# n dimensional ball!?

Hello, guys! It’s me! Warotan!

Japanese version

Let’s calculate the volume of an $$n$$ dimensional ball!
We use it in so many field (statistical mechanics, for example)

## What form does the quantum have??

The volume of $$n$$ dimensional ball.
Can you anticipate the form?

To begin with, because it is very difficult to start from the $$n$$ dimensional, we do with a 3 dimensional one.

As we already think about it in this article, the volume of a 3 dimensional ball with radius $$r$$ is

$$\begin{eqnarray}V_{3}=\frac{4}{3}\pi r^{3}.\end{eqnarray}$$

It is proportional to $$r^3$$.

Here, The area of a circle with radius $$r$$ is

$$\begin{eqnarray}V_{2}=\pi r^{2}.\end{eqnarray}$$

It is proportional to $$r^2$$.
The area in 2 dimensional plane corresponds to the volume in 3 dimensional space.
“Volume” shows the quantity of the biggest shape in the “space”.
Therefore, an area in a plane can be thought as a general “volume”.

The area of a circle (2 dimension) is proportional to $$r^2$$, and the volume of a ball (3 dimensional) is to $$r^3$$.
So, we expect that the volume of $$n$$ dimensional ball is proportional to $$r^n$$.

Assuming

$$\begin{eqnarray}V_{n}= A_{n} r^{n},\end{eqnarray}$$

$$A_{n}$$ is a constant independent of $$r$$.

## The volume of an $$n$$ dimensional ball

We will use a gamma function (English version coming soon)

$$\begin{eqnarray}\Gamma\left(\frac{1}{2}\right) &=& \int_{-\infty}^{\infty}{\rm e}^{-x^2}{\rm d}x \\ &=& \pi^{1/2}\end{eqnarray}$$

to express the volume.
It is an integration of a gaussian.

Here, let us consider the integration of a product of $$n$$ gaussians with different variables like

$$\begin{eqnarray}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}{\rm e}^{-x_{1}^{2}}{\rm e}^{-x_{2}^{2}}\cdots{\rm e}^{-x_{n}^{2}}{\rm d}x_{1}{\rm d}x_{2}\cdots{\rm d}x_{n} &=& \left(\int_{-\infty}^{\infty}{\rm e}^{-x^2}{\rm d}x\right)^{2} \\ &=& \pi^{n/2}\end{eqnarray}$$

On the other hand, we can calculate this integration in a different way.
First, the $$n$$ dimensional volume element $${\rm d}x_{1}{\rm d}x_{2}\cdots{\rm d}x_{n}$$ can be written as $${\rm d}V_{n}$$.
Next, the product of $$n$$ gaussians can be written as $${\rm e}^{-\left(x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}\right)}$$;

$$\begin{eqnarray} {\rm d}x_{1}{\rm d}x_{2}\cdots{\rm d}x_{n} &\rightarrow& {\rm d}V_{n} \\ {\rm e}^{-x_{1}^{2}}{\rm e}^{-x_{2}^{2}}\cdots{\rm e}^{-x_{n}^{2}} &\rightarrow& {\rm e}^{-\left(x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}\right)}. \end{eqnarray}$$

Therefore,

$$\begin{eqnarray} \pi^{n/2} = \int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}{\rm e}^{-\left(x_{1}^2+\cdots +x_{n}^2\right)}{\rm d}V_{n}.\end{eqnarray}$$

The exponent of the integrand is the square of the distance from the origin to the point $$\left(x_{1},x_{2},\cdots ,x_{n}\right)$$ in Euclid space.
Using the assumption we made above, $$V_{n}\sim r^n$$, we can rewrite it as

$$\begin{eqnarray} \pi^{n/2} &=& \int_{0}^{\infty}{\rm e}^{-r^2}nA_{n}r^{n-1}{\rm d}r \\ &=& nA_{n}\int_{0}^{\infty}{\rm e}^{-r^2}r^{n-1}{\rm d}r \end{eqnarray}$$

Yes! It is a gamma function!

$$\begin{eqnarray} \Gamma\left(z\right) &=& \int_{0}^{\infty}{\rm e}^{-t}t^{z-1}{\rm d}t \\ &=& 2\int_{0}^{\infty}{\rm e}^{-\xi^2}\xi^{2z-1}{\rm d}\xi \end{eqnarray}$$

They are the same if we put $$z=\frac{n}{2}$$.
The above equation can be written as

$$\begin{eqnarray} \pi^{n/2} = A_{n}\frac{n}{2}\Gamma\left(\frac{n}{2}\right) \end{eqnarray}$$

The $$A_{n}$$ is

$$\begin{eqnarray} A_{n} &=& \frac{\pi^{n/2}}{\frac{n}{2}\Gamma\left(\frac{n}{2}\right)} \\ &=& \frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)} \end{eqnarray}$$

In the end, the volume of an $$n$$ dimensional ball is

$$\begin{eqnarray} V_{n}=\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)}r^n \end{eqnarray}$$

The gamma function shows again!

## Summary

Now, that’s all what I wanted to write.

This time, we calculated the volume of an $$n$$ dimensional ball.
It is a common quantity in math or physics (and so on).
Seeing or hearing $$n$$ dim-ball, you will get uneasy but get accustom to it!