Hello folks,(adsbygoogle = window.adsbygoogle || []).push({});

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:

[itex]

{\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

[/itex]

whereais the scale factor for the usual spacial dimensions andbthe same for the extra dimensions, their amount being some integerd. [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 thatK=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 meansk=0.

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

[itex]

R_{a0a0} = a \ddot{a}

[/itex]

where the indexa(not that clevery chosen, if you ask me) is any of the indices in the usual spacial dimensions, that isa=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 settingK=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

[itex]

\gamma = \left(

\begin{array}{cc}

1 & 0 \\

0 & 1

\end{array}

\right)

[/itex]

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

[itex]

R_{1010} = a \ddot{a}

[/itex]

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

[itex]

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

[/itex]

and

[itex]

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

[/itex]

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 useddiag(-1,1,1,1,1,1)as the metric). And guess what? I get theexact same resultas 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 ony, 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 shereallymean 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.

regards,

/W

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

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