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Einstein tensor in the FLRW frame

  1. Aug 23, 2014 #1

    andrewkirk

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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.
     
  2. jcsd
  3. Aug 23, 2014 #2

    andrewkirk

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    Einstein tensor in the FLRW frame - Part 2

    Einstein tensor in the FLRW frame - Part 2 of 2

    NOTE - this is continued from Part 1, which needs to be read first.

    Now we calculate the Ricci tensor components. To do this, in each case we first need to calculate the Riemann tensor components, using:
    \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*}

    Now calculate the components of the Ricci tensor component ##R_{00}##:

    \begin{align*}
    R^0_{000}&={\Gamma^0}_{00,0}-{\Gamma^0}_{00,0}
    +{\Gamma^0}_{\sigma0}{\Gamma^\sigma}_{00}
    -{\Gamma^0}_{\sigma0}{\Gamma^\sigma}_{00}\\
    &=0\\
    \end{align*}
    For ##i>1## we have:
    \begin{align*}
    R^i_{0i0}&={\Gamma^i}_{00,i}-{\Gamma^i}_{0i,0}
    +{\Gamma^i}_{\sigma i}{\Gamma^\sigma}_{00}
    -{\Gamma^i}_{\sigma 0}{\Gamma^\sigma}_{0i}\\
    &=(0)_{,i}-\frac{\partial}{\partial t}(\frac{\dot{R}}{R})
    +{\Gamma^i}_{\sigma i}\times 0
    -{\Gamma^i}_{i0}{\Gamma^i}_{0i}\\
    &=-(\frac{\ddot{R}R-\dot{R}^2}{R^2})
    -(\frac{\dot{R}}{R})^2\\
    &=-\frac{\ddot{R}R}{R^2}\\
    \end{align*}

    Hence ##R_{00}=-3\frac{\ddot{R}R}{R^2}##.

    Now calculate the components of the Ricci tensor component ##R_{11}##:

    \begin{align*}
    R^0_{101}&={\Gamma^0}_{11,0}-{\Gamma^0}_{10,1}
    +{\Gamma^0}_{s0}{\Gamma^s}_{11}
    -{\Gamma^0}_{s1}{\Gamma^s}_{10}\\
    &=\frac{\partial}{\partial t}(\frac{\dot{R}}{R}g_{11})-(0)_{,1}
    +0\times{\Gamma^s}_{11}
    -{\Gamma^0}_{11}{\Gamma^1}_{10}\\
    &=\frac{\partial}{\partial t}(\frac{\dot{R}}{R}\frac{R^2}{1-kr^2})-0+0
    -(\frac{\dot{R}}{R}\frac{R^2}{1-kr^2})(\frac{\dot{R}}{R})\\
    &=\frac{1}{1-kr^2}\Big(\frac{\partial}{\partial t}(\dot{R}R)
    -\dot{R}^2\Big)\\
    &=\frac{1}{1-kr^2}\Big((\ddot{R}R+\dot{R}^2)
    -\dot{R}^2\Big)\\
    &=\frac{\ddot{R}R}{1-kr^2}\\
    \end{align*}
    \begin{align*}
    R^1_{111}&={\Gamma^1}_{11,1}-{\Gamma^1}_{11,1}
    +{\Gamma^1}_{\sigma1}{\Gamma^\sigma}_{11}
    -{\Gamma^1}_{\sigma1}{\Gamma^\sigma}_{11}\\
    &=0\\
    \end{align*}
    If ##i=2## or ##i=3## we have:
    \begin{align*}
    R^i_{1i1}&={\Gamma^i}_{11,i}-{\Gamma^i}_{1i,1}
    +{\Gamma^i}_{si}{\Gamma^s}_{11}
    -{\Gamma^i}_{s1}{\Gamma^s}_{1i}\\
    &=(0)_{,i}-\frac{\partial}{\partial r}(\frac{1}{r})
    +\Big({\Gamma^i}_{0i}{\Gamma^0}_{11}+{\Gamma^i}_{1i}{\Gamma^1}_{11}\Big)
    -{\Gamma^i}_{i1}{\Gamma^i}_{1i}\\
    &=0-(-\frac{1}{r^2})
    +\Big((\frac{\dot{R}}{R})(\frac{\dot{R}}{R}g_{11})+(\frac{1}{r})(\frac{kr}{1-kr^2})\Big)
    -(\frac{1}{r})^2\\
    &=\Big((\frac{\dot{R}}{R})^2\frac{R^2}{1-kr^2}+\frac{k}{1-kr^2}\Big)\\
    &=\frac{\dot{R}^2+k}{1-kr^2}\\
    \end{align*}

