Chapter 2: Equation of State

Introduction The Local Thermodynamic Equilibrium The Distribution Function Black Body Radiation Fermi-Dirac EoS The Complete Degenerate Gas Application to White Dwarfs Temperature Effects Ideal Gas The Saha Equation “Almost Perfect” EoS Adiabatic Exponents and Other Derivatives Outline

Introduction

The Local Thermodynamic Equilibrium

The Distribution Function

Black Body Radiation

Fermi-Dirac EoS The Complete Degenerate Gas Application to White Dwarfs Temperature Effects

Ideal Gas

The Saha Equation

“Almost Perfect” EoS

Adiabatic Exponents and Other Derivatives The EoS, together with the thermodynamic equation, allows to study how the stellar material properties react to the heat, changing density, etc.

Introduction

Goal of the Chapter: 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. The EoS, together with the thermodynamic equation, allows to study how the stellar material properties react to the heat, changing density, etc. Thermodynamics

Thermodynamics is deﬁned as the branch of science that deals with the relationship between heat and other forms of energy, such as work. The Laws of Thermodynamics: I First law: Energy can be neither created nor destroyed. This is a version of the law of conservation of energy, adapted for (isolated) thermodynamic systems. I Second law: In an isolated system, natural processes are spontaneous when they lead to an increase in disorder, or entropy, ﬁnally reaching an equilibrium. I Third law: The entropy of a system at absolute zero is a constant, determined only by the degeneracy of the ground state. For a given set of macroscopic variables, the entropy measures the degree to which the probability of the system is spread out over different possible microstates, according to the fundamental postulate in statistical mechanics: the occupation of any microstate is assumed to be equally probable.

Entropy

The most general interpretation of entropy is as a measure of our uncertainty about a system. The equilibrium state of a system maximizes the entropy because we have lost all information about the initial conditions except for the conserved variables; maximizing the entropy maximizes our ignorance about the details of the system. For a given set of macroscopic variables, the entropy measures the degree to which the probability of the system is spread out over different possible microstates, according to the fundamental postulate in statistical mechanics: the occupation of any microstate is assumed to be equally probable. The thermodynamic relation

The fundamental thermodynamic relation with chemical reactions: X dQ = TdS = dE + PdV − µi dNi i

where µi is the chemical potential of the ith species

∂E µi = . ∂Ni S,V

Chemical reactions determine the species number density Ni , which may have the units of number per gram of material; i.e., Ni = ni /ρ. Outline

Introduction

The Local Thermodynamic Equilibrium

The Distribution Function

Black Body Radiation

Fermi-Dirac EoS The Complete Degenerate Gas Application to White Dwarfs Temperature Effects

Ideal Gas

The Saha Equation

“Almost Perfect” EoS

Adiabatic Exponents and Other Derivatives The equilibrium of chemical reactions means X µi dNi = 0, i

appropriate for a star, since the mean free path of particles is generally short, compared to the variation scales of the −1 thermodynamic quantities; e.g., λph = (κρ) is on order of 1 cm ( except for stellar photospheres). The time scale to establish the LTE is also typically short enough, except for nuclear reactions, compared to the evolutionary time scale of a star with the exception of various explosions.

The Local Thermodynamic Equilibrium

LTE means that in a sufﬁciently small volume at any position in a star, the complete thermodynamic (mechanical, thermal, and chemical) equilibrium is very nearly true. The time scale to establish the LTE is also typically short enough, except for nuclear reactions, compared to the evolutionary time scale of a star with the exception of various explosions.

The Local Thermodynamic Equilibrium

LTE means that in a sufﬁciently small volume at any position in a star, the complete thermodynamic (mechanical, thermal, and chemical) equilibrium is very nearly true. The equilibrium of chemical reactions means X µi dNi = 0, i

+ − 0 H + e ↔ H + χH

where χH = 13.6 eV is the ionization potential from the ground state of H. For simplicity, we have assumed that H has only one bound level and that the gas is pure hydrogen.

Let us ﬁrst consider a classic black body cavity ﬁlled with radiation in thermodynamic equilibrium with the wall. The equilibrium of a chemical reaction with photons can be written as X µi dNi + µγ dNγ = 0 i

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. So we can neglect the photons here in the discussion of the equilibrium of a chemical reaction. Let us ﬁrst consider a classic black body cavity ﬁlled with radiation in thermodynamic equilibrium with the wall. The equilibrium of a chemical reaction with photons can be written as X µi dNi + µγ dNγ = 0 i

+ − 0 H + e ↔ H + χH where χH = 13.6 eV is the ionization potential from the ground state of H. For simplicity, we have assumed that H has only one bound level and that the gas is pure hydrogen. Clearly in such an equilibrium, Ni is closely linked to µi and depends on T and ρ or equivalent thermodynamic quantities, as well as on a catalog of the possible reactions.

