Size of the Observable Universe: Blog 32, Worksheet 10.1, Problem 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?

Light is both the fastest known phenomena in the universe and our way of observing the universe. All we can see is determined by how quickly light can reach our eyes, so we are limited to a particle horizon determined by the speed of 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.
$$ds^2 = -c^2dt^2 + a^2(t) [\frac{dr^2}{1-kr^2} + r^2 (d \theta + sin^2( \theta) d \theta ]$$ With the conditions for light and comoving radii this becomes $$0= -c^2dt^2 + a^2(t) [\frac{dr^2}{1-kr^2} + 0 ]$$ $$\frac{c^2dt^2}{ a^2(t)} = \frac{dr^2}{1-kr^2}$$

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 =  $t_0$). (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 $t_0$.) $$\frac{c^2dt^2}{ a^2(t)} = \frac{dr^2}{1-kr^2}$$ In a flat universe, this simplifies to $$\frac{c}{ a(t)} dt = dr$$ The integral is $$\int_0^{t_0} \frac{c}{ a(t)} dt = \int_0^{r_H} dr$$ We know from a previous blog post that $a(t) \propto  t^{\frac{2}{3}}$ so the integral becomes $$\int_0^{t_0} \beta  \frac{c}{  \left( \frac{t}{t_0}\right)^{\frac{2}{3}}} dt = \int_0^{r_H} dr$$ where beta is a constant. So we have $$r_H = 3 \beta c t_0$$

Correction: we can solve this problem more accurately by replacing the relation $a(t) \propto  t^{\frac{2}{3}}$ with the more accurate $a(t) \propto \left( \frac{ t}{t_0} \right)^{\frac{2}{3}}$ in this way we can redefine beta in terms of $a_0 \text{ and } t_0$ where $a_0$ is 1 so our final answer cleans up to just be $$r_H = 3  c t_0$$