Assumptions – most of the

• Spherical symmetry Broken by e.g., , rotation, magnetic fields, explosion, instabilities Makes equations a lot easier. Also facilitates Lecture 4 the use of Lagrangian ( shell) coordinates and Time Scales • Homogeneous composition at birth

• Isolation frequently assumed Glatzmaier and Krumholz 3 and 4

Prialnik 2 • Pols 2 When not forming or exploding

Uniqueness Hydrostatic Equilibrium One of the basic tenets of is the Consider the acting upon a spherical mass shell Russell-Vogt Theorem, which states that the mass dm = 4π r 2 dr ρ and chemical composition of a , and in particular The shell is attracted to the center of the star by a how the chemical composition varies within the star, per unit area uniquely determine its radius, , and internal −Gm(r)dm −Gm(r) ρ Fgrav = = dr structure, as well as its subsequent evolution. (4πr 2 )r 2 r 2 where m(r) is the mass interior to the radius r M dm A consequence of the theorem is that it is possible It is supported by the to uniquely describe all of the parameters for a star gradient. The pressure P+dP simply from its location in the Hertzsprung-Russell on its bottom is smaller P Diagram. There is no proof for the theorem, and in fact, than on its top. dP is negative it fails in some instances. For example if the star has rotation or if small changes in initial conditions cause m(r) FP = P(r) i area−P(r + dr) i area large variations in outcome (chaos). or F dP P = dr area dr Hydrostatic Equilibrium Examples of hydrostatic equilibrium: 1)Consider, for example, an isothermal composed If these forces are unbalanced there will be an acceleration of ideal , P = nkT (T= constant; n is the number ). per unit area Let the atmosphere rest on the surface of a spherical mass M F F 2 F grav − P dm 4πr ρdr with radius R. R and M are both constant. h is the height = = r = 2 r Area Area Area 4πr above the surface: ⎛ Gm(r)ρ dP ⎞ dr r = dr (ρ ) − 2 + ⎝⎜ r dr ⎠⎟ dP GMρ = − 2 r = R + h h << r −Gm(r) 1 dP dr r r = − dr dh r 2 ρ dr = N kT dρ GMρ ρN r will be non-zero in the case of stellar explosions A = − = − gρ P = nkT n = A T const. µ dh R2 µ or dynamical collpase, but in general it is very small ρ dρ µg h µgh = − dh ⇒ ln ρ −ln ρ = − in compared with the right hand side, so ∫ ρ N kT ∫ 0 N kT ρ0 A 0 A dP Gm(r)ρ ⎛ ⎞ h N kT = − 2 ρ A dr r ln⎜ ⎟ = − H = = density scale height ⎝ ρ ⎠ H µg where m(r) is the mass interior to the radius r. This is 0 h h − − kT called the equation of hydrostatic equilibrium. It is the ρ =ρ e H equivalently n = n e H H = 0 0 g most basic of the equations.

Examples: Examples: 3) Stellar photospheres 2) Or water in the ocean (density constant; incompressible) As we will disuss next , a beam of light going through dP GMρ a medium with opacity, κ (cm2 g−1), suffers attenuation = − 2 r = R − h h << r dr r dI = − κρI ⇒ I = I e−κρr ≡ I e−τ dr = − dh h = depth below surface dr 0 0 dP GMρ where τ is the "optical depth". The photosphere is defined − = − = − gρ dh R2 by place where, integrating inwards, τ ≈1 (a value of 2/3 is also sometimes used). P = gρh e.g., at one km depth Consider hydrostatic equilibrium in a stellar atmosphere =(980)(1)(105 ) = 9.8×107 dyne cm−2 where M and R can assumed to be constant = 98bars dP GM = − 2 ρ = − gρ dr R Multiplying by κ and dividing by ρ

