# Defintion of Curvature

maverick280857
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

$$6\pi \frac{C_{exp}-C_{meas}}{C_{exp}A_{exp}}$$

The situation is as follows:

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

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

Any thoughts?

Thanks.
Cheers
Vivek.

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Mentz114
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.

maverick280857
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 $6\pi$ factor. How does all that come in?

pmb_phy
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

$$6\pi \frac{C_{exp}-C_{meas}}{C_{exp}A_{exp}}$$

The situation is as follows:

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

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

Mentz114
maverick280857 , if you want a derivation, check out Wiki on 'Gaussian curvature'. There's even an expression in terms of Christoffel symbols.

maverick280857
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.

maverick280857
Ok, so I guess its a special case of the expression given there, for 2 dimensions.