What Are the Steps to Solve Einstein Equations for This Metric?

[unscientific]

Homework Statement

[/B]
(a) Find the christoffel symbols
(b) Find the einstein equations
(c) Find A and B
(d) Comment on this metric

Homework Equations

\Gamma_{\alpha\beta}^\mu \frac{1}{2} g^{\mu v} \left( \partial_\alpha g_{\beta v} + \partial_\beta g_{\alpha v} - \partial_\mu g_{\alpha \beta} \right)

R_{v \beta} = \partial_\mu \Gamma_{\beta v}^\mu - \partial_\beta \Gamma_{\mu v}^\mu + \Gamma_{\mu \epsilon}^\mu \Gamma_{v \beta}^\epsilon - \Gamma_{\epsilon \beta}^\mu \Gamma_{v \mu}^\epsilon

The Attempt at a Solution

Part(a)[/B]
After some math, I found the christoffel symbols to be:
$\Gamma_{11}^0 = \frac{A A^{'}}{c^2}$
$\Gamma_{22}^0 = \frac{B B^{'}}{c^2}$
$\Gamma_{33}^0 = \frac{B B^{'}}{c^2}$
$\Gamma_{01}^1 = \frac{A^{'}}{A}$
$\Gamma_{02}^2 = \frac{B^{'}}{B}$
$\Gamma_{03}^3 = \frac{B^{'}}{B}$

Part (b)
Now brace yourselves for the ricci tensors...
R_{00} = -\partial_0 \left( \Gamma_{01}^1 + \Gamma_{02}^2 + \Gamma_{03}^3 \right) - \Gamma_{10}^1 \Gamma_{01}^1 - 2\Gamma_{20}^2 \Gamma_{02}^2
R_{00} = -\frac{A^{''}}{A} - 2 \frac{B^{''}}{B}

By symmetry, $R_{01} = R_{02} = R_{03} = R_{12} = R_{13} = R_{23} = 0$.

Now to find the $11$ component:
R_{11} = \partial_0 \Gamma_{11}^0 + \Gamma_{11}^0 \left( \Gamma_{10}^1 + \Gamma_{20}^2 + \Gamma_{30}^3 \right) - \Gamma_{11}^0 \Gamma_{10}^1 - \Gamma_{01}^1 \Gamma_{11}^0
= \partial_0 \Gamma_{11}^0 + 2 \Gamma_{11}^0 \Gamma_{20}^2 - \Gamma_{11}^0 \Gamma_{10}^1
R_{11} = \frac{A A^{''}}{c^2} + 2 \left( \frac{A}{B} \right) \frac{A^{'} B^{'}}{c^2}

By symmetry, to find $22$ and $33$ components, we swap $A$ with $B$:
R_{22} = R_{33} = \frac{B B^{''}}{c^2} + 2 \left( \frac{B}{A} \right) \frac{A^{'} B^{'}}{c^2}The einstein field equations are given by:
G^{\alpha \beta} = \frac{8 \pi G}{c^4} T^{\alpha \beta} - \Lambda g^{\alpha \beta}

Thus, the simultaneous equations we seek are:
G^{00} = \frac{8 \pi G}{c^4} T^{00}
For $\mu, v \neq 0$ we have
R_{\mu v} = 0
So we simply equate $R_11 = 0$, $R_22 = R_{33} = 0$.However, the equations don't match..

 

[unscientific]

bumpp

 

[unscientific]

bumpp

 

[unscientific]

bumpp

 

[unscientific]

bumpp

 

[unscientific]

bumpp

 

[unscientific]

bumpp

 

[unscientific]

bumpp

 

[unscientific]

bumpp on part (b)/(c)

 

[unscientific]

Would appreciate help with my "ricci-nightmare"

 

[unscientific]

Anyone managed to get a different result for the ricci tensors yet?

 

[unscientific]

anyone else had a go with the ricci tensors?

 

[unscientific]

tried again, still didn't get the required ricci tensors.

 

[thierrykauf]

Here is what I find for one term for Ricci

 

[unscientific]

Here is what I find for one term for Ricci

I think the term is not appearing, do you mind posting it again?