κ dP dP dP g = κ = − g or = − κρ dr dτ dτ κ This was first noted by K. Schwartzschild in 1906 Solar photosphere Integrating inwards and assuming g and κ are constant (the latter is As we learned last time, the solar photosphere is not such a good approximation) largely neutral. H and He are not ionized much at all -4 g (~10 for H). The electrons come mostly from elements P ≈ photosphere κ like Na, Ca, K, etc. that are very rare. Consequently the GM ion pressure overwhelmingly dominates at the solar For the g =  = 2.7×104 R2 photosphere (radiation pressure is negligible). This is  So if κ  1, P ≈ 0.027 bars. not the case if one goes deeper in the star. photosphere The electron pressure at the photosphere, though small, − κ is due to the H ion gives the electron density which enters into the Saha and varies rapidly with equation and determines the spectrum. This is in fact but is of order unity (between 0.1 how it is determined. and 1 in the region of interest). Actually P ~ 105 dyne cm−2 0.1 bar =

Chapter 2: The Photosphere 29

Table 2-1: The Holweger-Müller Model Atmosphere7 Lagrangian coordinates Optical Temper- Electron Height8 Depth ature Pressure Pressure Density Opacity -2 -2) -3 (km) (τ5000) (°K) (dynes cm ) (dynes cm (g cm ) (κ5000) The hydrostatic equilibrium equation 550 5.0×10-5 4306 5.20×102 5.14×10-2 1.90×10-9 0.0033 dP Gm(r)ρ 507 1.0×10-4 4368 8.54×102 8.31×10-2 3.07×10-9 0.0048 = − 2 441 3.2×10-4 4475 1.75×103 1.68×10-1 6.13×10-9 0.0084 dr r 404 6.3×10-4 4530 2.61×103 2.48×10-1 9.04×10-9 0.012 can also be expressed with the mass as the

3 -1 -8 366 0.0013 4592 3.86×10 3.64×10 1.32×10 0.016 independent variable by the substitution dm= 4πr 2 ρ dr 304 0.0040 4682 7.35×103 6.76×10-1 2.47×10-8 0.027 dP dr Gm(r) 1 254 0.010 4782 1.23×104 1.12 4.03×10-8 0.040 ρ Holweger and Muller = − 2 2 Solar , 1974 202 0.025 4917 2.04×104 1.92 6.52×10-8 0.061 dr dm r 4πr ρ 176 0.040 5005 2.63×104 2.54 8.26×10-8 0.075 149 0.063 5113 4 3.42 -7 0.092 3.39×10 1.04×10 dP Gm(r) 121 0.10 5236 4.37×104 4.68 1.31×10-7 0.11 = − 4 94 0.16 5357 5.61×104 6.43 1.64×10-7 0.14 dm 4πr 66 0.25 5527 7.16×104 9.38 2.03×10-7 0.19 29 0.50 5963 9.88×104 22.7 2.60×10-7 0.34 This is the "Lagrangian" form of the equation. We will find 0 1.0 6533 5 73.3 -7 0.80 1.25×10 3.00×10 that all of our stellar structure equations can have either -34 3.2 7672 1.59×e+005 551 3.24×10-7 3.7 -75 16 8700 2.00×e+005 2.37×103 3.57×10-7 12 an Eulerian form or a Lagrangian form.

7Holweger, H. and Müller, E. A. “The Photospheric Barium Spectrum: Solar Abundance and Collision Broadening of Ba II Lines by Hydrogen”, Solar Physics 39, pg 19-30 (1974). Extra points have been cubic spline interpolated by J. E. Ross. The optical properties (such as the optical depth and the opacity) of a model atmosphere are, obviously, very important, and will be considered later. See table C-4 for complete details of the Holweger-Müller model atmosphere including all depth points used. 8The height scale is not arbitrary. The base of the photosphere (height = 0 km) is chosen to be at

standard optical depth of one (i.e. τ5000Å = 1 ). Merits of using Lagrangian coordinates • In fact the merit of Lagrangian coordinates is so great for spherically symmetric problems • Material interfaces are preserved if part or that all 1D stellar evolution codes are writen all of the star expands or contracts. Avoids in Lagrangian coordinates “advection”. • On the other hand almost all multi-D stellar codes • Avoids artificial mixing of of composition are written in “Eulerian” coordinates. and transport of energy

• In a stellar code can place the zones “where the action is”, e.g., at high density in the center of the star

• Handles large expansion and contraction (e.g., to a without regridding.