The ionization-recombination reaction can then be written as

1H+ + 1e− − 1H0 = 0

And in general, a reaction can be written in such a symbolic form: X νi Ci = 0, i where Ci is the names of the participating particle (except for photons). Clearly, dNi ∝ νi . Thus X µi νi = 0 i

This is the equation of chemical equilibrium. The ionization-recombination reaction can then be written as

1H+ + 1e− − 1H0 = 0

This is the equation of chemical equilibrium. Clearly in such an equilibrium, Ni is closely linked to µi and depends on T and ρ or equivalent thermodynamic quantities, as well as on a catalog of the possible reactions. Outline

Introduction

The Local Thermodynamic Equilibrium

The Distribution Function

Black Body Radiation

Fermi-Dirac EoS The Complete Degenerate Gas Application to White Dwarfs Temperature Effects

Ideal Gas

The Saha Equation

“Almost Perfect” EoS

Adiabatic Exponents and Other Derivatives and “+” or “−” is for Fermi-Dirac or Bose-Einstein particles. The number of states in the phase space, accounting for the degeneracy of the state gj , is

g d 3pd 3r j . h3

The Distribution Function

For a particular species of elementary nature in the LTE, the occupation number at a certain quantum state in the coordinate-momentum phase space is

1 F(j, E(p)) = , exp[−µ + Ej + E(p)]/kT ± 1

where Ej is the internal energy state j referred to some reference energy level, E(p) is the kinetic energy as a function of the momentum p, E(p) = (p2c2 + m2c4)1/2 − mc2, µ is the chemical potential of the species and is to be determined, The number of states in the phase space, accounting for the degeneracy of the state gj , is

g d 3pd 3r j . h3

The Distribution Function

For a particular species of elementary nature in the LTE, the occupation number at a certain quantum state in the coordinate-momentum phase space is

1 F(j, E(p)) = , exp[−µ + Ej + E(p)]/kT ± 1

where Ej is the internal energy state j referred to some reference energy level, E(p) is the kinetic energy as a function of the momentum p, E(p) = (p2c2 + m2c4)1/2 − mc2, µ is the chemical potential of the species and is to be determined, and “+” or “−” is for Fermi-Dirac or Bose-Einstein particles. The Distribution Function

For a particular species of elementary nature in the LTE, the occupation number at a certain quantum state in the coordinate-momentum phase space is

1 F(j, E(p)) = , exp[−µ + Ej + E(p)]/kT ± 1

where Ej is the internal energy state j referred to some reference energy level, E(p) is the kinetic energy as a function of the momentum p, E(p) = (p2c2 + m2c4)1/2 − mc2, µ is the chemical potential of the species and is to be determined, and “+” or “−” is for Fermi-Dirac or Bose-Einstein particles. The number of states in the phase space, accounting for the degeneracy of the state gj , is

g d 3pd 3r j . h3 So the phase-space density can be deﬁned as

1 X gj ( ) = . n p 3 h exp[−µ + Ej + E(p)]/kT ± 1 j

We can then calculate the physical space density n, kinetic pressure P, and internal energy E as Z n = n(p)4πp2dp, p which gives the connection between n, T , and µ.

1 Z P = n(p)vp4πp2dp, 3 p ∂E(p) where v = . ∂p Z E = n(p)E(p)4πp2dp p Outline

Introduction

The Local Thermodynamic Equilibrium

The Distribution Function

Black Body Radiation

Fermi-Dirac EoS The Complete Degenerate Gas Application to White Dwarfs Temperature Effects

Ideal Gas

The Saha Equation

“Almost Perfect” EoS

Adiabatic Exponents and Other Derivatives Black Body Radiation

Photons are mass-less bosons of unit spin and µ = 0, as discussed earlier. Since they travel at c (hence have no rest frame), they have only two states (two spin orientations or polarizations). Hence,

8π Z ∞ p2dp = ∝ 3. nγ 3 T h 0 exp(pc/kT ) − 1

aT 4 Similarly, one can easily get P = and E = aT 4 = 3P , rad 3 rad rad where a is the radiation constant. The nice thing about the LTE is that all these quantities depend only on T . So radiation follows a γ-law EoS P = (γ − 1)E with γ = 4/3. The speciﬁc energy density per unit frequency (ν) is

8πhν3 1 u = erg cm−3Hz−1. ν c3 exp(hν/kT ) − 1 and the frequency-dependent Planck function c B (T ) = u . ν 4π ν The integrated Planck function is B(x) = x3/[exp(x) − 1]. Z σ 4 −2 −1 −1 The maximum is at B(T ) = Bν (T )dν = T erg cm s sr . π x = hν/kT = 2.82. where σ = ca/4 is the Stefan-Boltzmann constant. Outline

Introduction

The Local Thermodynamic Equilibrium

The Distribution Function

Black Body Radiation

Fermi-Dirac EoS The Complete Degenerate Gas Application to White Dwarfs Temperature Effects

Ideal Gas

The Saha Equation

“Almost Perfect” EoS

Adiabatic Exponents and Other Derivatives Fermi-Dirac EoS

Elections, protons, and neutrons are fermions. They all have spin one half. Hence g = 2.

