Defintion of Curvature

  • #1
maverick280857
1,789
4
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:

Answers and Replies

  • #2
Mentz114
5,432
292
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.
 
  • #3
maverick280857
1,789
4
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?
 
  • #4
pmb_phy
2,952
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.
That relation is teh Gaussian Curvature for a two dimensional surface. Gaussian curvature is not defined for spaces higher than two.

Pete
 
  • #5
Mentz114
5,432
292
maverick280857 , if you want a derivation, check out Wiki on 'Gaussian curvature'. There's even an expression in terms of Christoffel symbols.
 
  • #6
maverick280857
1,789
4
maverick280857 , if you want a derivation, check out Wiki on 'Gaussian curvature'. There's even an expression in terms of Christoffel symbols.

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.
 
  • #7
maverick280857
1,789
4
Ok, so I guess its a special case of the expression given there, for 2 dimensions.
 

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