Friedmann equations and different types of universes

Question

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?

Answer 1

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 $$0 = \frac{2M}{a_E^3} - \frac{1}{a_E^2} + \frac{\Lambda_E}{3} $$