The occupation number is

1 F(E) = , exp ([E(p) − µ]/kT ) + 1

p 2 where E(p) = mc2 (1 + )1/2 − 1 and mc p p 2−1/2 v(p) = 1 + . m mc When T → 0, the integral tends to be either The shaded area and the zero or unity, depending on whether the kinetic dashed line show how F(E) energy E(p) < µ or E(p) > µ. We call this is changed by raising the temperature from kT = 0 to critical kinetic energy, EF = µ, as the “Fermi energy”, though we are yet to determine µ. EF /20. The Complete Degenerate Gas When T = 0, fermions ﬁll the energy range, 0 ≤ E ≤ EF . Correspondingly, we have pF , as deﬁned by p E = mc2 (1 + x 2)1/2 − 1, where x = F . F F F mc The number density can be calculated as

−3 8π Z pF 8π h = 2 = 3, n 3 p dp xF h 0 3 mc h where (= 2.4 × 10−10 cm for electrons) is the Compton mc wavelength. For the complete degenerate electron gas (i.e., n = ne = ρ/(µema)),

ρ 3 = BxF , µe −3 8πma h 5 −3 where B = = 9.7 × 10 g cm and µe is electron 3 mec mean molecular weight. ρ Once is known, we can derive xF , and hence pF and EF . µe The pressure and internal energy density can be calculated as

8π m4c5 Z xF x 4 P = dx, 3 2 1/2 3 h 0 (1 + x ) and −3 h Z xF E = 8π mc2 x 2[(1 + x 2)1/2 − 1]dx, mc 0 5 5/3 which have limiting forms of P ∝ E ∝ xF ∝ ρ for xF 1 and 4 4/3 P ∝ E ∝ xF ∝ ρ for xF 1. It is easy to show E/P = 3/2 (hence γ = 5/3, assuming a γ-law EoS, P = (γ − 1)E) for xF 1 and E/P = 3 (γ = 4/3) for xF 1. xF ∼ 1 or pF ∼ mec corresponds to the demarcation between non-relativistic and relativistic mechanics. The corresponding density is ∼ 106 g cm−3, 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 xF ≈ 0.35. Thus neutron stars are non-relativistic. The completely degenerate gas in a The equation of state for non-relativistic limit acts like a monatomic completely degenerate ideal gas whereas, in the extreme relativistic electrons. The transition limit, it behaves like a photon gas. between the non-relativistic to the extremely relativistic cases is smooth, but takes place at densities around 6 3 ρtr ≈ 10 µe g cm . Application to White Dwarfs The above P(x(ρ)) relation together with the hydrostatic equation forms a second-order differential equation (e.g., ρ vs. Mr ). Its solution would give the structure of a WD (see Chapter 7); The WD radius can be approximately expressed as

" #1/2 2 5/3 M −1/3 M 4/3 R ≈ 0.013R 1 − , µe M MCh

where the Chandrasekhar limiting mass is

2 2 MCh = 1.4M . µe

When M → MCh, R → 0! The interpretation is that an electron degenerate object (of ﬁxed µe) cannot have the mass exceeding the Chandrasekhar limit. The EoS is too soft! Increasing density and pressure cannot halt the collapse because the relativistic limit has already been reached. A similar limit (∼ 3M ) can be found for a neutron star mass. Temperature Effects

Suppose the temperature is low — on some scale yet to be determined — but not zero. Fermions near the surface of the sea, within a depth of ∼ kT , may ﬁnd themselves elevated into energies greater an EF .

The net effect is that the occupation distribution smooths out to higher energies. Apparently, the criterion for the transition between degeneracy and near- or degeneracy is EF ∼ kT . The criterion gives a ρ/µe − T relation that roughly separates the two regimes.

For a non-relativistic electron gas 2 2 (EF = mec xF /2), ρ = 6.0 × 10−9T 3/2 g cm−3. µe

2 For a relativistic electron gas (EF = mec xF ), Figure: The location of the ρ −24 3 −3 = 4.6 × 10 T g cm . center of the present-day sun µe in these coordinates is indicated by the sign. Outline

Introduction

The Local Thermodynamic Equilibrium

The Distribution Function

Black Body Radiation

Fermi-Dirac EoS The Complete Degenerate Gas Application to White Dwarfs Temperature Effects

Ideal Gas

The Saha Equation

“Almost Perfect” EoS

Adiabatic Exponents and Other Derivatives Taking logarithmic differentials of the above equation on both sides and noticing

∞ Z 2 e−p /(2mkT )p2dp = (2πmkT )3/2/(4π) (2) 0 yield the distribution function for a Maxwell-Boltzmann ideal gas:

dn(p) 4π 2 = e−p /2mkT p2dp. n (2πmkT )3/2

From this equation, it is easy to get the pressure P and to prove P = nkT , which is true even if the particles are relativistic.

