Astronomy 112: The Physics of Stars Class 10 Notes: Applications and Extensions of Polytropes In the last class we saw that polytropes are a simple approximation to a full solution to the stellar structure equations, which, despite their simplicity, yield important physical insight. This is particularly true for certain types of stars, such as white dwarfs and very low mass stars. In today’s class we will explore further extensions and applications of simple polytropic models, which we can in turn use to generate our first realistic models of typical main sequence stars. I. The Binding Energy of Polytropes We will begin by showing that any realistic model must have n < 5. Of course we already determined that solutions to the Lane-Emden equation reach Θ = 0 at finite ξ only for n < 5, which already hints that n ≥ 5 is a problem. Nonetheless, we have not shown that, as a matter of physical principle, models of this sort are unacceptable for stars. To demonstrate this, we will prove a generally useful result about the energy content of polytropic stars. As a preliminary to this, we will write down the polytropic relation (n+1)/n P = KP ρ in a slightly different form. Consider a polytropic star, and imagine moving down within it to the point where the pressure is larger by a small amount dP . The corre- sponding change in density dρ obeys n + 1 dP = K ρ1/ndρ. P n Similarly, since P = K ρ1/n, ρ P it follows that P ! K 1 dP ! d = P ρ(1−n)/ndρ = ρ n n + 1 ρ We can regard this relation as telling us how much the ratio of pressure to density changes when we move through a star by an amount such that the pressure alone changes by dP . With this out of the way, we now compute the gravitational potential energy of a polytropic star of total mass M and radius R. For convenience, since we’ll be changing variables many times, we will write limits of integration as s for surface or c for center, 1 indicating that, whatever variable we’re using, it is to be evaluated at the surface or the center. The potential energy is Z s Gm Ω = − dm c r 1 Z s G = − d(m2) 2 c r "Gm2 #s 1 Z s Gm2 = − − 2 dr 2r c 2 c r GM 2 1 Z s Gm2 = − − dr 2R 2 c r2 In the third step, we integrated by parts. Next, we use the equation of hydrostatic balance, dP/dr = −Gmρ/r2, or dP = −(Gm/r)(ρ/r) dr to replace Gm/r in the integral, and then make use of the result we just derived: GM 2 1 Z s dP Ω = − + m 2R 2 c ρ GM 2 n + 1 Z s P ! = − + m d . 2R 2 c ρ Now we integrate by parts one more time: GM 2 "n + 1 P #s n + 1 Z s P Ω = − + m − dm 2R 2 ρ c 2 c ρ GM 2 n + 1 Z s P = − − 4πr2ρ dr 2R 2 c ρ GM 2 n + 1 Z s 4π = − − P d(r3). 2R 2 c 3 In the second step, the term in brackets vanishes because m = 0 at the center and P/ρ = 0 at the surface. Finally, one last integration by parts, followed by another application of hydrostatic balance: GM 2 n + 1 4π s n + 1 Z s Ω = − − P r3 + 4πr3 dP 2R 2 3 c 6 c GM 2 n + 1 Z s Gm = − − 4πr3 ρ dr 2R 6 c r2 GM 2 n + 1 Z s Gm = − − dm 2R 6 c r As before, the term in brackets vanished in the second step because r = 0 at the center and P = 0 at the surface. 2 Notice that we have gotten back to where we started: the integral on the right hand side is just −Ω. Thus we have shown that GM 2 n + 1 Ω = − + Ω 2R 6 3 GM 2 Ω = − . 5 − n R Given this result, one can get the total energy simply by applying the virial theorem. If the star consists of monotonic ideal gas, we have 1 3 GM 2 E = Ω + U = Ω = − . 2 10 − 2n R This is a handy formula to know, but it also shows us why only polytropic models with n < 5 are acceptable models for stars. At n = 5, the total energy of the polytrope becomes infinite, and for n > 5 the energy E > 0, meaning the object is unbound. Thus only models with n < 5 represent gravitationally bound objects with finite binding energy. II. Radiation Pressure and the Eddington Limit Polytropes are very useful, but, since they separate the hydrostatic balance of a star (as embodied by the first two stellar structure equations) from its energy balance (as embodied by the second two), they have limitations. In particular, they cannot by themselves tell us anything about how energy flows through