Notes
  • Notes
  • 恒星结构与演化
    • Chapter 7. Equation of State
    • Chapter 3. Virial Theorem
    • Chapter 11. Main Sequence
    • Chapter 4. Energy Conservation
    • Chapter 12. Post-Main Sequence
    • Chapter 2. Hydrostatic Equilibrium
    • Chapter 6. Convection
    • Chapter 9. Nuclear Reactions
    • Chapter 10 Polytrope
    • Chapter 8. Opacity
    • Chapter 14. Protostar
    • Chapter 13. Star Formation
    • Chapter 5. Energy Transport
  • 天体光谱学
    • Chapter 6 气体星云光谱
    • Chapter 5 磁场中的光谱
    • Chapter 7 X-射线光谱
    • Chapter 3 碱金属原子
    • Chapter 1 光谱基础知识
    • Chapter 9 分子光谱
    • Chapter 4 复杂原子
    • Chapter 2 氢原子光谱
  • 物理宇宙学基础
    • Chapter 2 Newtonian Cosmology
    • Chapter 1 Introduction
    • Chapter 5* Monochromatic Flux, K-correction
    • Chapter 9 Dark Matter
    • Chapter 10 Recombination and CMB
    • Chapter 8 Primordial Nucleosynthesis
    • Chapter 7 Thermal History of the Universe
    • Chapter 6 Supernova cosmology
    • Chapter 5 Redshifts and Distances
    • Chapter 4 World Models
    • Chapter 3 Relativistic Cosmology
  • 数理统计
    • Chapter 6. Confidence Sets (Intervals) 置信区间
    • Chapter 1. Data Reduction 数据压缩
    • Chapter 7. Two Sample Comparisons 两个样本的比较
    • Chapter 3. Decision Theory 统计决策
    • Chapter 4. Asymptotic Theory 渐近理论
    • Chapter 5. Hypothesis Testing 假设检验
    • Chapter 9. Linear Models 线性模型
    • Chapter 10 Model Selection 模型选择
    • Chapter 2. Estimation 估计
    • Chapter 11 Mathematical Foundation in Causal Inference 因果推断中的数理基础
    • Chapter 8. Analysis of Variance 方差分析
  • 天体物理动力学
    • Week8: Orbits
    • Week7: Orbits
    • Week6: Orbits
    • Week5: Orbits
    • Week4: Orbits
    • Week3: Potential Theory
    • Week2
    • Week1
  • 天体物理吸积过程
    • Chapter 4. Spherically Symmetric Flow
    • Chapter 2. Fluid Dynamics
    • Chapter 5. Accretion Disk Theory
    • Chapter 3. Compressible Fluid
  • 天文技术与方法
    • Chapter1-7
  • 理论天体物理
    • Chapter 6 生长曲线的理论和应用
    • Chapter 5 线吸收系数
    • Chapter 4 吸收线内的辐射转移
    • Chapter 3 恒星大气模型和恒星连续光谱
    • Chapter 2 恒星大气的连续不透明度
    • Chapter 1 恒星大气辐射理论基础
  • 常微分方程
    • 线性微分方程组
    • 高阶微分方程
    • 奇解
    • 存在和唯一性定理
    • 初等积分法
    • 基本概念
  • 天体物理观测实验
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On this page
  • Lane-Emden Equation
  • Analytical Solutions
  • Total Mass
  • Gravitational Potential Energy
  • The Isothermal Case
  • Mass-Radius Relation
  • $n = 1, P\propto\rho^2$ - Neutron Star
  • $n = 3/2, P\propto\rho^{5/3}$ - Fully Convective Star
  • $n = 3, P\propto\rho^{4/3}$ - Massive White Dwarfs/Extremely Massive Stars
  • Chandrasekhar Mass
  • Extremely Massive Star
  • Core Solution and Envelope Solution
  1. 恒星结构与演化

Chapter 10 Polytrope

In reality, the solution and structure of stars are very complex. However, the EoS can often be approximated with a simple power-law.

P=Kργ,γ=1+1nP=K\rho^\gamma,\quad\gamma=1+\frac 1nP=Kργ,γ=1+n1​

where $n$ is the polytropic index.

Lane-Emden Equation

Starting from the 1-D hydrostatic equation, we will derive the stellar profile given a polytropic index $n$.

