Chapter 2 Newtonian Cosmology

vA=H0rA, vB=H0rB\vec v_A=H_0 \vec r_A,\ \vec v_B=H_0 \vec r_B
  • Recession velocity if B as seen by A is

    vBA=vBvA=H0(rBrA)\vec v_{BA}=\vec v_B-\vec v_A=H_0(\vec r_B-\vec r_A)

    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)

    rBA=a(t)xBA\vec r_{BA} = a(t)\cdot\vec x_{BA}

    $a(t)$ 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

    U=T+V=12mr˙2GMmr=12mr˙24π3Gρr2mU=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^2m

Friedmann equation

(a˙a)2=8πG3ρkc2a2, 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}
  • $k$ 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

    v=H0r, r=axv=r˙r^=a˙arHa˙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}

    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

    H2=8πG3ρkc2a2H^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}

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

    ρc=3H028πG\rho_c=\frac{3H_0^2}{8\pi G}
  • In thermodynamics, the first law points that

    dE+PdV=TdS\text{d} E+P\text dV=T\text d S

    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

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

    ρ˙+3a˙a(ρ+Pc2)=0\dot\rho+3\frac{\dot a}{a}\left(\rho+\frac{P}{c^2}\right)=0
    • 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,

    PP(ρ)P\equiv P(\rho)
    • 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

    2a˙aaa¨a˙2a2=8πG3ρ˙+2kc2a3a˙a¨a(a¨a)2=8πG3ρkc2a32\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}
  • Acceleration equation

    a¨a=4πG3(ρ+3Pc2)\frac{\ddot a}{a}=-\frac{4\pi G}{3}\left(\rho+\frac{3P}{c^2}\right)
  • Analytical solution

    • $k=0$

      • Dust, $P=0$

        ρ˙+3a˙aρ=0ddt(ρa3)=0\dot\rho+3\frac{\dot a}{a}\rho=0\Rightarrow\frac{\text{d}}{\text{d}t}(\rho a^3)=0
        ρ=ρ0a3a˙2=8πGρ031aa(t)=(tt0)2/3, ρ(t)=ρ0t02t2H(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}

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

      • Radiation, $P=\rho c^2/3$

        ρ˙+4a˙aρ=0ddt(ρa4)=0a(t)=(tt0)1/2, ρ(t)=ρ0t02t2H(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}
      • Radiation dominated universe

        a(t)t1/2, ρradt2, ρdusta3t3/2a(t)\propto t^{1/2},\ \rho_{rad}\propto t^{-2},\ \rho_{dust}\propto a^{-3}\propto t^{-3/2}

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

      • Matter dominated universe

        a(t)t2/3, ρdustt2, ρdusta4t8/3a(t)\propto t^{2/3},\ \rho_{dust}\propto t^{-2},\ \rho_{dust}\propto a^{-4}\propto t^{-8/3}

        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

    μ=0rmax4πr2n0(L04πr2)dr=n0L0rmaxrmaxμ\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\infty

    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)

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