Ricci-Tensor from Riemann in higher dimensional flat space

  • Thread starter wildemar
  • Start date
20
0
This is driving me nuts.
(I originally posted this to the coursework section, but in thinking about this, I felt that it might not be the right place (this is for a term paper, not really any ongoing coursework, so there). Hope I'm not imposing ... I feel quite embarrassed on this one, since it should be simple. Enough of my babbling.)

Mission statement
The problem deals with an extension to the Robertson-Walker Metric for a flat space. To the usual 3 spatial dimensions with scale factor a(t), d more are appended with another scale factor b(t).

I thus have the following line-element:
[itex]
\mathrm{d}s^2 = - \mathrm{d}t^2 + a(t)^2 \sum_{\alpha=1}^{3} (\mathrm{d} x^\alpha)^2+ b(t)^2 \sum_{\mu=1}^{3} (\mathrm{d} x^\mu)^2
[/itex]

I need to calculate the Ricci-Tensor for this.

My attempt at a solution
For the given metric, I get the following Riemann (we can assume that this is right, Maple says so, but feel free to correct me (the paper I'm using as a reference has positive signs for the first two components listed):
[itex]
R_{0 \alpha 0 \alpha} = -a \ddot{a} \quad\quad
R_{0 \mu 0 \mu} = -b \ddot{b} \quad\quad
R_{\alpha\beta\alpha\beta} = a^2 \dot{a}^2 \quad\quad
R_{\mu\nu\mu\nu} = b^2 \dot{b}^2 \quad\quad
R_{\alpha\mu\alpha\mu} = a \dot{a} b \dot{b}
[/itex]
(Note that alpha and beta run over the usual 3 spatial dimensions while mu and nu run over the d additional ones. All Latin indices run over all dimensions available, including time.)

I then said (no summation convention here)
[tex]
\begin{align*}
R_{ab}
&= \sum_{c,d=0}^{4+d} g^{cd} R_{cadb}
= \sum_{c=0}^{4+d} g^{cc} R_{cacb} \\
&= (-1) \cdot R_{0a0b}
+ \sum_{\alpha = 1}^{3} a^{-2} R_{\alpha a \alpha b}
+ \sum_{\mu = 4}^{3+d} b^{-2} R_{\mu a \mu b} \\
&= (-1) \cdot R_{0 a 0 b}
+ \quad 3 \, \frac{R_{\alpha a \alpha b}}{a^2} \quad\
+ \quad d \, \frac{R_{\mu a \mu b}}{b^2}
\end{align*}
[/tex]
where the Greek indices in the last line are symbolic in the way that you can use any value that is in line with the above convention as all those components are equal.

So with this I get
[tex]
\begin{align*}
R_{00} &= -3\frac{\ddot{a}}{a} - d \frac{\ddot{b}}{b} \\
R_{\alpha\alpha} &= a \ddot{a} + 3 \dot{a}^2 + d \frac{a \dot{a} \dot{b}}{b} \\
R_{\mu\mu} &= b \ddot{b} + 3 \frac{\dot{a} b \dot{b}}{a} + d \dot{b}^2
\end{align*}
[/tex]


The Problem
Both Maple and the paper I'm working through say that in the second line it should read 2 instead of 3 and in the third d-1 instead of d (I apparently can't use any additional latex in this post, so you'll have to use your imaginations).
That is, in the last two lines, the "pure" spatial terms (those that have no mixed indices of normal and higher dimensions) occur on time less than what I calculate.

I can't for the life of me find the error I'm making. This isn't a complicated calculation at all, but I'm completely stumped.
 
20
0
Uhm, anyone?

If what I'm writing is unclear, or even if you think what I do is correct, I'd really like to know. Sorry to bother you.

Maybe this is more appropriate in the GR section? If so, can I move the thread?
 
Last edited:

nrqed

Science Advisor
Homework Helper
Gold Member
3,540
181
Uhm, anyone?

If what I'm writing is unclear, or even if you think what I do is correct, I'd really like to know. Sorry to bother you.

Maybe this is more appropriate in the GR section? If so, can I move the thread?
I just saw your post. I have been doing tons of calculations similar to these for calculations in brane world models (and I do everything by hand) so I will check your question tomorrow (it's too late now, I need to go to sleep)

Regards
 

nrqed

Science Advisor
Homework Helper
Gold Member
3,540
181
Uhm, anyone?

If what I'm writing is unclear, or even if you think what I do is correct, I'd really like to know. Sorry to bother you.

Maybe this is more appropriate in the GR section? If so, can I move the thread?
Hi,


I only had time to look at your calculation this afternoon, sorry about the delay.


The reason is this: The components of the Riemann tensor are zero whenever the four indices are identical. So [tex] R_{1111} = R_{2222} = R_{3333} .... = 0 [/tex]

Therefore you have to be careful. When you give the expressions for [tex] R_{\alpha \beta \alpha \beta} [/tex] for example, it is valid only if [tex] \alpha \neq \beta [/tex].


Now, look at when you calculate the Ricci tensor with two spatial indices. In that case, you cannot write

[tex] \sum_{\alpha=1}^3 a^{-2} R_{\alpha a \alpha b} = 3 \frac{R_{\alpha a \alpha b}}{a^2} [/tex]

It's clear if you write it out. For example consider [tex] R_{11} [/tex]
We should get

[tex] R_{11} = \frac{1}{a^2} \bigl( R_{2121} + R_{3131} \bigr) = 2 \dot{a}^2 [/tex]

i.e., the term [tex] R_{1111} [/tex] must not be included.

The same problem arises when you find R with two [tex] \mu [/tex] indices. In that case, the sum over the mu that you wrote should not include the term [tex] R_{\mu \mu \mu \mu} [/tex] so the factor in front of the sum is (d-1), not d.


I hope this helps. It is tricky. And since most people just use Maple, they don't have to worry about this kind of subtleties. Since I have done tons of calculations by hand, I have had to deal with this kind of tricky issues very often and had nobody to help. I hope that my hours of struggle will have helped you out a bit!


Patrick
 
20
0
*slaps forehead*

Man! Thanks Patrick. That was it, of course. Gee, there it was, right before my eyes and I didn't see. You're the man now, dog! :)

Really, I can not express how grateful I am. Now I can finally start doing some serious calculations again.

regards,
/W
 

Related Threads for: Ricci-Tensor from Riemann in higher dimensional flat space

Replies
5
Views
6K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
3K
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
3
Views
2K
Replies
3
Views
884
  • Last Post
Replies
7
Views
10K
Top