# Geometry of the Universe varying in time?

1. Nov 13, 2008

### Gio83

In the expression of the Robertson-Walker metric for an expanding universe, the parameter k is allowed to be 1, -1, 0 if the universe is respectively closed, open or flat. But Einstein says that the density of the matter content of the universe determines its geometry: can we think about a time-dependence of k during the evolution of the Universe? I've never found such an approach in books, so I wanted to know if someone ever heard about that.

2. Nov 13, 2008

### George Jones

Staff Emeritus
k doesn't change if the universe is homogeneous and isotropic, which seems to be reasonably close to reality for the observable universe on larges scales. Note that this does not imply that density of the universe is constant in time.

3. Nov 13, 2008

### wolram

This now seems to be a rally point, i do not have the evidence , but how doe's this image reality?

4. Nov 13, 2008

### Gio83

Yes, the density of matter is a decreasing function of time due to the expansion of space, right?
But I was wondering if this decrease of density could affect the geometry of the universe, changing the value of k. Say, during a phase of high density the universe could be described by a closed geometry, while in a phase after the expansion, when the density is minor, it could be described by a flat or open geometry.
Or maybe the decrease of density doesn't affect the geometry as long as the density distribution itself is homogeneous? Is that what you mean?

(sorry if make mistakes writing in english...)

5. Nov 13, 2008

### marcus

I'd like to throw in some background information, Gio---hopefully not getting in the way of George's response.

Spatial geometry is not determined merely by density alone, but by a ratio Omega which tells how the actual density at a given moment compares with the critical density at that moment.

You can learn to calculate what that critical density is, at the present time, if you want. There is a simple formula for it, involving the square of the Hubble parameter H---this H is currently estimated at around 71 km/s per Megaparsec. Please say if you want the formula for critical density.

The actual density and the critical density are both gradually changing. But their ratio stays in the same range (either always less than 1, or always equal 1, or always bigger than 1) and so the basic feature of geometry, the "k", stays unchanged as the standard model universe evolves.
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Gio, assuming it is not a nuisance to have the extra information. Here is the formula for calculating the critical density:

3H2/(8 pi G)

You know what H is (namely 71 km/s per Megaparsec) and you know pi=3.14, and you know Newton's constant G. So should be a cinch to calculate it. And a neat way is to use the Google calculator which has things like pi and G built in.
You just put this into the box and press return:

3(71 km/s per Mpc)^2/(8 pi G)

That will calculate something for you and give you the answer in kilograms per cubic meter. I prefer to have the answer as an energy density, so what I prefer to put into the Google calculator is:

3c^2(71 km/s per Mpc)^2/(8 pi G) in joules/km^3

Then it gives me the answer as an energy density in joules per cubic kilometer. You can see I inserted a c^2 to change it from mass density to equivalent energy density. Google calculator is flexible: it will give you the answer in the units you specify. It is fun to use. And it is a great timesaver because it already knows many constants, like the mass of the sun, the charge of the electron, Newton G, Boltzmann k, etc etc. You seldom have to look anything up in a handbook.

So this critical density that we just calculated is what you have to compare the REAL density of the universe to, if you want to know the spatial geometry.

Last edited: Nov 13, 2008
6. Nov 15, 2008

### Gio83

thanks a lot, all of you!