Einstein tensor in the FLRW frame

In summary, this conversation discusses the development of a formula for the ##G^{00}## component of the Einstein tensor in the FLRW coordinate system for a homogeneous and isotropic spacetime. FLRW coordinates are used, where the metric tensor components are all zero except for specific values. The Riemann tensor equation is also introduced and Christoffel symbols are calculated to determine the non-zero symbols in this system.
  • #1
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
4,116
1,712
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.
 
Physics news on Phys.org
  • #2
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.
 

1. What is the Einstein tensor in the FLRW frame?

The Einstein tensor in the FLRW (Friedmann–Lemaître–Robertson–Walker) frame is a mathematical expression used in general relativity to describe the curvature of spacetime. It is a combination of the Ricci tensor and the scalar curvature, and is used to represent the gravitational field in the FLRW metric, which is a solution to Einstein's field equations.

2. How is the Einstein tensor related to the FLRW metric?

The Einstein tensor is directly related to the FLRW metric through the Einstein field equations, which relate the curvature of spacetime to the distribution of matter and energy within it. The FLRW metric is a solution to these equations, and the Einstein tensor describes the curvature of spacetime in this specific frame.

3. What is the significance of the Einstein tensor in the FLRW frame?

The Einstein tensor is significant in the FLRW frame because it is used to describe the gravitational field in this particular solution to Einstein's field equations. It allows us to understand the curvature of spacetime and how it is affected by the distribution of matter and energy in the universe.

4. How does the Einstein tensor in the FLRW frame relate to the expansion of the universe?

The Einstein tensor in the FLRW frame is directly related to the expansion of the universe through the FLRW metric. This metric describes the expansion of the universe and the rate at which it is expanding. The Einstein tensor is used to calculate the curvature of spacetime in this expanding universe.

5. Can the Einstein tensor in the FLRW frame be applied to other cosmological models?

Yes, the Einstein tensor in the FLRW frame can be applied to other cosmological models because it is a general mathematical expression that describes the curvature of spacetime. However, it may need to be modified or adapted to fit different models, as the FLRW metric is specific to the expanding universe and may not accurately describe other cosmological scenarios.

Similar threads

  • Special and General Relativity
Replies
2
Views
117
  • Special and General Relativity
Replies
11
Views
1K
  • Special and General Relativity
Replies
2
Views
533
  • Special and General Relativity
Replies
6
Views
1K
  • Special and General Relativity
Replies
2
Views
1K
  • Special and General Relativity
Replies
5
Views
1K
  • Special and General Relativity
Replies
1
Views
806
  • Special and General Relativity
Replies
4
Views
3K
  • Special and General Relativity
Replies
16
Views
2K
  • Special and General Relativity
Replies
2
Views
785
Back
Top