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Home , Equation of state

The Equation of State

The equation of state is the function that relates the to the density, molecular weight, and at any place in the . Since it is a solely an internal property of the , it can, in principle, be computed once externally, and used via a lookup table, i.e., P = P (ρ, µ, T ). The net pressure can be divided into three components, pressure from radiation, pressure from ions, and pressure from electrons. The latter may not obey the law due to the effects of degeneracy. Radiation Pressure

Normally, the pressure due to radiation in is small. How- ever, in the centers of massive stars, the generation is large enough to make the term substantial. To compute the radia- tion pressure, recall that from kinetic theory, the pressure of an isotropic gas comes from the transfer of , i.e.,

1 P = n(p) · v · pd3p 3 Z Since the energy density of a photon gas is

4π 2hν3 1 u dν = dν (7.1.1) ν c c2 ehν/kT − 1 then the number density of photons, n = uν /hν, in terms of the momentum, p = hν/c, is

8π p2 n(p) dp = dp h3 ehν/kT − 1

The pressure from the photon gas is therefore

1 1 8πc p3 Prad = n(p) cpdp = dp 3 3 h3 ehν/kT − 1 Z Z or, letting x = pc/kT (i.e., p = kT x/c),

4 ∞ 8πc kT x3 Prad = dx (7.1.2) 3h3 c ex − 1 µ ¶ Z0 The integral in (7.1.2) can be evaluated analytically (although I wouldn’t recommend it), and is

∞ x3 π4 dx = ex − 1 15 Z0 Thus 5 4 4 8π k 4 aT Prad = T = (7.1.3) 45c3h3 3 µ ¶ where a is the radiation constant (a = 4σ/c = 7.56471 × 10−15 ergs cm−3 ◦K−4). Ion Pressure

Under almost all circumstances, the pressure from ions can be evaluated through the ρ Pion = kT (7.2.1) µima The exceptions are Ion . At extremely high densities and low tempera- tures, Coulomb interactions between ions can become important. The ions are then constrained to move within a lattice, and their properties change. This should happen when the thermal energy of each ion becomes less than the inter-ion Coulomb energy, i.e., 3 (Ze)2 kT ≈ 2 rion 3 If we note that the per ion, V = 4/3 πrion = 1/nion, then the ratio of Coulomb energy to thermal energy is 2 1/3 2 1/3 uC (Ze) 2 4 −3 Z nion Γ = = πnion = 1.8 × 10 (7.2.2) C u kT 3 3 T t µ ¶ (Actually, crystallization begins to become important when ΓC ∼ 100.) The temperature is thus 3 2 −1/3 1/3 Tm ∼ 2.3 × 10 Z µ ρ (7.2.3) Except for evolved white dwarfs, stellar are always greater than the melting temperature.

Ion Degeneracy. Because the mass of an ion is ∼> 2000 times that of an electron, the momentum vectors associated with ions are much larger. (In the case of protons, by a factor of mp/me.) Thus, the space available for ions is much greater than for p electrons, and ion degeneracy is not a problem. Note also that α-particles are not Fermions, and thus degeneracy rules do not apply to compact helium cores. Electron Pressure

For main sequence stars, the pressure from electrons obeys the ideal gas law. But for more evolved stars, electron degeneracy may be important. To understand how and why degeneracy works, first compare the Maxwellian velocity distribution

4πp2 p2 f(p) dpdV = n exp − dpdV (7.3.1) e (2πm kT )3/2 2m kT e µ e ¶ with the Pauli exclusion principle. According to Pauli, each elec- tron occupies a region of phase space

