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

[wildemar]

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" .

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]

where a is the scale factor for the usual spatial 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:

[itex]
R_{a0a0} = a \ddot{a}
[/itex]

where the index a (not that clevery chosen, if you ask me) is any of the indices in the usual spatial 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

[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 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.
regards,
/W

 

[cristo]

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

[tex]ds^2=-dt^2+a^2(t)g_{IJ}dx^Idx^J+b^2(t)\gamma_{ij}dx^idx^j[/tex]

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

[tex]ds^2=-dt^2+a^2(t)\delta_{IJ}dx^Idx^J+b^2(t)\delta_{ij}dx^idx^j[/tex].

If you use the connection coefficients derived from this metric, does it produce the results in the paper?

 

[wildemar]

Holy Moly, that was quick!

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

[tex]ds^2=-dt^2+a^2(t)g_{IJ}dx^Idx^J+b^2(t)\gamma_{ij}dx^idx^j[/tex]

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].

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.

Now, invoking flatness in both spatial parts simply gives the new line element

[tex]ds^2=-dt^2+a^2(t)\delta_{IJ}dx^Idx^J+b^2(t)\delta_{ij}dx^idx^j[/tex].

If you use the connection coefficients derived from this metric, does it produce the results in the paper?

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. :)

regards
/W

 

[cristo]

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".

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!

 

[wildemar]

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.

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.