PHYS 633: Introduction to Stellar Astrophysics Spring Semester 2006 Rich Townsend (Rhdt@Bartol.Udel.Edu)

PHYS 633: Introduction to Stellar Astrophysics Spring Semester 2006 Rich Townsend ([email protected])

Governing Equations In the foregoing analysis, we have seen how a star usually maintains a state of hydrostatic equilibrium. We have seen how the energy created in the star through nuclear reactions, or released through contraction, makes its way out in the form of the star’s luminosity. We have seen how the physical processes responsible for transporting this energy can either be radiation or convection. And, finally, we have seen how nuclear reactions within the star, and transport processes such as convective mixing, can alter the stars chemical composition. These four key pieces of physics lead to 3 + I of the differential equations governing stellar structure (here, I is the number of elements whose evolution we are following). Let’s quickly review these equations. First, we have the most general (spherically-symmetric) form for the equation governing momentum conservation, ∂P Gm 1 ∂2r = − − . (1) ∂m 4πr2 4πr2 ∂t2 If the second term on the right-hand side vanishes, then we recover the equation of hydrostatic equilibrium, expressed in Lagrangian form (i.e., derivatives with respect to the mass coordinate m, rather than the radial coordinate r). If this term does not vanish at some point in the star, then the mass shells at that point will being to expand or contract, with an acceleration given by ∂2r/∂t2. Second, we have the most general form for the energy conservation equation, ∂l ∂T δ ∂P = − − c + . (2) ∂m ν P ∂t ρ ∂t The first and second terms on the right-hand side come from nuclear energy gen- eration and neutrino losses, respectively. The third and fourth terms together compose what we have previously termed the gravothermal energy release rate, gi, P ∂ρ ∂u ∂T δ ∂P ≡ − = −c + ; (3) gi ρ2 ∂t ∂t P ∂t ρ ∂t they give a net positive contribution to the luminosity gradient when the star contracts. Third, we have the equation(s) of energy transport. These allow us to de- termine the physical temperature gradient ∇ ≡ ∂ ln T/∂ ln P at each point in the star, from which we can obtain the ‘real’ temperature gradient as ∂T GmT = − ∇. (4) ∂m 4πr4P

1 In radiative regions, the physical gradient is equal to the radiative gradient ∇rad, 3 κρH 3 κlP ∇ ≡ ∇ ≡ P F = . (5) rad rad 4ac T 4 16πacG mT 4

If ∇rad exceeds the adiabatic temperature gradient ∇rad, convection will set in, and we need instead to calculate ∇ from a convective model — usually, mixing-length theory. Finally, we have the I − 1 equations describing the change in the mass fractions describing the star’s chemical composition,   ∂Xi mi X i X i X ji X ik =  pjirji − pjirik = − . (6) ∂t ρ qji qik j k j k

The ﬁnal equation governing the chemical composition is the closure relation

X ∂Xi = 0. (7) ∂t i To describe mixing in convective regions, the above equations are replaced by

Z mb dX¯i ∂Xi dma dmb = dm − Xi,a(Xi,a − X¯i) + (Xi,b − X¯i). (8) dt ma ∂t dt dt There is one final differential equation we must add to the ones above — one based not so much on deep physics, and more on the definition of the mass coordinate m. This is of course the mass-radius relation ∂r 1 = (9) ∂m 4πr2ρ This equation, plus eqns. (1), (2), (4), (6) and (5), together form the 4 + I differential equations governing stellar structure. To solve them, we’re going to need informationr regarding the stellar material — for instance, its equation of state, thermodynamic coefficients, opacity, nuclear rates, and so on. We’re also going to need boundary conditions, to ensure that we obtain only those solutions that are physically-realistic. First, however, let’s step back and have a look over the equations. As Kippen- hahn & Weigert discussion in their §9, certain subsystems can be distinguished in the equations. For instance, the mechanical equations, (1) and (9), can be solved if the density can be expressed solely as a function of the pressure. One possible pressure-density relation is the polytropic relation, P = Kρ1+1/n; sub- stituting this into the mechanical equations leads to the Lane-Emden equation. Another pair of subsystems can be formed by splitting the chemical composition equations from the others. This split is highlighted in the title of Kippen- hahn & Weigert: ‘Stellar structure and Evolution’ indicates two seperate topics, and it is not far from the truth to view the chemical composition equations as relating to evolution, while the remaining equations relate to stellar structure.

