- #1
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Einstein tensor in the FLRW frame - Part 1 of 2
This note develops a formula for the ##G^{00}## component of the Einstein tensor in the FLRW coordinate system for a homogeneous and isotropic spacetime.
We use the convention that tensor indices ##i, j## or ##k## are used only for spatial dimensions (indices 1, 2 and 3), whereas any other pronumeral (whether Greek or Roman) in an index position can stand for any of 0, 1, 2 and 3.
We use FLRW coordinates for this, in which the metric tensor components are all zero except for:
\begin{align*}
g_{00}=-1,\ \ g_{11}=\frac{R^2}{1-kr^2},\ \ g_{22}=R^2r^2,\ \ g_{33}=R^2r^2sin^2\theta
\end{align*}
Here ##R## is the scale factor, which depends only on ##t##.
Since \textbf{g} is diagonal in the FLRW coordinates, so is its inverse, and ##g^{aa}=\frac{1}{g_{aa}}##.
The equation for the Riemann tensor in terms of partial derivatives of Christoffel symbols is as follows (see for example Schutz, equation 6.63):
\begin{align*}
R^\alpha_{\beta\mu\nu}&={\Gamma^\alpha}_{\beta\nu,\mu}-{\Gamma^\alpha}_{\beta\mu,\nu}
+{\Gamma^\alpha}_{\sigma\mu}{\Gamma^\sigma}_{\beta\nu}
-{\Gamma^\alpha}_{\sigma\nu}{\Gamma^\sigma}_{\beta\mu}
\end{align*}
First we calculate all Christoffel symbols, making plentiful use of the identity ##\Gamma^a_{bc}=\Gamma^a_{cb}##:
\begin{align*}
\Gamma^0_{ab}&=\frac{1}{2}g^{0s}(g_{as,b}+g_{bs,a}-g_{ab,s})\\
&=\frac{1}{2}g^{00}(g_{a0,b}+g_{b0,a}-g_{ab,0})\\
&=\frac{1}{2}g^{00}(g_{00,b}+g_{00,a}-\delta^a_b g_{aa,0})\\
&=\frac{1}{2}\delta^a_b g_{aa,0}\textrm{ [since }g_{00}\textrm{ is constant at -1]}\\
\end{align*}
So all off-diagonal elements ##\Gamma^0_{ab}## for ##a\neq b## are ##0##.
Next note that for all ##j>0##:
\begin{align*}
g_{jj,0}&=\frac{\partial}{\partial t}(g_{jj})\\
&=(\frac{g_{jj}}{R^2})\frac{\partial}{\partial t}(R^2)\textrm{ [as }\frac{g_{jj}}{R^2}\textrm{ is independent of }t\textrm{]}\\
&=(\frac{g_{jj}}{R^2})2R\dot{R}\\
&=2\frac{\dot{R}}{R}g_{jj}\\
\end{align*}
so that ##\Gamma^0_{jj}=\frac{\dot{R}}{R}g_{jj}##.
Also we have ##\Gamma^0_{00}=\frac{\partial}{\partial t}(-1)=0##.
\begin{align*}
\Gamma^j_{0k}&=\frac{1}{2}g^{js}(g_{0s,k}+g_{ks,0}-g_{0k,s})\\
&=\frac{1}{2}g^{jj}(g_{0j,k}+g_{kj,0}-g_{00,j}\delta^0_k)\\
&=\frac{1}{2}g^{jj}(g_{00,k}\delta^0_j+g_{jj,0}\delta^k_j+0\times\delta^0_k)\\
&=\frac{1}{2}g^{jj}(0\times\delta^0_j+g_{jj,0}\delta^k_j)\\
&=\frac{1}{2}g^{jj}g_{jj,0}\delta^k_j\\
&=g^{jj}\frac{\dot{R}}{R}g_{jj}\delta^k_j\\
&=\frac{\dot{R}}{R}\delta^k_j\\
\end{align*}
\begin{align*}
\Gamma^1_{jk} &=\frac{1}{2}g^{1s}(g_{js,k}+g_{ks,j}-g_{jk,s})\\
&=\frac{1}{2}g^{11}(g_{j1,k}+g_{k1,j}-g_{jk,1})\\
&=\frac{1}{2}g^{11}(g_{11,k}\delta^1_j+g_{11,j}\delta^1_k-g_{jj,1}\delta^k_j)\\
\end{align*}
Hence
\begin{align*}
\Gamma^1_{11}&=\frac{1}{2}g^{11}(g_{11,1}+g_{11,1}-g_{11,1})\\
&=\frac{1}{2}\frac{(1-kr^2)}{R^2}\frac{\partial}{\partial r}\frac{R^2}{1-kr^2}\\
&=\frac{1}{2}\frac{(1-kr^2)}{R^2}\Big(-\frac{(-2kr)R^2}{(1-kr^2)^2}\Big)\\
&=\frac{kr}{1-kr^2}\\
\end{align*}
And
\begin{align*}
\Gamma^1_{jk} (j\neq k))&=\frac{1}{2}g^{11}(g_{11,k}\delta^1_j+g_{11,j}\delta^1_k-g_{jj,1}\times 0)\\
&=\frac{1}{2}g^{11}(g_{11,k}\delta^1_j+g_{11,j}\delta^1_k)\\
&=0+0
\end{align*}
because if ##j=1## and ##j\neq k## then ##k>1## so ##g_{11,k}=0## and the same applies if we swap ##j## and ##k##.
