# S^3 sphere metric-unit vector

1. Jun 22, 2014

### ChrisVer

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?

Last edited by a moderator: Jun 22, 2014
2. Jun 22, 2014

### micromass

Staff Emeritus
LaTeX hint: use \sin and \cos.

3. Jun 22, 2014

### ChrisVer

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.

4. Jun 22, 2014

### ChrisVer

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.

5. Jun 23, 2014

### Matterwave

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?

6. Jun 24, 2014

### ChrisVer

latin indices= spatial indices= 1,2,3...