without numerically solving for θ[ζ], so your answer will just be interms of the solution to the Lane-Emden equation).Since we know thatgiven these two relations we can find( ) ρ(r) 1/nΘ =ξ = r ρaρ(r) = Θ n ρ cr 2 = a 2 ξ 2 a = R dr = Rdξand from the Lane-Amden equation we knowddξ(ξ 2 dΘdξ)= −ξ 2 Θ ngiven these following relationships we find thatZ 1 Z 1(M = −04πR 3 ξ 2 Θ n ρ c dξ = −4πR 3 dρ c ξ 2 dΘ )dξ0 dξ dξthus we find that a n is given byand we can finally show thata n = − 1 3Z 10ddξρ c = 3M4πR 3 a n(ξ 2 dΘ )dξdξ(b). Show that the central pressure of a polytrope can be written asP c = 4πGρ2 ca 2n+1where a ≠ a n is the constant (with units of length) defined in lecture (note that the polytropic relationP = Kρ γ can be used to write K = P c ρ −γc . Use this result and (a) to derive an expression for the centralpressure of a polytropic model of the form( GM2)P c =R 4 c n15

where c n is again a dimensionless function that you should write down. Also show that the centralpressure of a polytrope can be written asP c = d n GM 2/3 ρ 4/3cwhere d n depends on a n and c n . The values of a n ,d n , and d n can be determined by numerically solving theLane-Emden equation. The most useful cases for our purposes are γ = 4/3 (n = 3) and γ = 5/3 (n = 3/2)polytropes. For n = 1.5, a n = 5.99 and c n = 0.77 while for n = 3, a n = 54.183 and c n = 11.05. We will usethese quite a bit during this course. Note how, as mentioned in class, the results for the central pressure anddensity of polytropes above are very similar to what you would get from an order of magnitude estimate,except that for polytropes we get an exact correct numerical factor given by a n , c n and d n .We know thatand alsoa =given these relationships we can now find[K = P c ρ −γca 2 4πGn+1 = Kρ1/n−1 cbut since we know that γ = 1/n+1 we findand finally we findP c ρ −γc ρ1/n−1 c(n+1)Kρc1/n−1 ] 1/24πG= P c ρ −γc ρc1/n−1= P c ρc−(1+1/n) ρc1/n−1 = P c ρ −2cusing this results we can now deriveP c = a2 4πGρ 2 cn+1which becomesP c = a2 4πGρ 2 cn+1P c = a2 4πGa 2 nn+1looking at these two expressions we can see that= GM2R 4 c n( ) 3M 24πR 3 = GM2R 4 c nc n = 9a2 a 2 n4πR 2 1n+1now looking atand from (b) we findP c = d n GM 2/3 ρ 4/3cP c = a2 4πGρc 2/3 ρc4/3 ( )= a2 4πG an 3M 2/3n+1 n+1 4πR 3 ρc4/316

which becomes( )P c = GM 2/3 ρc4/3 4π 4/3c n 9 −2/3a nand after some fun algebra, which will be omitted here we findd n = c n( 4π3a n) 4/3(c). What are the values of d n for n = 3 and 1.5 polytropes, respectively?using the above result we findn = 3 a n = 5.99 c n = 0.77 d n = 0.477n = 3/2 a n = 54.183 c n = 11.05 d n = 0.363(d). Use your expressions for the central pressure and density to give an expression for the centraltemperature of a polytrope. Assume gas pressure dominates.to find these expression we will assume that gas pressure dominates, i.eP c = ρ c¯m k bT cthusT c = P c ¯m= d nGM 2/3 ρc1/3 ¯mρ c k b k band skipping some algebra we find that the central temperature is given byT c = d nGMRk bso all of the expressions can be written as( ) 1/3 3an m p4π 2T c = d nGMRk b( ) 1/3 3an m p4π 2P c = GM2R 4 c n ρ c = 3M4πR 3 a n(e). Calculate the central temperature, pressure, and density for γ = 4/3 (n = 3) and γ = 5/3 (n = 3/2)polytropes for M = M sun and R = R sun (i.e., for the sun). Assume fully ionized hydrogen for simplicity.Which polytrope better approximates the true interior temperature, pressure, and density of the sun? Canyou explain physically why this is the case?using the above expressions for temperature, density and pressure we find forγ = 5 3 a n = 5.99 c n = 0.77 d n = 0.477we findand forρ c ≈ 8.44 g/cm 3 P c ≈ 8.67 × 10 15 T c ≈ 6.2 × 10 6 Kγ = 4 3 a n = 54.183 c n = 11.05 d n = 0.36317

we findρ c ≈ 77.26 g/cm 3 P c ≈ 1.24 × 10 15 T c ≈ 1.02 × 10 7 Kthus we can see that the γ = 4/3 polytrope best represents the values observed in the sun, this is mainlydue to the fact that the center of the sun is radiative and not convective. Since we now know thatP ∝ ρ 4/3 radiativeP ∝ ρ 5/3 convectiveProblem # 4Consider 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 e f f ≈ constant. Butstars with M > 0.3M sun do not end up fully convective on the main sequence and so must go througha phase of KH contraction in which energy transport is dominated by photons. Assume throughout thisproblem that gas pressure dominates and that free-free ab- sorption dominates the opacity (because thetemperature is lower during KH contraction than on the main sequence, free-free absorption tends tobe even more important). Motivated by HW #2 Problem 1, assume that the luminosity of a star inwhich photons carry the energy out and the opacity is dominated by free-free absorption is given byL ≈ 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 andmass 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 massM). Do the luminosity and effective temperature increase or decrease as the star contracts?since we know thatthus we know thatP g > P rκ = κ f f( ) M 11/2 ( ) R −1/2L f f ≈ L sun ≈ L rad ≈ − 1 GM 2 dRM sun R sun 2 R 2 dtsince we are doing proportionalities we findso we find that the radius scales asplugging this intothusand to find the temperatureM 2 RR 2 t T ∝ M11/2 R −1/2R ∝ 1M 7 t 2L ∝ M 11/2 R −1/2 ∝ M 9 tL ∝ M 9 tL ∝ R 2 T 4e f f( ) L 1/4T e f f ∝R 2 ∝ M 23/4 t 5/418

thusT e f f ∝ M 23/4 t 5/4We can see that as time and mass increase the luminosity increases as well as the effective temperature.(b). Estimate the radius of a star (in R sun ) of a given mass M (in M sun ) at the time when energy transportby photons takes over from convection during the KH phase. At what luminosity does this occur (again asa function of mass M)? Assume based on lecture that the luminosity of a fully convective star isL ≈ 0.2L sun (M/M sun ) 4/7 (R/R sun ) 2if we set the free-free luminosity equal to the convective luminosity we findL sun( MM sun) 11/2 ( RR sun) −1/2≈ 0.2L sun (M/M sun ) 4/7 (R/R sun ) 2some algebra yieldsR ≈ 5 5/2 ( MM sun) 69/35R sun(c). Sketch the paths of 1M sun pre-stellar gas clouds during their KH contraction phase in the HRdiagram. Include both the convective and radiative parts of their evolution and the correct transition pointbetween the two based on b). Be sure to properly label your axes (L in L sun and T e f f in K). Note that onthe main sequence a 3 M sun star has L ≈ 40L sun and T e f f ≈ 10000 K (you know the values for the sun).The KH contraction phase ends when the star contracts to the point where its luminosity and temperaturehave these values.since we now have a relationship for the radius we can find what the luminosity is byL ≈ 0.2L sun( MM sun) 4/7 ( RR sun) 2≈ 5 2/5 · 0.2L sun( MM sun) 4/7 ( MM sun) 128/35which yieldsand a plot is given byL(M sun) ≈ 0.724L sun19

Figure 1: We plot the path that a 1 solar mass star would trace when moving from the Hayashi trackto the main sequence.Problem set 5Problem # 1In lecture we discussed the slow, nearly hydrostatic, contraction of pre-stellar gas clouds as they approachthe main sequence - Kelvin Helmholz (KH) contraction.(a). Argue that, for KH contraction to occur, the timescale for KH contraction t KH must be longer thanthe gravitational free-fall time of the cloud, t f f ≈ 1/ √ G〈ρ〉, where 〈ρ〉 is the mean density of the cloud.What happens if t KH < t f f ?Since we know thatt f f ≈1√G〈ρ〉t KH ≈[ ( ) ]M1/23R sun(2 × 10 7 yrs)M sun R20

froma a purely physical argument we know that things cannot fall faster than gravity can pull it. Thust KH ≫ t f fand 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 relationshipsof time in both free-fall and Kelvin-Helmholtz contraction we can see that as R ↓ that t f f ↓ thus theonly 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 , thecloud undergoes KH contraction according to your criterion from a). Recall that the luminosity of a fullyconvective star is L ≈ 0.2L 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 M 1/2 3R√ sun=(2 × 10 7 yrs)G〈ρ〉 M sun R cre-arranging this expression for R c we find( ) M 1/2R c =R sun (G〈ρ〉) 1/6 (2 × 10 7 yrs) 1/3M sunand lettingwe findif we put the mass off the sun we getSince we know thatwe can see that〈ρ〉 = 3M4πR 3 c≈ M R 3 c4/9 R2/3 sun(2 × 10 7 yrs) 1/3 G 1/9R c ≈ MMsun1/3R c ≈ 4.79 × 10 10 cmt f f ∝ R 3/2 t KH ∝ 1 R 3R ↑ t KH ↓ t f f ↑R ↓ t KH ↑ t f f ↓≈ M 4/9 7.6 × 10 −3from these two relationships we can see that for KH contraction to occur that t KH ≫ t f f and also thatR < 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 byT c = d ( ) 1/3nGM 3an m pRk b 4π 221

and since we know that fully convective stars have a polytropic index of γ = 5/3. Knowing this wefindn = 3 a n = 5.99 d n = 0.477thusT c ≈ 2.16 × 10 −15 M R c≈ 2.79 × 10 −15 M 5/9and if we want the temperature of a collapsing gas cloud with respect to M sun we getT c ≈ 1.9 × 10 4 KProblem # 2The globular cluster M13 in Hercules contains about 0.5 million stars with an average mass of abouthalf the solar mass. Use Jeans criteria to check whether this cluster could have formed in the early universejust 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 equationusing the values given we findand the mass of M13 isand we can see thatM j =( ) 3/2 kb T 3/2√Gm p ρM j ≈ 1.37 × 10 42 gM 13 ≈ 0.5 × 10 6 · M sun ≈ 9.95 × 10 38 gM j ≫ M 13which means that this cluster could not have formed in the early universe. Things only collapse if themass is greater than the Jeans massProblem # 3The binding energy per nucleon for 56 Fe is 8.8 MeV per nucleon. Estimate the energy released perkilogram of matter by the sequnce of reactions which fuse hydrogen to iron.We know that the enery released will be given asE tot =and the number of nucleons are given bythusN nucleon = M m p≈ 1 kgm pE bnucleon × N nucleonE tot ≈ 5.27 × 10 27 MeV≈ 5.98 × 10 26 nucleon(b). Consider two hypothetical stars of the same mass M and the same luminosity L (that is constant intime). The stars are initially pure hydrogen. In star A, fusion proceeds until the entire star is converted into22

