Curvature tensor of sphere radius R

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Discussion Overview

The discussion revolves around finding the curvature tensor of a sphere with radius R. Participants explore the mathematical formulation and coordinate representation necessary for this calculation, focusing on spherical coordinates and the metric tensor.

Discussion Character

  • Technical explanation

Main Points Raised

  • One participant seeks guidance on how to begin calculating the curvature tensor for a sphere of radius R.
  • Another participant clarifies that the discussion pertains to a three-dimensional sphere and provides the spherical coordinate transformations for x, y, and z.
  • The second participant outlines the process of calculating differentials and the metric tensor, ultimately leading to the expression for the metric tensor in spherical coordinates.
  • A later reply expresses gratitude and indicates that the initial poster understands the next steps, suggesting they may seek further assistance if problems arise.

Areas of Agreement / Disagreement

The discussion does not present any explicit disagreements, but it reflects a progression from inquiry to a clearer understanding of the topic. No consensus is established regarding the curvature tensor itself, as the initial poster has yet to perform the calculations.

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hello! I need to find curvature tensor of sphere of R radius. How can I start? thanks!
 
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hello
 
Are you talking about the sphere of radius R in three dimensions?

Start by writing out x, y, and z in spherical coordinates with \rho taken as the constant R:

x= Rcos(\theta)sin(\phi)
y= Rsin(\theta)sin(\phi)
z= R cos(\phi)

Calculate the differentials:
dx= - R sin(\theta)sin(\phi)d\theta+ Rcos(\theta)cos(\phi)d\phi
dy= R cos(\theta)sin(\phi)d\theta+ Rsin(\theta)cos(\phi)d\phi
dz= -R sin(\phi)d\phi

Find ds^2= dx^2+ dy^2+ dz^2 in terms of spherical coordinates:
dx^2= R^2 sin^2(\theta)sin^2(\phi)d\theta^2- 2R^2sin(\theta)cos(\theta)sin(\phi)cos(\phi)d\theta d\phi+ R^2cos^2(\theta)cos^2(\phi)d\phi^2
dy^2= R^2cos^2(\theta)sin^2(\phi)d\theta^2+ 2R^2sin(\theta)cos(\theta)sin(\phi)cos(\phi)d\thetad\phi+ R^2sin^2(\theta)cos^2(\phi)d\phi^2
dz^2= R^2 sin^2(\phi)d\phi^2

Adding those
ds^2= R^2sin^2(\phi)d\theta^2+ R^2 d\phi^2
which gives us the metric tensor:


g_{ij}= \begin{pmatrix}R^2sin^2(\phi) & 0 \\ 0 & R^2 \end{pmatrix}

You can calculate g^{ij}, the Clebsh-Gordon coefficients, and the curvature tensor from that.
 
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thanks you! i finally know what to do;] i going to try to do this, i ask if get problems
 

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