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

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

<p>PHYS 633: Introduction to Stellar Astrophysics </p><p>Spring Semester 2006 <br>Rich Townsend ([email protected]) </p><p>Governing Equations </p><p>In the foregoing analysis, we have seen how a star usually maintains a state of hydrostatic equilibrium.&nbsp;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.&nbsp;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. <br>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).&nbsp;Let’s quickly review these equations.&nbsp;First, we have the most general (spherically-symmetric) form for the equation governing momentum conservation, </p><p></p><ul style="display: flex;"><li style="flex:1">∂P </li><li style="flex:1">Gm </li></ul><p>4πr<sup style="top: -0.2398em;">2 </sup></p><p>1</p><p>∂<sup style="top: -0.3013em;">2</sup>r <br>4πr<sup style="top: -0.2398em;">2 </sup>∂t<sup style="top: -0.2398em;">2 </sup></p><p>= − </p><p>−</p><p>.</p><p>(1) </p><p>∂m </p><p>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 ∂<sup style="top: -0.3013em;">2</sup>r/∂t<sup style="top: -0.3013em;">2</sup>. <br>Second, we have the most general form for the energy conservation equation, </p><p></p><ul style="display: flex;"><li style="flex:1">∂l </li><li style="flex:1">∂T </li></ul><p>∂t δ ∂P ρ ∂t <br>= ꢀ − ꢀ<sub style="top: 0.1246em;">ν </sub>− c<sub style="top: 0.1246em;">P </sub></p><p>+</p><p>.</p><p>(2) </p><p>∂m </p><p>The first and second terms on the right-hand side come from nuclear energy generation and neutrino losses, respectively.&nbsp;The third and fourth terms together compose what we have previously termed the gravothermal energy release rate, </p><p>ꢀ<sub style="top: 0.1245em;">gi</sub>, </p><p>P ∂ρ&nbsp;∂u ρ<sup style="top: -0.2398em;">2 </sup>∂t </p><ul style="display: flex;"><li style="flex:1">∂T </li><li style="flex:1">δ ∂P </li></ul><p></p><p>ꢀ<sub style="top: 0.1245em;">gi </sub></p><p></p><ul style="display: flex;"><li style="flex:1">≡</li><li style="flex:1">−</li></ul><p></p><p>= −c<sub style="top: 0.1245em;">P </sub></p><p>∂t </p><p></p><ul style="display: flex;"><li style="flex:1">+</li><li style="flex:1">;</li><li style="flex:1">(3) </li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">∂t </li><li style="flex:1">ρ ∂t </li></ul><p></p><p>they give a net positive contribution to the luminosity gradient when the star contracts. <br>Third, we have the equation(s) of energy transport.&nbsp;These allow us to determine the physical temperature gradient ∇ ≡ ∂ ln T/∂ ln P at each point in the star, from which we can obtain the ‘real’ temperature gradient as </p><p></p><ul style="display: flex;"><li style="flex:1">∂T </li><li style="flex:1">GmT </li></ul><p>4πr<sup style="top: -0.2398em;">4</sup>P </p><p>= − </p><p>∇. </p><p>(4) </p><p>∂m </p><p>1<br>In radiative regions, the physical gradient is equal to the radiative gradient ∇<sub style="top: 0.1245em;">rad</sub>, </p><p>3 κρH<sub style="top: 0.1246em;">P </sub><br>4ac T<sup style="top: -0.2398em;">4 </sup></p><p>3</p><p>κlP <br>16πacG mT<sup style="top: -0.2398em;">4 </sup></p><p>∇</p><p>≡ ∇<sub style="top: 0.1246em;">rad </sub></p><p>≡</p><p>F = <br>.</p><p>(5) </p><p>rad </p><p>If ∇<sub style="top: 0.1245em;">rad </sub>exceeds the adiabatic temperature gradient ∇<sub style="top: 0.1245em;">rad</sub>, convection will set in, and we need instead to calculate ∇ from a convective model — usually, mixing-length theory. <br>Finally, we have the I − 1 equations describing the change in the mass fractions describing the star’s chemical composition, </p><p><br></p><ul style="display: flex;"><li style="flex:1">X</li><li style="flex:1">X</li><li style="flex:1">X</li><li style="flex:1">X</li></ul><p></p><p>∂X<sub style="top: 0.1246em;">i </sub><br>∂t </p><p>m<sub style="top: 0.1246em;">i </sub></p><p>ρ</p><p>ꢀ<sub style="top: 0.1246em;">ji </sub>q<sub style="top: 0.1245em;">ji </sub><br>ꢀ<sub style="top: 0.1245em;">ik </sub>q<sub style="top: 0.1245em;">ik </sub></p><p>=</p><p>p<sup style="top: -0.3428em;">i</sup><sub style="top: 0.2053em;">ji</sub>r<sub style="top: 0.1245em;">ji </sub></p><p>−</p><p>p<sup style="top: -0.3428em;">i</sup><sub style="top: 0.2053em;">ji</sub>r<sub style="top: 0.1245em;">ik </sub></p><p>=</p><p>−</p><p>.