1ρdPdr=−Gmr2⇒ddr(r21ρdPdr)=−Gdmdr=−4πGρr2\frac1\rho\frac{\text dP}{\text dr}=-\frac{Gm}{r^2}\Rightarrow\frac{\text d}{\text dr}\left(r^2\frac1\rho\frac{\text dP}{\text dr}\right)=-G\frac{\text dm}{\text dr}=-4\pi G\rho r^2ρ1​drdP​=−r2Gm​⇒drd​(r2ρ1​drdP​)=−Gdrdm​=−4πGρr2

Here we define a dimensionless parameter $\theta$ so that

ρ≡ρcθn,P≡Pcθn+1\rho\equiv\rho_c\theta^n,\quad P\equiv P_c\theta^{n+1}ρ≡ρc​θn,P≡Pc​θn+1

In this 1-D system, $\theta$ depends solely on the radius $r$. In this way

(n+1)Pc4πGρc21r2ddr(r2dθdr)=−θn\frac{(n+1)P_c}{4\pi G\rho_c^2}\frac1{r^2}\frac{\text d}{\text dr}\left(r^2\frac{\text d\theta}{\text dr}\right)=-\theta^n4πGρc2​(n+1)Pc​​r21​drd​(r2drdθ​)=−θn

If we further define

a≡[(n+1)Pc4πGρc2]1/2,ξ=raa\equiv\left[\frac{(n+1)P_c}{4\pi G\rho_c^2}\right]^{1/2},\quad \xi=\frac raa≡[4πGρc2​(n+1)Pc​​]1/2,ξ=ar​

Then we obtain the Lane-Emden equation

1ξ2ddξ(ξ2dθdξ)=−θn\frac1{\xi^2}\frac{\text d}{\text d\xi}\left(\xi^2\frac{\text d\theta}{\text d\xi}\right)=-\theta^nξ21​dξd​(ξ2dξdθ​)=−θn

This is a second-order odinary differential equation (ODE). We need two boundary conditions to determine $\theta(\xi)$. Here we impose the so-called regularity conditions

θ(0)=1,θ′(0)=0\theta(0)=1,\quad \theta'(0)=0θ(0)=1,θ′(0)=0

The first condition is trivial. The second one denotes that at the center, the pressure/density gradient is zero. Then for each $n$, this initial value problem of Cauchy has a unique solution. But only when $n<5$ will the solution $\theta(\xi)$ has a zero point $\xi_1$.

Analytical Solutions

  • $n=0$

    θ0=1−ξ26\theta_0=1-\frac{\xi^2}6θ0​=1−6ξ2​
  • $n=1$

    θ1=sin⁡ξξ\theta_1=\frac{\sin\xi}\xiθ1​=ξsinξ​
  • $n=5$

    θ5=(1+ξ23)−1/2\theta_5=\left(1+\frac{\xi^2}3\right)^{-1/2}θ5​=(1+3ξ2​)−1/2
  • For any other $n$, there is no analytical solution, including the three most important $n$ for astrophysics

    • $n=3/2\iff P\propto\rho^{5/3}$ - adiabatic ideal gas, or non-relativistic degenerate electron gas

    • $n=3\iff P\propto\rho^{4/3}$ - radiation dominated region, or relativistic degenerate electron gas

    • $n\to\infty\iff P\propto\rho$ - isothermal (here we have to redefine $\theta$, and apply the Emden-Chandrasekhar equation)

Total Mass

M=m(aξ1)=4πρca3∫0ξ1θnξ2dξ=−4πρca3∫0ξ1ddξ(ξ2dθdξ)dξ=−4πρca3ξ12dθdξ∣ξ1\begin{align*} M&=m(a\xi_1)=4\pi\rho_ca^3\int_0^{\xi_1}\theta^n\xi^2\text d\xi\\ &=-4\pi\rho_ca^3\int_0^{\xi_1}\frac{\text d}{\text d\xi}\left(\xi^2\frac{\text d\theta}{\text d\xi}\right)\text d\xi\\ &=-4\pi\rho_ca^3\xi_1^2\frac{\text d\theta}{\text d\xi}\Bigg|_{\xi_1} \end{align*}M​=m(aξ1​)=4πρc​a3∫0ξ1​​θnξ2dξ=−4πρc​a3∫0ξ1​​dξd​(ξ2dξdθ​)dξ=−4πρc​a3ξ12​dξdθ​​ξ1​​​

Since $M(R)=\frac43\pi R^3\cdot\bar\rho$, the average density is related to the central density