And
\begin{align*}
\Gamma^1_{22}&=\frac{1}{2}g^{1s}(g_{2s,2}+g_{2s,2}-g_{22,s})\\
&=\frac{1}{2}g^{11}(g_{21,2}+g_{21,2}-g_{22,1})\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}g_{22,1}\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}\frac{\partial}{\partial r}(R^2r^2)\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}(2rR^2)\\
&=-r(1-kr^2)\\
\end{align*}
And
\begin{align*}
\Gamma^1_{33}&=\frac{1}{2}g^{1s}(g_{3s,3}+g_{3s,3}-g_{33,s})\\
&=\frac{1}{2}g^{11}(g_{31,3}+g_{31,3}-g_{33,1})\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}g_{33,1}\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}\frac{\partial}{\partial r}(R^2r^2sin^2\theta)\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}(2rR^2sin^2\theta)\\
&=-r(1-kr^2)sin^2\theta\\
\end{align*}
Now do ##\Gamma^2_{jk}##.
\begin{align*}
\Gamma^2_{jk} &=\frac{1}{2}g^{2s}(g_{js,k}+g_{ks,j}-g_{jk,s})\\
&=\frac{1}{2}g^{22}(g_{j2,k}+g_{k2,j}-g_{jk,2})\\
&=\frac{1}{2}g^{22}(g_{22,k}\delta^2_j+g_{22,j}\delta^2_k-g_{jj,2}\delta^k_j)\\
\end{align*}
Hence
\begin{align*}
\Gamma^2_{11} &=\frac{1}{2}g^{22}(g_{22,1}\delta^2_1+g_{22,1}\delta^2_1-g_{11,2}\delta^1_1)\\
&=\frac{1}{2}g^{22}(0+0-0)\\
&=0
\end{align*}
\begin{align*}
\Gamma^2_{22} &=\frac{1}{2}g^{22}(g_{22,2}+g_{22,2}-g_{22,2})\\
&=\frac{1}{2}\frac{1}{R^2r^2}\frac{\partial}{\partial \theta}(R^2r^2)\\
&=0\\
\end{align*}
\begin{align*}
\Gamma^2_{33} &=\frac{1}{2}g^{22}(g_{22,3}\delta^2_3+g_{22,3}\delta^2_3-g_{33,2}\delta^3_3)\\
&=\frac{1}{2}\frac{1}{R^2r^2}(g_{22,3}\times 0+g_{22,3}\times 0-\frac{\partial}{\partial \theta}(R^2r^2 sin^2\theta))\\
&=-\frac{1}{2}\frac{1}{R^2r^2}(2R^2r^2sin\theta cos\theta)\\
&=-sin\theta cos\theta\\
\end{align*}
Schutz puts an erroneous minus sign in front of this.
