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The ratio $\frac{\text{perimeter}}{\text{diameter}}$ of circles on the surface of constant curvature $0$ is constantly $3.14\dots$ and is called $\pi$.

  • Is this ratio a constant for every other surface of constant curvature $\kappa$?

  • How then (by which formula) does this constant depend on $\kappa$?

2 Answers2

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On a 2-dimensional sphere of radius $R$, the discs would correspond to the caps on a cone with vertex at the center of the sphere (with opening given by an angle $0<\phi<\pi$). Since the caps are circles of radius $R\sin \phi$ and the diameter would be the length of the arc of the great circle (which has opening $2\phi$), the ratio would be \begin{equation}\frac{c}{d}=\frac{2\pi R\sin \phi}{2R\phi}=\frac{\pi\sin \phi}{\phi}. \end{equation}

AppliedSide
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The ratio is not constant on surfaces of non-zero curvature. Let $d(x)$ be the ratio of the perimeter of a circle of diameter $x$ by the diameter. If the curvature is negative then this is a strictly increasing function while if the curvature is positive it is a strictly decreasing function. It's quite easy to see that with particular models for such constant curvature spaces and directly computing. For instance, for circles on a sphere this is quite straightforward: the perimeter of the circle, as the diameter increases, reaches a maximum and then starts shrinking.

I don't know of an exact formula, so will be looking forward to other answers.

Ittay Weiss
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