Astronomy 112: the Physics of Stars Class 9 Notes: Polytropes with Our

Astronomy 112: The Physics of Stars

Class 9 Notes: Polytropes

With our discussion of nuclear reaction rates last time, we have mostly completed our survey of the microphysical properties of stellar matter – its pressure, how energy ﬂows through it, and how it generates energy from nuclear reactions. For the next few weeks we will be using those microphysical models to begin to make our ﬁrst models of stars.

I. The Stellar Structure Equations To begin we will collect the various equations we have developed thus far to describe the behavior of material in stars. As always, we consider a shell of material of mass dm and thickness dr, which is at a distance r from the center of the star and has a mass m interior to it. The shell has density ρ, pressure P , temperature T , and opacity κ. The radiation flux passing through it is F , and the shell generates energy via nuclear reactions at a rate per unit mass q. We will write down the equations describing this shell, under the assumption that the star is in both hydrostatic and thermal equilibrium, so we can drop all time derivatives describing change in position, energy, etc. Since we are assuming equilibrium, for now we will also assume that the composition is fixed, so that we know X, Y , Z, and any other quantities we need that describe the chemical makeup of the gas. The first equation is just the definition of the density for the shell, which says that ρ = dm/dV . Writing this in Eulerian or Lagrangian form (i.e. with either radius or mass as the independent variable), we have dm dr 1 = 4πr2ρ = . dr dm 4πr2ρ The second equation is the equation of hydrostatic balance, which we can also write in either Eulerian or Lagrangian form: dP Gm dP Gm = − ρ = − . dr r2 dm 4πr4 This just equates the change in gradient in pressure with the force of gravity. The third equation is the equation describing how the temperature changes with position within a star as a result of radiative diffusion: dT 3 κρ F dT 3 κ F = − = − . dr 4ac T 3 4πr2 dm 4ac T 3 (4πr2)2 Finally, we have the equation of energy conservation, which for material in equilibrium just equates the change in energy flux across a shell to the rate at which nuclear reactions generate power within it: dF dF = 4πr2ρq = q. dr dm 1 These are four coupled, non-linear ordinary differential equations. As we discussed a few weeks ago, by themselves they are not a complete system, because by themselves they contain more unknowns than equations. The unknowns appearing are ρ, T , P , F , κ, and q. If we adopt Eulerian coordinates, then m is also an unknown, and r is the independent variable. For Lagrangian coordinates, r is also an unknown, and m is the independent variable. Thus we have seven unknowns, but only four equations. We therefore need three more equations, and that is what we have spent the last few weeks providing. The pressure depends on density and temperature via

R 1 4 P = ρT + Pe + aT . µI 3 The first term is the ion pressure, which we have written assuming that ions are non-degenerate, which they are except in neutron stars. The last term is the radiation pressure. The middle term, the electron pressure, takes a form that depends on whether the electrons are degenerate or not, but which is a known function of ρ and T . The opacity and nuclear energy generate rate, we have seen, are in general quite complicated functions. However, we have also seen that they can be approximated reasonably well as powerlaws: a b m n κ = κ0ρ T q = q0ρ T . Regardless of whether we make the powerlaw approximation or not, we now know how to compute κ and q from ρ and T . Thus we have written down three more equations involving the unknowns. We are therefore up to seven equations for our seven unknowns, which is sufficient to fully specify the system. The only thing missing is boundary conditions, since differential equations produce constants of integration that must be determined by boundary conditions. Since there are four differential equations, we need four boundary conditions. Three of them are obvious. In Lagrangian coordinates, we have r = 0 and F = 0 at m = 0, and P = 0 at m = M. In words, the first condition says that the innermost mass element must reside at radius r = 0, and it must have zero flux (F = 0) entering it from below. The third condition says that the pressure falls to zero at the boundary of the star, m = M. The fourth condition is slightly more complicated. The simplest approach is to set T = 0 at the star’s surface. This is actually not a terrible approximation, since the temperature at the surface is very low compared to that in the interior. A better approach is to specify the relation between flux and temperature at the stellar surface as F = 4πr2σT 4. An even better approach is to make a detailed model of a stellar atmosphere and figure out how the flux through it depends on its temperature and pressure, and use that as a boundary condition for the stellar model. The set of seven equations and four boundary conditions we have written down now fully specifies the structure of a star. Solving those equations, however, is another matter entirely. There is no general method for solving sets of coupled non-linear

2 differential-algrebraic equations subject to boundary conditions specified at two points. There is every reason to believe that such equations cannot, in general, be solved in closed form. Today the standard approach is to hand the problem to a computer. A computer can integrate the equations and find solutions to any desired level of accuracy, and this problem is sufficiently simple that the calculations will run on an ordinary desktop machine in a matter of seconds. However, in the days when people first approached these problems, there were no such things as computers. Instead, people were forced to come up with analytic approximations, and it turns out that one can understand a great deal about the behavior of stars using such approximations. (It turns out that a significant fraction of being a good physicist consists in the ability to come up with good approximations for intractable differential equations.)