    Hence ##R_{11}=\frac{\ddot{R}R+2(\dot{R}^2+k)}{1-kr^2}##.

    Now calculate the components of the Ricci tensor component ##R_{22}##:

    \begin{align*}
    R^0_{202}&={\Gamma^0}_{22,0}-{\Gamma^0}_{20,2}+{\Gamma^0}_{\sigma0}{\Gamma^\sigma}_{22}
    -{\Gamma^0}_{\sigma2}{\Gamma^\sigma}_{20}\\
    &=\frac{\partial}{\partial t}(\frac{\dot{R}}{R}g_{22})-(0)_{,2}
    +0\times{\Gamma^\sigma}_{22}
    -{\Gamma^0}_{22}{\Gamma^2}_{20}\\
    &=\frac{\partial}{\partial t}((\frac{\dot{R}}{R})(R^2r^2))
    -(\frac{\dot{R}}{R}g_{22})(\frac{\dot{R}}{R})\\
    &=\frac{\partial}{\partial t}((\frac{\dot{R}}{R})(R^2r^2))
    -(\frac{\dot{R}}{R})^2(R^2r^2))\\
    &=\frac{\partial}{\partial t}(\dot{R}Rr^2)-\dot{R}^2r^2\\
    &=(\ddot{R}R+\dot{R}^2)r^2-\dot{R}^2r^2\\
    &=\ddot{R}Rr^2\\
    \end{align*}
    \begin{align*}
    R^1_{212}&={\Gamma^1}_{22,1}-{\Gamma^1}_{21,2}+{\Gamma^1}_{\sigma1}{\Gamma^\sigma}_{22}
    -{\Gamma^1}_{\sigma2}{\Gamma^\sigma}_{21}\\
    &=\frac{\partial}{\partial r}(-r(1-kr^2))-(0)_{,2}
    +({\Gamma^1}_{01}{\Gamma^0}_{22}
    +{\Gamma^1}_{11}{\Gamma^1}_{22})
    -{\Gamma^1}_{22}{\Gamma^2}_{21}\\
    &=(-1+3kr^2)
    +\Big(\frac{\dot{R}}{R}(\frac{\dot{R}}{R}g_{22})
    +\frac{kr}{1-kr^2}(-r(1-kr^2))\Big)
    -(-r(1-kr^2))\frac{1}{r}\\
    &=-1+3kr^2
    +\frac{\dot{R}}{R}(\frac{\dot{R}}{R}(R^2r^2))
    -kr^2
    +1-kr^2\\
    &=kr^2+\dot{R}^2r^2
    \end{align*}
    \begin{align*}
    R^2_{222}&={\Gamma^2}_{22,2}-{\Gamma^2}_{22,2}
    +{\Gamma^2}_{\sigma2}{\Gamma^\sigma}_{22}
    -{\Gamma^2}_{\sigma2}{\Gamma^\sigma}_{22}\\
    &=0\\
    \end{align*}
    \begin{align*}
    R^3_{232}&={\Gamma^3}_{22,3}-{\Gamma^3}_{23,2}
    +{\Gamma^3}_{\sigma3}{\Gamma^\sigma}_{22}
    -{\Gamma^3}_{\sigma2}{\Gamma^\sigma}_{23}\\
    &=(0)_{,3}-\frac{\partial}{\partial\theta}(cot\theta)
    +({\Gamma^3}_{13}{\Gamma^1}_{22}+{\Gamma^3}_{03}{\Gamma^0}_{22})
    -({\Gamma^3}_{32}{\Gamma^3}_{23})\\
    &=-(-\frac{1}{sin^2\theta})
    +\Big((\frac{1}{r})(-r(1-kr^2))+(\frac{\dot{R}}{R})(\frac{\dot{R}}{R}g_{22})\Big)
    -(cot\theta)(cot\theta)\\
    &=\frac{1}{sin^2\theta}
    +\Big(-(1-kr^2)+(\frac{\dot{R}}{R})^2(R^2r^2))\Big)
    -cot^2\theta\\
    &=(cosec^2\theta-cot^2\theta) -(1-kr^2)+(\dot{R}^2r^2)\\
    &=(1) -1+kr^2+\dot{R}^2r^2\\
    &=kr^2+\dot{R}^2r^2\\
    \end{align*}

