# Least Action

Nontrivializing triviality..and vice versa.

## Coulomb’s Law in (1+2) dimensions

The idea for this post came from an unsolved problem in Tony Zee’s book on QFT. We begin with the ‘mutual’ energy:

$E = -\frac{1}{2}\int \frac{d^2 k}{(2\pi)^2}\frac{e^{i\vec{k}\cdot(\vec{x}_1-\vec{x}_2)}}{k^2 + m^2}$

Define $\boldsymbol{r} = \boldsymbol{x}_1-\boldsymbol{x}_2$. Transforming to cylindrical polar coordinates, this is

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

The $\theta$ integral produces a Bessel Function of order 0, i.e. $2\pi J_{0}(kr)$, so that

$E = -\frac{1}{4\pi}\int_{0}^{\infty}\frac{k J_{0}(kr)}{k^2 + m^2}dk$

I’m lazy, so I used Mathematica to get the final result, which is

$E = -\frac{1}{4\pi}K_{0}(mr)$

where $K_{0}$ denotes the modified Bessel function of the second kind, of order 0. A plot of $K_{0}(mr)$ as a function of $mr$ is shown below.

The graph in red is that of the usual 1/r potential. So, in 2 spatial dimensions, the potential energy actually falls off faster than 1/r.

Written by Vivek

October 18, 2010 at 23:29

Posted in Uncategorized

### 2 Responses

1. It’s great to see another engineer following Zee’s book! 😀

The potential energy only falls of faster than 1/r due to the mass of the exchanged particle (it’s analogous to the exponential factor in the 3D Yukawa potential). If you do the integral for m = 0 using r = 1 as the reference point for the energy, the result is the usual logarithmic potential.

Best wishes!

mchouza

December 19, 2010 at 11:39

• Hi, thanks for your comment. Have a look at a sequel to this post as well.

Vivek

December 20, 2010 at 23:19