Ideal Gas The Boltzmann distribution for an ideal gas is characterized by µ/kT −1, i.e., the occupation number is very low. For simplicity, we assume that the gas particles are non-relativistic with E = p2/2m, v = p/m. The number density of the gas is then

p→∞ 4π Z 2 = (µ−p /2m)/kT 2 . n 3 g e p dp (1) h 0 Ideal Gas The Boltzmann distribution for an ideal gas is characterized by µ/kT −1, i.e., the occupation number is very low. For simplicity, we assume that the gas particles are non-relativistic with E = p2/2m, v = p/m. The number density of the gas is then

p→∞ 4π Z 2 = (µ−p /2m)/kT 2 . n 3 g e p dp (1) h 0

Taking logarithmic differentials of the above equation on both sides and noticing

∞ Z 2 e−p /(2mkT )p2dp = (2πmkT )3/2/(4π) (2) 0 yield the distribution function for a Maxwell-Boltzmann ideal gas:

dn(p) 4π 2 = e−p /2mkT p2dp. n (2πmkT )3/2

From this equation, it is easy to get the pressure P and to prove P = nkT , which is true even if the particles are relativistic. Now consider particles (e.g., ions) with multiple energy levels, which are populated or depopulated by photon absorption or emission, for example. We need to have the internal energy level Ej restored in Eqs. 1 and 3. But because the photon chemical potential is zero, µ1 = µ2 for any two levels. Then we have n1 g1 = e−(E1−E2)/kT , n2 g2 which is the Boltzmann population distribution.

Furthermore, the combination of Eqs. 1 and 2 yields

nh3 eµ/kT = (3) g(2πmkT )3/2 Then we have n1 g1 = e−(E1−E2)/kT , n2 g2 which is the Boltzmann population distribution.

Furthermore, the combination of Eqs. 1 and 2 yields

nh3 eµ/kT = (3) g(2πmkT )3/2

Now consider particles (e.g., ions) with multiple energy levels, which are populated or depopulated by photon absorption or emission, for example. We need to have the internal energy level Ej restored in Eqs. 1 and 3. But because the photon chemical potential is zero, µ1 = µ2 for any two levels. Furthermore, the combination of Eqs. 1 and 2 yields

nh3 eµ/kT = (3) g(2πmkT )3/2

Now consider particles (e.g., ions) with multiple energy levels, which are populated or depopulated by photon absorption or emission, for example. We need to have the internal energy level Ej restored in Eqs. 1 and 3. But because the photon chemical potential is zero, µ1 = µ2 for any two levels. Then we have n1 g1 = e−(E1−E2)/kT , n2 g2 which is the Boltzmann population distribution. Outline

Introduction

The Local Thermodynamic Equilibrium

The Distribution Function

Black Body Radiation

Fermi-Dirac EoS The Complete Degenerate Gas Application to White Dwarfs Temperature Effects

Ideal Gas

The Saha Equation

“Almost Perfect” EoS

Adiabatic Exponents and Other Derivatives To avoid the double counting, we may choose g− = 2 and g+ = 1 (or vise verse) as well as g0 = 2. Then from Eq. 3, we can obtain

3/2 3/2 2[2πm kT ] − [2πm kT ] + n = e eµ /kT ; n+ = p eµ /kT ; e h3 h3

3/2 2[2π(me + mp)kT ] 0 n0 = eµ /kT eχH /kT . h3

The Saha Equation

The LTE and hence the chemical equilibrium also allow us to determine the number densities of individual species. Again consider the above ionization-recombination reaction:

+ − 0 H + e ↔ H + χH .

We establish the reference energy levels for all species by taking the + − zero energy as the just-ionized to be the H + e state. Thus E0 is zero for both electrons and protons, whereas = −χH for neutral hydrogen at the ground state. Then from Eq. 3, we can obtain

3/2 3/2 2[2πm kT ] − [2πm kT ] + n = e eµ /kT ; n+ = p eµ /kT ; e h3 h3

3/2 2[2π(me + mp)kT ] 0 n0 = eµ /kT eχH /kT . h3

The Saha Equation

The LTE and hence the chemical equilibrium also allow us to determine the number densities of individual species. Again consider the above ionization-recombination reaction:

+ − 0 H + e ↔ H + χH .

The LTE and hence the chemical equilibrium also allow us to determine the number densities of individual species. Again consider the above ionization-recombination reaction:

+ − 0 H + e ↔ H + χH .

3/2 3/2 2[2πm kT ] − [2πm kT ] + n = e eµ /kT ; n+ = p eµ /kT ; e h3 h3

3/2 2[2π(me + mp)kT ] 0 n0 = eµ /kT eχH /kT . h3 + The electrical neutrality gives ne = n . Using y to parametrize the n+ fraction of all hydrogen that is ionized, i.e., y = (where n n = n0 + n+ is a constant, independent of density), the equation then becomes 2 3/2 y 1 2πm kT H = e e−χ /kT . (1 − y) n h2

The dominant factor here is the exponential, while the dependence on the density is weak. The characteristic temperature for ionization-recombination is ∼ 1.6 × 104 K , or roughly y ≈ 1/2 when χ/kT ∼ 10.