    Hence ##R_{22}=\ddot{R}Rr^2+2r^2(k+\dot{R}^2)##.

    Now calculate the components of the Ricci tensor component ##R_{33}##:

    \begin{align*}
    R^0_{303}&={\Gamma^0}_{33,0}-{\Gamma^0}_{30,3}+{\Gamma^0}_{\sigma0}{\Gamma^\sigma}_{33}
    -{\Gamma^0}_{\sigma3}{\Gamma^\sigma}_{30}\\
    &=\frac{\partial}{\partial t}(\frac{\dot{R}}{R}g_{33})-(0)_{,3}
    +0\times{\Gamma^\sigma}_{33}
    -{\Gamma^0}_{33}{\Gamma^3}_{30}\\
    &=\frac{\partial}{\partial t}(\frac{\dot{R}}{R}(R^2r^2sin^2\theta))
    -(\frac{\dot{R}}{R}g_{33})(\frac{\dot{R}}{R})\\
    &=(r^2sin^2\theta)\frac{\partial}{\partial t}(\dot{R}R)
    -(\frac{\dot{R}}{R})^2(R^2r^2sin^2\theta))\\
    &=(r^2sin^2\theta)(\ddot{R}R+\dot{R}^2)
    -\dot{R}^2r^2sin^2\theta\\
    &=r^2sin^2\theta\ddot{R}R
    \end{align*}
    \begin{align*}
    R^1_{313}&={\Gamma^1}_{33,1}-{\Gamma^1}_{31,3}+{\Gamma^1}_{\sigma1}{\Gamma^\sigma}_{33}
    -{\Gamma^1}_{\sigma3}{\Gamma^\sigma}_{31}\\
    &=\frac{\partial}{\partial r}(-r(1-kr^2)sin^2\theta)
    -(0)_{,3}
    +({\Gamma^1}_{01}{\Gamma^0}_{33}+{\Gamma^1}_{11}{\Gamma^1}_{33})
    -{\Gamma^1}_{33}{\Gamma^3}_{31}\\
    &=(-1+3kr^2)sin^2\theta
    +\Big((\frac{\dot{R}}{R})(\frac{\dot{R}}{R}g_{33})+(\frac{kr}{1-kr^2})(-r(1-kr^2)sin^2\theta)\Big)
    -(-r(1-kr^2)sin^2\theta)(\frac{1}{r})\\
    &=(-1+3kr^2)sin^2\theta
    +\Big((\frac{\dot{R}}{R})(\frac{\dot{R}}{R}(R^2r^2sin^2\theta))-kr^2sin^2\theta\Big)
    +(1-kr^2)sin^2\theta\\
    &=sin^2\theta\Big(-1+3kr^2
    +\dot{R}^2r^2-kr^2
    +1-kr^2\Big)\\
    &=r^2sin^2\theta(k+\dot{R}^2)\\
    \end{align*}
    \begin{align*}
    R^2_{323}&={\Gamma^2}_{33,2}-{\Gamma^2}_{32,3}
    +{\Gamma^2}_{\sigma2}{\Gamma^\sigma}_{33}
    -{\Gamma^2}_{\sigma3}{\Gamma^\sigma}_{32}\\
    &=\frac{\partial}{\partial\theta}(-sin\theta cos\theta)-(0)_{,3}
    +({\Gamma^2}_{02}{\Gamma^0}_{33}+{\Gamma^2}_{12}{\Gamma^1}_{33})
    -{\Gamma^2}_{33}{\Gamma^3}_{32}\\
    &=(sin^2\theta-cos^2\theta)-0
    +\Big((\frac{\dot{R}}{R})(\frac{\dot{R}}{R}g_{33})+(\frac{1}{r})(-r(1-kr^2)sin^2\theta)\Big)
    -(-sin\theta cos\theta)cot\theta\\
    &=(sin^2\theta-cos^2\theta)
    +\Big((\frac{\dot{R}}{R})^2(R^2r^2sin^2\theta)-(1-kr^2)sin^2\theta\Big)+cos^2\theta\\
    &=sin^2\theta
    +sin^2\theta\Big(\dot{R}^2r^2-1+kr^2)\Big)\\
    &=r^2sin^2\theta(k+\dot{R}^2)\\
    \end{align*}
    \begin{align*}
    R^3_{333}&={\Gamma^3}_{33,3}-{\Gamma^3}_{33,3}
    +{\Gamma^3}_{\sigma3}{\Gamma^\sigma}_{33}
    -{\Gamma^3}_{\sigma3}{\Gamma^\sigma}_{33}\\
    &=0
    \end{align*}

