PX282 - D3 - equations of state

\begin{matrix} \frac{d M_{r}}{d r} = 4 π r^{2} ρ \\ \frac{d P}{d r} = - ρ \frac{G M_{r}}{r^{2}} \\ \frac{d L}{d r} = 4 π r^{2} ϵ_{N} ρ \\ \frac{d T}{d r} = - \frac{3 κ ρ}{16 a c π r^{2} T^{3}} L \end{matrix}

the standard equation of state

\begin{matrix} (1) & P = n k_{B} T \end{matrix}

\begin{matrix} (1.1) & P = \frac{ρ}{μ m_{H}} k_{B} T \end{matrix}

where, $μ =$ the average mass per particle, and $m_{H} =$ mass of hydrogen

in the sun, the mass fractions of hydrogen, $X_{⊙} ≃ 0.7$ , helium, $Y_{⊙} ≃ 0.28$ , and everything else ('metals'), $Z_{⊙} ≃ 0.02$
- for a fully ionized hydrogen, $μ = \frac{1}{2}$ (2 particles per $m_{H} : 1 p, 1 e$ )
- for $H e : μ = \frac{4 m_{H}}{3 p a r t i c l e s} = \frac{4}{3}$ (1 nucleus and 2 electrons)
- for $Z :_{6}^{12} C,_{8}^{16} O \to μ ≃ 2$
- overall: $μ_{⊙} ≃ 0.6$

\begin{matrix} (2) & P = \frac{1}{3} a T^{4} \end{matrix}

where, $a = \frac{4 σ}{c}$

for an ideal gas, $T \to 0$ as $P \to 0$
consequently, the average particle speed, and hence the average momentum also tend to zero
from the uncertainty principle:

Δ x Δ p ≃ ℏ

so, as $Δ p \to 0, Δ x \to \infty$ , which is impossible
hence, there is a minimum $Δ p$
this is relevant for neutron stars and white dwarfs
a gas is degenerate when the average thermal energy is less than the fermi energy:

\frac{3}{2} k_{B} T < ϵ_{p} \propto ρ^{2 / 3}

T ρ^{\frac{2}{3}} < 1261 K m^{2} k g^{- 2 / 3}

P \propto n_{e}^{5 / 3}

P \propto n_{e}^{4 / 3}

for the sun: $T ρ^{2 / 3} = 5300 > 1251$
$P_{d e g e n ⊙} = 2.4 \times 10^{15} N m^{- 2}$
$P_{i d e a l ⊙} = n k_{B} T = 1.4 \times 10^{16} N m^{- 2}$
$\frac{P_{i d e a l}}{P_{d e g e n}} = 5.8$
$P_{r a d} = 1.5 \times 10^{13} N m^{- 2}$
for a white dwarf:
$R ≃ R_{\oplus}$ ,
$M ≃ M_{⊙}$ ,
$T ≃ 10^{7} K$ ,
$ρ ≃ 1.8 \times 10^{9} k g m^{- 3}$ ,
$μ ≃ 2 mass units / e^{-}$
$n_{e} ≃ \frac{ρ}{μ m_{H}} = 5.4 \times 10^{3} 5 m^{- 3}$
$∴ P_{d e g e n, n o n - r e l} ≃ 8.4 \times 10^{21} N m^{- 2}$ and $P_{d e g e n, r e l} ≃ 1.1 \times 10^{22} N m^{- 2}$