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A vector function can be decomposed to form a curl free and divergence
free parts:
[tex]\vec{f}(\vec{r})=\vec{f_{\parallel}}(\vec{r'})+\vec{f_{\perp}}(\vec{r'})[/tex]
where
[tex]\vec{f_{\parallel}}(\vec{r'}) = - \vec{\nabla} \left( \frac{1}{4
\pi} \int d^3 r' \frac{\vec{\nabla'} \cdot
\vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} \right)[/tex]
and
[tex]\vec{f_{\perp}}(\vec{r'}) = \vec{\nabla} \times \left( \frac{1}{4
\pi} \int d^3 r' \frac{\vec{\nabla'} \times
\vec{f}(\vev{r'})}{|\vec{r}-\vec{r'}|}[/tex]
I am trying to take the Fourier transform of
[itex]\vec{f_{\parallel}}(\vec{r'})[/itex] and
[itex]\vec{f_{\perp}}(\vec{r})[/itex]
I am starting at [itex]\vec{f_{\parallel}}(\vec{r'})[/itex]. We know
that the Fourier transform is given by:
[tex] \vec{f}(\vec{k}) = \int_{-\infty}^{\infty} d^3r e^{- i \vec{k}
\cdot \vec{r}} \vec{f}(\vec{r}) [/tex]
[tex] \vec{f}(\vec{r}) = \frac{1}{(2 \pi)^3} \int_{-\infty}^{\infty}
d^3k e^{- i \vec{k} \cdot \vec{r}} \vec{f}(\vec{k}) [/tex]
I'm not exactly sure where to begin. If I just plug and chug , we'd have:
[tex] \vec{f}(\vec{k}) = \int_{-\infty}^{\infty} d^3r e^{- i \vec{k}
\cdot \vec{r}} \vec{f}(\vec{r}) [/tex]
[tex] \vec{f}(\vec{k}) = \int_{-\infty}^{\infty} e^{- i \vec{k} \cdot
\vec{r}} - \vec{\nabla} \left( \frac{1}{4 \pi} \int
\frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3
r' \right) d^3r [/tex]
I just do not see a simple way of tacking this problem. Any thoughts
would be appreciated.
free parts:
[tex]\vec{f}(\vec{r})=\vec{f_{\parallel}}(\vec{r'})+\vec{f_{\perp}}(\vec{r'})[/tex]
where
[tex]\vec{f_{\parallel}}(\vec{r'}) = - \vec{\nabla} \left( \frac{1}{4
\pi} \int d^3 r' \frac{\vec{\nabla'} \cdot
\vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} \right)[/tex]
and
[tex]\vec{f_{\perp}}(\vec{r'}) = \vec{\nabla} \times \left( \frac{1}{4
\pi} \int d^3 r' \frac{\vec{\nabla'} \times
\vec{f}(\vev{r'})}{|\vec{r}-\vec{r'}|}[/tex]
I am trying to take the Fourier transform of
[itex]\vec{f_{\parallel}}(\vec{r'})[/itex] and
[itex]\vec{f_{\perp}}(\vec{r})[/itex]
I am starting at [itex]\vec{f_{\parallel}}(\vec{r'})[/itex]. We know
that the Fourier transform is given by:
[tex] \vec{f}(\vec{k}) = \int_{-\infty}^{\infty} d^3r e^{- i \vec{k}
\cdot \vec{r}} \vec{f}(\vec{r}) [/tex]
[tex] \vec{f}(\vec{r}) = \frac{1}{(2 \pi)^3} \int_{-\infty}^{\infty}
d^3k e^{- i \vec{k} \cdot \vec{r}} \vec{f}(\vec{k}) [/tex]
I'm not exactly sure where to begin. If I just plug and chug , we'd have:
[tex] \vec{f}(\vec{k}) = \int_{-\infty}^{\infty} d^3r e^{- i \vec{k}
\cdot \vec{r}} \vec{f}(\vec{r}) [/tex]
[tex] \vec{f}(\vec{k}) = \int_{-\infty}^{\infty} e^{- i \vec{k} \cdot
\vec{r}} - \vec{\nabla} \left( \frac{1}{4 \pi} \int
\frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3
r' \right) d^3r [/tex]
I just do not see a simple way of tacking this problem. Any thoughts
would be appreciated.