This Lagrangian form of the hydrostatic equilibrium equation can be integrated to obtain the central pressure: A better but still very approximate result comes from P surf M Gm(r) dm assuming constant density (remember how bad this is dP = P −P = − 3 3 ∫ surf cent ∫ 4πr 4 for the sun!) m(r)= 4/3 πr ρ  4/3 πr ρ (off by 2 !) Pcent 0 0 and since P 0 4/3 4/3 surf = ⎛ 3m(r)⎞ m(r) ⎛ 4πρ ⎞ r 4 = ⇒ = 0 m(r) M Gm(r) dm ⎜ 4πρ ⎟ r 4 ⎜ 3m(r)⎟ P = ⎝ 0 ⎠ ⎝ ⎠ cent ∫ 4 0 4πr 4/3 4/3 −G ⎛ 4πρ ⎞ M dm G ⎛ 4πρ ⎞ 3M 2/3 To go further one would need a description of how P ≈ 0 = 0 cent 4π ⎜ 3 ⎟ ∫ m1/3(r) 4π ⎜ 3 ⎟ 2 m(r) actually varies with r. Soon we will attempt that . ⎝ ⎠ 0 ⎝ ⎠ 4/3 4/3 Some interesting limits can be obtained already though. ⎛ 4πρ ⎞ ⎛ M ⎞ but 0 = so For example, r is always less than R, the radius of ⎝⎜ 3 ⎠⎟ ⎝⎜ R3 ⎠⎟ 1 1 2 the star, so > and 3GM GMρ0 3M r 4 R4 P = or since ρ is assumed = cent 8πR4 2R 0 4πR3 G M GM 2 P > m(r) dm = This is 3 bigger but still too small cent 4 R4 ∫ 8 R4 π 0 π The sun's average density is 1.4 g cm−3 but its Hydrodynamical Time Scale central density is about 160 g cm−3. We took r=0 to get hydrostatic equilibrium. It is also Interestingly, most stars are supported primarily by interesting to consider other limiting cases where r ρN kT pressure (TBD), P= A ,so for the constant density µ is finite and the or is negligible. (constant composition, ideal gas) case The former case would correspond to gravitational ρ N kT GMρ 0 A = 0 µ 2R collapse, e.g., a cloud collapsing to form a star or galaxy. so the density cancels and one has The latter might characterize an explosion. GMµ M Tcent = ∝ 2NAkR R For a typical value of =0.59 µ Gm(r) 1 dP ⎛ M / M ⎞ − 6   T = 6.8×10 K r = 2 − cent ⎜ R / R ⎟ ⎝  ⎠ r ρ dr This of course is still a lower bound. The actual present value dP 2 for the sun is twice as large, but the calculation is really for If →0, then r=- Gm(r) / r = g(r) a homogeneous (zero age) sun. A similar scaling (M/R) dr will be obtained for the average temperature using the which is the equation for free fall.

Virial Theorem

Free fall time scale

If g(r) is (unrealistically) taken to be a constant = g(R), the collapse time scale is given by Sometimes in the literature one sees instead 1/2 1 3 gτ 2 ≈R and for the whole star, R R ⎛ 3R ⎞ ff 2 τ ff = = =⎜ 3 ⎟ 2 3 vesc 2GM / R ⎝ 8πGR ρ ⎠ GMτ ff 4πρR ≈ R but M = (exact) 1/2 2R2 3 ⎛ 3 ⎞ 2 1340 / sec 4πGρRτ = ⎜ ⎟ = ρ ff ≈ R ⎝ 8πGρ ⎠ 6 3 And for the density, which changes logrithmically τ 2 = ff 2πGρ 3 times as fast as the density So 1/2 1 ⎛ 3 ⎞ 1 2680 s τ = = 446 / ρ sec ff ⎜ ⎟ τ ff ≈ = 3 ⎝ 8πGρ ⎠ 2π ρ Gρ 3 Clearly this is an overestimate since g(r) actually increases −1/2 In any case all go as ρ and are about 1000 s for ρ =1 during the collapse. Explosion Time Scale Examples

