Least Action

Nontrivializing triviality..and vice versa.

Coulomb’s Law in (1+2) dimensions

with 2 comments

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.

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Written by Vivek

October 18, 2010 at 23:29

Posted in Uncategorized

2 Responses

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  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


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