ρˉρc=−3ξ1θ′(ξ1)\frac{\bar\rho}{\rho_c}=-\frac{3}{\xi_1}\theta'(\xi_1)ρc​ρˉ​​=−ξ1​3​θ′(ξ1​)
  • Revisit the central temperature of the Sun

    The solar structure is approximated with an $n=3$ polytropic EoS. For $n=3$, $\xi_1=6.8968, \theta'(\xi_1)=-0.0424$, then

    ρcρˉ=54.182⇒ρc=75 g/cm3\frac{\rho_c}{\bar\rho}=54.182\Rightarrow \rho_c=75\text{ g/cm}^3ρˉ​ρc​​=54.182⇒ρc​=75 g/cm3

    The central pressure is given by

    a2=(R⊙ξ1)2=(n+1)Pc4πGρc⇒Pc=1.2×1017 erg/cm3a^2=\left(\frac{R_\odot}{\xi_1}\right)^2=\frac{(n+1)P_c}{4\pi G\rho_c}\Rightarrow P_c=1.2\times10^{17}\text{ erg/cm}^3a2=(ξ1​R⊙​​)2=4πGρc​(n+1)Pc​​⇒Pc​=1.2×1017 erg/cm3

    Finally

    Tc=μmpPcρkB∼1.2×107 KT_c=\frac{\mu m_\text pP_c}{\rho k_B}\sim1.2\times10^7\text{ K}Tc​=ρkB​μmp​Pc​​∼1.2×107 K

Gravitational Potential Energy

The gravitational potential energy of a mass element $\text dm$ is

dEg=−Gm(r)rdm=d[−Gm2(r)2r]−Gm2(r)2r2dr\text dE_\text g=-\frac{Gm(r)}{r}\text dm=\text d\left[-\frac{Gm^2(r)}{2r}\right]-\frac{Gm^2(r)}{2r^2}\text drdEg​=−rGm(r)​dm=d[−2rGm2(r)​]−2r2Gm2(r)​dr

From hydrostatic equation

−Gm(r)r2=1ρdPdr=(n+1)ddr(Pρ)-\frac{Gm(r)}{r^2}=\frac1\rho\frac{\text dP}{\text dr}=(n+1)\frac{\text d}{\text dr}\left(\frac P\rho\right)−r2Gm(r)​=ρ1​drdP​=(n+1)drd​(ρP​)

Then

dEg=d[−Gm2(r)2r]+12(n+1)m(r)ddr(Pρ)dr\text dE_\text g=\text d\left[-\frac{Gm^2(r)}{2r}\right]+\frac12(n+1)m(r)\frac{\text d}{\text dr}\left(\frac P\rho\right)\text drdEg​=d[−2rGm2(r)​]+21​(n+1)m(r)drd​(ρP​)dr

The total gravitational potential energy is

Eg=[−Gm2(r)2r]0R+12(n+1)∫0Rm(r)ddr(Pρ)dr=−GM2R+12(n+1){∫0Rddr[m(r)Pρ]dr−∫0MPρdm}=−GM2R+12(n+1){[m(aξ)Pcρcθ(ξ)]0ξ1−∫0MPρdm}=−GM2R−12(n+1)∫0MPρdm=−GM2R+16(n+1)Eg\begin{align*} E_\text g&=\left[-\frac{Gm^2(r)}{2r}\right]_0^R+\frac12(n+1)\int_0^R m(r)\frac{\text d}{\text dr}\left(\frac P\rho\right)\text dr\\ &=-\frac{GM}{2R}+\frac12(n+1)\left\{\int_0^R\frac{\text d}{\text dr}\left[m(r)\frac P\rho\right]\text dr -\int_0^M\frac P\rho\text dm\right\}\\ &=-\frac{GM}{2R}+\frac12(n+1)\left\{\left[m(a\xi)\frac {P_c}{\rho_c}\theta(\xi)\right]_0^{\xi_1} -\int_0^M\frac P\rho\text dm\right\}\\ &=-\frac{GM}{2R}-\frac12(n+1)\int_0^M\frac P\rho\text dm\\ &=-\frac{GM}{2R}+\frac16(n+1)E_\text g \end{align*}Eg​​=[−2rGm2(r)​]0R​+21​(n+1)∫0R​m(r)drd​(ρP​)dr=−2RGM​+21​(n+1){∫0R​drd​[m(r)ρP​]dr−∫0M​ρP​dm}=−2RGM​+21​(n+1){[m(aξ)ρc​Pc​​θ(ξ)]0ξ1​​−∫0M​ρP​dm}=−2RGM​−21​(n+1)∫0M​ρP​dm=−2RGM​+61​(n+1)Eg​​
⇒Eg=−35−nGM2R,n<5\Rightarrow E_\text{g}=-\frac{3}{5-n}\frac{GM^2}R,\quad n<5⇒Eg​=−5−n3​RGM2​,n<5

