Chapter 2: Equation of State - PDF Free Download

Introduction Chapter 2: Equation of State 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

Introduction Goal of the Chapter: derive the equation of state (or the mutual dependencies among local thermodynamic quantities such as P, T, ρ, and N i ), 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: 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. Second law: In an isolated system, natural processes are spontaneous when they lead to an increase in disorder, or entropy, finally reaching an equilibrium. Third law: The entropy of a system at absolute zero is a constant, determined only by the degeneracy of the ground state.

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: dq = TdS = de + PdV i µ i dn i where µ i is the chemical potential of the ith species ( ) E µ i =. N i Chemical reactions determine the species number density N i, which may have the units of number per gram of material; i.e., N i = n i /ρ. S,V

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 µ i dn i = 0, i appropriate for a star, since the mean free path of particles is generally short, compared to the variation scales of the thermodynamic quantities; e.g., λ ph = (κρ) 1 is on order of 1 cm ( except for stellar photospheres).

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 µ i dn i + µ γ dn γ = 0 i Since photon number is not strictly conserved, in general dn γ 0. While dn i = 0 for the cavity wall (as a whole), we have µ γ = 0. So we can neglect the photons here in the discussion of the equilibrium of a chemical reaction. As an example, let s consider the ionization-recombination reaction: H + + e H 0 + χ H where χ H = 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.

The ionization-recombination reaction can then be written as 1H + + 1e 1H 0 = 0 And in general, a reaction can be written in such a symbolic form: ν i C i = 0, i where C i is the names of the participating particle (except for photons). Clearly, dn i ν i. Thus µ i ν i = 0 i This is the equation of chemical equilibrium. Clearly in such an equilibrium, N i is closely linked to µ i and depends on T and ρ or equivalent thermodynamic quantities, as well as on a catalog of the possible reactions.

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 F(j, E(p)) = 1 exp[ µ + E j + E(p)]/kT ± 1, where E j 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) = (p 2 c 2 + m 2 c 4 ) 1/2 mc 2, µ is the chemical potential of the species and is to be determined,

So the phase-space density can be defined as n(p) = 1 h 3 j g j exp[ µ + E j + E(p)]/kT ± 1. We can then calculate the physical space density n, kinetic pressure P, and internal energy E as n = n(p)4πp 2 dp, which gives the connection between n, T, and µ. P = 1 n(p)vp4πp 2 dp, 3 where v = E(p) p. E = n(p)e(p)4πp 2 dp p p p

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, n γ = 8π p 2 dp h 3 0 exp(pc/kt ) 1 T 3. Similarly, one can easily get P rad = at 4 3 and E rad = at 4 = 3P 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 u ν = 8πhν3 c 3 1 exp(hν/kt ) 1 erg cm 3 Hz 1. and the frequency-dependent Planck function B ν (T ) = c 4π u ν. The integrated Planck function is B(T ) = B ν (T )dν = σ π T 4 erg cm 2 s 1 sr 1. B(x) = x 3 /[exp(x) 1]. The maximum is at x = hν/kt = 2.82. where σ = ca/4 is the Stefan-Boltzmann constant.

Fermi-Dirac EoS Elections, protons, and neutrons are fermions. They all have spin one half. Hence g = 2. The occupation number is F(E) = 1 exp ([E(p) µ]/kt ) + 1, ( p ) 2) where E(p) = mc [(1 2 + 1/2 1] mc [ ( p ) 2 ] 1/2 1 +. mc and v(p) = p m When T 0, the integral tends to be either zero or unity, depending on whether the kinetic energy E(p) < µ or E(p) > µ. We call this critical kinetic energy, E F = µ, as the Fermi energy, though we are yet to determine µ. The shaded area and the dashed line show how F (E) is changed by raising the temperature from kt = 0 to E F /20.