A related time scale is the explosive time scale. a)The sun ρ = 1.4 g cm−3 Say g suddenly went to zero. An approximate expansion τ = 2680./ 1.4 =1260 s =38 time scale for the resulting expansion would be HD If for some reason hydrostatic equilibrium were lost ⎛ dP ⎞ 1⎛ dP ⎞ r =4πr2 = R ~1/2 rτ 2 in the sun, this would be the time needed to restore it. If the ⎜ dm⎟ ⎜ dr ⎟ exp ⎝ ⎠ ρ ⎝ ⎠ 2R pressure of the sun were abruptly increased by a substantial 2R P r~ ~ τ 2 factor (~2), this would be the time for the sun to explode 2 ρR exp τ exp 1/2 ⎛ ρ ⎞ 6 8 −3 ~R R/c b) A ρ ~10 −10 g cm τ exp ⎜ ⎟ ≈ sound ⎝ P ⎠ 6 τ HD = 2680./ 10 =0.26 − 2.6 s Usually the two terms in the "hydrostatic equilibrium" equation ~ Time scale for white dwarfs to vibrate (can't be pulsars) for r are comparable, even in an explosion andτ ff ≈τ exp We shall ~Time for the iron core of a massive star to just use 2680 s / ρ for both. collapse to a neutron star

Examples

c) A neutron star ρ ~1015 g cm−3

15 τ HD = 2680./ 10 =0.084 ms ~ Time for a neutron star to readjust its structureafter Generally the hydrodynamical, aka free fall, core bounce aka explosion time scale is the shortest of all the relevant time scales and stars, except in d) A red giant - solar mass, 1013 cm ⇒ ρ ~5 ×10−7 g cm−3 their earliest stages of formation and last explosive

−7 5 stages, are in tight hydrostatic equilibrium. τ HD = 2680./ 5 ×10 = 4×10 s ≈5 days

~Time for the shock to cross a supergiant star making a Type IIp supernova ~Typical Cepheid time scale Some definitions Gravitational binding energy for a dm sphere of constant The total gravitational binding energy of a star of m 3 mass M, is 4πr ρ If ρ =constant = ρ , m(r)= 0 dm= 4πr 2ρ dr 0 3 0 M Gm GM 2 Ω = − dm = − α α ~ 1 (see next page) ∫ r R M R 2 0 Gm(r) 4πGr ρ Ω= − dm= − 0 4πr 2ρ dr ∫ ∫ 0 0 r 0 3 Similarly, the total internal energy is the integral 16π 2Gρ 2 R 16π 2Gρ 2R5 over the star's mass of its internal energy per gram, u. = − 0 ∫ r 4 dr = − 0 3 N kT 3 P 3 0 15 For an ideal gas u = A (TBD) = 2 2 µ 2 ρ 3 2 3G ⎛ 4πR ρ0 ⎞ 3GM = − ⎜ ⎟ = − M 5R ⎝ 3 ⎠ 5R U = ∫ udm 0 3 M P i.e., α =3/5 for the case of constant density. In the = dm for an ideal gas 2 ∫ ρ 0 general case it will be larger.

Similarly The Stellar Virial Theorem The total kinetic energy is The Lagrangian version of the hydrostatic equilibrium equation is dP Gm(r) M 2 r (m) 1 = − 4 T = ∫ dm Mv 2 if the velocity is the same everywhere dm 4πr 0 2 2 4 Multiply each side by the volume inside radius r, V= πr 3, The total nuclear power is 3 and integrate M ⎛ 4 ⎞ ⎛ Gm(r)⎞ 1 Gm(r)dm V dP(r) r 3 dm L = ε(m) dm where ε is the energy generation rate =⎜ π ⎟ ⎜ − 4 ⎟ = − nuc ∫ ⎝ 3 ⎠ ⎝ 4πr ⎠ 3 r 0 P(r ) m(r ) −1 −1 1 Gm'dm' in erg g s from the nuclear reactions ∫ V dP = − ∫ P 3 0 r ε is a function of T, ρ, and composition cent and the luminosity of the star The integral on the right hand side is just the total gravitational potential energy in the star interior to r, Ω(r). M The left hand side can be integrated by parts out to radius r < R

