Metric tensor in spherical coordinates

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The discussion focuses on the metric tensor in spherical coordinates, particularly the Schwarzschild metric and its implications for stress-energy tensors. It clarifies that the Schwarzschild metric describes the gravitational field outside a spherical, non-rotating mass, and that the stress-energy tensor components should be zero in this vacuum solution. Confusion arises regarding the Ricci tensor and Riemann tensor, with participants noting that while the Ricci tensor is zero, the Riemann tensor can be non-zero, indicating curvature due to mass. The conversation also touches on the approximation of planets as perfect fluids for simplicity in calculations, despite their actual complexity. Overall, the thread emphasizes the distinctions between different metrics and tensors in general relativity.
  • #101
stevendaryl,

One of the components of the Riemann tensor in Schwarzschild solution is

R^{r}_{\theta \theta r}{} = \dfrac{M}{R}

What does this component mean in real sense? Susskind told that, here, two \theta \theta indexes (which are downstairs) have something to do with two vectors which define a plane at some point in space. What other two r-r represent? Basically, Riemann tensor takes as input 4 vectors and outputs the curvature value, right?

Thank you.

Is this a true picture of basis vectors? http://www.springerimages.com/Images/RSS/1-10.1007_978-1-4614-0706-5_6-0
 
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  • #102
GRstudent said:
R^{r}_{\theta \theta r}{} = \dfrac{M}{R}

What does this component mean in real sense?
I am not completely certain that I have this right, but I believe that it means that if you move a dr vector around a closed differential d\theta, d\theta loop then it will change by a differential amount M/R in the dr direction.
 
  • #103
^
This makes some sense.
 
  • #104
DaleSpam said:
I am not completely certain that I have this right, but I believe that it means that if you move a dr vector around a closed differential d\theta, d\theta loop then it will change by a differential amount M/R in the dr direction.

I'm not sure about the convention for the order of the indices, but if you have a loop (parallelogram) in which the two sides are parallel (both d\theta), then the loop has area 0, and so the parallel transport gives 0.

I think that the correct interpretation is this: Make a parallelogram with one side equal to dr\ e_r and another side equal to d\theta \ e_\theta. March the vector V = e_\theta around the parallelogram to get a new vector V' that will be slightly different from e_\theta. Then letting \delta V = V' - V,

\delta V^r = R^r_{\theta \theta r} dr d\theta
 
  • #105
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