Closed Universe - FRW Equation

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1. Mar 1, 2015

unscientific

1. The problem statement, all variables and given/known data

(a) Show that the equations satisfy FRW equations.
(b) Show the metric when $\eta$ is taken as time

2. Relevant equations

3. The attempt at a solution

The FRW equation is:
$$3 \left( \frac{\dot a}{a} \right)^2 = 8\pi G \rho$$

Using $\frac{da}{dt} = \frac{da}{d\eta} \frac{d\eta}{dt}$:

$$\dot a = \frac{\frac{1}{\eta_*} sin \left( \frac{\eta}{\eta_*}\right)}{1 - cos \left( \frac{\eta}{\eta_*} \right)}$$
$$\dot a = \frac{C sin \left( \frac{\eta}{\eta_*} \right)}{\eta_* a}$$

The LHS is then

$$3 \left( \frac{\dot a}{a} \right)^2 = 3 \left[ \frac{C^2 sin^2 (\frac{\eta}{\eta_*})}{\eta_*^2 a^4} \right]$$

Not sure how to show this equals RHS..

2. Mar 1, 2015

Dick

In a closed universe the FRW equation has a spatial curvature term with a 'k' in it. You omitted that.

Last edited: Mar 1, 2015
3. Mar 2, 2015

unscientific

FRW equation is given by:
$$3 \left( \frac{\dot a}{a} \right)^2 + \frac{kc^2}{a^2} = 8\pi G \rho$$

Curvature parameter is given by $\Omega = \frac{8\pi \rho G}{3H_0^2}$.

$$\frac{C^2 sin^2 (\frac{\eta}{\eta_*})}{\eta_*^2 a^4} + \frac{kc^2}{a^2} = \frac{\Omega}{H_0^2}$$

At $t=0$, $\frac{\eta_0}{\eta_*} = sin (\frac{\eta_0}{\eta_*})$, so $a_0 = C\left[ 1 - \sqrt{1 - (\frac{\eta_0}{\eta_*})^2} \right]$.

Rearranging, $(\frac{\eta_0}{\eta_*})^2 = (\frac{a_0}{c})(1 - \frac{a_0}{c})$.

How am I to relate $\eta$ to $\eta_0$?

Last edited: Mar 2, 2015
4. Mar 2, 2015

unscientific

I'm just confused at this stage, shouldn't $a_0 =1$? I have a feeling this problem is much simpler than it seems..Would appreciate some help

Last edited: Mar 2, 2015
5. Mar 2, 2015

unscientific

5th Attempt

$$\left( \frac{\dot a}{a}\right)^2 = \frac{8\pi G \rho}{3} - \frac{kc^2}{a^2}$$

Using $\rho = \Omega \rho_c = \Omega \cdot \frac{3H^2}{8\pi G}$:

$$\left( \frac{\dot a}{a}\right)^2 = \Omega H^2 - \frac{kc^2}{a^2}$$

$$\dot {a}^2 = \Omega \dot {a}^2 - kc^2$$

$$\frac{da}{dt} = c \sqrt{\frac{k}{\Omega -1}}$$

But there is no factor of $C$ on the LHS, as $\frac{da}{dt} = \frac{sin(\frac{\eta}{\eta_*})}{\eta_* \left[1-cos(\frac{\eta}{\eta_*})\right]}$