3 dV dp = (∆x ∆px) (∆y ∆py)(∆z ∆pz) ∼ h and only two electrons (spin-up, spin-down) can occupy the same region of phase space at the same time. Therefore, since the vol- ume of phase space available to electrons with momenta between p and p+dp is 4πp2dp, the maximum phase-space density of these electrons is 2 (4πp2) f(p) dpdV ≤ dpdV (7.3.2) h3 For high densities, the number of electrons at a given momentum state may exceed the exclusion condition. The electrons are then forced into higher momentum states, until the Fermi momentum, pf is reached, i.e.,

pf 8πp2 8π n dV = dpdV = p3 dV (7.3.3) e h3 3h3 f Z0 This equation defines the Fermi momentum. It is the minimum momentum required to fit ne dV electrons into their lowest states. COMPLETELY DEGENERATE, NON-RELATIVISTIC GAS The equation of state for a completely degenerate non-relativistic 2 gas (i.e., one in which pf ¿ mec ) is fairly straightforward. When degenerate, the number of electrons with momenta between p and p + dp is 8πp2/h3 for p ≤ p f(p)= f (7.3.4) 0 for p > p ½ f To calculate the pressure (the flux of momenta through a unit surface), you need to know two things: 1) The flux of electrons hitting the surface, and 2) How much momentum is transferred by each electron. The former quantity can be calculated in a manner similar to dif- fusion. Recall that, for diffusion, we said that the flux of particles with velocity v diffusing across a boundary from one direction is 1 F1 = n − v − (3.1.1) 6 z l z l where l is the particle mean free path and v is the particle veloc- ity. If the mean free path is small, the total number of electrons crossing the boundary from both directions is 1 1 1 F1 + F2 = n − v − + n + v + ≈ nv 6 z l z l 6 z l z l 3 If each electron carries an amount of momentum, p, then the pressure at the boundary from electrons with velocity v is 1 P = n(v) v p 3 and the total pressure is ∞ ∞ 1 1 P = n(v) pvdv = n(p) pvdp (7.3.5) 3 3 Z0 Z0 For a non-relativistic electron gas, v = p/me, so for complete degeneracy 1 pf 8πp2 p 8π pf 8π P = p dp = p4dp = p5 e 3 h3 m 3m h3 15m h3 f Z0 µ e ¶ e Z0 e If we then substitute for pf using (7.3.3) 1 3 3h3 ρ / p = · (7.3.6) f 8π µ m µ e a ¶ and 2 3 5 3 h2 3 / ρ / P = e 20m m πm µ e a µ a ¶ µ e ¶ 5 3 ρ / = 1.0036 × 1013 cgs (7.3.7) µ µ e ¶ Note that there is no temperature dependence in this equation. The distribution of electrons in phase space is entirely determined by the Fermi exclusion principle, leaving pressure proportional only to density. Since there must be a smooth transition between the degenerate and non-degenerate states, we can estimate the density where this occurs by simply equating the two expressions from pressure 2 3 5 3 ρ h2 3 / ρ / P = kT = trans µm 20m m πm µ a e a µ a ¶ µ e ¶ Degeneracy therefore becomes important where 3 2 20m k / πm ρ = e a µ5/2 T 3/2 µ 3h3 e µ ¶ ³ ´ 3/2 T − − = 23.9 µ5/2 µ 3/2 g cm 3 (7.3.8) 106 e µ ¶ COMPLETELY DEGENERATE, RELATIVISTIC GAS If the density is extremely high, the velocities that the electrons must have to obey the Pauli exclusion rule will be relativistic. In this case, we must substitute for velocity using m v p = e (7.3.9) 1 − v2/c2 p The integral for pressure is then slightly more complicated

1 pf 8πp2 p/m P = p e dp (7.3.10) e 3 2 2 2 3 0 h 1+ p /(m c ) Z Ã e ! p If we let ξ = p/(mec), and x = pf /(mec), then the integral can be re-written as

8πm4c5 x ξ4 P = e dξ e 3h3 (1 + ξ2)1/2 Z0 This integral is analytic:

x 4 ξ dξ 1 − = x (2x2 − 3) (x2 + 1)1/2 + 3sinh 1 x (1 + ξ2)1/2 8 Z0 1 2 2 1 2 2 1 = x (2x − 3) (x + 1) / +3ln{x +(x + 1) 2 } 8

Following Chandrasekhar’s Principles of , if we let − f(x)= x (2x2 − 3) (x2 + 1)1/2 + 3sinh 1 x (7.3.11) then πm4c5 P = e f(x) (7.3.12) e 3h3 Note that if the gas is extremely relativisitic, pf À mec, thus x À 1, and f(x) → 2x4. In this limiting case