Notice that for the LTE, µ− + µ+ − µ0 = 0 and [memp/(me + mp)] ≈ me, we have the Saha equation:

+ 3/2 n n 2πm kT H e = e e−χ /kT . n0 h2 The dominant factor here is the exponential, while the dependence on the density is weak. The characteristic temperature for ionization-recombination is ∼ 1.6 × 104 K , or roughly y ≈ 1/2 when χ/kT ∼ 10.

Notice that for the LTE, µ− + µ+ − µ0 = 0 and [memp/(me + mp)] ≈ me, we have the Saha equation:

+ 3/2 n n 2πm kT H e = e e−χ /kT . n0 h2

+ The electrical neutrality gives ne = n . Using y to parametrize the n+ fraction of all hydrogen that is ionized, i.e., y = (where n n = n0 + n+ is a constant, independent of density), the equation then becomes 2 3/2 y 1 2πm kT H = e e−χ /kT . (1 − y) n h2 Notice that for the LTE, µ− + µ+ − µ0 = 0 and [memp/(me + mp)] ≈ me, we have the Saha equation:

+ 3/2 n n 2πm kT H e = e e−χ /kT . n0 h2

3kT E = (1 + y)n + ynχ erg g−1, 2 H including the ionization energy. P and E clearly depend only on the temperature and density. For real calculation, of course, all species, energy levels, and reactions must be considered. For example, one also has the ionization-recombination for Helium at high temperatures. The presence of such zones of ionization has profound consequences for the structure of a star. Later we will further use similar Saha equations for some fast nuclear reactions under certain states where the thermodynamic equilibrium can be assumed.

The thermal pressure can be expressed as

+ 0 P = (ne + n + n )kT = (1 + y)nρkT . For real calculation, of course, all species, energy levels, and reactions must be considered. For example, one also has the ionization-recombination for Helium at high temperatures. The presence of such zones of ionization has profound consequences for the structure of a star. Later we will further use similar Saha equations for some fast nuclear reactions under certain states where the thermodynamic equilibrium can be assumed.

The thermal pressure can be expressed as

+ 0 P = (ne + n + n )kT = (1 + y)nρkT .

The speciﬁc internal energy to be used later is

3kT E = (1 + y)n + ynχ erg g−1, 2 H including the ionization energy. P and E clearly depend only on the temperature and density. Later we will further use similar Saha equations for some fast nuclear reactions under certain states where the thermodynamic equilibrium can be assumed.

The thermal pressure can be expressed as

+ 0 P = (ne + n + n )kT = (1 + y)nρkT .

The speciﬁc internal energy to be used later is

+ 0 P = (ne + n + n )kT = (1 + y)nρkT .

The speciﬁc internal energy to be used later is

Introduction

The Local Thermodynamic Equilibrium

The Distribution Function

Black Body Radiation

Fermi-Dirac EoS The Complete Degenerate Gas Application to White Dwarfs Temperature Effects

Ideal Gas

The Saha Equation

“Almost Perfect” EoS

Adiabatic Exponents and Other Derivatives For example, the demarcation between the radiation and ideal gas 4 pressures (aT /3 = ρkT /µmA) for completely ionized hydrogen gas is ρ = 1.5 × 10−23T 3 g cm−3. At ρ ∼ 1 g cm−3, such gas can also get pressure-ionized because the 1/3 Wigner-Seitz radius a ∼ [3/(4πnI )] is comparable to the ﬁrst Bohr orbit. A measure of the interaction energy between ions is the Coulomb Z 2e2 potential. The ratio Γ ≡ characterizes the importance of the C akT Coulomb effects relative to the thermal agitation. When Γc is much greater than 1, the gas settles down into a crystal.

“Almost Perfect” EoS

Sometimes, interactions need to be accounted for that modify the above “perfect” results. In addition, a stellar EoS might consist of many components with radiation, Maxwell-Boltzmann, and degenerate gases competing in importance. At ρ ∼ 1 g cm−3, such gas can also get pressure-ionized because the 1/3 Wigner-Seitz radius a ∼ [3/(4πnI )] is comparable to the ﬁrst Bohr orbit. A measure of the interaction energy between ions is the Coulomb Z 2e2 potential. The ratio Γ ≡ characterizes the importance of the C akT Coulomb effects relative to the thermal agitation. When Γc is much greater than 1, the gas settles down into a crystal.

