- #1
wildemar
- 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.
(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.