n dimensional ball!?

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

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What form does the quantum have??
The volume of an \(n\) dimensional ball
Summary

What form does the quantum have??

The volume of \(n\) dimensional ball.
Can you anticipate the form?

Let’s think about it.
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

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

We have already seen the form.

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!

Thank you for reading and please spread this blog if you like.

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