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Robertson-Walker metric in higher dimensions (and problematic Riemann tensor)

  1. Aug 22, 2008 #1
    Hello folks,

    this is going to be a bit longish, but please bear with me, I'm going nuts over this.

    For a term paper I am working through a paper on higher dimensional spacetimes by Andrew, Bolen and Middleton. You can http://arxiv.org/abs/0708.0373" [Broken].

    My problem/confusion is in calculating the Riemann tensor from the given metric (basically a Robertson-Walker metric with additional dimensions), which reads:

    {\mathrm{d}s}^2 = -{\mathrm{d}t}^2 + a^2(t) \left[ \frac{{\mathrm{d}r}^2}{1-K r^2} + r^2 \left( {\mathrm{d}\theta}^2 + \sin^2\theta {\mathrm{d}\phi}^2 \right) \right] + b^2(t) \gamma_{m n}(y) {\mathrm{d}y}^m {\mathrm{d}y}^n

    where a is the scale factor for the usual spacial dimensions and b the same for the extra dimensions, their amount being some integer d. [itex]\gamma_{m n}[/itex] is the part of the metric in the extra dimensions.

    The assumptions are that both parts of the dimensions are flat. This means of course that K=0. And for the (of course maximally symmetric) Riemann tensor of the extra dimensions, [itex]R_{abcd} = k (\gamma_{ac}\gamma_{bd} - \gamma_{ad}\gamma_{bc})[/itex], this means k=0.

    They now go on to compute the components of the Riemann tensor, one of which reads:

    R_{a0a0} = a \ddot{a}

    where the index a (not that clevery chosen, if you ask me) is any of the indices in the usual spacial dimensions, that is a=1,2,3.

    OK, this is what they do. I hope you've been following so far.

    Now I'm trying to get these results myself. My professor said that the flatness of the extra dimensions means that [itex]\gamma_{m n}[/itex] is "essentially the unit matrix". Oh well, since I didn't (and still don't) have anything else to go on, I chose to believe and use that notion, as well as setting K=0.
    (In order to be able to calculate something at all, I added two extra dimensions to the usual four. That means that I set

    \gamma = \left(
    1 & 0 \\
    0 & 1

    So with this ansatz I calculated the Riemann components, but I found that while

    R_{1010} = a \ddot{a}

    as in the paper, for the second spatial index I get

    R_{2020} = r^2 a \ddot{a}


    R_{3030} = r^2 \sin^2\theta \ a \ddot{a}

    for the third. I think you can see the pattern here.

    So, guided by this pattern and as a form of sanity check I dumped the Robertson-Walker inspired metric in favor of a Minkowski metric for the first 4 dimensions while leaving [itex]\gamma[/itex] as it was (in other words, I used diag(-1,1,1,1,1,1) as the metric). And guess what? I get the exact same result as they do in the paper, as was to be expected by this point.

    I hope you can see why I am confused by this. It would seem to me that they used a Robertson-Walker metric to get the Riemann tensor for a Minkowski metric. What is it that I'm missing here?

    And need I say this everything gets infinitely more complicated when I leave [itex]\gamma[/itex] undefined and actually make it dependent on y, as they say in the original definition of the metric (see above)? I just don't know how I would impose the "flatness" of the additional space as a restriction on [itex]\gamma[/itex]. I'm completely stumped there.

    In the same vein: When my teacher said that a flat spacetime essentially means a unit matrix I didn't think much about it, but it doesn't make a whole lot of sense to me, considering that the (spatial part of the) Robertson-Walker metric is certainly not the unit matrix, not even when you have a flat space (K=0). So why did she say that, or rather, what did she really mean by that? (She's away for some time now, so I can't ask her).

    OK, this is my long story.
    Any ideas, I could really use them. Thanks in advance.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 22, 2008 #2


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    I've not read this properly, so could be wrong, but I would guess that the problem is the interpretation where they say "the index a runs from 1,...,3." I bet they've switched back to cartesian type coordinates. That is, the original line element can be written


    where I've written [itex]g_{IJ}[/itex] to be the spatial part of FRW, and I,J run over 1,2,3; i,j run over 4,...,n for the n dimensional extra spatial metric [itex]\gamma_{ij}[/itex]. Now, invoking flatness in both spatial parts simply gives the new line element


    If you use the connection coefficients derived from this metric, does it produce the results in the paper?
  4. Aug 22, 2008 #3
    Holy Moly, that was quick!

    I think they did. And part of my problem thus far was probably that I assumed that the Robertson Walker metric had to be in spherical coordinates, because of the isotropy of space (D'Inverno, whom I've used to learn GR/cosmology, used this argument for introducing spherical coordinates). It never occurred to me that the spherical part is (or rather: can be) just chosen for convenience.

    I actually did that, as described in my post. But I can't blame you for missing it. :) I actually got the correct results with that, but didn't understand why. I guess I do now though. I'm currently sort of feeling my way in GR and there are still a lot of things that I'm not sure are "OK to do".

    So thanks for answering all my questions in a batch. Turns out that all I needed was some confirmation. :)

  5. Aug 22, 2008 #4


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    Actually, it looks like that you did was to take a Minkowski-esqe metric, the diag(-1111111), instead of the above metric with the functions a and b in.

    Anyway, I'm glad you got the answer!
  6. Aug 22, 2008 #5
    The advice from one of my professors rings as true as ever: I need to be more precise. I of course did what you said, not what I said i did. *sigh*

    OK then, thanks again.
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