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 thatL He = E HeL Fe = E Fet He t Feand since we know that these two luminosities are theoretically equalwhich givesthus we see thatE He t Fe = E Fe t Het He = E HeE Fet Fe = 6.48.5 t Fe ≈ 0.72t Fet He < t Fewe can see that the time for all of the hydrogen to fuse into helium is less then the time for all of thehydrogen 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 (themean thermal energy at the center of the sun)? Estimate the probability that the protons tunnel through theCoulomb barrier trying to keep them apart. Answer the same two questions for two 4 He nuclei and for aproton and a 4 He nucleus with the same energy of 2 keV.The classical distance of closest approach is given byr c = e2 Z 1 Z 2E 0≈ 7.2 × 10 −11 cmWe know that the propbability for a particle-particle interaction is given bywhereP = e − ( EGE) 1/2E G = 2π 2 α 2 Z 2 1 Z2 2 (m rc 2 )is the Gamow energy. For a proton-proton interaction we findE G ≈ .493 MeVthus the probability is given asfor a He-He interaction we findP ≈ 1.51 × 10 −7r c = e2 Z 1 Z 2E 0≈ 2.8 × 10 −10 cmand the probabilty is given byE G = 32π 2 α 2 2m p c 2 ≈ 31.6 MeVP ≈ 2.50 × 10 −5523

for the proton-He interaction we findr c = e2 Z 1 Z 2E 0≈ 1.44 × 10 −10 cmthus the probability is given byE G ≈ 3.15 MeVP ≈ 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, andiii) 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 thesecorrespond to?Since we know thatfor the He-He interaction we findfor the proton-He interaction we findE =E G(lnP) 2E He−He ≈ 0.125 MeVE p−He ≈ .013 MeVfor the carbon-carbon interaction the Gamow energy is given byE G ≈ 2592π 2 α 2 6m p c 2 ≈ 7.7 GeVand thuswe know thatE c−c ≈ 31 MeVT ≈ E k bsoT He−He ≈ 1.45 × 10 9 KT p−He ≈ 1.5 × 10 8 KT c−c ≈ 3.6 × 10 11 KProblem # 5Calculations of nuclear reaction rates are done in the center of mass (COM) frame, so it is useful toremember a few results about the COM. Consider two particles of mass m 1 and m 2 with positions x 1 andx 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 bycom = m 1r 1 + m 2 r 2m 1 + m 224

so the velocity would bev com = m 1v 1 + m 2 v 2m 1 + m 2(b). What are the velocities of each of the two particles in the COM reference frame (i.e., in the framefor which the COM is at the origin)?We know that the relative velocities are given byv rel−1v rel−2= v 1 − v com= v 2 − v coma bit of algebra yieldsv rel−1 =v rel−2 =m 2m 1 + m 2(v 1 − v 2 )m 1m 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 ofthe reduced mass moving at the relative velocity, as claimed in class.We know that the total kinetic energy is given byK tot= 1 2 m 1v 2 rel−1 + 1 2 m 2v 2 rel−2m 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 ) 2where m r is the reduced mass.K tot = 1 2 m r(v 1 − v 2 ) 2Problem set 7Problem # 1 The Main Sequence for Fully Convective StarsIn this problem we will determine the main sequence for fully convective low mass stars. We showedin lecture that fully convective stars have T e f f ≈ 4000(L/L sun ) 1/102 (M/M sun ) 7/51 K (I actually derived acoefficient of 2600 K in lecture but commented that more detailed calculations get something similar butwith the coefficient closer to the value of 4000 K used here). We can also write this result as25

L ≈ 0.2(M/M sun ) 4/7 (R/R) 2 sunL sun ≡ LconvI called this luminosity L conv since it is derived from the properties of energy transport alone (convectiveinterior + radiative atmosphere with H − opacity). The luminosity of a star is also given byL f usion = 4πr 2 ρε(T,ρ)drwhere ε is due to the proton-proton chain for low mass stars (this was given in lecture). As discussed inclass, the main sequence is determined by the requirement that the energy escaping the star (in this case byconvection) 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 e f f ) (the HRdiagram) for fully convective stars, like we did for other examples in lecture. Approximate ε ∝ ρT β withan appropriate choice of β (recall that low mass stars will have somewhat lower central temperatures thanthe sun, closer to ≃ ×10 6 K, as you will see in part b).We know thatwe can find what β is byL conv ∝ M 4/7 R 2L f us ∝ R 3 ε(ρ,T) ∝ R 3 ρ 2 T ββ = − 2 ( ) 1/33 + EG4kTgiven that we know what the temperature is and also what E G for p-p reactionE G ≈ 500 keVT ≈ 5 × 10 6 Kwe find thatwe also knowβ = 5.92 ≈ 6.0ρ ∝ M R 3we know that in steady statethus we can findL f usion = L convM 4/7 R 2 ∝ R 3 ( M2R 6 )T 6c ⇒ T 6 ∝ M −5/21 R 5/6we know from the Virial temperature, assuming gas pressure dominatesT ∝ M Rthus we findT c ∝ M 25/77knowing this we can now findR ∝ M 52/77with this and the relationship for the convective luminosity we findL ∝ M 4/7 R 2 ∝ M 148/7726

with this we can now find what the effective temperature as a function of mass is, i.ethuswhich yieldsL ∝ R 2 T 4e f fT 4e f f ∝ L R 2 ∝ M4/7T e f f ∝ M 1/7to find what the luminosity as a function of the effective temperature is (HR diagram)M ∝ T 7e f fwhich yieldsL ∝ T 148/11e f fIn a) you just determined a scaling relation between stars of different mass, but not the absolute valuesof L, T e f f , etc. In class, we did the latter by scaling to the sun. Note, however, that it is not reasonableto estimate the properties of low mass stars by scaling from the properties of the sun, since the sun is nota 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 scalingarguments by using the fact that fully convective stars are n = 3/2 polytropes. It turns out that for apolytrope, in equation (1) can be Taylor expanded near the center to yieldL f usion ≃2.4ε cM(3+β) 3/2where I have again approximated ε ∝ ρT β and where ε c is evaluated at the center of the star. I am notasking you to prove equation (2). You will have to trust me. Note that for a typical value of β for the ppchain, equation (2) says that L f usion ≃ 0.1ε c M . This makes sense because fusion only takes place at thecenter 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 ofthe star T c , central density ρ c , and pp energy generation at the center of the star ε c in terms of the massM and radius R. Assume X = 0.7 and µ = 0.6 (typical for stars just reaching the main sequence). Notethat you should give expressions for T c , ρ c , and ε c here, with constants and real units, not just scalingrelationships. So that the constants in front of your expressions are reasonable, please normalize M toM sun and R to R sun .The general expressions given by HW 4 problem #3 areT c = d nGMRk bwe found that for a n = 3/2 polytrope( ) 1/3 3anµm p P c = GM24πR 4 c n ρ c = 3M4πR 3 a na n = 5.99 c n = 0.77 d n = 0.47727

thus we find thatT c = 0.322 m pGM sunk b R sun( )M/MsunR/R sunρ c = 1.43 M ( )sun M/MsunR 3 sun (R/R sun ) 3plugging in all the constants yieldsT c ≈ 7.43 × 10 6 K( )M/MsunR/R sun( )M/Msunρ c ≈ 8.41(R/R sun ) 3we also knowε c = 5 × 10 5 ρ c X 2 T −2/3−1/3−15.7T7e 7we also know that we can approximate this asε c = AρT β7 X 2 = AρT 67 X 2setting this two expressions equal to each other we can find what A is, i.e lettingT ≈ 10 7 K we findA = 5 × 10 5 e −15.7 ≈ 0.076thus we findsubstituting T c and ρ c givesε c ≈ 0.076ρ c T 67 X 2 ≈ 0.037ρ c T 67( ) M 7 ( ) 9 Rsunε c ≈ 0.053M sun Rc) 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 e f f ) relations for fully convective stars. If you use the same β, your expressions hereshould be the same as in a) except that you should now be able to determine the absolute normalizationfor R(M), L(M), etc., i.e., you have determined the true luminosity and radius of a ful ly convective starfrom first principles. In doing this problem, remember that β is temperature dependent so make sure youcheck that your value of β is reasonable given the resulting central temperature that you calculate.Using the results from b and alsothus we know thatL f usion ≃2.4ε cM(3+β) 3/2 ≈ .09ε cM (β ≈ 6)L con = L f usion( ) M 4/7 ( ) R 2 ( ) M 7 ( ) 9 ( )Rsun M0.2L sun = 0.0047M sunM sun R sun M sun R M sunrearranging this we findwhich yield( RR sun) 11= 0.023 M sunL sun( MM sun) 52/7( )RM 52/77≈ 0.67R sun M sun28

to find for the temperature we can useusing our previous resulst givesto find for the luminosity we useplugging in for the radius we findfor the effective temperature we findwhich can be simplified toT c ≈ 7.43 × 10 6 K( )M/MsunR/R sun( ) M 25/77T c ≈ 1.1 × 10 7 KM sun( ) M 4/7 ( ) R 2L conv = 0.2L sunM sun R sun( ) M 148/77L conv = 0.09L sunM sun( ) M 148/77L = 4πR 2 σTe 4f f ⇒ 0.09L sun = 4πR 2 σTe 4f fM sun( ) M 148/77 ( ) M 104/770.09L sun = 1.56 × 10 18 Te 4f fM sun M sunthuswhich can also be written as( ) M 1/7T e f f = 3868 KM sun( )M7 Te f f=M sun 3868 KL(T e f f ) = 4.85 × 10 −50 L sun T 148/11e f fd) What are your predicted luminosities, radii, and effective temperatures for main sequence stars withM = 0.1 and 0.3M sun ? Compare your values to the values of L = 0.01L sun , R = 0.3R sun , and T e f f = 3450K for M = 0.3M sun and L = 10 −3 L sun , R = 0.11R sun , and T e f f = 3000 K for M = 0.1M sun that I found in agraduate textbook (based on detailed models).Given our relationships we find forM = 0.1M sun L = 1.1 × 10 −3 L sun R = 0.14R sun T e f f = 2783 Kand forM = 0.3M sun L = .009L sun R = 0.298R sun T e f f = 3256 K29