The Complete Degenerate Gas When T = 0, fermions fill the energy range, 0 E E F. Correspondingly, we have p F, as defined by ( pf E F = mc [ 2 (1 + xf 2)1/2 1 ] ), where x F =. mc The number density can be calculated as where n = 8π pf h 3 p 2 dp = 8π 0 3 ( ) 3 h xf 3 mc, h mc (= 2.4 10 10 cm for electrons) is the Compton wavelength. For the complete degenerate electron gas (i.e., n = n e = ρ/(µ e m a )), ρ µ e = Bx 3 F, where B = 8πm ( ) 3 a h = 9.7 10 5 g cm 3 and µ e is electron 3 m e c mean molecular weight. Once ρ is known, we can derive x F, and hence p F and E F. µ e

The pressure and internal energy density can be calculated as P = 8π 3 m 4 c 5 h 3 xf 0 x 4 dx, (1 + x 2 ) 1/2 and ( ) 3 h xf E = 8π mc 2 x 2 [(1 + x 2 ) 1/2 1]dx, mc 0 which have limiting forms of P E x 5 F ρ5/3 for x F 1 and P E x 4 F ρ4/3 for x F 1. It is easy to show E/P = 3/2 (hence γ = 5/3, assuming a γ-law EoS, P = (γ 1)E) for x F 1 and E/P = 3 (γ = 4/3) for x F 1. x F 1 or p F m e c corresponds to the demarcation between non-relativistic and relativistic mechanics.

The corresponding density is 10 6 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 10 14 g cm 3 ), it is easy to show x F 0.35. Thus neutron stars are non-relativistic. 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. The equation of state for completely degenerate electrons. The transition between the non-relativistic to the extremely relativistic cases is smooth, but takes place at densities around ρ tr 10 6 µ e g cm 3.

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. M r ). Its solution would give the structure of a WD (see Chapter 7); The WD radius can be approximately expressed as ( ) 5/3 ( ) 1/3 ( ) ] 4/3 1/2 2 M M R 0.013R [1, µ e M M Ch where the Chandrasekhar limiting mass is M Ch = 1.4M ( 2 µ e ) 2. When M M Ch, R 0! The interpretation is that an electron degenerate object (of fixed µ e ) 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 E F. 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 E F kt.

The criterion gives a ρ/µ e T relation that roughly separates the two regimes. For a non-relativistic electron gas (E F = m e c 2 x 2 F /2), ρ µ e = 6.0 10 9 T 3/2 g cm 3. For a relativistic electron gas (E F = m e c 2 x F ), ρ µ e = 4.6 10 24 T 3 g cm 3. Figure: The location of the center of the present-day sun in these coordinates is indicated by the sign.

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 = p 2 /2m, v = p/m. The number density of the gas is then n = 4π h 3 g p 0 e (µ p2 /2m)/kT p 2 dp. (1) Taking logarithmic differentials of the above equation on both sides and noticing 0 e p2 /(2mkT ) p 2 dp = (2πmkT ) 3/2 /(4π) (2) yield the distribution function for a Maxwell-Boltzmann ideal gas: dn(p) n = 4π 2 e p /2mkT p 2 dp. (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.

Furthermore, the combination of Eqs. 1 and 2 yields e µ/kt = nh 3 g(2πmkt ) 3/2 (3)

Furthermore, the combination of Eqs. 1 and 2 yields e µ/kt = nh 3 g(2πmkt ) 3/2 (3) 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 E j restored in Eqs. 1 and 3. But because the photon chemical potential is zero, µ 1 = µ 2 for any two levels.

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: H + + e H 0 + χ H. 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 E 0 is zero for both electrons and protons, whereas = χ H for neutral hydrogen at the ground state. To avoid the double counting, we may choose g = 2 and g + = 1 (or vise verse) as well as g 0 = 2.

Notice that for the LTE, µ + µ + µ 0 = 0 and [m e m p /(m e + m p )] m e, we have the Saha equation: n + ( ) 3/2 n e 2πme kt n 0 = h 2 e χh /kt.

Notice that for the LTE, µ + µ + µ 0 = 0 and [m e m p /(m e + m p )] m e, we have the Saha equation: n + ( ) 3/2 n e 2πme kt n 0 = h 2 e χh /kt. The electrical neutrality gives n e = n +. Using y to parametrize the fraction of all hydrogen that is ionized, i.e., y = n+ n (where n = n 0 + n + is a constant, independent of density), the equation then becomes y 2 (1 y) = 1 ( ) 3/2 2πme kt n h 2 e χh /kt. The dominant factor here is the exponential, while the dependence on the density is weak. The characteristic temperature for ionization-recombination is 1.6 10 4 K, or roughly y 1/2 when χ/kt 10.

The thermal pressure can be expressed as P = (n e + n + + n 0 )kt = (1 + y)nρkt.

The thermal pressure can be expressed as P = (n e + n + + n 0 )kt = (1 + y)nρkt. The specific internal energy to be used later is E = (1 + y)n 3kT 2 + ynχ H erg g 1, 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.

