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Defintion of Curvature

  1. May 14, 2008 #1
    Hi.

    I am reading "An Introduction to Modern Astrophysics" by Carroll and Ostlie, for a summer project. In section 27.3 (Relativistic Cosmology) the curvature of a sphere is given by

    [tex]6\pi \frac{C_{exp}-C_{meas}}{C_{exp}A_{exp}}[/tex]

    The situation is as follows:

    Consider a sphere of radius R. An ant is moving on the sphere at a fixed polar angle [itex]\theta[/itex]. The ant measures the circumference [itex]C_{meas} = 2\pi R\sin\theta[/itex] whereas the expected value of circumference is [itex]C_{exp} = 2\pi D[/itex] where [itex]D = R\theta[/itex]. The expected area of the circle is [itex]A_{exp} = \pi D^2[/itex].

    I am not sure how the above expression leads to the curvature of the sphere...

    Any thoughts?

    Thanks.
    Cheers
    Vivek.
     
    Last edited: May 14, 2008
  2. jcsd
  3. May 14, 2008 #2

    Mentz114

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    My understanding of this is that the ant walks in a circle, not a great circle. In this case the expression you quote gives the curvature of the sphere. Imagine the ant walking is a circle on a flat surface, then let the flat surface curve. The ratio of the diameter to the radius changes.
     
  4. May 15, 2008 #3
    Mentz114, thanks for your reply. I'm not clear about the way the curvature has been written in terms of the two circumferences and the area, and also the [itex]6\pi[/itex] factor. How does all that come in?
     
  5. May 15, 2008 #4
    That relation is teh Gaussian Curvature for a two dimensional surface. Gaussian curvature is not defined for spaces higher than two.

    Pete
     
  6. May 15, 2008 #5

    Mentz114

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    maverick280857 , if you want a derivation, check out Wiki on 'Gaussian curvature'. There's even an expression in terms of Christoffel symbols.
     
  7. May 16, 2008 #6
    Thanks, I'm still learning these things and I don't know much about them. The page you have referred to does not explain the origin of the particular definition of curvature used here. But I will keep looking.
     
  8. May 16, 2008 #7
    Ok, so I guess its a special case of the expression given there, for 2 dimensions.
     
  9. May 16, 2008 #8

    robphy

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