Problem # 2 Very Massive StarsConsider very massive stars with M ∼ 50 − 100M sun . Recall that I showed in lecture and you showedon HW 3, Problem # 1, that in such stars, radiation pressure due to photons (a relativistic particle) is moreimportant than gas pressure. Fusion is by the CNO cycle. Assume for now that energy is transportedprimarily 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 e f f ) (the HRdiagram) for very massive stars, like we did for other examples in lecture.Using radiative diffusion along withwe also knowthusand radiative diffusion sayswhich gives usP rad = 1 3 aT 4dP rdT = 4 3 aT 3dPdR = dP dTdT dR = 4 3 dTaT3 dR = −ρGM R 2dTdR ∝ ρMT 3 R 2LρR 2 T 3 ∝ dTdR ∝ ρMT 3 R 2L ∝ Musing the Virial theorem, where P rad dominates rather than P gas we findT 4ρ ∝ M R⇒ T c ∝ M1/2Rwhere the left hand term is from the radiation pressure, but since we know that is an energy densitywe must devide by the density to find what the energy is per particle. Now using the steady state forluminosity we findL ∝ MρT 18where we chose β = 18 as a more appropriate value rather than the value given for the sun β = 20, thisis motivated by the fact that more massive stars have somewhat higher temperatures, thus reducing β. WefindT 18 ∝ 1 ρ ∝ R3Mand using the result from Virial temperature we findthus we findand we also find for the central temperatureM 9 R3∝R18 MR ∝ M 10/21T c ∝ M 1/4230

and to find the effective temperature we knowsimplifying givesT 4e f f ∝ L R 2 ∝ M R 2 ∝ M1/21T e f f ∝ M 1/84and finally the luminosity as a function of T e f f is given byL(T e f f ) ∝ T 84e f fb) Estimate the fraction of the mass in the star that is undergoing convection (recall that fusion bythe CNO cycle is very concentrated at small radii because of the strong temperature dependence). Forcomparison, detailed calculations show that the fraction of the mass that undergoes “core” convectionincreases from 10 % at 2M sun to 75% at 60M sun .The condition for convection is given bysince we know thatd lnTd lnP ≈ 1 4P totP radL L r /LL Edd M r /M > γ − 1γγ = 4 3P tot ≈ P radgives uswhich simplyfies to1 L r M> 1 4 L Edd M r 4L r> M rL edd Mwe know that in the limit that M → 150M sun L r → L Edd ,M rM < 1which means that the fraction of the mass of the star that is undergoing convection approaches 1, which is100% of the mass is undergoing convection. Its a little strange that stars that are much less massive thanthe sun and the stars that are much more massive than the sun are both almost fully convective.n=3/2 polytropeThe previous case yielded a result for a n=3 polytrope, we find that forn = 3/2 polytrope γ = 5 3M rM < 5 L r8 L Eddand in the limit where M → 150M sunL r → L eddM rM < 5 831

This seems rather strange in the sense that stars that are approximately 60M sun have a convective corethat encompasses 75% of the mass, which means that the convective core decreases after M > 60M sun ?c) Calculate the main sequence lifetime of a very massive star as a function of its mass M . Be sure totake into account the results of b).We know that the main sequence lifetime of a star is given bywhereE tot = NQt MS ≈ E totL EddQ ≈ 7 MeVwhere that is the total energy per reaction, we also knowNN= M r= M m p m p= M r= 5 Mm p 8m p(n = 3 polytrope)(n = 3/2 polytrope)we also know thatL Edd = 4πcGMκ Tso we find the main-sequence lifetime to be given ast MSt MS≈≈κ T Qm p 4πcG5κ T Qm p 32πcG(n = 3 polytrope)(n = 3/2 polytrope)we know thatκ T ≈ 0.4 cm 2 /g Q ≈ 7 MeV ≈ 1.12 × 10 −5 ergsthus we find that the main-sequence lifetime for both types of polytropes are given by2.21 × 10 6 yr < t MS < 3.39 × 10 6 yrseems reasonable.Problem set 8Problem # 1 Fermi gas32

a) Above what density is a gas of room temperature fermions degenerate? Below what temperaturewould gas with the density of air be degenerate?asWe know that if the density of the gas is n g ≥ n Q where n Q is the quantum concentration, n Q is definedn Q ≡and for a gas at room temperature to be degeneratewhere we used( 2π ¯mkTh 2 ) 3/2(1)( ) 2π ¯mkT 3/2n g ≥ n Q =≈ 1.46 × 10 26 cm −3h 2T = 300K¯m = 28m pdue to the fact that air is mostly composed of N 2 . If we assume that the questio is only speaking aboutfree electrons we get( ) 2πme kT 3/2n g ≥ n Q =≈ 1.25 × 10 19 cm −3using the same temperature as before.h 2To find the temperature at which gas with a density of air would be degenerate can by using the aboveexpression, except now we must find what the density of air is at STP and use this, i.eand now using Equation 1 we findusing ¯m = 28m pn air = PkT = 2.52 × 1019 cm −3 = n QT = n2/3 Qh22π ¯mkT ≈ 9.2 × 10−3 Kb) Compare the relative importance of the thermal energy, the electrostatic (Coulomb) energy betweenelectrons 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 asWe can write the Coulomb energy asE th ≈ 3 kT ≈ .039 eV2E coul ≈ Z 2 e 2 1 rwe know( ρ¯m) −1/3r ∼ n −1/3 ∼33

thusE coul ≈ Z 2 e 2( ρ) 1/3≈ 12.43 keV¯musing ρ ≃ 10 g/cm 3 and ¯m ≈ 100m p . The Fermi energy can be written asE f =( ) 3 2/3h 2 ( ) 47ρ 2/3≈ 75 eV8π 2m e ¯mwhere we used ¯m ≈ 100m p and the 47 comes from the fact that there are 47 electrons in a silver atom.We can see thatE coul ≫ E f > E thfor room temperature silver.Problem # 2 Deuterium Fusion in Contracting ProtostarsSmall amounts of Deuterium are made in the Big Bang. D is destroyed in the interiors of stars viathe reaction p+D → 3 He+γ . The S value for D-burning is2.5 × 10 −4 keV-barn = 4 × 10 −37 erg cm 2 ,each reaction releases ≈5.5 MeV, and the cosmic abundance of D from the Big Bang is n D ≈ 2×10 −5 n H .Let’s focus on a low mass fully convective star undergoing KH contraction; such a star can be reasonablywell modeled as an n = 3/2 polytrope. Assume that the star has cosmic composition ( µ ≃ 0.6). Note thatin this problem, you should not use the approximation ε ∝ ρT β . Instead, you will need to keep the fullexpression for ε.a) What is the Gamow energy for D fusion? Write down the resulting thermally averaged cross-section〈σv〉 for D fusion.The Gamow energy can be written asm rE G = Z1 2 Z2 2 MeVm pusing Z 1 = Z 2 = 1 and m r = 2 3 m p we find the Gamow energy to beThe thermally averaged cross-section is given asE G ≈ .67 MeV〈σv〉 = 2.6S(E)E1/6 Gk 2/3 T 2/3 e −3(E G/4kT) 1/3using all the constants given and the Gamow energy we findin terms of M and R we find〈σv〉 =〈σv〉 =3.7 × 10−15K −2/3 T 2/3 cm 3s e−3742(K/T)1/33.7 × 10−15K −2/3 T 2/3 cm 3s e−3742(K/T)1/3b) In class we derived a quantitative model for the Kelvin-Helmholtz contraction of a low mass star asit approaches the main sequence. Use these results to calculate the local contraction time t c ≡ R/|dR/dt|34

as a function of the mass and radius of the star. This is the amount of time that a star of a given mass Mspends at a given radius R. Does the contraction time get shorter or longer as the star contracts?From lecture we derived the following relationshipL = 3 GM 2 ∣ ∣∣∣ dR7 R 2 dt ∣ = 0.2L sun(M/M sun ) 4/7 (R/R sun ) 2thus we finddRdt= 7 · 0.2 R 2 sunL sun3 GM 2 sun= 3.34 × 10 −5 cm/s(Msun) 2 ( ) R 2 ( ) M 4/7 ( ) R 2M R sun M sun R sun) 10/7 ( ) R 4M R sun(Msunand we can find the local contraction time to be( )( ) R 1M 10/7 ( ) 3t c = R sunR sun (dR/dt) = 6.7 × Rsun107 yrM sun RAs the star contracts the contraction time gets longer.c) What is the lifetime t D of a D nucleus at the center of the star in terms of the local density andtemperature (the lifetime is the average time before a D nucleus is destroyed by fusion into 3 He)? Use theproperties of n = 3/2 polytropes to write t D as a function of M and R. Does the D lifetime get shorter orlonger as the star contracts?We know that average lifetime of a deuteron is given bywhich gives usbut we also knowthus we findwe also knowthus we findt D = l v = 1n p σv〈t D 〉 = 1n p 〈σv〉 = µm pρ c 〈σv〉ρ c = 3M4πR 3 a n = 1.43 M R 3〈t D 〉 = µm pρ c 〈σv〉 = µm pR 31.43〈σv〉M = 7.019 × 10−25 g 1 R 3〈σv〉 MT c = d nGMRk b( 3an4π) 1/3µm p = 2.60 × 10 −16 cm g KM R〈t D 〉 = 7.019 × 10 −25 g 1 R 3g s= 1.89 × 10−10〈σv〉 M cm 3 T 2/3 3742(K/T)1/3 R3eM35

