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Metric tensor of the 4-sphere

  1. Sep 10, 2014 #1
    After my recent studies of the curvature of the 2- sphere, I would like to move on to Minkowski space. However, I can not seem to find the metric tensor of the 4 sphere on line, nor can I seem to think of the vector of transformation properties that I would use to derive the metric tensor of the 4 sphere.

    Could anyone please post the metric tensor components for the 4-sphere (or a link to a page that has it) along with the labels telling what row and column each element is on?

    Thank you very much.

    If you can not do this, can you at least tell me what vector is differentiated with respect to θ and ø in order to derive the tangential vectors that are multiplied together (via the dot product) to get the metric tensor for the 2 sphere? If I can derive the metric tensor for the 2 sphere, then I should be able to extend it to 3 and 4 dimensions.
     
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  3. Sep 10, 2014 #2

    ChrisVer

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    In order to find the 4-sphere, you have to make an embedding in 5D... As you do for the 2 Sphere which you embed into [itex]E^{3}[/itex]...
    For the 4D space, I have only seen people defining the 3 Sphere...
    In general a 4-sphere would need 4 parameters, or in other words you need to look at the mapping of the parameter space [itex](u,v,\psi,w)[/itex] to the [itex]\mathbb{R}^5[/itex]:
    [itex] X= X (u,v,\psi, w) [/itex]. In order for this to be on a sphere, the vector's components should be be contrained: [itex] X = x_{1} \hat{e}_{1}+x_{2} \hat{e}_{2}+x_{3} \hat{e}_{3}+x_{4} \hat{e}_{4}+x_{5} \hat{e}_{5}[/itex] with [itex]x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=1[/itex]
    Then the tangent vectors on the surface point P, will be given by the derivatives of [itex]X[/itex] wrt your parameters on that point...
    The metric then will be defined by [itex] \bar{g}_{ab}= X_{a} \cdot X_{b}[/itex]...

    How do you define the 3 sphere for example?
    The embedding is done in the 4D space...
    Then the coordinates of the map [itex]X^{\mu}[/itex] will be:
    [itex]x_{1} = \sin \theta \cos \phi \sin \psi[/itex]
    [itex]x_{2} = \sin \theta \sin \phi \sin \psi[/itex]
    [itex]x_{3} = \cos \theta \sin \psi[/itex]
    [itex]x_{4} = \cos \psi[/itex]
    These make the above equation hold:
    [itex] \sum_{i} x_{i}^2 =\sin^2 \theta \cos^2 \phi \sin^2 \psi+ \sin^2 \theta \sin^2 \phi \sin^2 \psi + \cos^2 \theta \sin^2 \psi + \cos^2 \psi= (\sin^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi + \cos^2 \theta )_{=1,S^2}\sin^2 \psi +\cos^2 \psi=\sin^2 \psi +\cos^2 \psi=1[/itex]

    Then the metric is given by:
    [itex] g_{ab}= X_{a} \cdot X_{b} [/itex]

    Where [itex]\cdot[/itex] stands for the inner product of your space [in minkowski space embedding this means the contraction of them with the minkowski metric tensor], and a,b in subscripts stand for the derivatives of [itex]X^{\mu}= (x_{1},x_2,x_3,x_4)[/itex] vectors with respect to a,b=1,2,3 ([itex]\theta, \phi, \psi [/itex]). Eg in full mode, the 11 or [itex]\theta \theta[/itex] component of the embedded metric will be:

    [itex] g_{\theta \theta} = \eta_{\mu \nu} \frac{\partial X^{\mu}}{\partial \theta} \frac{\partial X^{\nu}}{\partial \theta}[/itex]

    If I see everything correctly,if you want to move one step further and look at the 4Sphere embedded in 5D space [although i don't know the metric of that space but let's say it's N=diag(-1,1,1,1,1) which is the most natural choice of course ], then you can just add up one more coordinate [itex]x_{5}= \cos w[/itex], and multiply all the previous with [itex]\sin w[/itex] (similiar to what you did with the move from your 2-Sphere to the 3-Sphere...
    then they will add up to unity (because if you take out the [itex]sin^2 w[/itex] from the sum, the factor in front will correspond to the [itex]S^{3}[/itex] and it will give 1, and so you'll remain with [itex] sin^{2} w + cos^{2} w= 1[/itex]). Now to find the metric you just need to know derivatives (now my subscripts a,b run over 1,2,3,4 or [itex]\theta, \phi, \psi, w[/itex]) and making a product...
     
