Contraction of the Riemann Tensor with the Weak Field Metric

[JMedley]

I have started with the space-time metric in a weak gravitational field (with the assumption of low velocity):
$$ds^2=-(1+2\phi)dt^2+(1-2\phi)(dx^2+dy^2+dz^2)$$
Where $$\phi<<1$$ is the gravitational potential. Using the standard form for the Christoffel symbols have found:
$$\Gamma^0_{00}=\phi_{,0}, \Gamma^0_{0i}=\Gamma^0_{i0}=\phi_{,i}, \Gamma^0_{ij}=\delta_{ij}\phi_{,0}$$
$$\Gamma^i_{00}=\phi^{,i}, \Gamma^i_{0j}=\Gamma^i_{j0}=-\delta^i_j\phi_{,0}, \Gamma^i_{jk}=\delta_{jk}\phi^{,i}-\delta^i_j\phi_{,k}-\delta^i_k\phi_{,j}$$
Then combining derivatives of these to first order (ignoring products of Christoffel symbols) using:
$$R^\alpha_{\beta\mu\nu}=\Gamma^\alpha_{\beta\nu,\mu} - \Gamma^\alpha_{\beta\mu,\nu}$$
to get:
$$R^0_{i0j}=\delta_{ij}\phi_00-\phi_{ij}, R^i_{0j0}=\phi^{,i}_{,j}+\delta^i_j\phi_{,00}$$
$$R^i_{0jk}=-\delta^i_k\phi_{,0j}+\delta^i_j\phi_{0k}, R^i_{kj0}=\delta^i_j\phi_{0k} - \delta_{jk}\phi^{,i}_{,0}$$
$$R^i_{kjl}=-\delta^i_l\phi_{,jk}+\delta_{kl}\phi^{,i}_{,j}+ {\delta^i_j}\phi_{,kl}-\delta_{jk}\phi^{,i}_{,l}$$
(Where greek indices run from 0 to 3 and latin indices run from 1 to 3, and commas denote coordinate partial differentiation). And here is where I run into problems.. When I try to use $$R_{\alpha\beta}=R^\sigma_{\alpha\sigma\beta}$$ to contract these down to find the Ricci tensor. For example I get:
$$R_{00}=R^\sigma_{0\sigma 0}=\phi^{,i}_{,i}+\phi_{,00}$$
Which doesn't agree with the text I'm using which gives $$R_{00}=\nabla^2\phi +3\phi_{,00}$$
Can anybody spot where I'm going wrong? Many Thanks for any help.
Jack M

 

[Valerie Park]

First of all, you should be aware that the Ricci tensor is not the same as the Christoffel symbols. The Ricci tensor is a second rank tensor obtained by taking the trace of the Riemann tensor, which is a fourth rank tensor. In your example, you have correctly calculated the components of the Riemann tensor. However, when calculating the Ricci tensor, you need to take the trace of the Riemann tensor in order to obtain the Ricci tensor. To do this, you need to sum over the indices $\alpha$ and $\beta$ in the equation $R_{\alpha\beta}=R^\sigma_{\alpha\sigma\beta}$. This gives:$R_{00}=R^\sigma_{0\sigma 0}+R^\sigma_{1\sigma 1}+R^\sigma_{2\sigma 2}+R^\sigma_{3\sigma 3}$ $R_{00}=\phi^{,i}_{,i}+3\phi_{,00}$ which agrees with the text you are using. Hope this helps!