# Question #3993a

Mar 3, 2017

As the metric tensor ${g}_{\mu \nu}$ is a diagonal matrix, the inverse is easily obtained as,

${g}^{\mu \nu} = \mathrm{di} a g \left(1 , {\Phi}^{-} 2 , {\Phi}^{-} 2 , {\Phi}^{-} 2\right)$

Further, the Christoffel symbols are given as :

Where ${g}_{m k , l}$ means partial derivative of ${g}_{m k}$ with respect to ${x}^{l}$

So, for example,

${\Gamma}_{\mu \nu}^{0} = \frac{1}{2} {g}^{0 m} \left({g}_{m \mu , \nu} + {g}_{m \nu , \mu} - {g}_{\mu \nu , m}\right)$

here, following Einstein's notation, the sum is over m. As the metric here is general, you have to calculate and write all the derivative terms separately. IN general, the terms will look like: $\Phi \Phi ' \left({x}^{0}\right) , \Phi \Phi ' \left({x}^{1}\right) , \Phi \Phi ' \left({x}^{2}\right) \ldots e t c$

The list of terms is extremely long. Be careful.

Once you have the Christoffel symbols, to find the RIcci Tensor, you first calculate the RIemann Tensor which is obtained as:

${R}_{j l m}^{i} \setminus \equiv {\partial}_{l} {\Gamma}_{m j}^{i} - \setminus {\partial}_{m} \setminus {\Gamma}_{l j}^{i} + \setminus {\Gamma}_{m j}^{k} \setminus {\Gamma}_{l k}^{i} - \setminus {\Gamma}_{l j}^{k} \setminus {\Gamma}_{k m}^{i}$

lower the $i$ index in RIemann Tensor and then contract with the metric tensor, to find the Ricci Tensor as:

${R}_{j m} = {g}^{i l} {R}_{i j l m} \setminus$

The Ricci scalar is found by contracting the Ricci tensor with metric tensor once more:

$R = {g}^{j m} {R}_{j m}$