# Massive Scalar Field in 2+1 Dimensions

1. Sep 26, 2011

### xGAME-OVERx

1. The problem statement, all variables and given/known data

We wish to find, in 2+1 dimensions, the analogue of $E = - \frac{1}{4\pi r} e^{-mr}$ found in 3+1 dimensions. Here r is the spatial distance between two stationary disturbances in the field.

2. Relevant equations

In 3+1 we start from $E = - \int \frac{ d^3 k }{(2\pi)^3} \frac{1}{ {\bf{k}}^2 + m^2 } e^{ i {\bf{k}} \cdot ( {\bf{x}}_1 - {\bf{x}}_2 ) }$ where $\bf{k}$ is momentum, and $\bf{x}_i$ are the spatial locations of the two disturbances.

3. The attempt at a solution

I think in 2+1 we must use the equation $E = - \int \frac{ d^2 k }{(2\pi)^2} \frac{1}{ {\bf{k}}^2 + m^2 } e^{ i {\bf{k}} \cdot ( {\bf{x}}_1 - {\bf{x}}_2 ) }$. I begin by transforming to polar coordinates, i.e. $E = - \frac{1}{(2\pi)^2} \int_{0}^{\infty} dk \int_{0}^{2\pi} d\theta \frac{k}{ k^2 + m^2 } e^{ i k r \cos\theta }$.

However, I am not sure what to do with this. As far as I know the theta integral can't be done in this form, and the r integral extends only down to 0, preventing it from being amenable to countour integration methods.

I tried a common trick of writing:

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

Which just makes the integral worse (I think). Any pointers would be greatly appreciated.

Thanks
Scott

2. Sep 26, 2011

### mathfeel

Integration over $\theta$ gives you the Bessel function of the first kind.

This leave you with $E = - \frac{1}{2\pi} \int_0^{\infty} dk \frac{k}{k^2 + m^2} J_0 (kr)$

This I think is another Bessel function...Second Kind...I think. Look it up in a table.

Last edited: Sep 26, 2011
3. Sep 26, 2011

### xGAME-OVERx

Thank you for your reply, but that doesn't match any of the definitions I have seen for Bessel functions. May I ask which definition you are using?

Thanks
Scott

EDIT: Sorry, found it in Abramowitz & Stegun...

Last edited: Sep 26, 2011
4. Sep 26, 2011

### mathfeel

Rewrite the integral in term of a new variable $t = \cos\theta$. Then it conforms to the first integral representation here: http://functions.wolfram.com/Bessel-TypeFunctions/BesselJ/07/01/01/