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

  • Context: Graduate 
  • Thread starter Thread starter ChrisVer
  • Start date Start date
  • Tags Tags
    Sphere Vector
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
Science Advisor
Messages
3,372
Reaction score
465
We have the (I think FRW) metric in the coordinates
[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?
 
Last edited by a moderator:
Physics news on Phys.org
ChrisVer said:
[itex]n= ( cosψ , sinψ sinθ cosφ, sinψ sinθ sinφ, sinψ cosθ )[/itex]

LaTeX hint: use \sin and \cos.
 
  • Like
Likes   Reactions: 1 person
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 [itex]y^{0}=t[/itex] that:

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

And I let time flow, to [itex]t' = t + \delta t[/itex]

Then:

[itex]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}|[/itex]

Inserting the Hubble's parameter [itex]H= \frac{\dot{f}}{f}[/itex]

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

Or

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

Could I write with that:

[itex]\delta_{t} g_{ij} = 2 H(t) \delta t g_{ij}[/itex]

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

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 [itex]g_{00}=1[/itex] because it's constant.
 
ChrisVer said:
We have the (I think FRW) metric in the coordinates
[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?

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...