# Friedmann equations and different types of universes

I'm currently reading a book (in Danish) about the Friedmann equations and I have stumbled upon a third degree polynomial:

$$\Omega_\Lambda = \frac{4K_0^3}{27\Omega_0^2} = \frac{4(\Omega_0 + \Omega_\Lambda -1)^3}{27\Omega_0^2}$$

Where $$\Omega_0$$ is called the density parameter (defined as $$\frac{\rho}{p_c}$$) and $$\Omega_\Lambda=\frac{\Lambda}{3H_0^2}$$. Here $$H_0$$ is the Hubble parameter. $$K_0$$ is in my book called the current curvature of space.

According to the book if the above third degree polynomial is solved with respect to $$\Omega_\Lambda$$ then it's possible to plot $$\Omega_\Lambda$$ as a function of $$\Omega_0$$ and create a "map" of different types of universes (no Big Bang, run-away universes and Big Crunch universes). Here is the "map":

My question is: How is this third degree polynomial derived from the Friedmann equations and why does solving it make it possible to classify different universes?

• For me it's unclear what you mean with density parameter, $\Omega_0$. I guess that $\rho$ in your book is the total density (not counting curvature) ? And maybe you wanted to write the critical density $\rho_c$ (instead of $p_c$), defined as $3 H_0^2/ 8 \pi G$? Commented Mar 28, 2020 at 21:33
• You need to give us more background about the equation. What kind of universe is this. What is $\Omega_0$ ? Is it $\Omega_0 = \Omega_{\Lambda}$ or $\Omega_0 = \Omega_{\Lambda} + \Omega_m$ etc ? Is there an english version of the book that you can share the name for us ? Theres not much information that we can analyse Commented Mar 29, 2020 at 14:05

Usually one defines $$\Omega_0 = \frac{8\pi G\rho_0}{3H_0^2},\ \ \ \Omega_k = -\frac{k}{H_0^2a_0^2}, \ \ \ \Omega_\Lambda = -\frac{\Lambda}{2H_0^2}$$ Then the Friedmann equation can be written $$\dot a^2 = H_0^2(\Omega_0\frac{a_0^3}{a} + \Omega_k a_0^2 + \Omega_\Lambda a^2 ) .$$ At current time this gives the identity $$\Omega_0 + \Omega_k + \Omega_\Lambda \equiv 1 .$$
This accounts for the right hand equality, by cancelling factors and taking the cube root, but I don't know why the author wants to use $$K_0 = -\Omega_k$$
I can only think the left hand equality must be a typo. You can have $$K_0 = 0$$ without constraining $$\Omega_\Lambda$$. The "map" does not make a great deal of sense, because big bang models are possible for all values of the parameters. In this diagram showing general properties of the Friedmann models the critical value $$\Lambda_E$$ refers to Einstein's static solution for which $$\dot a = 0, a=a_E$$, constant, for which the Friedmann equation reduces to $$0 = \frac{2M}{a_E^3} - \frac{1}{a_E^2} + \frac{\Lambda_E}{3}$$