# Density parameter (cosmology) as a function of redshift

Hopefully I'm missing some very basic algebra for this question. In essence, I need to derive the following:

$$\Omega=\Omega_{1}\frac{(1+z)}{1+\Omega_1{z}}$$ Now, I proceeded from the Friedmann equations as follows for a matter dominated Universe (and for $\Lambda=0$ and $k\neq{0}$): $$\frac{H^{2}}{H_{0}}=\Omega_{m}a^{-3}+\Omega_{\kappa}a^{-2}$$ Knowing that $a=1/(1+z)$ I can re-arrange this to: $$\frac{H^{2}}{H_{0}}=\Omega_m(1+z)(1+z)^2+\Omega_{\kappa}(1+z)^2$$ $$\frac{H^{2}}{H_{0}}=(\Omega_m+\Omega_mz)(1+z)^2+\Omega_{\kappa}(1+z)^2$$ Introducing the fact that $1-\Omega_{m}=\Omega_{\kappa}$: $$\frac{H^{2}}{H_{0}}=(\Omega_m+\Omega_{\kappa}+\Omega_mz)(1+z)^2=(1+\Omega_mz)(1+z)^2$$ This is where I feel I am almost there but I've stumbled a little as to my next options. Any tips or advice would be brilliant!

• Well for starters you need to introduce $\Omega_1$ and $\Omega$ in there with some other equation(s). – Chris Feb 9 '16 at 2:21
• I assume by $\Omega_1$ you actually mean $\Omega_m$ (i.e. $\Omega_0$). You have derived how the Hubble parameter evolves. Now you need to use this to find how the density parameter evolves. THe density parameter is defined as $\Omega \equiv \rho/(3M_{\mathrm{pl}}^2H^2)$. – Philo Feb 9 '16 at 17:57

## 1 Answer

You almost have it. The last step is $$\Omega = \frac{\rho}{\rho_\text{c}} = \frac{\rho_0\,a^{-3}}{\rho_\text{c,0}}\frac{\rho_\text{c,0}}{\rho_\text{c}} = \Omega_m\,(1+z)^3\frac{H_0^2}{H^2},$$ where we used the critical density $$\rho_\text{c} = \frac{3H^2}{8\pi G},\qquad \rho_\text{c,0} = \frac{3H_0^2}{8\pi G}.$$ The result follows immediately from what you already derived.