Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ricci Tensor from Schwarzschild Metric

  1. Dec 30, 2008 #1
    Looking for the Schwarzschild Solution for this equation:

    [tex]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[/tex]


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

    From this can be calculated the co- and contra-varient metric tensors and Affinity:

    Ricci Tensor is:

    [tex]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}[/tex]

    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 Earth’s orbit and geodesics. Could
    someone steer me in the right direction.
    Last edited: Dec 31, 2008
  2. jcsd
  3. Dec 30, 2008 #2


    User Avatar
    Gold Member

    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.
  4. Dec 30, 2008 #3
    My calculations included the squared term just a typo:

    [tex]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}[/tex]
  5. Dec 30, 2008 #4


    User Avatar
    Gold Member

    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.
  6. Dec 30, 2008 #5
    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.

    [tex]R_{bc} = R^{a}_{bac}[/tex]

    produces the same result because of symmetry

    [tex]R_{bc} = R^{a}_{bca} = R^{a}_{bac}[/tex]

    What is the rank of this operation?
    Last edited: Dec 31, 2008
  7. Dec 31, 2008 #6


    User Avatar
    Gold Member

    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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Ricci Tensor from Schwarzschild Metric