This person is not a sailor. Sailing orthogonal to the wind, a "beam reach", is the fastest point of sail due to the lift of the sail.
I knew someone would make this comment. I love HN for this kind of pedantry when it's specific, accurate and doesn't dismiss the entire article for one inaccurate analogy.
π in Other Universes
111–113 of 113 posts
Re: π in Other Universes
#112Earlier quoted context omitted.
It blows my mind to think of mathematics/logic almost like a huge cellular automaton. “axioms” don’t necessarily correspond to “truth”, to me they’re arbitrary constraints that can give rise to complexity. And sometimes the resulting systems can be useful
The whole of the puzzles of cosmology actually all might be obvious if we had a few different fundamental theorems. But because we hit on some that almost work, and then build upon them a hole edifice of mathematics that is internally consistent and almost fits the universe we keep beating on it, not realizing that backing up a little and then driving forward again at a slightly different angle might yield a simpler,…
yet if we just tried, oh, making the unit circle a unit... ellipse... all of the epiphenomenal complexity that comes from remediating the pervasively accumulated 0.01% error in that fundamental assumption would instantly vanish.
Re: π in Other Universes
#113Earlier quoted context omitted.
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, t…
How 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
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.