    Last edited: Sep 10, 2014
  4. Sep 10, 2014 #3

    ChrisVer

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    About the S2, what's the parametric representation of the sphere?
    [itex] \vec{x} = \vec{x}(\theta, \phi)[/itex]
    with:
    [itex]\vec{x} (\theta, \phi)= x_{1}(\theta, \phi) \hat{e}_{1} + x_{2} (\theta, \phi) \hat{e}_{2} + x_{3}(x_{1}(\theta, \phi), x_{2}(\theta,\phi)) \hat{e}_{3}= (\cos \phi \sin \theta) \hat{e}_{1} + (\sin \phi \sin \theta) \hat{e}_{2} + \cos \theta \hat{e}_{3}[/itex]

    since [itex]x_{3} = \sqrt{1-x_1^2-x_2^2}= \sqrt{1-sin^2 \theta}[/itex]
     
  5. Sep 10, 2014 #4

    PeterDonis

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    What does the 4 sphere have to do with Minkowski spacetime? Minkowski spacetime is flat.
     
  6. Sep 10, 2014 #5

    ChrisVer

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    Can you help me clarify one thing about that?
    If you write the minkowski metric in spherical coordinates you have:

    [itex] ds^{2} = dt^{2} - dr^{2} - r^{2} d \theta^{2} - r^{2} \sin^{2} \theta d \phi ^{2} [/itex]
    doesn't there exist an [itex]\mathcal{S}^{2}[/itex] in that (the last two terms) (and [itex]\mathcal{S}^{2}[/itex] is not flat)?
     
    Last edited: Sep 10, 2014
  7. Sep 10, 2014 #6

    PeterDonis

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    The last two terms are telling you that, if you have a surface with constant ##t## and ##r##, it is a 2-sphere, and yes, such a surface is not flat. But that just means you can embed a non-flat 2-sphere in flat Minkowski spacetime. It doesn't mean Minkowski spacetime itself is not flat.

    (Note that you can run the same argument for 3-dimensional Euclidean space--just leave out the ##dt^2## part of the metric, and flip the sign so everything is positive. You'll have the same thing: you can embed a non-flat 2-sphere in flat Euclidean 3-space, but the Euclidean 3-space is still flat.)
     
  8. Sep 10, 2014 #7

    ChrisVer

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    Hmmm that sounds legit.. I think I asked a stupid question...
    I wish I could have a program to put the metric [itex]n=diag(1,-1, -r^2 , -r^2 \sin^2 \theta) [/itex] and see by my own eyes if it has zero Riemann Curvature... Because when I wrote the question I thought that [itex]R^{\theta}_{\phi \theta \phi} \ne 0[/itex] (because of the sphere). :)
    thanks
     
  9. Sep 10, 2014 #8

    Matterwave

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    Do you doubt that ##R=0## using the metric ##ds^2=-dt^2+dx^2+dy^2+dz^2##? Or do you doubt that ##R## is a legitimate tensor so that if it is 0 in one set of coordinates, then it is 0 in all sets of coordinates? Or do you doubt that the metric ##ds^2=-dt^2+dr^2+r^2(d\theta^2+sin^2(\theta)d\phi^2)## is really just the Minkowski metric with a change of variables?

    If none of those give you doubt, then you should not doubt that ##R=0## in spherical coordinates. :)
     
  10. Sep 11, 2014 #9

    PeterDonis

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    There are several programs that can do this. I like Maxima, because it's free and it runs on Linux:

    http://en.wikipedia.org/wiki/Maxima_(software [Broken])
     
    Last edited by a moderator: May 6, 2017
  11. Sep 11, 2014 #10

    ChrisVer

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    No, I don't doubt those of course and that's why it confused me and I asked... Because for the 2sphere, the [itex]R^{\theta}_{\phi \theta \phi} \ne 0 [/itex] ... It came in my mind that in the minkowski metric in the spherical coords as I wrote, the [itex]R^{2}_{323} (Mink) = R^{\theta}_{\phi \theta \phi}(S^2)[/itex],,,
    But it's OK now...back to the topic
     
    Last edited: Sep 11, 2014
  12. Sep 12, 2014 #11

    stevendaryl

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    Yes, the 2-sphere has nonzero curvature, although the 2-sphere metric is part of the [itex]R^3[/itex] metric in spherical coordinates, which is flat. The arbitrariness of breaking an [itex]N+1[/itex] dimensional space into an [itex]N[/itex] dimensional space [itex]\times[/itex] a [itex]1[/itex] dimensional space means that you can't judge the flatness of the [itex]N+1[/itex] dimensional part from the flatness of the [itex]N[/itex] dimensional subspace.
     
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