Here we have applied the virial theorem,

∫0MPρdm=−13Eg\int_0^M\frac P\rho\text dm=-\frac13E_\text{g}∫0M​ρP​dm=−31​Eg​

The Isothermal Case

When $P=K\rho\propto\rho$, the Lane-Emden equation is no longer valid. Here we define the gravitational potential $\Phi$ so that

1ρdPdr=−dΦdr=Kdln⁡ρdr⇒Kln⁡ρ(r)ρc=−Φ(r)+Φ(0)\frac1\rho\frac{\text dP}{\text dr}=-\frac{\text d\Phi}{\text dr}=K\frac{\text d\ln \rho}{\text dr}\Rightarrow K\ln\frac{\rho(r)}{\rho_c}=-\Phi(r)+\Phi(0)ρ1​drdP​=−drdΦ​=Kdrdlnρ​⇒Klnρc​ρ(r)​=−Φ(r)+Φ(0)

For convenience we set $\Phi(0)=0$, then

ρ(r)=ρcexp⁡[−Φ(r)K]\rho(r)=\rho_c\exp\left[-\frac{\Phi(r)}K\right]ρ(r)=ρc​exp[−KΦ(r)​]

In this way, we can calculate the mass distribution by solving the Poisson equation

∇2Φ=−4πGρ(r)⇒1r2ddr(r2dΦdr)=4πGρcexp⁡[−Φ(r)K]\nabla^2\Phi=-4\pi G\rho(r)\Rightarrow \frac1{r^2}\frac{\text d}{\text dr}\left(r^2\frac{\text d\Phi}{\text dr}\right)=4\pi G\rho_c\exp\left[-\frac{\Phi(r)}K\right]∇2Φ=−4πGρ(r)⇒r21​drd​(r2drdΦ​)=4πGρc​exp[−KΦ(r)​]

Define $\xi\equiv r/a$ and $\Phi\equiv K\theta$, where

a2≡4πGρcKa^2\equiv\frac{4\pi G\rho_c}{K}a2≡K4πGρc​​

We have derived the Emden-Chandrasekhar equation

1ξ2ddξ(ξ2dθdξ)=e−θ\frac1{\xi^2}\frac{\text d}{\text d\xi}\left(\xi^2\frac{\text d\theta}{\text d\xi}\right)=\text e^{-\theta}ξ21​dξd​(ξ2dξdθ​)=e−θ

With the regularity conditions, the unique solution has to be numerically solved. The solution $\theta(\xi)$ does not have a zero-point.

This solution is practical when a star is embedded in an envolope, so that at the stellar surface, there is a finite external pressure $P_\text{ext}$. With the regularity conditions and the external pressure, we can still solve a stellar radius $R$.

Mass-Radius Relation

For finite $n<5$ only

Now that

R=aξ1=[(n+1)Pc4πGρc]1/2ξ1=[(n+1)K4πG]1/2ρc1−n2nR=a\xi_1=\left[\frac{(n+1)P_c}{4\pi G\rho_c}\right]^{1/2}\xi_1= \left[\frac{(n+1)K}{4\pi G}\right]^{1/2}\rho_c^{\frac{1-n}{2n}}R=aξ1​=[4πGρc​(n+1)Pc​​]1/2ξ1​=[4πG(n+1)K​]1/2ρc2n1−n​​
M=−4πρca3ξ12dθdξ∣ξ1=−4πρc[(n+1)Pc4πGρc2]3/2ξ12dθdξ∣ξ1=−4π[(n+1)K4πG]3/2ξ12dθdξ∣ξ1ρc3−n2nM=-4\pi\rho_ca^3\xi_1^2\frac{\text d\theta}{\text d\xi}\Bigg|_{\xi_1}=-4\pi\rho_c\left[\frac{(n+1)P_c}{4\pi G\rho_c^2}\right]^{3/2}\xi_1^2\frac{\text d\theta}{\text d\xi}\Bigg|_{\xi_1}=-4\pi\left[\frac{(n+1)K}{4\pi G}\right]^{3/2}\xi_1^2\frac{\text d\theta}{\text d\xi}\Bigg|_{\xi_1} \rho_c^{\frac{3-n}{2n}}M=−4πρc​a3ξ12​dξdθ​​ξ1​​=−4πρc​[4πGρc2​(n+1)Pc​​]3/2ξ12​dξdθ​​ξ1​​=−4π[4πG(n+1)K​]3/2ξ12​dξdθ​​ξ1​​ρc2n3−n​​