written in terms of M and R we find( ) R 7/3 ( ) 1/3〈t D 〉 = 1.23 × 10 −6 Msunse 19.19((M sun/M)(R/R sun )) 1/3R sun Mthe deuteron lifetime gets shorter as the star contracts.d)For any mass M show that there is a critical radius R D at which t D = t c . This represents the radius(time) at which D starts to undergo significant fusion. Give the numerical value of R D for M = 0.03 and 0.1M sun . For each of these two cases, also determine the central temperature of the star T c and the D lifetimet D when R = R D . Does D fusion occur before or after the star reaches the main sequence?We know thatwhich yieldst D = t c( ) R 7/3 ( ) 1/3 ( )1.23 × 10 −6 Msunse 19.19((M sun/M)(R/R sun )) 1/3 M 10/7 ( ) 3= 6.7 × 10 7 RsunyrR sun MM sun Rand so we find(RDR sun) 16/3= 1.71 × 10 21 ( MM sun) 37/21e −19.19((M sun/M)(R/R sun )) 1/3we can solve this numerically to findR D = 0.44R sunM = 0.03M sunR D = 1.11Rto find the central temperature we can usewe findT c = 7.4 × 10 6 KM = 0.1M sun( )( )M RsunM sun R DT c ≈ 5.0 × 10 5 K M = 0.03M sunT c ≈ 6.67 × 10 5 K M = 0.1M sunand to solve for the deuteron lifetime we find( ) 7/3 ( ) 1/3〈t D 〉 = 1.23 × 10 −6 RD Msunse 19.19((M sun/M)(R/R sun )) 1/3R sun Mso we findt D ≈ 4.67 × 10 6 yr M = 0.03M sunt D ≈ 4.1 × 10 5 yr M = 0.1M sune) Can D fusion halt (at least temporarily) the KH contraction of the star? Explain your answer quantitatively.36

Since we know thatwe also know thatε = Qr dρ dρ d = m d n d r d = n dt DL = Mε = MQr dρ d= MQ2m p t Dsince we know that Q = 8.8 × 10 −6 ergs, we findL = M(8.8 × 10−6 ergs)2m p t Dso for M = .03M sun we findand for M = 0.1M sun we findL ≈ 5.37 × 10 35 ergs/sL ≈ 4.08 × 10 37 ergs/sthus we can see that for both of these stars deuteron fusion can stop the KH contraction temporarily.Problem # 3 The R(M) Relation for Degenerate ObjectsConsider an object supported entirely by the pressure of non-relativistic degenerate electrons. BecauseP = Kρ 5/3 such an object can be modeled (rigorously) as an n = 3/2 polytrope.K is a constant that dependson the electron mean molecular weight µ e .a) Use your results for how the central pressure P c and density ρ c of an n = 3/2 polytrope dependson the radius R and mass M of the object to derive the R(M) relation for degenerate objects (the radiusalso depends on µ e ). Note that you should give an expression with proper constants and not just a scalingrelationship. Normalize the mass M to M sun and the radius R to R sun (this should sound pretty familiar bynow).We know thatrearranging this equation for ρ c yields( )P deg = P c = h2 3 2/3 ( ) 5/3 ρc= d n GM 2/3 ρc4/35m e 8π µ e m pρ −1/3c( )= h2 3 2/3 ( ) 1 5/315m e 8π µm p d n GM 2/3from the last problem set we showed( )( ) M3 Rsunρ c = 8.41M sun Rusing this we find( )R= 0.04µ −5/3 M −1/3eR sun M sunb) Use a) to estimate the radius of Jupiter. How does your result compare to the correct value?Using part a) with µ e ≈ 1.17 which is the value given for the sun on Google and M = M J , we findR ∼ 0.30R sun37

c) The results you have derived in a) should show that as M → 0,R → ∞. This is not correct, however,because Coulomb interactions become important in the equation of state of low-mass objects (browndwarfs and planets). Estimate the density at which the Coulomb energy per particle becomes comparableto the Fermi energy. What mass and radius does this correspond to? Explain why this is a very roughestimate of the maximum radius of a degenerate object.thusIf we knowthenbut we know thatso we find the density to be given byn =to find the mass we can useand using R from part b) we findwhich can be simplified towhich can also be expressed asusing this we can now find the radius to beE f = E coul1 e 2 ( ) 3 2/34πε 0 r = h 2n 2/38π 2m e1r ∼ n1/3( )13 2/3e 2 n 1/3 h 2=n 2/34πε 0 8π 2m e( ) 3 ( ) 2me 3 24πε 0 h 2 e2 ≈ 6.15 × 10 28 m −38πρ c = nm p = 1.43 M R 3M sun (M/M sun )1.43(0.03R sun ) 3 (M/M sun ) −1 = nm p(M nmp (.03R sun ) 3 ) 1/2=≈ 5.7 × 10 −4M sun 1.43M sunM ≈ 190M earthR ≈ 0.361R sunProblem set 938

Problem # 1Use the chemical potential µ for a non-degenerate, non-relativistic gas (derived in class; also 2.21in Phillips) to show that in the limit n ≪ n Q (the non-degenerate limit), the full quantum mechanicaldistribution function reduces to the classical Maxwell-Boltzmann distribution function. A good check thatyou have things correct is that the QM dist. fcn you start with has some h ′ s in it (Planck’s constant), butthe classical dist. fcn you end up with should, of course, be independent of h.We know that the chemical potential is defined asµ= mc 2 + kT ln( ngn Q)(2)and that the quantum distribution function is defined asg/h 3n(p) =e (Ep−µ)/kT ± 1to show that in the classical regimee (E p−µ)/kT ≫ 1We can write Equation 1 asn Qn = 1 g e(mc2 −µ)/kTn ≪ n Q e (mc2 −µ)/kT ≫ 1thus the quantum distribution function can be written asWe also know thatand using equation 1 we findn(p) =n(p) =g/h3e (E p−µ)/kT( )E p = mc 2 + p2 2πmkT 3/22m n Q =h 2g/h 3e (mc2 +p 2 /2m−mc 2 −kT ln(n Q /n))/kT =gnh 3 e E/kT n Qwhich after some simplification reduces to the Classical Boltzman distribution function( ) 1 3/2n(p) = ne −E/kT2πmkTProblem # 2Consider a cloud of gas that has a total mass M . Assume that all of the gas in the cloud is convertedinto stars with the initial mass function given in class dN/dm ∝ m −α where α = 2.35 and where thisformula is valid between m = 0.5M sun and m = 150M sun . Note that dN/dm has units of number of starsper unit mass.39

a) What is the ratio of the number of stars formed with masses within dm ≃ m 1 of m 1 and masseswithin dm ≃ m 2 of m 2 ? What is the ratio of the number of 150M sun stars formed to the number of 0.5M sunstars formed?We know that the Initial Mass Function IMF is given aswe also know that for dm ≃ m 1 of m 1 we findwe also know that for dm ≃ m 2 of m 2 we findand the fraction is given bydNdm ∝ m−α ∝ m −2.35 α = 2.35dN(m 1 ) = m −α1 dm 1 ≈ m 1−α1dN(m 2 ) = m −α2 dm 2 ≈ m 1−α2dN(m 1 )dN(m 2 ) = (m1m 2) 1−α≈ 2208 m 1 = 0.5M sun m 2 = 150M sunb) Estimate the mass of a cloud M so that approximately one 150M sun star forms in the cloud. If thetemperature of the cloud at the time of formation was 10 K, what was the density of the gas out of whichthe cloud formed?From part a) we found thatN s ≈ 2208N bwhere N b is for stars that are the number of ∼ 150M sun and N s is for stars that have ∼ 0.5M sun and toget the total mass we must multiply the total number of small stars to the average mass of the stars, andfrom lecture we are told that〈M ∗ 〉 ≈ 0.5M sunM cluster = N s · 〈M ∗ 〉 ≈ 1104M sunto find the density of the gas in which this cloud formed we can use the Jean’s densityProblem # 3ρ J = 3 ( ) 3kT 34πM 2 ≈ 3.14 × 10 −22 kg m −32G ¯mA stellar atmosphere consists almost entirely of hydrogen. Assume that 50 % of the hydrogen moleculesare dissociated into atoms and that the pressure is 100 Pa. Given that the binding energy of the hydrogenmolecule is 4.48 eV, estimate the temperature. Set all degeneracies to 1. As the hint at the back of the booksuggests, you should derive the Saha equation for the dissociation of H 2 into hydrogen, i.e., the reactionγ+H 2 ←→ H + H .We know thatand the Saha equation givesγ+H 2 ←→ H + Hµ(H 2 ) ←→ 2µ(H)40

