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

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?

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I'm curious where this linearity assumption comes from. The Friedmann equations give you $R$ as a function of $t$ - I've never seen an assumed form for $R(t)$ be inserted into them. – Chris White Jul 29 '13 at 17:30
I chose the linearity assumption as it gives the simplest expanding cosmology. – John Eastmond Jul 29 '13 at 22:57
I don't understand what you're trying to do. As you said yourself, equation (1) isn't valid for $k=1$. Instead, the volume of a sphere is $4\pi R^3\int_0^\chi\sin^2\chi'\,\text{d}\chi'=\pi R^3(2\chi-\sin 2\chi)$. – Pulsar Jul 30 '13 at 2:29
So you're saying I can't redefine the k=1 problem as a flat space problem with half the density. – John Eastmond Jul 30 '13 at 2:37