we have

M∝R3−n1−n  ⟺  R∝M1−n3−nM\propto R^{\frac{3-n}{1-n}}\iff R\propto M^{\frac{1-n}{3-n}}M∝R1−n3−n​⟺R∝M3−n1−n​

$n = 1, P\propto\rho^2$ - Neutron Star

Here we find $R$ is independent of $M$. The figure below gives mass-radius relation calulated assuming different EoSs of neutron stars, and under a framework of general relativity. The Newtonian hydrostatic equation is replaced by the Tolman–Oppenheimer–Volkoff (TOV) equation. Even so, many of the mass-radius relations suggest that the radii are roughly fixed when the mass is high enough.

$n = 3/2, P\propto\rho^{5/3}$ - Fully Convective Star

Here $M\propto R^{-3}$. This is common for some relatively cool objects, including protostars and red giants.

Now consider a interacting binary composed of a giant star and a compact star. When the giant fills its Roche lobe, it starts to be stripped by the compact companion, marked as the onset of mass transfer. As its mass decreases, its radius however expands, leading to higher mass transfer rate. Finally the materials from the donor form a common envelope for the binary. In a common envelope binary system the envelope does not generally rotate at the same rate as the embedded binary system, thus the binary stars suffer friction and gradually lose orbital energy. Chances are that the envelope, with energy and angular momentum gained from the embedded binary, will be blown away, while the binary orbit keeps shrinking to finally form a compact binary with significant gravitational wave emission.

$n = 3, P\propto\rho^{4/3}$ - Massive White Dwarfs/Extremely Massive Stars

$M$ is independent of $R$.

Notes:

  • For a zero-temperature relativistic degenerate elelctron gas, $K$ is strictly a constant. Thus without any approximation, the object built up with such gas should have a constant mass.

  • For massive stars where radiation pressure dominates, we also know that $P=K\rho^{4/3}$, but $K$ is only roughly a constant. It depends on the ratio between gas pressure and the radiation pressure.

Chandrasekhar Mass

For stars composed of cold relativistic degenerate electron gas, such as massive white dwarfs, the EoS is given by

P=(3π)1/3hc8(μmp)4/3ρ4/3P=\left(\frac3\pi\right)^{1/3}\frac{hc}{8(\mu m_\text p)^{4/3}}\rho^{4/3}P=(π3​)1/38(μmp​)4/3hc​ρ4/3

This is the $n=3$ polytropic EoS with

K=(3π)1/3hc8(μmp)4/3K=\left(\frac3\pi\right)^{1/3}\frac{hc}{8(\mu m_\text p)^{4/3}}K=(π3​)1/38(μmp​)4/3hc​

The mass of such a star is

M=−4π[(n+1)K4πG]3/2ξ12dθdξ∣ξ1∝(KG)3/2∝(ℏcG)3/21(μmp)2≡1μ2(mPlmp)3mp\begin{align*} M&=-4\pi\left[\frac{(n+1)K}{4\pi G}\right]^{3/2}\xi_1^2\frac{\text d\theta}{\text d\xi}\Bigg|_{\xi_1}\\ &\propto\left(\frac KG\right)^{3/2}\propto\left(\frac{\hbar c}{G}\right)^{3/2}\frac1{(\mu m_\text p)^2}\\ &\equiv\frac1{\mu^2}\left(\frac{m_{Pl}}{m_\text p}\right)^3m_\text p \end{align*}M​=−4π[4πG(n+1)K​]3/2ξ12​dξdθ​​ξ1​​∝(GK​)3/2∝(Gℏc​)3/2(μmp​)21​≡μ21​(mp​mPl​​)3mp​​

where

mPl≡ℏcGm_{Pl}\equiv\sqrt{\frac{\hbar c}{G}}mPl​≡Gℏc​​

is the Planck mass. The precise result is

MCh=3μ2(mPlmp)3mp=1.46 M⊙M_{Ch}=\frac3{\mu^2}\left(\frac{m_{Pl}}{m_\text p}\right)^3m_\text p=1.46\ M_\odotMCh​=μ23​(mp​mPl​​)3mp​=1.46 M⊙​

There can be Super-Chandrasekhar white dwarfs. Strong magnetic field strengthens the pressure gradient within the star so that is can still be bound over the Chandrasekhar limit.

