Fourier Transform of Curl & Divergence Free Vector Function

[LocationX]

A vector function can be decomposed to form a curl free and divergence
free parts:

$$\vec{f}(\vec{r})=\vec{f_{\parallel}}(\vec{r'})+\vec{f_{\perp}}(\vec{r'})$$

where

$$\vec{f_{\parallel}}(\vec{r'}) = - \vec{\nabla} \left( \frac{1}{4 \pi} \int d^3 r' \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} \right)$$

and

$$\vec{f_{\perp}}(\vec{r'}) = \vec{\nabla} \times \left( \frac{1}{4 \pi} \int d^3 r' \frac{\vec{\nabla'} \times \vec{f}(\vev{r'})}{|\vec{r}-\vec{r'}|}$$

I am trying to take the Fourier transform of
$\vec{f_{\parallel}}(\vec{r'})$ and
$\vec{f_{\perp}}(\vec{r})$


I am starting at $\vec{f_{\parallel}}(\vec{r'})$. We know
that the Fourier transform is given by:

$$\vec{f}(\vec{k}) = \int_{-\infty}^{\infty} d^3r e^{- i \vec{k} \cdot \vec{r}} \vec{f}(\vec{r})$$


$$\vec{f}(\vec{r}) = \frac{1}{(2 \pi)^3} \int_{-\infty}^{\infty} d^3k e^{- i \vec{k} \cdot \vec{r}} \vec{f}(\vec{k})$$


I'm not exactly sure where to begin. If I just plug and chug , we'd have:

$$\vec{f}(\vec{k}) = \int_{-\infty}^{\infty} d^3r e^{- i \vec{k} \cdot \vec{r}} \vec{f}(\vec{r})$$

$$\vec{f}(\vec{k}) = \int_{-\infty}^{\infty} e^{- i \vec{k} \cdot \vec{r}} - \vec{\nabla} \left( \frac{1}{4 \pi} \int \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3 r' \right) d^3r$$


I just do not see a simple way of tacking this problem. Any thoughts
would be appreciated.

 

[Jhero]

Thank you for bringing up this interesting topic. I can provide some insights on how to approach this problem. The first step in solving this problem is to recognize that the Fourier transform is a linear operator, which means that it can be applied to each term separately. This will help us break down the problem into smaller, more manageable parts.

Let's start with \vec{f_{\parallel}}(\vec{r'}). Using the definition of the Fourier transform, we can write:

\vec{f_{\parallel}}(\vec{k}) = \int_{-\infty}^{\infty} d^3r' e^{- i \vec{k} \cdot \vec{r'}} \vec{f_{\parallel}}(\vec{r'})

Next, we can use the definition of \vec{f_{\parallel}}(\vec{r'}) to simplify the integral as follows:

\vec{f_{\parallel}}(\vec{k}) = -\vec{\nabla} \left( \frac{1}{4 \pi} \int_{-\infty}^{\infty} d^3r' e^{- i \vec{k} \cdot \vec{r'}} \int_{-\infty}^{\infty} \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3r \right)

Now, we can use the convolution theorem for Fourier transforms, which states that the Fourier transform of a convolution is the product of the individual Fourier transforms. In this case, we have a convolution between e^{- i \vec{k} \cdot \vec{r'}} and \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|}. Therefore, we can write:

\vec{f_{\parallel}}(\vec{k}) = -\vec{\nabla} \left( \frac{1}{4 \pi} \int_{-\infty}^{\infty} d^3r' e^{- i \vec{k} \cdot \vec{r'}} \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|}