TheMan112
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If the Ricci-scalar [tex]R[/tex] 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):
[tex]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)[/tex]
Where k = -1, 0 or +1 and a is constant.
Why [tex]\frac{1}{1-kr^2} dr^2[/tex] ?
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.
A homogenous and isotropic hypersurface (disregarding time) has by definition the following line element (due to spherical symmetry):
[tex]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)[/tex]
Where k = -1, 0 or +1 and a is constant.
Why [tex]\frac{1}{1-kr^2} dr^2[/tex] ?
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.