Ricci Tensor from Schwarzschild Metric

In summary, the conversation discusses the search for the Schwarzschild Solution for a given equation, which is a solution of the GR field equations with a zero Ricci tensor. The conversation also includes the calculation of the co- and contra-varient metric tensors, Affinity, and Ricci Tensor, as well as some potential errors and corrections. The rank of the operation is not specified.
  • #1
Philosophaie
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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]

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:
[tex]g_{ab}[/tex]
[tex]g^{ab}[/tex]
[tex]\Gamma^{c}_{ab}[/tex]

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:
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  • #2
I don't understand what this means

Looking for the Schwarzschild Solution for this equation,

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
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]
 
  • #4
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
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:
  • #6
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.

https://www.physicsforums.com/showthread.php?t=243517
 

What is the Ricci tensor?

The Ricci tensor is a mathematical object used to describe the curvature of space in the theory of general relativity.

How is the Ricci tensor calculated from the Schwarzschild metric?

The Ricci tensor can be calculated from the Schwarzschild metric using the Einstein field equations. This involves taking the second derivatives of the metric and plugging them into a specific formula.

What does the Ricci tensor tell us about the Schwarzschild metric?

The Ricci tensor provides us with information about the curvature of space in the presence of a massive object, as described by the Schwarzschild metric. It tells us how spacetime is bending and warping due to the mass of the object.

Why is the Ricci tensor important in general relativity?

The Ricci tensor is a crucial component in the Einstein field equations, which are the foundation of general relativity. It helps us understand the relationship between matter and the curvature of spacetime.

Can the Ricci tensor be used to solve the Einstein field equations for other metrics?

Yes, the Ricci tensor can be used to solve the Einstein field equations for any metric that describes the curvature of spacetime. However, the calculations may become more complex for more complicated metrics.

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