Hi! How can I calculate the Fourier transform of the potential generated by a point charge in the origin with the charge q?
All right!... but for semplicity sake may be it is better to start with the two-dimensional case ...
The potential function generated by a charge q in the origin is...
$\displaystyle \displaystyle V(x,y)= \frac{q}{4 \pi \varepsilon_{0}\ \sqrt{x^{2}+y^{2}}}$ (1)
The two-dimension Fourier tranform of V(x,y) by definition is...
$\displaystyle \displaystyle \mathcal{F} \{V(x,y)\}= \int_{- \infty}^{+ \infty} \int_{- \infty}^{+ \infty} V(x,y)\ e^{-j 2 \pi (u\ x + v\ y)}\ dx dy$ (2)
Now if we pass to polar coordinates the (2) becomes...
$\displaystyle \displaystyle \mathcal{F} \{V(r, \theta)\}= \frac{q}{4 \pi \varepsilon_{0}}\ \int_{0}^{\infty} \int_{- \pi}^{\pi} e^{-j 2 \pi (u\ r \cos \theta + v\ r\ \sin \theta )}\ dr d \theta $ (3)
Now it is easy to verify that the integral...
$\displaystyle \displaystyle \int_{- \pi}^{\pi} e^{-j 2 \pi (u\ r \cos \theta + v\ r\ \sin \theta )}\ d \theta $ (4)
... vanishes everiwhere with the only exception of the case $\displaystyle u=v=0$ where is $\displaystyle 2 \pi$ so that the final result is...
$\displaystyle \displaystyle \mathcal{F} \{V(x, y)\}= \frac{q}{2 \varepsilon_{0}}\ \delta(u,v)$ (5)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Thank you, but does the integral really vanish? If $\displaystyle r\sqrt{u^2+v^2} < 1/4$ the exponent over e will be in the interval $\displaystyle (-j\pi/2, +j\pi/2)$, so the real part of the integrand will be positive for all $\displaystyle \theta$ and then the integral can't vanish, can it?
I'm still wondering what the Fourier transform is; the delta function can't be the answer since it's the Fourier transform of 1 (or at least of a real constant). The Fourier transform of 1/r in three dimensions has to be some kind of standard transform, hasn't it?