L = F(m) dm where F(m) is the energy flux entering P(r ) r V (r ) V (r ) ∫ since V = 0 0 V dP =PV ⎤ − P dV =P(r)V(r) − P dV ∫ ⎦ ∫ ∫ at the center each spherical shell whose inner P 0 0 cent cent boundary is at m P(r ) r V (r ) V (r ) The Virial Theorem for ideal V dP =PV ⎤ − P dV = P(r)V(r) − P dV ∫ ⎦ ∫ ∫ M 2 1 Pcent 0 0 u dm =− Ω cent So for ideal gas ∫ 3 3 4πr 2ρ dr dm 0 Substituting dV = 4πr 2dr = = ρ ρ The left hand side is 2/3 of the total internal energy of V (r ) m(r ) P the star, hence ∫ P dV = ∫ dm so 0 0 ρ 2U = - Ω (Prialnik 2.26) (Ω is defined < 0) The internal energy is, in magnitude, 1/2 the binding energy m(r ) P 1 P(r)V(r) − dm = Ω(r) of the star. We shall see later that similar though somewhat ∫ 3 0 ρ different expressions exist for radiation and relativistic degeneracy This is true at any value of r, but pick r = R where pressure. This is the Virial Theorem. P(R)= 0, then Note that we can also define a mass averaged temperature M P 1 in the star - dm = Ωtot (Prialnik 2.23) ∫ ρ 3 M 0 for any EOS 1 P 2 T = T(m)dm but for an ideal gas u M ∫ = 0 ρ 3

The Virial temperature N kρT This is similar to the value obtained from hydrostatic equilibrium For an ideal gas, P= A (same as nkT but in terms of ρ) µ α but (about 0.2) instead of 0.5, so cooler, about M M M 3 3 P 3 NAkT U = dm dm but T dm T M 6 ∫ = ∫ ∫ = 2.6 x 10 for the sun. 2 0 ρ 2 0 µ 0 3 N kMT 1 = A = − Ω 2 µ 2 Note that the temperature of nearly homogeneous 2 stars like the sun is set by their bulk properties, M and R. 3MNAkT αGM = As the mass on the main sequence rises, R empirically µ R rises only as about M2/3 so the central temperature of where e.g., α =3/5 for a sphere of constant density massive stars rises gently with their mass. We shall αµG M see later that the density actually decreases. T = Virial temperature 3NAk R note again the inverse dependence of T on R The change in internal energy in a thin shell of mass, δm, during a small change of time, δ t, is then dm and δm used (aka First law of ) interchangably here δudm=δQ+δW Consider the changes of energy that can be experienced P is the pressure by a small (spherical) mass element m = 4 r2 r.     in the zone. ε is the where Q m t F(m) t F(m m) t is r << R; m << M. If the zone is sufficiently small, we δ = ε δ δ + δ − + δ δ energy generation the internal energy generation plus the net can also think of m as dm and it is customary to in the zone. do so In stellar evolution codes. The zone’s internal energy, accumulation or loss of energy from fluxes at its F is the flux of energy at u, can change as a consequence of: upper and lower boundaries and m or m+dm ⎛ 1⎞ δW = − PδV = − Pδ δm • energy flowing in or out of its upper or lower ⎝⎜ ρ ⎠⎟ boundary by radiation, conduction, or convection is the energy lost to work because the zone expands and pushes

on its boundaries or is compressed and gains energy. Note that • compression or expansion 2 2 V =4 r r and m = constant = 4 r r so δ π δ δ π ρδ nuclear reactions generating or absorbing energy ⎛ 1⎞ • δ V =δm δ ⎜ ⎟ ⎝ ρ ⎠