II. Polytropes

A. Definition and Motivation The first two stellar structure equations, describing the definition of density and hydrostatic equilibrium, are linked to the second two only via the relationship between pressure and temperature. If we can write the pressure in terms of the density alone, without reference to the temperature, then we can separate these two equations from the others and solve them by themselves. Solving two differential equations (plus one algebraic equation relating P and ρ is much easier than solving seven equations. As a first step in this strategy, we can combine the first two first-order ODEs into a single second-order ODE. To do so, we start with the equation of hydrostatic equilibrium and multiply by r2/ρ to obtain r2 dP = −Gm. ρ dr Next we differentiate both sides: d r2 dP ! dm = −G . dr ρ dr dr Finally, we substitute for dm/dr using the definition of density, dm/dr = 4πr2ρ. Doing so we obtain 1 d r2 dP ! = −4πGρ. r2 dr ρ dr This is just another form of the equation of hydrostatic balance, this time with the definition of density explicitly substituted in. Thus far everything we have done is exact. Now we make our approximation. We approximate that the pressure and density are related by a powerlaw

γP P = KP ρ .

3 Equations of state of this sort are called polytropes. For historical reasons, it is common to deﬁne 1 1 γP = 1 + or n = n γP − 1 where we call n the polytropic index. Before going any further, it is important to consider whether an equation of state like this is at all sensible. Why should a star ever obey such an equation of state? The answer to this question becomes clearer if we recall that, for an adiabatic gas, the equation of state reads

γa P = Kaρ ,

where Ka is the adiabatic constant and adiabatic index. It is important to understand the difference between this relation and the polytropic relation. The polytropic relation describes how the pressure changes with density inside as one moves through a star, while the adiabatic equation of state describes how a given gas shell would respond to being compressed. The constant Ka depends on the entropy of the gas in a given shell, so different shells in a star can have different values of Ka. If different shells have different Ka values, then as I move through the star the pressure will not vary as P ∝ ργ, because different shells will have different constants of proportionality. Thus a star can be described by a polytropic relation only if Ka is the same for every shell. While this condition might seem far-fetched, it is actually satisfied under a wide range of circumstances. One circumstance when it is satisfied is if a star is dominated by the pressure of degenerate electrons, since in that case we proved a few 0 5/3 0 4/3 classes back that Pe = K1(ρ/µe) for a non-relativistic gas, or Pe = K2(ρ/µe) . 0 0 The proportionality constants K1 and K2 depend only on constants of nature like h, c, the electron mass, and the proton mass, and thus do not vary within a star. For such stars, γP = 5/3 or 4/3, corresponding to n = 1.5 or n = 3, for the non-relativistic and highly-relativistic cases, respectively.

Another situation where Ka is constant is if a star is convective. As we will discuss in a week or so, under some circumstances the material in a star can be subject to an instability that causes it to move around in such a way as to enforce that the entropy is constant. In a region undergoing convection, Ka does not vary from shell to shell, and a polytropic equation of state is applicable. Signiﬁcant fractions of the mass in the Sun and similar stars are subject to convection, and for those parts of stars, a polytropic equation of state applies well, and is a good approximation for the star as a whole. In low mass stars where the gas is non- relativistic and radiation pressure is not signiﬁcant, γa = 5/3, so constant Ka means that γP = 5/3 as well, and n = 1.5. Because they apply in a broad range of situations, polytropic models turn out to be extremely useful.