 

[unscientific]

bumpp ricci

 

[thierrykauf]

Hold on. Here is what I find

 

[thierrykauf]

Here is what I find R_{00}=\Gamma^x_{x0}\Gamma^x_{x0} + \Gamma^x_{x0}\Gamma^y_{y0} + \Gamma^z_{z0}\Gamma^y_{y0}

 

[unscientific]

Here is what I find R_{00}=\Gamma^x_{x0}\Gamma^x_{x0} + \Gamma^x_{x0}\Gamma^y_{y0} + \Gamma^z_{z0}\Gamma^y_{y0}

Thanks a lot for replying. I'll give it a go later today and post my updated work.

 

[thierrykauf]

Please do. And let's see what you have.

 

[unscientific]

Please do. And let's see what you have.

My method is to go symbol by symbol and evaluate all possibilities for each symbol. Let's try to find $R_{tt}$. For the first symbol, since all christoffel symbols are functions of $t$, the first term $\partial_t \Gamma^t_{tt}=0$. Third term is zero as there is a $\Gamma^{\epsilon}_{tt}$ term.Then we have

R_{\nu \beta} = \partial_\mu \Gamma^\mu_{\beta \nu} - \partial_\beta \Gamma^\mu_{\mu \nu} + \Gamma^\mu_{\mu \epsilon} \Gamma^\epsilon_{\nu \beta} - \Gamma^\mu_{\epsilon \beta} \Gamma^\epsilon_{\nu \mu}

R_{tt} = -\partial_t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) - \Gamma^\mu_{\epsilon t} \Gamma^\epsilon_{t\mu}

Last term is non-zero only if $\mu = \epsilon = x,y,z$.

= -\partial_t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) - \left[ (\Gamma^x_{tx})^2 + (\Gamma^y_{ty})^2 + (\Gamma^z_{tz})^2 \right]

 

[unscientific]

bumpp

 

[unscientific]

bumpp on
R_{tt} = -\partial_t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) - \left[ (\Gamma^x_{tx})^2 + (\Gamma^y_{ty})^2 + (\Gamma^z_{tz})^2 \right]

 

[unscientific]

My method is to go symbol by symbol and evaluate all possibilities for each symbol. Let's try to find $R_{tt}$. For the first symbol, since all christoffel symbols are functions of $t$, the first term $\partial_t \Gamma^t_{tt}=0$. Third term is zero as there is a $\Gamma^{\epsilon}_{tt}$ term.Then we have

R_{\nu \beta} = \partial_\mu \Gamma^\mu_{\beta \nu} - \partial_\beta \Gamma^\mu_{\mu \nu} + \Gamma^\mu_{\mu \epsilon} \Gamma^\epsilon_{\nu \beta} - \Gamma^\mu_{\epsilon \beta} \Gamma^\epsilon_{\nu \mu}

R_{tt} = -\partial_t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) - \Gamma^\mu_{\epsilon t} \Gamma^\epsilon_{t\mu}

Last term is non-zero only if $\mu = \epsilon = x,y,z$.

= -\partial_t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) - \left[ (\Gamma^x_{tx})^2 + (\Gamma^y_{ty})^2 + (\Gamma^z_{tz})^2 \right]

bumpp

 

[unscientific]

Please do. And let's see what you have.

R_{tt} = -\partial_t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) - \left[ (\Gamma^x_{tx})^2 + (\Gamma^y_{ty})^2 + (\Gamma^z_{tz})^2 \right]

bumpp

 

[unscientific]

anyone else had a go with the ricci tensors?

 

[unscientific]

bump

 

[unscientific]

Please do. And let's see what you have.

Any new update?

 

[thierrykauf]

Hey, sorry for letting you down. Here is what I find for \tex{R_{00}

 

[thierrykauf]

Hi, here is what I found for the first Ricci component
R_{00} = \partial_l\Gamma^l_{00} - \partial_0\Gamma^l_{0l} + \Gamma^l_{00}\Gamma^m_{lm} - \Gamma^m_{0l}\Gamma^l_{0m}
R_{00} = -(\frac{A'}{A})' - 2(\frac{B'}{B})'- (\frac{A'}{A})^2 - 2 (\frac{B'}{B})^2
R_{00} = -(\frac{A''}{A}) - 2(\frac{B''}{B})

Repeat for the three others then remove the trace with subtracting the scalar curvature. These calculations are not difficult but they always stomp me when I'm not fully awake :)