du d ⎛ 1⎞ ∂F +P =ε − dt dt ⎝⎜ ρ ⎠⎟ ∂m δQ = ε δmδt +F(m)δt −F(m + δm) δt F is the flux of energy at du is the rate at which the internal energy in erg g−1 is and since m or m+δm dt ⎛ dF ⎞ changing, e.g., in a given zone δm of a (Lagrangian) F(m + δm)=F(m) + δm ⎝⎜ dm⎠⎟ stellar model ⎛ dF ⎞ δQ= ε − δm δt d ⎛ 1⎞ ⎜ dm⎟ P is the PdV work being done on or by the zone ⎝ ⎠ dt ⎝⎜ ρ ⎠⎟ ⎛ 1⎞ ⎛ dF ⎞ as it contracts ( PdV is negative) or expands so δuδm+ Pδ δm = ε − δm δt ρ ↑⇒ ⎜ ⎟ ⎜ ⎟ 1 1 ⎝ ρ ⎠ ⎝ dm⎠ (ρ ↓⇒PdV is positive). Units are erg g− s− dividing by δm and δ t and taking the limit as δ t →0 ε is the nuclear energy generation rate minus neutrino losses in erg g−1 s−1 du d ⎛ 1⎞ dF The energy conservation +P =ε − dF dt dt ⎝⎜ ρ ⎠⎟ dm equation aka “the first law dm is the difference in energy (per ) going out the of thermodynamics dm top of the zone (by , conduction, or convection) minus the energy coming in at the bottom. It is positive if more energy is leaving than entering. Thermal Equilibrium Examples: If a steady state is reached where in a given zone the internal energy (u), density (ρ), and pressure (P) are not changing • Main sequence stars are in thermal equilibrium very much, then the left hand side of the 1st law becomes Massive stars becoming red giants are not. More approximately zero and • energy is being generated than is leaving the dF ε = surface. The star’s envelope is expanding dm Energy flows into or out of the zone to accomodate what • White dwarfs are in a funny thermal equilibrium is released or absorbed by nuclear reactions (plus neutrinos). where nuc = o and PdV is zero, but u is decreasing U L in order to provide L ⇒ = If this condition exists through the entire star M M • Late stages of massive stellar evolution may also ε dm =L = dF = L approach a funny equilibrium where neutrino ∫ nuc ∫ 0 0 losses balance nuclear energy generation, at least then the star is said to be in thermal equilibrium. It is also in those portions of the star where burning is going possible for thermal equilibrium to exist in a subset of the star. on at a rapid rate

Continuing: Integrating the first law of thermodynamics

du d ⎛ 1⎞ dF d ⎛ 1⎞ d ⎛ dV ⎞ d ⎛ dV ⎞ +P =ε − = = dt dt ⎝⎜ ρ ⎠⎟ dm dt ⎝⎜ ρ ⎠⎟ dt ⎝⎜ dm⎠⎟ dm ⎝⎜ dt ⎠⎟

Integrate over mass: d ⎛ 2 dr ⎞ = ⎜ 4πr ⎟ M du M d ⎛ 1⎞ M dm ⎝ dt ⎠ dm + P dm dm F(M) F(0) ∫ ∫ ⎜ ⎟ = ∫ ε − + Integrate by parts: 0 dt 0 dt ⎝ ρ ⎠ 0 M M L L nd d ⎛ 1⎞ ⎛ 2 dr ⎞ = nuc − (2 P dm = Pd 4πr term) ∫ dt ⎜ ρ ⎟ ∫ ⎜ dt ⎟ Since m does not depend on t the leftmost term can be 0 ⎝ ⎠ 0 ⎝ ⎠ M M rewritten 2 dr ⎤ 2 dr dP = 4πr P ⎥ − 4πr dm M M dt ∫ dt dm du d ⎦0 0 dm udm U ∫ = ∫ = The first term is zero at r = 0 and M (where P = 0) so 0 dt dt 0 M d ⎛ 1⎞ M dr dP The second term takes some work, one can rewrite 1/ρ P dm = − 4πr 2 dm ∫ dt ⎝⎜ ρ ⎠⎟ ∫ dt dm (since dm = dV / ρ) 0 0 d ⎛ 1⎞ d ⎛ dV ⎞ d ⎛ dV ⎞ = = dt ⎝⎜ ρ ⎠⎟ dt ⎝⎜ dm⎠⎟ dm ⎝⎜ dt ⎠⎟

