How is the S^3 metric defined by a unit vector and coordinate changes?

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We have the (I think FRW) metric in the coordinates
y^{0}=t,~~y^{1}=\psi,~~y^{2}=\theta,~~y^{3}=\varphi

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

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

n= ( \cos\psi , \sin\psi \sin\theta\cos\varphi, \sin\psi \sin\theta \sin\varphi, \sin\psi \cos\theta )

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

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

So I would like to interpret this result... I need some confirmation of how I interpreted it :)
Suppose you have the vector n. The metric is then by the equation above, defined by how the n vector changes \partial n along the change of the i-th coordinate ∂y^{i}. As I wrote it, by the module of the velocity of n wrt y^{i}. Also tried to do a grid diagram which I think is correct for S^{2} of coordinates (\theta,\varphi), just imagining the generalization of it with a 3rd coordinate ##\psi##.
Finally the metric is scaling by the flow of time (or y^{0}-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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ChrisVer said:
n= ( cosψ , sinψ sinθ cosφ, sinψ sinθ sinφ, sinψ cosθ )

LaTeX hint: use \sin and \cos.
 
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Better now? Well the form of it doesn't make the big difference since my maths were correct and the interpretation has to do with the final equation, but thanks :) that way other people might understand it better.
 
Another question... I did it for my own fun... because I said about that thing with "images" of sphere, and I wanted to see how , by the flow of "time", the metric would change.

Suppose I have at time y^{0}=t that:

g_{ij}(t, y^{a})= - \frac{f^{2}(t)}{\alpha} |n_{,i} \cdot n_{,j}|

And I let time flow, to t' = t + \delta t

Then:

g_{ij}'=g_{ij}(t+\delta t, y^{a})= g_{ij}(t, y^{a}) + \delta t \frac{\partial g_{ij}(t, y^{a})}{\partial t}= g_{ij}(t, y^{a}) -\frac{2}{\alpha} \delta t f(t) \dot{f}(t) |n_{,i} \cdot n_{,j}| = g_{ij}(t, y^{a}) -\frac{2 \dot{f}(t)}{\alpha f(t)} \delta t f^{2}(t) |n_{,i} \cdot n_{,j}|

Inserting the Hubble's parameter H= \frac{\dot{f}}{f}

g_{ij}'=g_{ij}(t, y^{a}) + \delta t 2 H(t) g_{ij}(t, y^{a})

Or

g_{ij}(t+\delta t, y^{a})=(1+ 2H(t) \delta t) g_{ij}(t, y^{a})

Could I write with that:

\delta_{t} g_{ij} = 2 H(t) \delta t g_{ij}

Or equivalently:
g_{ij}(t', y^{a})= e^{2 H(t) (t'-t)} g_{ij} (t, y^{a})
?

Meaning that the Hubble's parameter is somewhat related to the generator of the translation of time for the spatial components of the metric?
However that's not true for the g_{00}=1 because it's constant.
 
ChrisVer said:
We have the (I think FRW) metric in the coordinates
y^{0}=t,~~y^{1}=\psi,~~y^{2}=\theta,~~y^{3}=\varphi

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

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

n= ( \cos\psi , \sin\psi \sin\theta\cos\varphi, \sin\psi \sin\theta \sin\varphi, \sin\psi \cos\theta )

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

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

So I would like to interpret this result... I need some confirmation of how I interpreted it :)
Suppose you have the vector n. The metric is then by the equation above, defined by how the n vector changes \partial n along the change of the i-th coordinate ∂y^{i}. As I wrote it, by the module of the velocity of n wrt y^{i}. Also tried to do a grid diagram which I think is correct for S^{2} of coordinates (\theta,\varphi), just imagining the generalization of it with a 3rd coordinate ##\psi##.
Finally the metric is scaling by the flow of time (or y^{0}-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?

You posted four different ##g_{00}##, so I think you've not specified the metric correctly.

Also, when you use the Latin indices ##i,j## do they go from 1-3 or from 0-3?
 
latin indices= spatial indices= 1,2,3...
 
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