And
\begin{align*}
\Gamma^2_{jk} (j\neq k))&=\frac{1}{2}g^{22}(g_{22,k}\delta^2_j+g_{22,j}\delta^2_k-g_{jj,2}\delta^k_j)\\
&=\frac{1}{2}g^{22}(g_{22,k}\delta^2_j+g_{22,j}\delta^2_k+0)\\
\end{align*}
Hence one of the lower indices must be ##2##, and the other must be ##1## or ##3##. This gives us at most the following two nonzero Christoffel symbols of this type (bearing in mind that the symbols are symmetrical over the lower two indices):
\begin{align*}
\Gamma^2_{12} &=\frac{1}{2}g^{22}g_{22,1}\\
&=\frac{1}{2}\frac{1}{R^2r^2}\frac{\partial}{\partial r}(R^2r^2)\\
&=\frac{1}{2}\frac{1}{R^2r^2}(2rR^2)\\
&=\frac{1}{r}\\
\textrm{And}&\\
\Gamma^2_{32} &=\frac{1}{2}g^{22}g_{22,3}\\
&=0\textrm{ by (III)}
\end{align*}
Now do ##\Gamma^3_{jk}##.
\begin{align*}
\Gamma^3_{jk} &=\frac{1}{2}g^{3s}(g_{js,k}+g_{ks,j}-g_{jk,s})\\
&=\frac{1}{2}g^{33}(g_{j3,k}+g_{k3,j}-g_{jk,3})\\
&=\frac{1}{2}g^{33}(g_{33,k}\delta^3_j+g_{33,j}\delta^3_k-0)\\
&\textrm{Because none of the tensor components depend on }\phi\\
&=\frac{1}{2}g^{33}(g_{33,k}\delta^3_j+g_{33,j}\delta^3_k)\\
\end{align*}
Hence
\begin{align*}
\Gamma^3_{11} &=\frac{1}{2}g^{33}(g_{33,k}\delta^3_1+g_{33,j}\delta^3_1)\\
&=\frac{1}{2}g^{33}(0+0)\\
&=0
\end{align*}
\begin{align*}
\Gamma^3_{22} &=\frac{1}{2}g^{22}(g_{22,2}+g_{22,2}-g_{22,2})\\
&=\frac{1}{2}\frac{1}{R^2r^2}\frac{\partial}{\partial \theta}(R^2r^2)\\
&=0\\
\end{align*}
\begin{align*}
\Gamma^3_{33} &=\frac{1}{2}g^{33}(g_{33,3}\delta^3_3+g_{33,3}\delta^3_3-g_{33,3}\delta^3_3)\\
&=\frac{1}{2}g^{33}g_{33,3}\\
&=\frac{\partial}{\partial\phi}(R^2r^2sin^2\theta)\textrm{ [by (III)]}\\
&=0\end{align*}
And
\begin{align*}
\Gamma^3_{jk} (j\neq k))&=\frac{1}{2}g^{33}(g_{33,k}\delta^3_j+g_{33,j}\delta^3_k)
\end{align*}
Hence one of the lower indices must be ##3##, and the other must be ##1## or ##2##. This gives us the following two nonzero Christoffel symbols of this type (bearing in mind that the symbols are symmetrical over the lower two indices):
\begin{align*}
\Gamma^3_{13} &=\frac{1}{2}g^{33}g_{33,1}\\
&=\frac{1}{2}\frac{1}{R^2r^2sin^2\theta}\frac{\partial}{\partial r}(R^2r^2sin^2\theta)\\
&=\frac{1}{2}\frac{1}{R^2r^2sin^2\theta}(2rR^2sin^2\theta)\\
&=\frac{1}{r}
\textrm{And}&\\
\Gamma^3_{23} &=\frac{1}{2}g^{33}g_{33,2}\\
&=\frac{1}{2}\frac{1}{R^2r^2sin^2\theta}\frac{\partial}{\partial \theta}(R^2r^2sin^2\theta)\\
&=\frac{1}{2}\frac{1}{R^2r^2sin^2\theta}(2R^2r^2sin\theta cos\theta)\\
&=cot\theta
\end{align*}
We can summarise these results as follows
\begin{align*}
\Gamma^0_{jj}&=\frac{\dot{R}}{R}g_{jj}\\
\Gamma^j_{0j}&=\frac{\dot{R}}{R}\\
\Gamma^1_{11}&=\frac{kr}{1-kr^2}\\
\Gamma^1_{22}&=-r(1-kr^2)\\
\Gamma^1_{33}&=-r(1-kr^2)sin^2\theta\\
\Gamma^2_{33} &=-sin\theta cos\theta\\
\Gamma^2_{12} &=\frac{1}{r}\\
\Gamma^3_{13} &=\frac{1}{r}\\
\Gamma^3_{23} &=cot\theta
\end{align*}
and all other Christoffel symbols are zero.
V.465.