Continuing: So all together    U+Ω +T = Lnuc −L M d ⎛ 1⎞ M dr dP P dm = − 4πr 2 dm which expresses succinctly (and obviously) the conservation ∫ dt ⎝⎜ ρ ⎠⎟ ∫ dt dm 0 0 of energy for the star. Power generated or lost on the right dP −Gm r But = − balances the change in internal, gravitational, and kinetic dm 4πr 4 4πr 2 energies on the left. M d ⎛ 1⎞ M Gm dr M dr P dm = dm+ r dm ∫ ⎜ ⎟ ∫ 2 ∫ Suppose the star is static so that T =0, and the Virial 0 dt ⎝ ρ ⎠ 0 r dt 0 dt Theorem also applies so U = -1/2  (for an ideal gas) dr d ⎛ r2 ⎞ Gm dr d ⎛ Gm⎞ Ω and r = 2 = − Then dt dt ⎝⎜ 2 ⎠⎟ r dt dt ⎝⎜ r ⎠⎟ So 1  Ω = Lnuc −L M M M 2 d ⎛ 1⎞ d Gm 1 d P dm = − dm+ r2 dm =Ω +T ∫ dt ⎝⎜ ρ ⎠⎟ dt ∫ r 2 dt ∫ 0 0 0 This is interesting.

1  Also interesting is the for U in the same Ω = Lnuc −L remember Ω is negative 2 situation. Then

If L = L then the star is in a state of balanced 1 • nuc U = - Ω = L − L power. Over long period of time it neither expands 2 nuc or contracts. It is in thermal and dynamic equilibrium In the absence of nuclear energy input, the star increases • If Lnuc > L, more energy is being generated by its internal energy (heat) as it radiates. The more a star, nuclear reactions than is being radiated.  & supported by ideal gas pressure, radiates, the hotter it becomes less negative. The star expands to absorb the excess gets. The heat capacity is defined by C(T) = Δq/ΔT where Δq is the energy flowing into or out of the matter. Here ΔT

• If Lnuc < L, the star is radiating more energy than is positive when Δq is negative. We say that stars have a nuclear reactions are producing. If possible, the "negative heat capacity". The origin of the energy is gravity. star makes up the deficiency by contracting to

a more tightly bound state.  becomes more negative The Kelvin Helmholtz time scale Kelvin Helmholtz time scale for the sun The time scale for adjusting to an imbalance in For the sun the constant density expression gives energy generation (historically Lnuc = 0) is the Kelvin 3GM 2 Helmholtz timescale τ ()=  = 2.95×1014 s =9.3 My KH 10R L Ω Ω   τ = ≈ KH Ω 2L but the actual sun is far from constant density and the actual value is closer to 30 My. We will obtain a better value when we study polytropes later. M Gm R Gm(r) Ω= − ∫ dm = − ∫ 4πr 2ρ(r) dr 0 r 0 r αGM 2 = − where α depends on ρ(r) R I

This is (roughly) the time the sun could shine without nuclear reactions at its present radius and luminosity. In fact, if the radius is allowed to change gravity could One could in principle explain the modern sun power the sun or any star much longer. Even keeping the without recourse to nuclear reactions by allowing luminosity the same the sun could, in principle, continue it to become extremely centrally condensed (e.g., a to contract until it became a white dwarf. Taking a black hole at the center). But this would be quite white dwarf radius of ~5000 km, could be increased τ KH inconsistent with to 1.3 Gy, but then the sun wouldnot look like it does (with the same luminosity it would radiate in the ultraviolet). • Stellar physics and known equations of state

Taking this to the absurd limit (not allowed by quantum • The solar neutrino experiments for a 1 M star), ~Mc2, for a black hole  Ω 13 τ KH ~10 . Gravity, natures weakest force, can, in the end provide more power than any other source including nuclear reactions.

