- #1
xGAME-OVERx
- 86
- 0
Homework Statement
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.
Homework 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.
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