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maverick280857

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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.

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.

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