I have a question about a linear FRW cosmology with [itex]k=+1[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

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 [itex]k=-1[/itex] (open),0 (flat) or +1 (closed) and [itex]R(t)[/itex] 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 [itex]k=-1[/itex] then [itex]\rho=0[/itex]. This is the empty Milne Universe with negative spatial curvature.

If [itex]k=0[/itex] 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 [itex]M[/itex] in a sphere of radius [itex]R[/itex]:

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

so that

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

Now I want to look at the case with [itex]k=+1[/itex]:

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

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 [itex]\rho^\prime = \rho/2[/itex] 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 [itex]R[/itex] will have a mass [itex]M/2[/itex] 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 [itex]k=+1[/itex] is unique in that the curvature term and the density term have exactly the same [itex]1/R^2[/itex] dependence. I thought therefore that maybe the curvature can be included with the density leaving an effectively flat cosmology.

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

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