“Almost Perfect” EoS

Sometimes, interactions need to be accounted for that modify the above “perfect” results. In addition, a stellar EoS might consist of many components with radiation, Maxwell-Boltzmann, and degenerate gases competing in importance. For example, the demarcation between the radiation and ideal gas 4 pressures (aT /3 = ρkT /µmA) for completely ionized hydrogen gas is ρ = 1.5 × 10−23T 3 g cm−3. A measure of the interaction energy between ions is the Coulomb Z 2e2 potential. The ratio Γ ≡ characterizes the importance of the C akT Coulomb effects relative to the thermal agitation. When Γc is much greater than 1, the gas settles down into a crystal.

“Almost Perfect” EoS

Sometimes, interactions need to be accounted for that modify the above “perfect” results. In addition, a stellar EoS might consist of many components with radiation, Maxwell-Boltzmann, and degenerate gases competing in importance. For example, the demarcation between the radiation and ideal gas 4 pressures (aT /3 = ρkT /µmA) for completely ionized hydrogen gas is ρ = 1.5 × 10−23T 3 g cm−3. At ρ ∼ 1 g cm−3, such gas can also get pressure-ionized because the 1/3 Wigner-Seitz radius a ∼ [3/(4πnI )] is comparable to the ﬁrst Bohr orbit. A measure of the interaction energy between ions is the Coulomb Z 2e2 potential. The ratio Γ ≡ characterizes the importance of the C akT Coulomb effects relative to the thermal agitation. When Γc is much greater than 1, the gas settles down into a crystal. In general, the EoS here is approximated as P ∝ T χT ρχρ , where the exponents typically need to be evaluated numerically.

“Almost Perfect” EoS

This composite ﬁgure shows how the ρ − T plane is broken up into regions dominated by pressure ionization, degeneracy, radiation, ideal gas, crystallization, and ionization-recombination. The gas is assumed to be pure hydrogen. “Almost Perfect” EoS

In general, the EoS here is approximated as P ∝ T χT ρχρ , where the exponents typically need to be evaluated numerically. Outline

Introduction

The Local Thermodynamic Equilibrium

The Distribution Function

Black Body Radiation

Fermi-Dirac EoS The Complete Degenerate Gas Application to White Dwarfs Temperature Effects

Ideal Gas

The Saha Equation

“Almost Perfect” EoS

Adiabatic Exponents and Other Derivatives Such derivatives include the speciﬁc heats (cρ and cp), power-law exponents of the EoS:

∂lnP ∂lnP χT = , χρ = ; ∂lnT ρ ∂lnρ T

and adiabatic exponents:

∂lnP Γ2 ∂lnP 1 ∂lnT Γ1 = , = = , Γ3 − 1 = . ∂lnρ S Γ2 − 1 ∂lnT S ∇S ∂lnρ S

Adiabatic Exponents and Other Derivatives

With the EoS, we can in principle construct a simpliﬁed stellar model. But to construct realistic ones and to evolve them, we need several thermodynamic derivatives, for example, to study the stability of a model and to determine weather or not the convection needs to be considered. power-law exponents of the EoS:

∂lnP ∂lnP χT = , χρ = ; ∂lnT ρ ∂lnρ T

and adiabatic exponents:

∂lnP Γ2 ∂lnP 1 ∂lnT Γ1 = , = = , Γ3 − 1 = . ∂lnρ S Γ2 − 1 ∂lnT S ∇S ∂lnρ S

Adiabatic Exponents and Other Derivatives

∂lnP Γ2 ∂lnP 1 ∂lnT Γ1 = , = = , Γ3 − 1 = . ∂lnρ S Γ2 − 1 ∂lnT S ∇S ∂lnρ S

Adiabatic Exponents and Other Derivatives

∂lnP ∂lnP χT = , χρ = ; ∂lnT ρ ∂lnρ T Adiabatic Exponents and Other Derivatives

∂lnP ∂lnP χT = , χρ = ; ∂lnT ρ ∂lnρ T

and adiabatic exponents:

∂lnP Γ2 ∂lnP 1 ∂lnT Γ1 = , = = , Γ3 − 1 = . ∂lnρ S Γ2 − 1 ∂lnT S ∇S ∂lnρ S Speciﬁc heats The thermodynamic equation can be expressed as:

dQ = TdS = dE + PdV .

where the chemical energy of particles has been absorbed into E. As ∂Q ∂E an example, we consider the speciﬁc heat, cρ = = ∂T ρ ∂T ρ for the above pure hydrogen gas. Recall

3kT E = (1 + y)n + ynχ erg g−1. 2 H We have ∂E ∂E ∂E ∂y = + ∂T ρ ∂T y ∂y T ∂T ρ " # 3kT 1 2 3 χ 1 ∂y = (1 + y)n + + H . 2 T 3 2 kT (1 + y) ∂T ρ Note that D(1) = D(0) = 0 and, for general 0 ≤ y ≤ 1, D(y) ≥ 0. Thus the term is greater than zero, meaning that an increase of the temperature, keeping the density ﬁxed, requires more energy when the reaction occurs than when it does not.

