Einstein's static universe concept

Question

This question is I think related to Einstein's static universe. I was reading an article on it that I am quoting here, and I have some very simple questions about it. My questions are interspersed.

Denoting $R(t)$ to be a scalar function of time which measures how much the Universe has stretched between $t_0$ and $t$. Einstein came up with the following equation:

We have the equation $$\ddot{R} = -\dfrac{4\pi G\rho}{3R^2}+\dfrac{\lambda R}{3}$$

The Jacobian Matrix $$J(R,S) = \begin{bmatrix}0& 1 \\ \dfrac{8\pi G\rho}{3R^3}+\dfrac{\lambda R}{3} & 0 \end{bmatrix}$$

Now we find the equilibrium point of $\ddot{R}$. With the assumption that $R \neq 0$ and that bearing in mind $$\frac{dR}{dt} = 0 = \frac{dS}{dt}$$

We solve the equation to get that the only equilibrium solution to be $$S = 0 ~~~\text{ and }~~~ R = \left(\frac{4\pi G \rho_0}{\lambda}\right)^{1/3}$$

The article goes on to say

We briefly mention that $R$ is a constant at the equilibrium so the size of the Universe never changes

Is it because $R$ being constant means $\dot{R}=0$ hence the velocity of galaxies moving is $0$?

It then says that furthermore we need $\lambda$ to be positive to make sense. Why is this so?

And now at the equilibrium, we look at the stability of the system: $$J(R,S)_{equilibrium} = \begin{bmatrix}0& 1 \\ \lambda & 0 \end{bmatrix}$$

Cleary, this is a saddle and hence the equilibrium solution is unstable

But how does unstableness cause the "static universe" theory to be rejected?

On Wikipedia I found that if the universe expands slightly, then the expansion releases vacuum energy, which causes yet more expansion. Does this mean a static universe cannot be like this?

Answer 1

The Einstein static universe assumes a matter dominated universe and a positive cosmological constant, implying a positive curvature (closed universe) as well.

To undestand how it was conceived, let us consider the Friedmann equations.
$(\frac{\dot a}{a})^2 = \frac{8 \pi}{3} \rho - \frac{K}{a^2} + \frac{\Lambda}{3}$ Eq. (1)
$\frac{\ddot a}{a} = \frac{-4 \pi}{3} (\rho + 3p) + \frac{\Lambda}{3}$ Eq. (2)
where:
$c = G = 1$ natural units
$a$ scale factor (dimensionless)
$\rho$ matter or radiation density
$K = 1 / R_0^2$ curvature constant
$R_0$ radius of curvature
$\Lambda$ cosmological constant
$p$ matter or radiation pressure

A static universe requires to have both $\dot a$ and $\ddot a$ nil. Moreover in a matter dominated universe the pressure is negligible.

In Eq. (2) by setting $\ddot a = 0$ and noting that $\rho = \rho_0 / a^3$, you get
$a = (4 \pi \rho_0 / \Lambda)^{1/3}$ scale factor at equilibrium
Being a meaningful scale factor positive, the cosmological constant is required to be positive as well.

In Eq. (1) to have $\dot a = 0$ you need a positive $K$, i.e. a positive curvature (closed universe).

The Einstein static universe is unstable as any slight change in either the cosmological constant, the matter density, or the spatial curvature will result in a universe that either expands and accelerates forever or re-collapses to a singularity.