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 pressures (at 4 /3 = ρkt /µm A ) for completely ionized hydrogen gas is ρ = 1.5 10 23 T 3 g cm 3. At ρ 1 g cm 3, such gas can also get pressure-ionized because the Wigner-Seitz radius a [3/(4πn I )] 1/3 is comparable to the first Bohr orbit. A measure of the interaction energy between ions is the Coulomb potential. The ratio Γ C Z 2 e 2 characterizes the importance of the akt Coulomb effects relative to the thermal agitation. When Γ c is much greater than 1, the gas settles down into a crystal.

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. In general, the EoS here is approximated as P T χ T ρ χ ρ, where the exponents typically need to be evaluated numerically.

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. Such derivatives include the specific heats (c ρ and c p ), power-law exponents of the EoS: χ T = ( lnp lnt ), χ ρ = ρ ( ) lnp ; lnρ T

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, c ρ = = T ρ T ρ for the above pure hydrogen gas. Recall We have ( ) E T ρ = E = (1 + y)n 3kT 2 + ynχ H erg g 1. ( ) E + T y = (1 + y)n 3kT 2 ( ) E y T [ 1 T + 2 3 ( ) y T ρ ( 3 2 + χ H kt ) ( ) ] 1 y. (1 + y) T ρ

Specific heats Now recall the Saha equation y 2 (1 y) = 1 ( ) 3/2 2πme kt n h 2 e χh /kt. ( Take a log of it and then take the partial derivative of T, we get 2 y + 1 ) ( ) y 3 1 y T = 2 + χh 1 kt T ( ) 1 y 3 (1 + y) T = D(y) 2 + χh 1 kt T, where D(y) = y(1 y) (2 y)(1 + y). [ Therefore, c ρ = (1 + y)n 3k 1 + D(y) 2 ( 3 2 3 2 + χ ) ] 2 H erg g 1 K 1, 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.

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),

Variable Conversions Using the above expressions and such mathematical relations as 2 Φ P T = 2 Φ T P, one can readily do various variable transformations of the derivative.

Variable Conversions Using the above expressions and such mathematical relations as 2 Φ P T = 2 Φ 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 P S ( S V ) S T = ( P T ) V, V ( ) S P T S = ( V T These relations are very useful to express the derivatives in certain desirable forms and to know their relationships. ) P P.

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 This relation can also be derived from ( ) V dv = dt + T by holding V constant, dv = 0. P T ( ) V dp P T

Let s see a couple of examples for the use of the conversions. Example 1: Various γs: ( ) lnp P = K ρ γ, where γ = Γ 1 =. lnρ S γ c P /c V. This γ = Γ 1 is good for ideal gas, but does not hold, when the radiation pressure is important, for example. γ in the γ-law EoS: P = (γ 1)ρE. Let s try to prove that the last γ equals to Γ 3 = 1 + means that we need to prove ( ) lnt. This lnρ S ( ) P = (Γ 3 1)ρ (4) E ρ

( ) E From de = TdS PdV, we have = T P ( ) ρ E T =. S ρ ( ) T Furthermore, = 1 ( ) P ρ S ρ 2. S ρ We can then express the left side of Eq. 4 as ) ) ) ( P E ρ = 1 T It is also easy to show ( P S Γ 3 1 = 1 ρ ρ = ρ2 T ( T ρ S ( ) P = P χ T. E ρ ρt c V ( ) S and P ρ = (Γ 3 1)ρ. Note that γ = Γ 3 in the γ-law EoS is generally not equal to γ c P /c V = 1 + χ T (Γ 3 1) = 1 + P χ 2 T. (5) χ ρ ρtc V χ ρ Try to prove this latter relation! Clearly, when χ T = 1 and χ ρ = 1 (for ideal gas), then this latter γ is also equal to Γ 3.

Example 2: Relation between c P and c ρ or c Vρ. Start by letting T and P be independent and write the specific heat content as and This gives dq = T dq = TdS = T dp = ( ) S dt + T T P While holding V constant, ( ) q = T T V [( ) S dt + T P ( ) P dt + T V ( ) S P T c p c V = T ( ) S + T T P ( S P ( ) ] S dp. P T ( ) P dv. V T [( ) P dt + T V ( ) S ( ) P P T T ) ( ) P T. T V ( ) ] P dv. V 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 we have T ( P c p c V = T V = VP T ) 1 T ( lnp lnv ( ) 2 P T V ( lnp lnt ) 1 T ) 2 V = P χ 2 T, ρt χ ρ 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.