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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  • Comoving coordinates
  • Friedmann equation
  • Olber's Paradox
  1. 物理宇宙学基础

Chapter 2 Newtonian Cosmology

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Last updated 4 years ago

v⃗A=H0r⃗A, v⃗B=H0r⃗B\vec v_A=H_0 \vec r_A,\ \vec v_B=H_0 \vec r_BvA​=H0​rA​, vB​=H0​rB​
  • Recession velocity if B as seen by A is

    Hubble expansion is a natural property of an expanding universe that obeys the cosmological principle

Comoving coordinates

  • coordinates that are carried along with the expansion

  • comoving with space distance $\vec r$, comoving distance $\vec x$ (constant)

    $a(t)$ is the scale factor - all we want to find out

  • Birkhoof's theorem: the net gravitational effect of a uniform external medium on a spherical cavity is zero, the net gravitational effect comes from matter internal to radius $r$ (as a mass point)

    total energy of a particle of mass $m$ at A, B, C, D

Friedmann equation

  • $k$ must be independent of $x$, therefore $U\propto x^2$

  • $k$ is a independent of $t$, for $U​$ is conserved

  • Thus $k$ is just a constant

    • $k>0\Rightarrow UT$ , collapse

    • $k<0\Rightarrow U<0, V<T$ , expand forever

    • $k=0\Rightarrow U=0$ , expansion terminates when $t\to\infty$

  • Hubble parameter

    which is a function of time, so called Hubble constant $H_0$ is the Hubble parameter at present

  • Using Hubble parameter we can rewrite Friedmann equation

    if $k=0$ , the critical density $\rho$ is then

  • In thermodynamics, the first law points that

    where $P$ is the pressure, $S$ is the entropy

  • Expanding volume $V$ of unit comoving radius $x=1 (r=a)$, $E=mc^2$, energy within the volume

  • Assuming a reservable expansion $\text dS=0​$ , we obtain the fluid equation

    • The first term - the dilution in the density because the volume has increased

    • The second term - the loss in energy because the pressure of the material has done work as the universe’s volume increased.

  • Equation of state

    There is a unique pressure associated with each density,

    • For non-relativistic matter, $P=0$ , dust

    • For highly-relativistic matter , $P=\rho c^2/3$ , radiation pressure of light

  • Time-derivation on both sides of the Friedmann equation

  • Acceleration equation

  • Analytical solution

    • $k=0$

      • Dust, $P=0$

        However, the current age of universe $t_0​$ obtained in this model is wrong

      • Radiation, $P=\rho c^2/3$

      • Radiation dominated universe

        Unstable, the radiation density decreases faster than the dust density, the universe will turn to matter dominated on day

      • Matter dominated universe

        Stable

Olber's Paradox

  • In a shell from $r​$ to $r+\mathrm d r​$, the total number of stars is $4\pi r^2n_0\mathrm d r​$, where $n_0​$ is the number density

  • Total intensity of the sky

    which is $10^{13}$ times brighter than it actually is

  • Thermodynamic problem

    • Infinite volume

    • Infinite time to reach thermodynamic equilibrium

    • BUT the universe is not homogeneous (not the temperature of CMB)