This note develops a formula for the ##G^{00}## component of the Einstein tensor in the FLRW coordinate system for a homogeneous and isotropic spacetime.
We use the convention that tensor indices ##i, j## or ##k## are used only for spatial dimensions (indices 1, 2 and 3), whereas any other pronumeral (whether Greek or Roman) in an index position can stand for any of 0, 1, 2 and 3.
We use FLRW coordinates for this, in which the metric tensor components are all zero except for:
\begin{align*}
g_{00}=-1,\ \ g_{11}=\frac{R^2}{1-kr^2},\ \ g_{22}=R^2r^2,\ \ g_{33}=R^2r^2sin^2\theta
\end{align*}
Here ##R## is the scale factor, which depends only on ##t##.
Since \textbf{g} is diagonal in the FLRW coordinates, so is its inverse, and ##g^{aa}=\frac{1}{g_{aa}}##.
The equation for the Riemann tensor in terms of partial derivatives of Christoffel symbols is as follows (see for example Schutz, equation 6.63):
\begin{align*}
R^\alpha_{\beta\mu\nu}&={\Gamma^\alpha}_{\beta\nu,\mu}-{\Gamma^\alpha}_{\beta\mu,\nu}
+{\Gamma^\alpha}_{\sigma\mu}{\Gamma^\sigma}_{\beta\nu}
-{\Gamma^\alpha}_{\sigma\nu}{\Gamma^\sigma}_{\beta\mu}
\end{align*}
First we calculate all Christoffel symbols, making plentiful use of the identity ##\Gamma^a_{bc}=\Gamma^a_{cb}##:
\begin{align*}
\Gamma^0_{ab}&=\frac{1}{2}g^{0s}(g_{as,b}+g_{bs,a}-g_{ab,s})\\
&=\frac{1}{2}g^{00}(g_{a0,b}+g_{b0,a}-g_{ab,0})\\
&=\frac{1}{2}g^{00}(g_{00,b}+g_{00,a}-\delta^a_b g_{aa,0})\\
&=\frac{1}{2}\delta^a_b g_{aa,0}\textrm{ [since }g_{00}\textrm{ is constant at -1]}\\
\end{align*}
So all off-diagonal elements ##\Gamma^0_{ab}## for ##a\neq b## are ##0##.
Next note that for all ##j>0##:
\begin{align*}
g_{jj,0}&=\frac{\partial}{\partial t}(g_{jj})\\
&=(\frac{g_{jj}}{R^2})\frac{\partial}{\partial t}(R^2)\textrm{ [as }\frac{g_{jj}}{R^2}\textrm{ is independent of }t\textrm{]}\\
&=(\frac{g_{jj}}{R^2})2R\dot{R}\\
&=2\frac{\dot{R}}{R}g_{jj}\\
\end{align*}
so that ##\Gamma^0_{jj}=\frac{\dot{R}}{R}g_{jj}##.
Also we have ##\Gamma^0_{00}=\frac{\partial}{\partial t}(-1)=0##.
\begin{align*}
\Gamma^j_{0k}&=\frac{1}{2}g^{js}(g_{0s,k}+g_{ks,0}-g_{0k,s})\\
&=\frac{1}{2}g^{jj}(g_{0j,k}+g_{kj,0}-g_{00,j}\delta^0_k)\\
&=\frac{1}{2}g^{jj}(g_{00,k}\delta^0_j+g_{jj,0}\delta^k_j+0\times\delta^0_k)\\
&=\frac{1}{2}g^{jj}(0\times\delta^0_j+g_{jj,0}\delta^k_j)\\
&=\frac{1}{2}g^{jj}g_{jj,0}\delta^k_j\\
&=g^{jj}\frac{\dot{R}}{R}g_{jj}\delta^k_j\\
&=\frac{\dot{R}}{R}\delta^k_j\\
\end{align*}
\begin{align*}
\Gamma^1_{jk} &=\frac{1}{2}g^{1s}(g_{js,k}+g_{ks,j}-g_{jk,s})\\
&=\frac{1}{2}g^{11}(g_{j1,k}+g_{k1,j}-g_{jk,1})\\
&=\frac{1}{2}g^{11}(g_{11,k}\delta^1_j+g_{11,j}\delta^1_k-g_{jj,1}\delta^k_j)\\
\end{align*}
Hence
\begin{align*}
\Gamma^1_{11}&=\frac{1}{2}g^{11}(g_{11,1}+g_{11,1}-g_{11,1})\\
&=\frac{1}{2}\frac{(1-kr^2)}{R^2}\frac{\partial}{\partial r}\frac{R^2}{1-kr^2}\\
&=\frac{1}{2}\frac{(1-kr^2)}{R^2}\Big(-\frac{(-2kr)R^2}{(1-kr^2)^2}\Big)\\
&=\frac{kr}{1-kr^2}\\
\end{align*}
And
\begin{align*}
\Gamma^1_{jk} (j\neq k))&=\frac{1}{2}g^{11}(g_{11,k}\delta^1_j+g_{11,j}\delta^1_k-g_{jj,1}\times 0)\\
&=\frac{1}{2}g^{11}(g_{11,k}\delta^1_j+g_{11,j}\delta^1_k)\\
&=0+0
\end{align*}
because if ##j=1## and ##j\neq k## then ##k>1## so ##g_{11,k}=0## and the same applies if we swap ##j## and ##k##.
