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There is no concept of the "background metric" here. Both the radius and the circumference are measured in the defined metric itself. Any metric that "pulls on
by azeemba 3y ago
There is no concept of the "background metric" here. Both the radius and the circumference are measured in the defined metric itself.
Any metric that "pulls on the origin" compared to Euclidean distance will have to do the mapping in a continuous way. This will basically result in both the radius and circumference being expanded in that metric.
Matter of fact, I linked an article that proves that for _all_ metrics, the value of π is always between 3 and 4 (inclusive). Unfortunately the article might have gotten the hug of death so here is an alternative link: https://www.researchgate.net/publication/353330827_Extremal_Values_of_Pi https://www.researchgate.net/publication/353330827_Extremal_...
- charlieyu1 3y agoHow is circumference defined? And I can think of a counterexample on a sphere, just using Euclidean distance on the surface. Consider a circle with centre at North Pole and radius being the distance from the North Pole to a point on the equator. For this circle it is easy to find out that pi=2
- Filligree 3y agoHmm. And if you keep increasing the radius, pi will shrink all the way to 0.
- azeemba 3y agoThanks for your example! I have been thinking about it. Your observation is correct and the surface of the sphere is a metric. The ratio of radius to circumference is not constant with that metric though so I feel like something should disqualify it. But I am not sure how. So I think your observation shows that we need a stronger constraint than just being a metric. Other commenters have hinted that you need a normed vector space but I am not sure if that's sufficient.