Just beyond the horizon, but no way we can see it

We are bounded in a nutshell of Infinite space: Blog Post #32, Worksheet # 10.1, Problem #3: Just beyond the horizon, but no way we can see it

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?

(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 \(d\theta = d\phi = 0\), find the differential equation in terms of the coordinates t and r only.

(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 \(t= t_0\)).

(Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to \(r_{horizon}\) and t from 0 to \(t_0\).)

(a) As we all know, every modern theory of the large scale universe is based on several constants which are assumed and tested repeatedly to check for the consistency of the value. This is the case of the speed of light, c, the constant of the universe Einstein used to create special and general relativity, and the consequences of this knowledge have allowed us to understand the deeper throws of the universe. So, in order to find the radius of the horizon of the universe, we can use the only object which has reached us from the earliest point in time we can see, photons. Light has simply gotten to us before anything else, so it is the best representation of the horizon we are going to have.

(b) Simply put, we are going to find how the base FWR Metric: \[ds^2 = -cdt^2 + a^2(t) \left[\frac{dr^2}{1-kr^2} + r^2 (d\theta^2 + \sin^2\theta d\phi^2)\right],\]  can become a differential equation once we set \(ds^2\) to 0: \[ cdt^2 = a^2(t) \left[\frac{dr^2}{1-kr^2} + r^2 (d\theta^2 + \sin^2\theta d\phi^2)\right],\] \[ cdt^2 = \frac{a^2(t) (dr^2)}{1-kr^2},\] \[ \frac{cdt^2 }{a^2(t)} = \frac{ dr^2}{1-kr^2}.\] After all this rearrangement of the terms initially found, with both derivatives of angles set to 0, the equation becomes a relatively simple expression to describe the path light travels.

 (c) Furthermore, by establishing a flat universe, we have a definitive value for k, 0 in this case, so the equation becomes:\[ \frac{cdt^2 }{a^2(t)} = \frac{ dr^2}{1-kr^2},\]  \[ \frac{cdt^2 }{a^2(t)} = { dr^2},\]  which can now be set up to be solved in a way similar to the Friedmann Equations, using a parameter they established to complete the problem. So, from the Friedmann equation for a Matter Dominated Universe(http://ay17-rcordova.blogspot.com/2015/11/cosmology-101-part-2.html), the relation of the universal scale factor and time is: \[a(t) = a_0 \left(\frac{t}{t_0}\right)^{2/3} ,\] which can be placed into our derived form of the FRW metric and get: \[ dr^2= \frac{cdt^2 }{ a_0^2 \left(\frac{t}{t_0}\right)^{4/3}} ,\] which can be rewritten and analyzed in a manner similar to problem #2 of this Worksheet:  \[\int_0^{r_{horizon}} dr= \int_0^{t_0} \frac{cdt }{ a_0\left(\frac{t}{t_0}\right)^{2/3}} ,\] and after some integration and separation of variables, as well as recognizing \(a_0 \) and \(t_0\) as constants, we solve and get: \[ r_{horizon}= \frac{c t_0^{2/3}}{ a_0} \cdot 3t_0^{1/3},\] \[ r_{horizon}=3 \frac{c }{ a_0} t_0,\] which is the description for the radius of horizon today.