And
\begin{align*}
\Gamma^1_{22}&=\frac{1}{2}g^{1s}(g_{2s,2}+g_{2s,2}-g_{22,s})\\
&=\frac{1}{2}g^{11}(g_{21,2}+g_{21,2}-g_{22,1})\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}g_{22,1}\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}\frac{\partial}{\partial r}(R^2r^2)\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}(2rR^2)\\
&=-r(1-kr^2)\\
\end{align*}
And
\begin{align*}
\Gamma^1_{33}&=\frac{1}{2}g^{1s}(g_{3s,3}+g_{3s,3}-g_{33,s})\\
&=\frac{1}{2}g^{11}(g_{31,3}+g_{31,3}-g_{33,1})\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}g_{33,1}\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}\frac{\partial}{\partial r}(R^2r^2sin^2\theta)\\
&=-\frac{1}{2}\frac{(1-kr^2)}{R^2}(2rR^2sin^2\theta)\\
&=-r(1-kr^2)sin^2\theta\\
\end{align*}
Now do ##\Gamma^2_{jk}##.
\begin{align*}
\Gamma^2_{jk} &=\frac{1}{2}g^{2s}(g_{js,k}+g_{ks,j}-g_{jk,s})\\
&=\frac{1}{2}g^{22}(g_{j2,k}+g_{k2,j}-g_{jk,2})\\
&=\frac{1}{2}g^{22}(g_{22,k}\delta^2_j+g_{22,j}\delta^2_k-g_{jj,2}\delta^k_j)\\
\end{align*}
Hence
\begin{align*}
\Gamma^2_{11} &=\frac{1}{2}g^{22}(g_{22,1}\delta^2_1+g_{22,1}\delta^2_1-g_{11,2}\delta^1_1)\\
&=\frac{1}{2}g^{22}(0+0-0)\\
&=0
\end{align*}
\begin{align*}
\Gamma^2_{22} &=\frac{1}{2}g^{22}(g_{22,2}+g_{22,2}-g_{22,2})\\
&=\frac{1}{2}\frac{1}{R^2r^2}\frac{\partial}{\partial \theta}(R^2r^2)\\
&=0\\
\end{align*}
\begin{align*}
\Gamma^2_{33} &=\frac{1}{2}g^{22}(g_{22,3}\delta^2_3+g_{22,3}\delta^2_3-g_{33,2}\delta^3_3)\\
&=\frac{1}{2}\frac{1}{R^2r^2}(g_{22,3}\times 0+g_{22,3}\times 0-\frac{\partial}{\partial \theta}(R^2r^2 sin^2\theta))\\
&=-\frac{1}{2}\frac{1}{R^2r^2}(2R^2r^2sin\theta cos\theta)\\
&=-sin\theta cos\theta\\
\end{align*}
Schutz puts an erroneous minus sign in front of this.
