## Saturday, November 28, 2015

### 35 | Worksheet 11.1, Problem 4: The Flatness Problem of the Big Bang Model

Despite the success of the Big Bang model mentioned in the lecture, there are also problems with it. These problems mostly have to do with the initial conditions for the Big Bang model, and provide motivations for scientists to seek deeper answers for the origin of the Big Bang. In this exercise, we study for one of these problems.

(a) Recall the Friedmann equation that we learned in a previous lecture:
$H^2 = \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2}$
The critical density of the universe $$\rho_c$$ is defined to be the value of $$\rho$$ at which the curvature parameter $$k=0$$. Namely,
$\rho_c (t) = \frac{3H(t)^2}{8\pi G}$
Rewrite the Friedmann equation in terms of the density parameter $$\Omega$$ defined as:
$\Omega (t) \equiv \frac{\rho (t)}{\rho_c (t)}$
and show that:
$1+ \frac{kc^2}{a^2 H^2} = \Omega$

Starting with the Friedmann equation given:
$H^2 = \frac{8\pi G}{3} \rho - \frac{kc^2}{a^2}$
We notice the first term on the right-hand side looks similar to $$\rho_c$$, which is:
$\rho_c = \frac{3H^2}{8\pi G}$
We can rewrite the Friedmann equation in terms of $$\rho_c and \rho$$:
$H^2 = \frac{\rho}{\rho_c}H^2 - \frac{kc^2}{a^2}$
Since the ratio of $$\rho$$ and $$\rho_c$$ is defined to be the density parameter $$\Omega$$, we can rewrite this as:
$H^2 = \Omega H^2 - \frac{kc^2}{a^2}$
Solving for $$\Omega$$, we get:
$\boxed{\Omega = 1 + \frac{kc^2}{H^2 a^2}}$

(b) Consider the epoch of our universe that is dominated by matter. We have already computed the time dependence of $$a$$ and $$H$$ in such a universe. Use this result and (a) to show the time dependence of $$\Omega$$.

In Worksheet 9.1, we found that the time dependence of the scale factor $$a$$ in such a universe is given by:
$a \propto t^{\frac{2}{3}}$
We also know that the Hubble constant is defined as:
$H = \frac{\dot{a}}{a}$
Since we've already found the time dependence of $$a$$, we can easily find the time dependence of $$H$$. Taking the derivative of $$a$$, we find that $$\dot{a} \propto t^{-1/3}$$, so $$H \propto t^{-1}$$. Above, we showed that:
$\Omega - 1 \propto a^{-2}H^{-2}$
Plugging in the time dependences found above, we find that
$\boxed{\Omega -1 \propto t^{2/3}}$

(c) Today, experiments have measured that the density parameter of our universe is within -0.005<$$\Omega -1$$<0.005, namely $$\Omega$$ is close to 1 today. Use your above result to determine the range of $$\Omega -1$$ at the time of CMB formation, using the fact that today the age of universe is about 13.7 billion years and the age of the universe when the CMB was formed was about 380,000 years.

We've shown how the density parameter is related to time, namely:
$\Omega -1 \propto t^{2/3}$
In order to solve for $$\Omega$$ we will say that $$\Omega - 1 = C t^{2/3}$$ where $$C$$ is a proportionality constant.

We know both the time and density parameter today, so we can use these values to find the proportionality constant. Since we know a range for the density parameter, we cannot use equal signs, but instead I will use arrows below:
$C t^{2/3} \Rightarrow \Omega - 1$
$C(13.7\times 10^9 \text{ yr})^{2/3} \Rightarrow 0.005$
$C \Rightarrow 8.73\times 10^{-10}$
Now we can use this constant to find the density parameter range during the time the CMB was formed:
$\Omega - 1 \Rightarrow (8.73\times 10^{-1-})(3.8\times 10^5 \text{ yr}) = 4.58\times 10^{-6}$
So, the range for the density parameter at the time of CMB formation was:
$\boxed{-4.6\times 10^{-6} < \Omega - 1< 4.6\times 10^{-6}}$
This means that $$\Omega$$ is very close to 1.