Density parameter against scale factor for perfect fluids

[spaghetti3451]

Homework Statement

This is a basic cosmology problem.

The Friedmann equations are

$\Big( \frac{\dot{a}}{a}\Big)^{2}+\frac{k}{a^{2}}=\frac{8\pi}{3m_{Pl}^{2}}\rho$ and $\Big( \frac{\ddot{a}}{a} \Big) = - \frac{4\pi}{3m_{Pl}^{2}}(\rho + 3p)$.

Using the density parameter $\Omega \equiv \frac{\rho}{\rho_{c}}=\frac{8\pi}{3m_{Pl}^{2}}\frac{\rho}{H^{2}}$, we can write the density parameter as $\Omega = 1 + \frac{k}{(aH)^{2}}$.

Furthermore, for perfect fluids, $p=\omega\rho$ so that the continuity equation $\dot{\rho}+3\Big(\frac{\dot{a}}{a}\Big)(\rho + p)=0$ for perfect fluids leads to $\rho \propto a^{-3(1+w)}$.

(a) Show that $\frac{d\Omega}{d\text{ln}a}=(1+3w)\Omega(\Omega -1)$.

(b) For matter and radiation, $1+3w>0$. Show that this implies that $\frac{d|\Omega -1|}{d\text{ln}a}>0$. What does this mean for a flat universe?

Homework Equations

The Attempt at a Solution

I have to substitute $\rho \propto a^{-3(1+w)}$ into the Freidmann equation $\Big( \frac{\dot{a}}{a}\Big)^{2}+\frac{k}{a^{2}}=\frac{8\pi}{3m_{Pl}^{2}}\rho$ and find an expression for H in terms of a and k.

Is this the correct approach?

 

[mark726]

Yes, that is the correct approach. By substituting the expression for ρ into the Friedmann equation, you can solve for H in terms of a and k. Then, by using the expression for H in terms of a and k, you can find the derivative of ρ with respect to ln(a) and plug it into the equation given in part (a). This will result in the desired expression for dΩ/dln(a).