B. The Lane-Emden Equation

4 γP (n+1)/n Having motivated the choice P = KP ρ = KP ρ for an equation of state, we can proceed to substitute it into the equation of hydrostatic balance. Doing so, after a little bit of simpliﬁcation we ﬁnd (n + 1)K 1 d r2 dρ! P = −ρ. 4πGn r2 dr ρ(n−1)/n dr

As a second-order equation, this ODE requires two boundary conditions. At the surface r = R, the pressure P (R) = 0, and since P ∝ ργP , one of our boundary conditions could be ρ(R) = 0. However, usually one instead sets the density at the center, ρ(0) = ρc, and then R is the radius at which ρ ﬁrst goes to zero. For the second boundary condition, recall that the original equation of hydrostatic equilibrium read dP/dr = −ρGm/r2. Unless the density becomes inﬁnite in the 3 2 center, m must vary as m ∝ ρcr near the center of the star, so m/r must vary as r. Thus dP/dr goes to zero at r = 0, and, for our polytropic equation of state, dρ/dr must therefore approach zero in the center as well. This is our second boundary condition: dρ/dr = 0 at r = 0.

Note that our equation now involves three constants: KP and n, which come from the equation of state, and R, which sets the total radius of the star. Since these are the only constants that appear (other than physical ones), these must fix the solution. In other words, for a given polytrope with a given choice of KP , n, and R, there is a single unique density profile ρ(r) which is in hydrostatic equilibrium. In fact, it is even simpler than that, as becomes clear if we make a change of variables. Let ρc be the density in the center of the star, and let us define the new variable Θ by ρ = Θn. ρc Note that Θ is a dimensionless number, and that it runs from 1 at the center of the star to 0 at the edge of the star. With this change of variable, we can re-write the equation of hydrostatic balance as " # ! (n + 1)KP 1 d 2 dΘ n (n−1)/n 2 r = −Θ . 4πGρc r dr dr The quantity in square brackets has units of length squared, and this suggests a second change of variables. We let " # 2 (n + 1)KP α = (n−1)/n 4πGρc and we then set r ξ = α so that ξ is also a dimensionless number. With this substitution, the equation becomes 1 d dΘ! ξ2 = −Θn. ξ2 dξ dξ

5 The two central boundary conditions are now Θ = 1 and dΘ/dξ = 0 at ξ = 0. The equation we have just derived is called the Lane-Emden Equation. Clearly the only constant in the equation is n. Before moving on to solve the equation, it is worth mentioning the procedure used to derive it, which is a very general and powerful one. This procedure is called non-dimensionalization. The basic idea is to take an equation relating physical quantities, like the hydrostatic balance equation we started with, and re-write all the physical quantities as dimensionless numbers times their characteristic values. n For example, we re-wrote the density as ρ = ρcΘ . We re-wrote the length as r = αξ. The advantage of this approach is that it allows us to factor out all the dimensional quantities and leave behind only a pure mathematical equation, and in the process we often discover that the underlying problem does not depend on the dimensional quantities. In this case, it was far from obvious that the structure of a polytrope didn’t depend on R, ρc, KP , etc. After all, those quantities appear in the hydrostatic balance equation. However, the non-dimensionalization procedure shows that they just act as multipliers on an underlying solution whose behavior depends only on n. Tricks like this come up all the time in the study of diﬀerential equations, and are well worth remembering if you plan to think about them at any point in the future. With that aside out of the way, consider the Lane-Emden equation itself. To get a sense of how it behaves, we can solve it for some chosen values of n. First consider n = 0, corresponding to the limit γP → ∞. In this case the equation is

d dΘ! ξ2 = −ξ2, dξ dξ which we can integrate to obtain

dΘ ξ3 ξ2 = − + C, dξ 3 where C is a constant of integration. Bringing the ξ2 to the other side and integrating again gives ξ2 C Θ = − − + D. 6 ξ Applying the boundary conditions that Θ = 1 and dΘ/dξ = 0 at ξ = 0, we immediately see that we must choose C = 0 and D = 1; so the solution is therefore ξ2 Θ = 1 − . 6 This√ is a function that decreases monotonically at ξ > 0, and reaches 0 at ξ = ξ1 = 6. Analytic solutions also exist for n = 1 and n = 5. Numerically integrating the equation for other values of n shows that, for n < 5, this is generic behavior: Θ decreases monotonically with ξ and reaches zero at

6 some ﬁnite value ξ1. As we will see in a moment, it is particularly useful to know 2 ξ1 and −ξ1 (dΘ/dξ)ξ1 , and these can be obtained trivially from the numerical solution. Some reference values are

2 n ξ1 −ξ1 (dΘ/dξ)ξ1 1.0 3.14 3.14 1.5 3.65 2.71 2.0 4.35 2.41 2.5 5.36 2.19 3.0 6.90 2.02

The radius at which ξ reaches zero is clearly the radius of the star, so

R = αξ1.