</p><p>(6) </p><p></p><ul style="display: flex;"><li style="flex:1">j</li><li style="flex:1">j</li></ul><p></p><ul style="display: flex;"><li style="flex:1">k</li><li style="flex:1">k</li></ul><p></p><p>The final equation governing the chemical composition is the closure relation </p><p>X</p><p>∂X<sub style="top: 0.1246em;">i </sub></p><p></p><ul style="display: flex;"><li style="flex:1">= 0. </li><li style="flex:1">(7) </li></ul><p></p><p>∂t </p><p>i</p><p>To describe mixing in convective regions, the above equations are replaced by </p><p>Z</p><p>m<sub style="top: 0.0881em;">b </sub></p><p>¯</p><p>dX<sub style="top: 0.1246em;">i </sub></p><p>∂X<sub style="top: 0.1246em;">i </sub><br>∂t </p><p>dm<sub style="top: 0.1246em;">a </sub></p><p>dt dm dt </p><p><sup style="top: -0.4371em;">b </sup>(X<sub style="top: 0.1246em;">i,b </sub>− X<sub style="top: 0.1246em;">i</sub>). </p><p>(8) </p><ul style="display: flex;"><li style="flex:1">¯</li><li style="flex:1">¯</li></ul><p>=</p><p>dm − </p><p>X<sub style="top: 0.1246em;">i,a</sub>(X<sub style="top: 0.1246em;">i,a </sub>− X<sub style="top: 0.1246em;">i</sub>) + </p><p>dt </p><p>m<sub style="top: 0.0831em;">a </sub></p><p>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 </p><p>∂r </p><p>1</p><ul style="display: flex;"><li style="flex:1">=</li><li style="flex:1">(9) </li></ul><p></p><p>∂m <br>4πr<sup style="top: -0.2398em;">2</sup>ρ </p><p>This equation, plus eqns. (1), (2), (4), (6) and (5), together form the 4&nbsp;+ I differential equations governing stellar structure.&nbsp;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.&nbsp;We’re also going to need boundary conditions, to ensure that we obtain only those solutions that are physically-realistic. <br>First, however, let’s step back and have a look over the equations.&nbsp;As Kippenhahn &amp; Weigert discussion in their §9, certain subsystems can be distinguished in the equations.&nbsp;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 = Kρ<sup style="top: -0.3012em;">1+1/n</sup>; substituting this into the mechanical equations leads to the Lane-Emden equation. <br>Another pair of subsystems can be formed by splitting the chemical composition equations from the others. This split is highlighted in the title of Kippenhahn &amp; 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. </p><p>2<br>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. <br>Let’s now look at the additional information we need to solve the differential equations. This comprises: </p><p>• The equation of state </p><p>ρ = ρ(P, T, X<sub style="top: 0.1245em;">i</sub>), </p><p>(10) which relates the density at each point to the temperature and pressure, <br>• Three thermodynamic coefficient equations, for instance </p><p>c<sub style="top: 0.1246em;">P </sub>= c<sub style="top: 0.1246em;">P </sub>(P, T, X<sub style="top: 0.1246em;">i</sub>), </p><p>(11) (12) (13) </p><p>Γ<sub style="top: 0.1246em;">1 </sub>= Γ<sub style="top: 0.1246em;">1</sub>(P, T, X<sub style="top: 0.1246em;">i</sub>), </p><p>∇</p><p>= ∇<sub style="top: 0.1246em;">rad</sub>(P, T, X<sub style="top: 0.1246em;">i</sub>), </p><p>rad </p><p>from which the 21 other possible thermodynamic coefficients can be derived, </p><p>• The equation for the Rosseland mean opacity (with conduction included), </p><p>κ = κ(P, T, X<sub style="top: 0.1246em;">i</sub>), </p><p>which tells us how easily radiation can flow through stellar material, <br>• The nuclear rate equations <br>(14) </p><p>r<sub style="top: 0.1246em;">ik </sub>= r<sub style="top: 0.1246em;">ik</sub>(P, T, X<sub style="top: 0.1246em;">i</sub>), </p><p>(15) (16) (17) </p><p>ꢀ = ꢀ(P, T, X<sub style="top: 0.1246em;">i</sub>), <br>ꢀ = ꢀ<sub style="top: 0.1246em;">ν</sub>(P, T, X<sub style="top: 0.1246em;">i</sub>) </p><p>3</p><p>Combining these with the differential equations, we are able to write the righthand sides of the differential equations purely in terms of r, P, T, l, and X<sub style="top: 0.1245em;">i </sub>— that is, 4+I unknowns. Since&nbsp;there are 4+I differential equations, we therefore have sufficient information to calculate a solution — at last! <br>Equations. (1) and (2) both contain time derivatives as well as space derivatives, marking them out as mixed differential equations.