Least Action

Nontrivializing triviality..and vice versa.

Coulomb’s Law in (n+1) dimensions

with 3 comments

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

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  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)}

    Using this and completing squares in the exponent I got

    \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.


    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 😛


    December 22, 2010 at 08:38

  3. mchouza

    December 22, 2010 at 08:49

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