# Least Action

Nontrivializing triviality..and vice versa.

## Coulomb’s Law in (n+1) dimensions

The pairwise interaction energy in $n$ spatial dimensions is given by

$E = -\int\frac{d^n k}{(2\pi)^{n}}\frac{e^{i\vec{k}\cdot\vec{r}}}{k^2+m^2}$

To evaluate this $n$-dimensional Fourier integral, we must employ hyperspherical coordinates. The volume element is

$dV = k^{n-1}\sin^{n-2}\phi_1\sin^{n-3}\phi_2\ldots\sin\phi_{n-2}\,dk d\phi_1\ldots d\phi_{n-1}$

The angles $\phi_{1}, \ldots, \phi_{n-2}$ range from $0$ to $\pi$ and $\phi_{n-1}$ ranges from $0$ to $2\pi$. Writing $\vec{k}\cdot\vec{r} = kr\cos\phi_1$, the integral can be written as

$E = -\frac{(2\pi)^{d} C}{(2\pi)^{n}}\int_{0}^{\infty}\frac{dk\,k^{n-1}}{k^2+m^2}\int_{0}^{\pi}d\phi_1\,e^{ikr\cos\phi_1}\sin^{n-2}\phi_1$

The constant $C$ is equal to the product of integrals of the form $\int_{0}^{\pi} \sin^{n-k}\phi\,d\phi$. The number of such integrals depends on the dimension $n$. In two and three dimensions, $C = 1$. In more than 3 dimensions, there are $n-3$ such terms, for $k = 2, 3, \ldots, n-2$ (the integral over $\phi_{n-1}$ produces a $2\pi$, which we’ve already factored out). The value of $d$ is $0$ for 2 dimensions, and is $1$ for 3 and higher dimensions.

The integral over $\phi_1$ produces a regularized confluent hypergeometric function $_{2}F_3(a_1,a_2;b_1,b_2;\alpha)$. Specifically, for $n \geq 2$, the integral over $\phi_1$ produces

$\frac{\sqrt{\pi}e^{-ikr}\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\left[_{2}F_{3}\left(\frac{n-1}{4},\frac{n+1}{4};\frac{1}{2},\frac{n-1}{2},\frac{n}{2};-k^2 r^2\right) + (ikr) _{2}F_{3}\left(\frac{n+1}{4},\frac{n+3}{4};\frac{3}{2},\frac{n+1}{2},\frac{n}{2};-k^2 r^2\right)\right]$

(As a check, for $n =2$, this becomes $\pi J_{0}(kr)$, which is what we obtained for 2 spatial dimensions in a previous post.)

The resulting integration over $k$ is rather tricky. At this point, I don’t know if its possible to do it by hand, so I am using Mathematica to perform it.

Edit: a few hours later..

So, it seems this integral does not have a closed form representation, or at least not one that Mathematica can find.

Written by Vivek

October 21, 2010 at 18:39

### 3 Responses

1. I’ve found a closed form solution for this problem, though I needed to use Google to find an integration “trick” that I had never seen before (seems somewhat specific to this problem):

$\displaystyle \frac{1}{\mathbf{k}^2+m^2} = \int_0^{+\infty} dx e^{-x(\mathbf{k}^2 + m^2)}$

$\displaystyle = -(2\pi)^{-\frac{D}{2}} r^{1-\frac{D}{2}} m^{\frac{D}{2}-1} K_{\frac{D}{2}-1}(mr)$,

matching the results that were previously obtained for D = 2 and D = 3.

mchouza

December 21, 2010 at 17:57

2. Okay, sounds interesting. I would like to see your full working, as I’m not too smart with the math 😛

Vivek

December 22, 2010 at 08:38

3. mchouza

December 22, 2010 at 08:49