ut we know that the chemical potential is given by( )( )µ(H) = m H c 2 nQ,H g H− kT lnµ(H 2 ) = m H2 c 2 nQ,K2 g H2− kT lnn H n H2wherethus we findwhich becomeswe are given thatthus we findbut we are given thatthusm H c 2 = m p c 2 + m e c 2 − χ Hm H2 c 2 = 2m p c 2 + 2m e c 2 − χ H2 − 2χ H( ) [ ( )]m H2 c 2 nQ,K2 g H2− kT ln = 2 m H c 2 nQ,H g H− kT lnn H2 n H− χ H 2kT = ln [ (nQ,H2g H2n H2)(nHn Q,H g H) 2]g H2 = g H = 1n Q,H2 ≈ n Q,H ≈( 2πmkTh 2 ) 3/2n H2= n (He −χ H h /kT 2) 3/22 = n H e −χ H 2/kTn H n Q 2πmkTn H = P3kTn H2n H= 1 212 = P ( h2) 3/2e −χ H 2/kT3kT 2πmkTthis can only be solved analyticaly, we find that the temperature is given byProblem # 4 Lines from HydrogenT ≈ 2260 KConsider a pure hydrogen gas. In this problem we will calculate the fraction of H atoms that havean electron in the n = 2 state (a result I plotted in class), and use that to understand some aspects of theobserved lines of H from stars. Recall that the energy levels of the H atom are given by E = −13.6/n 2 eVand the degeneracies are g n = 2n 2 .a) Use the Saha equation to solve for the fraction of hydrogen atoms that are ionized as a function oftemperature T . If n is the total number density of hydrogen atoms (both neutral and ionized) then whatwe are after is n p /n since an ionized hydrogen atom is just a proton. Your result for n p /n will dependon n (because, as discussed in class, the ionization of a gas depends weakly on density in addition to theprimary dependence on temperature). For densities appropriate to the photosphere of the sun, make a plotof n p /n as a function of temperature T . If you are familiar with graphing using IDL, Mathematica, etc.feel free to use that. Otherwise, you can just plug values into your calculator and make the plot by hand. Inyour calculation, assume that all of the neutral hydrogen atoms are in the n = 1 (ground) state. The reasonthis is an ok approximation is as follows. According to the reasoning in class, which you will confirmhere, Hydrogen is 1/2 ionized at T ≃ 1.5 × 10 4 K. At that temperature, nearly all of the neutral H atoms41

are in the ground state (check it if you don’t believe me!), so for temperatures at which H is largely neutral(T ≤ 1.5 × 10 4 K), it is reasonable to say that almost everything is in the ground state.We know from the Saha equationn e n pn H= g eg pg H( ) 2πmkT 3/2e −χ/kTh 2we knowthusand son e ∼ n p n = n H + n p χ ≈ −13.6 eV g eg pg H= 1n 2 pn H= αwhich becomes a quadratic equation of the formwith the solution of n p being( ) 2πmkT 3/2α ≡e −χ/kTh 2n 2 p = αn H = α(n − n p )n 2 p + αn p − αn = 0n p = −α ± √ α 2 + 4αn2and since we know that this must be a positive thus we will take the positive solutionand finally we are looking forthe plot is given byn p = −α+√ α 2 + 4αn2n pn = −α+√ α 2 + 4αn2n42

n p vs n total0.80.6n p /n0.40.20.05.0•10 3 1.0•10 4 1.5•10 4 2.0•10 4 2.5•10 4 3.0•10 4Temperaturewe can see that at a temperature of T ≃ 1.4 − 1.5 × 10 7 K roughly ∼ 50% of the hydrogen atoms areionized.b) Use your result from a) to calculate the fraction of all H atoms that have an electron in the n = 2state of hydrogen. If n 2 is the number density of atoms with electrons in the n = 2 state, then what weare after here is n 2 /n. You will need to use the Boltzmann factor in addition to your result from the Sahaequation in a). For densities appropriate to the photosphere of the sun, make a plot of n 2 /n as a functionof temperature T . If you are familiar with graphing using IDL, Mathematica, etc. feel free to use that.Otherwise, you can just plug values into your calculator and make the plot by hand.We know thatn Hn = n − n pnand from the Boltzman equation we know thatwe also knowandwhat we are looking for is= 1 − n pnn 2n 1= g 2g 1e −(E 2−E 1 )/kTn H = n 1 + n 2 → n 2 = n H − n 1 → n Hn 2= 1+ n 1n 2(n 2= 1+ n ) −11n H n 2n 2n = n (2 n Hn H n = 1+ n ) −1(11 − n )pn 2 n43

thus we find the fraction of all H atoms that have an electron in the n = 2 state of hydrogen given by(n 2n = 1+ n ) −1(11 − n )p=n 2 n(1+ g ) ()−11e (E 2−E 1 )/kT1 − −α+√ α 2 + 4αng 2 2nwhere α has been explicitily defined already. The plot is given byn 2 vs n total4•10 −53•10 −5n p /n2•10 −51•10 −501.0•10 4 1.5•10 4 2.0•10 4 2.5•10 4 3.0•10 4 3.5•10 4 4.0•10 4TemperatureWe can see that the fraction of hydrogen atomes in the energy state n = 2 peaks at ∼ 1.5 × 10 4 K.c) The Balmer lines of hydrogen are produced by transitions between the n = 2 states of Hydrogen andthe n = 3,4, .... states. What are the wavelengths of the H α (n = 2 → 3)and H β (n = 2 → 4) lines of H?Use your result from b) to explain why A stars show the most prominent H α lines of hydrogen (relative tomore massive stars such as O stars and less massive stars such as M stars).We know thatthus∆E = hν = hcλλ = hc∆E = hc(E 2 − E 1 )and13.6 evE = −n 2thus for the n = 2 → n = 3 transition we getλ =hc≈ 656.3 nm(3.4 − 1.51)eV44

and for the n = 2 → n = 4 transition we getλ =hc≈ 486.7 nm(3.4 − 0.85)eVFrom the plot given in part b) we can see that the fractional number of atoms in the n = 2 energy statepeaks at around 1.5 × 10 4 K, which is approximately the surface temperature of A stars, we can also seethat for O type stars that have surface temperatures much greater than 15,000 K that there are ≈ 0% ofhydrogen atoms in the n = 2 energy state, most of the atoms are already ionized. The situation is similarfor M stars that have surface temperatures that are much lower than 15,000 K. We can see that at thesetemperatures there are approximately 0 atoms with electrons in the n = 2 energy state.d) The Lyman lines of hydrogen are produced by transitions between the n = 1 states of Hydrogen andthe n = 2,3,4, .... states. What is the wavelength of the Ly α (n = 1 → 2) line of H? Roughly what fractionof H atoms have electrons in the ground (n = 1) state of H in the atmosphere of an M-star? Would youexpect to see prominent Ly α lines from an M-star? Why or why not?Usingwe findλ = hc∆E = hc(E 2 − E 1 )λ = hc∆E = hc≈ 121.6 nm(13.6 − 3.4)eVWe would not expect to see any L yα lines from M stars, even though all of the hydrogen atoms are inthe ground state, there is not enough thermal energy to excite the electrons from n = 1 to n = 2.Problem set 10Problem # 1Consider a gas with total mass density ρ and temperature T . Recall that the mean molecular weight µis defined by P ≡ ρkT/µm p where P is the total ideal gas pressure (ions and electrons), while the electronmean molecular weight µ e is defined by n e ≡ ρ/µ e m p .Since we know that the total pressure is given byP T = P I + P e = ρkTm p( 1µ I+ 1 µ e)= ρkTµm pand thusbut we know that1µ = 1 µ I+ 1 µ e1µ I= X A1= XZµ e A45

where X is the mass fraction of the species, Z is the number of electrons, and A is the atomic numberof the species, thus we findµ= 1 Aµ e = A(3)X 1+Z XZa) What are the values of µ and µ e if the gas consists ofi) ionized H,Since we know thatthenX = 1 A = 1 Z = 1µ= 1 2 µ e = 1ii) 75 % (by mass) ionized H and 25 % (by mass) ionized He,We have to treat this case seperately since we have two species contributing to the mean molecularweightX = 0.75 Y = 0.25 A = 4 Z = 2thusand so we find1= X µ I A + Y A = 1 13(X +Y) =A 16µ= 1627 µ e = 8 71µ e= Z A (X +Y) = 7 8iii) ionized He,Since this is a pure fully ionized gas we can use Equation 1 withX = 1 A = 4 Z = 2we findiv) ionized O,Using Equation 1 withwe findv) ionized FeUsing Equation 1 withwe findµ= 4 3 µ e = 2X = 1 A = 16 Z = 8µ= 169µ e = 2X = 1 A = 56 Z = 26µ= 5627µ e = 281346

) Which gas has the largest ideal gas pressure? Which gas has the largest electron degeneracy pressure?Assume that ρ and T are the same in all cases.Since we know that gas pressure goes asisP g ∝ 1 µthan the smallest µ will give us the highest pressure, thus the element that has the highest gas pressureµ= 1 2Hydrogen gasfor degeneracy pressure we knowisP d ∝ n 5/3e( ) 1 5/3∝µ ethus the gas that gives the highest degeneracy pressure is the one with the lowest value for µ e and thisProblem # 2 The Helium Main Sequenceµ e = 1 Hydrogen gasIn certain stages of stellar evolution, stars are largely composed of He and He fusion dominates thestellar luminosity. One can approximate such stars as lying on a He main sequence. In this problemwe will calculate the properties of the He main sequence assuming that a star is composed of pure He,that energy transport is via radiation, that electron scattering dominates the opacity, and that gas pressuredominates. The energy generation rate for He fusing to Carbon isε = 5 × 10 11 ρ 2 T −38exp(−44/T 8 ) ergs s −1 g −1and 7.65 MeV is released converting 3 He nuclei into 1 C nucleus. Note that throughout this problemyou should not just give scaling (proportionality) laws for the desired relations; you should also determinereasonable normalizations.a) Calculate the relationship between mass M and luminosity L for the He main sequence.For a star that has the given properties: energy transport is via radiation, electron scattering dominatesthe opacity (σ = σ T ), and that gas pressure dominates (P ∼ P g ) we find the luminosity given asL ∝ M 3 µ 4 µ eThis equation gives the evolution of the lumunosity on the MS as chemical composition changes. Wecan scale this to the sun to find( ) M 3 ( ) µ 4 ( )µeL = L sunM sun µ sun µ e−sunwe also know thatµ He = 4 3 µ e = 2 µ sun = 0.6 µ e−sun ≈ 1.14and so we find that the lumonosity for the Helium main sequence can be expressed asL ≈ 42.7L sun( MM sun) 347