2 In a sence, the chemical composition is a star’s memory; it records where a star has been during its past evolution, and dictates where it will go in the future. As we shall discuss later, armed solely with knowledge of the total mass of a star and its internal chemical composition distribution, it is in principle possible to reconstruct its full internal structure. Let’s now look at the additional information we need to solve the diﬀerential equations. This comprises: • The equation of state ρ = ρ(P,T,Xi), (10) which relates the density at each point to the temperature and pressure, • Three thermodynamic coeﬃcient equations, for instance

cP = cP (P,T,Xi), (11)

Γ1 = Γ1(P,T,Xi), (12)

∇rad = ∇rad(P,T,Xi), (13) from which the 21 other possible thermodynamic coeﬃcients can be de- rived, • The equation for the Rosseland mean opacity (with conduction included),

κ = κ(P,T,Xi), (14) which tells us how easily radiation can ﬂow through stellar material, • The nuclear rate equations

rik = rik(P,T,Xi), (15)

= (P,T,Xi), (16)

3 = ν (P,T,Xi) (17)

Combining these with the differential equations, we are able to write the right- hand sides of the differential equations purely in terms of r, P , T , l, and Xi — that is, 4+I unknowns. Since there are 4+I differential equations, we therefore have sufficient information to calculate a solution — at last! Equations. (1) and (2) both contain time derivatives as well as space derivatives, marking them out as mixed differential equations. Often, mixed equations can introduce quite a bit of difficulty in numerical solution techniques. However, in the present case there are a couple of simplifications we can make. Let’s first consider the momentum equation, (1). Suppose the pressure gradient term on the left-hand side could be magically set to zero (corresponding to ‘turning off’ the pressure supporting the star against gravity). The approximate timescale τdyn for the surface layers to collapse down to the center is given by the solution of GM R 0 = − 4 + 2 , (18) 4πR τdyn

3 which is r R τ ∼ . (19) dyn GM This ‘dynamical timescale’, which turns out to be around 20 minutes for the Sun, tells us how quickly a star will readjust itself into hydrostatic equilibrium, in response to a disturbance. It also tells us how important the time derivative term in eqn. (1) is. If the internal structure of a star is changing on a timescale much larger than τdyn, then we can neglect the time derivative, to give ∂P Gm = − . (20) ∂m 4πr2 Let’s do a similar thing for the energy equation. Suppose that the nuclear energy sources and neutrino losses went away, so that the star had to provide its luminosity from its gravitational and internal energy, ∂l ∂T δ ∂P = −c + . (21) ∂m P ∂t ρ ∂t On the left-hand side, we have approximately the energy being lost from the star per unit mass; and on the right hand side, we have approximately the rate- of-change of thermal and gravitational energy per unit mass supplying this loss. Hence, the star will run out of energy over a timescale L L τKH ∼ ∼ , (22) Eg Ei where we have used Eg and Ei before to denote the gravitational and thermal energy, respectively. This timescale is of course is the Kelvin-Helmholtz timescale. As with the dynamical timescale, the Kelvin-Helmholtz timescale tells us how quickly a star will respond to departures from thermal equilibrium (these departures coming, e.g., from losses through the surface that require a contraction of the star). If the internal structure of a star is changing on a timescale much larger than τKH, then we can neglect the time derivative in the energy equation, to give ∂l = − (23) ∂m ν Typically, a star evolves during nuclear burning over a nuclear timescale L L τn ∼ ∼ 2 , (24) En fMc where the mass-energy conversion fraction f is ∼ 10−4. This timescale is much longer than the dynamical and Kelvin-Helmholtz timescales, so the star re- mains in hydrostatic and thermal equilibrium during nuclear burning. Of course, when nuclear reactions cease, the star then evolves over the Kelvin-Helmholtz timescale (i.e., much more rapidly), until conditions are reached where fresh nuclear burning can occur. During this rapid evolution, the star is not in thermal equilibrium, and we cannot neglect the time derivative in the energy equation (2).