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S^3 sphere metric-unit vector

  1. Jun 22, 2014 #1


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    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: Jun 22, 2014
  2. jcsd
  3. Jun 22, 2014 #2
    LaTeX hint: use \sin and \cos.
  4. Jun 22, 2014 #3


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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.
  5. Jun 22, 2014 #4


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    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]


    [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]


    [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.
  6. Jun 23, 2014 #5


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    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?
  7. Jun 24, 2014 #6


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