) Estimate the core temperature of a 1 solar mass He star. You do not need to do the full integralL f usion = R dMε , but can approximate this as L f usion ∼ 0.1Mε(r = 0).In steady state we can express the lumunosity of energy transport be equal to the luminosity due tofusionL f usion ≈ L transportwhere we can use part a) and the approximation given to find42.7L sun( MM sun) 3= 0.1Mε = 0.1M5 × 10 11 ρ 2 T −38exp(−44/T 8 ) ergs s −1 g −1if we letwe findsolving this numerically yieldsM = M sun ρ ≈ 150 g/cm 3T 38 e44/T 8= 1.36 × 10 13T 8 ≈ 1.52T c ≈ 1.5 × 10 8 Kc) Given your result for T c for a 1 M sun star from b), calculate the power-law relation T c (M) by imposingthe steady state requirement that L f usion = L photons and using ε ∝ ρ α T β (where L photons is the energy carriedout of the star by photons from a).Since we know thatL photons ∝ M 3L f usion ∝ Mεwe findM 2 ∝ ε ∝ ρ α T βwhere in part b) we are givenα = 2 β = −3+ 44T 8≈ 26knowing that the density and temperature scale asρ ∝ M R 3T ∝ M Rgives( ) M 2 ( ) M 26M 2 ∝R 3 → R ∝ M 13/16Rusing this along with our expression for the temperature givesT ∝ M R ∝to scale to a one solar mass star, from part b) we findM ∝ M3/16M13/16 ( ) M 3/16T ≈ 1.5 × 10 8 KM sun48

d) Use your results above to determine the R(M)and T e f f (L)relations for the He main sequence. Thensketch the relative positions of the H & He main sequences in the HR diagram.From the Virial temperature we knowT c = Gµm pkwhere k is Boltzmans constant, this expression can be scaled to the sunT c = Gµm ( )( )p M sun M Rsunk R sun M sun Rusing the result from c) and plugging in all the constants along with µ= 4/3 givesand we find( ) M 3/16 ( )( )M1.5 × 10 8 K = 3.08 × 10 7 RsunKM sun M sun RMR( ) ( ) RM 13/16= 0.21R sun M sunTo find the relationship between the luminosity and the effective temperature we can usebut we know from our previous expressionL = 4πR 2 σT 4e f fR 2 = (0.21R sun ) 2 ( MM sun) 13/8plugging this in our expression of the luminosity givesL = 4πσ(0.21R sun ) 2 ( MM sun) 13/8T 4e f fnow we can use the result from a)( ) ( ) M L 1/3=M sun 42.6L sunusing this we getthus we find thatL sun( LL sun)The HR diagram for this star is given by( )= 4πσ(0.21R sun ) 2 L 13/24T 442.6L sun( ) L 11/96T e f f ≈ 2.1 × 10 4 KL sune f f49

which seems rather odd, We would expect this main sequence He buring star to be above the mainsequence line. This can be explained by our initial assumption that went into deriving this relationship.We assumed that this was a pure ball of He gas.e) At what mass does the luminosity of the star exceed the Eddington luminosity?We know that the Eddington luminosity is given byL Edd = 4πcGMκL f usion = 42.7L sun( MM sun) 3setting these two expressions equal to each other givesgiven the opacity defined asthus4πcGM sunκ( MM sun)= 42.7L sun( MM sun) 3κ = n eσ Tρn e =ρµ e m pκ = 1µ e m pσ T µ e = 2 κ = σ T2m p≈ 0.2 g/cmM ≥ 38.95M sunis the mass that will exceed the Eddington luminosityf ) What is the He main sequence lifetime as a function of stellar mass? Compare this to the correspondingH burning lifetime.50

We know that the main sequence lifetime is given byt = E LE = NQ = 0.1M (7.65MeV) = 0.1M sun M(7.65MeV )12m p 12m p M sunthust =0.1M sun12m p (42.7L sun )thus the main sequence lifetime is(MsunM) 2(7.65MeV) ≈ 2.34 × 10 7 yrst ≈ 2.34 × 10 7 yrs( ) 2 MsunM( ) 2 Msunthis is much much shorter than the H burning lifetime which is t ∼ 10 10 yrs for M ∼ 1M sun . Also canbe written ast He ∼ 0.2% the time of the main sequence Hydrogen burningProblem # 3 The Thin Shell InstabilityAs we discussed in lecture, during several phases of stellar evolution, fusion takes place in a thin shell.Consider such a shell located a distance R s from the center of a star. The mass interior to R s is M , themass of the shell itself is M shell and the thickness of the shell is H ≪ dR ≪ R, where H is the scale-heightat radius R s (recall that H is the distance over which the density, pressure, temperature, etc. change).a) Use hydrostatic equilibrium to show that the pressure at the base of the shell is given byHE giveswhich can be written in differential formbut we know thatP(R s ) ≃ GMM shell4πR 4 sdPdr = −ρGM R 2P(R s + dR) − P(R s )dR= −ρ GMR 2 sρ s = M s≈M sV s 4πR 2 sdRand since we know that H ≪ dR ≪ R (this comes from the definition of H) then P(R s + dR) ≪ P(R s )thanwhich simplifies toP(R s )dR= GMM s4πR 4 s dRP(R s ) = GMM s4πR 4 sb) Use your result in a), together with the strong temperature dependence of fusion reactions, to explainwhy fusion in a thin shell is unstable and will runaway, as in a bomb. Hint: How will P,ρ,T,and dR of theshell change if there is a small perturbation that increases the amount of fusion in the shell?51M

Since we know thatL ∝ Mρ 2 T −44/T 8ρ ∝ MdR 3If we apply a small pertubation that increases fusion then we know T ↑,ρ ↓, and dR ↑and since theenergy generation has such a high power temperature dependence we know that in order for this to be stablethan the density must decrease to compensate (assume constant pressure). But we can see that the densitydependence is a function of the radius and cannot decrease by 20 orders of magnitude to compensate, andthus this becomes a runaway reaction, i.e like a bomb.This unstable fusion occurs primarily when stars are on the asymptotic giant branch (fusion of He ina thin shell outside a C/O core) and may be part of the reason that such stars lose so much mass on theirway to becoming white dwarfs.Problem set 11Problem # 1Consider a 0.5 M sun WD. Approximate it as an n = 3/2 polytrope, reasonably appropriate since weare below the Chandrasekhar mass. Estimate the ratio of the energy transported by photons (radiativediffusion) to the energy transported by degenerate electrons (thermal conduction) at the center of the WD.Scale the central temperature of the WD to 10 8 K, an appropriate number for a newly formed WD. Assumethat the opacity is due to electron scattering. Show that the energy transported by electron conductiondominates that transported by photons.We know that the radiative flux for photons is given by( 4 acT 3 )F r = − ∇T = −κ r ∇T3 κ 0 ρwhere κ o is the opacity and κ r is the conductivity (any process that transports energy). We also know thatflux due to thermal conduction of degenerate electrons is given byF deg = −κ deg ∇Twhere κ deg is the conductivity due to degenerate electrons, thus the ratio of the energy transported byphotons (radiative diffusion) to the energy transported by degenerate electrons (thermal conduction) isgiven byF r= κ rF deg κ degwe have defined the degenerate electron conductivity to be( ) 3/2 EFκ deg ≃ κ cls = k2 h 3 T n ikT 32e 4 m 2 = k2 h 3 T ρ ce 32e 2 m 2 e µ im pn i = ρ cµ i m p52

and the radiative diffusion conductivity to beκ r = 4 acT 3= 4 acT 3= 4 acT 3 µ e m p3 κ o ρ c 3 n e σ T 3 ρ c σ TWe also know that ρ c (n = 3/2) polytrope equation is given bywhere the radius is given byρ c = 4M3πR 3 a n ≈ 1.43 M Ra n = 5.99( ) M −1/3(µe ) ( ) −5/3 m −1R ≈ 0.013R sun ≈ 0.016R sunM sun 2 m eif we use M = 0.5M sun and µ e = 2. The ratio can be written asκ rκ deg= 128acT 2 e 4 m 2 e m2 p µ eµ i3k 2 h 3 σ T ρ 2 c(4)we also knowwhere we find1µ = 1 + 1 n1 X i=µ i µ e µ i∑i=0Aµ i = 967µ e = 2n1 X i Z=µ e∑i=0Aassuming 50% C and 50 % O. If we use a temperature of T ≈ 10 8 K on Equation 1 we findκ rκ deg≈ 0.084⇒ κ deg ≈ 11.9κ rThus we can see that the energy transported by electron conduction dominates this process.Problem # 2Assume that stars are formed with the Salpeter initial mass function (dN/dM ∝ M −2.35 ) between 0.5and 150 M sun , that stars with M i < 8M sun become 0.5M sun WDs, that stars with 30M sun > M i > 8M sunbecome 1.4 M sun NSs, and that stars with M i > 30M sun become 7M sun BHs (the typical WD, NS, and BHmasses chosen here are well-motivated observationally). Assume further that all NSs and BHs are formedvia SN explosions.a) What fraction of stars undergo SN explosions at the end of their lives?We know that the Salpeter initial mass function is given asZdN = M −2.35 dMThus the fraction of the stars that undergo SN explosions would be given by the sum of the fractionsof the stars that become NS and BH, this is given bySN explosions =R 1508 M −2.35 dMR 1500.5 M−2.35 dM = 2.3%53

