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Massive Scalar Field in 2+1 Dimensions

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant 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.

    3. 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
     
  2. jcsd
  3. Sep 26, 2011 #2
    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: Sep 26, 2011
  4. Sep 26, 2011 #3
    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
  5. Sep 26, 2011 #4
    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/
     
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