Similarly, given a solution Θ(ξ), we can also compute the mass of the star:

Z R M = 4πr2ρ dr 0 Z ξ1 3 2 n = 4πα ρc ξ Θ dξ 0 ! Z ξ1 3 d 2 dΘ = −4πα ρc ξ dξ 0 dξ dξ dΘ! = −4πα3ρ ξ2 . c 1 dξ ξ1

In the third step we used the equation of hydrostatic balance to replace ξ2Θn. From a polytropic model, we can derive a bunch of other useful numbers and relationships. As one example, it is often convenient to know how centrally concentrated a star is, i.e. how much larger its central density is than its mean density. We deﬁne this quantity as ρ D ≡ c n ρ 4πR3 = ρ c 3M  ! −1 4π 3 3 2 dΘ = ρc (αξ1) −4πα ρcξ  3 1 dξ ξ1  −1 3 dΘ! = −   ξ1 dξ ξ1

A second useful relationship is between mass and radius. We start by expressing the central density ρc in terms of the other constants in the problem and our

7 length scale α: "(n + 1)K #n/(n−1) ρ = P . c 4πGα2 Next we substitute this into the equation for the mass:

"(n + 1)K #n/(n−1) dΘ! M = −4πα3 P ξ2 . 4πGα2 1 dξ ξ1

Finally, we substitute α = R/ξ1. Making the substitution and re-arranging, we arrive at " #n−1 !3−n n GM R [(n + 1)KP ] 2 = −ξ1 (dΘ/dξ)ξ1 ξ1 4πG Thus mass and radius are related by M ∼ R(n−3)/(n−1). A third useful expression is for the central pressure. From the equation of state (n+1)/n we have Pc = KP ρc . We can then use the mass-radius relation to solve for KP and then substitute it into the equation of state, which gives

1/n " #(n−1)/n !(3−n)/n (4πG) GM R (n+1)/n Pc = 2 ρc . n + 1 −ξ1 (dΘ/dξ)ξ1 ξ1

Then, using the centrally concentrated measure, Dn, to elliminate R, we get

1/3 2/3 4/3 Pc = (4π) BnGM ρc

where  −1  ! 2/3 2 dθ Bn = − (n + 1) ξ    1 dξ  ξ1 which is relatively independent of n. As we will see, all realistic models have n = 1.5 to n = 3. Thus we expect a nearly universal relation among the central pressure, central density and mass of stars.

C. The Chandrasekhar Mass and Relativistic Gasses Consider what this analysis of polytropes implies for stars where the pressure is dominated by electron degeneracy pressure. White dwarf stars are examples of such stars. To see why, consider what their position on the HR diagram and their masses tells us about them. Observations of white dwarfs in binary systems imply that they have masses comparable to the mass of the Sun. On the other hand, their extremely low luminosities, combined with surface temperatures that are not very diﬀerent from that of the Sun, implies that they must have very small 8 6 radii: r ∼ rE = 6 × 10 cm. The corresponding mean density is ρ ∼ 2 × 10 g cm−3. In contrast, we showed a few classes ago that electrons become degenerate 7 3/2 −3 at a density above ρ/µe ≈ 750(T/10 K) g cm . Thus, unless white dwarf

8 interiors are hotter than ∼ 109 K, which they do not appear to be, the gas must be degenerate.

0 5/3 As long as the gas is non-relativistic, this implies that its pressure is P = K1(ρ/µe) , 0 where K1 is a constant that depends only on fundamental constants. Thus a white dwarf is a polytrope, and γa = γP = 5/3 implies that n = 3/2. For a polytrope of index n, we have just shown that mass and radius are related by R ∝ M (n−1)/(n−3) = M −1/3. Thus more massive white dwarfs have smaller radii, with the radius falling as M −1/3. It is important to realize that this statement is not just true of a particular white dwarf, it is true of all white dwarfs. The constant of proportionality between 0 5/3 M and R depends only on KP and G, and for a degenerate gas KP = K1/µe . 0 For a given composition (ﬁxed µe), this is a universal constant of nature, since K1 depends only on fundamental constants. Thus all white dwarfs in the universe follow a common mass-radius relation that is imposed by quantum physics. However, this cannot continue forever, with ever more massive white dwarfs having smaller and smaller radii. The thing that breaks is that, as M increases and R shrinks, eventually the electrons in the white dwarf are forced by the Pauli exclusion principle to occupy higher and higher momenta. As a result, the gas becomes relativistic. The mean density varies as M ρ ∝ ∝ M 2, R3 and we showed a few weeks that a degenerate gas becomes relativistic when its density exceeds ρ ≈ 3 × 106 g cm−3. µe 6 −3 We got a density of 2 × 10 g cm for a mass of M and a radius of rE, so it doesn’t take much of an increase in mass to push the star above the relativistic limit.