) What fraction of stars will become WDs? NSs? BHs?The fraction of stars that become WD is given byThe fraction of NS is given byR 80.5M −2.35 dMWD stars = R 1500.5 M−2.35 dM = 97.7%and the fraction of BH is given byNS stars =BH =R 308 M−2.35 dMR 1500.5 M−2.35 dM = 1.97%R 15030 M−2.35 dMR 1500.5 M−2.35 dM = 0.35%c) Estimate the fraction of the mass of a stellar population that is returned to the interstellar medium(via stellar winds or explosions) after 10 Gyrs. You do not need to do a rigorous, accurate to manysignificant digits, calculation.We know that the total mass of a specific type of star can be calculated usingM = dNMdN = AM −2.35 dMwhere A is a normalization constantZ bM T = A M −1.35 dMawe also know that the total mass that is redistributed to the ISM can only come from the fraction of starsthat have M ≥ M sun and for WD this means that the total mass that is redistributed to the ISM (mass loss)is given byZ bM loss = A M −2.35 (M − M end )dMathis becomes[ ( 1 1M loss = A0.35 a 0.35 − 1 )b 0.35 − M (end 11.35 a 1.35 − 1 )]b 1.35we chose integration limits motivated by the knowledge that the sun M = M sun has an approximate lifetimeof 10 Gyr, thus stars wiuth this mass range are the only ones contributing to this enrichment of the ISM.The mass fraction is given byM f rac = M lossM totwhereUsingon Equation 2 andZ bM tot = A M −1.35 dM = A [ 1a .35 a .35 − 1 ]b .35a = 1M sun b = 8M suna = 0.5M sun54M end = 0.5M sunb = 8M sun(5)(6)

on Equation 3 givesM f rac (WD) ≈ 48%This tells us that approximately 48 % of the mass from these stars is given back to the ISM, since we knowthat this population comprises 97 % then the total contribution to the ISM will beFor NS we can sayUsing this in equations 2 and 3 we findf rac(WD) = 0.48 ∗ 97.7 ≈ 47%a = 8 b = 30 M end = 1.4M sunM f rac (NS) = 65%this tells us that approximately 65 % of this mass is given to the ISM and since we know that this populationcomprises 1.97 % then the total mass fraction contribution to the ISM will beand finally For BH we can saUsing this in equations 2 and 3 we findf rac(NS) = .65 ∗.0197 ≈ 1.2%a = 30 b = 150 M end = 7M sunM f rac (BH) = 87.5%this tells us that approximately 87.5 % of this mass is given to the ISM and since we know that thispopulation comprises 0.3 % then the total mass fraction contribution to the ISM will bef rac(NS) = .875 ∗.003 ≈ .2%Thus we can conclude that the total fraction of the mass that is redistributed to the ISM is given byf rac T ≈ 48.4%Problem # 3Consider a white dwarf with a mass of M = 0.5M sun and an effective temperature of 10 4 K.a) Estimate the radius, luminosity, central temperature, and age of the WD. You are free to use any ofthe results on WD cooling quoted in lecture.We know that the radius of a WD can be estimated using( ) M −1/3(µe ) ( ) −5/3 m −1R ≈ 0.013R sun ≈ 0.016R sun ≈ 1.11 × 10 9 cmM sun 2 m eThe luminosity can be calculated usingL = 4πσR 2 T 4e f f = 5.83 × 1030 erg s −1 = 2.3 × 10 −3 L sunThe central temperature is given by( ) LT c = 10 8 M 2/7sunK= 1.35 × 10 7 K5L sun M55

and finally the age can be calculated using( ) Lt = 10 6 M −5/7sunyrs= 1.49 × 10 8 yrs5L sun Mb) Estimate the thickness of the photosphere of the WD. What is the number density in the photosphere?Assume for simplicity that the opacity in the photosphere is approximately equal to the electronscattering opacity.We know that the thickness of the photosphere is given by the scale heighth = R2 kT¯mGM =R2 kTµ H m p GM ≈ 1.5 × 104 cm ≈ 1 × 10 −5 R W Dwe used µ H ≈ 1 since we assume that the atmosphere of the WD is comprised primarily of hydrogen. Weknow that the mean free path is equal to the scale height in the photosphere, thusthusl = 1nσ = hn = 1hσ T≈ 9.75 × 10 19 cm −3c) Use the Saha equation for the ionization of hydrogen to estimate the temperature at which hydrogenis 1/2 ionized at the surface of a WD. Is this larger or smaller than the temperature at which hydrogen is1/2 ionized for photospheric densities appropriate to MS stars?We know that the Saha equation can be written asn p n en H= g pg eg Hn Q,e e −χ/kT e f fwe know that in a gas of 1/2 ionized hydrogen n e = n p = n H and also g p = 1 g e = 2 g H = 2 and χ =13.6 eV. The quantum density is given asthus we can write the Saha equation asn H =n Q,e =( 2πme kTh 2 ) 3/2( 2πme kh 2 ) 3/2T 3/2e f f e−13.6/kT e f fwe know that for every hydrogen atom there are both a proton and electron, thus the hydrogen densityaccounts for 1/3 the total densityPlugging in all the constants yieldn H = n 3 = ( 2πme kh 2 ) 3/2T 3/2e f f e−13.6/kT e f f7.43 × 10 −5 T 3/2e f f e−1.57×105 /T e f f− 1 = 056

solving this numerically yieds a temperature ofT e f f ≈ 2.6 × 10 4 Kwhich is a higher than the 1/2 ionization temperature for the solar photosphere T ∼ 1.3 × 10 4 K. This isdue to the differences in the densities.Problem # 4Consider the late stages of evolution of a 25M sun star. Focus on the core which has a mass ∼ 1M sunand a radius ∼ 10 8 cm. The star’s photon luminosity is 3 × 10 5 L sun . You might find some of the numbersin Table 4.2 of Phillips useful.a) Estimate the temperature at which cooling by neutrinos (which are optically thin and leave the coredirectly) exceeds cooling by photons (which random walk out of the star). Use the expression for theneutrino luminosity from class. At what stage of nuclear fusion (H, He, C, O, Ne, Si, ....) does neutrinocooling become dominant?We know that the luminosity given by the neutrinos is given byand the luminosity of the photons is given by( ) 3L ν ≈ 10 12 T93 Rc10 −2 e −11.9/T 9L sunR sunL ph = 3 × 10 5 L sunwe can find the temperatures at which this are approximately equal by setting these expressions equal toeach other and solving it numerically, i.ethis simplifies tosolving this numerically yields( ) 3L ν = L ph 10 12 T93 Rc10 −2 e −11.9/T 9L sun = 3 × 10 5 L sunR sun3.3 × 10 6 T 39 e−11.9/T 9− 1 = 0T 9 ≈ .794⇒ T ≈ 7.93 × 10 8 KNeutrino cooling becomes important after Helium fusion and before Carbon fusion due to the temperaturesgiven in table 4.2 of Phillips for these reactions.b) If neutrino cooling were unimportant (and thus the photon luminosity determined the energy lost bythe star), estimate the time it would take the ≃ 1M sun core of the star to fuse from 20 Ne to 56 Fe. Comparethis to the true time of about 1.5 years (from Phillip’s Table) set by neutrino cooling. Assume that theluminosity of the star is independent of time and that fusion of heavy elements releases ≃ 0.7 MeV pernucleon (≃ 10 times less the fusion of H to He because the binding energies of heavy nuclei are closer toeach other.We know that the time is given byt = EL ph= NQL ph57

we know that the number of particles are given byand the energy released isthus we find that the time isN = M sunm pQ ≈ 0.7 MeV/nucleont = M sunQm p L ph≈ 3.6 × 10 4 yrsWe can see that neutrino cooling is very effective in “killing” a star. Due to the fact that the star wouldexist for much longer if it were not for those pesky neutrinos.Problem set 12Problem # 1The Energy needed to dissociate one 4 He nucleus into two neutrons and two protons is Q = 28.3 MeV.Derive an expression relating the numbers of 4 He nuclei, neutrons and protons coexisting at a temperatureT in an equilibrium set up by the reactionsγ+ 4 He ⇋ 2n+2pCalculate the temperature for 50 % dissociation when the density is 10 12 kg m −3 . [Note :This is a simpleexample of nuclear statistical equilibrium (NSE) discussed in class, i.e., the balance of nuclei determinedwhen nuclear reactions go both ways at high temperatures (because photons have enough energy to photodisintegratenuclei into their more basic constituents).] In addition to calculating the temperature for 50% dissociation of He, also show explicitly that at high temperatures, NSE favors the nuclei being brokenapart (n and p in this case) while at low temperatures it favors nuclei being bound (He in this case).We can use the Saha equation, which is given byµ(γ)+µ( 4 He) = µ(2n)+µ(2p)which can also be written as( )( )m He c 2 gHe n Q,He− kT ln = 2m n c 2 gn n 2 ( )Q,n− kT ln + 2m p c 2 gp n 2Q,p− kT lnn He n pn n58

earranging this equation yieldsc 2 (m He − 2m n − 2m p ) = kT lnwe are given the binding energy needed to dissociate a helium atomthuswe are also givenand the quantum concentration is defined asand if we assume for simplicityit follows that( (gHe ) ( ))n Q,He n 2 n n2 pn He (g n n Q,n ) 2 (g p n Q,p ) 2c 2 (m He − 2m n − 2m p ) = −28.3 MeVn 2 nn 2 ( ) ( )p gHe n Q,He=n He (g n g p ) 2 (n 2 e −28.4MeV/kTQ,n n2 Q,pn Q,n = n Q,p =g He = 1 g n = g p = 2n Q,A =( 2πmA kTh 2 ) 3/2( ) 2πmp kT 3/2 ( ) 8πmp kT 3/2h 2 n Q,He =h 2n 2 nn 2 pn Heinserting the expressions for the quantum density givesif we assume that this gas is 50% dissociated gives us= n4 Q,pn Q,Hee −28.4MeV/kTn 2 ( )n n2 p 2πmp kT 9/2= 16(8)n He h 2 e −28.4MeV/kT (7)n n = n p = 2 5 n n He = 1 5 nfor every helium nuclei there are two neutrons and two protons, giving a total of five particles. We alsoknow thatn = ρ¯m ¯m = m He + 2m n + 2m p≈ 8 5 5 m p n = 5 ρ8Using this along with plugging in all the constants into Equation 1 gives us7.84 × 10 21 T 9/212 e−0.328/T 12= 1 T 12 = T10 12 Ksolving this equation numerically yields a temperature ofm pT 12 = 0.0109T ≈ 1.09 × 10 10 K59

