# Contraction of the Riemann Tensor with the Weak Field Metric

1. Sep 20, 2011

### 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

Last edited: Sep 20, 2011
2. Sep 20, 2011

### Bill_K

Ri0j0,ijijϕ,00

When you contract this, don't forget that δii = 3.

3. Sep 20, 2011

### JMedley

Ok so that takes care of the factor 3, then how does $$\phi^{,i}_{,i}$$ correspond to $$\nabla^2\phi$$? Cheers for the help

Last edited: Sep 20, 2011
4. Sep 22, 2011

### millitiz

Because they are the same. I would assume that $\phi$ is a scalar (from what you showed) - then partial derivative is the same as the full derivative. And what you just wrote are exactly the same - just different way of expressing it. Hope it helps

5. Sep 22, 2011

### WannabeNewton

You have $\partial ^{i}\partial _{i}\phi = \delta ^{ij}\partial _{j}\partial _{i}\phi$ and this, in background flat 3 - space with a Cartesian chart is $\delta ^{xx}\partial ^{2}_{x}\phi + \delta ^{yy}\partial ^{2}_{y}\phi + \delta ^{zz}\partial^{2} _{z}\phi = \partial ^{2}_{x}\phi + \partial ^{2}_{y}\phi + \partial^{2} _{z}\phi = \triangledown ^{2}\phi$