Outline

4. Equation of State and Related Descriptions

4.5. Adiabatic Exponents and Specific Heats

As a star contracts, expands, or undergoes convective motions, it must concern itself with the release or entrainment of internal energy resulting from volume changes. The calculation of the necessary quantities is relatively straightforward for mixtures of ideal gasses, radiation, and fully degenerate components, as long as no reactions (e.g. ionizations) are taking place.

A. Fixed composition

Recall that the specific heat a constant x is defined as c_x ≡ (dQ /dT ) _x .
Useful x's are P, ρ, and specific volume V_ρ = 1/ρ .
The first law of thermodynamics for a reversible process is

dQ = dE + Pd(1/ρ) = dE + Pdρ /ρ²,

so c_Vρ = (dE /dT )_ρ.
But (ideal monatomic gas) E = 3N_AkT / 2μ = c_VρT.

Find c_P from c_P − c_Vρ = − T (∂P/∂T )_ρ² (∂P/∂V_ρ )_T⁻¹.

Choose an equation of state of the form P = P_o (ρ/ρ_o) ^β (T/T_o) ^α,
then c_P − c_Vρ = (P/ρT ) α²/β , = N_Ak/μ for ideal monatomic.

Later we will want the ratio of specific heats γ ≡ c_P/c_Vρ = 1 + (P/ρTc_Vρ ) α²/β.

Adiabatic exponents

Γ1 ≡	(∂ lnP /∂ lnρ)S	Intimately tied to dynamics through the sound speed
Γ2 /( Γ2 − 1) ≡	(∂ lnP /∂ lnT )S ≡ 1/∇adxx	Related to convective instability
Γ3 − 1 ≡	(∂ lnT /∂ lnρ )S	The relation between heat content and compression

In terms of the power-law equation of state:
Γ₃ − 1 = (P/ρT ) α / c_Vρ
Γ₁ = α ( Γ₃ − 1) + β
Γ₂ /(Γ₂ − 1) = α + β /( Γ₃ − 1)
γ = Γ₁ /β
Note: the 'γ' in the so-called γ-law equation of state is Γ₃, not the ratio of specific heats γ.

Ideal gas plus radiation
Assume P = (ρ/μ) N_AkT + aT ⁴/3 dyne/cm⁻².
Now β ≡ (∂ lnP /∂ lnρ)_T, which turns out to be P_g /P.
We also have α = 4 − 3β.
We may substitute these values into the formulae above to get

c_Vρ = (3N_AkT / 2μ) (8 − 7β) /β erg/g K,
Γ₃ − 1 = (2/3) (4 − 3β) / (8 − 7β)
Γ₁ = β + (4 − 3β) (Γ₃ − 1)
Γ₂ /(Γ₂ − 1) = (32 − 24β − 3β ²) / (8 − 6β)
γ = Γ₁ /β

B. Pure hydrogen ionization

H&K carry out the algebra for the specific heats and exponents in a gas consisting of pure hydrogen with a single-state neutral atom. If we define n as the total number density of hydrogen nuclei and y as the fraction of ionized nuclei n₊/n, the pressure and internal energy become

P = (1 + y) nkT
E = (n/ρ) [(1 + y) 3kT/2 + y χ],
χ = ionization energy.

After derivatives and algebra,

c_Vρ = (3/2) (n/ρ) k (1 + y) [1 + D (2/3) (3/2 + χ/kT )² ] erg / gK,
D ≡ y (1 − y) / (2 − y) (1 + y)
β = 1 − D,
α = 1 + D (3/2 + χ/kT ).

These expressions may be substituted into the formulae above for the Γ's.
Note: once we have expressed the equation of state as a power-law, the required first-order derivatives depend on the local instantaneous values of α and β, and not on their own variations.

Γ₃ − 1 = [2 + 2D (3/2 + χ/kT )] / [3 + 2D (3/2 + χ/kT )²]

We will not reproduce the other Γ's since these formulae are instructive only and do not represent any real situation with accuracy.
However, note that D(0) = D(1) = 0, and reaches a maximum of 1/9 when y = 1/2.
Thus Γ₃ (and all the other Γ's) remains near the ideal gas value 5/3 except when the ionization is partial. Typically, hydrogen becomes partially ionized when χ/kT is of order 10 (T = 10⁴ K). Under these circumstances, Γ₃ may drop to 1.15 or so. The other Γ's show similar decreases.