0 4/3 For a relativistic electron gas, we have shown that P = K2(ρ/µe) , where, as 0 0 with K1, the quantity K2 depends only on fundamental constants. Since such a gas has γa = 4/3 and ﬁxed Ka, it can be described by a polytropic equation of state with index n = 1/(γP − 1) = 3. The mass-radius relation has an interesting behavior near n = 3. We showed that it can be written " #n−1 !3−n n GM R [(n + 1)KP ] 2 = , −ξ1 (dΘ/dξ)ξ1 ξ1 4πG or equivalently

1 dΘ! [(n + 1)K ]n/(n−1) M = − ξ(n+1)/(n−1) P R(n−3)/(n−1). (4π)1/(n−1) 1 dξ Gn/(n−1) ξ1

9 Notice that, for n = 3, the R dependence disappears, and we ﬁnd

4 dΘ! K 3/2 M = −√ ξ2 P π 1 dξ G ξ1 A relativistic electron gas has

0 1/3 K2 3 hc KP = 4/3 = 4/3 , µe π 8(µemH) and plugging this into the mass we have just calculated gives   1/2 ! !3/2 3 1 2 dΘ hc 1 5.83 M = Mch = −ξ1  2 = 2 M . 32 π dξ G (mHµe) µ ξ1 e

Everything on the right-hand side except µe is a constant, which means that for a highly relativistic electron gas, there is only a single possible mass which can be in hydrostatic equilibrium. If we have a gas that is depleted of hydrogen, so X = 0, then µe ≈ 2, and we have Mch = 1.46 M . This quantity is called the Chandresekhar mass, after Subrahmanyan Chan- dresekhar, who first derived it. He did the calculation while on his first trip out of India, to start graduate school at Cambridge... at age 20... which goes to prove that the rest of us are idiots.... To understand the significance of this mass, consider what happens when a star has a mass near it. If the mass is slightly smaller than Mch, the star can respond by puffing out a little and adopting a larger radius. This reduces the density and makes the gas slightly less relativistic, so that n decreases slightly and the star can be in equilibrium.

On the other hand, if M > Mch, the gas is already relativistic, and there is no adjustment possible. If a Chandresekhar mass white dwarf gets just a little bit more massive, there simply is no equilibrium state that the star can possibly reach. Instead, it is forced to collapse on a dynamical timescale. The result is invariably a massive explosion. For a white dwarf composed mainly of carbon and oxygen, the carbon and oxygen undergo rapid nuclear burning to elements near the iron peak, and the result is essentially that a nuclear bomb goes oﬀ with a solar mass worth of fuel. The resulting explosion, known as a type Ia supernova, can easily outshine an entire galaxy.

D. Very Massive Stars The Chandrasekhar limit is one way that a star can get into trouble if it is supported by a relativistic gas. It is, however, not the only way. Massive stars are also close to being n = 3 polytropes, but instead of being supported by a relativistic gas of electrons, they are supported by a relativistic gas of photons – i.e. by radiation pressure.

10 Massive stars don’t explode while they are on the main sequence, but they still suffer from instabilities associated with being close to the instability line of n = 3. In massive stars, this instability tends to manifest as rapid mass loss and ejection of gas. The mechanism is not fully understood today, but observations make it clear that massive stars do suffer from instabilities. Perhaps the most spectacular example is the star η Carinae, a massive star that is reasonably close to Earth. In 1843, η Carinae (which is visible with the naked eye, despite being > 2 kpc away), suddenly brightened, briefly becoming the second- brightest star in the night sky. Subsequently it dimmed again, although it remains naked-eye visible. In modern times, observations using the Hubble Telescope show that η Carinae is surrounded by a nebula of gas, called the Homunculus Nebula, which was presumably ejected in the 1843 eruption. [Slide 1 – η Carinae and the Homunculus Nebula] The physical mechanism behind the explosion is still not very well understood, and it is an active area of research. Nonetheless, it seems clear that it is connected with the fact that η Carinae is supported primarily by radiation pressure, which puts it dangerously close to the n = 3 line of instability. We’ll discuss the nature of this instability more in a week or so.