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I FLRW metric derivation

  1. Dec 15, 2016 #1
    Why is it needed to consider a 4D Euclidean space to introduce the FLRW metric? Is it because with a fourth parameter, we can set the radius of the 4D sphere formed with the four parametres as constant?
     
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  3. Dec 15, 2016 #2

    vanhees71

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    Why do you need to consider a 4D Euclidean space to introduce the FLRW metric? It's a Lorentzian manifold, and you just ask for the spacetimes with maximally symmetric spaces.
     
  4. Dec 15, 2016 #3
    Thanks
     
  5. Dec 15, 2016 #4

    PeterDonis

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    We don't. The FLRW metric is locally Lorentzian, not Euclidean. Why do you think a 4D Euclidean space is needed?

    (Btw, vanhees71 was asking you the same question. He was not answering your question in the OP; he was telling you the question was based on a false assumption.)
     
  6. Dec 16, 2016 #5
    Ah :biggrin:

    Because I found two derivations where the author introduces such a space.
    One of them is here:
    http://www2.warwick.ac.uk/fac/sci/physics/current/teach/module_home/px436/notes/lecture20.pdf

    The another derivation I found in a book I've read.
     
  7. Dec 16, 2016 #6

    martinbn

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    But you need to read it more carefully. The 4D Euclidean space is introduced in order to define a 3D sphere, it is not the space-time in the FRWL model. That 3D sphere is the constant cosmological time surface. Think of the sphere through all time, that forms the space-time manifold, which is not Euclidean.
     
  8. Dec 16, 2016 #7

    vanhees71

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    As usual, Weinberg is much more straight forward. He just derives the maximally symmetric 3D spaces of constant curvature using symmetry. Symmetry is anyway the key to all of physics. So it's well invested time to study it also in the context of GR (keyword: Killing vectors).
     
  9. Dec 16, 2016 #8

    martinbn

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    In the section "Maximally symmetric spaces. Constructions", Weinberg constructs them the same way, via embedings in a ##n+1## dimensional Euclidean space.
     
  10. Dec 16, 2016 #9

    vanhees71

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    I guess you mean n+1-dimensional affine (flat) space. The negative-curvature case cannot be completely embedded in an Euclidean space. If I remember right, Weinberg discusses this within the mentioned chapter too.
     
  11. Dec 16, 2016 #10

    martinbn

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    It is the hyperboloid ##-x_0^2+x_1^2+\cdots+x_n^2=0##.

    My point was that he also does that. I assumed that when you said that his method was more straight forward you meant that he doesn't construct them the same way i.e. as hypersurfaces in an Euclidean space of one more dimension.
     
  12. Dec 16, 2016 #11

    vanhees71

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    Maybe I'm wrongly attributing the derivation to Weinberg (I've to check as soon as I'm back home), but hasn't he also given the derivation of finding systematically the spaces with maximal symmetry via the Killing equation?
     
  13. Dec 16, 2016 #12

    martinbn

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    Well, he shows that they have constant curvature and that the curvature and the signature of the metric uniquely determine the space (at least locally and up to an isometry). But to find them he says that since they are unique it suffices to construct them any way we want, and he does the same thing as in the lectures in the link above.
     
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