Blog #32, Worksheet #10.1, Problem #3, Observable Universe

3. Observable Universe. It is important to realize that in our Big Bang universe, at any given time, the size of the observable universe is finite. The limit of this observable part of the universe is called the horizon (or particle horizon to be precise, since there are other definitions of horizon.) In this problem, we’ll compute the horizon size in a matter dominated universe in co-moving coordinates. To compute the size of the horizon, let us compute how far the light can travel since the Big Bang. 

(a) First of all - Why do we use the light to figure out the horizon size? 
Well we classify the horizon as the edge of the observable universe and we can only observe as far as light has traveled. Thus if we want to calculate the size of the "observable" universe, we must use the fastest known observable element, which in our case is light.

(b) Light satisfies the statement that \(ds^2 = 0\). Using the FRW metric, write down the differential equation that describes the path light takes. We call this path the geodesic for a photon. Choosing convenient coordinates in which the light travels in the radial direction so that we can set dθ = dφ = 0, find the differential equation in terms of the coordinates t and r only. 

We begin with the general FRW metric given earlier in the worksheet and set ds^2 = 0 giving us \[c^2 dt^2 = a^2 (t) [\frac{dr^2}{1-kr^2} + r^2 (d \theta^2 + sin^2 \theta d \phi^2)] \]We simplify given the information that dθ = dφ = 0 and get \[c^2 dt^2 = \frac{a^2 (t) dr^2}{1-kr^2}\]which gives us an equation \[c^2 = \frac{a^2}{1-kr^2} (\frac{dr}{dt})^2\]

(c) Suppose we consider a flat universe. Let’s consider a matter dominated universe so that a(t) as a function of time is known (in the last worksheet). Find the radius of the horizon today (at t = t0). (Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to rhorizon and t from 0 to t0.)

We begin with the equation we just derived and put in the fact that for a flat universe, k = 0. Thus we are left with \[c^2 = a^2 (\frac{dr}{dt})^2 \rightarrow dr = \frac{c}{a(t)} dt \] In last week's worksheet we found that a(t) in modern times is proportional to \(t^{\frac{2}{3}}\). We then plug into our equation \[a(t) = a_0 (\frac{t}{t_0})^{2/3}\] \[dr = \frac{c t_0^{\frac{2}{3}}}{a_0 t^{\frac{2}{3}}}\]We then integrate both sides and get \[R = \frac{c}{a_0} t_0^{\frac{2}{3}} \int_{0}^{t_0} t^{- \frac{2}{3}} = \frac{c}{a_0} t_0^{\frac{2}{3}} \cdot 3 t_0^{\frac{1}{3}}\]Which gives us our final relationship of \[R = \frac{3c t_0}{a_0}\]