    Hence ##R_{33}=r^2sin^2\theta(\ddot{R}R+2(k+\dot{R}^2))##.

    So now we can calculate the Ricci scalar:
    \begin{align*}
    R&=R_{ab}g^{ab}\\
    &=R_{aa}g^{aa}\textrm{ [as }\textbf{g}\textrm{ is diagonal in the FLRW coordinates]}\\
    &=R_{00}g^{00}+R_{11}g^{11}+R_{22}g^{22}+R_{33}g^{33}\\
    &=(-3\frac{\ddot{R}R}{R^2})(-1)
    +\Big(\frac{\ddot{R}R+2(\dot{R}^2+k)}{1-kr^2}\Big)\frac{1-kr^2}{R^2}\\
    &+(\ddot{R}Rr^2+2r^2(k+\dot{R}^2))\frac{1}{R^2r^2}
    +(r^2sin^2\theta(\ddot{R}R+2(k+\dot{R}^2)))\frac{1}{R^2r^2sin^2\theta}\\
    \end{align*}
    \begin{align*}
    &=3\frac{\ddot{R}R}{R^2}
    +\frac{\ddot{R}R+2(\dot{R}^2+k)}{R^2}
    +\frac{(\ddot{R}R+2(k+\dot{R}^2))}{R^2}
    +\frac{(\ddot{R}R+2(k+\dot{R}^2))}{R^2}\\
    &=\frac{1}{R^2}\Big(3\ddot{R}R
    +(\ddot{R}R+2(k+\dot{R}^2))
    +(\ddot{R}R+2(k+\dot{R}^2))
    +(\ddot{R}R+2(k+\dot{R}^2))\Big)\\
    &=\frac{1}{R^2}\Big(6\ddot{R}R
    +6\dot{R}^2
    +6k\Big)\\
    &=\frac{6}{R^2}\Big(\ddot{R}R
    +\dot{R}^2
    +k\Big)\\
    \end{align*}

    Before we compute the Einstein tensor component we need to raise the indices of ##R_{00}## to get:
    \begin{align*}
    R^{00}&=g^{a0}g^{b0}R_{ab}\\
    &=g^{00}g^{00}R_{00}\textrm{ [as }\textbf{g}\textrm{ is diagonal in the FLRW coordinates]}\\
    &=(\frac{1}{-1})(\frac{1}{-1})R_{00}\\
    &=R_{00}\\
    &=-3\frac{\ddot{R}R}{R^2}\\
    \end{align*}

    Finally, the ##00## component of the Einstein tensor is:

    \begin{align*}
    G^{00}&=R^{00}-\frac{1}{2}Rg^{00}\\
    &=-3\frac{\ddot{R}R}{R^2}
    -\frac{1}{2}\frac{6}{R^2}\Big(\ddot{R}R+\dot{R}^2 +k\Big)(-1)\\
    &=\frac{3}{R^2}(\dot{R}^2 +k)\\
    \end{align*}

    V.421.
     
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