Speciﬁc heats

Now recall the Saha equation

2 3/2 y 1 2πm kT H = e e−χ /kT . (1 − y) n h2

Now recall the Saha equation

2 3/2 y 1 2πm kT H = e e−χ /kT . (1 − y) n h2

Take a log of it and then take the partial derivative of T , we get 2 1 ∂y 3 χH 1 + = + → y 1 − y ∂T 2 kT T 1 ∂y 3 χH 1 y(1 − y) = D(y) + , where D(y) = . (1 + y) ∂T 2 kT T (2 − y)(1 + y) " # 3k 2 3 χ 2 Therefore, c = (1 + y)n 1 + D(y) + H erg g−1 K−1, ρ 2 3 2 kT where the second term arises from the consideration of the ionization-recombination reaction. Note that D(1) = D(0) = 0 and, for general 0 ≤ y ≤ 1, D(y) ≥ 0. Thus the term is greater than zero, meaning that an increase of the temperature, keeping the density fixed, requires more energy when the reaction occurs than when it does not. Among the five variables in the first law of thermodynamics (E, S, P, T , and V ), only two are independent. The law can be expressed in various forms by replacing E with H = E + PV (enthalpy), F = E − TS (free energy), or Φ = E + PV − TS (the Gibbs free energy), and correspondingly replacing the independent variables S and V with S and P, T and V , or T and P, respectively. For example,

dE = TdS − PdV → dΦ = −SdT + VdP

Variable Conversions

In realistic (complicated) situation, the calculations of the derivatives need to be done numerically. Thus it is desirable to know the analytic conversions of the derivatives from a minimal set of calculations. The law can be expressed in various forms by replacing E with H = E + PV (enthalpy), F = E − TS (free energy), or Φ = E + PV − TS (the Gibbs free energy), and correspondingly replacing the independent variables S and V with S and P, T and V , or T and P, respectively. For example,

dE = TdS − PdV → dΦ = −SdT + VdP

Variable Conversions

In realistic (complicated) situation, the calculations of the derivatives need to be done numerically. Thus it is desirable to know the analytic conversions of the derivatives from a minimal set of calculations. Among the ﬁve variables in the ﬁrst law of thermodynamics (E, S, P, T , and V ), only two are independent. F = E − TS (free energy), or Φ = E + PV − TS (the Gibbs free energy), and correspondingly replacing the independent variables S and V with S and P, T and V , or T and P, respectively. For example,

dE = TdS − PdV → dΦ = −SdT + VdP

Variable Conversions

In realistic (complicated) situation, the calculations of the derivatives need to be done numerically. Thus it is desirable to know the analytic conversions of the derivatives from a minimal set of calculations. Among the ﬁve variables in the ﬁrst law of thermodynamics (E, S, P, T , and V ), only two are independent. The law can be expressed in various forms by replacing E with H = E + PV (enthalpy), or Φ = E + PV − TS (the Gibbs free energy), and correspondingly replacing the independent variables S and V with S and P, T and V , or T and P, respectively. For example,

dE = TdS − PdV → dΦ = −SdT + VdP

Variable Conversions

In realistic (complicated) situation, the calculations of the derivatives need to be done numerically. Thus it is desirable to know the analytic conversions of the derivatives from a minimal set of calculations. Among the ﬁve variables in the ﬁrst law of thermodynamics (E, S, P, T , and V ), only two are independent. The law can be expressed in various forms by replacing E with H = E + PV (enthalpy), F = E − TS (free energy), and correspondingly replacing the independent variables S and V with S and P, T and V , or T and P, respectively. For example,

dE = TdS − PdV → dΦ = −SdT + VdP

Variable Conversions

In realistic (complicated) situation, the calculations of the derivatives need to be done numerically. Thus it is desirable to know the analytic conversions of the derivatives from a minimal set of calculations. Among the ﬁve variables in the ﬁrst law of thermodynamics (E, S, P, T , and V ), only two are independent. The law can be expressed in various forms by replacing E with H = E + PV (enthalpy), F = E − TS (free energy), or Φ = E + PV − TS (the Gibbs free energy), For example,

dE = TdS − PdV → dΦ = −SdT + VdP

Variable Conversions

In realistic (complicated) situation, the calculations of the derivatives need to be done numerically. Thus it is desirable to know the analytic conversions of the derivatives from a minimal set of calculations. Among the ﬁve variables in the ﬁrst law of thermodynamics (E, S, P, T , and V ), only two are independent. The law can be expressed in various forms by replacing E with H = E + PV (enthalpy), F = E − TS (free energy), or Φ = E + PV − TS (the Gibbs free energy), and correspondingly replacing the independent variables S and V with S and P, T and V , or T and P, respectively. Variable Conversions

In realistic (complicated) situation, the calculations of the derivatives need to be done numerically. Thus it is desirable to know the analytic conversions of the derivatives from a minimal set of calculations. Among the ﬁve variables in the ﬁrst law of thermodynamics (E, S, P, T , and V ), only two are independent. The law can be expressed in various forms by replacing E with H = E + PV (enthalpy), F = E − TS (free energy), or Φ = E + PV − TS (the Gibbs free energy), and correspondingly replacing the independent variables S and V with S and P, T and V , or T and P, respectively. For example,

dE = TdS − PdV → dΦ = −SdT + VdP For example, one can easily drive the Maxwell’s relations (giving it a try!):

∂T ∂P ∂T ∂V = − , = , ∂V S ∂S V ∂P S ∂S P ∂S ∂P ∂S ∂V = , = − . ∂V T ∂T V ∂P T ∂T P

These relations are very useful to express the derivatives in certain desirable forms and to know their relationships.

