Here we discuss two effects producing pressure.
Ideal Gas
Above all, for ideal gas, the EoS gives
P gas = ρ k B T μ m p P_\text{gas}=\frac{\rho k_BT}{\mu m_\text{p}} P gas = μ m p ρ k B T Radiation Pressure
P rad = 1 3 a T 4 P_\text{rad}=\frac13aT^4 P rad = 3 1 a T 4 Let us consider gas-radiation mixed fluid
P = P gas + P rad = ρ k B T μ m p + 1 3 a T 4 P=P_\text{gas}+P_\text{rad}=\frac{\rho k_BT}{\mu m_\text{p}}+\frac13aT^4 P = P gas + P rad = μ m p ρ k B T + 3 1 a T 4 Define
β ≡ P gas P \beta\equiv\frac{P_\text{gas}}{P} β ≡ P P gas Let's check the $\beta$ value for different stars
Sun: $\rho_c\sim150$ g/cm$^3$, $T_c\sim10^7$ K, $\Rightarrow\beta\sim1$
Massive star ($20 M_\odot$): $\rho_c\sim5$ g/cm$^3$, $T_c\sim4\times10^7$ K, $\Rightarrow\beta\sim0.84$
Since the density
ρ = μ m p k B T ( P − a 3 T 4 ) = ρ ( P , T ) \rho=\frac{\mu m_\text{p}}{k_BT}\left(P-\frac a3T^4\right)=\rho(P,T) ρ = k B T μ m p ( P − 3 a T 4 ) = ρ ( P , T ) We can define two derivatives
α ≡ ( ∂ ln ρ ∂ ln P ) T = P P gas = 1 β , δ ≡ − ( ∂ ln ρ ∂ ln T ) P = P + 3 P rad P gas = 4 − 3 β β \alpha\equiv\left(\frac{\partial \ln \rho}{\partial \ln P}\right)_T=\frac{P}{P_\text{gas}}=\frac1\beta,\quad \delta\equiv-\left(\frac{\partial \ln \rho}{\partial \ln T}\right)_P=\frac{P+3P_\text{rad}}{P_\text{gas}}=\frac{4-3\beta}{\beta} α ≡ ( ∂ ln P ∂ ln ρ ) T = P gas P = β 1 , δ ≡ − ( ∂ ln T ∂ ln ρ ) P = P gas P + 3 P rad = β 4 − 3 β The specific internal energy is
e ( ρ , T ) = e gas + e rad = 3 2 k B μ m p T + 1 ρ a T 4 e(\rho,T)=e_\text{gas}+e_\text{rad}=\frac{3}{2}\frac{k_B}{\mu m_\text{p}}T+\frac1\rho aT^4 e ( ρ , T ) = e gas + e rad = 2 3 μ m p k B T + ρ 1 a T 4 Thus the specific heat capacity (when the pressure is a constant) is
c P ≡ ( d e d T ) P = ( ∂ e ∂ T ) P + P [ ∂ ( 1 / ρ ) ∂ T ] P = ( ∂ e ∂ T ) ρ + ( ∂ e ∂ ρ ) T ( ∂ ρ ∂ T ) P − P ρ 2 ( ∂ ρ ∂ T ) P = k B μ m p [ 3 2 + ( 4 + δ ) 3 P rad P gas ] + k B μ m p δ P P gas = k B μ m p [ 3 2 + ( 4 + 4 − 3 β β ) 3 ( 1 − β ) β + 4 − 3 β β 2 ] = k B μ m p [ 3 2 + 3 ( 4 + β ) ( 1 − β ) β 2 + 4 − 3 β β 2 ] \begin{align*}
c_P&\equiv\left(\frac{\text d e}{\text d T}\right)_P=\left(\frac{\partial e}{\partial T}\right)_P+P\left[\frac{\partial (1/\rho)}{\partial T}\right]_P\\
&=\left(\frac{\partial e}{\partial T}\right)_\rho+\left(\frac{\partial e}{\partial \rho}\right)_T\left(\frac{\partial \rho}{\partial T}\right)_P-\frac{P}{\rho^2}\left(\frac{\partial \rho}{\partial T}\right)_P\\