To show explicitly that at high temperatures, NSE favors the nuclei being broken apart (n and p in thiscase) while at low temperatures it favors nuclei being bound (He in this case) we must consider Equation1 along with assumingn = n He + n n + n p ≈ n He + 2n pwe can see thatputting this into Equation 1 givesn He ≈ n − 2n pn 2 nn 2 ( )p2πmp kT 9/2= 16(8)n − 2n p h 2 e −28.4MeV/kT (8)If we consider the case where T → ∞ (very high temperatures) we can see that Equation 2 goes to infinity,this only happens if the denominator is 0n − 2n p = n He = 0n = 1 2 n pthus there are no bound nucleus only protons and neutrons in equal numbers. If we now consider the casewhere T → 0 we can see that Equation 2 goes to 0 this can only happen inn p = n n = 0n = n Heand thus it favors nuclei being bound.Problem # 2Compare the total energy released by a 25 M sun star during (a) its pre-main sequence evolution (KHcontraction), (b) its time on the MS, (c) its post-main-sequence-evolution, and (d) the supernova explosionto form a neutron star.To calculate the total energy released during the pre-main sequence we can assume can just calculatethe total gravitational energy relesed from contraction in the star. We know that the energy for a boundsystem is given by the Virial theorem asis given byE ∼ − U 2∆E ≈ |E i | − |E F | ≈ |E F | ≈ − U 2 ∼ GM22R MSwhere R MS is given by the main sequence radius which is defined assince we know that the mass of this star isR MS ≈ R sun( MM sun) 6/7we find the radius to beM = 25M sunR = 15.78R sun60

and the gravitational energy, which is the total energy is∆E pre−MS = G(25M sun) 22 · 15.78R sun≈ 7.45 × 10 49 ergsTo calculate the total energy of a 25M sun star during the main sequence we need to multiply totalluminosity by the total time that the star spends on the main sequencewhere the luminosity is given by∆E = L MS t MSL = L sun( MM sun) 3.5= 7.8 × 10 4 L sunand the time a star spends o the main sequence is given asthis gives the total energy ast MS = 10 10 ( MM sun) −2.5yrs = 3.2 × 10 6 yrs∆E MS ≈ 3.08 × 10 52 ergswe can compare this to the value derived by nuclear energetics, we can say that the total energy is givenbywhere we know∆E ≈ 0.1NQ He = 0.1 M m pQ HeM = 25M sunQ He ≈ 7 MeV/nucleonWhere Q He is the nuclear binding energy per nucleon of helium. We find the energy to be given aswhich is almost them.as∆E ≈ 3.33 × 10 52 ergsTo find the total energy during the post main sequence stage we need to consider the energy to be given∆E = 0.5N∆Qwhere ∆Q is the difference in the binding energy of iron to the binding energy of helium and we assumedthat 50% of the star will undergo fusion. The binding energy per nucleon of iron and helium are Q Fe ≈8.78 MeV/nucleon and Q He ≈ 7 MeV/nucleon, thusN = 25M sunm p∆Q ≈ 1.78 MeV/nucleonand we find the total energy to be given as∆E post−MS ≈ 4.24 × 10 52 ergs61

To find the total energy released during the supernova explosion can be estimated using the sameequation as part a) except now the final radius is given by the radius of te neutron star.∆E ≈ GM22R NSwhere the mass that we will consider will be the mass of the coreM ≈ 1.4M sunR ≈ 10 kmthus we find the energy to be∆E SN ≈ 2.59 × 10 53 ergsthis energy release is much greater than all other processes.Problem # 3Consider an ideal degenerate gas of electrons, protons and neutrons, and the equilibrium establishedby the reactionsn → p+e − + ν¯e and e − + p → n+ν eAssume equal numbers of electrons and protons and assume that the density is so high that all the degenerateparticles are ultra-relativistic. Show that the number densities of the particles are in the ration e : n p : n n = 1 : 1 : 8Using the Saha equation, along with the knowledge that all of the particles are now reletavistic we findand we know that the Fermi momentum is given byµ(n)+µ(ν e ) = µ(p)+µ(e − )p F =and the Fermi energy for relativistic particles isthus we find( ) 1/3 3nehc+8π( ) 1/3 3nphc −8π( ) 3n 1/3h8πε F = p F c( ) 1/3 3nnhc = m n c 2 − m p c 2 ≈ 08πwhere we made the assumption that m n c 2 − m p c 2 ≈ 0. We are also told thatn e = n pthus we find2n 1/3p = n n n e = n p = 1 8 n nProblem # 462

Assume that a hot, bloated neutron star emits thermal neutrino radiation from a surface of radius R atan effective temperature equal to T E . Assume that three types of massless, or nearly massless, neutrinos,ν e ,ν ν ,ν τ and their antiparticles, are emmited in equal numbers, in thermal equilibrium with zero chemicalpotential. Show that the luminosity is given byL ν = 218 σT 4 E 4πR2where σ is Stefan’s constant. Find an expression for the average energy for a neutrino in this radiation.[Hint: Look back at Chapter 2 and reconsider Problem 2.5]If we refer to Philipps problem 2.5 we find that the energy density of fermions is given byand we know that the energy density of a photon isu F = 7 8 aT 4 (9)u p = aT 4 (10)the differences in these two expressions comes from solving the following two integralsn = 1 VZ ∞0[ ] kT 3 Z ∞ x 2N(p)dp = 8πhc 0 e −x ± 1 dxthis is the number density of particles with momentum p and p + dp, the ± is to differentiate betweenbosons and fermions. Since neutrinos are fermions. And the energy density is given asu = 1 VZ ∞0[ ] kT 3 Z ∞ x 3ε p N(p)dp = 8π kThc 0 e x ± 1 dxfrom Equationquations 3 and 4 we can see that the solution to the energy density for a fermion is given byu F = 7 8 aT 4this was using the assumptions that the polarization of the fermion is 2, but we know that the polarizationof neutrinos is 1. We also need to take into account the 6 different species of neutrinos, thus for neutrinoswe find that the energy density is given aswe also know that Stefan’s constant is defined asu ν = 6 72 8 aT 4 = 218 aT 4σ = ac4thus the energy density is now given asu ν = 21 48 c σT 4but we know that the flux due to the neutrinos is given byF ν = 218 σT 463

where the factor of 4/c was taking care of the fact the intensity radiated at a particular frequency is c/4times the photon energy density at this frequency. We know that the luminosity of neutrinos is given byL ν = F ν 4πR 2and we just derived the flux for neutrinos, thus the luminosity isProblem # 5L ν = 218 σT 4 E 4πR 2In this problem we will calculate the properties of the neutrinos emitted by a newly formed neutronstar (a “proto-NS”). The neutron star is formed during a supernova explosion and its gravitational bindingenergy E NS is released in the form of neutrinos on a timescale t KH , so that the neutrino luminosity ofthe NS is L = E NS /t KH . Assume that the NS has a mass of 1.4M sun and approximate it as an n = 3/2polytrope supported exclusively by neutron degeneracy pressure. The initial central temperature of the NSis ≃ 10 11 K.a) Calculate the radius and central density ρ c of the NS.We know that the relationship between the radius and the mass of a NS is given by( ) M −1/3R NS = 15 kmM sunand since we know that M = 1.4M sun we find the radius to be approximatelyR NS = 13.41 kmthe mass density can be found by using the central density of a n = 3/2 polytrop which is given asthus the central density isgiven the mass and radius we findρ c = 3M4πR 3 a na n = 5.99 n = 3/2 polytropeρ c ≈ 1.43 MR 3 NSρ c ≈ 1.65 × 10 15 g/cm −3b) Show that the neutrinos are initially degenerate in the core of the NS. Use this fact to estimate thetypical energy E ν of a neutrino in the core of the NS. The neutrinos are relativistic.wherewe know thatwe know thatE F (e) =E ν ≈ E F (e)( ) 1/3 3nehc8πn e ≈ 1 8 n n ≈ 1 864ρ cm p

thus we find the Fermi energy to beE ν ≈ E F (e) ≈to show that they are degenerate we just need to show( ) 1/3 3nnhc ≈ 303 MeV64πE ν ≥ E Twhere E T is the thermal energy, we find that for relativistic particles isE T ≈ 3kT ≈ 25.8 MeVthus we can see thatE ν ≥ E Tc) The cross section for neutrino’s interacting with matter isσ ν ≃ 10 −44 (Eνm e c 2 ) 2cm 2Estimate the optical depth τ = nσ ν R ≃ R/l ν of the NS to neutrinos, where l ν is the neutrino mean freepath in the core of the NS.Since we know that the optical depth is given byτ = nσ ν R NSwhere we have defined n to be the total number densitywe findn ≈ ρ cm p≈ 9.86 × 10 38 cm −3 R NS ≈ 13.41 km m e c 2 ≈ 0.508 MeVτ ≈ 4.71 × 10 6and we know that the neutrino mean free path is given byl ν = R NSτ≈ 0.284 cmd) The timescale t KH for the NS to radiate away its binding energy in neutrinos is the time for theneutrinos to random walk out of the NS. Use your result from c) to estimate the time t KH and the neutrinoluminosity L ν of the NS.we know that the time it takes for a neutrino to random walk out of the NS is given byt = R2 NSvl ν= R2 NScl νgiven the radius of the NS and the mean free path we find the time to bet ≈ 210 s65

we know that the luminosity is given asL ν = E νtusing the neutrino energy in part b) and the time we find that the neutrino luminosity isL ν ≈ 9.19 × 10 50 ergs/se) Look at Problem 4 Note Phillips’ hints at the back of the book for problems 6.3 and 2.5.f ) Use your results from d) and problem 4 to calculate the effective temperature of the neutrino radiation(in K) and the average energy of a neutrino emitted by the newly formed NS (in MeV). For comparisonto the results you have calculated in this problem, the observed timescale of neutrino emission was ≃ 10 sfor SN 1987A and the typical neutrino energy was ≃20 MeV.The effective temperature is given as( )8L 1/4νT E =21 · 4πR 2 ≈ 1.48 × 10 10 Kσthe energy of a neutrino can be calculated by knowing the energy density and the number density. Theenergy is given byE ν ≈ u νn νwhere the number density is given by Philips equation 2.42 with a modification coming fro the fact thatwe are dealing with fermionsn = bT 3 where b = 6 8πk31.803 ×2 h 3 c 3 = 45.48K−3 cm −3thus we know that the energy of a neutrino is given byE ν = 218aT ≈ 6.24 MeVb66