v⃗BA=v⃗B−v⃗A=H0(r⃗B−r⃗A)\vec v_{BA}=\vec v_B-\vec v_A=H_0(\vec r_B-\vec r_A)vBA​=vB​−vA​=H0​(rB​−rA​)
r⃗BA=a(t)⋅x⃗BA\vec r_{BA} = a(t)\cdot\vec x_{BA}rBA​=a(t)⋅xBA​
U=T+V=12mr˙2−GMmr=12mr˙2−4π3Gρr2mU=T+V=\frac{1}{2}m\dot{r}^2-\frac{GMm}{r}=\frac{1}{2}m\dot{r}^2-\frac{4\pi}{3}G\rho r^2mU=T+V=21​mr˙2−rGMm​=21​mr˙2−34π​Gρr2m
(a˙a)2=8πG3ρ−kc2a2, where kc2=−2Umx2\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2},\ \text{where }kc^2=-\frac{2U}{mx^2}(aa˙​)2=38πG​ρ−a2kc2​, where kc2=−mx22U​
v⃗=H0r⃗, r⃗=ax⃗⇒v⃗=∣r⃗˙∣r⃗^=a˙ar⃗⇒H≡a˙a\vec v=H_0\vec r,\ \vec r=a\vec x\\ \Rightarrow \vec v=|\dot{\vec r}|\hat{\vec r}=\frac{\dot a}{a}\vec r\\ \Rightarrow H\equiv \frac{\dot a}{a}v=H0​r, r=ax⇒v=∣r˙∣r^=aa˙​r⇒H≡aa˙​
H2=8πG3ρ−kc2a2H^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}H2=38πG​ρ−a2kc2​
ρc=3H028πG\rho_c=\frac{3H_0^2}{8\pi G}ρc​=8πG3H02​​
dE+PdV=TdS\text{d} E+P\text dV=T\text d SdE+PdV=TdS
E=4π3a3ρc2dEdt=4πa2ρc2dadt+4π3a3c2dρdtdVdt=4πa2dadtE=\frac{4\pi}{3}a^3\rho c^2\\ \frac{\text dE}{\text{d} t}=4\pi a^2\rho c^2\frac{\text da}{\text{d} t}+\frac{4\pi}{3} a^3 c^2\frac{\text d\rho}{\text{d} t}\\ \frac{\text dV}{\text{d} t}=4\pi a^2\frac{\text da}{\text{d} t}E=34π​a3ρc2dtdE​=4πa2ρc2dtda​+34π​a3c2dtdρ​dtdV​=4πa2dtda​
ρ˙+3a˙a(ρ+Pc2)=0\dot\rho+3\frac{\dot a}{a}\left(\rho+\frac{P}{c^2}\right)=0ρ˙​+3aa˙​(ρ+c2P​)=0
P≡P(ρ)P\equiv P(\rho)P≡P(ρ)
2a˙aaa¨−a˙2a2=8πG3ρ˙+2kc2a3a˙a¨a−(a¨a)2=8πG3ρ−kc2a32\frac{\dot a}{a}\frac{a\ddot a-\dot a^2}{a^2}=\frac{8\pi G}{3}\dot\rho+2\frac{kc^2}{a^3}\dot a\\ \frac{\ddot a}{a}-\left(\frac{\ddot a}{a}\right)^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^3}2aa˙​a2aa¨−a˙2​=38πG​ρ˙​+2a3kc2​a˙aa¨​−(aa¨​)2=38πG​ρ−a3kc2​
a¨a=−4πG3(ρ+3Pc2)\frac{\ddot a}{a}=-\frac{4\pi G}{3}\left(\rho+\frac{3P}{c^2}\right)aa¨​=−34πG​(ρ+c23P​)
ρ˙+3a˙aρ=0⇒ddt(ρa3)=0\dot\rho+3\frac{\dot a}{a}\rho=0\Rightarrow\frac{\text{d}}{\text{d}t}(\rho a^3)=0ρ˙​+3aa˙​ρ=0⇒dtd​(ρa3)=0
ρ=ρ0a3⇒a˙2=8πGρ031a⇒a(t)=(tt0)2/3, ρ(t)=ρ0t02t2⇒H(t)=a˙a=23t, t0=23H0\rho=\frac{\rho_0}{a^3}\Rightarrow\dot a^2=\frac{8\pi G\rho_0}{3}\frac{1}{a}\\\Rightarrow a(t)=\left(\frac{t}{t_0}\right)^{2/3},\ \rho(t)=\frac{\rho_0t_0^2}{t^2}\\ \Rightarrow H(t)=\frac{\dot a}{a}=\frac{2}{3t},\ t_0=\frac{2}{3H_0}ρ=a3ρ0​​⇒a˙2=38πGρ0​​a1​⇒a(t)=(t0​t​)2/3, ρ(t)=t2ρ0​t02​​⇒H(t)=aa˙​=3t2​, t0​=3H0​2​
ρ˙+4a˙aρ=0⇒ddt(ρa4)=0⇒a(t)=(tt0)1/2, ρ(t)=ρ0t02t2⇒H(t)=a˙a=12t<23t, t0=12H0\dot\rho+4\frac{\dot a}{a}\rho=0\Rightarrow\frac{\text{d}}{\text{d}t}(\rho a^4)=0\\ \Rightarrow a(t)=\left(\frac{t}{t_0}\right)^{1/2},\ \rho(t)=\frac{\rho_0t_0^2}{t^2}\\ \Rightarrow H(t)=\frac{\dot a}{a}=\frac{1}{2t}<\frac{2}{3t},\ t_0=\frac{1}{2H_0}ρ˙​+4aa˙​ρ=0⇒dtd​(ρa4)=0⇒a(t)=(t0​t​)1/2, ρ(t)=t2ρ0​t02​​⇒H(t)=aa˙​=2t1​<3t2​, t0​=2H0​1​
a(t)∝t1/2, ρrad∝t−2, ρdust∝a−3∝t−3/2a(t)\propto t^{1/2},\ \rho_{rad}\propto t^{-2},\ \rho_{dust}\propto a^{-3}\propto t^{-3/2}a(t)∝t1/2, ρrad​∝t−2, ρdust​∝a−3∝t−3/2
a(t)∝t2/3, ρdust∝t−2, ρdust∝a−4∝t−8/3a(t)\propto t^{2/3},\ \rho_{dust}\propto t^{-2},\ \rho_{dust}\propto a^{-4}\propto t^{-8/3}a(t)∝t2/3, ρdust​∝t−2, ρdust​∝a−4∝t−8/3
μ=∫0rmax4πr2n0(L04πr2)dr=n0L0rmaxrmax→∞⇒μ→∞\mu=\int_0^{r_{max}}4\pi r^2n_0\left(\frac{L_0}{4\pi r^2}\right)\mathrm d r=n_0L_0r_{max}\\ r_{max}\to\infty\Rightarrow \mu\to\inftyμ=∫0rmax​​4πr2n0​(4πr2L0​​)dr=n0​L0​rmax​rmax​→∞⇒μ→∞