&=\frac{k_B}{\mu m_\text{p}}\left[\frac{3}{2}+(4+\delta)\frac{3P_\text{rad}}{P_\text{gas}}\right]+\frac{k_B}{\mu m_\text{p}}\frac{\delta P}{P_\text{gas}}\\
&=\frac{k_B}{\mu m_\text{p}}\left[\frac{3}{2}+\left(4+\frac{4-3\beta}{\beta}\right)\frac{3(1-\beta)}{\beta}+\frac{4-3\beta}{\beta^2}\right]\\
&=\frac{k_B}{\mu m_\text{p}}\left[\frac{3}{2}+\frac{3(4+\beta)(1-\beta)}{\beta^2}+\frac{4-3\beta}{\beta^2}\right]
\end{align*} c P ≡ ( d T d e ) P = ( ∂ T ∂ e ) P + P [ ∂ T ∂ ( 1/ ρ ) ] P = ( ∂ T ∂ e ) ρ + ( ∂ ρ ∂ e ) T ( ∂ T ∂ ρ ) P − ρ 2 P ( ∂ T ∂ ρ ) P = μ m p k B [ 2 3 + ( 4 + δ ) P gas 3 P rad ] + μ m p k B P gas δ P = μ m p k B [ 2 3 + ( 4 + β 4 − 3 β ) β 3 ( 1 − β ) + β 2 4 − 3 β ] = μ m p k B [ 2 3 + β 2 3 ( 4 + β ) ( 1 − β ) + β 2 4 − 3 β ] When gas pressure dominates, $\beta=1$, thus
c P = 5 2 k B μ m p c_P=\frac52\frac{k_B}{\mu m_\text{p}} c P = 2 5 μ m p k B When radiation dominates, $\beta=0$, and $c_P$ diverges.
The adiabatic temperature gradient
∇ ad = δ ρ c P P T = k B μ m p c P δ β = [ 3 2 + 3 ( 4 + β ) ( 1 − β ) β 2 + 4 − 3 β β 2 ] − 1 4 − 3 β β 2 = 8 − 6 β 32 − 24 β − 3 β 2 \begin{align*}
\nabla_\text{ad}&=\frac{\delta}{\rho c_P}\frac PT=\frac{k_B}{\mu m_\text{p}c_P}\frac{\delta}{\beta}\\
&=\left[\frac{3}{2}+\frac{3(4+\beta)(1-\beta)}{\beta^2}+\frac{4-3\beta}{\beta^2}\right]^{-1}\frac{4-3\beta}{\beta^2}\\
&=\frac{8-6\beta}{32-24\beta-3\beta^2}
\end{align*} ∇ ad = ρ c P δ T P = μ m p c P k B β δ = [ 2 3 + β 2 3 ( 4 + β ) ( 1 − β ) + β 2 4 − 3 β ] − 1 β 2 4 − 3 β = 32 − 24 β − 3 β 2 8 − 6 β The heat index is
γ ≡ ( ∂ ln P ∂ ln ρ ) s = [ ( ∂ ln ρ ∂ ln P ) T + ( ∂ ln ρ ∂ ln T ) P ( ∂ ln T ∂ ln P ) s ] − 1 = 1 α − δ ∇ ad = 32 − 24 β − 3 β 3 24 − 21 β = { 5 / 3 , β = 1 4 / 3 + β / 6 + O ( β 2 ) , β → 0 \begin{align*}
\gamma&\equiv\left(\frac{\partial \ln P}{\partial\ln \rho}\right)_s=\left[\left(\frac{\partial \ln \rho}{\partial\ln P}\right)_T+\left(\frac{\partial \ln \rho}{\partial\ln T}\right)_P\left(\frac{\partial \ln T}{\partial\ln P}\right)_s\right]^{-1}\\
&=\frac1{\alpha-\delta\nabla_\text{ad}}\\
&=\frac{32-24\beta-3\beta^3}{24-21\beta}\\
&=\left\{
\begin{array}{l}
5/3,\quad \beta=1\\
4/3+\beta/6+\mathcal{O}(\beta^2),\quad \beta\to0\\
\end{array}
\right.
\end{align*} γ ≡ ( ∂ ln ρ ∂ ln P ) s = [ ( ∂ ln P ∂ ln ρ ) T + ( ∂ ln T ∂ ln ρ ) P ( ∂ ln P ∂ ln T ) s ] − 1 = α − δ ∇ ad 1 = 24 − 21 β 32 − 24 β − 3 β 3 = { 5/3 , β = 1 4/3 + β /6 + O ( β 2 ) , β → 0 This fact is extremely important for the stability of a star.
