Value of $\pi$ in the $L^3$ metric

Question

What is the semicircumference of a unit circle in $L^3$ space? Does it gave the same generalizations to other dimensions as $\pi$ does in $L^2$ space?

Answer 1

I presume that by "a unit circle in $L^3$ space" you're talking about the curve $|x|^3 + |y|^3 = 1$ in $\mathbb R^2$ with norm $\|(a,b)\| = (|a|^3 + |b|^3)^{1/3}$. The part of this "circle" in the first quadrant is parametrized as $R(t) = [ \cos(t)^{2/3}, \sin(t)^{2/3}]$, $0 \le t \le \pi/2$. Thus $dR/dt = [- \frac{\sin(t)}{\cos(t)^{1/3}}, \frac{\cos(t)}{\sin(t)^{1/3}}]$ and the arc length (using the given norm) is $$\int_{0}^{\pi/2} \|dR/dt\|\; dt = \frac{2}{3} \int_0^{\pi/2} \left( \frac{\sin(t)^3}{\cos(t)} + \frac{\cos(t)^3}{\sin(t)}\right)^{1/3}\; dt$$ I don't believe this can be done in "closed form", but numerically it is approximately $1.6298839965294975486$. The semicircumference is then twice this, or approximately $3.2597679930589950972$.

Answer 2

Take a look at https://www.youtube.com/watch?v=ineO1tIyPfM&t=12s The answer is somewhere the 6.30 mark