&nbsp;Often, mixed equations can introduce quite a bit of difficulty in numerical solution techniques.&nbsp;However, in the present case there are a couple of simplifications we can make. Let’s first consider the momentum equation, (1).&nbsp;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).&nbsp;The approximate timescale τ<sub style="top: 0.1246em;">dyn </sub>for the surface layers to collapse down to the center is given by the solution of </p><p>GM 4πR<sup style="top: -0.2398em;">4 </sup><br>R</p><p>τ<sub style="top: 0.2505em;">d</sub><sup style="top: -0.2859em;">2</sup><sub style="top: 0.2505em;">yn </sub></p><p></p><ul style="display: flex;"><li style="flex:1">0 = − </li><li style="flex:1">+</li></ul><p></p><p>,</p><p>(18) <br>3which is </p><p>r</p><p>R</p><p>τ<sub style="top: 0.1246em;">dyn </sub></p><p>∼</p><p>.</p><p>(19) </p><p>GM </p><p>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 τ<sub style="top: 0.1245em;">dyn</sub>, then we can neglect the time derivative, to give </p><p></p><ul style="display: flex;"><li style="flex:1">∂P </li><li style="flex:1">Gm </li></ul><p>4πr<sup style="top: -0.2398em;">2 </sup></p><p>= − </p><p>.</p><p>(20) </p><p>∂m </p><p>Let’s do a similar thing for the energy equation.&nbsp;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, </p><p></p><ul style="display: flex;"><li style="flex:1">∂l </li><li style="flex:1">∂T </li></ul><p>∂t δ ∂P ρ ∂t </p><p></p><ul style="display: flex;"><li style="flex:1">= −c<sub style="top: 0.1246em;">P </sub></li><li style="flex:1">+</li></ul><p></p><p>.</p><p>(21) </p><p>∂m </p><p>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 rateof-change of thermal and gravitational energy per unit mass supplying this loss. Hence, the star will run out of energy over a timescale </p><p></p><ul style="display: flex;"><li style="flex:1">L</li><li style="flex:1">L</li></ul><p></p><p>τ<sub style="top: 0.1245em;">KH </sub></p><p></p><ul style="display: flex;"><li style="flex:1">∼</li><li style="flex:1">∼</li></ul><p></p><p>,</p><p>(22) </p><p></p><ul style="display: flex;"><li style="flex:1">E<sub style="top: 0.1246em;">g </sub></li><li style="flex:1">E<sub style="top: 0.1246em;">i </sub></li></ul><p></p><p>where we have used E<sub style="top: 0.1245em;">g </sub>and E<sub style="top: 0.1245em;">i </sub>before to denote the gravitational and thermal energy, respectively.&nbsp;This timescale is of course is the Kelvin-Helmholtz timescale. As&nbsp;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).&nbsp;If the internal structure of a star is changing on a timescale much larger than τ<sub style="top: 0.1245em;">KH</sub>, then we can neglect the time derivative in the energy equation, to give </p><p>∂l <br>= ꢀ − ꢀ<sub style="top: 0.1245em;">ν </sub></p><p>(23) </p><p>∂m </p><p>Typically, a star evolves during nuclear burning over a nuclear timescale </p><p></p><ul style="display: flex;"><li style="flex:1">L</li><li style="flex:1">L</li></ul><p>fMc<sup style="top: -0.2398em;">2 </sup></p><p>τ<sub style="top: 0.1246em;">n </sub>∼ </p><p>∼</p><p>,</p><p>(24) </p><p>E<sub style="top: 0.1245em;">n </sub></p><p>where the mass-energy conversion fraction f is ∼ 10<sup style="top: -0.3012em;">−4</sup>. This timescale is much longer than the dynamical and Kelvin-Helmholtz timescales, so the star remains in hydrostatic and thermal equilibrium during nuclear burning.&nbsp;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). </p><p>4</p>

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