The universe is REALLY old and REALLY big

We are bounded in a nutshell of Infinite space: Blog Post #27, Worksheet # 8.1, Problem #3: The universe is REALLY old and REALLY big

3. It is not strictly correct to associate this ubiquitous distance-dependent redshift we observe with the velocity of the galaxies (at very large separations, Hubble’s Law gives ‘velocities’ that exceeds 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, as you explored in the opening thought experiment, 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, \(\dot{a} = da/dt \), 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\(\sim 12 Gyr)? What you will calculate is known as the Hubble Time.

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

(a) In this problem, we’ll be examining some of the first methods developed that correctly estimated the age, and size of the universe (from our perspective). Going back to one of the pioneers in the field of cosmology, Edwin Hubble established a set of relationships and constants which have largely determined the basis for all cosmology. One of these basics is the Hubble constants, a value found by analyzing the Red Shifts of many galaxies and comparing them all in their velocities and distance from Earth. The slope of this graphical comparison yields the Hubble Constant, which can be approximated to being \( 70 \frac{\frac{km}{s}}{Mpc} \). Knowing this, we can start to use calculus and integrate the above calculation for the Hubble Constant as a function of time when \(a(t) = a(t_0)\): \[H(t_0) = \frac{1}{a(t_0)} \frac{da}{dt}|_{t=0} , \] \[H(t_0) = \frac{da}{dt} , \] \[ dt = \frac{da}{ H(t) } , \] and now we integrate to find the total \(H(t)\) : \[\int_{t=0}^{t=present} dt = \int^{a(t_0)}_0 \frac{da}{ H(t) } , \] \[t = H_0 ^{-1} .\] And now, plugging in the compiled value for Hubble Time, we can convert it to an actual time by making the dimensional analysis work. So if \( H_0 = 70 \frac{\frac{km}{s}}{Mpc} , \) then we simply find the value of km in pc, and turn Mpc into pc, so the final dimensional analysis would be: \[ H_0 = 70 \frac{\frac{km}{s}}{Mpc} \cdot ( 3.24 \times 10^{-20} \frac{pc}{s} ) \cdot (1 \times 10^{-6} \frac{Mpc}{pc} ) ,\] which yields: \[H_0 = 2.27 \times 10^{-18} \frac{1}{s} ,\]  and once this is plugged into the equation for the age of the universe we found a just a bit ago, we now have: \[t = \frac{1}{2.27 \times 10^{-18}  s^{-1}} ,\] \[ t = 4.41 \times 10^{17} s\] and now we just convert this value into years so it looks like it makes sense: \[ t =  4.41 \times 10^{17} s \cdot (3600 \frac{s}{hour}) \cdot (24 \frac{hours}{day}) \cdot (365 \frac{days}{year}) ,\] \[t = 1.4 \times 10^{10} years ,\] a close approximation to the current definition for the age of the universe, which is 200 million years older than the oldest globular cluster we have observed.

(b) To find the Total Distance of the galaxy, it is as simple as multylping the age of the universe times the maximum speed of the universe, the speed of light. So by taking the speed of light and turning it into the distance it travels in one year: \[ c = 3 \times 10^8 \frac{m}{s} \cdot (3600 \frac{s}{hour}) \ cdot (24 \frac{hours}{day}) \cdot (365 \frac{days}{year}) , \] \[c = 9.46 \times 10^{15} \frac{m}{year} , \] and next we turn the speed of light into Mpc per year: \[ c = 9.46 \times 10^{15} \frac{m}{year} \cdot \frac{1}{3.241 \times 10^23 \frac{m}{Mpc} } , \] \[c = 2.77 \times 10^{-8} \frac{Mpc}{year} , \] and now finally multiply this by the age of the universe and we get: \[1.4 \times 10^{years} ~years ~\cdot 2.77 \times 10^{-8} \frac{Mpc}{year} = 3.87 \times 10^2 Mpc .\] However, this is not the total observable distance of the universe, rather it is only the radius of our line of sight, so to complete the distance, we would need to multiply this radius by 2 to get the total diameter of the sphere that we consider the observable universe, the Hubble Length. \[ Hubble ~Length= 2 \times  3.87 \times 10^2 Mpc ,\] \[Hubble ~Length = 7.7 \times 10^2 Mpc . \]