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Math Help - Fourier transform of potential from point charge

  1. #1
    Senior Member TriKri's Avatar
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    Fourier transform of potential from point charge

    Hi! How can I calculate the Fourier transform of the potential generated by a point charge in the origin with the charge q?
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  2. #2
    MHF Contributor chisigma's Avatar
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    How many dimensions?... one, two, three... or more?...

    Kind regards

    \chi \sigma
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  3. #3
    Senior Member TriKri's Avatar
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    Three dimensions!
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by TriKri View Post
    Three dimensions!
    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 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 \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 \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 \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 u=v=0 where is 2 \pi so that the final result is...

    \displaystyle \mathcal{F} \{V(x, y)\}= \frac{q}{2 \varepsilon_{0}}\ \delta(u,v) (5)

    Kind regards

    \chi \sigma
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  5. #5
    Senior Member TriKri's Avatar
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    Thank you, but does the integral really vanish? If r\sqrt{u^2+v^2} < 1/4 the exponent over e will be in the interval (-j\pi/2, +j\pi/2), so the real part of the integrand will be positive for all \theta and then the integral can't vanish, can it?
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  6. #6
    Senior Member TriKri's Avatar
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    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?
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