Massive Scalar Field in 2+1 Dimensions

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
xGAME-OVERx
Messages
86
Reaction score
0

Homework Statement



We wish to find, in 2+1 dimensions, the analogue of [itex]E = - \frac{1}{4\pi r} e^{-mr}[/itex] 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 [itex]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 ) }[/itex] where [itex]\bf{k}[/itex] is momentum, and [itex]\bf{x}_i[/itex] are the spatial locations of the two disturbances.

The Attempt at a Solution



I think in 2+1 we must use the equation [itex]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 ) }[/itex]. I begin by transforming to polar coordinates, i.e. [itex]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 }[/itex].

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:

[itex]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 }[/itex]

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

Thanks
Scott
 
Physics news on Phys.org
xGAME-OVERx said:
I think in 2+1 we must use the equation [itex]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 ) }[/itex]. I begin by transforming to polar coordinates, i.e. [itex]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 }[/itex].

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.

Integration over [itex]\theta[/itex] gives you the Bessel function of the first kind.

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

This I think is another Bessel function...Second Kind...I think. Look it up in a table.
 
Last edited:
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:
xGAME-OVERx said:
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...

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