this post was submitted on 17 May 2025
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[โ€“] bss03@infosec.pub 4 points 6 days ago (1 children)

Cause it's just a (n-1)-dimensional ball extruded along the remaining axis, or do all 3d shapes exist on (nearly) all 3d metrics?

[โ€“] Shardikprime@lemmy.world 3 points 3 days ago

Mostly because the actual pi values can vary in between non/euclidean geometries. Within extremely strong gravitational fields, spacetime becomes highly non euclidean, affecting the C/d ratio of an actual circle, so I'd wager this would affect pi as well