Deriving the line element in homogenous isotropic space

[TheMan112]

If the Ricci-scalar $$R$$ is constant for a given spatial hypersurface, then the curvature of that region should be homogenous and isotropic, right?

A homogenous and isotropic hypersurface (disregarding time) has by definition the following line element (due to spherical symmetry):

$$d\sigma^2 = a^2 \left(\frac{1}{1-kr^2} dr^2 + r^2(d \theta^2 + sin^2(\theta) d \Phi^2) \right)$$

Where k = -1, 0 or +1 and a is constant.

Why $$\frac{1}{1-kr^2} dr^2$$ ?

This is apparently very important as the value of k determines the evolution of the universe, but I don't know how to come to this line element.

 

[Mentz114]

The three cases come from three possible solutions of the Robertson-Walker metric. A good description is given here

http://www.jb.man.ac.uk/~jpl/cosmo/RW.html

and even Wiki on FLRW is not bad

en.wikipedia.org/wiki/Friedmann-Lemaître-Robertson-Walker_metric

I hope this helps, I don't know if you've seen this material before.

 

[Entropia]

The line element for a homogenous and isotropic space can be derived from the Friedmann equations, which describe the evolution of the universe. These equations take into account the energy density and pressure of the universe, as well as the curvature of space.

In a homogenous and isotropic space, the energy density and pressure are assumed to be constant and the curvature is also assumed to be homogenous and isotropic. This means that the Ricci-scalar R is constant for a given spatial hypersurface, as stated in the content.

Using the Friedmann equations, we can derive the line element for a homogenous and isotropic space. The equation for the Friedmann curvature constant, k, is given by:

k = -\frac{R}{6}a^2

Where R is the Ricci-scalar and a is the scale factor of the universe.

Substituting this into the line element, we get:

d\sigma^2 = a^2 \left(\frac{1}{1-\frac{R}{6}a^2r^2} dr^2 + r^2(d \theta^2 + sin^2(\theta) d \Phi^2) \right)

Since R is constant, we can simplify this to:

d\sigma^2 = a^2 \left(\frac{1}{1-kr^2} dr^2 + r^2(d \theta^2 + sin^2(\theta) d \Phi^2) \right)

Where k = -\frac{R}{6}a^2.

This explains why the term \frac{1}{1-kr^2} is present in the line element. It is determined by the value of the Friedmann curvature constant, which in turn is determined by the constant Ricci-scalar in a homogenous and isotropic space.

In summary, if the Ricci-scalar R is constant for a given spatial hypersurface, then the curvature of that region should be homogenous and isotropic, and the line element for such a space can be derived using the Friedmann equations. This line element is crucial in understanding the evolution of the universe and the effects of curvature on its structure.