## Saturday, November 21, 2015

### 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 universe is called the particle 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) Why do we use the light to figure out the horizon size?

We use the light, as there is no information in the universe that can travel faster than the speed of light. This is to say that if we want to see as far as possible into our universe, we must use light, as this information has traveled the farthest in the finite time since the Big Bang.

 (image: http://frigg.physastro.mnsu.edu)

(b) Light satisfies the statement that $$\mathrm{d}s^2 = 0$$. Using the Friedmann-Robertson-Walker metric, write down the differential equation that describes the path the light takes. We call this path the geodesic for a photon. Choosing convenient coordinates in which the light travels in the radial direction, we can set $$\mathrm{d}\theta = \mathrm{d}\phi = 0$$. Find the differential equation in terms of the coordinates $$t$$ and $$r$$ only.

The Friedmann-Robertson-Walker metric is given by:
$\mathrm{d}s^2 = -c^2 \mathrm{d}t^2 + a^2(t) \left[\frac{\mathrm{d}r^2}{1-kr^2} + r^2 (\mathrm{d}\theta^2 + \sin^2 \theta \mathrm{d}\phi^2)\right]$
Since we are dealing with light, we can set $$\mathrm{d}s^2 = 0$$. Also, we are considering light traveling in the radial direction, so $$\mathrm{d}\theta = \mathrm{d}\phi = 0$$. This simplifies the Friedmann-Robertson-Walker metric to:
$c^2 \mathrm{d}t^2 = \frac{a^2(t) \mathrm{d}r^2}{1-kr^2}$
Let's group the $$t$$-dependent terms on the left-hand side and the $$r$$-dependent terms on the right:
$\frac{c^2 \mathrm{d}t^2}{a^2(t)} = \left(\frac{1}{1-kr^2}\right)\mathrm{d}r^2$
Taking the square root of both sides gives:
$\boxed{\frac{c \mathrm{d}t}{a(t)} = \left(\frac{1}{1-kr^2}\right)^{\frac{1}{2}}\mathrm{d}r}$

(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. Find the radius of the horizon today ($$t = t_0$$).

In (b), we found the following differential equation:
$\frac{c \mathrm{d}t}{a(t)} = \left(\frac{1}{1-kr^2}\right)^{\frac{1}{2}}\mathrm{d}r$
For now, we will consider a flat universe, so $$k =0$$. This means that our expression becomes:
$\frac{c \mathrm{d}t}{a(t)} = \mathrm{d}r$
Now we can integrate both sides. We will integrate over time from $$0 \rightarrow t_0$$ and over the distance $$0 \rightarrow r_h$$, where $$r_h$$ is the radius of the particle horizon:
$\int_0^{t_0} \frac{c}{a(t)}\mathrm{d}t = \int_0^{r_h} \mathrm{d}r$
Our expression includes $$a(t)$$, which we need to consider when taking the integral. In a previous worksheet, we found that $$a \propto t^{2/3}$$ in a matter-dominated universe. Since this is the scenario we are considering in this problem, we will say $$a(t) = \frac{1}{A}t^{2/3}$$ where $$A$$ is a proportionality constant. Our expression is now:
$cA \int_0^{t_0} \frac{1}{t^{\frac{2}{3}}} = \int_0^{r_h}\mathrm{d}r$
Solving this integral gives us the radius of the particle horizon:
$\boxed{r_h = 3cAt_0^{\frac{1}{3}}}$