a star, and thus about stars’ luminosities. To put this into our models, we must re-insert the temperature- dependence. This brings back into the picture the third structure equation: dT 3 κρ F dT 3 κ F = − = − . dr 4ac T 3 4πr2 dm 4ac T 3 (4πr2)2 We care about the temperature is because of its relationship with the pressure, and in particular with the pressure of radiation. That is because the gas pressure follows 4 Pgas ∝ T , but radiation pressure has a much steeper dependence: Prad = aT /3. Thus at sufficiently high temperatures radiation pressure always dominates. You calculated when radiation pressure begins to dominate on your last homework. The strong dependence of Prad on T has important consequences for stellar structure. To see this, it is helpful to rewrite our last equation in terms of radiation pressure rather than temperature: dT 3 κρ F = − dr 4ac T 3 4πr2 4 dT κρ F aT 3 = − 3 dr c 4πr2 dP κρ F rad = − dr c 4πr2 3 Repeating the same trick using the Lagrangian form gives dP κ F rad = − . dm c (4πr2)2 Now consider what this implies for hydrostatic balance. Since P = Pgas + Prad, the equation of hydrostatic balance can be written as dP dP Gm rad + gas = −ρ dr dr r2 dP Gm dP rad = −ρ − gas dr r2 dr Since density and temperature always fall with radius within a star, dPgas/dr is always negative, so the term −dPgas/dr > 0. Thus we have dP Gm rad > −ρ dr r2 κρ F Gm − > −ρ c 4πr2 r2 ! 2 −1 ! 4πcGm 4 M 0.4 cm g F < = 3.2 × 10 L , κ M κ where we have normalized κ to the electron scattering opacity because that is usually the smallest possible opacity in a star, producing the maximum possible F . This result is known as the Eddington limit, named after Arthur Eddington, who first derived it. In his honor, the quantity on the right-hand side is called the Eddington luminosity: 4πcGm L = . Edd κ This result represents a fundamental limit on the luminosity of any object in hydro- static equilibrium. It applies to stars, but it applies equally well to any other type of astronomical system, and the Eddington limit is important for black holes, entire galaxies, and in many other contexts. This also lets us derive a useful relation describing how the ratio of radiation pressure to total pressure varies within a star. We have already shown that dP κρ F rad = − dr c 4πr2 dP Gm = −ρ , dr r2 and taking the ratio of these two equations gives dP κF F rad = = . dP 4πcGm LEdd 4 The Eddington limit also tells us something about nuclear energy generation at the center of a star, where m = 0 and F (m) = 0. We can Taylor expand F around m = 0, so F ≈ m(dF/dm). Inserting this into our inequality gives dF 4πcG = q < , dm c κ where we have noted that the final stellar structure equation asserts that dF/dm = q. Thus we have derived an upper limit on the rate of nuclear energy generation qc in the center of a star. There is an important subtlety here, which is important to notice. Our equation relating dT/dr and F is calculated assuming that F comes solely from radiation, i.e. that there are no other sources of energy transport in stars. This means that the F that appears in the limit includes only the radiative heat flow F , not any other sources of heat transport – and we will see next week that there is potentially another type of energy transport that is important in some stars. Although the radiative heat flow always obeys the limit we have derived, the total heat flow need not. This is obviously not much a limitation at the surface of a star, since there radiation is the only means by which energy can move around. III. The Eddington Model Considering the importance of radiation pressure leads to an important model, essen- tially the first quasi-realistic model for stellar structure. It was first derived by Arthur Eddington, and is therefore known as the Eddington model. The first step in this model is the equation we just derived for the relation between dP κF F rad = = . dP 4πcGm LEdd Eddington then assumes that the right hand side is constant. This might seem like a crazy assumption, but it turns out to be reasonably good. You will show that it holds for massive stars on your homework.