And
\begin{align*}
\Gamma^2_{jk} (j\neq k))&=\frac{1}{2}g^{22}(g_{22,k}\delta^2_j+g_{22,j}\delta^2_k-g_{jj,2}\delta^k_j)\\
&=\frac{1}{2}g^{22}(g_{22,k}\delta^2_j+g_{22,j}\delta^2_k+0)\\
\end{align*}
Hence one of the lower indices must be ##2##, and the other must be ##1## or ##3##. This gives us at most the following two nonzero Christoffel symbols of this type (bearing in mind that the symbols are symmetrical over the lower two indices):
\begin{align*}
\Gamma^2_{12} &=\frac{1}{2}g^{22}g_{22,1}\\
&=\frac{1}{2}\frac{1}{R^2r^2}\frac{\partial}{\partial r}(R^2r^2)\\
&=\frac{1}{2}\frac{1}{R^2r^2}(2rR^2)\\
&=\frac{1}{r}\\
\textrm{And}&\\
\Gamma^2_{32} &=\frac{1}{2}g^{22}g_{22,3}\\
&=0\textrm{ by (III)}
\end{align*}
Now do ##\Gamma^3_{jk}##.
\begin{align*}
\Gamma^3_{jk} &=\frac{1}{2}g^{3s}(g_{js,k}+g_{ks,j}-g_{jk,s})\\
&=\frac{1}{2}g^{33}(g_{j3,k}+g_{k3,j}-g_{jk,3})\\
&=\frac{1}{2}g^{33}(g_{33,k}\delta^3_j+g_{33,j}\delta^3_k-0)\\
&\textrm{Because none of the tensor components depend on }\phi\\
&=\frac{1}{2}g^{33}(g_{33,k}\delta^3_j+g_{33,j}\delta^3_k)\\
\end{align*}
Hence
\begin{align*}
\Gamma^3_{11} &=\frac{1}{2}g^{33}(g_{33,k}\delta^3_1+g_{33,j}\delta^3_1)\\
&=\frac{1}{2}g^{33}(0+0)\\
&=0
\end{align*}
\begin{align*}
\Gamma^3_{22} &=\frac{1}{2}g^{22}(g_{22,2}+g_{22,2}-g_{22,2})\\
&=\frac{1}{2}\frac{1}{R^2r^2}\frac{\partial}{\partial \theta}(R^2r^2)\\
&=0\\
\end{align*}
\begin{align*}
\Gamma^3_{33} &=\frac{1}{2}g^{33}(g_{33,3}\delta^3_3+g_{33,3}\delta^3_3-g_{33,3}\delta^3_3)\\
&=\frac{1}{2}g^{33}g_{33,3}\\
&=\frac{\partial}{\partial\phi}(R^2r^2sin^2\theta)\textrm{ [by (III)]}\\
&=0\end{align*}
And
\begin{align*}
\Gamma^3_{jk} (j\neq k))&=\frac{1}{2}g^{33}(g_{33,k}\delta^3_j+g_{33,j}\delta^3_k)
\end{align*}
Hence one of the lower indices must be ##3##, and the other must be ##1## or ##2##. This gives us the following two nonzero Christoffel symbols of this type (bearing in mind that the symbols are symmetrical over the lower two indices):
\begin{align*}
\Gamma^3_{13} &=\frac{1}{2}g^{33}g_{33,1}\\
&=\frac{1}{2}\frac{1}{R^2r^2sin^2\theta}\frac{\partial}{\partial r}(R^2r^2sin^2\theta)\\
&=\frac{1}{2}\frac{1}{R^2r^2sin^2\theta}(2rR^2sin^2\theta)\\
&=\frac{1}{r}
\textrm{And}&\\
\Gamma^3_{23} &=\frac{1}{2}g^{33}g_{33,2}\\
&=\frac{1}{2}\frac{1}{R^2r^2sin^2\theta}\frac{\partial}{\partial \theta}(R^2r^2sin^2\theta)\\
&=\frac{1}{2}\frac{1}{R^2r^2sin^2\theta}(2R^2r^2sin\theta cos\theta)\\
&=cot\theta
\end{align*}
We can summarise these results as follows
\begin{align*}
\Gamma^0_{jj}&=\frac{\dot{R}}{R}g_{jj}\\
\Gamma^j_{0j}&=\frac{\dot{R}}{R}\\
\Gamma^1_{11}&=\frac{kr}{1-kr^2}\\
\Gamma^1_{22}&=-r(1-kr^2)\\
\Gamma^1_{33}&=-r(1-kr^2)sin^2\theta\\
\Gamma^2_{33} &=-sin\theta cos\theta\\
\Gamma^2_{12} &=\frac{1}{r}\\
\Gamma^3_{13} &=\frac{1}{r}\\
\Gamma^3_{23} &=cot\theta
\end{align*}
and all other Christoffel symbols are zero.
V.465.