As the star contracts, so long as it remains approximately Other places the Kelvin Helmholtz time scale enters: an ideal gas, the Virial Theorem (or hydrostatic equilibrium) • – it is the time required for a implies: protostar to settle down on the main sequence Ω 3 GM 2 GMρ U=− ⇒ N kTM = α ; or P ∝ and ignite nuclear burning 2 2µ A 2R c R

M M • The time scale between major nuclear burning T ∝ but ρ ∝ 3 stages where degeneracy does not enter in, R R e.g., between helium depletion and carbon So T ∝ M2/3 ρ1/3 ignition in a massive star As the the star contracts, the temperature rises as the cube

root of the density and, for a given density, is less in stars of • The time a proto-neutron star requires to release its binding energy as neutrinos. During this smaller mass. time it can power a supernova We will make these argumentsmore quantitative later

The nuclear time scale

Evolution of the central As we shall see later nuclear fusion – up to the density and temperature of element iron is capable of releasing large energies 15 M and 25 M stars   per gram of fuel (though well short of mc2).

Kelvin Helmholtz Roughly the energy release during each phase is nuclear the fraction of the star that burns times Kelvin Helmholtz Fuel fraction mc2 nuclear -3 Kelvin Helmholtz Hydrogen 7 x 10 Helium 7 x 10-4 nuclear Carbon 1 x 10-4 3 x 10-4 Silicon 1 x 10-4

The nuclear time scale There are several things to notice here. One can also define a nuclear time scale First the energy yield is a small fraction of the rest Mq τ = nuc mass energy. It just comes from shuffling around nuc L the same neutrons and protons inside different but everywhere except the main sequence, one must nuclei. The number of neutrons plus protons is be quite cautious as to what to use for L and M because being conserved. No mass is being annihilated. more than one fuel may be burning at a give time and Less energy is produced once helium has been neutrinos can carry away appreciable energy. Also made. The binding energies of neutrons and only a fraction of the star burns, just that part that is hot protons in helium, carbon, oxygen and silicon enough. For the sun if 10% burns are similar, all about 8 MeV/nucleon. More energy is released by oxygen 0.1 0.007c 2 M ( ) (  ) burning simply because the oxygen abundance τ ≈ ( ) nuc L is about 5 times the carbon abundance.  = 10Gy We shall see later that the nuclear time scale can which is fortuitously quite close to correct. be greatly accelerated by neutrino losses

Random walk: Radiation diffusion time scale

In a random walk, how far are you from the origin in n steps, and how long did it take to get there?

=3 cm2 x = 3 cm 3 The thermal time scale

The average of s itself is zero

2   R ⎛ (6.9 ×1010 )2 ⎞ s is =  1.6 1012 sec τ dif () ≈ = 10 = × c ⎝⎜ (0.1)(3 ×10 ⎠⎟ or about 50,000 years. A more accurate value is 170,000 years

http://adsabs.harvard.edu/full/1992ApJ...401..759M

Ordering of Time scales In general, and especially on the main sequence τ << τ <τ <τ To summarize we have discussed 4 time scales HD therm KH nuc that characterize stellar evolution. There are however interesting places where this ordering • Hydrodynamical time – 446/1/2 to 2680/1/2 sec breaks down: – the time to adjust to and maintain hydrostatic τ ~τ ~τ during the late stages of massive equilibrium. Also the free fall or explosion time scale therm KH nuc star evolution Thermal – R2/lc – the time to establish thermal • τ HD ~τ nuc for SN Ia; for explosive equilibrium if diffusion dominates (it may not) nucleosynthesis in SN II

• Kelvin-Helmholtz – GM2/(2RL) – the time to adjust There are also other relevant time scales for e.g., structure when the luminosity changes. rotational mixing, angular transport, convective mixing, etc. Usually a phenomenon can • Nuclear – q M /L – the time required to fuse a nuc be characterized and its importance judged by examining given fuel to the next heavier one the relevant time scales.