Stellar Objects: Equation of State 1 Equation of State Our Goal here is to derive the equation of state (or the mutual dependencies among local thermodynamic quantities such as P,T,ρ, and Ni), not only for the classic ideal gas, but also for photons and fermions, for which quantum effects are important. The EoS together the thermodynamic equation allows to study how the stellar material react to the heat, changing density, etc. 1 The local thermodynamic equilibrium We do this under the local thermodynamic equilibrium (LTE) assump- tion. The LTE means that at any position in a star, complete thermodynamic equilibrium is very nearly true: e.g., the Boltzmann populations of ion energy levels are consistent with the local kinetic temperature of electrons. This assumption is appropriate for a star, since the mean free path of particles is short generally short, compare to the variation scales of the thermodynamic −1 quantities; e.g., λph = (κρ) is on order of 1 cm. Exception cases are the stellar photosphere. With such a short mean free path, how long does it take for a photon to diffuse out of a star? Random walk: 1-D random walk theory. 2 2 The mean of N steps is zero, but the dispersion is not, σ ( i xi)= Nσ . The time scale to establish the LTE is also typically short enough,P except for nuclear reactions, compared to the evolutionary time scale of a star with the exception of various explosions. 2 Chemical reactions Chemical reactions determine the species number density Ni may have the units of number per gram of material; i.e., Ni = ni/ρ. Stellar Objects: Equation of State 2 The thermodynamic equation with chemical reactions: dE = TdS − P dV + µidNi (1) i X where µi is the chemical potential associated with an ith species and clearly can be defined as ∂E µ = . (2) i ∂N i S,V The LTE requires that µidNi =0 (3) i X As an example, + − 0 H + e ↔ H + χH (4) where χH = 13.6 eV is the ionization potential from the ground state of H. For simplicity, we’ll assume that H has only one bound level and no other reactions that involve the above constituents and that the gas is pure hydro- gen. We will neglect the photons here since its chemical potential is zero. Consider a classic black body cavity filled with radiation in thermodynamic equilibrium with the wall. The above equation can be written as µidNi + µγdNγ =0 (5) i X Since photon number is not strictly conserved, in general dNγ =6 0. While dNi = 0 for the cavity wall (as a whole), we have µγ = 0. Now let us get back to ionization-recombination reaction, which can be writ- ten as 1H+ + 1e− − 1H0 =0 (6) And in general, a reaction can be written in such a symbolic form: νiCi = 0, (7) i X Stellar Objects: Equation of State 3 where Ci is the names of the participating particle (except for photons). Clearly, dN = νi dN . Since dN is arbitrary, we have i ν1 1 1 µνi =0 (8) i X This the equation of chemical equilibrium. Here µi, containing the informa- tion of Ni, depending on T and ρ or equivalent thermodynamic quantities as well as on a catalog of the possible reactions. When they are known, we should be able to find all Ni. For a perfect (or ideal) gas, as will be assumed here, it is not difficult to establish such connection. 3 The Distribution Function For a particular species of elementary nature (ions, photons, etc.) in the LTE, the occupation number in a certain quantum state, defined by the internal energy level j in the coordinate-momentum phase space is found from statistical mechanics to be 1 F (j,E(p)) = , (9) exp[−µ + Ej + E(p)]/kT ± 1 where Ej is the energy state j referred to the reference energy level, E(p) is the kinetic energy as a function of the momentum p, E(p)=(p2c2 + m2c4)1/2 − mc2. A “+” or “-” is used for Fermi-Dirac or Bose-Einstein parti- cles (fermions of half-integer spin or bosons of zero and whole integer spin), respectively. The number of states in the phase space, accounting for the degeneracy of the state gj, is g dpdr j . (10) h3 So the phase-space density can be defined as 1 g n(p)= j . (11) h3 exp[−µ + E + E(p)]/kT ± 1 j j X Stellar Objects: Equation of State 4 We can then calculate the physical space density n, kinetic