We revisit the virial theorem,
E tot = 3 γ − 4 3 ( γ − 1 ) E g E_\text{tot}=\frac{3\gamma-4}{3(\gamma-1)}E_\text g E tot = 3 ( γ − 1 ) 3 γ − 4 E g When $\beta\to0$, $E\text{tot}\to\beta E \text g/2$. Since $E_\text g0$, the star can exist (thanks to the fact that $\beta>4/3$).
We try to compress a star of the mass $M$ and discuss the stability afterwards.
The gravitational force and the pressure gradient force are
F g ∼ − G M R 2 , F P ∼ 1 ρ P R ∝ ρ γ − 1 R F_\text g\sim-\frac{GM}{R^2},\quad F_{P}\sim\frac1\rho\frac{P}{R}\propto\frac{\rho^{\gamma-1}}{R} F g ∼ − R 2 GM , F P ∼ ρ 1 R P ∝ R ρ γ − 1 Since $M\propto\rho R^3$ is fixed, $\rho\propto R^{-3}$, thus
F g ∝ − R − 2 , F P ∝ ρ γ − 1 R ∝ R − 3 γ + 2 ∝ R − 3 ( γ − 4 / 3 ) − 2 F_\text g\propto-R^{-2},\quad F_{P}\propto\frac{\rho^{\gamma-1}}{R}\propto R^{-3\gamma+2}\propto R^{-3(\gamma-4/3)-2} F g ∝ − R − 2 , F P ∝ R ρ γ − 1 ∝ R − 3 γ + 2 ∝ R − 3 ( γ − 4/3 ) − 2 Obviously, when $\gamma>4/3$, when $R$ is compressed, $F_\text{g}+F_P>0$, so the net force resists the compression. If $\gamma<4/3$, the star is dynamically unstable and the star would finally collapse.
Degenerate Electron Gas
For cold and dense gas, quantum effects cause degenerate pressure .
For zero-temperature Fermions such as electrons, in the phase space, they all lie beneath a Fermi momentum . The total number of electrons $N$ is given by
N = 2 ∫ d x d y d z d p x d p y d p z h 3 = 2 V h 3 ∫ 0 p F 4 π p 2 d p N=2\int\frac{\text dx\text dy\text dz\text dp_x\text dp_y\text dp_z}{h^3}=\frac{2V}{h^3}\int_0^{p_\text F}4\pi p^2\text dp N = 2 ∫ h 3 d x d y d z d p x d p y d p z = h 3 2 V ∫ 0 p F 4 π p 2 d p where the factor of 2 comes from the electron spin. Thus the number density $n_e$ is given by
n e = N V = 8 π p F 3 3 h 3 n_e=\frac NV=\frac{8\pi p_\text F^3}{3h^3} n e = V N = 3 h 3 8 π p F 3 Note that pressure is simply the surface integral of momentum flux crossing each surface element $\Omega_s$
P e = 1 4 π ∫ 2 π ∫ 0 ∞ f ( p ) u ( p ) p cos 2 θ d p d Ω s P_e=\frac1{4\pi}\int_{2\pi}\int_0^\infty f(p)u(p)p\cos^2\theta\text dp\text d\Omega_s P e = 4 π 1 ∫ 2 π ∫ 0 ∞ f ( p ) u ( p ) p cos 2 θ d p d Ω s where $f(p)$ is the distribution function
f ( p ) = { 8 π p 2 h 3 , p > p F 0 , p > p F f(p)=\left\{
\begin{array}{c}
\frac{8\pi p^2}{h^3},\quad p>p_\text F\\
0,\quad p>p_\text F
\end{array}
\right. f ( p ) = { h 3 8 π p 2 , p > p F 0 , p > p F thus
P e = 8 π 3 h 3 ∫ 0 p F p 3 u ( p ) d p P_e=\frac{8\pi}{3h^3}\int_0^{p_\text F}p^3u(p)\text dp P e = 3 h 3 8 π ∫ 0 p F p 3 u ( p ) d p Since special relativity gives
