Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\kappa$?

share|improve this question

2 Answers 2

up vote 5 down vote accepted

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}

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.