Ricci Tensor from Schwarzschild Metric

1. Dec 30, 2008

Philosophaie

Looking for the Schwarzschild Solution for this equation:

$$ds^2 = -A(r) / c^2 * dr^2 - r^2 / c^2 *(d\\theta^2 +(sin(\\theta))^2 *d\\phi^2) + B(r) * dt^2$$

where

A(r) = 1 / (1-2*m/r)
And
B(r) = (1-2*m/r)

From this can be calculated the co- and contra-varient metric tensors and Affinity:
$$g_{ab}$$
$$g^{ab}$$
$$\Gamma^{c}_{ab}$$

Ricci Tensor is:

$$R_{bc} = R^{a}_{bca} = \Gamma^{a}_{dc}*\Gamma^{d}_{ba} - \Gamma^{a}_{da}*\Gamma^{d}_{bc} + d/dx^{c} * \Gamma^{a}_{ba} - d/dx^{a} * \Gamma^{a}_{bc}$$

My solution is a 4x4 matrix with all zeros except on the diagonal.

My choices for A(r) and B(r) may not be correct for Earthâ€™s orbit and geodesics. Could
someone steer me in the right direction.

Last edited: Dec 31, 2008
2. Dec 30, 2008

Mentz114

I don't understand what this means

The Schwarzschild metric is a solution of the GR field equations which has a zero Ricci tensor.

Is your formula as you want it ? dphi should be squared, and it seems to be missing a '+' also.

3. Dec 30, 2008

Philosophaie

My calculations included the squared term just a typo:

$$ds^2 = -A(r) / c^{2} * dr^{2}-r^{2} / c^{2} *(d\theta^{2} + (sin(\theta))^{2} *d\phi^{2}) + B(r) * dt^{2}$$

4. Dec 30, 2008

Mentz114

OK. I think you've got A(r) and B(r) switched. All the Ricci components should be zero.
I've just seen another possible error. To contract the Riemann tensor to the Ricci, indexes 1 and 3 are summed over. You are showing contraction over the first and last.

5. Dec 30, 2008

Philosophaie

The Schwarzschild metric was taken directly from a textbook and confirmed from many websites. My calculations of the metric tensors and affinity are from the Schwarzschild metric. I switched the indices for the Ricci Tensor is summed from the Riemann 1st and 3rd indices and the Ricci tensor was not all zeros with nonzero on the diagonal.

$$R_{bc} = R^{a}_{bac}$$

produces the same result because of symmetry

$$R_{bc} = R^{a}_{bca} = R^{a}_{bac}$$

What is the rank of this operation?

Last edited: Dec 31, 2008
6. Dec 31, 2008

Mentz114

Hi,
please have a look at this thread, and the PDF file there which has the curvature results.

You aren't calculating the metric tensor - that is given. I assume you calculated the connections and used them to get the Riemann tensor.