We have the (I think FRW) metric in the coordinates(adsbygoogle = window.adsbygoogle || []).push({});

[tex] y^{0}=t,~~y^{1}=\psi,~~y^{2}=\theta,~~y^{3}=\varphi [/tex]

[tex] g_{00}=1,~~g_{00}= - \frac{f^{2}(t)}{\alpha} ,~~ g_{00}= - \frac{f^{2}(t)}{\alpha} \sin^{2}\psi ,~~g_{00}= - \frac{f^{2}(t)}{\alpha} \sin^{2}\psi sin^{2}\theta [/tex]

Suppose we have define a unit vector [itex]n \in \mathbb{R}^{4} [/itex] such that:

[tex] n= ( \cos\psi , \sin\psi \sin\theta\cos\varphi, \sin\psi \sin\theta \sin\varphi, \sin\psi \cos\theta ) [/tex]

So far I was able to show that (by doing the derivative calculations- is there any faster way one can work?)

[tex] g_{ij} = - \frac{f^{2}(t)}{\alpha} \sum_{A=1}^{4} \frac{\partial n^{A}}{∂y^{i}}\frac{\partial n^{A}}{\partial y^{j}} [/tex]

So I would like to interpret this result... I need some confirmation of how I interpreted it :)

Suppose you have the vector [itex]n[/itex]. The metric is then by the equation above, defined by how the [itex]n[/itex] vector changes [itex]\partial n[/itex] along the change of the i-th coordinate [itex]∂y^{i}[/itex]. As I wrote it, by the module of the velocity of [itex]n[/itex] wrt [itex]y^{i}[/itex]. Also tried to do a grid diagram which I think is correct for [itex]S^{2}[/itex] of coordinates [itex](\theta,\varphi)[/itex], just imagining the generalization of it with a 3rd coordinate ##\psi##.

Finally the metric is scaling by the flow of time (or [itex]y^{0}[/itex]-coord) so it's more like, as time passes, we get different images of a 3-sphere, each having its "grid" rescaled by some factor.

Is that correct? Do you think I'm missing something important?

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# S^3 sphere metric-unit vector

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