p = m e u 1 − u 2 / c 2 ⟺ u = p m e 2 + p 2 / c 2 p=\frac{m_eu}{\sqrt{1-{u^2}/{c^2}}}\iff u=\frac{p}{\sqrt{m_e^2+p^2/c^2}} p = 1 − u 2 / c 2 m e u ⟺ u = m e 2 + p 2 / c 2 p Let $\xi=p/mec, x=p \text F/m_ec$, we have
P e = 8 π c 5 m e 4 3 h 3 ∫ 0 x ξ 4 ( 1 + ξ 2 ) 1 / 2 d ξ P_e=\frac{8\pi c^5m_e^4}{3h^3}\int_0^x\frac{\xi^4}{\left(1+\xi^2\right)^{1/2}}\text d\xi P e = 3 h 3 8 π c 5 m e 4 ∫ 0 x ( 1 + ξ 2 ) 1/2 ξ 4 d ξ Define
f ( x ) = ∫ 0 x ξ 4 ( 1 + ξ 2 ) 1 / 2 d ξ = 1 8 { x ( 2 x 2 − 3 ) ( 1 + x 2 ) 1 / 2 + 3 ln [ x + ( 1 + x 2 ) 1 / 2 ] } f(x)=\int_0^x\frac{\xi^4}{\left(1+\xi^2\right)^{1/2}}\text d\xi=\frac18\left\{x\left(2x^2-3\right)\left(1+x^2\right)^{1/2}+3\ln\left[x+\left(1+x^2\right)^{1/2}\right]\right\} f ( x ) = ∫ 0 x ( 1 + ξ 2 ) 1/2 ξ 4 d ξ = 8 1 { x ( 2 x 2 − 3 ) ( 1 + x 2 ) 1/2 + 3 ln [ x + ( 1 + x 2 ) 1/2 ] } For non-relativistic case, $x\to 0$
f ( x ) ∼ ∫ 0 x ξ 4 d ξ = 1 5 x 5 f(x)\sim\int_0^x\xi^4\text d\xi=\frac15x^5 f ( x ) ∼ ∫ 0 x ξ 4 d ξ = 5 1 x 5 ⇒ P e = 1 20 ( 3 π ) 2 / 3 h 2 m e n e 5 / 3 ∼ 1 0 13 ( ρ μ e ) 5 / 3 g/cm/s − 2 \Rightarrow P_e=\frac1{20}\left(\frac3\pi\right)^{2/3}\frac{h^2}{m_e}n_e^{5/3}\sim10^{13}\left(\frac{\rho}{\mu_e}\right)^{5/3}\text{ g/cm/s}^{-2} ⇒ P e = 20 1 ( π 3 ) 2/3 m e h 2 n e 5/3 ∼ 1 0 13 ( μ e ρ ) 5/3 g/cm/s − 2 For extreme relativistic case, $x\to \infty$
f ( x ) ∼ ∫ 0 x ξ 3 d ξ = 1 4 x 4 f(x)\sim\int_0^x\xi^3\text d\xi=\frac14x^4 f ( x ) ∼ ∫ 0 x ξ 3 d ξ = 4 1 x 4 ⇒ P e = ( 3 π ) 1 / 3 h c 8 n e 4 / 3 ∼ 1 0 15 ( ρ μ e ) 4 / 3 g/cm/s − 2 \Rightarrow P_e=\left(\frac3\pi\right)^{1/3}\frac{hc}{8}n_e^{4/3}\sim10^{15}\left(\frac{\rho}{\mu_e}\right)^{4/3}\text{ g/cm/s}^{-2} ⇒ P e = ( π 3 ) 1/3 8 h c n e 4/3 ∼ 1 0 15 ( μ e ρ ) 4/3 g/cm/s − 2 EoS in a Star
In our discussion above, we see that
P gas ∝ ρ T P_\text{gas}\propto \rho T P gas ∝ ρT P rad ∝ T 4 P_\text{rad}\propto T^4 P rad ∝ T 4 P e ∝ ρ 4 / 3 − 5 / 3 P_e\propto\rho^{4/3-5/3} P e ∝ ρ 4/3 − 5/3 Therefore, radiation pressure dominates when $T$ is high and $\rho$ is low, while the degenerate pressure dominates in the exact opposite. So the $T-\rho$ plane can be divided into three
Notes:
Our Sun generally lies in the region of ideal gas.
Increase the mass - the stellar center may become radiation dominated (e.g., maassive stars);
Decrease the mass - the stellar center may fall into the electron degenerate region (e.g., brown dwarfs).
For white dwarfs in non-relativistic region, $E\text{tot}<0$, even if nuclear burning is lighted somewhere, the star itself is still stable - helium flash . For white dwarfs in relativistic regions, however, since $E \text{tot}$ is only slightly below 0, any nuclear fusion may lead to destructive explosions, such as Type Ia SN .
