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I was wondering if there is an intuitive explanation why the surface of an $n$-dimensional sphere is maximal at $n=6$ and for the volume $n=5$. I know that for the surface you have: $$A_n=\frac{2\pi^\frac{(n+1)}{2}}{\Gamma(\frac{n+1}{2})}$$

And for the volume
$$V_n=\frac{A_{n-1}}{n}$$

Finally you get
$$A_n(R)=A_nR^n$$ $$V_n(R)=V_nR^n$$
Where $R$ is the radius of the ball.
Plugging in the dimensions it's easy to show that $V_n$ is max for $n=5$ and the surface for $n=6$
Is there an intuitive way of explaining this?

Conifold
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Dominik Car
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  • Maybe, but I would like an explenation from a physicist rather than a mathematician due to the way one looks at space. – Dominik Car May 14 '17 at 18:42
  • According to Mathworld's Hypersphere article, "for the unit hypersphere, the hyper-surface area reaches a maximum and then decreases towards 0... the seven-dimensional unit hypersphere has maximum hyper-surface area". For the volume the maximum does occur in dimension 5, but that is also an artifact of choosing radius = 1, for other radii it occurs in other dimensions, see Richeson's explanation. In other words, there is nothing special about 5 and 7. – Conifold May 14 '17 at 20:33

1 Answers1

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Here is a naive intuitive explanation for the growth in volume of the unit $n$-sphere as $n$ increases.

Embed the $n$-sphere in an $n$-cube which has $2^n$ corners, $2n$ facets and volume 1. The space in the cube not occupied by the sphere is concentrated near the corners and the space in the cube occupied by the sphere is concentrated near the middle of the facets. For $n \leq 4$, the latter effect overrides the former. But for $n > 4$, the corners win.

Phill
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