pressure P , and internal energy E as n = n(p)4πp2dp, (12) Zp This gives the connection between n, T , and µ. 1 P = n(p)vp4πp2dp, (13) 3 Zp ∂E(p) where v = ∂p . E = n(p)E(p)4πp2dp (14) Zp 4 Black body Radiation Photons are mass-less bosons of unit spin. Since they travel at c, they have only two states (two spin orientations or polarization) for a given energy. Hence, 8π ∞ p2dp n = ∝ T 3. (15) γ h3 exp(pc/kT − 1) Z0 aT 4 4 Similarly, one can easily get Prad = 3 and Erad = aT = 3Prad, where a is the radiation constant. The nice thing about the LTE is that all these quantities depend only on T . Note that the above equation is a γ-law EoS P =(γ − 1)E with γ = 4/3. For later use, it is easy to define the energy density per unit frequency (ν) as 8πhν3 1 u dν = erg cm−3Hz−1Hz. (16) ν c3 exp(hν/kT − 1) and the frequency-dependent Planck function c B (T )= u , (17) ν 4π ν Stellar Objects: Equation of State 5 which peaks at hν = 2.8kT . The integrated Planck function is σ B(t)= B (T )dν = T 4 erg cm−2 s−1 sr−1. (18) ν π Z where σ is the Stefan-Boltzmann constant. Reading assignment: CH 3.5-.6 5 Fermi-Dirac EoS Elections, protons, and neutrons are fermions. They all have spin one half. 2 Hence g = 2. We choose an energy reference level of E0 = mc (no internal energy level; alternative choices would only lead to an additive constant in the definition of µ). The occupation number is 1 F (E)= , (19) exp ([E(p) − (µ − mc2)]/kT ) ± 1 −1/2 2 p 2 1/2 p p 2 where, in general, E(p)= mc (1 + mc ) − 1 and v(p)= m 1+ mc . h i h i 5.1 The Complete Degenerate Gas Now let us examine the behavior of the above integral when T → 0. The exponential tends either to zero or infinity, depending on whether the kinetic energy E(p) > (µ−mc2) or E(p) < (µ−mc2). We call this critical kinetic en- 2 ergy, EF =(µ−mc ), as the “Fermi energy”, though we are yet to determine µ (= the total energy (including the rest mass energy) of the most energetic particle of the system). Fermions fill the energy range, 0 ≤ E ≤ EF — “filled Fermi sea”, which is capped by EF . Correspondingly, we have pF , as 2 2 1/2 pF defined by EF = mc (1 + x ) − 1 , where x = mc is the dimensionless momentum. The number density can be calculated as − 8π pF 8π h 3 n = p2dp = x3, (20) h3 3 mc Z0 Stellar Objects: Equation of State 6 h −10 where mc is the Compton wavelength and is 2.4 × 10 cm for electrons and about 2000 smaller for neutrons (due to the mass difference). For the complete degenerate electron gas, we then have ρ = Bx3, (21) µe −3 where B = 8πma h = 9.7 × 105 g cm−3. Therefore, once ρ is known, 3 mc µe we can derive x, and hence p and E . F F The demarcation between relativistic and relativistic mechanics occur when 6 −3 pF ∼ mec or x ∼ 1. The corresponding density is ∼ 10 g cm , which, incidentally, is a typical central density of white dwarfs. For typical densities in a neutron star (comparable to the nuclear densities of ρ ≈ 2.7 × 1014 g cm−3, it is easy to show x ≈ 0.35. Thus the neutron stars are non-relativistic. The Pressure and internal energy density can be calculated as 8π m4c5 x x4 P = dx, (22) 3 h3 (1 + x2)1/2 Z0 and − h 3 x E = 8π mc2 x2[(1 + x2)1/2 − 1]dx, (23) mc Z0 which have limiting forms of P ∝ E ∝ x5 ∝ ρ5/3 for x ≪ 1 and P ∝ E ∝ x4 ∝ ρ4/3 for x ≫ 1. It is also easy to show E/P = 3/2(γ = 5/3) for x ≪ 1 and E/P = 3(γ = 4/3) for x ≫ 1. For a γ-law EoS, the completely degenerate gas in a non-relativistic limit acts like a monatomic ideal gas whereas, in the extreme relativistic limit, it behaves like a photon gas. 5.2 Application to White Dwarfs The above P (x(ρ)) relation together with the mass conservation equation can be combined to for a second-order different equation (e.g., ρ vs. Mr). Its solution would give the structure of a WD. Stellar Objects: Equation of State 7 But here, we will only show the M − R scaling via the virial theorem in the hydrostatic form 3(γ − 1)U = −Ω (remember 2K = 3P ), assuming a white dwarf with constant density and composition throughout. Of course, the constant density (hence constant pressure) assumption does not satisfy the hydrostatic equation. But this does not change the above scaling, quali- tatively.