Blog #27: WS8.1, #3

3. It is not strictly correct to associate this ubiquitous distance-dependent redshift we observe with the velocity of galaxies (at very large separations, Hubble's Law gives 'velocities' that exceed the speed of light and becomes poorly defined). What we have measured is the cosmological redshift, which is actually due to the overall expansion of the universe itself. This phenomenon is dubbed the Hubble Flow, and it is due to space itself being stretched in an expanding universe. 

Since everything seems to be getting away from us, you might be tempted to imagine we are located at the center of this expansion. But, in actuality, everything is rushing away from everything else, everywhere in the universe, in the same way. So an alien astronomer observing the motion of galaxies in its locality would arrive at the same conclusions we do. 

In cosmology, the scale factor, a(t), is a dimensionless parameter that characterizes the size of the universe and the amount of space in between grid points in the universe at time t. In the current epoch, \(t=t_0\) and \(a(t_0)\equiv 1\). a(t) is a function of time. It changes over time, and it was smaller in the past (since the universe is expanding). This means that two galaxies in the Hubble Flow separated by distance \(d_0=d(t_0)\) in the present were \(d(t)=a(t)d_0\) apart at time t. 

The Hubble Constant is also a function of time, and is defined so as to characterize the fractional rate of change of the scale factor: 
\[H(t)=\frac{1}{a(t)}\frac{da}{dt}|_t\]
and the Hubble Law is locally valid for any t: 
\[v=H(t)d\]
where v is the relative recessional velocity between two points and d the distance that separates them. 

a) Assume the rate of expansion, \(\dot{a}\equiv da/dt\), has been constant for all time. How long ago was the Big Bang (i.e. when a(t=0)=0)? How does this compare with the age of the oldest globular clusters (~12 Gyr)? What you will calculate is known as the Hubble Time. 

Since we know the rate of expansion is constant, we can figure out a as a function of t.
\(\frac{da}{dt}=c\)
\(\int_0^{a(t)}da=\int_0^tdt'\)
\(a(t)=ct\)
We can then plug this into the equation for the Hubble Constant, with \(\frac{da}{dt}\) simplified to a constant.
\(H(t)=\frac{1}{ct}c\)
\(H(t)=\frac{1}{t}\)
Now we just need to plug in the Hubble Constant (68.6 (km/s)/Mpc, calculated in #2) in for H(t) to calculate the Hubble Time.
\(68.6\frac{km/s}{Mpc}=\frac{1}{t}\)
\(t=0.0146\frac{Mpc}{km/s}\rightarrow14.28Gyr\)
The universe is about 2Gyr older than the oldest globular clusters.

b) What is the size of the observable universe? What you will calculate is known as the Hubble Length.

The radius of the observable universe should be the distance that light can travel over the Hubble Time, so we can just multiply the Hubble Time by the speed of light.
\(r=(0.0146\frac{Mpc}{km/s})(3.0*10^5km/s)=4380Mpc\)