Variable Conversions

Using the above expressions and such mathematical relations as

∂2Φ ∂2Φ = , ∂P∂T ∂T ∂P one can readily do various variable transformations of the derivative. Variable Conversions

Using the above expressions and such mathematical relations as

∂2Φ ∂2Φ = , ∂P∂T ∂T ∂P one can readily do various variable transformations of the derivative. For example, one can easily drive the Maxwell’s relations (giving it a try!):

∂T ∂P ∂T ∂V = − , = , ∂V S ∂S V ∂P S ∂S P ∂S ∂P ∂S ∂V = , = − . ∂V T ∂T V ∂P T ∂T P

These relations are very useful to express the derivatives in certain desirable forms and to know their relationships. Variable Conversions

For variable changes, one may ﬁnd useful such Jacobian transformation as

∂(V , P) ∂P ∂V ∂(V , T ) ∂T = = V ∂T ∂(T , P) ∂P P − ∂(V , T ) ∂V T This relation can also be derived from ∂V ∂V dV = dT + dP ∂T P ∂P T by holding V constant, dV = 0. Let’s see a couple of examples for the use of the conversions. Example 1: Various γs: γ ∂lnP I P = K ρ , where γ = Γ1 = . ∂lnρ S I γ ≡ cP /cV . This γ = Γ1 is good for ideal gas, but does not hold, when the radiation pressure is important, for example. I γ in the γ-law EoS: P = (γ − 1)ρE. ∂lnT Let’s try to prove that the last γ equals to Γ3 = 1 + . This ∂lnρ S means that we need to prove

∂P = (Γ3 − 1)ρ (4) ∂E ρ ∂E ∂S From dE = TdS − PdV , we have = T and ∂P ρ ∂P ρ ∂E T = . ∂S ρ ∂T 1 ∂P = Furthermore, 2 . ∂ρ S ρ ∂S ρ We can then express the left side of Eq. 4 as

∂P 1 ∂P ρ2 ∂T = = = (Γ3 − 1)ρ. ∂E ρ T ∂S ρ T ∂ρ S It is also easy to show 1 ∂P P χT Γ3 − 1 = = . ρ ∂E ρ ρT cV

Note that γ = Γ3 in the γ-law EoS is generally not equal to

2 χT P χT γ ≡ cP /cV = 1 + (Γ3 − 1) = 1 + . (5) χρ ρTcV χρ

Try to prove this latter relation! Clearly, when χT = 1 and χρ = 1 (for ideal gas), then this latter γ is also equal to Γ3. Example 2: Relation between cP and cρ or cVρ . Start by letting T and P be independent and write the speciﬁc heat content as

∂S ∂S dq = TdS = T dT + dP . ∂T P ∂P T and ∂P ∂P dP = dT + dV . ∂T V ∂V T This gives

∂S ∂S ∂P ∂P dq = T dT + T dT + dV . ∂T P ∂P T ∂T V ∂V T While holding V constant,

∂q ∂S ∂S ∂P = T + T ∂T V ∂T P ∂P T ∂T V ∂S ∂P cp − cV = −T . ∂P T ∂T V Now let’s express the right-hand side with the exponents of the E.o.S. Using one of the Maxwell’s relations

∂S ∂V = − ∂P T ∂T P and then the relation we have just derived:

∂P ∂V ∂T = V ∂T ∂P P − ∂V T we have

∂P −1 ∂P 2 cp − cV = −T ∂V T ∂T V VP ∂lnP −1 ∂lnP 2 P χ2 = − = T , T ∂lnV T ∂lnT V ρT χρ which is same as Eq. 5. Review Key Concepts: local thermodynamic equilibrium, chemical potential µ, chemical equilibrium, Fermi energy, Chandrasekhar limiting mass 1. What is an equation of state? 2. Why should the chemical potential of photon gas be equal to zero? 3. What is the degenerate pressure? 4. What is the ρ − T relation corresponding to the condition under which the degenerate and thermal energies are comparable? 5. What is the Saha equation? Under which circumstance can the equation be used? 6. In reality, what may be various interactions that may make an EoS imperfect? In addition, you should be able to derive such quantities as density, pressure, and thermal energy for photon, ideal (Maxwell-Boltzmann), and completely degenerate gases, starting from the basic occupation distribution (with different µ limits). Get familiar with derivations of various adiabatic exponents and other derivatives.