A Very Gentle Approach to Thinking About Hyperspheres

[bob012345]

TL;DR

A 2D problem is extended to 3D and beyond. Perhaps even to infinite dimensions!

Here is an interesting geometry problem given for general interest. In the figure, compute the fraction that is shaded. Then extend that to a sphere snugly packed in an octant of a larger sphere. That means the smaller sphere is tangent to each plane and to the larger sphere in the most compact way possible. Those who are brave can extend that further to four dimensions or higher and make some interesting observations about how things change at higher dimensions. Have fun!

 

[bob012345]

Here is the first part;

Let the center to center distance be $d$ and $r$ the smaller circle radius and $R$ the larger circle radius so $d+r=R$. We know $d^2=r^2+r^2$ or $d=\sqrt{2}r$. Solving for $r$ in terms of $R$ we get $$r=\frac{R}{\sqrt{2}+1}$$ therefore the fraction that is shaded is $$\frac{\pi r^2}{\frac{1}{4}\pi R^2}= 4(\frac{r}{R})^2= \frac{4}{(\sqrt{2}+1)^2}\approx 0.686$$

In the 3D case, $d=\sqrt{3}r$ giving $$r=\frac{R}{\sqrt{3}+1}$$ now we have the fraction of volume of a sphere to an octant which is $$\frac{\frac{4}{3} \pi r^3}{\frac{1}{8}\frac{4}{3}\pi R^3}= 8(\frac{r}{R})^3= \frac{8}{(\sqrt{3}+1)^3}\approx0.392$$

 

[bob012345]

Here is the next part;


In the case of 4D, $d=\sqrt{4}r=2r$ then $r=\frac{R}{3}$. In the general case $d=\sqrt{n}r$ and the ratio $$\frac{r}{R}=\frac{1}{\sqrt{n}+1}$$


The general equation for hypersphere volume is

$$V_n(R) = \frac{\pi^{n/2}}{\Gamma\!\left(\frac{n}{2} + 1\right)} \, R^n$$


Note that in higher dimensions the number of orthants (quadrants in 2D and octant in 3D) goes as $2^n$

Thus the fraction of the hypersphere of dimension $n$ to its orthant is $$\frac{2^n}{(\sqrt{n}+1)^n}$$

As $n$ gets larger, this fraction tends towards zero as does the ratio of $r$ to $R$ and a hyper cube volume in general.

 

[Ibix]

I got the same as #3.

Spoiler: Probably doesn't need spoiler tags

The values drop quite quickly with dimension. Here's a plot of the values from $n=2$ to $n=8$. The latter is the first time the ratio is less than 1%, and also the first time it's less than 1% of the $n=2$ value.

 

[bob012345]

Looking at a projection of the 3D case we can see how the ball is smaller for the same $R$.

 

[pbuk]

$$\frac{4}{(\sqrt{2}+1)^2}=0.686$$

I think that should be $\approx$

In the 3D case, $d=\sqrt{3}r$

Why?

Looking at a projection of the 3D case we can see how the ball is smaller for the same $R$.
View attachment 371412

I don't see how this diagram helps visualize how the smaller sphere touches the face of the octant.

 

[Hornbein]

Suppose any two volumes with one properly contained within the other. As the number of dimensions increases the relative volume of the smaller shape goes to zero. This is because with high dimensions almost all of the volume is at the surface.

 

[Hornbein]

Consider a sphere and cube with the same center and volume. As the number of dimensions increases the volume of their intersection goes to zero.

 

[robphy]

Possibly related and interesting:

3Blue1Brown (YouTube) - Thinking outside the 10-dimensional box
3Blue1Brown (YouTube) - The most beautiful formula not enough people understand
Stanford Math (YouTube) - Who cares about high-dimensional spheres? - Grant Sanderson (3Blue1Brown)

 

[bob012345]

I think that should be $\approx$



Why?



I don't see how this diagram helps visualize how the smaller sphere touches the face of the octant.

At the time I couldn’t figure out how to make Latex to do $\approx$. Now I do and I edited the above post. Thanks.

Regarding $d=\sqrt{3}r$, the vector from the origin to the center of the small sphere has x,y,z components each equal to $r$.

Regarding the graph, it just visualizes the smaller size.

 

[robphy]

Looking at a projection of the 3D case we can see how the ball is smaller for the same $R$.
View attachment 371412

desmos.com/3d/ky3el7b8v7

 

[pbuk]

Regarding $d=\sqrt{3}r$, the vector from the origin to the center of the small sphere has x,y,z components each equal to $r$.

Ah yes, of course - obvious once you point it out.