# I Changing coordinate systems

1. Nov 30, 2017

### Tio Barnabe

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 by a moderator: Nov 30, 2017
2. Nov 30, 2017

### eys_physics

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 $\vec{p}\cdot\vec{x}=px\cos\theta$.