Chapter 2 Newtonian Cosmology

Recession velocity if B as seen by A is
vBA=vB−vA=H0(rB−rA)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$a(t)$ 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=21mr˙2−rGMm=21mr˙2−34πGρr2m
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
v=H0r, r=ax⇒v=∣r˙∣r^=aa˙r⇒H≡aa˙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=38πGρ−a2kc2if $k=0$ , the critical density $\rho$ is then
ρc=8πG3H02In thermodynamics, the first law points that
dE+PdV=TdSwhere $P$ 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=34πa3ρc2dtdE=4πa2ρc2dtda+34πa3c2dtdρdtdV=4πa2dtdaAssuming a reservable expansion $\text dS=0$ , we obtain the fluid equation
ρ˙+3aa˙(ρ+c2P)=0The 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,
P≡P(ρ)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
2aa˙a2aa¨−a˙2=38πGρ˙+2a3kc2a˙aa¨−(aa¨)2=38πGρ−a3kc2Acceleration equation
aa¨=−34πG(ρ+c23P)Analytical solution
$k=0$
Dust, $P=0$
ρ˙+3aa˙ρ=0⇒dtd(ρa3)=0ρ=a3ρ0⇒a˙2=38πGρ0a1⇒a(t)=(t0t)2/3, ρ(t)=t2ρ0t02⇒H(t)=aa˙=3t2, t0=3H02However, the current age of universe $t_0$ obtained in this model is wrong
Radiation, $P=\rho c^2/3$
ρ˙+4aa˙ρ=0⇒dtd(ρa4)=0⇒a(t)=(t0t)1/2, ρ(t)=t2ρ0t02⇒H(t)=aa˙=2t1<3t2, t0=2H01Radiation dominated universe
a(t)∝t1/2, ρrad∝t−2, ρdust∝a−3∝t−3/2Unstable, 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, ρdust∝t−2, ρdust∝a−4∝t−8/3Stable
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(4πr2L0)dr=n0L0rmaxrmax→∞⇒μ→∞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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