4 πm4c5 p 2πc P = e 2 f = p4 e 3h3 m c 3h3 f µ e ¶

Substituting density for pf using (7.3.6), we get

1 3 4 3 hc 3 / ρ / P = e 8m πm µ µ a ¶ µ a ¶ µ e ¶ 4 3 ρ / = 1.2435 × 1015 cgs (7.3.13) µ µ e ¶ Again, the pressure is independent of temperture, but this time, the exponent is 4/3. The transition point between the relativisitic and non-relativistic cases occurs when pf ∼ mec. If we substitute mec into (7.3.6), we find that this occurs at

8πµ m − ρ> e a m3c3 = 9.7 × 105 g cm 3 (7.3.14) 3h3 e PARTIAL DEGENERACY In many cases, when both the temperature and density are high, partial degeneracy occurs. When this happens, the electron dis- tribution in phase space is neither Maxwellian

4πp2 p2 f M (p) dpdV = n exp − dpdV (7.3.1) e (2πm kT )3/2 2m kT e µ e ¶ nor degenerate

8πp2 f D(p)dpdV = dpdV p

8πp2 1 f P (p) dpdV = dpdV (7.3.15) h3 1+ eE/kT −ψ where ψ is the degeneracy parameter,

n h3 ψ = ln e (7.3.16) 2(2πm kT )3/2 ½ e ¾ and E is the kinetic energy of the electron

2 1/2 2 2 p 2 E = Etot − m c = m c 1+ − m c (7.3.17) e e m2c2 e µ e ¶

Note that this equation does have the correct form. If ne is small, P M then ψ ¿ 0, and f (p) → f (p). On the other hand, if ne is large, eE/kT −ψ ¿ 0, and f P (p) → f D(p). 3/2 Note also that ψ depends only on ne/T . With this general expression for the electron phase-space occupa- tion, the expressions for ne, Pe, and ue are ∞ 8π p2 ne = 3 E/kT −ψ dp (7.3.18) h 0 1+ e Z ∞ 8π p3v(p) Pe = 3 E/kT −ψ dp (7.3.19) 3h 0 1+ e Z∞ 8π p2 u = E dp (7.3.20) e h3 1+ eE/kT −ψ Z0 where in the non-relativisitic case, 1 2 p2 / p2 E = m c2 1+ − 1 ≈ e m2c2 2m (µ e ¶ ) e p v = ≈ p/me (7.3.21) 2 2 2 1/2 (me + p /c ) 2 For the non-relativistic case, if we let η = p /2mekT , then these integrals become ∞ 4π η1/2 n = (2m kT )3/2 dη (7.3.22) e h3 e 1+ eη−ψ Z0 5/2 ∞ 3/2 4π (2mekT ) η Pe = 3 η−ψ dη (7.3.23) 3h me 0 1+ e Z ∞ 2π (2m kT )5/2 η3/2 u = e dη (7.3.24) e h3 m 1+ eη−ψ e Z0 All the above integrals have the form ∞ ην F (ψ)= dη (7.3.25) ν 1+ eη−ψ Z0 These are called Fermi-Dirac integrals, and they are tabulated in McDougall & Stoner 1939, Phil. Trans. R. Soc. London, 237, 67 (so you don’t need to numerically integrate them each time). The General Equation of State

We get the general equation of state for stellar by summing the expressions for radiation pressure, ion pressure, and electron pressure. Thus

∞ aT 4 ρ 8π p3 v(p) P = + kT + dp (7.4.1) 3 µ m 3h3 1+ eE/kT −ψ i a Z0 ∞ 4π E1/2 ρ = (2m )3/2µ m dE (7.4.2) 3 e e a 1+ eE/kT −ψ Z0 Thus, we have implicit formulae for the equation of state. Given ρ and T , you can solve for ψ via (7.3.16), and then P . Similarly, the equation for each particle’s is

∞ aT 4 3 k 8π p2E(p) u = + T + dp (7.4.3) ρ 2 µ m h3ρ 1+ eE/kT −ψ i a Z0 Unfortunately, there is no way to use these equations to generate analytical expressions for the thermodynamic quantities cP , cV , α, δ, and ∇ad. Unless one of the terms dominates, these quantities must be computed numerically.