Metric tensor in spherical coordinates

  • #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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