Extremely Massive Star

Usually, $P\text{rad}\gg P\text{gas}$, and $\beta=P\text{gas}/P\ll1$, where $P=P\text{rad}+ P\text{gas}=\frac{\rho k_BT}{\mu m\text p}+\frac13aT^4$. Then we can derive

P=(3a)1/3(kBμmp)4/3(1−ββ4)1/3ρ4/3P=\left(\frac3a\right)^{1/3}\left(\frac{k_B}{\mu m_\text p}\right)^{4/3}\left(\frac{1-\beta}{\beta^4}\right)^{1/3}\rho^{4/3}P=(a3​)1/3(μmp​kB​​)4/3(β41−β​)1/3ρ4/3

$\beta$ is also a function of $P$, but let us assume it to be a constant in the entire stellar structure anyway.

On the other hand, the definition of $a$ requires

K=πGρc2/3R2ξ12≃4.1×1014(MM⊙)2/3 in cgs unitsK=\pi G\rho_c^{2/3}\frac{R^2}{\xi_1^2}\simeq4.1\times10^{14}\left(\frac{M} {M_\odot}\right)^{2/3}\text{ in cgs units}K=πGρc2/3​ξ12​R2​≃4.1×1014(M⊙​M​)2/3 in cgs units

Then we can in turn solve

1−βμ4β4≃3×10−3(MM⊙)2\frac{1-\beta}{\mu^4\beta^4}\simeq3\times10^{-3}\left(\frac{M} {M_\odot}\right)^2μ4β41−β​≃3×10−3(M⊙​M​)2

For $\mu=0.6, M=10^6M_\odot$, $\beta\simeq0.0086$.

Recall that in radiation dominated region, $\gamma$ is slightly above $4/3$. In the first order of $\beta$,

γ=43+β6≃1.33477\gamma=\frac43+\frac\beta6\simeq1.33477γ=34​+6β​≃1.33477

The modification is $\sim0.1\%$, so the deviation from $\gamma=4/3$ is really small.

Another interesting fact is that for massive stars, Newton gravity may not be applicable anymore. GR requires a slightly larger critical $\gamma$ for the onset of instability, say

γcri=43+αGMRc2\gamma_\text{cri}=\frac43+\alpha\frac{GM}{Rc^2}γcri​=34​+αRc2GM​

where $\alpha\sim\mathcal{O}(1)$. Increasing the stellar mass promotes the $\gamma$ index, while the critical $\gamma$ index decreases, until when $M\sim10^6M\odot$, $\gamma=\gamma\text{cri}$. So Theoretically, there cannot be stars more massive than $10^6M_\odot$.

  • When astronomers were debating the mechanism of quasars, some people had assumed that they were extremely massive stars. However to be as bright as they appeared to be, these imaginary stars should weight over $10^8M_\odot$. This assumption is thus ruled out.

Core Solution and Envelope Solution

So far we have only discussed the Cauchy problem with regularity conditions. But we can also set other conditions.

One example is we can fix $\xi_1$ and $\theta'(\xi_1)$. One fatal result is that when we numerically solve this equation by decreasing $\xi$, $\theta$ could be negative before $\xi$ reaches 0. These solutions are known as the collapse type. To get physical solution, we really need fine-tuning.

The outer initial conditions are really useful when we try to model the core and envelope for a giant star. We solve two different Lane-Emden equations, each with an $n$, under different initial conditions. Then with fine-tuning, these two solutions meet at a certain point $(\rho_0,r_0)$. We may add extra restrictions, such as same derivative at $(\rho_0,r_0)$, to obtain more realistic solutions.

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Last updated 4 years ago

This is known as the Chandrasekhar limit. Mathematically it is only the mass of extreme-relativistic white dwarfs. But in we have discussed the stability of a polytropic star. Only when $\gamma>4/3$ is the star gravitationally bound. Thus $M_{Ch}$ is also an upper bound of white dwarf mass, since the white dwarf gets increasingly relativistic with higher mass, and the $\gamma$ index approaches $4/3$.

Chapter 3