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FRW metric, parameter k characterizes space or space-time?

  1. Feb 28, 2015 #1
    This is probably a stupid question but does k=1,0,-1 correspond to closed,flat,open refer to space or space-times?

    Looking at a derivation what each geometrically represents is only done when talking about the spatial part of the FRW metric.

    As space can be flat and space-time still curved couldn't the , say k=0, space be flat, but space-time not flat.
    Why is it that we say k=0 gives a flat 'universe' etc.

    Thanks in advance
     
  2. jcsd
  3. Feb 28, 2015 #2
    First off, space and time are the same thing essentially. You can have one without the other, but then it's completely pointless.

    Secondly space-time is not necessarily flat, that's just the easiest way to express it. Space-time occupies all things, as a field.

    To elaborate on the need for each other. With only space, you just have a area of static objects, and since light can't move, you can't see any of them nor think. With only time, well you still have nothing as there's no room for anything.
     
  4. Feb 28, 2015 #3

    wabbit

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    Space.
    True.
    Right - Flat spacetime would mean no gravity, so pretty boring cosmology. Current observations suggest our (observable) universe is in fact close to spatially flat - but by no means is our spacetime expected to be flat.
     
  5. Feb 28, 2015 #4

    ChrisVer

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    The parameter [itex]k[/itex] corresponds to the spatial-curvature if you take a slice of the hypersurface of spacetime (i.e. [itex]dt=0[/itex]), and not the spacetime curvature.
     
  6. Feb 28, 2015 #5

    ChrisVer

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    I don't understand what you mean by that question. Looking at the FRW metric in the comoving coordinate form, you can see that in the case of k=0, you get:
    [itex]ds^2 = dt^2 - (dr^2 + r^2 d \theta^2 + r^2 \sin^2 \theta d \phi^2 )[/itex]
    The part in the parenthesis is just an Euclidean space (flat) written in spherical coordinates...If you make the well known transformation to Cartesian coordinates x,y,z you will have:
    [itex]ds^2 = dt^2 - dx^2 - dy^2 -dz^2 [/itex]
    Geodesics are just straight lines.

    That is not true for any other choice of k.
     
  7. Feb 28, 2015 #6
    So the books I'm looking at on dynamics of the universe for a given matter content, say dust/radiation before stating the solution says 'for k=0, 'the flat universe'...' 'for k=1, 'the closed universe'..'

    My question is the use of the word 'universe' , i interpret as space-time and not space being called flat.
     
  8. Feb 28, 2015 #7

    ChrisVer

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    No, it's the spatial part of the Universe. The names "closed/open/flat" have to do with the geodesic solutions for example of [itex]x^1[/itex] (being hyperbolic, linear or periodic).Whether you allow for time to be in the game, these solutions still hold- that's why some texts deal with the spatial FRW metric independently at first, derive the solutions, and then when they write the time coordinate , they say that these solutions still apply.
     
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