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I Changing coordinate systems

  1. Nov 30, 2017 #1
    I should evaluate ##\int d^3 p \ \exp(i \vec{p} \cdot \vec{x}) / \sqrt{|p| + m^2}## over all ##\mathbb{R}^3##. How can I do this in spherical coordinates? Since ##\vec{p}## is a position vector in ##\mathbb{R}^3##, our ##\vec{r}## of the spherical coordinates would be just equal to ##\vec{p}##, correct? (The same would be said of ##\vec{x}##, which we could call an ##\vec{r'}##.)

    So the integral above would be re-written as ##\int dp \ d\varphi \ d\theta \ p^2 \exp(i \vec{p} \cdot \vec{x}) / \sqrt{|p| + m^2}##. I suppose that the correct range for our "spherical variables" is ##p \in (-\infty, \infty)## and ##\varphi, \theta \in [0, 2 \pi]##. Now the integral would then become ##4 \pi \int dp \ p^2 \exp(i \vec{p} \cdot \vec{x}) / \sqrt{|p| + m^2}##. How do I evaluate it?

    EDIT: I don't need to give the final result, just to evaluate the last integral a little further.
     
    Last edited: Nov 30, 2017
  2. jcsd
  3. Nov 30, 2017 #2
    No, in spherical coordinates your volume element is $$dV=r^2\sin\theta d\theta d\phi,$$
    so you are missing a factor of ##sin\theta##. Also ##p## will be ##p=|\vec{p}|##, so the limits are ##p\in (0,\infty)##. Furthermore, your range for ##\theta## should be ##\theta\in [0,\pi]##. I also believe that in your initial expression the denominator should be ##\sqrt{|p|^2+m^2}##, if now ##m## is a mass. Notice, also that the scalar product depends on ##\theta##. You orientate the coordinate system so that ##\vec{p}\cdot\vec{x}=px\cos\theta##.
     
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