# Closed linear cosmology implies G M / R = c^2?

1. Jul 30, 2013

### johne1618

I have a question about a linear FRW cosmology with $k=+1$.

Assuming zero cosmological constant the first Friedmann equation can be written:

$$\left(\frac{\dot R}{R}\right)^2 + \frac{kc^2}{R^2}=\frac{8\pi G}{3}\rho$$

where scalar curvature $k=-1$ (open),0 (flat) or +1 (closed) and $R(t)$ is the radius of curvature of space.

Now I assume a linear cosmology so that:

$$R = c t.$$

Thus the Friedmann equation becomes:

$$\frac{c^2}{R^2} + \frac{kc^2}{R^2}=\frac{8\pi G}{3}\rho.$$

If $k=-1$ then $\rho=0$. This is the empty Milne Universe with negative spatial curvature.

If $k=0$ then we have a flat Universe with:

$$\frac{c^2}{R^2} = \frac{8 \pi G}{3} \rho.$$

Because the Universe is flat I can write an expression for the total mass $M$ in a sphere of radius $R$:

$$M = \frac{4}{3} \pi R^3 \rho\ \ \ \ \ \ \ \ \ \ \ \ (1)$$

so that

$$\frac{GM}{R}=\frac{c^2}{2}.$$

Now I want to look at the case with $k=+1$:

$$\frac{c^2}{R^2} + \frac{c^2}{R^2}=\frac{8\pi G}{3}\rho.$$

One could say that this case must have a positive spatial curvature and thus the equation (1) linking mass, density and volume of a sphere is not valid.

But I can define $\rho^\prime = \rho/2$ so that I get the equation:

$$\frac{c^2}{R^2} = \frac{8 \pi G}{3} \rho^\prime.$$

This is exactly the same as the flat Friedmann equation but with half the density. Therefore equation (1) is still valid. Thus a sphere of radius $R$ will have a mass $M/2$ so that, in this case, one would get the equation:

$$\frac{GM}{R}=c^2.$$

Is this correct?

I ask because the linear model with $k=+1$ is unique in that the curvature term and the density term have exactly the same $1/R^2$ dependence. I thought therefore that maybe the curvature can be included with the density leaving an effectively flat cosmology.